Thomas’ Calculus. Early Transcendentals [15, Global, in SI Units ed.] 9780137559893, 1292725907, 9781292725901, 9781292457628

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GLOBAL EDITION

THOMAS’ CALCULUS Early Transcendentals FIFTEENTH EDITION IN SI UNITS

Hass • Heil • Bogacki • Weir

'LJLWDO5HVRXUFHVIRU6WXGHQWV 0. (b) Find the tangent line to the curve y = x at x = 4.

Cancel h ≠ 0 and evaluate.

144

Chapter 3 Derivatives

Solution

Derivative of the Square Root Function 1 d x = , x > 0 dx 2 x

(a) We use the alternative formula to calculate f ′: f ′( x ) = lim z→ x

= lim z→ x

y y = 1x + 1 4

= lim z→ x

= lim

y = "x

(4, 2)

z→ x

1 0

z − x z−x z −

x

1 z +

x

=

1 . 2 x

The curve y = x and its tangent line at ( 4, 2 ). The tangent line’s slope is found by evaluating the derivative at x = 4 (Example 2).

The tangent line is the line through the point ( 4, 2 ) with slope 1 4 (Figure 3.5):

y =

A

Slope −1 B C

y = f (x)

Slope − 4 3 E D Slope 0

5

5

There are many ways to denote the derivative of a function y = f ( x ), where the independent variable is x and the dependent variable is y. Some common alternative notations for the derivative are ≈8

f ′( x ) = y′ =

≈ 4 x-units 10 15

x

(a) Slope

4

−2

f ′(a) =

E′

2 1 5 C′

D′ 10

15

x

We made the graph of y = f ′( x ) in (b) by plotting slopes from the graph of y = f ( x ) in (a). The vertical coordinate of B ′ is the slope at B, and so on. The slope at E is approximately 8 4 = 2. In (b) we see that the rate of change of f is negative for x between A′ and D′; the rate of change is positive for x to the right of D′.

dy dx

x=a

=

df dx

x=a

=

d f (x) dx

x=a

.

For instance, in Example 2, f ′(4) =

B′ Vertical coordinate −1 (b)

FIGURE 3.6

dy df d = = f ( x ) = D( f )( x ) = D x f ( x ). dx dx dx

The symbols d dx and D indicate the operation of differentiation. We read dy dx as “the derivative of y with respect to x,” and df dx and ( d dx ) f ( x ) as “the derivative of f with respect to x.” The “prime” notations y ′ and f ′ originate with Newton. The d dx notations are similar to those used by Leibniz. The symbol dy dx should not be regarded as a ratio; it simply denotes a derivative. To indicate the value of a derivative at a specified number x = a, we use the notation

y = f '(x)

3

−1

1 x + 1. 4

Notation

Slope 0

A'

1( x − 4) 4

y = 2+

y

0

Cancel and evaluate.

1 1 = . 4 2 4

f ′(4) =

FIGURE 3.5

10

1 1 = ( a − b )( a + b ) a2 − b2

( z − x )( z + x )

(b) The slope of the curve at x = 4 is

x

4

f (z) − f ( x ) z−x

d x dx

x=4

=

1 2 x

x=4

=

1 1 = . 4 2 4

Graphing the Derivative We can often make an approximate plot of the derivative of y = f ( x ) by estimating the slopes on the graph of f. That is, we plot the points ( x , f ′( x ) ) in the xy-plane and connect them with a curve that represents y = f ′( x ). EXAMPLE 3

Graph the derivative of the function y = f ( x ) in Figure 3.6a.

Solution We sketch the tangent lines to the graph of f at frequent intervals and use their slopes to estimate the values of f ′( x ) at these points. We plot the corresponding ( x , f ′( x ) ) pairs and connect them with a curve as sketched in Figure 3.6b.

3.2  The Derivative as a Function

145

What can we learn from the graph of y = f ′( x )? At a glance we can see 1. where the rate of change of f is positive, negative, or zero; 2. the rough size of the growth rate at any x; 3. where the rate of change itself is increasing or decreasing.

Differentiability on an Interval; One-Sided Derivatives

Slope = f(a + h) − f(a) lim + h h:0

Slope = f (b + h) − f (b) lim h h:0−

A function y = f ( x ) is differentiable on an open interval (finite or infinite) if it has a derivative at each point of the interval. It is differentiable on a closed interval [ a, b ] if it is differentiable on the interior ( a, b ) and if the limits f ( a + h ) − f ( a) h f ( b + h ) − f ( b) lim h → 0− h lim

h→ 0+

y = f (x)

a

a+h h>0

b+h h 0,

y y = 0x0 y′ = −1

d( x dx

)

=

d d (x) = (1 ⋅ x ) = 1. dx dx

x = x since x > 0,

d( mx + b ) = m dx

y′ = 1

To the left, when x < 0, 0

x y′ not defined at x = 0: right-hand derivative ≠ left-hand derivative

FIGURE 3.8 The function y = x is not differentiable at the origin where the graph has a “corner” (Example 4).

d( x dx

)

=

d d (−x ) = (−1 ⋅ x ) = −1 dx dx

x = − x since x < 0

(Figure 3.8). The two branches of the graph come together at an angle at the origin, forming a non-smooth corner. There is no derivative at the origin because the one-sided derivatives differ there: 0+h − 0 h = lim+ h→0 h→0 h h h h = h  when  h > 0 = lim+ h→0 h = lim+ 1 = 1

Right-hand derivative of  x  at zero = lim+

h→0

0+h − 0 h = lim− h→0 h→0 h h −h h = − h  when  h < 0 = lim− h→0 h = lim− −1 = −1.

Left-hand derivative of  x  at zero = lim−

h→0

146

Chapter 3 Derivatives

y

EXAMPLE 5

In Example 2 we found that for x > 0, 1 d x = . dx 2 x

2 y=

x

We apply the definition to examine whether the derivative exists at x = 0:

1

lim

h→ 0+

0

1

x

2

0+h− h

0

= lim+ h→0

1 = ∞. h

Since the (right-hand) limit is not finite, there is no derivative at x = 0. Since the slopes of the secant lines joining the origin to the points ( h, h ) on a graph of y = x approach ∞, the graph has a vertical tangent line at the origin. (See Figure 3.9 and Exercises 37 and 38 in Section 3.1.)

FIGURE 3.9 The square root function is not differentiable at x = 0, where the graph of the function has a vertical tangent line.

When Does a Function Not Have a Derivative at a Point? A function has a derivative at a point x 0 if the slopes of the secant lines through P ( x 0 , f ( x 0 ) ) and a nearby point Q on the graph approach a finite limit as Q approaches P. Thus differentiability is a “smoothness” condition on the graph of f. A function can fail to have a derivative at a point for many reasons, including the existence of points where the graph has

Q−

P P

P

Q− Q+

Q+

Q−

Q+

1. a corner, where the one-sided derivatives differ

2. a cusp, where the slope of PQ approaches ∞ from one side and −∞ from the other

3. a vertical tangent line, where the slope of PQ approaches ∞ from both sides or approaches −∞ from both sides (here, it approaches −∞) y y= Q−

P

P

5

x sin 1 , x 0,

x≠0 x=0

Q+ P

Q+ Q− Q−

4. a discontinuity (two examples shown)

x

Q+

5. wild oscillation

The last example shows a function that is continuous at x = 0, but whose graph oscillates wildly up and down as it approaches x = 0. The slopes of the secant lines through 0 oscillate between −1 and 1 as x approaches 0, and do not have a limit at x = 0.

147

3.2  The Derivative as a Function

Differentiable Functions Are Continuous A function is continuous at every point where it has a derivative.

THEOREM 1—Differentiable Implies Continuous

If f has a derivative at

x = c, then f is continuous at x = c. Proof Given that f ′(c) exists, we must show that lim f ( x ) = f (c), or, equivalently, x →c that lim f ( c + h ) = f (c). If h ≠ 0, then h→0

f ( c + h ) = f (c) + ( f ( c + h ) − f (c) ) = f (c) +

f ( c + h ) − f (c) ⋅ h. h

Add and subtract  f (c). Divide and multiply by h.

Now take limits as h → 0. By Theorem 1 of Section 2.2, lim f ( c + h ) = lim f (c) + lim

h→ 0

h→ 0

h→ 0

f ( c + h ) − f (c) ⋅ lim h h→ 0 h

= f (c) + f ′(c) ⋅ 0 = f (c) + 0 = f (c). Similar arguments with one-sided limits show that if f has a derivative from one side (right or left) at x = c, then f is continuous from that side at x = c. Theorem 1 says that if a function has a discontinuity at a point (for instance, a jump discontinuity), then it cannot be differentiable there. The greatest integer function y = ⎣ x ⎦ fails to be differentiable at every integer x = n (Example 4, Section 2.6). Caution The converse of Theorem 1 is false. A function need not have a derivative at a point where it is continuous, as we saw with the absolute value function in Example 4.

3.2

EXERCISES

Finding Derivative Functions and Values

Slopes and Tangent Lines

Using the definition, calculate the derivatives of the functions in Exercises 1–6. Then find the values of the derivatives as specified.

In Exercises 13–16, differentiate the functions and find the slope of the tangent line at the given value of the independent variable.

1. f ( x ) = 4 − x 2 ;

f ′(−3), f ′(0), f ′(1)

13. f ( x ) = x +

2. F ( x ) = ( x − 1) + 1; F ′(−1), F ′(0), F ′(2) 1 3. g(t ) = 2 ; g′(−1), g′(2), g′( 3 ) t 1− z 4. k ( z ) = ; k ′(−1), k ′(1), k ′( 2 ) 2z 2

15. s = t 3 − t 2 , t = −1

3θ ;

2s + 1; r ′(0), r ′(1), r ′(1 2 )

17. y = f ( x ) =

In Exercises 7–12, find the indicated derivatives. dy 7. dx

if

9.

ds dt

if

s =

11.

dp dq

if

p = q3 2

y = 2x 3 t 2t + 1

16. y =

x+3 , x = −2 1− x

In Exercises 17–18, differentiate the functions. Then find an equation of the tangent line at the indicated point on the graph of the function.

p′(1), p′(3), p′( 2 3 )

5. p(θ) = 6. r (s) =

1 9 , x = −3 14. k ( x ) = , x = 2 x 2+ x

8 , ( x , y ) = ( 6, 4 ) x−2

18. w = g( z ) = 1 +

dr 8. ds

if

r = s 3 − 2s 2 + 3

10.

dv dt

if

v = t−

12.

dz dw

if

z =

1 t

1 w2 − 1

4 − z , ( z , w ) = ( 3, 2 )

In Exercises 19–22, find the values of the derivatives. 19.

ds dt

t =−1

21.

dr dθ

θ= 0

if

s = 1 − 3t 2

20.

dy dx

x= 3

if

r =

2 4−θ

22.

dw dz

z=4

if if

y = 1− w = z+

1 x z

148

Chapter 3 Derivatives

31. Consider the function f graphed here. The domain of f is the interval [ −4, 6 ] and its graph is made of line segments joined end to end.

Using the Alternative Formula for Derivatives

Use the formula f ′( x ) = lim z→x

f (z) − f ( x ) z−x

y

23. f ( x ) =

1 x+2

(6, 2)

(0, 2)

to find the derivative of the functions in Exercises 23–26.

y = f (x)

24. f ( x ) = x 2 − 3 x + 4

x 25. g( x ) = x −1

0

(−4, 0)

26. g( x ) = 1 +

1

x (1, −2)

Graphs

Match the functions graphed in Exercises 27–30 with the derivatives graphed in the accompanying figures (a)–(d). y′

x

6

y′

(4, −2)

a. At which points of the domain interval is f ′ not defined? Give reasons for your answer. b. Graph the derivative of f. The graph should show a step function. 32. Recovering a function from its derivative a. Use the following information to graph the function f over the closed interval [ −2, 5 ].

x

0 x

0 (a)

(b)

y′

y′

i) The graph of f is made of closed line segments joined end to end. ii) The graph starts at the point (−2, 3 ). iii) The derivative of f is the step function in the figure shown here.

x

0

y′

x

0

y′ = f ′(x) 1

(c)

(d)

27.

28.

y

−2 y

x

0

29.

y = f4(x) x

0

3

5

x

b. Repeat part (a), assuming that the graph starts at (−2, 0 ) instead of (−2, 3 ). 33. Growth in the economy The graph in the accompanying figure shows the average annual percentage change y = f (t ) in the U.S. gross national product (GNP) for the years 2005–2011. Graph dy dt (where defined).

y

y = f3(x)

0

x

0

30.

y

1 −2

y = f 2 (x)

y = f1(x)

0

x

7% 6 5 4 3 2 1 0 2005 2006 2007 2008 2009 2010 2011

34. Fruit flies (Continuation of Example 4, Section 2.1.) Populations starting out in closed environments grow slowly at first, when there are relatively few members, then more rapidly as the number of reproducing individuals increases and resources are still abundant, then slowly again as the population reaches the carrying capacity of the environment.

149

3.2  The Derivative as a Function

a. Use the graphical technique of Example 3 to graph the derivative of the fruit fly population as a function of time (in days). The graph of the population is reproduced here.

c. Estimate the rate of change of home prices at the beginning of i) 2007

iii) 2014

d. During what year did home prices drop most rapidly and what is an estimate of this rate?

p

Number of flies

ii) 2010

e. During what year did home prices rise most rapidly and what is an estimate of this rate?

350 300 250 200 150 100 50

f. Use the graphical technique of Example 3 to graph the derivative of home price P versus time t. One-Sided Derivatives

0

10

20 30 Time (days)

40

t

50

Compute the right-hand and left-hand derivatives as limits to show that the functions in Exercises 37–40 are not differentiable at the point P. 38.

37. y

b. During what days does the population seem to be increasing fastest? Slowest?

y y = f (x)

y=x

y = 2x

y = f (x)

2

y=2 2

35. Temperature The given graph shows the temperature T in °C between 6 a.m. and 6 p.m.

P(1, 2)

y=x

T

1

Temperature (°C)

25

x

P(0, 0)

0

1

2

x

20

39.

15

40. y

10

y y = f (x)

5 0

3

6

9

12

6 A.M.

9 A.M.

12 NOON Time (h)

3 P.M.

6 P.M.

i) 7 a.m.

ii) 9 a.m.

iii) 2 p.m.

1

iv) 4 p.m.

b. At what time does the temperature increase most rapidly? Decrease most rapidly? What is the rate for each of those times? c. Use the graphical technique of Example 3 to graph the derivative of temperature T versus time t. 36. Average single-family home prices P (in thousands of dollars) in Sacramento, California, are shown in the accompanying figure from the beginning of 2006 through the end of 2015.

0

P(1, 1) y = "x 1

1

x

y=x

390 310

230

x

In Exercises 41–44, determine whether the piecewise-defined function is differentiable at x  0. ⎪⎧ 2 x − 1, x ≥ 0 41. f ( x ) = ⎪⎨ 2 ⎪⎪ x + 2 x + 7, x < 0 ⎪⎩ ⎧⎪ x 2 3 , x ≥ 0 42. g( x ) = ⎪⎨ 1 3 ⎪⎪ x , x < 0 ⎪⎩

Differentiability and Continuity on an Interval 2007

2009

2011

2013

2015

t

Each figure in Exercises 45–50 shows the graph of a function over a closed interval D. At what domain points does the function appear to be a. differentiable?

a. During what years did home prices decrease? increase?

b. continuous but not differentiable?

b. Estimate home prices at the end of

c. neither continuous nor differentiable?

i) 2007

1

y = 1x

⎪⎧ 2 x + tan x , x ≥ 0 43. f ( x ) = ⎪⎨ 2 ⎪⎪ x , x < 0 ⎪⎩ ⎧⎪ 2 x − x 3 − 1, x ≥ 0 ⎪⎪ 44. g( x ) = ⎨ ⎪⎪ x − 1 , x < 0 x +1 ⎪⎪⎩

P

150

y = 2x − 1

t

a. Estimate the rate of temperature change at the times

y = f (x) P(1, 1)

ii) 2012

iii) 2015

Give reasons for your answers.

150

Chapter 3 Derivatives

45.

y 2

46. y = f (x) D: −3 ≤ x ≤ 2

1 0 1

2

x

−2 −1

0

−1

−1

−2

−2

47.

y = f (x) D: −2 ≤ x ≤ 3

2

1 −3 −2 −1

y

48.

y

1

2

3

x

b. Show that

y

is differentiable at x = 0 and find f ′(0).

y = f (x) D: −2 ≤ x ≤ 3

T 61. Graph y = 1 ( 2 x ) in a window that has 0 ≤ x ≤ 2. Then, on the same screen, graph

3 −3 −2 −1 0 −1

1

2

3

−2 −1

1

0

y 4

3

x

y = f (x) D: −3 ≤ x ≤ 3

1

2

x

−3 −2 −1 0

1 2

3

x

Theory and Examples

In Exercises 51–54, a. Find the derivative f ′( x ) of the given function y = f ( x ). b. Graph y = f ( x ) and y = f ′( x ) side by side using separate sets of coordinate axes, and answer the following questions. c. For what values of x, if any, is f ′ positive? Zero? Negative? d. Over what intervals of x-values, if any, does the function y = f ( x ) increase as x increases? Decrease as x increases? How is this related to what you found in part (c)? (We will say more about this relationship in Section 4.3.) −x 2

52. y = −1 x

53. y = x 3 3

54. y = x 4 4

51. y =

for h = 1, 0.5, 0.1. Then try h = −1, − 0.5, − 0.1. Explain what is going on.

y =

(x

+ h )3 − x 3 h

for h = 2, 1, 0.2. Then try h = −2, −1, − 0.2. Explain what is going on.

2

1

0

2

x

T 62. Graph y = 3 x 2 in a window that has −2 ≤ x ≤ 2, 0 ≤ y ≤ 3. Then, on the same screen, graph

50.

y y = f (x) D: −1 ≤ x ≤ 2

−1

x+h − h

y =

1

−2

49.

2

x

60. a. Let f ( x ) be a function satisfying f ( x ) ≤ x 2 for −1 ≤ x ≤ 1. Show that f is differentiable at x = 0 and find f ′(0). ⎪⎧⎪ 2 1 x sin , x ≠ 0 x f ( x ) = ⎪⎨ ⎪⎪ x = 0 ⎪⎪⎩ 0,

y = f (x) D: −3 ≤ x ≤ 3 1

59. Limit of a quotient Suppose that functions g(t ) and h(t ) are defined for all values of t and g(0) = h(0) = 0. Can lim g (t ) h (t ) exist? If it does exist, must it equal zero? Give t→0 reasons for your answers.

55. Tangent line to a parabola Does the parabola y = 2 x 2 − 13 x + 5 have a tangent line whose slope is −1? If so, find an equation for the line and the point of tangency. If not, why not?

63. Derivative of y = x Graph the derivative of f ( x ) = x . Then graph y = ( x − 0 ) ( x − 0 ) = x x . What can you conclude? T 64. Weierstrass’s nowhere differentiable continuous function The sum of the first eight terms of the Weierstrass function ∞ f ( x ) = ∑ n = 0 ( 2 3 ) n cos( 9 n π x ) is g( x ) = cos( π x ) + ( 2 3 )1 cos( 9π x ) + ( 2 3 ) 2 cos( 9 2 π x ) + ( 2 3 )3 cos( 9 3 π x ) +  + ( 2 3 ) 7 cos( 9 7 π x ). Graph this sum. Zoom in several times. How wiggly and bumpy is this graph? Specify a viewing window in which the displayed portion of the graph is smooth. COMPUTER EXPLORATIONS

Use a CAS to perform the following steps for the functions in Exercises 65–70. a. Plot y = f ( x ) to see that function’s global behavior. b. Define the difference quotient q at a general point x, with general step size h.

56. Tangent line to y = x Does any tangent line to the curve y = x cross the x-axis at x = −1? If so, find an equation for the line and the point of tangency. If not, why not?

c. Take the limit as h → 0. What formula does this give?

57. Derivative of − f Does knowing that a function f ( x ) is differentiable at x = x 0 tell you anything about the differentiability of the function − f at x = x 0? Give reasons for your answer.

e. Substitute various values for x larger and smaller than x 0 into the formula obtained in part (c). Do the numbers make sense with your picture?

58. Derivative of multiples Does knowing that a function g(t ) is differentiable at t = 7 tell you anything about the differentiability of the function 3g at t = 7? Give reasons for your answer.

f. Graph the formula obtained in part (c). What does it mean when its values are negative? Zero? Positive? Does this make sense with your plot from part (a)? Give reasons for your answer.

d. Substitute the value x = x 0 and plot the function y = f ( x ) together with its tangent line at that point.

3.3  Differentiation Rules

x −1 , x 0 = −1 3x 2 + 1 69. f ( x )  sin 2 x , x 0  Q 2

65. f ( x ) = x 3 + x 2 − x , x 0 = 1 66. f ( x ) = x

68. f ( x ) =

+ x , x0 = 1 4x , x0 = 2 67. f ( x ) = 2 x +1 13

151

23

70. f ( x )  x 2 cos x , x 0  Q 4

3.3 Differentiation Rules This section introduces several rules that allow us to differentiate constant functions, power functions, polynomials, exponential functions, rational functions, and certain combinations of them, simply and directly, without having to take limits each time.

Powers, Multiples, Sums, and Differences A basic rule of differentiation is that the derivative of every constant function is zero. Derivative of a Constant Function

y

If f has the constant value f ( x )  c, then

c

(x, c)

0

x

(x + h, c)

df d  (c)  0. dx dx

y=c

h x+h

x

The rule ( d dx )(c) = 0 is another way to say that the values of constant functions never change and that the slope of a horizontal line is zero at every point.

FIGURE 3.10

Proof We apply the definition of the derivative to f ( x )  c, the function whose outputs have the constant value c (Figure 3.10). At every value of x, we find that f ′( x ) = lim

h→ 0

f ( x + h ) − f (x) c−c = lim = lim 0 = 0. h→ 0 h→ 0 h h

We now consider powers of x. From Section 3.1, we know that

( )

d 1 1 = − 2 , or dx x x

d −1 ( x ) = −x − 2 . dx

From Example 2 of the last section we also know that d ( x ) = 1 , or dx 2 x

d 12 1 ( x ) = x −1 2 . dx 2

These two examples illustrate a general rule for differentiating a power x n . We first prove the rule when n is a positive integer.

Derivative of a Positive Integer Power

If n is a positive integer, then d n x = nx n −1. dx

HISTORICAL BIOGRAPHY Richard Courant (1888–1972) Courant obtained a PhD from Göttingen in 1910. He founded Göttingen’s Mathematics Institute and was its director from 1920 until 1933. His research work focused on mathematical physics. To know more, visit the companion Website.

Proof of the Positive Integer Power Rule zn

−

xn

= (z −

x )( z n −1

+

z n−2  x

The formula + " + z x n −2 + x n −1 )

can be verified by multiplying out the right-hand side. Then, from the alternative formula for the definition of the derivative, f (z) − f ( x ) zn − xn = lim z→ x z→ x z − x z−x n 1 n 2 − − = lim ( z + z x + " + zx n −2 + x n −1 )

f ′( x ) = lim z→ x

= nx n −1.

n terms

152

Chapter 3 Derivatives

The Power Rule is actually valid for all real numbers n, not just for positive integers. We have seen examples for a negative integer and fractional power, but n could be an irrational number as well. Here we state the general version of the rule, but postpone its proof until Section 3.8.

Power Rule (General Version)

If n is any real number, then d n x = nx n −1, dx for all x where the powers x n and x n −1 are defined.

EXAMPLE 1

(b) x 2 3

(a) x 3

Applying the Power Rule Subtract 1 from the exponent and multiply the result by the original exponent.

Differentiate the following powers of x. (c) x

2

(d)

1 x4

(e) x −4 3

(f )

x 2+ π

Solution

(a) (b) (c) (d) (e) (f )

d 3 ( x ) = 3 x 3−1 = 3 x 2 dx d 23 2 2 ( x ) = x ( 2 3)−1 = x −1 3 dx 3 3 d ( x 2 ) = 2 x 2 −1 dx d 1 d −4 4 = ( x ) = −4 x −4 −1 = −4 x −5 = − 5 dx x 4 dx x d −4 3 4 4 (x ) = − x −( 4 3)−1 = − x −7 3 dx 3 3 d d 1+ ( π 2 ) π 1+( π 2 )−1 1 = (2 + π) x π x (x ) = 1+ ( x 2 + π ) = dx dx 2 2

( )

(

)

The next rule says that when a differentiable function is multiplied by a constant, its derivative is multiplied by the same constant.

Derivative Constant Multiple Rule

If u is a differentiable function of x, and c is a constant, then d du (cu) = c . dx dx

Proof cu ( x + h ) − cu( x ) d cu = lim h→ 0 dx h u ( x + h ) − u( x ) = c lim h→ 0 h du =c dx

Derivative definition with  f ( x ) = cu ( x ) Constant Multiple Rule for Limits u  is differentiable.

3.3  Differentiation Rules

y

EXAMPLE 2

y = 3x 2

(a) The derivative formula

Slope = 3(2x) = 6x Slope = 6(1) = 6

(1, 3)

3

y = x2 2

d (3 x 2 ) = 3 ⋅ 2 x = 6 x dx says that if we rescale the graph of y  x 2 by multiplying each y-coordinate by 3, then we multiply the slope at each point by 3 (Figure 3.11). (b) Negative of a function The derivative of the negative of a differentiable function u is the negative of the function’s derivative. The Constant Multiple Rule with c = −1 gives d d( d du (−u) = −1 ⋅ u ) = −1 ⋅ (u) = − . dx dx dx dx

Slope = 2x Slope = 2(1) = 2 (1, 1)

1

0

1

153

2

x

The graphs of y  x 2 and y  Tripling the y-coordinate triples the slope (Example 2).

The next rule says that the derivative of the sum of two differentiable functions is the sum of their derivatives.

FIGURE 3.11

3x 2 .

Derivative Sum Rule

If u and V are differentiable functions of x, then their sum u V is differentiable at every point where u and V are both differentiable. At such points, d( du dV . u + V) = + dx dx dx

Denoting Functions by u and V The functions we are working with when we need a differentiation formula are likely to be denoted by letters like f and g. We do not want to use these same letters when stating general differentiation rules, so instead we use letters like u and V that are not likely to be already in use.

Proof

We apply the definition of the derivative to f ( x ) = u( x ) + V( x ):

[ u ( x + h ) + V ( x + h ) ] − [ u( x ) + V( x ) ] d[ u( x ) + V( x ) ] = lim h→ 0 dx h V ( x + h ) − V( x ) ⎤ u ( x + h ) − u( x ) = lim ⎡⎢ + ⎥⎦ h→ 0 ⎣ h h V ( x + h ) − V( x ) u ( x + h ) − u( x ) du dV = lim + lim = + . h→ 0 h→ 0 h h dx dx

Combining the Sum Rule with the Constant Multiple Rule gives the Difference Rule, which says that the derivative of a difference of differentiable functions is the difference of their derivatives: d( d[ du ( ) d V du dV u − V) = u + ( − 1) V ] = + −1 = − . dx dx dx dx dx dx The Sum Rule also extends to finite sums of more than two functions. If u1 , u 2 , !, u n are differentiable at x, then so is u1 u 2 " u n , and du n du1 du 2 d (u + u2 + " + un ) = + +"+ . dx 1 dx dx dx A proof by mathematical induction for any finite number of terms is given in Appendix A.3.

154

Chapter 3 Derivatives

EXAMPLE 3

Solution

Find the derivative of the polynomial y = x 3 +

(

)

dy d 3 d 4 2 d d = x + x − (5 x ) + (1) dx dx dx 3 dx dx 8 4 = 3x 2 + ⋅ 2 x − 5 + 0 = 3x 2 + x − 5 3 3

4 2 x − 5 x + 1. 3

Sum and Difference Rules

We can differentiate any polynomial term by term, the way we differentiated the polynomial in Example 3. All polynomials are differentiable at all values of x. EXAMPLE 4 If so, where?

Does the curve y = x 4 − 2 x 2 + 2 have any horizontal tangent lines?

Solution The horizontal tangent lines, if any, occur where the slope dy dx is zero. We have y

y=

x4

−

2x 2

+2

dy d( 4 = x − 2 x 2 + 2 ) = 4 x 3 − 4 x. dx dx

(0, 2)

(−1, 1) −1

1

0

FIGURE 3.12

Now solve the equation

4x 3 − 4x = 0

(1, 1)

1

dy  0 for  x : dx

4 x ( x 2 − 1) = 0 x = 0, 1, −1.

x

The curve in Example 4 and its horizontal tangent lines.

The curve y = x 4 − 2 x 2 + 2 has horizontal tangent lines at x  0, 1, and 1. The corresponding points on the curve are ( 0, 2 ), (1, 1), and (−1, 1). See Figure 3.12.

Derivatives of Exponential Functions We briefly reviewed exponential functions in Section 1.4. When we apply the definition of the derivative to f ( x )  a x , we get d x a x+h − a x (a ) = lim h→ 0 dx h ax ⋅ ah − ax = lim h→ 0 h ah − 1 = lim a x ⋅ h→ 0 h h −1 a = a x ⋅ lim h→ 0 h h a −1 = lim ⋅ a x. h→ 0 h 

(

Derivative definition a x+h = a x ⋅ a h Factoring out  a x a x is constant as h → 0.

)

(1)

a fixed number L

Thus we see that the derivative of a x is a constant multiple L of a x . The constant L is a limit we have not encountered before. Note, however, that it equals the derivative of f ( x )  a x at x  0: f ′(0) = lim

h→ 0

ah − a0 ah − 1 = lim = L. h → 0 h h

3.3  Differentiation Rules

y

a = 3 a = e a = 2.5

a=2

L = 1.0 1.1 0.92

h y = a − 1, a > 0 h

0.69

The limit L is therefore the slope of the graph of f ( x )  a x where it crosses the y-axis. In Chapter 7, where we carefully develop the logarithmic and exponential functions, we prove that the limit L exists and has the value ln a. For now we investigate values of L by graphing the function y = ( a h − 1) h and studying its behavior as h approaches 0. Figure 3.13 shows the graphs of y = ( a h − 1) h for four different values of a. The limit L is approximately 0.69 if a  2, about 0.92 if a  2.5, and about 1.1 if a  3. It appears that the value of L is 1 at some number a chosen between 2.5 and 3. That number is given by a = e ≈ 2.718281828. With this choice of base we obtain the natural exponential function f ( x )  e x as in Section 1.4, and see that it satisfies the property eh − 1 =1 h→ 0 h

f ′(0) = lim

h

0

FIGURE 3.13 The position of the curve y = ( a h − 1) h , a > 0, varies continu-

ously with a. The limit L of y as h l 0 changes with different values of a. The number for which L  1 as h l 0 is the number e between a  2 and a  3.

155

(2)

because it is the exponential function whose graph has slope 1 when it crosses the y-axis. That the limit is 1 implies an important relationship between the natural exponential function e x and its derivative:

(

)

d x eh − 1 (e ) = lim ⋅ ex h→ 0 dx h = 1 ⋅ e x = e x.

Eq. (1) with  a = e Eq. (2)

Therefore the natural exponential function is its own derivative.

Derivative of the Natural Exponential Function

d x (e )  e x dx

EXAMPLE 5 Find an equation for a line that is tangent to the graph of y  e x and goes through the origin.

y 6

y=

ex

4 (a, e a )

2 −1

a

x

FIGURE 3.14 The line through the origin is tangent to the graph of y  e x when a  1 (Example 5).

Solution Since the line passes through the origin, its equation is of the form y  mx, where m is the slope. If it is tangent to the graph at the point ( a, e a ), the slope is m = ( e a − 0 ) ( a − 0 ). The slope of the natural exponential at x  a is e a . Because these slopes are the same, we then have that e a  e a a. It follows that a  1 and m  e, so the equation of the tangent line is y  ex. See Figure 3.14.

We might ask if there are functions other than the natural exponential function that are their own derivatives. The answer is that the only functions that satisfy the property that f ′( x ) = f ( x ) are functions that are constant multiples of the natural exponential function, f ( x ) = c ⋅ e x, c any constant. We prove this fact in Section 7.2. Note from the Constant Multiple Rule that indeed d( d x c ⋅ ex ) = c ⋅ (e ) = c ⋅ e x . dx dx

Products and Quotients While the derivative of the sum of two functions is the sum of their derivatives, the derivative of the product of two functions is not the product of their derivatives. For instance, d( d 2 x ⋅ x) = (x ) = 2x, dx dx

while

d d (x) ⋅ ( x ) = 1 ⋅ 1 = 1. dx dx

The derivative of a product of two functions is the sum of two products, as we now explain.

156

Chapter 3 Derivatives

Derivative Product Rule

If u and υ are differentiable at x, then so is their product uυ, and d dυ du (uυ) = u + υ. dx dx dx

The derivative of the product uυ is u times the derivative of υ plus the derivative of u times υ. In prime notation, ( uυ ) ′ = uυ′ + u′υ. In function notation, d [ f ( x ) g( x ) ] = f ( x ) g′( x ) + f ′( x ) g( x ), or ( fg)′ = fg′ + f ′ g. dx

EXAMPLE 6

Find the derivative of (a) y =

(3)

1( 2 x + e x ), (b) y = e 2 x. x

Solution

(a) We apply the Product Rule with u = 1 x and υ = x 2 + e x :

(

Picturing the Product Rule Suppose u( x ) and υ( x ) are positive and increase when x increases, and h > 0.

)

1 1 d ⎡ 1( 2 x + e x )⎤⎥ = ( 2 x + e x ) + − 2 ( x 2 + e x ) ⎦ dx ⎢⎣ x x x ex ex =2+ −1− 2 x x x e = 1 + ( x − 1)   2 . x (b)

d( ) dυ du uυ = u + υ and dx dx dx

( ) = − x1

d 1 dx x

2

d 2x d( x d x d x (e ) = e ⋅ ex ) = ex ⋅ (e ) + (e ) ⋅ e x = 2e x ⋅ e x = 2e 2 x dx dx dx dx

y(x + h) Δy

u(x + h) Δy

Proof of the Derivative Product Rule

y(x) u(x)y(x)

0

Δu y(x)

Δu u(x) u(x + h)

Then the change in the product uυ is the difference in areas of the larger and smaller “boxes,” which is the sum of the areas of the upper and right-hand reddish-shaded rectangles. That is, Δ( uυ ) = u ( x + h ) υ ( x + h ) − u( x )υ( x ) = u ( x + h )Δυ + Δuυ( x ). Division by h gives Δ( uυ ) Δυ Δu = u( x + h ) + υ( x ). h h h The limit as h → 0 + yields the Product Rule.

u ( x + h ) υ ( x + h ) − u( x )υ( x ) d (uυ) = lim h→ 0 dx h To change this fraction into an equivalent one that contains difference quotients for the derivatives of u and υ, we subtract and add u ( x + h ) υ( x ) in the numerator: u ( x + h ) υ ( x + h ) − u ( x + h ) υ( x ) + u ( x + h ) υ( x ) − u( x ) υ ( x ) d (uυ) = lim h→ 0 dx h ( x + h ) − υ( x ) ( x + h ) − u( x ) υ u ⎡ υ( x ) ⎤⎥ = lim ⎢ u ( x + h ) + h→ 0 ⎣ ⎦ h h ( x + h ) − υ( x ) ( x + h ) − u( x ) υ u = lim u ( x + h ) ⋅ lim + lim ⋅ υ( x ). h→ 0 h→ 0 h→ 0 h h As h approaches zero, u ( x + h ) approaches u( x ) because u, being differentiable at x, is continuous at x. The two fractions approach the values of d υ dx at x and du dx at x. Therefore, d dυ du (uυ) = u + υ. dx dx dx

3.3  Differentiation Rules

157

The derivative of the quotient of two functions is given by the Quotient Rule.

Derivative Quotient Rule

If u and υ are differentiable at x and if υ( x ) ≠ 0, then the quotient u υ is differentiable at x, and

( )

d u = dx υ

υ

du dυ −u dx dx . υ2

In function notation, g( x ) f ′( x ) − f ( x ) g′( x ) d ⎡ f (x) ⎤ . ⎢ ⎥ = ⎢ ⎥ dx ⎣ g( x ) ⎦ g 2 (x) Find the derivative of (a) y =

EXAMPLE 7

t2 − 1 , (b) y = e − x . t3 + 1

Solution

(a) We apply the Quotient Rule with u = t 2 − 1 and υ = t 3 + 1: dy ( t 3 + 1) ⋅ 2t − ( t 2 − 1) ⋅ 3t 2 = 2 dt ( t 3 + 1)

υ( du dt ) − u( d υ dt ) d u = dt υ υ2

( )

2t 4 + 2t − 3t 4 + 3t 2 2 ( t 3 + 1) −t 4 + 3t 2 + 2t = . 2 ( t 3 + 1) =

(b)

( )= e

d −x d 1 (e ) =   dx dx e x

x

⋅ 0 − 1 ⋅ ex −1 = x = −e − x (e x ) 2 e

Proof of the Derivative Quotient Rule u( x ) u( x + h ) − υ ( x + h ) υ( x ) d u = lim h→ 0 dx υ h υ( x ) u ( x + h ) − u( x ) υ ( x + h ) = lim h→ 0 h υ ( x + h ) υ( x )

( )

To change the last fraction into an equivalent one that contains the difference quotients for the derivatives of u and υ, we subtract and add υ ( x ) u ( x ) in the numerator. We then get υ( x ) u ( x + h ) − υ( x ) u( x ) + υ( x ) u( x ) − u( x ) υ ( x + h ) d u = lim h→ 0 dx υ h υ ( x + h ) υ( x )

( )

= lim

h→ 0

υ( x )

u ( x + h ) − u( x ) υ ( x + h ) − υ( x ) − u( x ) h h . υ ( x + h ) υ( x )

Taking the limits in the numerator and denominator now gives the Quotient Rule. Exercise 76 outlines another proof. The choice of which rules to use in solving a differentiation problem can make a difference in how much work you have to do. Here is an example.

158

Chapter 3 Derivatives

EXAMPLE 8

Find the derivative of y =

HISTORICAL BIOGRAPHY Maria Gaetana Agnesi (1718–1799) Agnesi was a well-published scientist by age 20 and an honorary faculty member of the University of Bologna by age 30. Today, Agnesi is remembered chiefly for a bellshaped curve called the “Witch of Agnesi.”

(x

− 1)( x 2 − 2 x ) . x4

Solution Using the Quotient Rule here will result in a complicated expression with many terms. Instead, use some algebra to simplify the expression. First expand the numerator and divide by x 4 :

y =

(x

− 1)( x 2 − 2 x ) x 3 − 3x 2 + 2 x = = x −1 − 3 x −2 + 2 x −3 . 4 x x4

Then use the Sum, Constant Multiple, and Power Rules: dy = −x − 2 − 3 ( − 2 ) x − 3 + 2 ( − 3 ) x − 4 dx 1 6 6 = − 2 + 3 − 4. x x x

To know more, visit the companion Website.

Second- and Higher-Order Derivatives If y = f ( x ) is a differentiable function, then its derivative f ′( x ) is also a function. If f ′ is also differentiable, then we can differentiate f ′ to get a new function of x denoted by f ′′. So f ′′ = ( f ′)′. The function f ′′ is called the second derivative of f because it is the derivative of the first derivative. It is written in several ways: f ′′( x ) =

( )

d 2y dy′ d dy = y′′ = D 2 ( f )( x ) = D x2 f ( x ). = = dx 2 dx dx dx

The symbol D 2 means that the operation of differentiation is performed twice. If y = x 6, then y′ = 6 x 5 and we have y″ =

How to Read the Symbols for Derivatives y′

“y prime”

y′′

“y double prime”

d 2y dx 2

“d squared y by dx squared”

y′′′

“y triple prime”

y (n)

“y super n”

dny dx n Dn

“d to the n of y by dx to the n” “d to the n”

dy′ d (6 x 5 ) = 30 x 4. = dx dx

Thus D 2 ( x 6 ) = 30 x 4 . If y ′′ is differentiable, its derivative, y′′′ = dy′′ dx = d 3 y dx 3, is the third derivative of y with respect to x. The names continue as you imagine, with y (n) =

d ny d ( n −1) y = = Dny dx dx n

denoting the nth derivative of y with respect to x for any positive integer n. We can interpret the second derivative as the rate of change of the slope of the tangent line to the graph of y = f ( x ) at each point. You will see in the next chapter that the second derivative reveals whether the graph bends upward or downward from the tangent line as we move off the point of tangency. In the next section, we interpret both the second and third derivatives in terms of motion along a straight line. EXAMPLE 9

The first four derivatives of y = x 3 − 3 x 2 + 2 are First derivative:

y ′ = 3x 2 − 6 x

Second derivative:

y ′′ = 6 x − 6

Third derivative:

y ′′′ = 6

Fourth derivative:

y (4) = 0.

All polynomial functions have derivatives of all orders. In this example, the fifth and later derivatives are all zero.

3.3  Differentiation Rules

EXERCISES

3.3

Derivative Calculations

49. w =

In Exercises 1–12, find the first and second derivatives. 1. y = −x 2 + 3 2. y = x 2 + x + 8 3. s =

−

5t 3

4. w =

3t 5

4x 3 5. y = − x + 2e x 3 1 7. w = 3z −2 − z

3z 7

−

7z 3

+

10. y = 4 − 2 x − x −3

1 5 − 3s 2 2s

(

12. r =

)

16. y = ( 1 + x 2 )   ( x 3 4 − x −3 )

x2 − 4 x + 0.5

21. υ = ( 1 − t )( 1 + t 2 )−1 s −1 s +1

23. f ( s ) =

22. w = ( 2 x − 7 )−1 ( x + 5 ) 24. u =

25. υ =

1+ x−4 x x

27. y =

1 28. y = ( x 2 − 1 )( x 2 + x + 1 )

5x + 1 2 x

26. r = 2

( 1θ + θ )

(x (x

+ 1 )( x + 2 ) − 1 )( x − 2 )

x 2 + 3e x 2e x − x

29. y = 2e −x + e 3 x

30. y =

31. y =   x 3 e x

32. w =   re −r 34. y = x

−3 5

35. s = 2t 3 2 + 3e 2

36. w =

1

37. y =

7

38. y =

39. r =

es s

33. y = x

94

+

e −2 x

x2 − xe

z 1.4 3

40. r = e θ

52. w = e z ( z − 1 )( z 2 + 1 )

u(0) = 5, u′(0) = −3, υ(0) = −1, υ′(0) = 2. Find the values of the following derivatives at x = 0. d (uυ) dx

b.

+π π + z

+ θ −π 2

c.

( )

d υ dx u

d.

d( 7υ − 2u ) dx

54. Suppose u and υ are differentiable functions of x and that u(1) = 2, u′(1) = 0, υ(1) = 5, υ′(1) = −1. Find the values of the following derivatives at x = 1. a.

d (uυ) dx

b.

( )

d u dx υ

c.

( )

d υ dx u

d.

d( 7υ − 2u ) dx

Slopes and Tangent Lines

c. Tangent lines having specified slope Find equations for the tangent lines to the curve at the points where the slope of the curve is 8. 56. a. Horizontal tangent lines Find equations for the horizontal tangent lines to the curve y = x 3 − 3 x − 2. Also find equations for the lines that are perpendicular to these tangent lines at the points of tangency. b. Smallest slope What is the smallest slope on the curve? At what point on the curve does the curve have this slope? Find an equation for the line that is perpendicular to the curve’s tangent line at this point.

y y = 24x x +1

x 9.6 + 2e 1.3 2

( )

d u dx υ

57. Find the tangent lines to Newton’s serpentine (graphed here) at the origin and the point ( 1, 2 ).

32

( θ1

q2 + 3 ( q − 1)3 + ( q + 1)3

b. Smallest slope What is the smallest slope on the curve? At what point on the curve does the curve have this slope?

t2 − 1 2 t +t−2

20. f (t ) =

50. p =

)

55. a. Normal line to a curve Find an equation for the line perpendicular to the tangent line to the curve y = x 3 − 4 x + 1 at the point ( 2, 1 ).

Find the derivatives of the functions in Exercises 17–40. 2x + 5 4 − 3x 17. y = 18. z = 3x − 2 3x 2 + x 19. g( x ) =

(

53. Suppose u and υ are functions of x that are differentiable at x = 0 and that

a.

12 4 1 − 3 + 4 θ θ θ In Exercises 13–16, find y ′ (a) by applying the Product Rule and (b) by multiplying the factors to produce a sum of simpler terms to differentiate. 13. y = ( 3 − x 2 )( x 3 − x + 1 ) 14. y = ( 2 x + 3 )( 5 x 2 − 4 x ) 1 15. y = ( x 2 + 1 ) x + 5 + x

( 1 +3z3z ) 3 − z

51. w = 3z 2 e 2 z 21z 2

x3 x2 6. y = + + e −x 3 2 4 8. s = −2t −1 + 2 t

9. y = 6 x 2 − 10 x − 5 x −2 11. r =

159

)

Find the derivatives of all orders of the functions in Exercises 41–44. x4 3 x5 41. y = − x2 − x 42. y = 2 2 120 43. y = ( x − 1 )( x + 2 )( x + 3 ) 44. y = ( 4 x 2 + 3 )( 2 − x ) x Find the first and second derivatives of the functions in Exercises 45–52. x3 + 7 t 2 + 5t − 1 45. y = 46. s = x t2 2 2 ( θ − 1 )( θ + θ + 1 ) ( x + x )( x 2 − x + 1 ) 47. r = 48. u = 3 x4 θ

(1, 2) 2 1 0

1 2 3 4

x

58. Find the tangent line to the Witch of Agnesi (graphed here) at the point ( 2, 1 ). y y= 2 1 0

8 x2 + 4 (2, 1)

1 2 3

x

160

Chapter 3 Derivatives

59. Quadratic tangent to identity function The curve y = ax 2 + bx + c passes through the point ( 1, 2 ) and is tangent to the line y = x at the origin. Find a, b, and c. 60. Quadratics having a common tangent line The curves y = x 2 + ax + b and y = cx − x 2 have a common tangent line at the point (1, 0). Find a, b, and c. 61. Find all points ( x , y ) on the graph of f ( x ) = 3 x 2 − 4 x with tangent lines parallel to the line y = 8 x + 5. 62. Find all points ( x , y ) on the graph of g( x ) = 13 x 3 − 32 x 2 + 1 with tangent lines parallel to the line 8 x − 2 y = 1. 63. Find all points ( x , y ) on the graph of y = x ( x − 2 ) with tangent lines perpendicular to the line y = 2 x + 3. 64. Find all points ( x , y ) on the graph of f ( x ) = x 2 with tangent lines passing through the point ( 3, 8 ). y 10

f (x) = x 2

6 (x, y) 2 2

4

P ( x ) = a n x n + a n−1 x n−1 +  + a 2 x 2 + a1 x + a 0 , where a n ≠ 0. Find P ′( x ). 74. The body’s reaction to medicine The reaction of the body to a dose of medicine can sometimes be represented by an equation of the form C M R = M2 − , 2 3 where C is a positive constant and M is the amount of medicine absorbed in the blood. If the reaction is a change in blood pressure, R is measured in millimeters of mercury. If the reaction is a change in temperature, R is measured in degrees, and so on. Find dR dM. This derivative, as a function of M, is called the sensitivity of the body to the medicine. In Section 4.6, we will see how to find the amount of medicine to which the body is most sensitive.

(

(3, 8)

−2

71. Find the value of a that makes the following function differentiable for all x-values. ⎧⎪ ax , if  x < 0 g( x ) = ⎪⎨ 2 ⎪⎪ x − 3 x , if  x ≥ 0 ⎪⎩ 72. Find the values of a and b that make the following function differentiable for all x-values. ⎧⎪ ax + b, x > −1 f ( x ) = ⎪⎨ 2 ⎪⎪ bx − 3, x ≤ −1 ⎪⎩ 73. The general polynomial of degree n has the form

x

65. Assume that functions f and g are differentiable with f (1) = 2, f ′(1) = −3, g(1) = 4, and g′(1) = −2. Find the equation of the line tangent to the graph of F ( x ) = f ( x ) g( x ) at x = 1. 66. Assume that functions f and g are differentiable with f (2) = 3, f ′(2) = −1, g(2) = −4, and g′(2) = 1. Find an equation of the line perpendicular to the line tangent to the graph of f (x) + 3 F (x) = at x = 2. x − g( x ) 67. a. Find an equation for the line that is tangent to the curve y = x 3 − x at the point ( −1, 0 ). T b. Graph the curve and tangent line together. The tangent line intersects the curve at another point. Use Zoom and Trace to estimate the point’s coordinates. T c. Confirm your estimates of the coordinates of the second intersection point by solving the equations for the curve and tangent line simultaneously.

)

75. Suppose that the function υ in the Derivative Product Rule has a constant value c. What does the Derivative Product Rule then say? What does this say about the Derivative Constant Multiple Rule? 76. The Reciprocal Rule a. The Reciprocal Rule says that at any point where the function υ( x ) is differentiable and different from zero,

( )

d 1 1 dυ . = − 2 dx υ υ dx Show that the Reciprocal Rule is a special case of the Derivative Quotient Rule. b. Show that the Reciprocal Rule and the Derivative Product Rule together imply the Derivative Quotient Rule. 77. Generalizing the Product Rule The Derivative Product Rule gives the formula d dυ du (uυ) = u + υ dx dx dx

68. a. Find an equation for the line that is tangent to the curve y = x 3 − 6 x 2 + 5 x at the origin.

for the derivative of the product uυ of two differentiable functions of x.

T b. Graph the curve and tangent line together. The tangent line intersects the curve at another point. Use Zoom and Trace to estimate the point’s coordinates.

a. What is the analogous formula for the derivative of the product uυ w of three differentiable functions of x?

T c. Confirm your estimates of the coordinates of the second intersection point by solving the equations for the curve and tangent line simultaneously. Theory and Examples

For Exercises 69 and 70, evaluate each limit by first converting each to a derivative at a particular x-value. 69. lim

x →1

x 50 − 1 x −1

x2 9 − 1 x →−1 x + 1

70. lim

b. What is the formula for the derivative of the product u1u 2 u 3u 4 of four differentiable functions of x? c. What is the formula for the derivative of a product u1u 2u 3  u n of a finite number n of differentiable functions of x? 78. Power Rule for negative integers Use the Derivative Quotient Rule to prove the Power Rule for negative integers, that is, d −m ( x ) = −mx −m −1 dx where m is a positive integer.

3.4  The Derivative as a Rate of Change

79. Cylinder pressure If gas in a cylinder is maintained at a constant temperature T, the pressure P is related to the volume V by a formula of the form P =

nRT an 2 − 2 , V − nb V

161

80. The best quantity to order One of the formulas for inventory management says that the average weekly cost of ordering, paying for, and holding merchandise is hq km A(q) = + cm + , q 2

in which a, b, n, and R are constants. Find dP dV . (See accompanying figure.)

where q is the quantity you order when things run low (shoes, TVs, brooms, or whatever the item might be); k is the cost of placing an order (the same, no matter how often you order); c is the cost of one item (a constant); m is the number of items sold each week (a constant); and h is the weekly holding cost per item (a constant that takes into account things such as space, utilities, insurance, and security). Find dA dq and d 2 A dq 2 .

3.4 The Derivative as a Rate of Change In this section we study applications where derivatives model the rates at which things change. It is natural to think of a quantity changing with respect to time, but other variables can be treated in the same way. For example, an economist may want to study how the cost of producing steel varies with the number of tons produced, or an engineer may want to know how the power output of a generator varies with its temperature.

Instantaneous Rates of Change If we interpret the difference quotient ( f ( x + h ) − f ( x ) ) h as the average rate of change in f over the interval from x to x + h, we can interpret its limit as h → 0 as the instantaneous rate at which f is changing at the point x. This gives an important interpretation of the derivative.

DEFINITION

The instantaneous rate of change of f with respect to x at x 0 is

the derivative f ′( x 0 ) = lim

h→ 0

f ( x 0 + h ) − f (x 0 ) , h

provided the limit exists.

Thus, instantaneous rates are limits of average rates. It is conventional to use the word instantaneous even when x does not represent time. The word is, however, frequently omitted. When we say rate of change, we mean instantaneous rate of change. EXAMPLE 1

The area A of a circle is related to its diameter by the equation A =

π 2 D . 4

How fast does the area change with respect to the diameter when the diameter is 10 m?

162

Chapter 3 Derivatives

Solution The rate of change of the area with respect to the diameter is

dA Q QD . = ⋅ 2D = dD 4 2 When D  10 m, the area is changing with respect to the diameter at the rate of ( Q 2 )10 = 5Q m 2 m ≈ 15.71 m 2 m.

Motion Along a Line: Displacement, Velocity, Speed, Acceleration, and Jerk Suppose that an object (or body, considered as a whole mass) is moving along a coordinate line (an s-axis), usually horizontal or vertical, so that we know its position s on that line as a function of time t: s  f (t ). The displacement of the object over the time interval from t to t + Δt (Figure 3.15) is Position at time t …

Δs

and at time t + Δt s + Δs = f (t + Δt)

s = f(t)

Δs = f ( t + Δt ) − f (t ), s

and the average velocity of the object over that time interval is

FIGURE 3.15

The positions of a body moving along a coordinate line at time t and shortly later at time t + Δt. Here the coordinate line is horizontal.

s

V aV =

displacement f ( t + Δt ) − f ( t ) Δs . = = travel time Δt Δt

To find the body’s velocity at the exact instant t, we take the limit of the average velocity over the interval from t to t + Δt as %t approaches zero. This limit is the derivative of f with respect to t.

Velocity (instantaneous velocity) is the derivative of position with respect to time. If a body’s position at time t is s  f (t ), then the body’s velocity at time t is

DEFINITION s = f (t)

V(t ) =

ds > 0 dt

t

0 (a) s increasing: positive slope so moving upward s s = f (t) ds < 0 dt

t

0 (b) s decreasing: negative slope so moving downward

FIGURE 3.16 For motion s  f (t ) along a straight line (the vertical axis), V  ds dt is (a) positive when s increases and (b) negative when s decreases.

f ( t + Δt ) − f (t ) ds = lim . Δt → 0 dt Δt

Besides telling how fast an object is moving along the horizontal line in Figure 3.15, its velocity tells the direction of motion. When the object is moving forward (s increasing), the velocity is positive; when the object is moving backward (s decreasing), the velocity is negative. If the coordinate line is vertical, the object moves upward for positive velocity and downward for negative velocity. The blue curves in Figure 3.16 represent position along the line over time; they do not portray the path of motion, which lies along the vertical s-axis. If we drive to a friend’s house and back at 50 kmh, say, the speedometer will show 50 on the way over but it will not show 50 on the way back, even though our distance from home is decreasing. The speedometer always shows speed, which is the absolute value of velocity. Speed measures the rate of progress regardless of direction.

DEFINITION

Speed is the absolute value of velocity. Speed  V(t ) 

ds dt

EXAMPLE 2 Figure 3.17 shows the graph of the velocity V = f ′(t ) of a particle moving along a horizontal line as in Figure 3.15 (as opposed to the graph of a position function s  f (t ), such as in Figure 3.16). In the graph of the velocity function, it’s not the

163

3.4  The Derivative as a Rate of Change

y MOVES FORWARD

FORWARD AGAIN

(y > 0)

(y > 0)

Velocity y = f ′(t) Speeds up

Steady (y = const)

Slows down

Speeds up Stands still (y = 0)

0

1

2

3

4

5

6

7

t (s)

Greatest speed

Speeds up

Slows down

MOVES BACKWARD

(y < 0)

FIGURE 3.17 The velocity graph of a particle moving along a horizontal line, discussed in Example 2.

slope of the curve that tells us whether the particle is moving forward or backward along the line (which is not shown in the figure), but rather the sign of the velocity. Figure 3.17 shows that the particle moves forward for the first 3 s (when the velocity is positive), moves backward for the next 2 s (the velocity is negative), stands motionless for a full second, and then moves forward again. The particle is speeding up when its positive velocity increases during the first second, moves at a steady speed during the next second, and then slows down as the velocity decreases to zero during the third second. It stops for an instant at t  3 s (when the velocity is zero) and reverses direction as the velocity starts to become negative. The particle is now moving backward and gaining in speed until t  4 s, at which time it achieves its greatest speed during its backward motion. Continuing its backward motion at time t  4, the particle starts to slow down again until it finally stops at time t  5 (when the velocity is once again zero). The particle now remains motionless for one full second, and then moves forward again at t  6 s, speeding up during the final second of the forward motion indicated in the velocity graph. HISTORICAL BIOGRAPHY Bernard Bolzano (1781–1848) Bolzano was born in Prague, Czechoslovakia. He studied at the University of Prague, where he took courses in philosophy, physics, and mathematics. To know more, visit the companion Website.

The rate at which a body’s velocity changes is the body’s acceleration. The acceleration measures how quickly the body picks up or loses speed. In Chapter 12 we will study motion in the plane and in space, where acceleration of an object may also lead to a change in direction. A sudden change in acceleration is called a jerk. When a ride in a car or a bus is jerky, it is not that the accelerations involved are necessarily large but that the changes in acceleration are abrupt. DEFINITIONS Acceleration is the derivative of velocity with respect to time. If a body’s position at time t is s  f (t ), then the body’s acceleration at time t is

dV d 2s  . dt dt 2 Jerk is the derivative of acceleration with respect to time: a(t ) 

j (t ) 

da d 3s  3. dt dt

164

Chapter 3 Derivatives

Near the surface of Earth, all bodies fall with the same constant acceleration. Galileo’s experiments with free fall (see Section 2.1) lead to the equation s 

1 2 gt , 2

where s is the distance fallen and g is the acceleration due to Earth’s gravity. This equation holds in a vacuum, where there is no air resistance, and closely models the fall of dense, heavy objects, such as rocks or steel tools, for the first few seconds of their fall, before the effects of air resistance are significant. The value of g in the equation s = (1 2 ) gt 2 depends on the units used to measure t and s. With t in seconds (the usual unit), the value of g determined by measurement at sea level is approximately 9.8 m s 2 (meters per second squared) in metric units. (This gravitational constant depends on the distance from Earth’s center of mass, and is slightly lower on top of Mt. Everest, for example.) The jerk associated with the constant acceleration of gravity ( g = 9.8 m s 2 ) is zero: j 

d ( g)  0. dt

An object does not exhibit jerkiness during free fall.

t (seconds) t=0

s (meters)

t=1

5

0

10

EXAMPLE 3 Figure 3.18 shows the free fall of a heavy ball bearing released from rest at time t  0 s. (a) How many meters does the ball fall in the first 3 s? (b) What are its velocity, speed, and acceleration when t  3?

15 t=2

20 25

Solution

(a) The metric free-fall equation is s  4.9t 2. During the first 3 s, the ball falls s(3) = 4.9( 3 ) 2 = 44.1 m.

30 35

(b) At any time t, velocity is the derivative of position:

40 t=3

45

V(t ) = s′(t ) =

d( 4.9t 2 ) = 9.8t. dt

At t  3, the velocity is FIGURE 3.18

A ball bearing falling from rest (Example 3).

V(3)  29.4 m s in the downward (increasing s) direction. The speed at t  3 is speed  V(3)  29.4 m s. The acceleration at any time t is a(t ) = V′(t ) = s ′′(t ) = 9.8 m s 2. At t  3, the acceleration is 9.8 m s 2. EXAMPLE 4 A dynamite blast blows a heavy rock straight up with a launch velocity of 49 m s (176.4 km h) (Figure 3.19a). It reaches a height of s  = 49t − 4.9t 2 m after t seconds. (a) How high does the rock go? (b) What are the velocity and speed of the rock when it is 78.4 m above the ground on the way up? On the way down? (c) What is the acceleration of the rock at any time t during its flight (after the blast)? (d) When does the rock hit the ground again?

3.4  The Derivative as a Rate of Change

Solution

s

Height (m)

165

smax

y=0

78.4

t=?

(a) In the coordinate system we have chosen, s measures height from the ground up, so the velocity is positive on the way up and negative on the way down. The instant the rock is at its highest point is the one instant during the flight when the velocity is 0. To find the maximum height, all we need to do is to find when υ = 0 and evaluate s at this time. At any time t during the rock’s motion, its velocity is ds d = ( 49t − 4.9t 2 ) = 49 − 9.8t  m s. dt dt The velocity is zero when 49 − 9.8t = 0 or t = 5 s. υ =

The rock’s height at t = 5 s is s max = s ( 5 ) = 49( 5 ) − 4.9( 5 ) 2 = 245 − 122.5 = 122.5 m.

s=0 (a)

See Figure 3.19b.

s, y 122.5

(b) To find the rock’s velocity at 78.4 m on the way up and again on the way down, we first find the two values of t for which

s = 49t − 4.9t 2

s(t ) = 49t − 4.9t 2 = 78.4. To solve this equation, we write

49

0 −49

5

10

t

4.9t 2 − 49t + 78.4 = 0 4.9( t 2 − 10t + 16 ) = 0 (t

y = ds = 49 − 9.8t dt (b)

FIGURE 3.19

(a) The rock in Example 4. (b) The graphs of s and υ as functions of time; s is largest when υ = ds dt = 0. The graph of s is not the path of the rock: It is a plot of height versus time. The slope of the plot is the rock’s velocity, graphed here as a straight line.

− 2 )( t − 8 ) = 0 t = 2s, t = 8s.

The rock is 78.4 m above the ground 2 s after the explosion and again 8 s after the explosion. The rock’s velocities at these times are υ(2) = 49 − 9.8(2) = 49 − 19.6 = 29.4 m s. υ(8) = 49 − 9.8(8) = 49 − 78.4 = −29.4 m s. At both instants, the rock’s speed is 29.4 m/s. Since υ(2) > 0, the rock is moving upward (s is increasing) at t = 2 s; it is moving downward (s is decreasing) at t = 8 because υ(8) < 0. (c) At any time during its flight following the explosion, the rock’s acceleration is a constant a =

dv d (49 − 9.8t ) = −9.8 m s 2 . = dt dt

The acceleration is always downward and is the effect of gravity on the rock. As the rock rises, it slows down; as it falls, it speeds up. (d) The rock hits the ground at the positive time t for which s = 0. The equation 49t − 4.9t 2 = 0 factors to give 4.9t (10 − t) = 0, so it has solutions t = 0 and t = 10. At t = 0, the blast occurred and the rock was thrown upward. It returns to the ground 10 s later.

Derivatives in Economics and Biology Economists have a specialized vocabulary for rates of change and derivatives. They call them marginals. In a manufacturing operation, the cost of production c( x ) is a function of x, the number of units produced. The marginal cost of production is the rate of change of cost with respect to level of production, so it is dc dx.

166

Chapter 3 Derivatives

Cost y (dollars) Slope = marginal cost

y = c (x)

Suppose that c( x ) represents the dollars needed to produce x tons of steel in one week. It costs more to produce x + h tons per week, and the cost difference, divided by h, is the average cost of producing each additional ton: average cost of each of the additional c ( x + h ) − c( x ) = h  tons of steel produced. h

x+h x Production (tons/week)

0

The limit of this ratio as h → 0 is the marginal cost of producing more steel per week when the current weekly production is x tons (Figure 3.20):

x

FIGURE 3.20 Weekly steel production: c( x ) is the cost of producing x tons per week. The cost of producing an additional h tons is c ( x + h ) − c( x ).

c ( x + h ) − c( x ) dc = lim = marginal cost of production. h→ 0 dx h Sometimes the marginal cost of production is loosely defined to be the extra cost of producing one additional unit: c ( x + 1) − c ( x ) Δc = , Δx 1

y

which is approximated by the value of dc dx at x. This approximation is acceptable if the slope of the graph of c does not change quickly near x. Then the difference quotient will be close to its limit dc dx, which is the rise in the tangent line if Δx = 1 (Figure 3.21). The approximation often works well for large values of x. Economists often represent a total cost function by a cubic polynomial

y = c(x)

Δc

dc dx

c( x ) = α x 3 + β x 2 + γ x + δ , where δ represents fixed costs, such as rent, heat, equipment capitalization, and management costs. The other terms represent variable costs, such as the costs of raw materials, taxes, and labor. Fixed costs are independent of the number of units produced, whereas variable costs depend on the quantity produced. A cubic polynomial is usually adequate to capture the cost behavior on a realistic quantity interval.

Δx = 1

0

x

FIGURE 3.21

x+1

x

The marginal cost dc dx is approximately the extra cost Δc of producing Δx = 1 more unit.

EXAMPLE 5

Suppose that it costs c( x ) = x 3 − 6 x 2 + 15 x

dollars to produce x radiators when 8 to 30 radiators are produced and that r ( x ) = x 3 − 3 x 2 + 12 x gives the dollar revenue from selling x radiators. Your shop currently produces 10 radiators a day. About how much extra will it cost to produce one more radiator a day, and what is your estimated increase in revenue and increase in profit for selling 11 radiators a day? Solution The cost of producing one more radiator a day when 10 are produced is about c′(10): d( 3 c′( x ) = x − 6 x 2 + 15 x ) = 3 x 2 − 12 x + 15 dx c′(10) = 3(100 ) − 12(10 ) + 15 = 195.

The additional cost will be about $195. The marginal revenue is r ′( x ) =

d( 3 x − 3 x 2 + 12 x ) = 3 x 2 − 6 x + 12. dx

The marginal revenue function estimates the increase in revenue that will result from selling one additional unit. If you currently sell 10 radiators a day, you can expect your revenue to increase by about r ′(10) = 3(100 ) − 6(10 ) + 12 = $252 if you increase sales to 11 radiators a day. The estimated increase in profit is obtained by subtracting the increased cost of $195 from the increased revenue, leading to an estimated profit increase of $252 − $195 = $57.

3.4  The Derivative as a Rate of Change

167

EXAMPLE 6 Marginal rates frequently arise in discussions of tax rates. If your marginal income tax rate is 28% and your income increases by $1000, you can expect to pay an extra $280 in taxes. This does not mean that you pay 28% of your entire income in taxes. It just means that at your current income level I, the rate of increase of taxes T with respect to income is dT dI = 0.28. You will pay $0.28 in taxes out of every extra dollar you earn. As your income increases, you may land in a higher tax bracket, and your marginal rate will increase.

y 1

Sensitivity to Change y = 2p − p 2

0

1

When a small change in x produces a large change in the value of a function f ( x ), we say that the function is sensitive to changes in x. The derivative f ′( x ) is a measure of this sensitivity. The function is more sensitive when f ′( x ) is larger (when the slope of the graph of f is steeper).

p

(a)

EXAMPLE 7 Genetic Data and Sensitivity to Change The Austrian monk Gregor Johann Mendel (1822–1884), working with garden peas and other plants, provided the first scientific explanation of hybridization. His careful records showed that if p (a number between 0 and 1) is the frequency of the gene for smooth skin in peas (dominant) and (1 − p ) is the frequency of the gene for wrinkled skin in peas, then the proportion of smooth-skin peas in the next generation will be

dydp 2

dy = 2 − 2p dp

y = 2 p(1 − p ) + p 2 = 2 p − p 2. 0

1

p

(b)

FIGURE 3.22

(a) The graph of y = 2 p − p 2, describing the proportion of smooth-skin peas in the next generation. (b) The graph of dy dp (Example 7).

EXERCISES

The graph of y versus p in Figure 3.22a suggests that the value of y is more sensitive to a change in p when p is small than when p is large. Indeed, this fact is borne out by the derivative graph in Figure 3.22b, which shows that dy dp is close to 2 when p is near 0 and close to 0 when p is near 1. The implication for genetics is that introducing a few more smooth skin genes into a population where the frequency of wrinkled-skin peas is large will have a more dramatic effect on later generations than will a similar increase when the population has a large proportion of smooth-skin peas.

3.4

Motion Along a Coordinate Line

Exercises 1–6 give the positions s = f (t ) of a body moving on a coordinate line, with s in meters and t in seconds. a. Find the body’s displacement and average velocity for the given time interval. b. Find the body’s speed and acceleration at the endpoints of the interval. c. When, if ever, during the interval does the body change direction? 1. s = t 2 − 3t + 2, 0 ≤ t ≤ 2 2. s = 6t − t 2 , 0 ≤ t ≤ 6 3. s = −t 3 + 3t 2 − 3t , 0 ≤ t ≤ 3 4. s = ( t 4 4 ) − t 3 + t 2 , 0 ≤ t ≤ 3 5. s =

25 5 − , 1≤ t ≤ 5 t2 t

6. s =

25 , −4 ≤ t ≤ 0 t+5

7. Particle motion At time t, the position of a body moving along the s-axis is s = t 3 − 6t 2 + 9t m.

8. Particle motion At time t ≥ 0, the velocity of a body moving along the horizontal s-axis is υ = t 2 − 4 t + 3. a. Find the body’s acceleration each time the velocity is zero. b. When is the body moving forward? Backward? c. When is the body’s velocity increasing? Decreasing? Free-Fall Applications

9. Free fall on Mars and Jupiter The equations for free fall at the surfaces of Mars and Jupiter (s in meters, t in seconds) are s = 1.86t 2 on Mars and s = 11.44t 2 on Jupiter. How long does it take a rock falling from rest to reach a velocity of 27.8 m s (about 100 km h) on each planet? 10. Lunar projectile motion A rock thrown vertically upward from the surface of the moon at a velocity of 24 m s (about 86 km h) reaches a height of s = 24 t − 0.8t 2 m in t s. a. Find the rock’s velocity and acceleration at time t. (The acceleration in this case is the acceleration of gravity on the moon.) b. How long does it take the rock to reach its highest point?

a. Find the body’s acceleration each time the velocity is zero.

c. How high does the rock go?

b. Find the body’s speed each time the acceleration is zero.

d. How long does it take the rock to reach half its maximum height?

c. Find the total distance traveled by the body from t = 0 to t = 2.

e. How long is the rock aloft?

168

Chapter 3 Derivatives

11. Finding g on a small airless planet Explorers on a small airless planet used a spring gun to launch a ball bearing vertically upward from the surface at a launch velocity of 15 m s. Because the acceleration of gravity at the planet’s surface was g s m s 2, the explorers expected the ball bearing to reach a height of s = 15t − (1 2 ) g s t 2 m t s later. The ball bearing reached its maximum height 20 s after being launched. What was the value of g s?

b. When (approximately) is the body moving at a constant speed? c. Graph the body’s speed for 0 b t b 10. d. Graph the acceleration, where defined. 16. A particle P moves on the number line shown in part (a) of the accompanying figure. Part (b) shows the position of P as a function of time t. P

12. Speeding bullet A 45-caliber bullet shot straight up from the surface of the moon would reach a height of s = 250t − 0.8t 2 m after t seconds. On Earth, in the absence of air, its height would be s = 250t − 4.9t 2 m after t seconds. How long will the bullet be aloft in each case? How high will the bullet go?

(a) s (cm)

13. Free fall from the Tower of Pisa Had Galileo dropped a cannonball from the Tower of Pisa, 56 m above the ground, the ball’s height above the ground t seconds into the fall would have been s = 56 − 4.9t 2. a. What would have been the ball’s velocity, speed, and acceleration at time t? b. About how long would it have taken the ball to hit the ground? c. What would have been the ball’s velocity at the moment of impact? 14. Galileo’s free-fall formula Galileo developed a formula for a body’s velocity during free fall by rolling balls from rest down increasingly steep inclined planks and looking for a limiting formula that would predict a ball’s behavior when the plank was vertical and the ball fell freely; see part (a) of the accompanying figure. He found that, for any given angle of the plank, the ball’s velocity t seconds into motion was a constant multiple of t. That is, the velocity was given by a formula of the form V  kt. The value of the constant k depended on the inclination of the plank. In modern notation—part (b) of the figure—with distance in meters and time in seconds, what Galileo determined by experiment was that, for any given angle R, the ball’s velocity t s into the roll was υ = 9.8( sin θ ) t m s.

s (cm)

0

s = f (t)

2 0

1

2

3

4

5

6

t (s)

−2 (6, −4)

−4 (b)

a. When is P moving to the left? Moving to the right? Standing still? b. Graph the particle’s velocity and speed (where defined). 17. Launching a rocket When a model rocket is launched, the propellant burns for a few seconds, accelerating the rocket upward. After burnout, the rocket coasts upward for a while and then begins to fall. A small explosive charge pops out a parachute shortly after the rocket starts down. The parachute slows the rocket to keep it from breaking when it lands. The figure here shows velocity data from the flight of the model rocket. Use the data to answer the following. a. How fast was the rocket climbing when the engine stopped? b. For how many seconds did the engine burn?

Free-fall position

60

?

u (b)

(a)

a. What is the equation for the ball’s velocity during free fall? b. Building on your work in part (a), what constant acceleration does a freely falling body experience near the surface of Earth? 15. The accompanying figure shows the velocity V  ds dt  f (t ) ( m s ) of a body moving along a coordinate line. y (m/s)

0

y = f (t)

2

4

6

30 15 0 −15 −30 0

Understanding Motion from Graphs

3

Velocity (ms)

45

2

4 6 8 10 Time after launch (s)

12

c. When did the rocket reach its highest point? What was its velocity then? d. When did the parachute pop out? How fast was the rocket falling then?

8 10

t (s)

−3

a. When does the body reverse direction?

e. How long did the rocket fall before the parachute opened? f. When was the rocket’s acceleration greatest? g. When was the acceleration constant? What was its value then (to the nearest integer)?

3.4  The Derivative as a Rate of Change

18. The accompanying figure shows the velocity V  f (t ) of a particle moving on a horizontal coordinate line. y y = f(t) 0

t (s)

1 2 3 4 5 6 7 8 9

169

b. Find the marginal cost when 100 washing machines are produced. c. Show that the marginal cost when 100 washing machines are produced is approximately the cost of producing one more washing machine after the first 100 have been made, by calculating the latter cost directly. 22. Marginal revenue Suppose that the revenue from selling x washing machines is

(

r ( x ) = 20,000 1 −

1 x

)

dollars. a. When does the particle move forward? Move backward? Speed up? Slow down? b. When is the particle’s acceleration positive? Negative? Zero? c. When does the particle move at its greatest speed? d. When does the particle stand still for more than an instant? 19. The graphs in the accompanying figure show the position s, velocity V  ds dt, and acceleration a  d 2 s dt 2 of a body moving along a coordinate line as functions of time t. Which graph is which? Give reasons for your answers. y A

B

a. Find the marginal revenue when 100 machines are produced. b. Use the function r a( x ) to estimate the increase in revenue that will result from increasing production from 100 machines a week to 101 machines a week. c. Find the limit of r a( x ) as x → ∞. How would you interpret this number? Additional Applications

23. Bacterium population When a bactericide was added to a nutrient broth in which bacteria were growing, the bacterium population continued to grow for a while, but then stopped growing and began to decline. The size of the population at time t (hours) was b = 10 6 + 10 4 t − 10 3 t 2. Find the growth rates at a. t  0 hours.

C

b. t  5 hours. c. t  10 hours. t

0

20. The graphs in the accompanying figure show the position s, the velocity V  ds dt, and the acceleration a  d 2 s dt 2 of a body moving along a coordinate line as functions of time t. Which graph is which? Give reasons for your answers.

24. Body surface area A typical male’s body surface area S in square meters is often modeled by the formula S  601 wh, where h is the height in centimeters, and w the weight in kilograms, of the person. Find the rate of change of body surface area with respect to weight for males of constant height h  180 cm. Does S increase more rapidly with respect to weight at lower or higher body weights? Explain. T 25. Draining a tank It takes 12 hours to drain a storage tank by opening the valve at the bottom. The depth y of fluid in the tank t hours after the valve is opened is given by the formula

(

y

y = 6 1−

A

t 12

) m. 2

a. Find the rate dy dt ( m h ) at which the tank is draining at time t. b. When is the fluid level in the tank falling fastest? Slowest? What are the values of dy dt at these times? 0

t B

C

Economics

21. Marginal cost Suppose that the dollar cost of producing x washing machines is c( x ) = 2000 + 100 x − 0.1x 2. a. Find the average cost per machine of producing the first 100 washing machines.

c. Graph y and dy dt together and discuss the behavior of y in relation to the signs and values of dy dt. 26. Draining a tank The number of liters of water in a tank t minutes after the tank has started to drain is Q(t ) = 200 ( 30 − t ) 2. How fast is the water running out at the end of 10 min? What is the average rate at which the water flows out during the first 10 min? 27. Vehicular stopping distance Based on data from the U.S. Bureau of Public Roads, a model for the total stopping distance of a moving car in terms of its speed is s = 0.21V + 0.0063V 2 , where s is measured in meters and V is measured in km h. The linear term 0.21V models the distance the car travels during the time the

170

Chapter 3 Derivatives

is the exit velocity of a particle of lava, its height t seconds later will be s = V 0 t − 4.9t 2 m. Begin by finding the time at which ds dt  0. Neglect air resistance.)

driver perceives a need to stop until the brakes are applied, and the quadratic term 0.00636V 2 models the additional braking distance once they are applied. Find ds d V at V  50 and V  100 km h, and interpret the meaning of the derivative. 28. Inflating a balloon The volume V = ( 4 3 ) Qr 3 of a spherical balloon changes with the radius. a. At what rate ( m 3 m ) does the volume change with respect to the radius when r  2 m ? b. By approximately how much does the volume increase when the radius changes from 2 to 2.2 m?

Analyzing Motion Using Graphs

T Exercises 31–34 give the position function s  f (t ) of an object moving along the s-axis as a function of time t. Graph f together with the velocity function V(t ) = ds dt = f ′(t ) and the acceleration function a(t ) = d 2 s dt 2 = f ″(t ). Comment on the object’s behavior in relation to the signs and values of y and a. Include in your commentary such topics as the following:

29. Airplane takeoff Suppose that the distance an aircraft travels along a runway before takeoff is given by D = (10 9 ) t 2 , where D is measured in meters from the starting point and t is measured in seconds from the time the brakes are released. The aircraft will become airborne when its speed reaches 200 km h. How long will it take to become airborne, and what distance will it travel in that time?

a. When is the object momentarily at rest?

30. Volcanic lava fountains Although the November 1959 Kilauea Iki eruption on the island of Hawaii began with a line of fountains along the wall of the crater, activity was later confined to a single vent in the crater’s floor, which at one point shot lava 580 m straight into the air (a Hawaiian record). What was the lava’s exit velocity in meters per second? In kilometers per hour? (Hint: If V 0

f. When is it farthest from the axis origin?

b. When does it move to the left (down) or to the right (up)? c. When does it change direction? d. When does it speed up and slow down? e. When is it moving fastest (highest speed)? Slowest? 31. s = 60t − 4.9t 2 , 0 ≤ t ≤ 12.5 (a heavy object fired straight up from Earth’s surface at 60 m s) 32. s = t 2 − 3t + 2, 0 ≤ t ≤ 5 33. s = t 3 − 6t 2 + 7t , 0 ≤ t ≤ 4 34. s = 4 − 7t + 6t 2 − t 3 , 0 ≤ t ≤ 4

3.5 Derivatives of Trigonometric Functions Many phenomena of nature are approximately periodic (electromagnetic fields, heart rhythms, tides, weather). The derivatives of sines and cosines play a key role in describing periodic changes. This section shows how to differentiate the six basic trigonometric functions.

Derivative of the Sine Function To calculate the derivative of f ( x )  sin x , for x measured in radians, we combine the limits in Example 5a and Theorem 6 in Section 2.4 with the angle sum identity for the sine function (see Figure 1.46): sin ( x + h ) = sin x cos h + cos x sin h. If f ( x )  sin x , then sin ( x + h ) − sin x f ( x + h ) − f (x) = lim h→ 0 h→ 0 h h ( sin x cos h + cos x sin h ) − sin x lim h→ 0 h sin x ( cos h − 1) + cos x sin h lim h→ 0 h cos h − 1 sin h lim sin x ⋅ + lim cos x ⋅ h→ 0 h→ 0 h h cos h − 1 sin h sin x ⋅ lim + cos x ⋅ lim h →0 h → 0  h  h 

f ′( x ) = lim = = = =

(

)

limit 0

(

= sin x ⋅ 0 + cos x ⋅ 1 = cos x.

Derivative definition Identity for sin ( x + h )

)

limit 1

The derivative of the sine function is the cosine function:

d ( sin x ) = cos x. dx

Example 5a and Theorem 6, Section 2.4

3.5  Derivatives of Trigonometric Functions

171

EXAMPLE 1 We find derivatives of a difference, a product, and a quotient, each of which involves the sine function. (a) y = x 2 − sin x :

dy d = 2 x − ( sin x ) dx dx = 2 x − cos x

(b) y = e x sin x:

dy d d x = e x ( sin x ) + e sin x dx dx dx = e x cos x + e x sin x

Difference Rule

(

)

Product Rule

= e x ( cos x + sin x ) d ( sin x ) − sin x ⋅ 1 dx x2 x cos x − sin x = x2

sin x (c) y = : x

dy = dx

x⋅

Quotient Rule

Derivative of the Cosine Function With the help of the angle sum formula for the cosine function (see Figure 1.46), cos( x + h ) = cos x cos h − sin x sin h, we can compute the limit of the difference quotient: cos( x + h ) − cos x d ( cos x ) = lim h→ 0 dx h ( cos x cos h − sin x sin h ) − cos x = lim h→ 0 h cos x ( cos h − 1) − sin x sin h = lim h→ 0 h cos h − 1 sin h = lim cos x ⋅ − lim sin x ⋅ h→ 0 h→ 0 h h cos h − 1 sin h = cos x ⋅ lim − sin x ⋅ lim h→ 0 h→ 0 h h = cos x ⋅ 0 − sin x ⋅ 1

(

)

(

= − sin x.

Derivative definition Cosine angle sum identity

) Example 5a and Theorem 6, Section 2.4

The derivative of the cosine function is the negative of the sine function:

d ( cos x ) = −sin x. dx

y y = cos x

1 −p

p

0 −1 y′

x

y′ = −sin x

1 −p

0 −1

p

x

The curve y′ = − sin x as the graph of the slopes of the tangent lines to the curve y = cos x.

FIGURE 3.23

Figure 3.23 shows a way to visualize this result by graphing the slopes of the tangent lines to the curve y = cos x. EXAMPLE 2 functions.

We find derivatives of the cosine function in combinations with other

(a) y = 5e x + cos x: dy d d = (5e x ) + ( cos x ) dx dx dx = 5e x − sin x

Sum Rule

172

Chapter 3 Derivatives

(b) y = sin x cos x:

(

)

dy d d = sin x ( cos x ) + ( sin x ) cos x dx dx dx = ( sin x )( − sin x ) + ( cos x )( cos x )

Product Rule

= cos 2 x − sin 2 x (c) y =

cos x : 1 − sin x

dy = dx

(1 − sin x )  

d d ( cos x ) − cos x   (1 − sin x ) dx dx (1 − sin x ) 2

=

(1 − sin x )(− sin x ) − ( cos x )( 0 − cos x ) (1 − sin x ) 2

=

1 − sin x (1 − sin x ) 2

=

1 1 − sin x

Quotient Rule

sin 2 x + cos 2 x = 1

Simple Harmonic Motion Simple harmonic motion models the motion of an object or weight bobbing freely up and down on the end of a spring, with no resistance. The motion is periodic and repeats indefinitely, so we represent it using trigonometric functions. The next example models motion with no opposing forces (such as friction). −5

0

Rest position

5

Position at t=0

EXAMPLE 3 A weight hanging from a spring (Figure 3.24) is stretched down 5 units beyond its rest position and released at time t = 0 to bob up and down. Its position at any later time t is s = 5 cos t. What are its velocity and acceleration at time t? Solution We have

s

Position:

FIGURE 3.24

A weight hanging from a vertical spring and then displaced oscillates above and below its rest position (Example 3).

Velocity: Acceleration:

s = 5 cos t ds υ = = dt dυ a = = dt

d ( 5 cos t ) = −5 sin t dt d (−5 sin t ) = −5 cos t. dt

Notice how much we can learn from these equations: s, y 5

y = −5 sin t

0

p 2

p

s = 5 cos t

3p 2

2p 5p 2

t

−5

FIGURE 3.25 The graphs of the position and velocity of the weight in Example 3.

1. As time passes, the weight moves down and up between s = −5 and s = 5 on the s-axis. The amplitude of the motion is 5. The period of the motion is 2π, the period of the cosine function. 2. The velocity υ = −5 sin t attains its greatest magnitude, 5, when cos t = 0, as the graphs in Figure 3.25 show. Hence, the speed of the weight, υ = 5 sin t , is greatest when cos t = 0, that is, when s = 0 (the rest position). The speed of the weight is zero when sin t = 0. This occurs when s = 5 cos t = ±5, at the endpoints of the interval of motion. At these points the weight reverses direction. 3. The weight is acted on by the spring and by gravity. When the weight is below the rest position, the combined forces pull it up, and when it is above the rest position, they pull it down. The weight’s acceleration is always proportional to the negative of its displacement. This property of springs is called Hooke’s Law, and is studied further in Section 6.5.

3.5  Derivatives of Trigonometric Functions

173

4. The acceleration, a = −5 cos t , is zero only at the rest position, where cos t = 0 and the force of gravity and the force from the spring balance each other. When the weight is anywhere else, the two forces are unequal, and acceleration is nonzero. The acceleration is greatest in magnitude at the points farthest from the rest position, where cos t = ±1. EXAMPLE 4

The jerk associated with the simple harmonic motion in Example 3 is j =

da d = (−5 cos t ) = 5 sin t. dt dt

It has its greatest magnitude when sin t = ±1, not at the extremes of the displacement but at the rest position, where the acceleration changes direction and sign.

Derivatives of the Other Basic Trigonometric Functions Because sin x and cos x are differentiable functions of x, it follows from the Quotient Rule that the related functions tan x =

sin x , cos x

cot x =

cos x , sin x

sec x =

1 , cos x

and

csc x =

1 sin x

are differentiable at every value of x at which they are defined. Their derivatives are given by the following formulas. Notice the negative signs in the derivative formulas for the cofunctions.

The derivatives of the other trigonometric functions:

d ( tan x ) = sec 2 x dx d ( sec x ) = sec x tan x dx

d ( cot x ) = − csc 2 x dx d ( csc x ) = − csc x cot x dx

To show a typical calculation, we find the derivative of the tangent function. The other derivations are left for Exercise 54.

EXAMPLE 5

Find d ( tan x ) dx .

Solution We use the Derivative Quotient Rule to calculate the derivative:

d d ⎛⎜ sin x ⎞⎟ ( tan x ) = ⎟= ⎜ dx dx ⎜⎝ cos x ⎟⎠

cos x

d d ( sin x ) − sin x ( cos x ) dx dx cos 2 x

cos x cos x − sin x ( −sin x ) cos 2 x cos 2 x + sin 2 x = cos 2 x 1 = = sec 2 x. cos 2 x =

Quotient Rule

174

Chapter 3 Derivatives

Find y″  if y = sec x.

EXAMPLE 6

Solution Finding the second derivative involves a combination of trigonometric derivatives.

y = sec x y′ = sec x tan x

Derivative rule for secant function

d ( sec x tan x ) dx d d = sec x ( tan x ) + tan x ( sec x ) dx dx = ( sec x )( sec 2 x ) + ( tan x )( sec x tan x )

y″ =

=

EXERCISES

sec 3

x + sec x

Derivatives

x

31. p =

1. y = −10 x + 3 cos x

2. y =

3. y = x 2 cos x

4. y =

5. y = csc x − 4 x +

7 ex

7. f ( x ) = sin x tan x 9. y = xe −x sec x cot x 1 + cot x

1 4 + 13. y = cos x tan x

3 + 5 sin x x x sec x + 3

6. y = x 2 cot x −

1 x2

cos x 8. g( x ) = sin 2 x 10. y = ( sin x + cos x )sec x cos x 1 + sin x cos x x 14. y = + x cos x 12. y =

15. y = ( sec x + tan x )( sec x − tan x ) 17. f ( x ) = x 3 sin x cos x

18. g( x ) = ( 2 − x ) tan 2 x

In Exercises 19–22, find ds dt . 19. s = tan t − e −t 1 + csc t 1 − csc t

20. s = t 2 − sec t + 5e t 22. s =

sin t 1 − cos t

In Exercises 23–26, find dr dθ . 23. r = 4 − θ 2 sin θ

24. r = θ sin θ + cos θ

25. r = sec θ csc θ

26. r = (1 + sec θ )sin θ

1 cot q

sin q + cos q 29. p = cos q

32. p =

3q + tan q q sec q

33. Find y ′′ if a. y = csc x. 34. Find

y (4)

=

d4

b. y = sec x. y

a. y = −2 sin x.

dx 4

if b. y = 9 cos x.

Tangent Lines

In Exercises 35–38, graph the curves over the given intervals, together with their tangent lines at the given values of x. Label each curve and tangent line with its equation. 35. y = sin x , −3π 2 ≤ x ≤ 2π x = −π , 0, 3π 2 36. y = tan x , −π 2 < x < π 2 37. y = sec x , −π 2 < x < π 2 x = −π 3, π 4 38. y = 1 + cos x , − 3π 2 ≤ x ≤ 2π x = −π 3, 3π 2 T Do the graphs of the functions in Exercises 39–44 have any horizontal tangent lines in the interval 0 ≤ x ≤ 2π ? If so, where? If not, why not? Visualize your findings by graphing the functions with a grapher. 39. y = x + sin x

40. y = 2 x + sin x

41. y = x − cot x

42. y = x + 2 cos x

In Exercises 27–32, find dp dq. 27. p = 5 +

q sin q q2 − 1

x = −π 3, 0, π 3

16. y = x 2 cos x − 2 x sin x − 2 cos x

21. s =

Derivative rules

3.5

In Exercises 1–18, find dy dx.

11. y =

tan 2

Derivative Product Rule

28. p = (1 + csc q ) cos q tan q 30. p = 1 + tan q

43. y =

sec x 3 + sec x

44. y =

cos x 3 − 4 sin x

45. Find all points on the curve y = tan x , −π 2 < x < π 2, where the tangent line is parallel to the line y = 2 x. Sketch the curve and tangent lines together, labeling each with its equation.

3.5  Derivatives of Trigonometric Functions

46. Find all points on the curve y = cot x , 0 < x < Q , where the tangent line is parallel to the line y = −x. Sketch the curve and tangent lines together, labeling each with its equation.

175

where x is measured in centimeters and t is measured in seconds. See the accompanying figure.

In Exercises 47 and 48, find an equation for (a) the tangent line to the curve at P and (b) the horizontal tangent line to the curve at Q. y

47.

y

48.

Q

−10

p P a , 2b 2

2

0 1

p P a , 4b 4

4

0

p 2 2 y = 4 + cot x − 2csc x 1

Equilibrium position at x = 0

10 x

x Q

p1 4

0

2

a. Find the spring’s displacement when t  0, t  Q 3, and t  3Q 4. 3

x

y = 1 + " 2 csc x + cot x

b. Find the spring’s velocity when t  0, t  Q 3, and t  3Q 4. 56. Assume that a particle’s position on the x-axis is given by x = 3 cos t + 4 sin t ,

Theory and Examples

The equations in Exercises 49 and 50 give the position s  f (t ) of a body moving on a coordinate line (s in meters, t in seconds). Find the body’s velocity, speed, acceleration, and jerk at time t  Q 4 s.

where x is measured in meters and t is measured in seconds.

49. s = 2 − 2 sin t

b. Find the particle’s velocity when t  0, t  Q 2, and t  Q.

50. s = sin t + cos t 51. Is there a value of c that will make ⎧⎪ sin 2 3 x ⎪⎪ , x ≠ 0 f ( x ) = ⎪⎨ x 2 ⎪⎪ c, x = 0 ⎪⎩⎪ continuous at x  0? Give reasons for your answer. 52. Is there a value of b that will make ⎪⎧ x + b, x < 0 g( x ) = ⎪⎨ ⎪⎪ cos x , x ≥ 0 ⎩ continuous at x  0? Differentiable at x  0? Give reasons for your answers. 53. By computing the first few derivatives and looking for a pattern, find the following derivatives. d 999 a. ( cos x ) dx 999 c.

d 110 b. ( sin x − 3 cos x ) dx 110

d 73 ( x sin x ) dx 73

54. Derive the formula for the derivative with respect to x of a. sec x.

b. csc x.

c. cot x.

55. A weight is attached to a spring and reaches its equilibrium position ( x = 0 ). It is then set in motion resulting in a displacement of x  10 cos t ,

a. Find the particle’s position when t  0, t  Q 2, and t  Q.

T 57. Graph y  cos x for −Q ≤ x ≤ 2Q. On the same screen, graph y =

sin ( x + h ) − sin x h

for h  1, 0.5, 0.3, and 0.1. Then, in a new window, try h = −1, −0.5, and 0.3. What happens as h → 0 +? As h → 0 −? What phenomenon is being illustrated here? T 58. Graph y = − sin x for −Q ≤ x ≤ 2Q. On the same screen, graph y =

cos( x + h ) − cos x h

for h  1, 0.5, 0.3, and 0.1. Then, in a new window, try h = −1, −0.5, and 0.3. What happens as h → 0 +? As h → 0 −? What phenomenon is being illustrated here? T 59. Centered difference quotients The centered difference quotient f ( x + h) − f ( x − h) 2h is used to approximate f a( x ) in numerical work because (1) its limit as h l 0 equals f a( x ) when f a( x ) exists, and (2) it usually gives a better approximation of f a( x ) for a given value of h than the difference quotient f ( x + h ) − f (x) . h

176

Chapter 3 Derivatives

over the interval [ −π , 2π ] for h = 1, 0.5, and 0.3. Compare the results with those obtained in Exercise 58 for the same values of h.

See the accompanying figure. y Slope = f ′(x) C

B

60. A caution about centered difference quotients (Continuation of Exercise 59.) The quotient

f (x + h) − f (x) Slope = h

f ( x + h) − f ( x − h) 2h

A Slope =

f (x + h) − f (x − h) 2h

may have a limit as h → 0 when f has no derivative at x. As a case in point, take f ( x ) = x and calculate

y = f(x)

lim h

h

x−h

0

h→ 0

x

x+h

x

a. To see how rapidly the centered difference quotient for f ( x ) = sin x converges to f ′( x ) = cos x , graph y = cos x together with y =

sin ( x + h ) − sin ( x − h ) 2h

over the interval [ −π , 2π ] for h = 1, 0.5, and 0.3. Compare the results with those obtained in Exercise 57 for the same values of h. b. To see how rapidly the centered difference quotient for f ( x ) = cos x converges to f ′( x ) = − sin x , graph y = − sin x together with y =

0+h − 0−h . 2h

As you will see, the limit exists even though f ( x ) = x has no derivative at x = 0. Moral: Before using a centered difference quotient, be sure the derivative exists. 61. Slopes on the graph of the tangent function Graph y = tan x and its derivative together on (−π 2, π 2 ). Does the graph of the tangent function appear to have a smallest slope? A largest slope? Is the slope ever negative? Give reasons for your answers. 62. Exploring ( sin kx ) x Graph y = ( sin x ) x , y = ( sin 2 x ) x , and y = ( sin 4 x ) x together over the interval −2 ≤ x ≤ 2. Where does each graph appear to cross the y-axis? Do the graphs really intersect the axis? What would you expect the graphs of y = ( sin 5 x ) x and y = ( sin(−3 x ) ) x to do as x → 0? Why? What about the graph of y = ( sin kx ) x for other values of k? Give reasons for your answers.

cos ( x + h ) − cos ( x − h ) 2h

3.6 The Chain Rule How do we differentiate F ( x ) = sin ( x 2 − 4 )? This function is the composition f  g of two functions y = f (u) = sin u and u = g( x ) = x 2 − 4 that we know how to differentiate. The answer, given by the Chain Rule, says that the derivative is the product of the derivatives of f and g. We develop the rule in this section. 2

3 1

C: y turns B: u turns A: x turns

FIGURE 3.26 When gear A makes x turns, gear B makes u turns and gear C makes y turns. By comparing circumferences or counting teeth, we see that y = u 2 (C turns one-half turn for each B turn) and u = 3 x (B turns three times for A’s one), so y = 3 x 2. Thus, dy dx = 3 2 = (1 2 )( 3 ) = ( dy du )( du dx ).

Derivative of a Composite Function The function y =

1 3 1 x = (3x ) is the composition of the functions y = u and u = 3 x. 2 2 2

We have dy 3 = , dx 2 Since

dy 1 = , 2 du

and

3 1 = ⋅ 3, we see in this case that 2 2 dy dy du = ⋅ . dx du dx

du = 3. dx

3.6  The Chain Rule

177

If we think of the derivative as a rate of change, this relationship is intuitively reasonable. If y = f (u) changes half as fast as u, and u = g( x ) changes three times as fast as x, then we expect y to change 3 2 times as fast as x. This effect is much like that of a multiple gear train (Figure 3.26). Let’s look at another example.

EXAMPLE 1

The function y = ( 3 x 2 + 1) 2

is obtained by composing the functions y = f (u) = u 2 and u = g( x ) = 3 x 2 + 1. Calculating derivatives, we see that dy du ⋅ = 2u ⋅ 6 x du dx = 2( 3 x 2 + 1) ⋅ 6 x

Substitute for  u.

= 36 x 3 + 12 x. Calculating the derivative from the expanded formula ( 3 x 2 + 1) 2 = 9 x 4 + 6 x 2 + 1 gives the same result: dy d( 4 = 9 x + 6 x 2 + 1) = 36 x 3 + 12 x. dx dx The derivative of the composite function f ( g( x )) at x is the derivative of f at g( x ) times the derivative of g at x. This is known as the Chain Rule (Figure 3.27).

Composition f ˚ g Rate of change at x is f ′(g(x)) · g′(x).

x

g

f

Rate of change at x is g′(x).

Rate of change at g(x) is f ′( g(x)).

u = g(x)

y = f (u) = f (g(x))

FIGURE 3.27 Rates of change multiply: The derivative of f  g at x is the derivative of f at g( x ) times the derivative of g at x.

If f (u) is differentiable at the point u = g( x ) and g( x ) is differentiable at x, then the composite function ( f  g )( x ) = f ( g( x )) is differentiable at x, and

THEOREM 2—The Chain Rule

( f  g )′( x ) = f ′( g( x )) ⋅ g′( x ). In Leibniz’s notation, if y = f (u) and u = g( x ), then dy dy du = ⋅ , dx du dx where dy du is evaluated at u = g( x ).

178

Chapter 3 Derivatives

A Proof of One Case of the Chain Rule: by Δx, so that

Let Δu be the change in u when x changes

Δu = g ( x + Δx ) − g( x ). Then the corresponding change in y is Δy = f ( u + Δu ) − f (u). If Δu ≠ 0, we can write the fraction Δy Δx as the product Δy Δy Δu = ⋅ Δx Δu Δx

(1)

and take the limit as Δx → 0: Δy dy = lim Δx → 0 Δ x dx Δy Δu = lim ⋅ Δx → 0 Δ u Δx Δy Δu = lim ⋅ lim Δx → 0 Δ u Δx → 0 Δ x

(Note that  Δu → 0 as  Δx → 0 since  g  is continuous.)

Δy Δu ⋅ lim Δ x → 0 Δu Δx dy du = ⋅ . du dx = lim

Δu → 0

The problem with this argument is that if the function g( x ) oscillates rapidly near x, then Δu can be zero even when Δx ≠ 0, so the cancelation of Δu in Equation (1) would be invalid. A complete proof requires a different approach that avoids this problem, and we give one such proof in Section 3.11. EXAMPLE 2 An object moves along the x-axis so that its position at any time t ≥ 0 is given by x (t ) = cos( t 2 + 1). Find the velocity of the object as a function of t. Solution We know that the velocity is dx dt. In this instance, x is a composition of two functions: x = cos(u) and u = t 2 + 1. We have

dx = − sin(u) du du = 2 t. dt

x = cos (u ) u = t2 + 1

By the Chain Rule, dx dx du = ⋅ dt du dt = − sin(u) ⋅ 2t = − sin ( t 2 + 1) ⋅ 2t = −2t sin ( t 2 + 1).

Ways to Write the Chain Rule

( f  g )′( x ) = f ′( g( x )) ⋅ g′( x ) dy dy du = ⋅ dx du dx dy = f ′( g( x )) ⋅ g′( x ) dx d du f (u) = f ′(u) dx dx

“Outside-Inside” Rule A difficulty with the Leibniz notation is that it doesn’t state specifically where the derivatives in the Chain Rule are supposed to be evaluated. So it sometimes helps to write the Chain Rule using functional notation. If y = f ( g( x )), then dy = f ′( g( x )) ⋅ g′( x ). dx

3.6  The Chain Rule

179

In words, differentiate the “outside” function f and evaluate this derivative at the “inside” function g( x ) left alone; then multiply by the derivative of the “inside function.” EXAMPLE 3

Differentiate sin ( x 2 + e x ) with respect to x.

Solution We apply the Chain Rule directly and find

d 2 + e x ) = cos ( x 2 + e x ) ⋅ ( 2 x + e x ). sin ( x    dx inside

EXAMPLE 4

derivative of the inside

inside left alone

Differentiate y  e cos x .

Solution Here the inside function is u  g( x )  cos x and the outside function is the exponential function f ( x )  e x . Applying the Chain Rule, we get

dy d cos x d (e ) = e cos x ( cos x ) = e cos x ( − sin x ) = −e cos x sin x. = dx dx dx Generalizing Example 4, we see that the Chain Rule gives the formula d u du e  eu . dx dx For example, d kx d (e ) = e kx ⋅ ( kx ) = ke kx , dx dx

for any constant  k

and d x2 ( e ) = e x 2 ⋅ d ( x 2 ) = 2 xe x 2. dx dx

Repeated Use of the Chain Rule We sometimes have to use the Chain Rule two or more times to find a derivative. HISTORICAL BIOGRAPHY Johann Bernoulli (1667–1748) Johann Bernoulli was born in Switzerland and attended the University of Basel. His doctoral dissertation was in mathematics despite its medical title, which was used to hide his mathematical work from his father who wanted Johann to become a doctor. To know more, visit the companion Website.

EXAMPLE 5

Find the derivative of g(t ) = tan ( 5 − sin 2t ).

Solution Notice here that the tangent is a function of 5  sin 2t , whereas the sine is a function of 2t, which is itself a function of t. Therefore, by the Chain Rule, d g′(t ) = tan ( 5 − sin 2t ) dt

= sec 2 ( 5 − sin 2t ) ⋅

Derivative of tan u  with

d ( 5 − sin 2t ) dt

(

= sec 2 ( 5 − sin 2t ) ⋅ 0 − cos 2t ⋅

d (2t ) dt

u = 5 − sin 2 t

)

Derivative of −sin u with u = 2t

= sec 2 ( 5 − sin 2t ) ⋅ ( − cos 2t ) ⋅ 2 = −2( cos 2t )sec 2 ( 5 − sin 2t ).

The Chain Rule with Powers of a Function If n is any real number and f is a power function, f (u)  u n , the Power Rule tells us that f ′(u) = nu n −1. If u is a differentiable function of x, then we can use the Chain Rule to extend this to the Power Chain Rule: d n du (u ) = nu n −1 . dx dx

d (u n ) = nu n −1 du

180

Chapter 3 Derivatives

EXAMPLE 6 The Power Chain Rule simplifies computing the derivative of a power of an expression. (a)

d 7 6 d ( 5 x 3 − x 4 ) = 7( 5 x 3 − x 4 ) ( 5x 3 − x 4 ) dx dx

Power Chain Rule with u = 5x 3 − x 4 , n = 7

= 7( 5 x 3 − x 4 ) 6 (15 x 2 − 4 x 3 ) (b)

(

)

d 1 d( 3 x − 2 )−1 = dx 3 x − 2 dx = −1( 3 x − 2 )−2

d( 3x − 2 ) dx

Power Chain Rule with u = 3 x − 2, n = −1

= −1( 3 x − 2 )−2  ( 3 ) 3 =− ( 3x − 2 )2 In part (b) we could also find the derivative with the Quotient Rule. d( 5 ) d Power Chain Rule with  u = sin x , n = 5, (c) sin x = 5 sin 4 x ⋅ sin x n because  sin n x  means  ( sin x ) ,   n ≠ −1 dx dx = 5 sin 4 x cos x (d)

d (e dx

3 x +1

)=e

3 x +1

=e

3 x +1

=

d ( 3x + 1 ) dx 1 ⋅ ( 3 x + 1)−1 2 ⋅ 3 2 ⋅

3 e 2 3x + 1

Power Chain Rule with u = 3 x + 1, n = 1 2

3 x +1

EXAMPLE 7 In Example 4 of Section 3.2 we saw that the absolute value function y = x is not differentiable at x = 0. However, the function is differentiable at all other real numbers, as we now show. Since x = x 2 , we can derive the following formula, which gives an alternative to the more direct analysis seen before. d( x dx

Derivative of the Absolute Value Function

d( x dx

)

)

x , x ≠ 0 x ⎧⎪ 1, x > 0 = ⎪⎨ ⎪⎪ −1, x < 0 ⎩ =

EXAMPLE 8 is positive.

d x2 dx d 2 1 = ⋅ (x ) 2 x 2 dx 1 = ⋅ 2x 2x x = , x ≠ 0. x =

Power Chain Rule with u = x 2, n = 1 2, x ≠ 0 x2 = x

Show that the slope of every line tangent to the curve y = 1 (1 − 2 x )3

Solution We find the derivative:

dy d( = 1 − 2 x )−3 dx dx d( 1 − 2x ) dx ⋅ (−2 )

= −3(1 − 2 x )−4 ⋅ = −3(1 − 2 x )−4 6 = . (1 − 2 x ) 4

Power Chain Rule with  u = (1 − 2 x ) , n = −3

At any point ( x , y ) on the curve, the denominator is nonzero, and the slope of the tangent line is dy 6 , = (1 − 2 x ) 4 dx which is the quotient of two positive numbers.

181

3.6  The Chain Rule

EXAMPLE 9 The formulas for the derivatives of both sin x and cos x were obtained under the assumption that x is measured in radians, not degrees. The Chain Rule gives us new insight into the difference between the two. Since 180 ° = π radians, x ° = π x 180 radians, where x° is the size of the angle measured in degrees. By the Chain Rule,

( )

( )

d d πx π πx π sin( x °) = sin cos cos( x °). = = dx dx 180 180 180 180 See Figure 3.28. Similarly, the derivative of cos( x °) is −( π 180 ) sin( x °). The factor π 180 would propagate with repeated differentiation, showing an advantage for the use of radian measure in computations. y

y = sin(x°) = sin px 180

1

x

180

y = sin x

FIGURE 3.28 The function sin( x °) oscillates only π 180 times as often as sin x oscillates. Its maximum slope is π 180 at x = 0 (Example 9).

EXERCISES

3.6

Derivative Calculations

In Exercises 1–8, given y = f (u) and u = g( x ), dy dx = f ′( g( x )) g′( x ). 1. y = 6u − 9, u = (1 2 ) x 4 2. y = 2u 3 , u = 8 x − 1

find

3. y = sin u, u = 3 x + 1

4. y = cos u, u = e −x

5. y =

6. y = sin u, u = x − cos x

u , u = sin x

7. y = tan u, u = π x 2

8. y = − sec u, u =

1 + 7x x

In Exercises 9–22, write the function in the form y = f (u) and u = g( x ). Then find dy dx as a function of x. 9. y = ( 2 x + 1) 5 −7

( 7x ) x 1 13. y = ( + x− ) 8 x 11. y = 1 −

10. y = ( 4 − 3 x ) 9 −10 ⎛ x ⎞ 12. y = ⎜⎜ − 1⎟⎟⎟ ⎝ 2 ⎠

2

4

14. y =

3x 2 − 4 x + 6

(

15. y = sec ( tan x )

16. y = cot π −

17. y = tan 3 x

18. y = 5 cos −4 x

19. y =

20. y = e

e −5 x

)

( )

28. r = 6( sec θ − tan θ )3 2

cos −2

(

)

34. y = ( 2 x − 5 )−1 ( x 2 − 5 x )6

35. y = xe −x + e x 3

36. y = (1 + 2 x ) e −2 x

37. y = ( x 2 − 2 x + 2 ) e

⎛ sin θ ⎞⎟ 2 43. f (θ) = ⎜⎜ ⎜⎝ 1 + cos θ ⎟⎟⎠

38. y = ( 9 x 2 − 6 x + 2 ) e x 3 1 40. k ( x ) = x 2 sec x tan 3 x 42. g( x ) = ( x + 7 )4 1 + sin 3t −1 44. g(t ) = 3 − 2t

45. r = sin(θ 2 ) cos(2θ)

46. r = sec θ tan

5x 2

39. h( x ) = x tan ( 2 x ) + 7 41. f ( x ) =

47. q = sin

7 + x sec x

(

t t +1

( )

(

)

48. q = cot

)

( sint t )

( 1θ )

50. y = θ 3e −2θ cos 5θ 52. y = sec 2 πt

−4

1 x 29. y = x+ x x 30. y = sin −5 x − cos 3 x x 3 1( 1 −1 31. y = 3 x − 2 )6 + 4 − 18 2x 2 x 2 sin 4

4

33. y = ( 4 x + 3 ) 4 ( x + 1)−3

51. y = sin 2 ( πt − 2 )

Find the derivatives of the functions in Exercises 23–50. 23. p = 3 − t 24. q = 3 2r − r 2 3πt 3πt 4 4 25. s = sin 3t + cos 5t 26. s = sin + cos 3π 5π 2 2 27. r = ( csc θ + cot θ )−1

)

In Exercises 51–70, find dy dt .

x +x2)

( )

(

1 2 +1 8 x

49. y = cos ( e −θ 2 )

2x 3

22. y = e ( 4

21. y = e 5− 7 x

1 x

32. y = ( 5 − 2 x )−3 +

53. y = (1 + cos 2t )

54. y = (1 + cot ( t 2 ))−2

55. y = ( t tan t )10

56. y = ( t −3 4 sin t )

57. y = e cos 2( πt −1) t2 59. y = 3 t − 4t

58. y = ( e sin ( t 2 ) ) 3t − 4 −5 60. y = 5t + 2 t 62. y = cos 5 sin 3 1 64. y = (1 + cos 2 ( 7t ))3 6

(

)

3

(

3

61. y = sin ( cos ( 2t − 5 ))

(

63. y = 1 + tan 4

43

( 12t ))

3

)

(

( ))

182

Chapter 3 Derivatives

65. y =

1 + cos ( t 2 )

66. y = 4 sin ( 1 +

67. y = tan 2 ( sin 3 t )

68. y = cos 4 ( sec 2 3t )

69. y = 3t ( 2t 2 − 5 ) 4

70. y =

3t +

t)

2+

1−t

Second Derivatives

Find y ′′ in Exercises 71–78. 1 3 71. y = 1 + x 1 73. y = cot ( 3 x − 1) 9 75. y = x ( 2 x + 1) 4

(

)

77. y = e x 2 + 5 x

c.

f (x) , x =1 g( x ) + 1

d. f ( g( x )), x = 0 f. ( x 11 + f ( x ) )−2 , x = 1

e. g( f ( x )), x = 0 72. y = (1 −

−1

x)

g. f ( x + g( x ) ) , x = 0 91. Find ds dt when θ = 3π 2 if s = cos θ and dθ dt = 5.

( )

x 74. y = 9 tan 3 76. y = x 2 ( x 3 − 1) 5 78. y = sin ( x 2e x )

For each of the following functions, solve both f ′( x ) = 0 and f ′′( x ) = 0 for x. 79. f ( x ) = x ( x − 4 )3 80. f ( x ) = sec 2 x − 2 tan x for 0 ≤ x ≤ 2π

92. Find dy dt when x = 1 if y = x 2 + 7 x − 5 and dx dt = 1 3. Theory and Examples

What happens if you can write a function as a composition in different ways? Do you get the same derivative each time? The Chain Rule says you should. Try it with the functions in Exercises 93 and 94. 93. Find dy dx if y = x by using the Chain Rule with y as a composition of a. y = ( u 5 ) + 7 and u = 5 x − 35 b. y = 1 + (1 u ) and u = 1 ( x − 1).

Finding Derivative Values

In Exercises 81–86, find the value of ( f  g )′ at the given value of x. 81. f (u) = u 5 + 1, u = g( x ) = x , x = 1 1 1 , x = −1 82. f (u) = 1 − , u = g( x ) = 1− x u πu 83. f (u) = cot , u = g( x ) = 5 x , x = 1 10 1 , u = g( x ) = π x , x = 1 4 84. f (u) = u + cos 2 u 2u 85. f (u) = 2 , u = g( x ) = 10 x 2 + x + 1, x = 0 u +1 u −1 2 1 86. f (u) = , u = g( x ) = 2 − 1, x = −1 u+1 x 87. Assume that f ′(3) = −1, g′(2) = 5, g(2) = 3, and y = f ( g( x )). What is y ′ at x = 2?

(

Find the derivatives with respect to x of the following combinations at the given value of x. a. 5 f ( x ) − g( x ), x = 1 b. f ( x ) g 3 ( x ), x = 0

)

88. If r = sin( f (t )), f (0) = π 3, and f ′(0) = 4, then what is dr dt at t = 0? 89. Suppose that functions f and g and their derivatives with respect to x have the following values at x = 2 and x = 3. x

f ( x)

g( x )

f ′( x )

g ′( x )

2

8

   2

13

−3

3

3

−4

2π

   5

Find the derivatives with respect to x of the following combinations at the given value of x. a. 2 f ( x ), x = 2 b. f ( x ) + g( x ), x = 3 c. f ( x ) ⋅ g( x ), x = 3

d. f ( x ) g( x ), x = 2

e. f ( g( x )), x = 2

f.

f ( x ), x = 2

g. 1 g 2 ( x ), x = 3

h.

f 2 ( x ) + g 2 ( x ), x = 2

90. Suppose that the functions f and g and their derivatives with respect to x have the following values at x = 0 and x = 1. x

f ( x)

g( x )

f ′( x )

g ′( x )

0

1

   1

      5

   1 3

1

3

−4

−1 3

−8 3

94. Find dy dx if y = x 3 2 by using the Chain Rule with y as a composition of a. y = u 3 and u = x b. y =

u

and u = x 3 .

95. Find the tangent line to y = (( x − 1) ( x + 1)) 2 at x = 0. 96. Find the tangent line to y =

x 2 − x + 7 at x = 2.

97. a. Find the tangent line to the curve y = 2 tan ( π x 4 ) at x = 1. b. Slopes on a tangent curve What is the smallest value the slope of the curve can ever have on the interval −2 < x < 2? Give reasons for your answer. 98. Slopes on sine curves a. Find equations for the tangent lines to the curves y = sin 2 x and y = − sin ( x 2 ) at the origin. Is there anything special about how the tangent lines are related? Give reasons for your answer. b. Can anything be said about the tangent lines to the curves y = sin mx and y = − sin ( x m ) at the origin ( m a constant ≠ 0 )? Give reasons for your answer. c. For a given m, what are the largest values the slopes of the curves y = sin mx and y = − sin ( x m ) can ever have? Give reasons for your answer. d. The function y = sin x completes one period on the interval [ 0, 2π ], the function y = sin 2 x completes two periods, the function y = sin ( x 2 ) completes half a period, and so on. Is there any relation between the number of periods y = sin mx completes on [ 0, 2π ] and the slope of the curve y = sin mx at the origin? Give reasons for your answer. 99. Running machinery too fast Suppose that a piston is moving straight up and down and that its position at time t seconds is s = A cos( 2πbt ), with A and b positive. The value of A is the amplitude of the motion, and b is the frequency (number of times the piston moves up and down each second). What effect does doubling the frequency have on the piston’s velocity, acceleration, and jerk? (Once you find out, you will know why some machinery breaks when you run it too fast.)

3.6  The Chain Rule

100. Temperatures in Fairbanks, Alaska The graph in the accompanying figure shows the average Celsius temperature in Fairbanks, Alaska, during a typical 365-day year. The equation that approximates the temperature on day x is 2Q ( y = 20 sin ⎡⎢ x − 101)⎤⎥ − 4 ⎣ 365 ⎦ a. On what day is the temperature increasing the fastest? b. About how many degrees per day is the temperature increasing when it is increasing at its fastest? y

Temperature (°C)

15 5 −5 −15

T 107. The derivative of sin 2x Graph the function y  2 cos 2 x for −2 ≤ x ≤ 3.5. Then, on the same screen, graph

T 108. The derivative of cos ( x 2 ) Graph y = −2 x sin ( x 2 ) for −2 ≤ x ≤ 3. Then, on the same screen, graph

. ... .... ............ ..... .

. .... ....

x

102. Constant acceleration Suppose that the velocity of a falling body is V  k s m s (k a constant) at the instant the body has fallen s meters from its starting point. Show that the body’s acceleration is constant. 103. Falling meteorite The velocity of a heavy meteorite entering Earth’s atmosphere is inversely proportional to s when it is s km from Earth’s center. Show that the meteorite’s acceleration is inversely proportional to s 2 . 104. Particle acceleration A particle moves along the x-axis with velocity dx dt  f ( x ). Show that the particle’s acceleration is f ( x ) f a( x ). 105. Temperature and the period of a pendulum For oscillations of small amplitude (short swings), we may safely model the relationship between the period T and the length L of a simple pendulum with the equation L , g

where g is the constant acceleration of gravity at the pendulum’s location. If we measure g in centimeters per second squared, we measure L in centimeters and T in seconds. If the pendulum is made of metal, its length will vary with temperature, either increasing or decreasing at a rate that is roughly proportional to L. In symbols, with u being temperature and k the proportionality constant, dL  kL. du Assuming this to be the case, show that the rate at which the period changes with respect to temperature is kT 2. 106. Chain Rule Suppose that f ( x )  x 2 and g( x )  x . Then the compositions

( f D g )( x ) = x

2

= x2

cos(( x + h ) 2 ) − cos( x 2 ) h

for h  1.0, 0.7, and 0.3. Experiment with other values of h. What do you see happening as h l 0? Explain this behavior. Using the Chain Rule, show that the Power Rule ( d dx ) x n = nx n−1 holds for the functions x n in Exercises 109 and 110. 109. x 1 4 

and ( g D f )( x ) = x 2 = x 2

110. x 3 4 

x

x x

111. Consider the function ⎧⎪ ⎪ x sin 1 , x > 0 x f ( x ) = ⎪⎨ ⎪⎪ 0, x ≤ 0 ⎪⎩⎪

( )

101. Particle motion The position of a particle moving along a coordinate line is s = 1 + 4 t , with s in meters and t in seconds. Find the particle’s velocity and acceleration at t  6 s.

T  2Q

sin 2( x + h ) − sin 2 x h

for h  1.0, 0.5, and 0.2. Experiment with other values of h, including negative values. What do you see happening as h l 0? Explain this behavior.

y =

Ja n Fe b M ar A pr M ay Ju n Ju l A ug Se p O ct N ov D ec Ja n Fe b M ar

−25

.. ......

... .... ....

. .. .. ..

....... .... . ........ .. . ... . . .... . .... .. ... ... ... ... ... ... ..

are both differentiable at x  0 even though g itself is not differentiable at x  0. Does this contradict the Chain Rule? Explain.

y =

and is graphed in the accompanying figure.

183

a. Show that f is continuous at x  0. b. Determine f a for x v 0. c. Show that f is not differentiable at x  0. 112. Consider the function ⎧⎪ 2 ⎪ x cos 2 , x ≠ 0 x f ( x ) = ⎪⎨ ⎪⎪ 0, x = 0 ⎪⎪⎩

( )

a. Show that f is continuous at x  0. b. Determine f a for x v 0. c. Show that f is not differentiable at x  0. d. Show that f a is not continuous at x  0. 113. Verify each of the following statements. a. If f is even, then f a is odd. b. If f is odd, then f a is even. COMPUTER EXPLORATIONS Trigonometric Polynomials

114. As the accompanying figure shows, the trigonometric “polynomial” s = f (t ) = 0.78540 − 0.63662 cos 2t − 0.07074 cos 6t − 0.02546 cos 10 t − 0.01299 cos 14 t gives a good approximation of the sawtooth function s  g(t ) on the interval [ −Q , Q ]. How well does the derivative of f approximate the derivative of g at the points where dg dt is defined? To find out, carry out the following steps. a. Graph dg dt (where defined) over [ −Q , Q ]. b. Find df dt .

184

Chapter 3 Derivatives

graphed in the accompanying figure approximates the step function s = k (t ) shown there. Yet the derivative of h is nothing like the derivative of k.

c. Graph df dt . Where does the approximation of dg dt by df dt seem to be best? Least good? Approximations by trigonometric polynomials are important in the theories of heat and oscillation, but we must not expect too much of them, as we see in the next exercise. s

s

s = g(t) s = f(t)

p 2

p

0

s = h(t)

1 −p

−p

s = k(t)

t

p − 2

p 2

0

p

t

−1

a. Graph dk dt (where defined) over [ −π , π ].

115. (Continuation of Exercise 114.) In Exercise 114, the trigonometric polynomial f (t ) that approximated the sawtooth function g(t ) on [ −π , π ] had a derivative that approximated the derivative of the sawtooth function. It is possible, however, for a trigonometric polynomial to approximate a function in a reasonable way without its derivative approximating the function’s derivative at all well. As a case in point, the trigonometric “polynomial”

b. Find dh dt . c. Graph dh dt to see how badly the graph fits the graph of dk dt. Comment on what you see.

s = h(t ) = 1.2732 sin 2t + 0.4244 sin 6t + 0.25465 sin 10 t + 0.18189 sin 14 t + 0.14147 sin 18t

3.7 Implicit Differentiation y

5

Most of the functions we have dealt with so far have been described by an equation of the form y = f ( x ) that expresses y explicitly in terms of the variable x. We have learned rules for differentiating functions defined in this way. A different situation occurs when we encounter equations like

y = f1(x) (x 0, y 1) A

x 3 + y 3 − 9 xy = 0,

x 3 + y 3 − 9xy = 0 y = f2(x)

(x 0, y 2) x0

0

(x 0, y3)

5

x

y 2 − x = 0,

or

x 2 + y 2 − 25 = 0.

(See Figures 3.29, 3.30, and 3.31.) Each of these equations defines an implicit relation between the variables x and y, meaning that a value of x may determine more than one value of y, even though we do not have a simple formula for the y-values. In some cases we may be able to solve such an equation for y as an explicit function (or even several functions) of x. When we cannot put an equation F ( x , y ) = 0 in the form y = f ( x ) to differentiate it in the usual way, we may still be able to find dy dx by implicit differentiation. This section describes the technique.

y = f3 (x)

Implicitly Defined Functions FIGURE 3.29

The curve x 3 + y 3 − 9 xy = 0 is not the graph of any one function of x. The curve can, however, be divided into separate arcs that are the graphs of functions of x. This particular curve, called a folium, dates to Descartes in 1638.

We begin with examples involving familiar equations that we can solve for y as a function of x and then calculate dy dx in the usual way. Then we differentiate the equations implicitly, and find the derivative. We will see that the two methods give the same answer. Following the examples, we summarize the steps involved in the new method. In the examples and exercises, it is always assumed that the given equation determines y implicitly as a differentiable function of x so that dy dx exists. EXAMPLE 1

Find dy dx if y 2 = x.

Solution The equation y 2 = x defines two differentiable functions of x that we can actually find, namely y 1 = x and y 2 = − x (Figure 3.30). We know how to calculate the derivative of each of these for x > 0:

dy1 1 = dx 2 x

and

dy 2 1 = − . dx 2 x

3.7  Implicit Differentiation

y2 = x

y

Slope =

1 1 = 2y1 2 "x

y1 = "x

P(x, "x )

But suppose that we knew only that the equation y 2 = x defined y as one or more differentiable functions of x for x > 0 without knowing exactly what these functions were. Can we still find dy dx ? The answer is yes. To find dy dx , we simply differentiate both sides of the equation y 2 = x with respect to x, treating y = f ( x ) as a differentiable function of x: y2 = x

x

0

2y

Q(x, − "x ) Slope =

1 1 =− 2y 2 2 "x

185

y 2 = − "x

dy =1 dx dy 1 = . dx 2y

The Chain Rule gives 

dy d d 2 [ f ( x ) ] 2 = 2 f ( x ) f ′( x ) = 2 y . (y ) = dx dx dx

This one formula gives the derivatives we calculated for both explicit solutions y 1 =

The equation y 2 − x = 0, and y 2 = − x: or = x as it is usually written, defines two differentiable functions of x on the dy1 1 1 = = interval x > 0. Example 1 shows how dx 2 y1 2 x to find the derivatives of these functions without solving the equation y 2 = x for y. FIGURE 3.30

x

y2

EXAMPLE 2

y1 = "25 − x 2

0

5

(3, −4) y2 = −"25 −

x2

dy 2 1 1 1 = = = − . 2y2 dx 2(− x ) 2 x

Find the slope of the circle x 2 + y 2 = 25 at the point ( 3, −4 ).

Solution The circle is not the graph of a single function of x. Rather, it is the combined graphs of two differentiable functions, y 1 = 25 − x 2 and y 2 = − 25 − x 2 (Figure 3.31). The point ( 3, −4 ) lies on the graph of y 2 , so we can find the slope by calculating the derivative directly, using the Power Chain Rule:

y

−5

and

Slope = − xy = 3 4

FIGURE 3.31 The circle combines the graphs of two functions. The graph of y 2 is the lower semicircle and passes through ( 3, −4 ).

x

dy 2 dx

x =3

= −

−2 x 2 25 − x 2

x =3

= −

−6 3 = . 4 2 25 − 9

(

)

1 2 d −( 25 − x 2 ) = dx 1 2 − 1 (− 2 x ) − ( 25 − x 2 ) 2

We can solve this problem more easily by differentiating the given equation of the circle implicitly with respect to x: d 2 d 2 d (x ) + (y ) = (25) dx dx dx dy 2x + 2 y = 0 dx dy x = − . dx y The slope at ( 3, −4 ) is −

x y

= − ( 3, − 4 )

See Example 1.

3 3 = . 4 −4

Notice that unlike the slope formula for dy 2 dx, which applies only to points below the x-axis, the formula dy dx = −x y applies everywhere the circle has a slope—that is, at all circle points ( x , y ) where y ≠ 0. Notice also that the derivative involves both variables x and y, not just the independent variable x. To calculate the derivatives of other implicitly defined functions, we proceed as in Examples 1 and 2: We treat y as a differentiable implicit function of x and apply the usual rules to differentiate both sides of the defining equation.

Implicit Differentiation

1. Differentiate both sides of the equation with respect to x, treating y as a differentiable function of x. 2. Collect the terms with dy dx on one side of the equation and solve for dy dx .

186

Chapter 3 Derivatives

y

EXAMPLE 3 y2

4

=

x2 +

Find dy dx if y 2 = x 2 + sin xy (Figure 3.32).

sin xy

Solution We differentiate the equation implicitly.

y 2 = x 2 + sin xy

2

−4

−2

0

2

4

x

−2 −4

FIGURE 3.32

The graph of the equation in Example 3.

d 2 (y ) dx dy 2y dx dy 2y dx dy dy 2y − ( cos xy ) x dx dx dy ( 2 y − x cos xy ) dx dy dx

(

d 2 d (x ) + ( sin xy ) dx dx d = 2 x + ( cos xy ) ( xy) dx

Differentiate both sides with respect to x …

=

(

= 2 x + ( cos xy ) y + x

… treating y as a function of x and using the Chain Rule.

dy dx

) = 2x + ( cos xy ) y

)

Treat xy as a product. Collect terms with dy dx .

= 2 x + y cos xy =

2 x + y cos xy 2 y − x cos xy

Solve for dy dx .

Notice that the formula for dy dx applies everywhere that the implicitly defined curve has a slope. Notice again that the derivative involves both variables x and y, not just the independent variable x.

Derivatives of Higher Order Implicit differentiation can also be used to find higher derivatives. EXAMPLE 4

Find d 2 y dx 2 if 2 x 3 − 3 y 2 = 8.

Solution To start, we differentiate both sides of the equation with respect to x in order to find y′ = dy dx .

d( 3 d( ) 2x − 3y 2 ) = 8 dx dx 6 x 2 − 6 yy′ = 0 y′ =

x2 , y

Treat  y  as a function of  x.

when y ≠ 0

Solve for  y′.

We now apply the Quotient Rule to find y ″.

Light ray

Tangent line Curve of lens surface

Normal line A P

Point of entry

y″ =

2 xy − x 2 y′ d ⎛⎜ x 2 ⎞⎟ x2 2x = = − 2 ⋅ y′ ⎟ ⎜ ⎟ 2 dx ⎝ y ⎠ y y y

Finally, we substitute y′ = x 2 y to express y ′′ in terms of x and y. y″ =

2x x2⎛x2 ⎞ 2x x4 − 3, − 2 ⎜⎜ ⎟⎟⎟ = y y ⎝ y ⎠ y y

when y ≠ 0

B

Lenses, Tangent Lines, and Normal Lines

FIGURE 3.33 The profile of a lens, showing the bending (refraction) of a ray of light as it passes through the lens surface.

In the law that describes how light changes direction as it enters a lens, the important angles are the angles the light makes with the line perpendicular to the surface of the lens at the point of entry (angles A and B in Figure 3.33). This line is called the normal line to the surface at the point of entry. In a profile view of a lens like the one in Figure 3.33, the normal line is the line perpendicular (also said to be orthogonal) to the tangent line of the profile curve at the point of entry.

3.7  Implicit Differentiation

y

EXAMPLE 5 Show that the point ( 2, 4 ) lies on the curve x 3 + y 3 − 9 xy = 0. Then find the tangent line and normal line to the curve there (Figure 3.34).

ne t li n e ng pe m Ta slo

Solution The point ( 2, 4 ) lies on the curve because its coordinates satisfy the equation given for the curve: 2 3 + 4 3 − 9( 2 )( 4 ) = 8 + 64 − 72 = 0. To find the slope of the curve at ( 2, 4 ), we first use implicit differentiation to find a formula for dy dx :

e lin al /m rm −1 pe slo

No

4

0

x 3 + y 3 − 9 xy = 0

+

y3

d 3 d 3 d (x ) + ( y ) − ( 9 xy ) dx dx dx dy dy dx −9 x + y 3x 2 + 3 y 2 dx dx dx dy ( 3y 2 − 9x ) + 3x 2 − 9 y dx dy 3( y 2 − 3 x ) dx dy dx

x

2 x3

187

(

− 9xy = 0

FIGURE 3.34 Example 5 shows how to find equations for the tangent line and normal line to the folium of Descartes at ( 2, 4 ).

=

d (0) dx

)=0

Differentiate both sides with respect to x. Treat  xy  as a product and  y  as a function of  x.

= 0 = 9 y − 3x 2 =

3y − x 2 . y 2 − 3x

Solve for  dy dx .

We then evaluate the derivative at ( x , y ) = ( 2, 4 ): dy dx

( 2, 4 )

=

3y − x 2 y 2 − 3x

= ( 2, 4 )

3( 4 ) − 2 2 8 4 = = . 4 2 − 3( 2 ) 10 5

The tangent line at ( 2, 4 ) is the line through ( 2, 4 ) with slope 4 5: 4( x − 2) 5 4 12 y = x+ . 5 5 y = 4+

The normal line to the curve at ( 2, 4 ) is the line perpendicular to the tangent line there, the line through ( 2, 4 ) with slope 5 4: 5( x − 2) 4 5 13 y = − x+ . 4 2 y = 4−

EXERCISES

3.7

Differentiating Implicitly

Find dr dR in Exercises 17–20.

Use implicit differentiation to find dy dx in Exercises 1–16. 1. x 2 y + xy 2 = 6

2. x 3 + y 3 = 18 xy

3. 2 xy + y 2 = x + y

4. x 3 − xy + y 3 = 1

5. x 2 ( x − y ) = x 2 − y 2

6. ( 3 xy + 7 ) = 6 y

2

x −1 x+1 9. x  sec y 7. y 2 =

11. x + tan ( xy) = 0 ⎛1⎞ 13. y sin ⎜⎜ ⎟⎟⎟ = 1 − xy ⎝ y⎠ 15.

e 2x

Slopes of two nonvertical perpendicular lines are negative reciprocals of each other (see Appendix A.4).

= sin ( x + 3 y )

17. R 1 2 + r 1 2 = 1 19. sin(rR) 

2

18. r − 2 R =

1 2

3 23 4 R + R3 4 2 3

20. cos r + cot R = e rR

2x − y x + 3y 10. xy  cot ( xy)

In Exercises 21–28, use implicit differentiation to find dy dx and then d 2 y dx 2 . Write the solutions in terms of x and y only.

12. x 4 + sin y = x 3 y 2

21. x 2 + y 2 = 1

22. x 2 3 + y 2 3 = 1

23. y 2 = e

24. y 2 − 2 x = 1 − 2 y

8. x 3 =

14. x cos( 2 x + 3 y ) = y sin x 16.

e x 2y

= 2x + 2y

Second Derivatives

x2

+ 2x

25. 2 y = x − y 27. 3 + sin y = y −

26. xy + y 2 = 1 x3

28. sin y = x cos y − 2

188

Chapter 3 Derivatives

29. If x 3 + y 3 = 16, find the value of d 2 y dx 2 at the point (2, 2). 30. If xy + y 2 = 1, find the value of d 2 y dx 2 at the point ( 0, −1).

47. The devil’s curve (Gabriel Cramer, 1750) Find the slopes of the devil’s curve y 4 − 4 y 2 = x 4 − 9 x 2 at the four indicated points. y

In Exercises 31 and 32, find the slope of the curve at the given points.

y 4 − 4y 2 = x 4 − 9x 2

31. y 2 + x 2 = y 4 − 2 x at (−2, 1) and (−2, −1) 32. ( x 2 + y 2 ) 2 = ( x − y ) 2

at (1, 0 ) and (1, −1)

(−3, 2)

2

(3, 2)

Slopes, Tangent Lines, and Normal Lines

In Exercises 33–42, verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given point. 33.

x2

+ xy −

34.

x2

+

y2

y2

x 3 (3, −2)

−2

= 1, ( 2, 3 )

= 25, ( 3, −4 )

35. x 2 y 2 = 9, (−1, 3 ) 36. y 2 − 2 x − 4 y − 1 = 0, (−2, 1) 37. 6 x 2 + 3 xy + 2 y 2 + 17 y − 6 = 0, (−1, 0 ) 38. x 2 −

−3 (−3, −2)

3xy + 2 y 2 = 5, ( 3, 2 )

39. 2 xy + π sin y = 2π , (1, π 2 ) 40. x sin 2 y = y cos 2 x , ( π 4 , π 2 )

Theory and Examples

41. y = 2 sin ( π x − y ), (1, 0 ) 42. x 2 cos 2 y − sin y = 0, ( 0, π ) 43. Parallel tangent lines Find the two points where the curve x 2 + xy + y 2 = 7 crosses the x-axis, and show that the tangent lines to the curve at these points are parallel. What is the common slope of these tangent lines? 44. Normal lines parallel to a line Find the normal lines to the curve xy + 2 x − y = 0 that are parallel to the line 2 x + y = 0. 45. The eight curve Find the slopes of the curve y 4 = y 2 − x 2 at the two points shown here. y a" 3 , " 3b 4 2

1

48. The folium of Descartes (See Figure 3.29) a. Find the slope of the folium of Descartes x 3 + y 3 − 9 xy = 0 at the points ( 4, 2 ) and ( 2, 4 ). b. At what point other than the origin does the folium have a horizontal tangent line? c. Find the coordinates of the point A in Figure 3.29 where the folium has a vertical tangent line.

49. Intersecting normal line The line that is normal to the curve x 2 + 2 xy − 3 y 2 = 0 at (1, 1) intersects the curve at what other point? 50. Power rule for rational exponents Let p and q be integers with q > 0. If y = x p q , differentiate the equivalent equation y q = x p implicitly and show that, for y ≠ 0, p d pq x = x ( p q )−1. q dx 51. Normal lines to a parabola Show that if it is possible to draw three normal lines from the point ( a, 0 ) to the parabola x = y 2 shown in the accompanying diagram, then a must be greater than 1 2. One of the normal lines is the x-axis. For what value of a are the other two normal lines perpendicular? y x = y2

a" 3 , 1b 4 2 x

0 y4 = y2 − x2

0

x

(a, 0)

−1

46. The cissoid of Diocles (from about 200 b.c.) Find equations for the tangent line and normal line to the cissoid of Diocles y 2 ( 2 − x ) = x 3 at (1, 1). y

52. Is there anything special about the tangent lines to the curves y 2 = x 3 and 2 x 2 + 3 y 2 = 5 at the points (1, ±1)? Give reasons for your answer. y

y 2(2 − x) = x 3

y2 = x3 2x 2 + 3y 2 = 5

1

0

(1, 1)

(1, 1)

1

x

x

0 (1, −1)

3.8  Derivatives of Inverse Functions and Logarithms

53. Verify that the following pairs of curves meet orthogonally. a. x 2 + y 2 = 4, x 2 = 3 y 2 1 2 y 3 54. The graph of y 2  x 3 is called a semicubical parabola and is shown in the accompanying figure. Determine the constant b so 1 that the line y = − x + b meets this graph orthogonally. 3 b. x = 1 − y 2 , x =

y

y2

=

x3

189

58. Use the formula in Exercise 57 to find dy dx if a. y = ( sin −1 x ) 2 1 b. y = sin −1 . x

( )

COMPUTER EXPLORATIONS

Use a CAS to perform the following steps in Exercises 59–66. a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point P satisfies the equation. b. Using implicit differentiation, find a formula for the derivative dy dx and evaluate it at the given point P.

1 y=− x+b 3 0

c. Use the slope found in part (b) to find an equation for the tangent line to the curve at P. Then plot the implicit curve and tangent line together on a single graph.

x

59. x 3 − xy + y 3 = 7, P ( 2, 1)

In Exercises 55 and 56, find both dy dx (treating y as a differentiable function of x) and dx dy (treating x as a differentiable function of y). How do dy dx and dx dy seem to be related? 55. xy 3 + x 2 y = 6 56.

x3

+

y2

=

sin 2

y

57. Derivative of arcsine Assume that y = sin −1 x is a differentiable function of x. By differentiating the equation x  sin y implicitly, show that dy dx = 1 1 − x 2 .

60. x 5 + y 3 x + yx 2 + y 4 = 4, P (1, 1) 2+ x 61. y 2 + y = , P ( 0, 1) 1− x 62. y 3 + cos xy = x 2 , P (1, 0 ) 63. x + tan

( xy ) = 2, P (1, Q4 )

(

)

Q 64. xy 3 + tan ( x + y ) = 1, P ,  0 4 65. 2 y 2 + ( xy )1 3 = x 2 + 2, P (1, 1) 66. x 1 + 2 y + y = x 2 , P (1, 0 )

3.8 Derivatives of Inverse Functions and Logarithms In Section 1.5 we saw how the inverse of a function undoes, or inverts, the effect of that function. We defined there the natural logarithm function f −1 ( x ) = ln x as the inverse of the natural exponential function f ( x )  e x . This is one of the most important functioninverse pairs in mathematics and science. We learned how to differentiate the exponential function in Section 3.3. Here we develop a rule for differentiating the inverse of a differentiable function, and we apply the rule to find the derivative of the natural logarithm function. y

y = 2x − 2

y=x

y= 1x+1 2 1 −2

1

x

−2

FIGURE 3.35

Graphing a line and its inverse together shows the graphs’ symmetry with respect to the line y  x. The slopes are reciprocals of each other.

Derivatives of Inverses of Differentiable Functions We calculated the inverse of the function f ( x ) = (1 2 ) x + 1 to be f −1 ( x ) = 2 x − 2 in Example 3 of Section 1.5. Figure 3.35 shows the graphs of both functions. If we calculate their derivatives, we see that

(

)

d d 1 1 f (x) = x +1 = 2 dx dx 2 d −1 d( f (x) = 2 x − 2 ) = 2. dx dx The derivatives are reciprocals of one another, so the slope of one line is the reciprocal of the slope of its inverse line. (See Figure 3.35.) This is not a special case. Reflecting any nonhorizontal or nonvertical line across the line y  x always inverts the line’s slope. If the original line has slope m v 0 , the reflected line has slope 1 m .

190

Chapter 3 Derivatives

y y = f (x) Slope f ′(a) = y=x

p q

b = f (a)

p q

Slope ( f –1)′(b) =

1 q 1 = p = p f ′(a) q

p

y = f –1(x)

q a = f –1(b) a

x

b

The slopes are reciprocals: ( f −1 ) ′ (b) =

1 1 or ( f −1 ) ′ (b) = f ′(a) f ′( f −1 (b))

FIGURE 3.36 The graphs of inverse functions have reciprocal slopes at corresponding points.

The reciprocal relationship between the slopes of f and f −1 holds for other functions as well, but we must be careful to compare slopes at corresponding points. If the slope of y = f ( x ) at the point ( a, f (a) ) is f ′(a) and f ′(a) ≠ 0, then the slope of y = f −1 ( x ) at the point ( f (a), a ) is the reciprocal 1 f ′(a) (Figure 3.36). If we set b = f (a), then ( f −1 )′(b) =

1 1 = . f ′(a) f ′( f −1 (b))

If y = f ( x ) has a horizontal tangent line at (a, f (a)), then the inverse function f −1 has a vertical tangent line at ( f (a), a), so the slope is undefined and f −1 is not differentiable at f (a). Theorem 3 gives the conditions under which f −1 is differentiable in its domain (which is the same as the range of f ).

THEOREM 3—The Derivative Rule for Inverses

If f has an interval I as domain and f ′( x ) exists and is never zero on I, then f −1 is differentiable at every point in its domain (the range of f ). The value of ( f −1 )′ at a point b in the domain of f −1 is the reciprocal of the value of f ′ at the point a = f −1 (b): ( f −1 )′(b) =

1 f ′( f −1 (b) )

(1)

or df −1 dx

x=b

=

1 df dx

.

x = f −1 ( b )

Theorem 3 makes two assertions. The first of these has to do with the conditions under which f −1 is differentiable; the second assertion is a formula for the derivative of f −1

3.8  Derivatives of Inverse Functions and Logarithms

191

when it exists. While we omit the proof of the first assertion, the second one is proved in the following way: f ( f −1 ( x )) = x

Inverse function relationship

d f ( f −1 ( x )) = 1 dx d −1 f ′( f −1 ( x )) · f (x) = 1 dx 1 d −1 . f (x) = dx f ′( f −1 ( x ))

y = x 2, x > 0

Slope 4 (2, 4)

3

( f −1 )′( x ) = Slope 1– 4

2

(4, 2)

1

2

3

Solve for the derivative.

1 f ′( f

−1 ( x ))

1 2( f −1 ( x )) 1 . = 2( x ) =

y = "x

1 0

Chain Rule

EXAMPLE 1 The function f ( x ) = x 2 , x > 0 and its inverse f −1 ( x ) = x have derivatives f ′( x ) = 2 x and ( f −1 )′( x ) = 1 ( 2 x ). Let’s verify that Theorem 3 gives the same formula for the derivative of f −1 ( x ):

y

4

Differentiate both sides.

f ′( x ) = 2 x  with  x  replaced by  f −1 ( x ) f −1 ( x ) =

x

x

4

Theorem 3 gives a derivative that agrees with the known derivative of the square root function. Let’s examine Theorem 3 at a specific point. We pick x = 2 (the number a) and f (2) = 4 (the value b). Theorem 3 says that the derivative of f at 2, which is f ′(2) = 4, and the derivative of f −1 at f (2), which is ( f −1 )′(4), are reciprocals. It states that

FIGURE 3.37

The derivative of f −1 ( x ) = x at the point (4, 2) is the reciprocal of the derivative of f ( x ) = x 2 at ( 2, 4 ) (Example 1).

( f −1 )′(4) =

1 1 1 = = f ′( f −1 (4) ) f ′(2) 2x

x=2

=

1 . 4

See Figure 3.37. y 6

(2, 6)

y = x3 − 2 Slope 3x 2 = 3(2)2 = 12

Reciprocal slope: 1 12 (6, 2)

−2

0

6

x

We will use the procedure illustrated in Example 1 to calculate formulas for the derivatives of many inverse functions throughout this chapter. Equation (1) sometimes enables us to find specific values of df −1 dx without knowing a formula for f −1 . EXAMPLE 2 Let f ( x ) = x 3 − 2, x > 0. Find the value of df −1 dx at x = 6 = f (2) without finding a formula for f −1 ( x ). See Figure 3.38. Solution We apply Theorem 3 to obtain the value of the derivative of f −1 at x = 6:

df dx

−2

FIGURE 3.38

The derivative of − 2 at x = 2 tells us the f (x) = derivative of f −1 at x = 6 (Example 2). x3

df −1 dx

x=2

x = f ( 2)

= 3x 2 =

x=2

1 df dx

x=2

= 12 =

1 . 12

Eq. (1)

192

Chapter 3 Derivatives

Derivative of the Natural Logarithm Function Since we know that the exponential function f ( x ) = e x is differentiable everywhere, we can apply Theorem 3 to find the derivative of its inverse f −1 ( x ) = ln x : 1 f ′( f −1 ( x )) 1 = f −1 ( x ) e 1 = ln x e 1 = . x

( f −1 )′( x ) =

Theorem 3 f ′(u ) = e u x>0 Inverse function relationship

Alternative Derivation Instead of applying Theorem 3 directly, we can find the derivative of y = ln x using implicit differentiation, as follows:

y = ln x ey

= x

d y d (e ) = (x) dx dx dy ey =1 dx dy 1 1 = y = . dx e x

x>0 Inverse function relationship Differentiate implicitly. Chain Rule ey = x

No matter which derivation we use, the derivative of y = ln x with respect to x is

d 1 ln x = , x > 0. dx x

(2)

The Chain Rule extends this formula to positive differentiable functions u( x ): d 1 du ln u = , dx u dx EXAMPLE 3

u > 0.

We use Equations (2) and (3) to find derivatives.

d 1 d 1 1 ln 2 x = (2 x ) = ⋅2 = , x > 0 dx 2 x dx 2x x 2 (b) Equation (3) with u = x + 3 gives

(a)

d ( 2 1 1 2x d ln x + 3 ) = 2 · ( x 2 + 3) = 2 · 2x = 2 . dx x +3 x + 3 dx x +3 (c) Using the Chain Rule and Equation (2), we find 4 ( ln x )3 d d 1 , x > 0. ( ln x ) 4 = 4 ( ln x )3 · ( ln x ) = 4 ( ln x )3 · = dx dx x x

(3)

193

3.8  Derivatives of Inverse Functions and Logarithms

(d) Equation (3) with u  x gives an important derivative: d d du ln x = ln u · dx du dx 1 x = · u x 1 x · = x x x = 2 x 1 = . x

Derivative of ln x 1 d ln x = , x ≠ 0 x dx

u = x ,x ≠ 0 d dx

(x) =

x x

(Example 7, Section 3.6)

Substitute for  u.

So 1 x is the derivative of ln x on the domain x  0, and the derivative of ln(x ) on the domain x  0. Notice from Example 3a that the function y  ln 2 x has the same derivative as the function y  ln x. This is true of y  ln bx for any constant b, provided that bx  0 : d 1 d 1 1 ln bx  · (bx )  (b)  . dx bx dx bx x

d 1 ln bx = , bx > 0 dx x

(4)

EXAMPLE 4 A line with slope m passes through the origin and is tangent to the graph of y  ln x. What is the value of m? Solution Suppose the point of tangency occurs at the unknown point x = a > 0. Then we know that the point ( a, ln a ) lies on the graph and that the tangent line at that point has slope m  1 a (Figure 3.39). Since the tangent line passes through the origin, its slope is ln a − 0 ln a m = = . a−0 a Setting these two formulas for m equal to each other, we have

y

2 1

0

(a, ln a)

1 2 y = ln x

ln a 1  a a ln a  1

1 Slope = a 3

4

5

x

e ln a  e 1 a  e

FIGURE 3.39 The tangent line meets the curve at some point ( a, ln a ), where the slope of the curve is 1 a (Example 4).

m 

1 . e

The Derivatives of a x and log a x We start with the equation a x = e ln( a x ) = e x ln a , a > 0, which was seen in Section 1.5, where it was used to define the function a x : d x d x ln a a = e dx dx d = e x ln a · ( x ln a ) dx = a x ln a. That is, if a  0, then

ax

d u du e = eu dx dx ln a  is a constant.

is differentiable and d x a  a x ln a. dx

(5)

194

Chapter 3 Derivatives

This equation shows why e x is the preferred exponential function in calculus. If a  e, then ln a  1 and the derivative of a x simplifies to d x e  e x ln e  e x . dx

ln e  1

If a  0 and u is a differentiable function of x, then by the Chain Rule, a u is a differentiable function of x and d u du a  a u ln a . dx dx EXAMPLE 5

(6)

Here are some derivatives of general exponential functions.

d x 3  3 x ln 3 dx d −x d (b) 3 = 3 − x ( ln 3 ) (−x ) = −3 − x ln 3 dx dx d sin x d (c) 3 = 3 sin x ( ln 3 ) ( sin x ) = 3 sin x ( ln 3 ) cos x dx dx d d (d) sin(3 x ) = cos(3 x ) 3 x = cos(3 x ) ⋅ 3 x ln 3 dx dx (a)

Eq. (6)  with  a  3, u  x Eq. (6)  with  a = 3, u =  x Eq. (6)  with  u = sin x Chain Rule and Eq. ( 5 )

In Section 3.3 we looked at the derivative f a(0) for the exponential functions f ( x )  a x at various values of the base a. The number f a(0) is the limit, lim ( a h − 1) h , and gives the h→ 0

slope of the graph of a x when it crosses the y-axis at the point ( 0, 1). We now see from Equation (5) that the value of this slope is lim

h→ 0

ah − 1 = ln a. h

(7)

In particular, when a  e we obtain lim

h→ 0

eh − 1 = ln e = 1. h

However, we have not fully justified that these limits actually exist. While all of the arguments given in deriving the derivatives of the exponential and logarithmic functions are correct, they do assume the existence of these limits. In Chapter 7 we will give another development of the theory of logarithmic and exponential functions which fully justifies that both limits do in fact exist and have the values derived above. To find the derivative of log a x for an arbitrary base ( a > 0,   a ≠ 1), we use the change-of-base formula for logarithms (reviewed in Section 1.5) to express log a x in terms of natural logarithms: log a x 

ln x . ln a

Then we take derivatives d d ⎛⎜ ln x ⎞⎟ log a x = ⎟ ⎜ dx dx ⎜⎝ ln a ⎟⎠ =

1 d ⋅ ln x ln a dx

=

1 1 ⋅ , ln a x

Differentiate both sides.

ln a  is a constant.

195

3.8  Derivatives of Inverse Functions and Logarithms

which yields d 1 log a x = dx x ln a

a > 0, a ≠ 1.

(8)

If u is a differentiable function of x and u  0, the Chain Rule gives a more general formula: d 1 du log a u = dx u ln a dx

a > 0, a ≠ 1.

(9)

Logarithmic Differentiation The derivatives of positive functions given by formulas that involve products, quotients, and powers can often be found more quickly if we take the natural logarithm of both sides before differentiating. This enables us to use the laws of logarithms to simplify the formulas before differentiating. The process, called logarithmic differentiation, is illustrated in the next example. EXAMPLE 6

Find dy dx if y =

( x 2 + 1)( x + 3 )1 2

x −1

,

x > 1.

Solution We take the natural logarithm of both sides and simplify the result with the algebraic properties of logarithms from Theorem 1 in Section 1.5: ( x 2 + 1)( x + 3 )1 2

ln y = ln

= ln (

(x2

x −1 + 1)( x + 3 )1 2 ) − ln ( x − 1)

= ln ( x 2 + 1) + ln ( x + 3 ) = ln ( x 2 + 1) +

Rule 2

− ln ( x − 1)

Rule 1

1 ( ln x + 3 ) − ln ( x − 1). 2

Rule 4

12

We then take derivatives of both sides with respect to x, using Equation (3): 1 1 1 1 dy 1 . = 2 ⋅ 2x + ⋅ − y dx x +1 2 x+3 x −1 Next we solve for dy dx :

(

)

dy 2x 1 1 . = y 2 + − dx x +1 2x + 6 x −1 Finally, we substitute for y:

(

)

dy ( x 2 + 1)( x + 3 )1 2 2x 1 1 = + − . dx x −1 x2 + 1 2x + 6 x −1 The computation in Example 6 would be much longer if we used the product, quotient, and power rules.

Irrational Exponents and the Power Rule (General Version) The natural logarithm and the exponential function will be defined precisely in Chapter 7. We can use the exponential function to define the general exponential function, which enables us to raise any positive number to any real power n, rational or irrational. That is, we can define the power function y  x n for any exponent n.

196

Chapter 3 Derivatives

DEFINITION

For any x > 0 and for any real number n, x n = e n ln x.

Because the logarithm and exponential functions are inverses of each other, the definition gives ln x n = n ln x , for all real numbers  n. That is, the rule for taking the natural logarithm of a power holds for all real exponents n, not just for rational exponents. The definition of the power function also enables us to establish the derivative Power Rule for any real power n, as stated in Section 3.3.

General Power Rule for Derivatives

For x > 0 and any real number n, d n x = nx n −1. dx If x ≤ 0, then the formula holds whenever the derivative, x n , and x n −1 all exist.

Proof

Differentiating x n with respect to x gives d n d n ln x x = e dx dx d = e n ln x ⋅ ( n ln x ) dx n = xn ⋅ x

Definition of  x n ,   x > 0 Chain Rule for  e u Definition and derivative of  ln x

= nx n −1.

xn ⋅

1 = x n −1 x

In short, whenever x > 0, d n x = nx n −1. dx For x < 0 , if y = x n , y′, and x n −1 all exist, then ln y = ln x

n

= n ln x .

Using implicit differentiation (which assumes the existence of the derivative y′) and Example 3(c), we have y′ n = . y x Solving for the derivative, we find that y′ = n

y xn = n = nx n −1. x x

y = xn

It can be shown directly from the definition of the derivative that the derivative equals 0 when x = 0 and n > 1 (see Exercise 107). This completes the proof of the general version of the Power Rule for all values of x.

3.8  Derivatives of Inverse Functions and Logarithms

197

Differentiate f ( x )  x x , x  0.

EXAMPLE 7

Solution The Power Rule tells us how to differentiate a function of the form x a , where a is a fixed real number. However, the exponent in x x is not a fixed constant, so we cannot use the Power Rule to differentiate x x . Equation (5) tells us how to differentiate a x when the base a is constant. We cannot use that equation either, because the base x in x x is not constant. Instead, to find the derivative of this function, we note that f ( x )  x x  e x ln x , so differentiation gives

d x ln x (e ) dx d = e x ln x ( x ln x ) dx

f ′( x ) =

(

= e x ln x ln x + x ·

The base e  is a constant. d u e , u = x ln x dx

1 x

)

Product Rule

= x x ( ln x + 1).

x > 0

We can also find the derivative of y  x x using logarithmic differentiation, assuming ya exists.

The Number e Expressed as a Limit In Section 1.4 we defined the number e as the base value for which the exponential function y  a x has slope 1 when it crosses the y-axis at ( 0, 1). Thus e is the constant that satisfies the equation eh − 1 = ln e = 1. h→ 0 h lim

Slope equals ln e  from Eq. ( 7 ) .

We now prove that e can be calculated as a certain limit. THEOREM 4—The Number e as a Limit

The number e can be calculated as

the limit e = lim (1 + x )1 x. x→0

Proof If f ( x )  ln x , then f ′( x ) = 1 x , so f ′(1) = 1. But, by the definition of derivative, f (1 + h ) − f (1) f (1 + x ) − f (1) = lim x→0 h x ln (1 + x ) − ln 1 1 = lim = lim ln (1 + x ) x→0 x→0 x x = lim ln (1 + x )1 x = ln ⎡⎢ lim (1 + x )1 x ⎤⎥ . ⎣ x→0 ⎦ x→0

f ′(1) = lim

h→ 0

y e 3 2

y = (1 + x)1x

ln is continuous,  Theorem 9 in Chapter 2

Because f ′(1) = 1, we have ln ⎡⎢ lim (1 + x )1 x ⎤⎥ = 1. ⎣ x→0 ⎦

1

0

ln 1 = 0

x

Therefore, exponentiating both sides, we get lim (1 + x )1 x = e. x→0

FIGURE 3.40

The number e is the limit of the function graphed here as x l 0.

See Figure 3.40. Approximating the limit in Theorem 4 by taking x very small gives approximations to e. Its value is e x 2.718281828459045 to 15 decimal places.

198

Chapter 3 Derivatives

3.8

EXERCISES

Assuming the inverse function f −1 is differentiable, find the slope of f −1 ( x ) at

Derivatives of Inverse Functions

In Exercises 1–4: a. Find f −1 ( x ). b. Graph f and f

b. x = 2

a. x = 1 −1 together.

12. The accompanying figure shows the graph of the function g.

c. Evaluate df dx at x = a and df −1 dx at x = f (a) to show that ( df −1 dx ) x = f ( a ) = 1 ( df dx ) x = a .

y

1. f ( x ) = 2 x + 3, a = −1 2. f ( x ) =

3

3. f ( x ) = 5 − 4 x , a = 1 2 5. a. Show that f ( x ) = x 3 and g( x ) = another.

3

c. Find the slopes of the tangent lines to the graphs of f and g at (1, 1) and (−1, −1) (four tangent lines in all). d. What lines are tangent to the curves at the origin? 6. a. Show that h( x ) = x 3 4 and k ( x ) = (4 x ) 1 3 are inverses of one another.

0

c. Find the slopes of the tangent lines to the graphs at h and k at ( 2, 2 ) and (−2, −2 ). d. What lines are tangent to the curves at the origin? 7. Let f ( x ) = x 3 − 3 x 2 − 1,   x ≥ 2. Find the value of df −1 dx at the point x = −1 = f (3). 8. Let f ( x ) = x 2 − 4 x − 5,   x > 2. Find the value of df −1 dx at the point x = 0 = f (5). 9. Suppose that the differentiable function y = f ( x ) has an inverse and that the graph of f passes through the point ( 2, 4 ) and has a slope of 1 3 there. Find the value of df −1 dx at x = 4. 10. Suppose that the differentiable function y = g( x ) has an inverse and that the graph of g passes through the origin with slope 2. Find the slope of the graph of g −1 at the origin. 11. The accompanying figure shows the graph of the function f. y

(3, 4) Slope = 4

4

(2, 3) Slope = 1/3

(1, 2) Slope = 3 (0, 4/3) Slope = 1/4 2

3

(2, 1) Slope = −3 1

2

3

x

4

b. x = 2

a. x = 1

c. x = 3

13. Suppose that the function f and its derivative with respect to x have the following values at x = 0, 1, 2, 3, and 4. x

0

1

2

3

4

f (x)

3

6

0

1

2

f ′( x )

43

5

4

12

17

Assuming the inverse function f −1 is differentiable, find the slope of f −1 ( x ) at b. x = 2

a. x = 1

c. x = 3

14. Suppose that the function g and its derivative with respect to x have the following values at x = 0, 1, 2, 3, and 4. 0

1

2

3

4

g( x )

−4

−1

1

2

3

g′( x )

3

2

54

23

15

x

Assuming the inverse function g −1 is differentiable, find the slope of g −1 ( x ) at a. x = 1

b. x = 2

c. x = 3

Derivatives of Logarithms

y = f (x)

1

(4, 3) (1, 2) Slope = −1/3 Slope = −1/2

Assuming the inverse function g −1 is differentiable, find the slope of g −1 ( x ) at

b. Graph h and k over an x-interval large enough to show the graphs intersecting at ( 2, 2 ) and (−2, −2 ). Be sure the picture shows the required symmetry about the line y = x.

3

(0, 7/3) Slope = −1/6

1

x are inverses of one

b. Graph f and g over an x-interval large enough to show the graphs intersecting at (1, 1) and (−1, −1). Be sure the picture shows the required symmetry about the line y = x.

0

(3, 4) Slope = −4

2

4. f ( x ) = 2 x 2 , x ≥ 0, a = 5

−1

y = g(x)

4

x +2 1 , a = 2 1− x

2 (−1, 1) Slope = 1/2 1

c. x = 3

x

In Exercises 15–44, find the derivative of y with respect to x, t, or θ, as appropriate. 1 16. y = 15. y = ln 3 x + x ln 3 x 18. y = ln ( t 3 2 ) + t 17. y = ln (t 2 ) 3 19. y = ln 20. y = ln ( sin x ) x 21. y = ln ( θ + 1) − e θ 22. y = ( cos θ ) ln ( 2θ + 2 ) 24. y = ( ln x )3

23. y = ln x 3 25. y = t ( ln t )

2

26. y = t ln t

199

3.8  Derivatives of Inverse Functions and Logarithms

x4 x4 ln x − 4 16 ln t 29. y = t ln x 31. y = 1 + ln x

30. y =

33. y = ln ( ln x )

34. y = ln ( ln ( ln x ))

27. y =

28. y = ( x 2 ln x ) 4 t ln t x ln x 32. y = 1 + ln x

38. y =

ln t

⎛ ( x 2 + 1) 5 ⎟⎞ 43. y = ln ⎜⎜ ⎟ ⎝ 1 − x ⎟⎠

44. y = ln

+

47. y =

49. y = ( sin θ ) θ + 3 51. y = t ( t + 1)( t + 2 ) 53. y = 55. y = 57. y =

θ+5 θ cos θ x x2 + 1 + 1) 2 3

(x 3

x ( x − 2) x2 + 1

46. y =

(x2

48. y =

1 t ( t + 1)

+

1)( x

−

1) 2

50. y = ( tan θ ) 2θ + 1 1 52. y = t ( t + 1)( t + 2 ) θ sin θ 54. y = sec θ (x

56. y = 58. y =

+ 1)10 + 1) 5

( 2x 3

x ( x + 1)( x − 2 )

( x 2 + 1)( 2 x + 3 )

In Exercises 59–70, find the derivative of y with respect to x, t, or θ, as appropriate. 59. y = ln ( cos 2 θ )

60. y = ln ( 3θe −θ )

61. y = ln ( 3te −t )

62. y = ln ( 2e −t sin t )

( 1 +e e ) θ

θ

65. y = e ( cos t + ln t )

⎛ θ ⎞⎟ 64. y = ln ⎜⎜ ⎝ 1 + θ ⎟⎟⎠ 66. y = e sin t ( ln t 2 + 1)

In Exercises 67–70, find dy dx. 67. ln y = e y sin x

68. ln xy = e x + y

69. x y = y x

70. tan y = e x + ln x

In Exercises 71–92, find the derivative of y with respect to the given independent variable. 71. y = 2 x

72. y = 3 −x

73. y = 5

74. y = 2

75. y =

s

xπ

77. y = log 2 5θ

87. y = log 5 e x

88. y = log 2

ln 5

( sineθ 2cos θ ) θ

θ

( 2 xx e+ 1 ) 2 2

90. y = 3 log 8 ( log 2 t ) ln 2

92. y = t log 3 ( e ( sin t )( ln 3) )

)

In Exercises 93–104, use logarithmic differentiation or the method in Example 7 to find the derivative of y with respect to the given independent variable.

(s 2 )

76. y = t 1− e 78. y = log 3 (1 + θ ln 3 )

99. y = x

94. y = x ( x +1)

t

97. y = ( sin x )

Finding Derivatives

63. y = ln

86. y = log 7

95. y = ( t )

In Exercises 45–58, use logarithmic differentiation to find the derivative of y with respect to the given independent variable. t t +1

( 3x7+x 2 )

84. y = log 5

85. y = θ sin ( log 7 θ )

93. y = ( x + 1) x

+ 1) 5 ( x + 2 ) 20 (x

Logarithmic Differentiation

45. y =

(( xx +− 11) )

Powers with Variable Bases and Exponents

⎛ sin θ cos θ ⎞⎟ 42. y = ln ⎜⎜ ⎜⎝ 1 + 2 ln θ ⎟⎟⎠

1)

82. y = log 3 r ⋅ log 9 r ln 3

91. y = log 2 ( 8t

41. y = ln ( sec ( ln θ ))

x( x

81. y = log 2 r ⋅ log 4 r

89. y = 3 log 2 t

1 1+ x ln 2 1− x

40. y =

80. y = log 25 e x − log 5 x

83. y = log 3

35. y = θ ( sin ( ln θ ) + cos( ln θ )) 36. y = ln ( sec θ + tan θ ) 1 37. y = ln x x +1 1 + ln t 39. y = 1 − ln t

79. y = log 4 x + log 4 x 2

96. y = t x

98. y = x

t sin x

100. y = ( ln x ) ln x

ln x

101. y x = x 3 y

102. x sin y = ln y

103. x = y xy

104. e y = y ln x

Theory and Applications

105. If we write g( x ) for f −1 ( x ), Equation (1) can be written as 1 g′( f (a)) = , or g′( f (a)) · f ′(a) = 1. f ′(a) If we then write x for a, we get g′( f ( x )) · f ′( x ) = 1. The latter equation may remind you of the Chain Rule, and indeed there is a connection. Assume that f and g are differentiable functions that are inverses of one another, so that ( g  f )( x ) = x. Differentiate both sides of this equation with respect to x, using the Chain Rule to express ( g  f )′( x ) as a product of derivatives of g and f. What do you find? (This is not a proof of Theorem 3 because we assume here the theorem’s conclusion that g = f −1 is differentiable.) x n = e x for any x > 0. 106. Show that lim 1 + n →∞ n 107. If f ( x ) = x n ,  n > 1, show from the definition of the derivative that f ′(0) = 0.

(

)

108. Using mathematical induction, show that for n > 1, ( n − 1)! dn ln x = (−1) n−1 . dx n xn

COMPUTER EXPLORATIONS

In Exercises 109–116, you will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS: a. Plot the function y = f ( x ) together with its derivative over the given interval. Explain why you know that f is one-to-one over the interval. b. Solve the equation y = f ( x ) for x as a function of y, and name the resulting inverse function g.

200

Chapter 3 Derivatives

c. Find an equation for the tangent line to f at the specified point ( x 0 , f ( x 0 ) ).

112. y =

d. Find an equation for the tangent line to g at the point ( f ( x 0 ),   x 0 ) located symmetrically across the 45 n line y  x (which is the graph of the identity function). Use Theorem 3 to find the slope of this tangent line.

113. y = x 3 − 3 x 2 − 1, 2 ≤ x ≤ 5, x 0 =

x2

x3 , −1 ≤ x ≤ 1, x 0 = 1 2 +1

114. y = 2 − x − x 3 , −2 ≤ x ≤ 2, x 0

27 10 3 = 2

115. y = e x , −3 ≤ x ≤ 5, x 0 = 1 Q Q 116. y = sin x , − ≤ x ≤ , x 0 = 1 2 2

e. Plot the functions f and g, the identity, the two tangent lines, and the line segment joining the points ( x 0 ,   f ( x 0 ) ) and ( f ( x 0 ),   x 0 ). Discuss the symmetries you see across the main diagonal (the line y  x). 2 ≤ x ≤ 4, x 0 = 3 109. y = 3 x − 2, 3 3x + 2 110. y = , −2 ≤ x ≤ 2, x 0 = 1 2 2 x − 11 4x 111. y = 2 , −1 ≤ x ≤ 1, x 0 = 1 2 x +1

In Exercises 117 and 118, repeat the steps above to solve for the functions y  f ( x ) and x = f −1 ( y) defined implicitly by the given equations over the interval. 117. y 1 3 − 1 = ( x + 2 )3 , −5 ≤ x ≤ 5, x 0 = − 3 2 118. cos y = x 1 5 , 0 ≤ x ≤ 1, x 0 = 1 2

3.9 Inverse Trigonometric Functions We introduced the six basic inverse trigonometric functions in Section 1.5 but focused there on the arcsine and arccosine functions. Here we complete the study of how all six basic inverse trigonometric functions are defined, graphed, and evaluated, and how their derivatives are computed.

Inverses of tan x, cot x, sec x , and csc x The graphs of these four basic inverse trigonometric functions are shown in Figure 3.41. We obtain these graphs by reflecting the graphs of the restricted trigonometric functions (as discussed in Section 1.5) through the line y  x. Let’s take a closer look at the arctangent, arccotangent, arcsecant, and arccosecant functions. Domain: −∞ < x < ∞ Range: − p < y < p 2 2 y p 2 −2 −1 −

p 2

y p

Domain: x ≤ −1 or x ≥ 1 Range: − p ≤ y ≤ p , y ≠ 0 2 2 y p 2

p

y = arctan x x 1 2

p 2 −2 −1

y = arccot x

1 (b)

(a)

FIGURE 3.41

Domain: x ≤ −1 or x ≥ 1 Range: 0 ≤ y ≤ p, y ≠ p 2 y

Domain: −∞ < x < ∞ 0 −1 be differentiable for all values of x? b. Discuss the geometry of the resulting graph of g. 21. Odd differentiable functions Is there anything special about the derivative of an odd differentiable function of x? Give reasons for your answer. 22. Even differentiable functions Is there anything special about the derivative of an even differentiable function of x? Give reasons for your answer.

Show that the derivative f a( x ) exists at every value of x and that f ′( x ) = f ( x ). 29. The generalized product rule Use mathematical induction to prove that if y  u1u 2 "u n is a finite product of differentiable functions, then y is differentiable on their common domain, and du du1 du dy u "u n + u1 2 "u n + " + u1u 2 "u n −1 n . = dx dx 2 dx dx 30. Leibniz’s rule for higher-order derivatives of products Leibniz’s rule for higher-order derivatives of products of differentiable functions says that d 2 ( uV ) d 2u du d V d 2V a. = + u 2. V+2 2 2 dx dx dx dx dx d 3 ( uV ) d 3u d 2u d V du d 2 V d 3V b. = +3 + u 3. V+3 2 3 3 2 dx dx dx dx dx dx dx

234

Chapter 3 Derivatives

n( n n −1 ) c. d unV = d un V + n d n−u1 d V + " dx dx dx dx n ( n − 1) " ( n − k + 1) d n− k u d k V + k! dx n− k dx k d nV +"+ u n. dx

The equations in parts (a) and (b) are special cases of the equation in part (c). Derive the equation in part (c) by mathematical induction, using

32. The melting ice cube Assume that an ice cube retains its cubical shape as it melts. If we call its edge length s, its volume is V  s 3 and its surface area is 6s 2 . We assume that V and s are differentiable functions of time t. We assume also that the cube’s volume decreases at a rate that is proportional to its surface area. (This latter assumption seems reasonable enough when we think that the melting takes place at the surface: Changing the amount of surface changes the amount of ice exposed to melt.) In mathematical terms,

⎛ m ⎞⎟ ⎛⎜ m ⎟⎞ m! m! ⎜⎜ ⎟ + ⎜ ⎟= + . ⎜⎝ k ⎟⎠ ⎜⎝ k + 1⎟⎟⎠ k !( m − k )! ( k + 1)!( m − k − 1)! 31. The period of a clock pendulum The period T of a clock pendulum (time for one full swing and back) is given by the formula T 2  4 Q 2 L g , where T is measured in seconds, g  9.8 m s 2, and L, the length of the pendulum, is measured in meters. Find approximately a. the length of a clock pendulum whose period is T  1 s. b. the change dT in T if the pendulum in part (a) is lengthened 0.01 m.

dV = −k ( 6s 2 ), dt

k > 0.

The minus sign indicates that the volume is decreasing. We assume that the proportionality factor k is constant. (It probably depends on many things, such as the relative humidity of the surrounding air, the air temperature, and the incidence or absence of sunlight, to name only a few.) Assume a particular set of conditions in which the cube lost 1 4 of its volume during the first hour, and assume that the volume is V0 when t  0. How long will it take the ice cube to melt?

c. the amount the clock gains or loses in a day as a result of the period’s changing by the amount dT found in part (b).

CHAPTER 3

Technology Application Projects

Mathematica/Maple Projects

Projects can be found at www.pearsonglobaleditions.com or within MyLab Math. • Convergence of Secant Slopes to the Derivative Function You will visualize the secant line between successive points on a curve and observe what happens as the distance between them becomes small. The function, sample points, and secant lines are plotted on a single graph, while a second graph compares the slopes of the secant lines with the derivative function. • Derivatives, Slopes, Tangent Lines, and Making Movies Parts I–III. You will visualize the derivative at a point, the linearization of a function, and the derivative of a function. You will learn how to plot the function and selected tangent lines on the same graph. Part IV (Plotting Many Tangent Lines) Part V (Making Movies). Parts IV and V of the module can be used to animate tangent lines as one moves along the graph of a function. • Convergence of Secant Slopes to the Derivative Function You will visualize right-hand and left-hand derivatives. • Motion Along a Straight Line: Position l Velocity l Acceleration Observe dramatic animated visualizations of the derivative relations among the position, velocity, and acceleration functions. Figures in the text can be animated.

4

Applications of Derivatives OVERVIEW One of the most important applications of the derivative is its use as a tool for finding the optimal (best) solutions to problems. For example, what are the height and diameter of the cylinder of largest volume that can be inscribed in a given sphere? What are the dimensions of the strongest rectangular wooden beam that can be cut from a cylindrical log of given diameter? How many items should a manufacturer produce to maximize profit? In this chapter we apply derivatives to find extreme values of functions, to determine and analyze the shapes of graphs, and to solve equations numerically. We also investigate how to recover a function from its derivative. The key to many of these applications is the Mean Value Theorem, which connects the derivative and the average change of a function.

4.1 Extreme Values of Functions on Closed Intervals This section shows how to locate and identify extreme (maximum or minimum) values of a function from its derivative. Once we can do this, we can solve a variety of optimization problems (see Section 4.6). The domains of the functions we consider are intervals or unions of separate intervals.

DEFINITIONS Let f be a function with domain D. Then f has an absolute maximum value on D at a point c if

f ( x ) ≤ f (c)

for all  x  in D

and an absolute minimum value on D at c if y y = cos x

−

f ( x ) ≥ f (c)

for all  x  in D.

1 y = sin x

p 2

0

p 2

x

−1

FIGURE 4.1 Absolute extrema for the sine and cosine functions on [ −π 2, π 2 ]. These values can depend on the domain of a function.

Maximum and minimum values are called extreme values of the function f . Absolute maxima or minima are also referred to as global maxima or minima. For example, on the closed interval [ −π 2, π 2 ] the function f ( x ) = cos x takes on an absolute maximum value of 1 (once) and an absolute minimum value of 0 (twice). On the same interval, the function g( x ) = sin x takes on a maximum value of 1 and a minimum value of −1 (Figure 4.1). Functions defined by the same equation or formula can have different extrema (maximum or minimum values), depending on the domain. A function might not have a maximum or minimum if the domain is unbounded or fails to contain an endpoint. We see this in the following example.

235

236

Chapter 4 Applications of Derivatives

EXAMPLE 1 The absolute extrema of the following functions on their domains can be seen in Figure 4.2. Each function has the same defining equation, y  x 2 , but the domains vary.

y = x2

Function rule

Domain D

Absolute extrema on D

(a) y  x 2

(−∞, ∞ )

No absolute maximum Absolute minimum of 0 at x  0

(b) y  x 2

[ 0, 2 ]

Absolute maximum of 4 at x  2 Absolute minimum of 0 at x  0

(c) y  x 2

( 0, 2 ]

Absolute maximum of 4 at x  2 No absolute minimum

(d) y  x 2

( 0, 2 )

No absolute extrema

y = x2

y

D = (−∞, ∞)

FIGURE 4.2

HISTORICAL BIOGRAPHY Daniel Bernoulli (1700–1789) Daniel Bernoulli was the second son of mathematician Johann Bernoulli. In 1724, Bernoulli published his Exercitationes mathematicae that attracted considerable attention. To know more, visit the companion Website.

x

2 (b) abs max and min

y = x2

y

D = (0, 2]

D = [0, 2]

2 (a) abs min only

y = x2

y

x

D = (0, 2)

2 (c) abs max only

y

x

2 (d) no max or min

x

Graphs for Example 1.

Some of the functions in Example 1 do not have a maximum or a minimum value. The following theorem asserts that a function which is continuous over (or on) a finite closed interval [ a, b ] has an absolute maximum and an absolute minimum value on the interval. We look for these extreme values when we graph a function.

THEOREM 1—The Extreme Value Theorem

If f is continuous on a closed interval [ a, b ], then f attains both an absolute maximum value M and an absolute minimum value m in [ a, b ]. That is, there are numbers x 1 and x 2 in [ a, b ] with f ( x 1 )  m, f (x 2 )  M , and m b f ( x ) b M for all x in [ a, b ].

The proof of the Extreme Value Theorem requires a detailed knowledge of the real number system (see Appendix A.9) and we will not give it here. Figure 4.3 illustrates possible locations for the absolute extrema of a continuous function on a closed interval [ a, b ]. As we observed for the function y  cos x , it is possible that an absolute minimum (or absolute maximum) may occur at two or more different points of the interval. The requirements in Theorem 1 that the interval be closed and finite, and that the function be continuous, are essential. Without them, the conclusion of the theorem need not hold. Example 1 shows that an absolute extreme value may not exist if the interval fails to be both closed and finite. The exponential function y = e x over ( −∞, ∞ ) shows that

237

4.1  Extreme Values of Functions on Closed Intervals

(x2, M) y = f (x)

y = f (x)

M

M x1

a

x2

b

0m0

m

x

b

a

x

Maximum and minimum at endpoints

(x1, m) Maximum and minimum at interior points

y = f (x) y = f (x)

M

m

m a

y No largest value

1 Smallest value

b

a

x1

b

x

Minimum at interior point, maximum at endpoint

FIGURE 4.3 Some possibilities for a continuous function’s maximum and minimum on a closed interval [ a, b ].

y=x 0≤ x 0. Likewise, f has a local maximum at an interior point x = c if f ( x ) ≤ f (c) for all x in some open interval ( c − δ , c + δ ) , δ > 0, and a local maximum at the endpoint x = b if f ( x ) ≤ f (b) for all x in some half-open interval ( b − δ , b ] , δ > 0. The inequalities are reversed for local minimum values. In Figure 4.5, the function f has local maxima at c and d and local minima at a, e, and b. Local extrema are also called relative extrema. Some functions can have infinitely many local extrema, even over a finite interval. One example is the function f ( x ) = sin (1 x ) on the interval ( 0, 1 ]. (We graphed this function in Figure 2.41.) An absolute maximum is also a local maximum. Being the largest value overall, it is also the largest value in its immediate neighborhood. Hence, a list of all local maxima will

238

Chapter 4 Applications of Derivatives

Absolute maximum No greater value of f anywhere. Also a local maximum.

Local maximum No greater value of f nearby.

Local minimum No smaller value of f nearby.

y = f (x)

Absolute minimum No smaller value of f anywhere. Also a local minimum.

Local minimum No smaller value of f nearby. a

FIGURE 4.5

c

e

d

b

x

How to identify types of maxima and minima for a function with domain

a ≤ x ≤ b.

automatically include the absolute maximum if there is one. Similarly, a list of all local minima will include the absolute minimum if there is one.

Finding Extrema The next theorem explains why we usually need to investigate only a few values to find a function’s extrema.

THEOREM 2—The First Derivative Theorem for Local Extreme Values

If f has a local maximum or minimum value at an interior point c of its domain, and if f ′ is defined at c, then f ′(c) = 0.

Local maximum value

Proof To prove that f ′(c) is zero at a local extremum, we show first that f ′(c) cannot be positive and second that f ′(c) cannot be negative. The only number that is neither positive nor negative is zero, so that is what f ′(c) must be. To begin, suppose that f has a local maximum value at x = c (Figure 4.6) so that f ( x ) − f (c) ≤ 0 for all values of x near enough to c. Since c is an interior point of f ’s domain, f ′(c) is defined by the two-sided limit

y = f (x)

lim Secant slopes ≥ 0 (never negative)

x

Secant slopes ≤ 0 (never positive)

c

x

x→c

This means that the right-hand and left-hand limits both exist at x = c and equal f ′(c). When we examine these limits separately, we find that

x

FIGURE 4.6 A curve with a local maximum value. The slope at c, simultaneously the limit of nonpositive numbers and nonnegative numbers, is zero.

f ( x ) − f (c) . x−c

f ′(c) = lim+

f ( x ) − f (c) ≤ 0. x−c

Because  x − c > 0 and  f ( x ) ≤ f (c)

(1)

f ′(c) = lim−

f ( x ) − f (c) ≥ 0. x−c

Because  x − c < 0 and  f ( x ) ≤ f (c)

(2)

x →c

Similarly, x →c

Together, Equations (1) and (2) imply f ′(c) = 0. This proves the theorem for local maximum values. To prove it for local minimum values, we simply use f ( x ) ≥ f (c), which reverses the inequalities in Formulas (1) and (2).

4.1  Extreme Values of Functions on Closed Intervals

y

Theorem 2 says that a function’s first derivative is always zero at an interior point where the function has a local extreme value and the derivative is defined. If we recall that all the domains we consider are intervals or unions of separate intervals, the only places where a function f can possibly have an extreme value (local or global) are

y = x3 1

−1

0

1

239

x

−1

1. interior points where f ′ = 0, 2. interior points where f ′ is undefined, 3. endpoints of an interval in the domain of f .

At x = c and x = e in Fig. 4.5 At x = d in Fig. 4.5 At x = a and x = b in Fig. 4.5

The following definition helps us to summarize these results. (a)

DEFINITION An interior point of the domain of a function f where f ′ is zero or

y

undefined is a critical point of f .

1 y = x13 −1

0

1

x

−1 (b)

FIGURE 4.7 Critical points without extreme values. (a) y′ = 3 x 2 is 0 at x = 0, but y = x 3 has no extremum there. (b) y′ = (1 3 ) x −2 3 is undefined at x = 0, but y = x 1 3 has no extremum there.

Thus the only domain points where a continuous function on a closed and finite interval can assume extreme values are critical points and endpoints. However, be careful not to misinterpret what is being said here. A function may have a critical point at x = c without having a local extreme value there. For instance, both of the functions y = x 3  and y = x 1 3 have critical points at the origin, but neither function has a local extreme value at the origin. Instead, each function has a point of inflection there (see Figure 4.7). We define and explore inflection points in Section 4.4. If the interval is not closed or not finite (such as a < x < b or a < x < ∞), we have seen that absolute extrema need not exist. However, many problems that ask for extreme values call for finding the absolute extrema of a continuous function on a closed and finite interval. Theorem 1 assures us that such values exist; Theorem 2 tells us that they are taken on only at critical points and endpoints. Often we can simply list these points and calculate the corresponding function values to find what the largest and smallest values are, and where they are located. Finding the Absolute Extrema of a Continuous Function f on a Finite Closed Interval

1. Find all critical points of f on the interval. 2. Evaluate f at all critical points and endpoints. 3. Take the largest and smallest of these values.

EXAMPLE 2 [ −2, 1 ].

Find the absolute maximum and minimum values of f ( x ) = x 2 on

Solution The function is differentiable over its entire domain, so the only critical point occurs where f ′( x ) = 2 x = 0, namely x = 0. We need to check the function’s values at x = 0 and at the endpoints x = −2 and x = 1:

Critical point value: Endpoint values:

f (0) = 0 f (−2) = 4 f (1) = 1.

The function has an absolute maximum value of 4 at x = −2 and an absolute minimum value of 0 at x = 0. EXAMPLE 3 Find the absolute maximum f ( x ) = 10 x ( 2 − ln x ) on the interval [ 1, e 2 ].

and

minimum

values

of

240

Chapter 4 Applications of Derivatives

Solution Figure 4.8 suggests that f has its absolute maximum value near x = 3 and its absolute minimum value of 0 at x = e 2 . Let’s verify this observation. We evaluate the function at the critical points and endpoints and take the largest and smallest of the resulting values. The first derivative is

y 30

(e, 10e)

25 20 15

(1, 20)

10

f ′( x ) = 10 ( 2 − ln x ) − 10 x

5 0

(e 2, 0) 1

2

3

4

5

6

7

x

8

FIGURE 4.8

The extreme values of f ( x ) = 10 x ( 2 − ln x ) on [ 1, e 2 ] occur at x = e and x = e 2 (Example 3).

( 1x ) = 10(1 − ln x ).

The only critical point in the domain [ 1, e 2 ] is the point x = e, where ln x = 1. The values of f at this one critical point and at the endpoints are Critical point value:

f (e) = 10e ( 2 − ln e ) = 10e

Endpoint values:

f (1) = 10 ( 2 − ln 1) = 20 f (e 2 ) = 10e 2 ( 2 − 2 ln e ) = 0.

We can see from this list that the function’s absolute maximum value is 10e ≈ 27.2; it occurs at the critical interior point x = e. The absolute minimum value is 0 and occurs at the right endpoint x = e 2 .

EXAMPLE 4 Find the absolute maximum and minimum values of f ( x ) = x 2 3 on the interval [ −2, 3 ]. Solution We evaluate the function at the critical points and endpoints and take the largest and smallest of the resulting values. The first derivative, y

Local maximum 2

f ′( x ) =

y = x 23, −2 ≤ x ≤ 3 Absolute maximum; also a local maximum

1 −2

−1

0

FIGURE 4.9

1 2 3 Absolute minimum; also a local minimum

The extreme values of f ( x ) = x on [ −2, 3 ] occur at x = 0 and x = 3 (Example 4). 23

has no zeros but is undefined at the interior point x = 0. The values of f at this one critical point and at the endpoints are Critical point value:

x

2 −1 3 2 x = 3 , 3 3 x

Endpoint values:

f (0) = 0 f (−2) = (−2 ) 2 3 =

3

f (3) = ( 3 ) 2 3 =

9.

3

4

We can see from this list that the function’s absolute maximum value is 3 9 ≈ 2.08, and it occurs at the right endpoint x = 3. The absolute minimum value is 0, and it occurs at the interior point x = 0 where the graph has a cusp (Figure 4.9). Theorem 1 leads to a method for finding the absolute maxima and absolute minima of a differentiable function on a finite closed interval. On more general domains, such as ( 0, 1), [ 2, 5 ) , [ 1, ∞ ), and (−∞,   ∞ ) , absolute maxima and minima may or may not exist. To determine if they exist, and to locate them when they do, we will develop methods to sketch the graph of a differentiable function. With knowledge of the asymptotes of the function, as well as the local maxima and minima, we can deduce the locations of the absolute maxima and minima, if any. For now we can find the absolute maxima and the absolute minima of a function on a finite closed interval by comparing the values of the function at its critical points and at the endpoints of the interval. For a differentiable function on a closed and finite interval [ a, b ], these are the only points where the extrema have the potential to occur.

241

4.1  Extreme Values of Functions on Closed Intervals

4.1

EXERCISES

Finding Extrema from Graphs

In Exercises 11–14, match the table with a graph.

In Exercises 1–6, determine from the graph whether the function has any absolute extreme values on [ a, b ]. Then explain how your answer is consistent with Theorem 1.

11. x

1.

2.

y

y y = f (x)

y = h(x)

f ′( x )

a

0

a

0

b

0

b

0

c

5

c

−5

f ′( x )

13. x a

0

3.

c1

c2

b

x

4.

y

a

0

c

a

5.

c

b

does not exist

a

does not exist

b

0

b

does not exist

c

−2

c

−1.7

y = h(x)

x

a

0

6.

y

c

a

x

b

b c

(a)

b

x

b

c

0

a

c

8.

y

b

y

9.

1

−2

10.

y

0

⎧⎪ −x , 0 ≤ x 0,

with s in meters and t in seconds. Find the body’s maximum height. 70. Peak alternating current Suppose that at any given time t (in seconds) the current i (in amperes) in an alternating current circuit is i = 2 cos t + 2 sin t. What is the peak current for this circuit (largest magnitude)? T Graph the functions in Exercises 71–74. Then find the extreme values of the function on the interval and say where they occur.

In Exercises 75–82, you will use a CAS to help find the absolute extrema of the given function over the specified closed interval. Perform the following steps. a. Plot the function over the interval to see its general behavior there. b. Find the interior points where f ′ = 0. (In some exercises, you may have to use the numerical equation solver to approximate a solution.) You may want to plot f ′ as well. c. Find the interior points where f ′ does not exist. d. Evaluate the function at all points found in parts (b) and (c) and at the endpoints of the interval. e. Find the function’s absolute extreme values on the interval and identify where they occur. 75. f ( x ) = x 4 − 8 x 2 + 4 x + 2, [ −20 25, 64 25 ] 76. f ( x ) = −x 4 + 4 x 3 − 4 x + 1, [ −3 4 , 3 ] 77. f ( x ) = x 2 3 ( 3 − x ) , [ −2, 2 ] 78. f ( x ) = 2 + 2 x − 3 x 2 3 , [ −1, 10 3 ]

73. h( x ) = x + 2 − x − 3 , −∞ < x < ∞

x + cos x , [ 0, 2π ] 1 80. f ( x ) = x − sin x + , [ 0, 2π ] 2 81. f ( x ) = π x 2e −3 x 2, [ 0, 5 ]

74. k ( x ) = x + 1 + x − 3 , −∞ < x < ∞

82. f ( x ) = ln ( 2 x + x sin x ) , [ 1, 15 ]

79. f ( x ) =

71. f ( x ) = x − 2 + x + 3 , − 5 ≤ x ≤ 5

34

72. g( x ) = x − 1 − x − 5 , − 2 ≤ x ≤ 7

4.2 The Mean Value Theorem We know that constant functions have zero derivatives, but could there be a more complicated function whose derivative is always zero? If two functions have identical derivatives over an interval, how are the functions related? We answer these and other questions in this chapter by applying the Mean Value Theorem. First we introduce a special case, known as Rolle’s Theorem, which is used to prove the Mean Value Theorem.

y f ′(c) = 0 y = f (x)

a

0

c

Rolle’s Theorem As suggested by its graph, if a differentiable function crosses a horizontal line at two different points, there is at least one point between them where the tangent to the graph is horizontal and the derivative is zero (Figure 4.10). We now state and prove this result.

x

b

(a) y f ′(c1 ) = 0

0

a

c1

f ′(c3 ) = 0 f ′(c2 ) = 0

c2

THEOREM 3—Rolle’s Theorem

Suppose that y = f ( x ) is continuous over the closed interval [ a, b ] and differentiable at every point of its interior (a, b). If f (a) = f (b), then there is at least one number c in (a, b) at which f ′(c) = 0.

y = f (x)

c3

b

x

(b)

FIGURE 4.10

Rolle’s Theorem says that a differentiable curve has at least one horizontal tangent between any two points where it crosses a horizontal line. It may have just one (a), or it may have more (b).

Proof Being continuous, f assumes absolute maximum and minimum values on [ a, b ] by Theorem 1. These can occur only 1. at interior points where f ′ is zero, 2. at interior points where f ′ does not exist, 3. at endpoints of the interval, in this case a and b.

244

Chapter 4 Applications of Derivatives

HISTORICAL BIOGRAPHY Michel Rolle (1652–1719) French mathematician Michel Rolle was largely self-educated in mathematics. He worked as an accountant and studied algebra and the Diophantine equations whenever he found time.

By hypothesis, f has a derivative at every interior point. That rules out possibility (2), leaving us with interior points where f ′ = 0 and with the two endpoints a and b. If either the maximum or the minimum occurs at a point c between a and b, then f ′(c) = 0 by Theorem 2 in Section 4.1, and we have found a point for Rolle’s Theorem. If both the absolute maximum and the absolute minimum occur at the endpoints, then because f (a)  f (b) it must be the case that f is a constant function with f ( x )  f (a)  f (b) for every x ∈ [ a, b ]. Therefore, f ′( x ) = 0 and the point c can be taken anywhere in the interior (a, b). The hypotheses of Theorem 3 are essential. If they fail at even one point, the graph may not have a horizontal tangent (Figure 4.11).

To know more, visit the companion Website.

y

y

y

y = f (x)

a

y = f (x)

b

x

(a) Discontinuous at an endpoint of [a, b]

a

x0 b

(b) Discontinuous at an interior point of [a, b]

y = f(x)

x

a

x0

b

x

(c) Continuous on [a, b] but not differentiable at an interior point

FIGURE 4.11

There may be no horizontal tangent line if the hypotheses of Rolle’s Theorem do not hold.

Rolle’s Theorem may be combined with the Intermediate Value Theorem to show when there is only one real solution of an equation f ( x )  0, as we illustrate in the next example.

EXAMPLE 1

Show that the equation

y

x 3 + 3x + 1 = 0

(1, 5)

has exactly one real solution.

y = x 3 + 3x + 1

Solution We define the continuous function

f ( x ) = x 3 + 3 x + 1.

1 −1

0

1

x

(−1, −3)

FIGURE 4.12 The only real zero of the polynomial y = x 3 + 3 x + 1 is the one shown here where the curve crosses the x-axis between 1 and 0 (Example 1).

Since f (−1) = −3 and f (0)  1, the Intermediate Value Theorem tells us that the graph of f crosses the x-axis somewhere in the open interval (−1, 0 ). (See Figure 4.12.) Now, if there were even two points x  a and x  b where f ( x ) was zero, Rolle’s Theorem would guarantee the existence of a point x  c between them where f a was zero. However, the derivative f ′( x ) = 3 x 2 + 3 is never zero (because it is always positive). Therefore, f has no more than one zero. We will use Rolle’s Theorem to prove the Mean Value Theorem.

245

4.2  The Mean Value Theorem

Tangent line parallel to secant line

y

Slope f ′(c)

The Mean Value Theorem, which was first stated by Joseph-Louis Lagrange, is a slanted version of Rolle’s Theorem (Figure 4.13). The Mean Value Theorem guarantees that there is a point where the tangent line is parallel to the secant line that joins A and B.

B Slope

The Mean Value Theorem

f (b) − f (a) b−a

A 0

a y = f(x)

c

x

b

FIGURE 4.13 Geometrically, the Mean Value Theorem says that somewhere between a and b the curve has at least one tangent line parallel to the secant line that joins A and B. HISTORICAL BIOGRAPHY Joseph-Louis Lagrange (1736–1813) Lagrange was born in Turin, Italy. He enjoyed studying mathematics, despite his father’s wish that he study law. Lagrange’s mathematical contributions began as early as 1754 with the discovery of the calculus of variations and continued with applications to mechanics in 1756. To know more, visit the companion Website.

y = f(x)

B(b, f (b))

THEOREM 4—The Mean Value Theorem

Suppose y  f ( x ) is continuous over a closed interval [ a, b ] and differentiable on the interval’s interior (a, b). Then there is at least one point c in (a, b) at which f (b) − f (a) = f ′(c). b−a

Proof We picture the graph of f and draw a line through the points A ( a, f (a) ) and B ( b, f (b) ). (See Figure 4.14.) The secant line is the graph of the function g( x ) = f (a) +

f (b) − f (a) ( x − a) b−a

(2)

(point-slope equation). The vertical difference between the graphs of f and g at x is

A(a, f(a))

h( x ) = f ( x ) − g( x ) = f ( x ) − f (a) −

a

b

x

FIGURE 4.14 The graph of f and the secant line AB over the interval [ a, b ]. y = f(x)

B h(x)

y = g(x)

A

(1)

a

FIGURE 4.15

x

b

The secant line AB is the graph of the function g( x ). The function h( x ) = f ( x ) − g( x ) gives the vertical distance between the graphs of f and g at x.

(3)

Figure 4.15 shows the graphs of f , g, and h together. The function h satisfies the hypotheses of Rolle’s Theorem on [ a, b ]. It is continuous on [ a, b ] and differentiable on (a, b) because both f and g are. Also, h(a)  h(b)  0 because the graphs of f and g both pass through A and B. Therefore h′(c) = 0 at some point c ∈ ( a, b ). This is the point we want for Equation (1) in the theorem. To verify Equation (1), we differentiate both sides of Equation (3) with respect to x and then set x  c: h′( x ) = f ′( x ) −

f (b) − f (a) b−a

Derivative of Eq.  ( 3)

h′(c) = f ′(c) −

f (b) − f (a) b−a

Evaluated at  x = c

0 = f ′(c) −

f (b) − f (a) b−a

h ′(c ) = 0

h(x) = f (x) − g(x) x

f (b) − f (a) ( x − a ). b−a

f ′(c) =

f (b) − f (a) , b−a

which is what we set out to prove.

Rearranged

246

Chapter 4 Applications of Derivatives

The hypotheses of the Mean Value Theorem do not require f to be differentiable at either a or b. One-sided continuity at a and b is enough (Figure 4.16).

y = "1 − x 2, −1 ≤ x ≤ 1 y 1

−1

0

x

1

FIGURE 4.16

The function f ( x ) = 1 − x 2 satisfies the hypotheses (and conclusion) of the Mean Value Theorem on [ −1, 1 ] even though f is not differentiable at 1 and 1.

y B(2, 4)

4

EXAMPLE 2 The function f ( x )  x 2 (Figure 4.17) is continuous for 0 b x b 2 and differentiable for 0  x  2. Since f (0)  0 and f (2)  4, the Mean Value Theorem says that at some point c in the interval, the derivative f ′( x ) = 2 x must have the value ( 4 − 0 ) ( 2 − 0 ) = 2. In this case we can identify c by solving the equation 2c  2 to get c  1. However, it is not always easy to find c algebraically, even though we know it always exists.

A Physical Interpretation We can think of the number ( f (b) − f (a) ) ( b − a ) as the average change in f over [ a, b ] and can view f a(c) as an instantaneous change. Then the Mean Value Theorem says that the instantaneous change at some interior point is equal to the average change over the entire interval. EXAMPLE 3 If a car accelerating from zero takes 8 s to go 176 m, its average velocity for the 8-s interval is 176 8  22 m s. The Mean Value Theorem says that at some point during the acceleration, the speedometer must read exactly 79.2 km h ( 22 m s ) (Figure 4.18).

3 y = x2 2

Mathematical Consequences 1

(1, 1)

1

A(0, 0)

x

2

FIGURE 4.17 As we find in Example 1, c  1 is where the tangent line is parallel to the secant line.

s

Distance (m)

200

s = f (t) (8, 176)

160

At the beginning of the section, we asked what kind of function has a zero derivative over an interval. The first corollary of the Mean Value Theorem provides the answer that only constant functions have zero derivatives. If f ′( x ) = 0 at each point x of an open interval ( a, b ), then f ( x )  C for all x ∈ ( a, b ) , where C is a constant.

COROLLARY 1

Proof We want to show that f has a constant value on the interval ( a, b ). We do so by showing that if x 1 and x 2 are any two points in ( a, b ) with x 1  x 2, then f ( x 1 )  f ( x 2 ). Now f satisfies the hypotheses of the Mean Value Theorem on [ x 1 , x 2 ]: It is differentiable at every point of [ x 1 , x 2 ] and hence continuous at every point as well. Therefore, f (x 2 ) − f (x1) = f ′(c) x 2 − x1

120 80 40 0

FIGURE 4.18

At this point, the car’s speed was 79.2 km/h. t

5 Time (s)

Distance versus elapsed time for the car in Example 3.

at some point c between x 1 and x 2 . Since f ′ = 0 throughout ( a, b ), this equation implies successively that f (x 2 ) − f (x1) = 0, x 2 − x1

f ( x 2 ) − f ( x 1 ) = 0,

and

f ( x 1 ) = f ( x 2 ).

At the beginning of this section, we also asked about the relationship between two functions that have identical derivatives over an interval. The next corollary tells us that their values on the interval have a constant difference. COROLLARY 2 If f ′( x ) = g′( x ) at each point x in an open interval ( a, b ), then there exists a constant C such that f ( x ) = g( x ) + C for all x ∈ ( a, b ). That is, f  g is a constant function on ( a, b ).

4.2  The Mean Value Theorem

y

y = x2 + C

Proof

C=2

h′( x ) = f ′( x ) − g′( x ) = 0.

C=0

Thus, h( x ) = C on ( a, b ) by Corollary 1. That is, f ( x ) − g( x ) = C on ( a, b ), so f ( x ) = g( x ) + C.

C = −2

1 0

At each point x ∈ ( a, b ) the derivative of the difference function h = f − g is

C=1

C = −1 2

247

x

−1

Corollaries 1 and 2 are also true if the open interval ( a, b ) fails to be finite. That is, they remain true if the interval is ( a, ∞ ) , (−∞, b ) ,  or  ( −∞, ∞ ). Corollary 2 will play an important role when we discuss antiderivatives in Section 4.8. It tells us, for instance, that since the derivative of f ( x ) = x 2 on (−∞, ∞ ) is 2x, any other function with derivative 2x on (−∞, ∞ ) must have the formula x 2 + C for some value of C (Figure 4.19).

−2

FIGURE 4.19

From a geometric point of view, Corollary 2 of the Mean Value Theorem says that the graphs of functions with identical derivatives on an interval can differ only by a vertical shift. The graphs of the functions with derivative 2x are the parabolas y = x 2 + C, shown here for several values of C.

EXAMPLE 4 Find the function f ( x ) whose derivative is sin x and whose graph passes through the point ( 0, 2 ). Solution Since the derivative of g( x ) = − cos x is g′( x ) = sin x, we see that f and g have the same derivative. Corollary 2 then says that f ( x ) = − cos x + C for some constant C. Since the graph of f passes through the point ( 0, 2 ), the value of C is determined from the condition that f (0) = 2:

f (0) = − cos(0) + C = 2,

C = 3.

so

The function is f ( x ) = − cos x + 3.

Finding Velocity and Position from Acceleration We can use Corollary 2 to find the velocity and position functions of an object moving along a vertical line. Assume the object or body is falling freely from rest with acceleration 9.8 m sec 2. We assume the position s(t ) of the body is measured positive downward from the rest position (so the vertical coordinate line points downward, in the direction of the motion, with the rest position at 0). We know that the velocity υ(t ) is some function whose derivative is 9.8. We also know that the derivative of g(t ) = 9.8t is 9.8. By Corollary 2, υ(t ) = 9.8t + C for some constant C. Since the body falls from rest, υ(0) = 0. Thus 9.8 ( 0 ) + C = 0,

and

C = 0.

The velocity function must be υ(t ) = 9.8t. What about the position function s(t )? We know that s(t ) is some function whose derivative is 9.8t. We also know that the derivative of f (t ) = 4.9t 2 is 9.8t. By Corollary 2, s(t ) = 4.9t 2 + C for some constant C. Since s(0) = 0, 4.9 ( 0 ) 2 + C = 0,

and

C = 0.

The position function is s(t ) = 4.9t 2 until the body hits the ground. The ability to find functions from their rates of change is one of the very powerful tools of calculus. As we will see, it lies at the heart of the mathematical developments in Chapter 5.

248

Chapter 4 Applications of Derivatives

EXERCISES

4.2 iii) y = x 3 − 3 x 2 + 4 = ( x + 1)( x − 2 ) 2

Checking the Mean Value Theorem

Find the value or values of c that satisfy the equation

iv) y = x 3 − 33 x 2 + 216 x = x ( x − 9 )( x − 24 )

f (b) − f (a) = f ′(c) b−a in the conclusion of the Mean Value Theorem for the functions and intervals in Exercises 1–8. 1. f ( x ) =

+ 2x 1 3. f ( x ) = x + , x 5. f ( x ) = arcsin x , 7. f ( x ) =

x2

x3

−

x 2,

− 1, [ 0, 1 ] ⎡ 1 , 2⎤ ⎢⎣ 2 ⎥⎦ [ −1, 1 ]

, [ 0, 1 ]

2. f ( x ) = x

23

4. f ( x ) =

x − 1, [ 1, 3 ]

6. f ( x ) = ln ( x − 1) , [ 2, 4 ]

[ −1, 2 ]

⎧⎪ x 3, −2 ≤ x ≤ 0 8. g( x ) = ⎪⎨ 2 ⎪⎪⎩ x , 0 < x ≤ 2 Which of the functions in Exercises 9–14 satisfy the hypotheses of the Mean Value Theorem on the given interval, and which do not? Give reasons for your answers. , [ −1, 8 ]

9. f ( x ) = x

23

10. f ( x ) = x

45

11. f ( x ) =

x (1 − x ), [ 0, 1 ]

, [ 0, 1 ]

⎧⎪ sin x , −π ≤ x < 0 ⎪ 12. f ( x ) = ⎪⎨ x ⎪⎪ x = 0 ⎪⎩ 0, −2 ≤ x ≤ −1 ⎧⎪ x 2 − x , 13. f ( x ) = ⎪⎨ 2 ⎩⎪⎪ 2 x − 3 x − 3, −1 < x ≤ 0 0 ≤ x ≤ 2 ⎧⎪ 2 x − 3, 14. f ( x ) = ⎨ 2 ⎩⎪⎪ 6 x − x − 7, 2 < x ≤ 3 15. The function ⎧⎪ x , 0 ≤ x < 1 f (x) = ⎨ ⎩⎪⎪ 0, x = 1 is zero at x = 0 and x = 1 and differentiable on ( 0, 1), but its derivative on ( 0, 1) is never zero. How can this be? Doesn’t Rolle’s Theorem say the derivative has to be zero somewhere in ( 0, 1)? Give reasons for your answer. 16. For what values of a, m, and b does the function ⎪⎧⎪ 3, x = 0 ⎪ f ( x ) = ⎪⎨ −x 2 + 3 x + a, 0 < x < 1 ⎪⎪ 1≤ x ≤ 2 ⎪⎪⎩ mx + b, satisfy the hypotheses of the Mean Value Theorem on the interval [ 0, 2 ]? Roots (Zeros)

17. a. Plot the zeros of each polynomial on a line together with the zeros of its first derivative. i) y = x 2 − 4 ii) y = x 2 + 8 x + 15

b. Use Rolle’s Theorem to prove that between every two zeros of x n + a n −1 x n −1 +  + a1 x + a 0 there lies a zero of nx n −1 + ( n − 1) a n −1 x n − 2 +  + a1 . 18. Suppose that f ′′ is continuous on [ a, b ] and that f has three zeros in the interval. Show that f ′′ has at least one zero in ( a, b ). Generalize this result. 19. Show that if f ′′ > 0 throughout an interval [ a, b ], then f ′ has at most one zero in [ a, b ]. What if f ′′ < 0 throughout [ a, b ] instead? 20. Show that a cubic polynomial can have at most three real zeros. Show that the functions in Exercises 21–28 have exactly one zero in the given interval. 21. f ( x ) = x 4 + 3 x + 1, [ −2, −1 ] 4 22. f ( x ) = x 3 + 2 + 7, (−∞, 0 ) x t + 1 + t − 4, ( 0, ∞ ) 1 24. g(t ) = + 1 + t − 3.1, (−1, 1) 1−t θ 25. r(θ) = θ + sin 2 − 8, (−∞, ∞ ) 3

23. g(t ) =

( )

26. r(θ) = 2θ − cos 2 θ +

2, (−∞, ∞ )

27. r(θ) = sec 2 θ − cos(2θ) − 1, ( 0, π 2 ) 28. r(θ) = 3 tan θ − cot θ − θ, ( 0, π 2 ) Finding Functions from Derivatives

29. Suppose that f (−1) = 3 and that f ′( x ) = 0 for all x. Must f ( x ) = 3 for all x? Give reasons for your answer. 30. Suppose that f (0) = 5 and that f ′( x ) = 2 for all x. Must f ( x ) = 2 x + 5 for all x? Give reasons for your answer. 31. Suppose that f ′( x ) = 2 x for all x. Find f (2) if a. f (0) = 0

b. f (1) = 0

c. f (−2) = 3.

32. What can be said about functions whose derivatives are constant? Give reasons for your answer. In Exercises 33–38, find all possible functions with the given derivative. 33. a. y ′ = x

b. y ′ = x 2

c. y ′ = x 3

34. a. y ′ = 2 x

b. y ′ = 2 x − 1 1 b. y ′ = 1 − 2 x 1 b. y ′ = x t b. y′ = cos 2

c. y ′ = 3 x 2 + 2 x − 1 1 c. y ′ = 5 + 2 x 1 c. y ′ = 4 x − x t c. y′ = sin 2t + cos 2

b. y ′ =

c. y′ =

1 x2 1 36. a. y ′ = 2 x 35. a. y ′ = −

37. a. y′ = sin 2t 38. a. y′ = sec 2 θ

θ

θ − sec 2 θ

In Exercises 39–42, find the function with the given derivative whose graph passes through the point P. 39. f ′( x ) = 2 x − 1, P ( 0, 0 ) 1 40. g ′( x ) = 2 + 2 x , P (−1, 1) x

4.2  The Mean Value Theorem

( 32 )

60. Rolle’s Theorem

41. f ′( x ) = e 2 x , P 0,

42. r ′(t ) = sec t tan t − 1, P ( 0, 0 ) Finding Position from Velocity or Acceleration

Exercises 43–46 give the velocity V  ds dt and initial position of an object moving along a coordinate line. Find the object’s position at time t. 43. V = 9.8t + 5, s(0) = 10

44. V = 32t − 2, s(0.5) = 4

45. υ  sin πt , s(0)  0

2 2t 46. υ  cos , s(π 2 )  1 π π

Exercises 47–50 give the acceleration a  d 2 s dt 2, initial velocity, and initial position of an object moving on a coordinate line. Find the object’s position at time t. 47. a  e t , V(0)  20, s(0)  5 48. a = 9.8, V(0) = −3, s(0) = 0 49. a = −4 sin 2t , V(0) = 2, s(0) = −3 50. a =

249

9 3t cos , υ(0) = 0, s(0) = −1 π2 π

Applications

51. Temperature change It took 14 s for a mercury thermometer to rise from −19 °C to 100 nC when it was taken from a freezer and placed in boiling water. Show that somewhere along the way, the mercury was rising at the rate of 8.5 nC s. 52. A trucker handed in a ticket at a tollbooth showing that in 2 hours she had covered 230 km on a toll road with speed limit 100 kmh. The trucker was cited for speeding. Why?

a. Construct a polynomial f ( x ) that has zeros at x = −2, −1, 0, 1, and 2. b. Graph f and its derivative f a together. How is what you see related to Rolle’s Theorem? c. Do g( x )  sin x and its derivative g a illustrate the same phenomenon as f and f a ? 61. Unique solution Assume that f is continuous on [ a, b ] and differentiable on ( a, b ). Also assume that f (a) and f (b) have opposite signs and that f ′ ≠ 0 between a and b. Show that f ( x )  0 exactly once between a and b. 62. Parallel tangent line Assume that f and g are differentiable on [ a, b ] and that f (a)  g(a) and f (b)  g(b). Show that there is at least one point between a and b where the tangent lines to the graphs of f and g are parallel or the same line. Illustrate with a sketch. 63. Suppose that f ′( x ) ≤ 1 f (4) − f (1) ≤ 3.

for

1 b x b 4.

Show

that

64. Suppose that 0 < f ′( x ) < 1 2 for all x-values. Show that f (−1)  < f (1) < 2 + f (−1). 65. Show that cos x − 1 ≤ x for all x-values. (Hint: Consider f (t )  cos t on the closed interval with the endpoints 0 and x.) 66. Show that for any numbers a and b, the sine inequality sin b − sin a ≤ b − a is true. 67. If the graphs of two differentiable functions f ( x ) and g( x ) start at the same point in the plane and the functions have the same rate of change at every point, do the graphs have to be identical? Give reasons for your answer.

53. Classical accounts tell us that a 170-oar trireme (ancient Greek or Roman warship) once covered 184 sea miles in 24 hours. Explain why at some point during this feat the trireme’s speed exceeded 7.5 knots (sea or nautical miles per hour).

68. If f ( w) − f ( x ) ≤ w − x for all values w and x and f is a differentiable function, show that −1 ≤ f ′( x ) ≤ 1 for all x-values.

54. A marathoner ran the 42-km New York City Marathon in 2.2 hours. Show that at least twice the marathoner was running at exactly 18 kmh, assuming the initial and final speeds are zero.

69. Assume that f is differentiable on a b x b b and that f (b)  f (a). Show that f a is negative at some point between a and b.

55. Show that at some instant during a 2-hour automobile trip the car’s speedometer reading will equal the average speed for the trip.

70. Let f be a function defined on an interval [ a, b ]. What conditions could you place on f to guarantee that

56. Free fall on the moon On our moon, the acceleration of gravity is 1.6 m s 2 . If a rock is dropped into a crevasse, how fast will it be going just before it hits bottom 30 s later? Theory and Examples

57. The geometric mean of a and b The geometric mean of two positive numbers a and b is the number ab. Show that the value of c in the conclusion of the Mean Value Theorem for f ( x )  1 x on an interval of positive numbers [ a, b ]  is  c = ab . 58. The arithmetic mean of a and b The arithmetic mean of two numbers a and b is the number ( a + b ) 2. Show that the value of c in the conclusion of the Mean Value Theorem for f ( x )  x 2 on any interval [ a, b ]  is  c = ( a + b ) 2. T 59. Graph the function f ( x ) = sin x sin ( x + 2 ) − sin 2 ( x + 1). What does the graph do? Why does the function behave this way? Give reasons for your answers.

min   f ′ ≤

f (b) − f (a) ≤ max   f ′, b−a

where min   f a and max   f a refer to the minimum and maximum values of f a on [ a, b ]? Give reasons for your answers. T 71. Use the inequalities in Exercise 70 to estimate f (0.1) if f ′( x ) = 1 (1 + x 4 cos x ) for 0 b x b 0.1 and f (0)  1. T 72. Use the inequalities in Exercise 70 to estimate f (0.1) if f ′( x ) = 1 (1 − x 4 ) for 0 b x b 0.1 and f (0)  2. 73. Let f be differentiable at every value of x and suppose that f (1)  1, that f ′ < 0 on  (−∞, 1) , and that f ′ > 0 on  (1, ∞ ). a. Show that f ( x ) p 1 for all x. b. Must f ′(1) = 0? Explain. 74. Let f ( x ) = px 2 + qx + r be a quadratic function defined on a closed interval [ a, b ]. Show that there is exactly one point c in ( a, b ) at which f satisfies the conclusion of the Mean Value Theorem.

250

Chapter 4 Applications of Derivatives

4.3 Monotonic Functions and the First Derivative Test In sketching the graph of a differentiable function, it is useful to know where it increases (rises from left to right) and where it decreases (falls from left to right) over an interval. This section gives a test to determine where it increases and where it decreases. We also show how to test the critical points of a function to identify whether local extreme values are present.

Increasing Functions and Decreasing Functions As another corollary to the Mean Value Theorem, we show that functions with positive derivatives are increasing functions and functions with negative derivatives are decreasing functions. A function that is either increasing on an interval or decreasing on an interval is said to be monotonic on the interval.

COROLLARY 3

( a, b ).

Suppose that f is continuous on [ a, b ] and differentiable on

If  f ′( x ) > 0 at each point  x ∈ ( a, b ),  then f  is increasing on  [ a, b ]. If  f ′( x ) < 0 at each point  x ∈ ( a, b ),  then  f  is decreasing on  [ a, b ].

Proof Let x 1 and x 2 be any two points in [ a, b ] with x 1 < x 2 . The Mean Value Theorem applied to f on [ x 1 , x 2 ] says that f ( x 2 ) − f ( x 1 ) = f ′(c)( x 2 − x 1 ) for some c between x 1 and x 2 . The sign of the right-hand side of this equation is the same as the sign of f ′(c) because x 2 − x 1 is positive. Therefore, f ( x 2 ) > f ( x 1 ) if f ′ is positive on ( a, b ) and f ( x 2 ) < f ( x 1 ) if f ′ is negative on ( a, b ). Corollary 3 tells us that f ( x ) = x is increasing on the interval [ 0, b ] for any b > 0 because f ′( x ) = 1 ( 2 x ) is positive on ( 0, b ). The derivative does not exist at x = 0, but Corollary 3 still applies. The corollary is also valid for open and for infinite intervals, so f ( x ) = x is also increasing on ( 0, 1), on ( 0, ∞ ) , and on [ 0, ∞ ). To find the intervals where a function f is increasing or decreasing, we first find all of the critical points of f . If a < b are two critical points for f , and if the derivative f ′ is continuous but never zero on the interval ( a, b ), then by the Intermediate Value Theorem applied to f ′ , the derivative must be everywhere positive on ( a, b ), or everywhere negative there. One way we can determine the sign of f ′ on ( a, b ) is simply by evaluating the derivative at a single point c in ( a, b ). If f ′(c) > 0, then f ′( x ) > 0 for all x in ( a, b ) so f is increasing on [ a, b ] by Corollary 3; if f ′(c) < 0, then f is decreasing on [ a, b ]. It doesn’t matter which point c we choose in ( a, b ) since the sign of f ′(c) is the same for all choices. Usually we pick c to be a point where it is easy to evaluate f ′(c). The next example illustrates how we use this procedure. EXAMPLE 1 Find the critical points of f ( x ) = x 3 − 12 x − 5 and identify the open intervals on which f is increasing and those on which f is decreasing. Solution The function f is everywhere continuous and differentiable. The first derivative

f ′( x ) = 3 x 2 − 12 = 3( x 2 − 4 ) = 3( x + 2 )( x − 2 )

4.3  Monotonic Functions and the First Derivative Test

y y = x3 − 12x − 5 20 (−2, 11) 10

−4 −3 −2 −1 0

1

2

3

4

x

−10 −20

is zero at x = −2 and x  2. These critical points subdivide the domain of f to create nonoverlapping open intervals (−∞, −2 ) , (−2, 2 ) , and ( 2, ∞ ) on which f a is either positive or negative. We determine the sign of f a by evaluating f a at a convenient point in each subinterval. We evaluate f a at x = −3 in the first interval, x  0 in the second interval, and x  3 in the third since f a is relatively easy to compute at these points. The behavior of f is determined by then applying Corollary 3 to each subinterval. The results are summarized in the following table, and the graph of f is given in Figure 4.20.

Interval

−∞ < x < −2

−2 < x < 2

2 < x < ∞

f ′(−3) = 15

f ′(0) = −12

f ′(3) = 15







increasing

decreasing

increasing

f a evaluated

(2, −21)

Sign of f a

FIGURE 4.20

The function f ( x ) = x 3 − 12 x − 5 is monotonic on three separate intervals (Example 1).

251

Behavior of f

x −3

−2

−1

0

1

2

3

We used “strict” less-than inequalities to identify the intervals in the summary table for Example 1, since open intervals were specified. Corollary 3 says that we could use b inequalities as well. That is, the function f in the example is increasing on −∞ < x ≤ −2, decreasing on −2 ≤ x ≤ 2, and increasing on 2 ≤ x < ∞. We do not talk about whether a function is increasing or decreasing at a single point.

HISTORICAL BIOGRAPHY Edmund Halley (1656–1742) Halley, a British biologist, geologist, sea captain, astronomer, and mathematician, encouraged Newton to write the Principia. Despite all of Halley’s accomplishments, he is known today as the man who calculated the orbit of the comet of 1682. To know more, visit the companion Website.

First Derivative Test for Local Extrema In Figure 4.21, at the points where f has a minimum value, f ′ < 0 immediately to the left and f ′ > 0 immediately to the right. (If the point is an endpoint, there is only one side to consider.) Thus, the function is decreasing on the left of the minimum value and it is increasing on its right. Similarly, at the points where f has a maximum value, f ′ > 0 immediately to the left and f ′ < 0 immediately to the right. Thus, the function is increasing on the left of the maximum value and decreasing on its right. In summary, at a local extreme point, the sign of f a( x ) changes. Absolute max f ′ undefined Local max f′ = 0 No extremum f′ = 0 f′ > 0

y = f(x)

No extremum f′ = 0

f′ > 0

f′ < 0

f′ < 0

f′ < 0 Local min

Local min f′ = 0

f′ > 0 Absolute min a

c1

c2

c3

c4

c5

b

x

FIGURE 4.21

The critical points of a function locate where it is increasing and where it is decreasing. The first derivative changes sign at a critical point where a local extremum occurs.

These observations lead to a test for the presence and nature of local extreme values of differentiable functions.

252

Chapter 4 Applications of Derivatives

First Derivative Test for Local Extrema

Suppose that c is a critical point of a continuous function f , and that f is differentiable at every point in some interval containing c except possibly at c itself. Moving across this interval from left to right, 1. if f ′ changes from negative to positive at c, then f has a local minimum at c; 2. if f ′ changes from positive to negative at c, then f has a local maximum at c; 3. if f ′ does not change sign at c (that is, f ′ is positive on both sides of c or negative on both sides), then f has no local extremum at c.

The test for local extrema at endpoints is similar, but there is only one side to consider in determining whether f is increasing or decreasing, based on the sign of f ′ . Proof of the First Derivative Test Part (1). Since the sign of f ′ changes from negative to positive at c, there are numbers a and b such that a < c < b, f ′ < 0 on ( a, c ), and f ′ > 0 on ( c, b ). If x ∈ ( a, c ) , then f (c) < f ( x ) because f ′ < 0 implies that f is decreasing on [ a, c ]. If x ∈ ( c, b ) , then f (c) < f ( x ) because f ′ > 0 implies that f is increasing on [ c, b ]. Therefore, f ( x ) ≥ f (c) for every x ∈ ( a, b ). By definition, f has a local minimum at c. Parts (2) and (3) are proved similarly. EXAMPLE 2

Find the critical points of f (x) = x 1 3 ( x − 4 ) = x 4 3 − 4 x 1 3.

Identify the open intervals on which f is increasing and those on which it is decreasing. Find the function’s local and absolute extreme values. Solution The function f is continuous at all x since it is the product of two continuous functions, x 1 3 and ( x − 4 ). The first derivative,

d 43 4 4 ( x − 4 x 1 3 ) = x 1 3 − x −2 3 dx 3 3 ( x − 1) 4 −2 3 ( 4 x − 1) = , = x 3 3x 2 3

f ′( x ) =

is zero at x = 1 and undefined at x = 0. There are no endpoints in the domain, so the critical points x = 0 and x = 1 are the only places where f might have an extreme value. The critical points partition the x-axis into open intervals on which f ′ is either positive or negative. The sign pattern of f ′ reveals the behavior of f between and at the critical points, as summarized in the following table. Interval Sign of f ′ Behavior of f

x < 0

0 < x 1

−

−

+

decreasing

decreasing

increasing x

−1

0

1

2

Corollary 3 to the Mean Value Theorem implies that f decreases on (−∞, 0 ) , decreases on ( 0, 1), and increases on (1, ∞ ). The First Derivative Test for Local Extrema tells us that f does not have an extreme value at x = 0 ( f ′ does not change sign) and that f has a local minimum at x = 1 ( f ′ changes from negative to positive).

253

4.3  Monotonic Functions and the First Derivative Test

y

The value of the local minimum is f (1) = 11 3 (1 − 4 ) = −3. This is also an absolute minimum since f is decreasing on (−∞, 1) and increasing on (1, ∞ ). Figure 4.22 shows this value in relation to the function’s graph. Note that lim f ′( x ) = −∞, so the graph of f has a vertical tangent line at the origin.

4 y = x13(x − 4) 2

x→0

1 −1 0 −1

1

2

3

EXAMPLE 3

x

4

f (x) = ( x 2 − 3)e x .

−2 −3

Find the critical points of

Identify the open intervals on which f is increasing and those on which it is decreasing. Find the function’s local and absolute extreme values.

(1, −3)

FIGURE 4.22

The function f ( x ) = x 1 3 ( x − 4 ) decreases when x < 1 and increases when x > 1 (Example 2).

Solution The function f is continuous and differentiable for all real numbers, so the critical points occur only at the zeros of f ′. Using the Derivative Product Rule, we find the derivative

d( 2 d x e + x − 3) ⋅ e x dx dx = ( x 2 − 3 ) e x + (2 x )e x = ( x 2 + 2 x − 3)e x .

f ′( x ) = ( x 2 − 3 ) ⋅

Since e x is never zero, the first derivative is zero if and only if y

x 2 + 2x − 3 = 0 ( x + 3 )( x − 1) = 0.

y = (x 2 − 3)e x

4

The zeros x = −3 and x = 1 partition the x-axis into open intervals as follows.

3 2 1 −5 −4 −3 −2 −1 −1

1

2

3

x

−2

Interval Sign of f ′ Behavior of f

−3

x < −3

−3 < x < 1

1< x

+

−

+

decreasing

increasing

increasing

x −4

−4

−3

−2

−1

0

1

2

3

−5 −6

We can see from the table that there is a local maximum (about 0.299) at x = −3 and a local minimum (about −5.437) at x = 1. The local minimum value is also an absolute minimum because f ( x ) > 0 for x > 3 . There is no absolute maximum. The function increases on (−∞, −3 ) and (1, ∞ ) and decreases on (−3, 1). Figure 4.23 shows the graph.

FIGURE 4.23

The graph of f ( x ) = ( x 2 − 3 ) e x (Example 3).

4.3

EXERCISES

Analyzing Functions from Derivatives

Answer the following questions about the functions whose derivatives are given in Exercises 1–14:

9.

a. What are the critical points of f ? b. On what open intervals is f increasing or decreasing? c. At what points, if any, does f assume local maximum or minimum values? 1. f ′( x ) = x ( x − 1)

2. f ′( x ) = ( x − 1)( x + 2 )

4. f ′( x ) = ( x − 1) 2 ( x + 2 ) 2 3. f ′( x ) = ( x − 1) ( x + 2 ) − x 5. f ′( x ) = ( x − 1) e 6. f ′( x ) = ( x − 7 )( x + 1)( x + 5 ) x 2 ( x − 1) , x ≠ −2 7. f ′( x ) = x+2 2

− 2 )( x + 4 ) , x ≠ −1, 3 + 1)( x − 3 ) 4 6 , x ≠ 0 10. f ′( x ) = 3 − f ′( x ) = 1 − 2 , x ≠ 0 x x f ′( x ) = x −1 3 ( x + 2 ) 12. f ′( x ) = x −1 2 ( x − 3 ) f ′( x ) = ( sin x − 1)( 2 cos x + 1) , 0 ≤ x ≤ 2π f ′( x ) = ( sin x + cos x )( sin x − cos x ) , 0 ≤ x ≤ 2π

8. f ′( x ) =

11. 13. 14.

(x

(x

Identifying Extrema

In Exercises 15–18: a. Find the open intervals on which the function is increasing and those on which it is decreasing. b. Identify the function’s local and absolute extreme values, if any, saying where they occur.

254

Chapter 4 Applications of Derivatives

y

15.

y

16.

2

2

y = f (x)

1 −3 −2 −1 −1

1

2

3

−3 −2 −1 −1

−2

2

3

2

−3 −2 −1 −1

1

2

3

a. Find the local extrema of each function on the given interval, and say where they occur.

1

−3 −2 −1 −1

−2

T b. Graph the function and its derivative together. Comment on the behavior of f in relation to the signs and values of f ′.

2

y = f (x) x

1

2

3

a. Find the open intervals on which the function is increasing and those on which it is decreasing. b. Identify the function’s local extreme values, if any, saying where they occur. 19. g(t ) = −t 2 − 3t + 3

20. g(t ) = −3t 2 + 9t + 5

21. h( x ) = −x 3 + 2 x 2

22. h( x ) = 2 x 3 − 18 x

4θ 3

24. f (θ) = 6θ − θ 3

25. f (r ) = 3r 3 + 16r

26. h(r ) = ( r + 7 )3

27. f ( x ) = x 4 − 8 x 2 + 16 3 29. H (t ) = t 4 − t 6 2 31. f ( x ) = x − 6 x − 1

28. g( x ) = x 4 − 4 x 3 + 4 x 2

−

59. f ( x ) = sin 2 x , 0 ≤ x ≤ π 60. f ( x ) = sin x − cos x , 0 ≤ x ≤ 2π 3 cos x + sin x , −π 62. f ( x ) = −2 x + tan x , 2 x x 63. f ( x ) = − 2 sin , 0 ≤ 2 2 64. f ( x ) = −2 cos x − cos 2 x ,

32. g( x ) = 4 x − 34. g( x ) =

x2

y

67.

39. h( x ) = x 1 3 ( x 2 − 4 )

40. k ( x ) = x 2 3 ( x 2 − 4 )

41. f ( x ) = e 2 x + e −x

42. f ( x ) = e

43. f ( x ) = x ln x

44. f ( x ) = x 2 ln x

45. g( x ) = x ( ln x ) 2

46. g( x ) = x 2 − 2 x − 4 ln x

x

−4

T b. Graph the function over the given domain. Which of the extreme values, if any, are absolute?

−2

x3 − 2x 2 + 4 x, 0 ≤ x < ∞ 3 54. k ( x ) = x 3 + 3 x 2 + 3 x + 1, −∞ < x ≤ 0

53. h( x ) =

4

x

−4

−2

0 −2

−4

−4

y

2

4

x

y

70.

2

2 y = f 9(x)

1

− 6 x − 9, −4 ≤ x < ∞

52. f (t ) = t 3 − 3t 2 , −∞ < t ≤ 3

2

2

−2

69.

− 4 x + 4, 1 ≤ x < ∞

51. f (t ) = 12t − t 3 , −3 ≤ t < ∞

0

f9

f9

b. Either use the graph to determine which intervals f is positive on and which intervals f is negative on, or explain why this information cannot be determined from the graph.

48. f ( x ) = ( x + 1) 2, −∞ < x ≤ 0 50. g( x ) =

4

a. Either use the graph to determine which intervals f is increasing on and which intervals f is decreasing on, or explain why this information cannot be determined from the graph.

a. Identify the function’s local extreme values in the given domain, and say where they occur.

−x 2

y

In Exercises 69 and 70, the graph of f ′ is given. Assume that f has domain (−2, 2 ).

In Exercises 47–58:

49. g( x ) =

−π ≤ x ≤ π

68.

2

+3

5−x x3 36. f ( x ) = 3x 2 + 1 38. g( x ) = x 2 3 ( x + 5 )

x2

x ≤ 2π

4

x2

47. f ( x ) = 2 x − x 2 , −∞ < x ≤ 2

0 ≤ x ≤ 2π π < x < 2

65. f ( x ) = csc 2 x − 2 cot x , 0 < x < π −π π < x < 66. f ( x ) = sec 2 x − 2 tan x , 2 2 In Exercises 67 and 68, the graph of f ′ is given. Assume that f is continuous, and determine the x-values corresponding to local minima and local maxima.

30. K (t ) = 15t 3 − t 5

33. g( x ) = x 8 − x2 − 3 35. f ( x ) = , x ≠ 2 x−2 37. f ( x ) = x 1 3 ( x + 8 ) x2

x

61. f ( x ) =

−2

In Exercises 19–46:

23. f (θ) =

x 2 − 2 x − 3, 3 ≤ x < ∞ x−2 57. g( x ) = 2 , 0 ≤ x 0 for  x < 1 and f ′( x ) < 0 for  x > 1; b. f ′( x ) < 0 for  x < 1 and  f ′( x ) > 0 for  x > 1; c. f ′( x ) > 0 for  x ≠ 1; d. f ′( x ) < 0 for  x ≠ 1. 74. Sketch the graph of a differentiable function y = f ( x ) that has

79. Determine the values of constants a and b so that f ( x ) = ax 2 + bx has an absolute maximum at the point (1, 2 ). 80. Determine the values of constants a, b, c, and d so that f ( x ) = ax 3 + bx 2 + cx + d has a local maximum at the point ( 0, 0 ) and a local minimum at the point (1, −1). 81. Locate and identify the absolute extreme values of a. ln (cos x) on [ −π 4 , π 3 ], b. cos (ln x) on [ 1 2, 2 ]. 82. a. Prove that f ( x ) = x − ln x is increasing for x > 1. b. Using part (a), show that ln x < x if x > 1. 83. Find the absolute maximum and the absolute minimum values of f ( x ) = e x − 2 x on [ 0, 1 ]. 84. Where does the periodic function f ( x ) = 2e sin ( x 2 ) take on its extreme values and what are these values? y

a. a local minimum at (1, 1) and a local maximum at ( 3, 3 ); b. a local maximum at (1, 1) and a local minimum at ( 3, 3 );

y = 2e sin (x2)

c. local maxima at (1, 1) and ( 3, 3 ); d. local minima at (1, 1) and ( 3, 3 ). 75. Sketch the graph of a continuous function y = g( x ) such that a. g(2) = 2, 0 < g′ < 1 for  x < 2, g′( x ) → 1− as x → 2 − , −1 < g′ < 0 for  x > 2, and  g′( x ) → −1+ as x → 2 + ; b. g(2) = 2, g′ < 0 for  x < 2, g′( x ) → −∞ as x → 2 − , g ′ > 0 for  x > 2, and g ′( x ) → ∞ as  x → 2 +. 76. Sketch the graph of a continuous function y = h( x ) such that

255

x

0

85. Find the absolute maximum value of f ( x ) = x 2 ln (1 x ) and say where it occurs. 86. a. Prove that e x ≥ 1 + x if x ≥ 0. b. Use the result in part (a) to show that 1 2 x . 2

a. h(0) = 0, −2 ≤ h( x ) ≤ 2 for all  x , h′( x ) → ∞ as  x → 0 − , and h ′( x ) → ∞ as  x → 0 + ;

ex ≥ 1 + x +

b. h(0) = 0, −2 ≤ h( x ) ≤ 0 for all  x , h′( x ) → ∞ as  x → 0 − , and h ′( x ) → −∞ as  x → 0 +.

87. Show that increasing functions and decreasing functions are oneto-one. That is, show that for any x 1 and x 2 in I, x 2 ≠ x 1 implies f ( x 2 ) ≠ f ( x 1 ).

77. Discuss the extreme-value behavior of the function f ( x ) = x sin (1 x ) , x ≠ 0 . How many critical points does this function have? Where are they located on the x-axis? Does f have an absolute minimum? An absolute maximum? (See Exercise 49 in Section 2.3.) 78. Find the open intervals on which the function f ( x ) = ax 2 + bx + c, a ≠ 0, is increasing and those on which it is decreasing. Describe the reasoning behind your answer.

Use the results of Exercise 87 to show that the functions in Exercises 88–92 have inverses over their domains. Find a formula for df −1 dx using Theorem 3, Section 3.8. 88. f ( x ) = (1 3 ) x + ( 5 6 )

89. f ( x ) = 27 x 3

90. f ( x ) = 1 − 8 x 3

91. f ( x ) = (1 − x )3

92. f ( x ) = x 5 3

4.4 Concavity and Curve Sketching We have seen how the first derivative tells us where a function is increasing, where it is decreasing, and whether a local maximum or local minimum occurs at a critical point. In this section we see that the second derivative gives us information about how the graph of a differentiable function bends or turns. With this knowledge about the first and second derivatives, coupled with our previous understanding of symmetry and asymptotic behavior studied in Sections 1.1 and 2.5, we can now draw an accurate graph of a function. By organizing all of these ideas into a coherent procedure, we give a method for sketching graphs and revealing visually the key features of functions. Identifying and knowing the locations of these features is of major importance in mathematics and its applications to science and engineering, especially in the graphical analysis and interpretation of data. When the domain of a function is not a finite closed interval, sketching a graph helps to determine whether absolute maxima or absolute minima exist and, if they do exist, where they are located.

256

Chapter 4 Applications of Derivatives

y

Concavity

CA

VE

UP

y=x

3

CO

N W

f ′ increases x

0

CO NC AV E

DO

f ′ decreases

N

The graph of f ( x ) = x 3 is concave down on (−∞, 0 ) and concave up on ( 0, ∞ ) (Example 1a). FIGURE 4.24

y 4

y = x2

U VE

CO NC AV E

2 P

−1

The graph of a differentiable function y = f ( x ) is

(a) concave up on an open interval I if f ′ is increasing on I; (b) concave down on an open interval I if f ′ is decreasing on I.

UP

CA CON

−2

DEFINITION

A function whose graph is concave up is also often called convex. If y = f ( x ) has a second derivative, we can apply Corollary 3 of the Mean Value Theorem to the first derivative function. We conclude that f ′ increases if f ′′ > 0 on I, and decreases if f ′′ < 0.

3

y″ > 0

As you can see in Figure 4.24, the curve y = x 3 rises as x increases, but the portions defined on the intervals (−∞, 0 ) and ( 0, ∞ ) turn in different ways. As we approach the origin from the left along the curve, the curve turns to our right and falls below its tangent lines. The slopes of the tangent lines are decreasing on the interval (−∞, 0 ). As we move away from the origin along the curve to the right, the curve turns to our left and rises above its tangent lines. The slopes of the tangent lines are increasing on the interval ( 0, ∞ ). This turning or bending behavior defines the concavity of the curve.

1

The Second Derivative Test for Concavity

Let y = f ( x ) be twice-differentiable on an interval I.

y″ > 0

0

1

x

2

The graph of f ( x ) = x 2 is concave up on every interval (Example 1b).

1. If f ′′ > 0 on I, the graph of f over I is concave up. 2. If f ′′ < 0 on I, the graph of f over I is concave down.

FIGURE 4.25

EXAMPLE 1

y 4

y = 3 + sin x

(a) The curve y = x 3 (Figure 4.24) is concave down on (−∞, 0 ) , where y′′ = 6 x < 0, and concave up on ( 0, ∞ ) , where y′′ = 6 x > 0. (b) The curve y = x 2 (Figure 4.25) is concave up on (−∞, ∞ ) because its second derivative y′′ = 2 is always positive.

(p, 3)

3 2 1 0 −1

If y = f ( x ) is twice-differentiable, we will use the notations f ′′ and y′′ interchangeably when denoting the second derivative.

p

2p

x

y″ = −sin x

Using the sign of y ′′ to determine the concavity of y (Example 2). FIGURE 4.26

EXAMPLE 2

Determine the concavity of y = 3 + sin x  on  [ 0, 2π ].

Solution The first derivative of y = 3 + sin x is y′ = cos x , and the second derivative is y′′ = − sin x. The graph of y = 3 + sin x is concave down on ( 0, π ) , where y′′ = − sin x is negative. It is concave up on ( π , 2π ) , where y′′ = − sin x is positive (Figure 4.26).

Points of Inflection The curve y = 3 + sin x in Example 2 changes concavity at the point ( π, 3 ). Since the first derivative y′ = cos x exists for all x, we see that the curve has a tangent line of slope −1 at the point ( π, 3 ). This point is called a point of inflection of the curve. Notice from Figure 4.26 that the graph crosses its tangent line at this point and that the second derivative y′′ = − sin x has value 0 when x = π. In general, we have the following definition.

4.4  Concavity and Curve Sketching

257

DEFINITION A point ( c, f (c) ) where the graph of a function has a tangent line

and where the concavity changes is a point of inflection.

y 3 2

y = x 3 − 3x 2 + 2 Concave down

We observed that the second derivative of f ( x ) = 3 + sin x is equal to zero at the inflection point ( π, 3 ). Generally, if the second derivative exists at a point of inflection ( c, f (c) ), then f ′′(c) = 0. This follows immediately from the Intermediate Value Theorem whenever f ′′ is continuous over an interval containing x = c because the second derivative changes sign moving across this interval. Even if the continuity assumption is dropped, it is still true that f ′′(c) = 0, provided the second derivative exists (although a more advanced argument is required in this noncontinuous case). Since a tangent line must exist at the point of inflection, either the first derivative f ′(c) exists (is finite) or the graph has a vertical tangent line at the point. At a vertical tangent, neither the first nor second derivative exists. In summary, one of two things can happen at a point of inflection.

1 −1

0 −1 −2

1

2

x

3

Point of inflection

EXAMPLE 3

Concave up

Solution We start by computing the first and second derivatives.

f ′( x ) = 3 x 2 − 6 x ,

y = x53

1 0

1 Point of inflection

–1

f ′′( x ) = 6 x − 6.

To determine concavity, we look at the sign of the second derivative f ′′( x ) = 6 x − 6. The sign is negative when x < 1, is 0 at x = 1, and is positive when x > 1. It follows that the graph of f is concave down on (−∞, 1) , is concave up on (1, ∞ ) , and has an inflection point at the point (1, 0 ) where the concavity changes. The graph of f is shown in Figure 4.27. Notice that we did not need to know the shape of this graph ahead of time in order to determine its concavity.

y

–1

Determine the concavity and find the inflection points of the function f ( x ) = x 3 − 3 x 2 + 2.

FIGURE 4.27 The concavity of the graph of f changes from concave down to concave up at the inflection point (Example 3).

2

At a point of inflection ( c, f (c) ), either f ′′(c) = 0 or f ′′(c) fails to exist.

x

The next example illustrates that a function can have a point of inflection where the first derivative exists but the second derivative fails to exist.

–2

FIGURE 4.28 The graph of f ( x ) = x has a horizontal tangent at the origin where the concavity changes, although f ′′ does not exist at x = 0 (Example 4). 53

y y = x4

EXAMPLE 4 The graph of f ( x ) = x 5 3 has a horizontal tangent at the origin because f ′( x ) = ( 5 3 ) x 2 3 = 0 when x = 0. However, the second derivative, f ′′( x ) =

(

)

d 5 23 10 −1 3 x x , = 9 dx 3

fails to exist at x = 0. Nevertheless, f ′′( x ) < 0 for x < 0 and f ′′( x ) > 0 for x > 0, so the second derivative changes sign at x = 0 and there is a point of inflection at the origin. The graph is shown in Figure 4.28. The following example shows that an inflection point need not occur even though both derivatives exist and f ′′ = 0.

2 1 y″ = 0 −1

0

1

x

The graph of y = x 4 has no inflection point at the origin, even though y ′′ = 0 there (Example 5).

FIGURE 4.29

EXAMPLE 5 The curve y = x 4 has no inflection point at x = 0 (Figure 4.29). Even though the second derivative y ″ = 12 x 2 is zero there, it does not change sign. The curve is concave up everywhere. In the next example, a point of inflection occurs at a vertical tangent to the curve where neither the first nor the second derivative exists.

258

Chapter 4 Applications of Derivatives

EXAMPLE 6 The graph of y = x 1 3 has a point of inflection at the origin because the second derivative is positive for x < 0 and negative for x > 0:

y Point of inflection

y = x13 0

y′′ =

x

(

)

d 1 −2 3 2 d2 13 (x ) = x = − x −5 3 . dx 3 9 dx 2

However, both y′ = x −2 3 3 and y″ fail to exist at x = 0, and there is a vertical tangent there. See Figure 4.30. FIGURE 4.30 A point of inflection where y ′ and y″ fail to exist (Example 6).

Caution Example 4 in Section 4.1 (Figure 4.9) shows that the function f ( x ) = x 2 3 does not have a second derivative at x = 0 and does not have a point of inflection there (there is no change in concavity at x = 0). Combined with the behavior of the function in Example 6 above, we see that when the second derivative does not exist at x = c, an inflection point may or may not occur there. So we need to be careful about interpreting functional behavior whenever first or second derivatives fail to exist at a point. At such points the graph can have vertical tangent lines, corners, cusps, or various discontinuities. To study the motion of an object moving along a line as a function of time, we often are interested in knowing when the object’s acceleration, given by the second derivative, is positive or negative. The points of inflection on the graph of the object’s position function reveal where the acceleration changes sign. EXAMPLE 7 A particle is moving along a horizontal coordinate line (positive to the right) with position function s(t ) = 2t 3 − 14 t 2 + 22t − 5,

t ≥ 0.

Find the velocity and acceleration, and describe the motion of the particle. Solution The velocity is

υ(t ) = s′(t ) = 6t 2 − 28t + 22 = 2( t − 1)( 3t − 11) , and the acceleration is a(t ) = υ′(t ) = s ′′(t ) = 12t − 28 = 4 ( 3t − 7 ). When the function s(t ) is increasing, the particle is moving to the right; when s(t ) is decreasing, the particle is moving to the left. Notice that the first derivative ( υ = s′ ) is zero at the critical points t = 1 and t = 11 3. Interval

0 < t 0, then f has a local minimum at x = c. 3. If f ′(c) = 0 and f ′′(c) = 0, then the test fails. The function f may have a local maximum, a local minimum, or neither.

f ′ = 0, f ″ < 0 1 local max

f ′ = 0, f ″ > 0 1 local min

Proof Part (1). If f ′′(c) < 0, then f ′′( x ) < 0 on some open interval I containing the point c, since f ′′ is continuous. Therefore, f ′ is decreasing on I. Since f ′(c) = 0, the sign of f ′ changes from positive to negative at c, so f has a local maximum at c by the First Derivative Test. The proof of Part (2) is similar. For Part (3), consider the three functions y = x 4 , y = −x 4 , and y = x 3 . For each function, the first and second derivatives are zero at x = 0. Yet the function y = x 4 has a local minimum there, y = −x 4 has a local maximum, and y = x 3 is increasing in any open interval containing x = 0 (having neither a maximum nor a minimum there). Thus the test fails. This test requires us to know f ′′ only at c itself and not in an interval about c. This makes the test easy to apply. That’s the good news. The bad news is that the test is inconclusive if f ′′ = 0 or if f ′′ does not exist at x = c. When this happens, use the First Derivative Test for local extreme values. Together f ′ and f ′′ tell us the shape of the function’s graph—that is, where the critical points are located and what happens at a critical point, where the function is increasing and where it is decreasing, and how the curve is turning or bending as indicated by its concavity. We use this information to sketch a graph of the function that captures its key features. EXAMPLE 8

Sketch a graph of the function f ( x ) = x 4 − 4 x 3 + 10

using the following steps. (a) Identify where the extrema of f occur. (b) Find the intervals on which f is increasing and the intervals on which f is decreasing. (c) Find where the graph of f is concave up and where it is concave down. (d) Sketch the general shape of the graph for f . (e) Plot some specific points, such as local maximum and minimum points, points of inflection, and intercepts. Then sketch the curve.

260

Chapter 4 Applications of Derivatives

Solution The function f is continuous since f ′( x ) = 4 x 3 − 12 x 2 exists. The domain of f is (−∞, ∞ ) , and the domain of f ′ is also (−∞, ∞ ). Thus, the critical points of f occur only at the zeros of f ′. Since

f ′( x ) = 4 x 3 − 12 x 2 = 4 x 2 ( x − 3 ) , the first derivative is zero at x = 0 and x = 3. We use these critical points to define intervals where f is increasing or decreasing. Interval

x < 0

0 < x < 3

3 < x

−

−

+

decreasing

decreasing

increasing

Sign of f ′ Behavior of f

(a) Using the First Derivative Test for local extrema and the table above, we see that there is no extremum at x = 0 and a local minimum at x = 3. (b) Using the table above, we see that f is decreasing on (−∞, 0 ] and [ 0, 3 ], and increasing on [ 3, ∞ ). (c) f ′′( x ) = 12 x 2 − 24 x = 12 x ( x − 2 ) is zero at x = 0 and x = 2. We use these points to define intervals where the graph of f is concave up or concave down. Interval

x < 0

0 < x < 2

2 < x

+

−

+

concave up

concave down

concave up

Sign of f ′ Behavior of f

We see that the graph of f is concave up on the intervals (−∞, 0 ) and ( 2, ∞ ) , and concave down on ( 0, 2 ). (d) Summarizing the information in the last two tables, we obtain the following.

y y = x 4 − 4x 3 + 10

x < 0

0 < x < 2

2 < x < 3

3 < x

decreasing concave up

decreasing concave down

decreasing concave up

increasing concave up

20

The general shape of the curve is shown in the accompanying figure.

15 (0, 10) Inflection 10 point 5 −1

0 −5 −10

1 Inflection point

2

3

4

(2, −6)

−15 −20

FIGURE 4.31

(3, −17) Local minimum

The graph of f ( x ) = x 4 − 4 x 3 + 10 (Example 8).

x

decr

decr

decr

incr

conc up

conc down

conc up

conc up

0

2

3

infl point

infl point

local min

General shape

(e) Plot the curve’s intercepts (if possible) and the points where y′ and y′′ are zero. Indicate any local extreme values and inflection points. Use the general shape as a guide to sketch the curve. (Plot additional points as needed.) Figure 4.31 shows the graph of f . The steps in Example 8 give a procedure for graphing the key features of a function. Asymptotes were defined and discussed in Section 2.5. We can find them for many classes

4.4  Concavity and Curve Sketching

261

of functions (including rational functions), and the methods in the next section give tools to help find them for even more general functions.

Procedure for Graphing y = f ( x )

1. Identify the domain of f and any symmetries the curve may have. 2. Find the derivatives y′ and y′′. 3. Find the critical points of f , if any, and identify the function’s behavior at each one. 4. Find where the curve is increasing and where it is decreasing. 5. Find the points of inflection, if any occur, and determine the concavity of the curve. 6. Identify any asymptotes that may exist. 7. Plot key points, such as the intercepts and the points found in Steps 3–5, and sketch the curve together with any asymptotes that exist.

EXAMPLE 9

Sketch the graph of f ( x ) =

+ 1) 2 . 1 + x2

(x

Solution

1. The domain of f is (−∞, ∞ ) and there are no symmetries about either axis or the origin (Section 1.1). 2. Find f ′ and f ′′. f (x) = f ′( x ) = = f ′′( x ) = =

+ 1) 2 1 + x2

x -intercept at  x = − 1, y-intercept at  y = 1

(x

( 1 + x 2 ) ⋅ 2 ( x + 1) − ( x + 1) 2 ⋅ 2 x 2 (1 + x 2 )

2 (1 − x 2 ) 2 (1 + x 2 )

Critical points:  x = − 1, x = 1

2

(1 + x 2 ) ⋅ 2 ( −2 x ) − 2 (1 − x 2 )[ 2 (1 + x 2 ) ⋅ 2 x ] 4 (1 + x 2 )

4 x ( x 2 − 3) 3 (1 + x 2 )

After some algebra, including canceling the common factor (1 + x 2 )

3. Behavior at critical points. The critical points occur only at x = ±1 where f ′( x ) = 0 (Step 2) since f ′ exists everywhere over the domain of f . At x = −1, f ′′(−1) = 1 > 0, yielding a relative minimum by the Second Derivative Test. At x = 1, f ′′(1) = −1 < 0, yielding a relative maximum by the Second Derivative test. 4. Increasing and decreasing. We see that on the interval (−∞, − 1) the derivative f ′( x ) < 0, and the curve is decreasing. On the interval (−1, 1) , f ′( x ) > 0 and the curve is increasing; it is decreasing on (1, ∞ ) where f ′( x ) < 0 again. 5. Inflection points. Notice that the denominator of the second derivative (Step 2) is always positive. The second derivative f ′′ is zero when x = − 3, 0, and 3. The second derivative changes sign at each of these points: negative on (−∞, − 3 ) , positive on (− 3, 0 ) , negative on ( 0, 3 ) , and positive again on ( 3, ∞ ). Thus each point is a point of inflection. The curve is concave down on the interval (−∞, − 3 ) , concave up on (− 3, 0 ) , concave down on ( 0, 3 ) , and concave up again on ( 3, ∞ ).

262

Chapter 4 Applications of Derivatives

y 2

(1, 2)

Point of inflection where x = " 3

6. Asymptotes. Expanding the numerator of f ( x ) and then dividing both numerator and denominator by x 2 gives + 1) 2 x 2 + 2x + 1 = 2 1+ x 1 + x2 1 + ( 2 x ) + (1 x 2 ) . = (1 x 2 ) + 1

f (x) =

y=1

1 Horizontal asymptote −1 Point of inflection where x = − " 3

x

1

FIGURE 4.32

The graph of + 1) 2 y = (Example 9). 1 + x2 (x

(x

Expanding numerator Dividing by x 2

We see that f ( x ) → 1 as x → ∞ and that f ( x ) → 1 as x → −∞. Thus, the line y = 1 is a horizontal asymptote. Since the function is continuous on (−∞, ∞ ) , there are no vertical asymptotes. 7. The graph of f is sketched in Figure 4.32. Notice how the graph is concave down as it approaches the horizontal asymptote y = 1 as x → −∞, and concave up in its approach to y = 1 as x → ∞.

Sketch the graph of f ( x ) =

EXAMPLE 10

x2 + 4 . 2x

Solution

1. The domain of f is all nonzero real numbers. There are no intercepts because neither x nor f ( x ) can be zero. Since f (−x ) = − f ( x ), we note that f is an odd function, so the graph of f is symmetric about the origin. 2. We calculate the derivatives of the function, but we first rewrite it in order to simplify our computations:

y 4 2 −4

−2 (−2, −2)

0

2 y= x +4 2x

(2, 2) y= x 2 2

4

x

−2 −4

x2 + 4 FIGURE 4.33 The graph of y = 2x (Example 10).

f (x) =

x2 + 4 x 2 = + 2x 2 x

Function simplified for differentiation

f ′( x ) =

1 2 x2 − 4 − 2 = 2 x 2x 2

Combine fractions to solve easily  f ′( x ) = 0.

f ′′( x ) =

4 x3

Exists throughout the entire domain of  f

3. The critical points occur at x = ±2 where f ′( x ) = 0. Since f ′′(−2) < 0 and f ′′(2) > 0, we see from the Second Derivative Test that a relative maximum occurs at x = −2 with f (−2) = −2, and a relative minimum occurs at x = 2 with f (2) = 2. 4. On the interval (−∞, −2 ) the derivative f ′ is positive because x 2 − 4 > 0 so the graph is increasing; on the interval (−2, 0 ) the derivative is negative and the graph is decreasing. Similarly, the graph is decreasing on the interval ( 0, 2 ) and increasing on ( 2, ∞ ). 5. There are no points of inflection because f ′′( x ) < 0 whenever x < 0, f ′′( x ) > 0 whenever x > 0, and f ′′ exists everywhere and is never zero throughout the domain of f . The graph is concave down on the interval (−∞, 0 ) and concave up on the interval ( 0, ∞ ). 6. From the rewritten formula for f ( x ), we see that lim

x → 0+

( 2x + 2x ) = +∞

and

lim

x → 0−

( 2x + 2x ) = −∞,

so the y-axis is a vertical asymptote. Also, as x → ∞ or as x → −∞, the graph of f ( x ) approaches the line y = x 2. Thus y = x 2 is an oblique asymptote. 7. The graph of f is sketched in Figure 4.33.

4.4  Concavity and Curve Sketching

263

Sketch the graph of f ( x ) = e 2 x .

EXAMPLE 11

Solution The domain of f is ( −∞, 0 ) ∪ ( 0, ∞ ) and there are no symmetries about either axis or the origin. The derivatives of f are

(

f ′( x ) = e 2 x −

)

2 2e 2 x = − 2 2 x x

and f ′′( x ) = − y 5

y = e 2x

4 3 Inflection 2 point 1 −2

−1

y=1 0

1

2

x

3

FIGURE 4.34 The graph of y = e 2 x has a point of inflection at (−1, e −2 ). The line y = 1 is a horizontal asymptote and x = 0 is a vertical asymptote (Example 11).

x 2 ( 2e 2 x )(−2 x 2 ) − 2e 2 x (2 x ) 4 e 2 x (1 + x ) . = x4 x4

Both derivatives exist everywhere over the domain of f . Moreover, since e 2 x and x 2 are both positive for all x ≠ 0, we see that f ′ < 0 everywhere over the domain and the graph is decreasing on the intervals (−∞, 0 ) and ( 0, ∞ ). Examining the second derivative, we see that f ′′( x ) = 0 at x = −1. Since e 2 x > 0 and x 4 > 0 , we have f ′′ < 0 for x < −1 and f ′′ > 0 for x > −1, x ≠ 0. Since f ′′ changes sign, the point (−1, e −2 ) is a point of inflection. The curve is concave down on the interval (−∞, −1) and concave up over (−1, 0 ) ∪ ( 0, ∞ ). From Example 7, Section 2.5, we see that lim− f ( x ) = 0 . As x → 0 +, we see that x→0 2 x → ∞, so lim+ f ( x ) = ∞ and the y-axis is a vertical asymptote. Also, as x → −∞ x→0

or  x → ∞, 2 x → 0 and so lim f ( x ) = lim f ( x ) = e 0 = 1. Therefore, y = 1 is a x →−∞ x →∞ horizontal asymptote. There are no absolute extrema, since f never takes on the value 0 and has no absolute maximum. The graph of f is sketched in Figure 4.34.

EXAMPLE 12

Sketch the graph of f ( x ) = cos x −

2 x over 0 ≤ x ≤ 2π. 2

Solution The derivatives of f are

f ′( x ) = − sin x − y 4 2

y = cos x − π 2

!2 x 2

5π 3π 7π 4 2 4 2π

0 −2 −4

inflection points

FIGURE 4.35 The graph of the function in Example 12.

x

2 2

and

f ′′( x ) = − cos x.

Both derivatives exist everywhere over the interval ( 0, 2π ). Within that open interval, the first derivative is zero when sin x = − 2 2 , so the critical points are x = 5π 4 and x = 7π 4. Since f ′′ ( 5π 4 ) = − cos ( 5π 4 ) = 2 2 > 0, the function has a local minimum value of f ( 5π 4 ) ≈ −3.48 (evaluated with a calculator) by the Second Derivative Test. Also, f ′′ ( 7π 4 ) = − cos ( 7π 4 ) = − 2 2 < 0, so the function has a local maximum value of f ( 7π 4 ) ≈ −3.18. Examining the second derivative, we find that f ′′ = 0 when x = π 2 or x = 3π 2. Since f ′′ changes sign at these two points, we conclude that ( π 2 , f ( π 2 )) ≈ ( π 2 , −1.11) and ( 3π 2 , f ( 3π 2 )) ≈ ( 3π 2 , −3.33 ) are points of inflection. Finally, we evaluate f at the endpoints of the interval to find f (0) = 1 and f (2π) ≈ −3.44. Therefore, the values f (0) = 1 and f ( 5π 4 ) ≈ −3.48 are the absolute maximum and absolute minimum values of f over the closed interval [ 0, 2π ]. The graph of f is sketched in Figure 4.35.

Graphical Behavior of Functions from Derivatives As we saw in Examples 8–12, we can learn much about a twice-differentiable function y = f ( x ) by examining its first derivative. We can find where the function’s graph rises and falls and where any local extrema are located. We can differentiate y′ to learn how the graph bends as it passes over the intervals of rise and fall. Together with information about the function’s asymptotes and its value at some key points, such as intercepts, this information about

264

Chapter 4 Applications of Derivatives

the derivatives helps us determine the shape of the function’s graph. The following figure summarizes how the first derivative and second derivative affect the shape of a graph. y = f (x)

y = f (x)

Differentiable 1 smooth, connected; graph may rise and fall

y′ > 0 1 rises from left to right; may be wavy

or

y = f (x)

y′ < 0 1 falls from left to right; may be wavy

−

or +

y″ > 0 1 concave up throughout; no waves; graph may rise or fall or both

+

−

or

−

y″ < 0 1 concave down throughout; no waves; graph may rise or fall or both

+

y′ changes sign 1 graph has local maximum or local minimum

EXERCISES

y′ = 0 and y″ < 0 at a point; graph has local maximum

y′ = 0 and y″ > 0 at a point; graph has local minimum

4.4

Analyzing Functions from Graphs

Identify the inflection points and local maxima and minima of the functions graphed in Exercises 1–8. Identify the open intervals on which the functions are differentiable and the graphs are concave up and concave down. 3 2 1. y = x − x − 2x + 1

3 y

4 2. y = x − 2x2 + 4

3

2

5. y = x + sin 2x, − 2p ≤ x ≤ 2p 3

4

3

2p 3

0

2p 3

y

x

0

3. y =

3 2 23 ( x − 1) 4

0

x

0

8. y = 2 cos x − " 2 x, −p ≤ x ≤ 3p 2 y

x

−p

0

3p 2

x

NOT TO SCALE

4. y = 9 x13(x 2 − 7) 14

y

y

0

Graphing Functions

x

In Exercises 9–70, graph the function using appropriate methods from the graphing procedures presented just before Example 9, identifying the coordinates of any local extreme points and inflection points. Then find coordinates of absolute extreme points, if any. 9. y = x 2 − 4 x + 3

0

y

x

7. y = sin 0 x 0 , −2p ≤ x ≤ 2p

x

6. y = tan x − 4x, − p < x < p 2 2

y

−

y

0

y″ changes sign at an inflection point

x

10. y = 6 − 2 x − x 2

11. y = x 3 − 3 x + 3

12. y = x ( 6 − 2 x ) 2

13. y = −2 x 3 + 6 x 2 − 3

14. y = 1 − 9 x − 6 x 2 − x 3

15. y = ( x − 2 )3 + 1

16. y = 1 − ( x + 1)3

4.4  Concavity and Curve Sketching

17. y = x 4 − 2 x 2 = x 2 ( x 2 − 2 ) 18. y =

−x 4

+

−4 =

6x 2

x 2 (6

Sketching the General Shape, Knowing y ′

−

x2)

−4

19. y = 4 x 3 − x 4 = x 3 ( 4 − x )

20. y = x 4 + 2 x 3 = x 3 ( x + 2 )

21. y = x 5 − 5 x 4 = x 4 ( x − 5 )

22. y = x

2x 2 + x − 1 23. y = x2 − 1

x 2 − 49 24. y = 2 x + 5 x − 14

x4 + 1 25. y = x2

x2 − 4 26. y = 2x

27. y =

1 x2 − 1

29. y = −

x2 − 2 x2 − 1

( 2x − 5 )

28. y =

x2 x2 − 1

30. y =

x2 − 4 x2 − 2

4

31. y =

x2 x +1

32. y = −

x2 − 4 x +1

33. y =

x2 − x + 1 x −1

34. y = −

x2 − x + 1 x −1

35. y =

x 3 − 3x 2 + 3x − 1 x2 + x − 2

36. y =

x3 + x − 2 x − x2

37. y =

x x2 − 1

38. y =

4x ( Newton’s x 2 + 4 serpentine )

8 ( Agnesi’s 39. y = 2 x + 4 witch )

x

40. y =

x2 + 1

Each of Exercises 71–92 gives the first derivative of a continuous function y = f ( x ). Find y ′′ and then use Steps 2–4 of the graphing procedure described in this section to sketch the general shape of the graph of f . 71. y ′ = 2 + x − x 2

72. y ′ = x 2 − x − 6

73. y ′ = x ( x − 3 )

74. y ′ = x 2 ( 2 − x )

2

75. y ′ = x ( x 2 − 12 )

76. y ′ = ( x − 1) 2 ( 2 x + 3 )

77. y ′ = ( 8 x − 5 x 2 )( 4 − x ) 2 78. y ′ = ( x 2 − 2 x )( x − 5 ) 2 π π 79. y′ = sec 2 x , − < x < 2 2 π π 80. y′ = tan x , − < x < 2 2 θ θ 81. y ′ = cot , 0 < θ < 2π 82. y ′ = csc 2 , 0 < θ < 2π 2 2 π π 2 83. y ′ = tan θ − 1, − < θ < 2 2 84. y ′ = 1 − cot 2 θ, 0 < θ < π 85. y′ = cos t , 0 ≤ t ≤ 2π

86. y′ = sin t , 0 ≤ t ≤ 2π

87. y ′ = ( x + 1)−2 3

88. y ′ = ( x − 2 )−1 3

89. y ′ = x −2 3 ( x − 1)

90. y ′ = x −4 5 ( x + 1)

⎪⎧−2 x , x ≤ 0 91. y′ = 2 x = ⎨ x > 0 ⎩⎪⎪ 2 x ,

42. y = x − sin x , 0 ≤ x ≤ 2π

2 ⎪⎧−x , x ≤ 0 92. y′ = ⎪⎨ 2 ⎪⎪⎩ x , x > 0

43. y =

Sketching y from Graphs of y ′ and y ′′

41. y = x + sin x , 0 ≤ x ≤ 2π 3x − 2 cos x , 0 ≤ x ≤ 2π

Each of Exercises 93–96 shows the graphs of the first and second derivatives of a function y = f ( x ). Copy the picture and add to it a sketch of the approximate graph of f , given that the graph passes through the point P.

−π π 4 < x < 44. y = x − tan x , 2 2 3 45. y = sin x cos x , 0 ≤ x ≤ π 46. y = cos x + 47. y = x

3 sin x , 0 ≤ x ≤ 2π

49. y = 2 x − 3 x 2 3 5 51. y = x 2 3 −x 2 53. y = x 8 − x 2

(

55. y =

)

54. y = ( 2 −

x 2 )3 2

56. y = x 2 +

2 x

x2 − 3 x−2 8x 59. y = 2 x +4

5 60. y = 4 x +5

61. y = x 2 − 1

62. y = x 2 − 2 x

63. y =

⎪⎧ −x , x = ⎪⎨ ⎪⎪ x , ⎪⎩

64. y =

x−4 x 9 − x2

y = f ′(x)

P

52. y = x 2 3 ( x − 5 )

16 − x 2

58. y =

3

x3 + 1

x

x P

y = f ″(x)

95. y P

y = f ′(x)

x

0 y = f ″(x)

x < 0 x ≥ 0 96. y 66. y =

y

94.

y = f ′(x)

50. y = 5 x 2 5 − 2 x

57. y =

65. y =

y

93.

48. y = x 2 5

15

265

y = f ′(x)

x2 1− x

67. y = ln ( 3 − x 2 )

68. y = ( ln x ) 2

69. y = ln ( cos x )

70. y =

1 ex = − x 1+e 1 + ex

x

0 y = f ″(x) P

y = f ″(x)

266

Chapter 4 Applications of Derivatives

Theory and Examples

97. The accompanying figure shows a portion of the graph of a twice-differentiable function y = f ( x ). At each of the five labeled points, classify y ′ and y ′′ as positive, negative, or zero.

102. Sketch the graph of a twice-differentiable function y = f ( x ) that passes through the points (−3, −2 ) , (−2, 0 ) , ( 0, 1), (1, 2 ), and ( 2, 3 ) and whose first two derivatives have the following sign patterns.

y

2 y = f (x) R

P

1

1

y9:

S

−3

2

0

x

2

T

Q

y 0:

1

2 −2

1 0

2 x

1

x

0

98. Sketch a smooth connected curve y = f ( x ) with f (−2) = 8,

f ′(2) = f ′(−2) = 0,

  f (0) = 4,

f ′( x ) < 0 for

x < 2,

  f (2) = 0,

f ′′( x ) < 0 for

x < 0,

f ′( x ) > 0 for

x > 2,

1 4

4 < x < 6 6

7

x > 6

y

x > 2

4

x

−4

−2 −

0

−2

−2

−4

−4

2

4

x

y ′ > 0,

y ′′ > 0

y ′ > 0,

y ′′ = 0

In Exercises 105 and 106, the graph of f ′ is given. Determine x-values corresponding to local minima, local maxima, and inflection points for the graph of f .

y ′ > 0,

y ′′ < 0

105.

y ′ = 0,

y ′′ < 0

y ′ < 0,

y ′′ < 0

Derivatives y ′′ = 0

y′ > 0,

y″ > 0

0

y ′ > 0,

y ′′ = 0

y ′ > 0,

y ′′ < 0

y ′ = 0,

y ′′ < 0

y ′ < 0,

y ′′ < 0

y ′ < 0,

y ′′ = 0

y ′ < 0,

y ′′ > 0

y′ = 0,

y″ > 0

y′ > 0,

y″ > 0

3 2

1 < x < 2 2

2

y ′′ > 0

y ′ = 0,

0 < x < 1 1

0

y ′ = 0,

−1

−1 < x < 0 0

−2

2

y ′′ > 0

y ′′ < 0

−2 < x < −1 −1

−4

f9

y ′ < 0,

y ′ > 0,

x < −2 −2

f9

2

x > 0.

100. Sketch the graph of a twice-differentiable function y = f ( x ) that passes through the points (−2, 2 ), (−1, 1), ( 0, 0 ), (1, 1), and ( 2, 2 ) and whose first two derivatives have the following sign patterns. + − + − y′ : 0 2 −2 − + − y ′′ : 1 −1 101. Sketch the graph of a twice-differentiable function y = f ( x ) with the following properties. Label coordinates where possible. x

4

Derivatives

y

2 < x < 4 4

y

104.

4

f ′′( x ) > 0 for

x < 2 2

y

103.

99. Sketch the graph of a twice-differentiable function y = f ( x ) with the following properties. Label coordinates where possible. x

In Exercises 103 and 104, the graph of f ′ is given. Determine x-values corresponding to inflection points for the graph of f .

0

y

y

106. 4

4 2 −4

−2

0

2

f9

2

f9 4

x

−44

−2 −

0

−2

−2

−4

−4

2

4

x

107. A function f ( x ) has domain (−2, 2 ). The graph below is a plot of the derivative of f , not a plot of f itself. In other words, this is a graph of y = f ′( x ). Either use this graph to determine on which intervals the graph of f is concave up and on which intervals the graph of f is concave down, or explain why this information cannot be determined from the graph. y 2 y = f 9(x) 1 -2

-1

0 -1 -2

1

2

x

4.4  Concavity and Curve Sketching

108. A function f ( x ) has domain (−2, 2 ). The graph below is a plot of the second derivative of f , not a plot of f itself. In other words, this is a graph of y = f ′′( x ).

267

112. The accompanying graph shows the monthly revenue of the Widget Corporation for the past 12 years. During approximately what time intervals was the marginal revenue increasing? Decreasing?

y

y

2 1 -2

-1

0

y = r(t)

y = f 0(x) 1

2

x

-1

0

-2

b. Either use the graph above to determine on which intervals f is increasing and on which intervals f is decreasing, or explain why this information cannot be determined from the graph. Motion Along a Line The graphs in Exercises 109 and 110 show the position s = f (t ) of an object moving up and down on a coordinate line. (a) When is the object moving away from the origin? Toward the origin? At approximately what times is the (b) velocity equal to zero? (c) Acceleration equal to zero? (d) When is the acceleration positive? Negative? s

t

y ′ = ( x − 1) 2 ( x − 2 ). At what points, if any, does the graph of f have a local minimum, local maximum, or point of inflection? (Hint: Draw the sign pattern for y ′.) 114. Suppose the derivative of the function y = f ( x ) is y ′ = ( x − 1) 2 ( x − 2 )( x − 4 ). At what points, if any, does the graph of f have a local minimum, local maximum, or point of inflection? 115. For x > 0, sketch a curve y = f ( x ) that has f (1) = 0 and f ′( x ) = 1 x . Can anything be said about the concavity of such a curve? Give reasons for your answer.

Displacement

116. Can anything be said about the graph of a function y = f ( x ) that has a continuous second derivative that is never zero? Give reasons for your answer. s = f (t)

0

5

10 Time (s)

15

t

117. If b, c, and d are constants, for what value of b will the curve y = x 3 + bx 2 + cx + d have a point of inflection at x = 1? Give reasons for your answer. 118. Parabolas a. Find the coordinates of the vertex of the parabola y = ax 2 + bx + c, a ≠ 0.

s

b. When is the parabola concave up? Concave down? Give reasons for your answers.

Displacement

110.

10

113. Suppose the derivative of the function y = f ( x ) is

a. Either use the graph above to determine on which intervals the graph of f is concave up and on which intervals the graph of f is concave down and the inflection points of f , or explain why this information cannot be determined from the graph.

109.

5

119. Quadratic curves What can you say about the inflection points of a quadratic curve y = ax 2 + bx + c, a ≠ 0? Give reasons for your answer.

s = f (t)

5

0

10 Time (s)

15

t

120. Cubic curves What can you say about the inflection points of a cubic curve y = ax 3 + bx 2 + cx + d , a ≠ 0? Give reasons for your answer. 121. Suppose that the second derivative of the function y = f ( x ) is

111. Marginal cost The accompanying graph shows the hypothetical cost c = f ( x ) of manufacturing x items. At approximately what production level does the marginal cost change from decreasing to increasing? c

y ′′ = ( x + 1)( x − 2 ). For what x-values does the graph of f have an inflection point? 122. Suppose that the second derivative of the function y = f ( x ) is y ″ = x 2 ( x − 2 )3 ( x + 3 ). For what x-values does the graph of f have an inflection point?

c = f (x) Cost

123. Find the values of constants a, b, and c such that the graph of y = ax 3 + bx 2 + cx has a local maximum at x = 3, local minimum at x = −1, and inflection point at (1, 11). 20 40 60 80 100 120 Thousands of units produced

x

124. Find the values of constants a, b, and c such that the graph of y = ( x 2 + a ) ( bx + c ) has a local minimum at x = 3 and a local maximum at (−1, −2 ).

268

Chapter 4 Applications of Derivatives

COMPUTER EXPLORATIONS

In Exercises 125–128, find the inflection points (if any) on the graph of the function and the coordinates of the points on the graph where the function has a local maximum or local minimum value. Then graph the function in a region large enough to show all these points simultaneously. Add to your picture the graphs of the function’s first and second derivatives. How are the values at which these graphs intersect the x-axis related to the graph of the function? In what other ways are the graphs of the derivatives related to the graph of the function? 125. y = x 5 − 5 x 4 − 240

126. y = x 3 − 12 x 2

4 5 x + 16 x 2 − 25 5 x4 x3 128. y = − − 4 x 2 + 12 x + 20 4 3 129. Graph f ( x ) = 2 x 4 − 4 x 2 + 1 and its first two derivatives together. Comment on the behavior of f in relation to the signs and values of f a and f aa. 127. y =

130. Graph f ( x )  x cos x and its second derivative together for 0 b x b 2Q. Comment on the behavior of the graph of f in relation to the signs and values of f aa.

4.5 Indeterminate Forms and L’Hôpital’s Rule Consider the four limits x3 = lim x 2 = 0, x→0 x x 1 lim = lim 2 = ∞, x→0 x 3 x→0 x x lim = lim 1 = 1, x→0 x x→0 x 1 1 lim = lim = . x → 0 2x x→0 2 2 In each case both the numerator and the denominator approach zero as x l 0, even though these limits ultimately lead to completely different results: 0, d, 1, and 1 2. We say the expression f (x) lim when lim f ( x ) = 0 and lim g( x ) = 0 (1) x → a g( x ) x→a x→a involves an indeterminate form 0/0. The expression “ 0 0 ” has the form of a number, but it is not a meaningful quantity. Stating that both the numerator and the denominator approach zero does not provide sufficient information to obtain the limit of the ratio. We have to examine the behavior of the expression in more detail by performing algebraic manipulation or by applying methods that we will introduce in this section. Other forms exhibit behavior similar to Equation (1). For instance, if both the numerator and the denominator approach +∞ or −∞, then the limit of the ratio leads to an indeterminate form d d. Additional indeterminate forms we consider in this section are ∞ ⋅ 0, ∞ − ∞, 1d , 0 0 , and d 0. Their purpose is to summarize the behavior of certain types of limits. John (Johann) Bernoulli discovered a rule for using derivatives to calculate limits of fractions whose numerators and denominators both approach zero or ±∞. The rule is known today as l’Hôpital’s Rule, after Guillaume de l’Hôpital. He was a French nobleman who wrote the first introductory differential calculus text, where the rule first appeared in print. Limits involving transcendental functions often require some use of this rule. lim

x→0

HISTORICAL BIOGRAPHY Guillaume François Antoine de l’Hôpital

(1661–1704) In the late 1600s, John Fernoulle discovered a rule for calculating limits of fractions whose numerators and denominators both approach zero. Today the rule is known as l’Hôpital’s rule. To know more, visit the companion Website. Johann Bernoulli

Indeterminate Form 0 0 It is important to understand that the notation “0 0” is not intended to imply numerically dividing 0 by 0. Instead, the indeterminate form 0 0 refers to a limit of a ratio of two functions, each of which approaches zero. L’Hôpital’s rule can help us evaluate such limits.

(1667–1748) Johann Bernoulli was born in Switzerland and attended the University of Basel. His doctoral dissertation was in mathematics despite its medical title, which was used to hide his mathematical work from his father who wanted Johann to become a doctor. To know more, visit the companion Website.

THEOREM 6—L’Hôpital’s Rule

Suppose that lim f ( x ) = lim g( x ) = 0 , that x→a

x→a

f and g are differentiable on an open interval I containing a, and that g′( x ) ≠ 0 on I  if  x v a. Then f (x) f ′( x ) lim = lim , x → a g( x ) x → a g′( x ) assuming that the limit on the right side of this equation exists.

4.5  Indeterminate Forms and L’Hôpital’s Rule

269

We give a proof of Theorem 6 at the end of this section. Theorem 6 also applies if x → ±∞ or when f ′( x ) g′( x ) → ±∞, but we will not prove this. Caution

To apply l’Hôpital’s Rule to f g, divide the derivative of f by the derivative of g. Do not make the mistake of taking the derivative of f g. The quotient to use is f ′ g ′, not ( f g ) ′.

EXAMPLE 1 The following limits involve 0 0 indeterminate forms, so we apply l’Hôpital’s Rule. In some cases, it must be applied repeatedly. 3 x − sin x x 3 − cos x          = lim x→0 1 3 − cos 0          = = 2 1

(a) lim

x→0

The numerator and the denominator are both approaching 0; apply l’Hôpital’s Rule.   Not  0 0 Limit is found.

1 1+ x −1 2 1 + x = 1 = lim (b) lim x→0 x→0 x 1 2 (c) lim

x→0

(d) lim

x→0

1+ x −1− x 2 x2 (1 2 )(1 + x )−1 2 − 1 2 = lim x→0 2x −(1 4 )(1 + x )−3 2 1 = lim = − x→0 2 8 x − sin x x3

0 0

1 − cos x 3x 2 sin x = lim x →0 6x cos x 1 = lim = x→0 6 6 = lim

x→0

(e) lim

x →∞

0 0

; apply l’Hôpital’s Rule.

0 Still  ; apply l’Hôpital’s Rule again. 0 0 Not  ; limit is found. 0

; apply l’Hôpital’s Rule.

0 Still  ; apply l’Hôpital’s Rule again. 0 0 Still  ; apply l’Hôpital’s Rule again. 0 0 Not  ; limit is found. 0

ln (1 + 1 x ) sin (1 x )

0 0

1 1+1x          = lim x →∞ −(1 x 2 ) cos (1 x ) −(1 x 2 )

; apply l’Hôpital’s Rule.

0 Still  ; simplification is easier than applying 0 l’Hôpital’s Rule.

1 1 1+1x 1 0 =1 +          = lim = x →∞ cos (1 x ) cos 0

Not

0 0

; limit is found.

Here is a summary of the procedure we followed in Example 1.

Using L’Hôpital’s Rule

To find lim

x→a

f (x) g( x )

by l’Hôpital’s Rule, we continue to differentiate f and g, so long as we still get the form 0 0 as x → a. But as soon as one or the other of these derivatives no longer approaches zero, we stop differentiating. L’Hôpital’s Rule does not apply when either the numerator or the denominator has a finite nonzero limit.

270

Chapter 4 Applications of Derivatives

EXAMPLE 2

Be careful to apply l’Hôpital’s Rule correctly: 1 − cos x x + x2 sin x = lim x →0 1 + 2x

lim

x→0

0 0 Not

0 0

It is tempting to try to apply l’Hôpital’s Rule again, which would result in lim

x→0

cos x 1 = , 2 2

but this is not the correct limit. l’Hôpital’s Rule can be applied only to limits that give indeterminate forms, and lim ( sin x ) (1 + 2 x ) does not give an indeterminate form. x→0

Instead, this limit is 0 1 = 0, and the correct answer for the original limit is 0. L’Hôpital’s Rule applies to one-sided limits as well. EXAMPLE 3

In this example the one-sided limits are different.

sin x x2

(a) lim+ x→0

0 0

cos x = lim+ = ∞ x→0 2x

Positive for  x > 0

sin x x2

(b) lim− x→0

0 0

cos x = lim− = −∞ x→0 2x

Negative for  x < 0

Indeterminate Forms ∞ ∞ , ∞ ⋅ 0, ∞ − ∞ Recall that ∞ and +∞ mean the same thing.

Sometimes when we try to evaluate a limit as x → a, we get an indeterminate form like ∞ ∞ , ∞ ⋅ 0, or ∞ − ∞, instead of 0 0. We first consider the form ∞ ∞. More advanced treatments of calculus prove that l’Hôpital’s Rule applies to the indeterminate form ∞ ∞, as well as to 0 0. If f ( x ) → ±∞ and g( x ) → ±∞ as x → a, then lim

x→a

f (x) f ′( x ) , = lim x → a g ′( x ) g( x )

provided the limit on the right exists or approaches ∞ or  −∞. In the notation x → a, a may be either finite or infinite. Moreover, x → a may be replaced by the one-sided limits x → a + or x → a − . EXAMPLE 4

Find the limits of these ∞ ∞ forms:

sec x x → π 2 1 + tan x

(a) lim

ln x x →∞ 2 x

(b) lim

ex . x →∞ x 2

(c) lim

Solution

(a) The numerator and denominator are discontinuous at x = π 2 , so we investigate the one-sided limits there. To apply l’Hôpital’s Rule, we can choose I to be any open interval with x = π 2 as an endpoint.

4.5  Indeterminate Forms and L’Hôpital’s Rule

lim

−

x →( π 2 )

sec x 1 + tan x

=

∞

from left, apply l’Hôpital’s Rule

∞

sec x tan x = sec 2 x

lim

x →( π 2 )−

lim

x →( π 2 )−

271

sin x = 1

The right-hand limit is 1 also, with (−∞ ) (−∞ ) as the indeterminate form. Therefore, the two-sided limit is equal to 1. ln x 1x 1 = lim = lim = 0 x →∞ 1 x →∞ 2 x x x

(b) lim

x →∞

ex

ex

1 x 1

x

=

x x

=

1 x

ex

= lim = lim = ∞ x →∞ 2 x x →∞ 2 x2 Next we turn our attention to the indeterminate forms ∞ ⋅ 0 and ∞ − ∞. Sometimes these forms can be handled by using algebra to convert them to a 0 0 or ∞ ∞ form. Here again, we do not mean to suggest that ∞ ⋅ 0 or ∞ − ∞ is a number. They are only notations for functional behaviors when considering limits. Here are examples of how we might work with these indeterminate forms. (c) lim

x →∞

Find the limits of these ∞ ⋅ 0 forms:

EXAMPLE 5

(

x →∞

(b) lim+ x→0

Solution

(

(a) lim x sin x →∞

)

1 x x ln x

(a) lim x sin

sin (1 x ) 1 = lim x →∞ 1x x

)

∞ ⋅ 0 converted to

( cos (1 x ))(−1 x 2 ) x →∞ −1 x 2

= lim

(

= lim cos x →∞

0 0

 

L’Hôpital’s Rule applied

)

1 =1 x

(See Example 6b in Section 2.5 for an alternative method to solve this problem.) (b) lim+ x→0

x  ln x = lim+ x→0

= lim+ x→0

ln x 1 x

∞ ⋅ 0 converted to  ∞ ∞  

1x −(1 2 ) x 3 / 2

l’Hôpital’s Rule applied

= lim+ ( −2 x ) = 0 x→0

EXAMPLE 6

Find the limit of this ∞ − ∞ form: ⎛ 1 1⎞ lim ⎜⎜ − ⎟⎟⎟. x→0 ⎜ x⎠ ⎝ sin x

Solution If x → 0 + , then sin x → 0 + and

1 1 − → ∞ − ∞. sin x x Similarly, if x → 0 − , then sin x → 0 − and 1 1 − → −∞ − (−∞ ) = −∞ + ∞. sin x x

272

Chapter 4 Applications of Derivatives

Neither form reveals what happens in the limit. To find out, we first combine the fractions: x − sin x 1 1 . − = sin x x x sin x

Common denominator is  x sin x .

Then we apply l’Hôpital’s Rule to the result: ⎛ 1 x − sin x 1⎞ lim ⎜⎜ − ⎟⎟⎟ = lim ⎜ x → 0 ⎝ sin x x ⎠ x → 0 x sin x = lim

x→0

0 0

1 − cos x sin x + x cos x

Still 

0 0

sin x 0 = = 0. x → 0 2 cos x − x sin x 2

= lim

Indeterminate Powers Limits that lead to the indeterminate forms 1∞ , 0 0 , and ∞ 0 can sometimes be handled by first taking the logarithm of the function. We use l’Hôpital’s Rule to find the limit of the logarithm expression and then exponentiate the result to find the original function limit. This procedure is justified by the continuity of the exponential function and Theorem 9 in Section 2.6, and it is formulated as follows. (The formula is also valid for one-sided limits.) If lim ln f ( x ) = L , then x→a

lim f ( x ) = lim e ln f ( x ) = e L . x→a

x→a

Here a may be either finite or infinite. EXAMPLE 7

Apply l’Hôpital’s Rule to show that lim+ (1 + x )1 x = e . x→0

Solution The limit leads to the indeterminate form 1∞ . We let f ( x ) = (1 + x )1 x and find lim+ ln f ( x ). Since x→0

ln f ( x ) = ln (1 + x )1 x =

1 ( ln 1 + x ) , x

l’Hôpital’s Rule now applies to give ln (1 + x ) x 1 1 + x = lim+ x→0 1 1 = = 1. 1

lim ln f ( x ) = lim+

x → 0+

x→0

0 0

L’Hôpital’s Rule applied

Therefore, lim+ (1 + x )1 x = lim+ f ( x ) = lim+ e ln f ( x ) = e 1 = e. x→0

EXAMPLE 8

x→0

x→0

Find lim x 1 x. x →∞

Solution The limit leads to the indeterminate form ∞ 0. We let f ( x ) = x 1 x and find lim ln f ( x ). Since x →∞ ln x ln f ( x ) = ln x 1 x = , x

4.5  Indeterminate Forms and L’Hôpital’s Rule

273

l’Hôpital’s Rule gives ln x x →∞ x 1x = lim x →∞ 1 0 = = 0. 1

lim ln f ( x ) = lim

x →∞

∞ ∞ L’Hôpital’s Rule applied

Therefore, lim x 1 x = lim f ( x ) = lim e ln f ( x ) = e 0 = 1. x →∞

x →∞

x →∞

Proof of L’Hôpital’s Rule y y = f ′(a)(x − a) f (x) y = g′(a)(x − a) g(x) 0

a

FIGURE 4.36 The two functions in l’Hôpital’s Rule, graphed with their linear approximations at x  a.

HISTORICAL BIOGRAPHY Augustin-Louis Cauchy

(1789–1857) Cauchy was born in Paris the year the French revolution began. He was the first to define fully the ideas of convergence and absolute convergence of infinite series. His classic works Cours d’analyse (Course on Analysis, 1821) and Résumé des leçons … sur le calcul infinitésimal (1823) were his greatest contributions to calculus. To know more, visit the companion Website.

When g( x )  x , Theorem 7 is the Mean Value Theorem.

x

Before we prove l’Hôpital’s Rule, we consider a special case to provide some geometric insight for its reasonableness. Consider the two functions f ( x ) and g( x ) having continuous derivatives and satisfying f (a)  g(a)  0, g′(a) ≠ 0. The graphs of f ( x ) and g( x ), together with their linearizations y = f ′(a)( x − a ) and y = g′(a)( x − a ), are shown in Figure 4.36. We know that near x  a, the linearizations provide good approximations to the functions. In fact, f ( x ) = f ′(a)( x − a ) + F1 ( x − a ) and

g( x ) = g′(a)( x − a ) + F 2 ( x − a ) ,

where F1 l 0 and F 2 l 0 as x l a. So, as Figure 4.36 suggests, lim

x→a

f ′(a)( x − a ) + F1 ( x − a ) f (x) = lim x → a g ′(a) ( x − a ) + F 2 ( x − a ) g( x ) = lim

f ′(a) + F1 f ′(a) = g′(a) + F 2 g′(a)

g′( a ) ≠ 0

= lim

f ′( x ) , g′( x )

Continuous derivatives

x→a

x→a

as asserted by l’Hôpital’s Rule. We now proceed to a proof of the rule based on the more general assumptions stated in Theorem 6, which do not require that g′(a) ≠ 0 or that the two functions have continuous derivatives. The proof of l’Hôpital’s Rule is based on Cauchy’s Mean Value Theorem, an extension of the Mean Value Theorem that involves two functions instead of one. We prove Cauchy’s Theorem first and then show how it leads to l’Hôpital’s Rule.

THEOREM 7—Cauchy’s Mean Value Theorem

Suppose functions f and g are continuous on [ a, b ] and differentiable throughout ( a, b ) and also suppose g′( x ) ≠ 0 throughout ( a, b ). Then there exists a number c in ( a, b ) at which f ′(c) f (b) − f (a) . = g(b) − g(a) g′(c)

Proof We apply the Mean Value Theorem of Section 4.2 twice. First we use it to show that g(a) v g(b). For if g(b) did equal g(a), then the Mean Value Theorem would give g′(c) =

g(b) − g(a) = 0 b−a

for some c between a and b, which cannot happen because g′( x ) ≠ 0 in ( a, b ).

274

Chapter 4 Applications of Derivatives

We next apply the Mean Value Theorem to the function F ( x ) = f ( x ) − f (a) −

f (b) − f (a) [ g( x ) − g(a) ]. g(b) − g(a)

This function is continuous and differentiable where f and g are, and F (b)  F (a)  0. Therefore, there is a number c between a and b for which F ′(c) = 0. When expressed in terms of f and g, this equation becomes F ′(c) = f ′(c) −

f (b) − f (a) [ g′(c) ] = 0 g(b) − g(a)

so that f ′(c) f (b) − f (a) . = g(b) − g(a) g′(c)

y

Slope =

f ′(c) g′(c) B (g(b), f (b))

P

Slope =

A 0

f (b) − f (a) g(b) − g(a)

(g(a), f (a))

Cauchy’s Mean Value Theorem has a geometric interpretation for a general winding curve C in the plane joining the two points A  ( g(a), f (a)) and B  ( g(b), f (b)). In Chapter 10 you will learn how to describe general curves such as C, along with their tangent lines. There is at least one point P on the curve for which the tangent to the curve at P is parallel to the secant line joining the points A and B. The slope of that tangent line turns out to be the quotient f a g a evaluated at the number c in the interval ( a, b ) , which is the left-hand side of the equation in Theorem 7. Because the slope of the secant line joining A and B is

x

FIGURE 4.37 There is at least one point P on the curve C for which the slope of the tangent line to the curve at P is the same as the slope of the secant line joining the points A( g(a), f (a)) and B( g(b), f (b)).

f (b)  f (a) , g(b)  g(a) the equation in Cauchy’s Mean Value Theorem says that the slope of the tangent line equals the slope of the secant line. This geometric interpretation is shown in Figure 4.37. Notice from the figure that it is possible for more than one point on the curve C to have a tangent line that is parallel to the secant line joining A and B. Proof of l’Hôpital’s Rule

Since lim f ( x ) = lim g( x ) = 0 and both f and g are x→a

x→a

differentiable at a, they must also be continuous at a, hence f (a)  g(a)  0. We first establish the limit equation for the case x → a + . The method needs almost no change to apply to x → a − , and the combination of these two cases establishes the result. Suppose that x lies in the interval to the right of a. Then g′( x ) ≠ 0, and we can apply Cauchy’s Mean Value Theorem to the closed interval from a to x. This step produces a number c between a and x such that f ′(c) f ( x ) − f (a) . = g( x ) − g(a) g′(c) But f (a)  g(a)  0, so f ′(c) f (x) = . g( x ) g′(c) As x approaches a, c approaches a because it always lies between a and x. Therefore, lim

x → a+

f (x) f ′(c) f ′( x ) = lim+ , = lim+ c → a g ′(c) x → a g ′( x ) g( x )

which establishes l’Hôpital’s Rule for the case where x approaches a from above. The case where x approaches a from below is proved by applying Cauchy’s Mean Value Theorem to the closed interval [ x , a ], x  a.

4.5  Indeterminate Forms and L’Hôpital’s Rule

275

4.5

EXERCISES

Finding Limits in Two Ways

In Exercises 1–6, use l’Hôpital’s Rule to evaluate the limit. Then evaluate the limit using a method studied in Chapter 2. sin 5 x x +2 1. lim 2 2. lim x →−2 x − 4 x→0 x 5 x 2 − 3x x3 − 1 3. lim 4. lim 3 x →∞ 7 x 2 + 1 x →1 4 x − x − 3 1 − cos x 2 x 2 + 3x 5. lim 6. lim 3 x→0 x →∞ x + x + 1 x2

45. lim

cos θ − 1 eθ − θ − 1

46. lim

47. lim

et + t 2 et − t

48. lim x 2 e −x

49. lim

x − sin x x tan x

50. lim

θ − sin θ cos θ θ→ 0 tan θ − θ

52. lim

θ→0

t →∞

x→0

h→ 0

e h − (1 + h ) h2

x →∞

( e x − 1)

2

x sin x

x→0

sin 3 x − 3 x + x 2 x→0 sin x sin 2 x

51. lim

Applying l’Hôpital’s Rule

Use l’Hôpital’s rule to find the limits in Exercises 7–52. x−2 x 2 − 25 8. lim 7. lim 2 x→2 x − 4 x →−5 x + 5 − 4 t + 15 t →−3 t 2 − t − 12 5x 3 − 2x 11. lim x →∞ 7 x 3 + 3 sin t 2 13. lim t→0 t 8x 2 15. lim x → 0 cos x − 1

+3 t →−1 4 t 3 − t + 3 x − 8x 2 12. lim x →∞ 12 x 2 + 5 x sin 5t 14. lim t→0 2t sin x − x 16. lim x→0 x3

2θ − π 17. lim θ → π 2 cos ( 2π − θ ) 1 sin θ − 2 19. lim π θ→π 6 θ− 6 1 − sin θ 21. lim θ → π 2 1 + cos 2θ

3θ + π 18. lim θ →−π 3 sin ( θ + ( π 3 ))

t3

9. lim

23. lim

x→0

x2 ln ( sec x )

t (1 − cos t ) 25. lim t→0 t − sin t 27.

lim

x →( π 2 )−

(

)

π x− sec x 2

3 sin θ − 1 29. lim θ→0 θ x2 x 31. lim x x→0 2 − 1 ln ( x + 1) 33. lim x →∞ log 2 x 35. lim+ x→0

ln ( x 2 + 2 x ) ln x

10. lim

3t 3

tan θ − 1 20. lim π θ→π 4 θ− 4 x −1 22. lim x →1 ln x − sin π x ln ( csc x ) 24. lim 2 x → π 2 ( x − ( π 2 ))

lim

x →( π 2 )−

(

)

π − x tan x 2

(1 2 ) θ − 1 30. lim θ→0 θ 3x − 1 32. lim x x→0 2 − 1 log 2 x 34. lim x →∞ log 3 ( x + 3 ) 36. lim+ x→0

ln ( e x − 1) ln x

5 y + 25 − 5 37. lim y→ 0 y 39. lim ( ln 2 x − ln ( x + 1))

ay + a 2 − a 38. lim , a > 0 y→ 0 y 40. lim+ ( ln x − ln sin x )

( ln x ) 2 41. lim+ x → 0 ln ( sin x ) ⎛ 1 1 ⎞⎟ − 43. lim+ ⎜⎜ ⎟ x →1 ⎜ ⎝ x − 1 ln x ⎟⎠

⎛ 3x + 1 1 ⎞⎟ − 42. lim+ ⎜⎜ ⎟ x→0 ⎜ sin x ⎟⎠ ⎝ x

x →∞

x→0

44. lim+ ( csc x − cot x + cos x ) x→0

Find the limits in Exercises 53–68. 53. lim+ x 1 (1− x )

54. lim+ x 1 ( x −1)

x →1

x →1

1x

55. lim ( ln x )

56. lim+ ( ln x )1

57. lim+ x −1 ln x

58. lim x 1 ln x

x →∞

( x −e )

x →e

x→0

x →∞

1 ( 2 ln x )

59. lim (1 + 2 x )

60. lim ( e x + x )1 x

61. lim+ x x

62. lim+ 1 +

x →∞

x→0

x→0

63. lim

x →∞

x2

65. lim+ x→0

(

x→0

( xx +− 12 )

x

64. lim

x →∞

(

1 x

)

x

( xx ++21) 2

1x

66. lim+ x ( ln x ) 2

ln x

π 67. lim+ x tan −x x→0 2

x→0

)

68. lim+ sin x ⋅ ln x x→0

Theory and Applications

L’Hôpital’s Rule does not help with the limits in Exercises 69–76. Try it—you just keep on cycling. Find the limits some other way. 9x + 1 x +1

70. lim+

sec x tan x

72. lim+

cot x csc x

73. lim

2x − 3x 3x + 4 x

74. lim

2x + 4x 5x − 2x

75. lim

e x2 xe x

76. lim+

x e −1 x

69. lim

x →∞

t sin t 26. lim t → 0 1 − cos t 28.

Indeterminate Powers and Products

71.

lim

x →( π 2 )−

x →∞

x →∞

x sin x

x→0

x→0

x →−∞

x→0

77. Which one is correct, and which one is wrong? Give reasons for your answers. 1 1 0 x−3 x−3 a. lim 2 b. lim 2 = lim = = = 0 x →3 x − 3 x →3 2x x →3 x − 3 6 6 78. Which one is correct, and which one is wrong? Give reasons for your answers. a. lim

x→0

x 2 − 2x 2x − 2 = lim x → 0 2 x − cos x x 2 − sin x = lim

x→0

b. lim

x→0

2 2 = =1 2 + sin x 2+0

x 2 − 2x 2x − 2 −2 = lim = = 2 x → 0 2 x − cos x x 2 − sin x 0 −1

276

Chapter 4 Applications of Derivatives

79. Only one of these calculations is correct. Which one? Why are the others wrong? Give reasons for your answers.

a. Use l’Hôpital’s Rule to show that

x→0

ln x −∞ c. lim+ x ln x = lim+ = = −1 x→0 x → 0 (1 x ) ∞ ln x d. lim+ x ln x = lim+ x→0 x → 0 (1 x )

f (x) = 1 +

a. f ( x ) = x ,

g( x ) = x 2 ,

( a, b ) = (−2, 0 )

b. f ( x ) = x ,

g( x ) = x 2 ,

( a, b )  arbitrary ( a, b ) = ( 0, 3 )

g( x ) = x 2 ,

81. Continuous extension Find a value of c that makes the function ⎧⎪ 9 x − 3 sin 3 x , x ≠ 0 ⎪ f ( x ) = ⎪⎨ 5x 3 ⎪⎪ x = 0 ⎪⎩ c, continuous at x = 0. Explain why your value of c works. 82. For what values of a and b is x→0

( tanx 2x + xa 3

2

)

sin bx = 0? x

+

T 83. ∞ − ∞ Form a. Estimate the value of lim ( x −

x →∞

by graphing f ( x ) = x − interval of x-values.

x2 + x)

x 2 + x over a suitably large

b. Now confirm your estimate by finding the limit with l’Hôpital’s Rule. As the first step, multiply f ( x ) by the fraction ( x + x 2 + x ) ( x + x 2 + x ) and simplify the new numerator. 84. Find lim

x →∞

(

x2 + 1 −

x ).

x →1

2 x 2 − ( 3 x + 1) x + 2 x −1

by graphing. Then confirm your estimate with l’Hôpital’s Rule. 86. This exercise explores the difference between

(

lim 1 +

x →∞

1 x2

)

x

and

(

x

= e.

x

(

and g( x ) = 1 +

1 x

)

x

c. Confirm your estimate of lim f ( x ) by calculating it with x →∞ l’Hôpital’s Rule. 87. Show that

1 lim 1 + x →∞ x

)

x

= e.

(

lim 1 +

k →∞

r k

)

k

= er.

88. Given that x > 0, find the maximum value, if any, of a. x 1 x b. x 1 x

2 n

c. x 1 x (n a positive integer) d. Show that lim x 1 x x →∞

n

= 1 for every positive integer n.

89. Use limits to find horizontal asymptotes for each function. 1 a. y = x tan x 3x + e 2 x b. y = 2x + e 3x ⎧⎪ e −1 x 2 , x ≠ 0 90. Find f ′(0) for f ( x ) = ⎪⎨ ⎪⎪ 0, x = 0. ⎪⎩

( )

x T 91. The continuous extension of ( sin x ) to [ 0, π ]

a. Graph f ( x ) = ( sin x ) x on the interval 0 ≤ x ≤ π . What value would you assign to f to make it continuous at x = 0? b. Verify your conclusion in part (a) by finding lim+ f ( x ) with x→0 l’Hôpital’s Rule. c. Returning to the graph, estimate the maximum value of f on [ 0, π ]. About where is max f taken on? d. Sharpen your estimate in part (c) by graphing f ′ in the same window to see where its graph crosses the x-axis. To simplify your work, you might want to delete the exponential factor from the expression for f ′ and graph just the factor that has a zero. tan x T 92. The function ( sin x )

T 85. 0 0 Form Estimate the value of lim

)

together for x ≥ 0. How does the behavior of f compare with that of g? Estimate the value of lim f ( x ).

80. Find all values of c that satisfy the conclusion of Cauchy’s Mean Value Theorem for the given functions and interval.

lim

)

1 x

x →∞

(1 x ) = lim+ (−x ) = 0 x→0 (−1 x 2 )

c. f ( x ) = x 3 3 − 4 x ,

1 x2

(

x→0

x→0

(

T b. Graph

b. lim+ x ln x = 0 ⋅ (−∞ ) = −∞

= lim+

x →∞

lim 1 +

a. lim+ x ln x = 0 ⋅ (−∞ ) = 0

(Continuation of Exercise 91)

a. Graph f ( x ) = ( sin x ) tan x on the interval −7 ≤ x ≤ 7. How do you account for the gaps in the graph? How wide are the gaps? b. Now graph f on the interval 0 ≤ x ≤ π . The function is not defined at x = π 2, but the graph has no break at this point. What is going on? What value does the graph appear to give for f at x = π 2? (Hint: Use l’Hôpital’s Rule to find lim f as x → ( π 2 )− and x → ( π 2 )+ .) c. Continuing with the graphs in part (b), find max f and min f as accurately as you can and estimate the values of x at which they are taken on.

4.6  Applied Optimization

277

4.6 Applied Optimization What are the dimensions of a rectangle with fixed perimeter having maximum area? What are the dimensions for the least expensive cylindrical can of a given volume? How many items should be produced for the most profitable production run? Each of these questions asks for the best, or optimal, value of a given function. In this section we use derivatives to solve a variety of optimization problems in mathematics, physics, economics, and business. x

Solving Applied Optimization Problems

1. Read the problem. Read the problem until you understand it. What is given? What is the unknown quantity to be optimized (maximized or minimized)?

12

2. Introduce variables. List every relevant relation in the problem as an equation. In most problems it is helpful to draw a picture.

x x

x

12

3. Write an equation for the unknown quantity. Express the quantity to be optimized as a function of a single variable. This may require considerable manipulation.

(a)

4. Test the critical points and endpoints in the domain of the function found in the previous step. Use what you know about the shape of the function’s graph. Use the first and second derivatives to identify and classify the function's critical points.

x

12 − 2x 12 12 − 2x x

x

(b)

FIGURE 4.38 An open box made by cutting the corners from a square sheet of tin. What size corners maximize the box’s volume (Example 1)?

EXAMPLE 1 An open-top box is to be made by cutting small congruent squares from the corners of a 12-cm-by-12-cm sheet of tin and bending up the sides. How large should the squares cut from the corners be to make the box hold as much as possible? Solution We start with a picture (Figure 4.38). In the figure, the corner squares are x cm on a side. The volume of the box is a function of this variable:

V ( x ) = x (12 − 2 x ) 2 = 144 x − 48 x 2 + 4 x 3 .

Since the sides of the sheet of tin are only 12 cm long, x ≤ 6 and the domain of V is the interval 0 ≤ x ≤ 6. A graph of V (Figure 4.39) suggests a minimum value of 0 at x 0 and x 6 and a maximum near x 2. To learn more, we examine the first derivative of V with respect to x:

Maximum y

Volume 0

dV = 144 − 96 x + 12 x 2 = 12 (12 − 8 x + x 2 ) = 12 ( 2 − x )( 6 − x ). dx

y = x(12 − 2x)2, 0≤x≤6

min

min 2

V = hlw

6

x

NOT TO SCALE

FIGURE 4.39 The volume of the box in Figure 4.38 graphed as a function of x.

Of the two zeros, x = 2 and x = 6, only x = 2 lies in the interior of the function’s domain and makes the critical-point list. The values of V at this one critical point and two endpoints are Critical-point value: V (2) = 128 Endpoint values: V (0) = 0, V (6) = 0. The maximum volume is 128 cm 3 . The cutout squares should be 2 cm on a side.

278

Chapter 4 Applications of Derivatives

EXAMPLE 2 You have been asked to design a one-liter can shaped like a right circular cylinder (Figure 4.40). What dimensions will use the least material?

2r

Solution Volume of can: If r and h are measured in centimeters, then the volume of the can in cubic centimeters is h

πr 2 h = 1000. πr 2 + 2 πrh A = 2

Surface area of can:

circular ends

FIGURE 4.40

Example 2 shows that this one-liter can uses the least material when h = 2r.

1 liter = 1000 cm 3

cylindrical wall

How can we interpret the phrase “least material”? For a first approximation we can ignore the thickness of the material and the waste in manufacturing. Then we ask for dimensions r and h that make the total surface area as small as possible, while satisfying the constraint πr 2 h = 1000 cm 3 . To express the surface area as a function of one variable, we solve for one of the variables in πr 2 h = 1000 and substitute that expression into the surface area formula. Solving for h is easier: h =

1000 . πr 2

Thus, A = 2πr 2 + 2πrh = 2πr 2 + 2πr = 2πr 2 +

( 1000 ) πr 2

2000 . r

Our goal is to find a value of r > 0 that minimizes the value of A. Since A is differentiable on r > 0, an interval with no endpoints, it can have a minimum value only where its first derivative is zero. dA 2000 = 4 πr − 2 dr r 2000 0 = 4 πr − 2 r 3 4 πr = 2000 Tall and thin

Short and wide

r =

3

A

Short and wide can

min

FIGURE 4.41

Solve for  r .

d2A 4000 = 4π + dr 2 r3

2000 , r > 0 A = 2pr2 + —— r

3

Multiply by  r 2 .

What happens at r = 3 500 π ? The second derivative

Tall and thin can

0

500 ≈ 5.42 π

Set  dA dr = 0.

500 p

r

The graph of A = 2πr 2 + 2000 r is concave up.

is positive throughout the domain of A. The graph is therefore everywhere concave up, and the value of A at r = 3 500 π is an absolute minimum. See Figure 4.41. The corresponding value of h (after a little algebra) is h =

1000 500 = 23 = 2r. 2 πr π

The one-liter can that uses the least material has height equal to twice the radius, here with r ≈ 5.42 cm and h ≈ 10.84 cm.

4.6  Applied Optimization

279

Examples from Mathematics and Physics EXAMPLE 3 A rectangle is to be inscribed in a semicircle of radius 2. What is the largest area the rectangle can have, and what are its dimensions? Solution Let ( x , 4 − x 2 ) be the coordinates of the upper right corner of the rectangle obtained by placing the circle and rectangle in the coordinate plane (Figure 4.42). The length, height, and area of the rectangle can then be expressed in terms of the position x of the lower right-hand corner:

y x 2 + y2 = 4 Qx, "4 − x 2 R

Length: 2 x , 2 −2 −x

0

x 2

x

 Height:  4  x 2 ,

Area: 2 x 4  x 2 .

Notice that the values of x are to be found in the interval 0 b x b 2, where the selected corner of the rectangle lies. Our goal is to find the absolute maximum value of the function A( x ) = 2 x 4 − x 2

FIGURE 4.42 The rectangle inscribed in the semicircle in Example 3.

on the domain [ 0, 2 ]. The derivative dA = dx

−2 x 2 + 2 4 − x2 4 − x2

is not defined when x  2 and is equal to zero when −2 x 2 + 2 4 − x2 = 0 4 − x2 −2 x 2 + 2 ( 4 − x 2 ) = 0 8 − 4x 2 = 0 x2 = 2 x = ± 2. Of the two zeros, x  2 and x = − 2, only x  2 lies in the interior of A’s domain and makes the critical-point list. The values of A at the endpoints and at this one critical point are Critical-point value: A( 2 ) = 2 2 4 − 2 = 4 Endpoint values:

A(0) = 0,

A(2) = 0.

The area has a maximum value of 4 when the rectangle is 4 − x 2 = 2 x  2 2 units long.

HISTORICAL BIOGRAPHY Willebrord Snell van Royen

(1580–1626) Snell was born in Leiden, Holland. Snell developed an important result involving the measure of light refraction as it travels into different media. While he never published the result, Descartes did so ten years after Snell’s death, and today it is known as Snell’s law. To know more, visit the companion Website.

2 units high and

EXAMPLE 4 The speed of light depends on the medium through which it travels, and is generally slower in denser media. Fermat’s principle in optics states that light travels from one point to another along a path for which the time of travel is a minimum. Describe the path that a ray of light will follow in going from a point A in a medium where the speed of light is c1 to a point B in a second medium where the speed of light is c 2 . Solution Since light traveling from A to B follows the quickest route, we look for a path that will minimize the travel time. We assume that A and B lie in the xy-plane and that the line separating the two media is the x-axis (Figure 4.43). We place A at coordinates ( 0, a ) and B at coordinates ( d , −b ) in the xy-plane. In a uniform medium, where the speed of light remains constant, “shortest time” means “shortest path,” and the ray of light will follow a straight line. Thus the path from A

280

Chapter 4 Applications of Derivatives

y A a

u1

to B will consist of a line segment from A to a boundary point P, followed by another line segment from P to B. Distance traveled equals rate times time, so

Angle of incidence u1 P

x

0

Medium 2

b

u2 d−x

distance . rate

Time =

Medium 1

d Angle of refraction

x

From Figure 4.43, the time required for light to travel from A to P is t1 =

B

AP = c1

a2 + x2 . c1

From P to B, the time is

FIGURE 4.43

A light ray refracted (deflected from its path) as it passes from one medium to a denser medium (Example 4).

t2 =

PB = c2

b 2 + ( d − x )2 . c2

The time from A to B is the sum of these: t = t1 + t 2 =

a2 + x2 + c1

b 2 + ( d − x )2 . c2

This equation expresses t as a differentiable function of x whose domain is [ 0, d ]. We want to find the absolute minimum value of t on this closed interval. We find the derivative dt x d−x = − dx c1 a 2 + x 2 c 2 b 2 + ( d − x )2 dtdx negative

0

dtdx zero

− − − − − +++++++++ x0

FIGURE 4.44

Example 4.

and observe that it is continuous. In terms of the angles θ1 and θ 2 in Figure 4.43,

dtdx positive

sin θ1 sin θ 2 dt = − . dx c1 c2

x d

The function t has a negative derivative at x = 0 and a positive derivative at x = d. Since

The sign pattern of dt dx in dt dx is continuous over the interval [ 0, d ], by the Intermediate Value Theorem for continu-

ous functions (Section 2.6), there is a point x 0 ∈ [ 0, d ] where dt dx = 0 (Figure 4.44). There is only one such point because dt dx is an increasing function of x (Exercise 70). At this unique point we then have sin θ1 sin θ 2 = . c1 c2 This equation is Snell’s Law or the Law of Refraction, and it is an important principle in the theory of optics. It describes the path the ray of light follows.

Examples from Economics Suppose that r ( x ) = the revenue from selling  x  items c( x ) = the cost of producing the  x  items p( x ) = r ( x ) − c( x ) = the profit from producing and selling x  items. Although x is usually an integer in many applications, we can learn about the behavior of these functions by defining them for all nonzero real numbers and by assuming they are differentiable functions. Economists use the terms marginal revenue, marginal cost, and marginal profit to name the derivatives r ′( x ), c′( x ), and p′( x ) of the revenue, cost, and profit functions. Let’s consider the relationship of the profit p to these derivatives. If r ( x ) and c( x ) are differentiable for x in some interval of production possibilities, and if p( x ) = r ( x ) − c( x ) has a maximum value there, it occurs at a critical point of p( x ) or at an endpoint of the interval. If it occurs at a critical point, then p′( x ) = r ′( x )  − c′( x ) = 0 and we see that r ′( x ) = c′( x ). In economic terms, this last equation means that

4.6  Applied Optimization

281

At a production level yielding maximum profit, marginal revenue equals marginal cost (Figure 4.45).

y

Dollars

Cost c(x)

Revenue r(x) Break-even point B

Maximum profit, c′(x) = r ′(x)

Local maximum for loss (minimum profit), c′(x) = r′(x) x Items produced

0

FIGURE 4.45 The graph of a typical cost function starts concave down and later turns concave up. It crosses the revenue curve at the break-even point B. To the left of B, the company operates at a loss. To the right, the company operates at a profit, with the maximum profit occurring where c′( x ) = r ′( x ). Farther to the right, cost exceeds revenue (perhaps because of a combination of rising labor and material costs, and market saturation) and production levels become unprofitable again.

EXAMPLE 5 Suppose that r ( x ) = 9 x and c( x ) = x 3 − 6 x 2 + 15 x are the revenue and the cost functions, given in millions of dollars, where x represents millions of MP3 players produced. Is there a production level that maximizes profit? If so, what is it? y

Solution Notice that r ′( x ) = 9 and c′( x ) = 3 x 2 − 12 x + 15.

3 x 2 − 12 x + 15 = 9

c(x) = x 3 − 6x2 + 15x

− 12 x + 6 = 0 The two solutions of the quadratic equation are 12 − 72 = 2− 6 12 + 72 x2 = = 2+ 6 x1 =

r(x) = 9x Maximum for profit

Local maximum for loss 0 2 − "2

2

2 + "2

NOT TO SCALE

FIGURE 4.46 The cost and revenue curves for Example 5.

Set  c′( x ) = r ′( x ).

3x 2

x

2 ≈ 0.586

and

2 ≈ 3.414.

The possible production levels for maximum profit are x ≈ 0.586 million MP3 players or x ≈ 3.414 million. The second derivative of p( x ) = r ( x ) − c( x ) is p′′( x ) = −c′′( x ) since r ′′( x ) is everywhere zero. Thus, p′′( x ) = 6( 2 − x ) , which is negative at x = 2 + 2 and positive at x = 2 − 2. By the Second Derivative Test, a maximum profit occurs at about x = 3.414 (where revenue exceeds costs) and maximum loss occurs at about x = 0.586. The graphs of r ( x ) and c( x ) are shown in Figure 4.46. EXAMPLE 6 A cabinetmaker uses cherry wood to produce 5 desks each day. Each delivery of one container of wood is $5000, whereas the storage of that material is $10 per day per unit stored, where a unit is the amount of material needed by her to produce 1 desk. How much material should be ordered each time, and how often should the material be delivered, to minimize her average daily cost in the production cycle between deliveries? Solution If she asks for a delivery every x days, then she must order 5x units to have enough material for that delivery cycle. The average amount in storage is approximately

282 y

Chapter 4 Applications of Derivatives

c(x) =

one-half of the delivery amount, or 5 x 2. Thus, the cost of delivery and storage for each cycle is approximately Cost per cycle = delivery costs + storage costs 5x Cost per cycle = 5000 10 + ⋅ ⋅ N Nx N 2 N delivery number of storage cost

5000 x + 25x y = 25x

( )

Cost

cost

days stored

per day

We compute the average daily cost c( x ) by dividing the cost per cycle by the number of days x in the cycle (see Figure 4.47). 5000 c( x ) = + 25 x , x > 0. x As x l 0 and as x → ∞, the average daily cost becomes large. So we expect a minimum to exist, but where? Our goal is to determine the number of days x between deliveries that provides the absolute minimum cost. We find the critical points by determining where the derivative is equal to zero:

5000 y= x x

min x value Cycle length

average amount stored

FIGURE 4.47 The average daily cost c( x ) is the sum of a hyperbola and a linear function (Example 6).

500 + 25 = 0 x2 x = ± 200 ≈ ±14.14.

c′( x ) = −

Of the two critical points, only 200 lies in the domain of c( x ). The critical point value of the average daily cost is 5000 c ( 200 ) = + 25 200 = 500 2 ≈ $707.11. 200 We note that c( x ) is defined over the open interval ( 0, ∞ ) with c′′( x ) = 10000 x 3 > 0. Thus, an absolute minimum exists at x = 200 ≈ 14.14 days. The cabinetmaker should schedule a delivery of 5 (14 ) = 70 units of wood every 14 days.

EXERCISES

4.6

Mathematical Applications

Whenever you are maximizing or minimizing a function of a single variable, we urge you to graph it over the domain that is appropriate to the problem you are solving. The graph will provide insight before you calculate and will furnish a visual context for understanding your answer. 1. Minimizing perimeter What is the smallest perimeter possible for a rectangle whose area is 16 cm 2, and what are its dimensions? 2. Show that among all rectangles with an 8-m perimeter, the one with largest area is a square. 3. The figure shows a rectangle inscribed in an isosceles right triangle whose hypotenuse is 2 units long. y

c. What is the largest area the rectangle can have, and what are its dimensions? 4. A rectangle has its base on the x-axis and its upper two vertices on the parabola y = 12 − x 2 . What is the largest area the rectangle can have, and what are its dimensions? 5. You are planning to make an open rectangular box from a 24-cm-by45-cm piece of cardboard by cutting congruent squares from the corners and folding up the sides. What are the dimensions of the box of largest volume you can make this way, and what is its volume?

7. The best fencing plan A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a single-strand electric fence. With 800 m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions?

P(x, ?)

A 0

b. Express the area of the rectangle in terms of x.

6. You are planning to close off a corner of the first quadrant with a line segment 20 units long running from ( a, 0 )  to  ( 0, b ). Show that the area of the triangle enclosed by the segment is largest when a  b.

B

−1

a. Express the y-coordinate of P in terms of x. (Hint: Write an equation for the line AB.)

x

1

x

8. The shortest fence A 216 m 2 rectangular pea patch is to be enclosed by a fence and divided into two equal parts by another fence parallel to one of the sides. What dimensions for the outer rectangle will require the smallest total length of fence? How much fence will be needed?

283

4.6  Applied Optimization

removed from the other corners so that the tabs can be folded to form a rectangular box with lid. x NOT TO SCALE

9. Designing a tank Your iron works has contracted to design and build a 4 m 3, square-based, open-top, rectangular steel holding tank for a paper company. The tank is to be made by welding thin stainless steel plates together along their edges. As the production engineer, your job is to find dimensions for the base and height that will make the tank weigh as little as possible. a. What dimensions do you tell the shop to use? b. Briefly describe how you took weight into account. 10. Catching rainwater A 20 m 3 open-top rectangular tank with a square base x m on a side and y m deep is to be built with its top flush with the ground to catch runoff water. The costs associated with the tank involve not only the material from which the tank is made but also an excavation charge proportional to the product xy. a. If the total cost is c = 5( x 2 + 4 xy ) + 10 xy,

x

x

x

30 cm

Base

Lid

x

x x

x 45 cm

a. Write a formula V ( x ) for the volume of the box. b. Find the domain of V for the problem situation, and graph V over this domain. c. Use a graphical method to find the maximum volume and the value of x that gives it. d. Confirm your result in part (c) analytically.

what values of x and y will minimize it? b. Give a possible scenario for the cost function in part (a). 11. Designing a poster You are designing a rectangular poster to contain 312.5 cm 2 of printing with a 10-cm margin at the top and bottom and a 5-cm margin at each side. What overall dimensions will minimize the amount of paper used? 12. Find the volume of the largest right circular cone that can be inscribed in a sphere of radius 3.

T 17. Designing a suitcase A 60-cm-by-90-cm sheet of cardboard is folded in half to form a 60-cm-by-45-cm rectangle as shown in the accompanying figure. Then four congruent squares of side length x are cut from the corners of the folded rectangle. The sheet is unfolded, and the six tabs are folded up to form a box with sides and a lid. a. Write a formula V ( x ) for the volume of the box. b. Find the domain of V for the problem situation and graph V over this domain. c. Use a graphical method to find the maximum volume and the value of x that gives it.

3

d. Confirm your result in part (c) analytically. e. Find a value of x that yields a volume of 17, 500 cm 3 .

3

y

f. Write a paragraph describing the issues that arise in part (b).

x

x

x

x

13. Two sides of a triangle have lengths a and b, and the angle between them is R. What value of R will maximize the triangle’s area? (Hint: A = (1 2 ) ab sin R.)

60 cm

60 cm x

14. Designing a can What are the dimensions of the lightest open-top right circular cylindrical can that will hold a volume of 1000 cm 3 ? Compare the result here with the result in Example 2. cm 3

right circular 15. Designing a can You are designing a 1000 cylindrical can whose manufacture will take waste into account. There is no waste in cutting the aluminum for the side, but the top and bottom of radius r will be cut from squares that measure 2r units on a side. The total amount of aluminum used up by the can will therefore be A =

8r 2

x

x x

90 cm

x

45 cm

The sheet is then unfolded.

60 cm

Base

+ 2Qrh

rather than the A = 2Qr 2 + 2Qrh in Example 2. In Example 2, the ratio of h to r for the most economical can was 2 to 1. What is the ratio for the most economical can now? T 16. Designing a box with a lid A piece of cardboard measures 30 cm by 45 cm. Two equal squares are removed from the corners of a 30-cm side as shown in the figure. Two equal rectangles are

90 cm

T 18. A rectangle is to be inscribed under the arch of the curve y = 4 cos ( 0.5 x ) from x = −Q to x  Q. What are the dimensions of the rectangle with largest area, and what is the largest area?

284

Chapter 4 Applications of Derivatives

19. Find the dimensions of a right circular cylinder of maximum volume that can be inscribed in a sphere of radius 10 cm. What is the maximum volume?

be used if the volume is fixed and the cost of construction is to be kept to a minimum. Neglect the thickness of the silo and waste in construction.

20. a. A certain Postal Service will accept a box for domestic shipment only if the sum of its length and girth (distance around) does not exceed 276 cm. What dimensions will give a box with a square end the largest possible volume?

24. The trough in the figure is to be made to the dimensions shown. Only the angle R can be varied. What value of R will maximize the trough’s volume?

Girth = distance around here

1′ u

u 1′ 1′

Length

20′

Square end

T b. Graph the volume of a 276-cm box (length plus girth equals 276 cm) as a function of its length, and compare what you see with your answer in part (a). 21. (Continuation of Exercise 20)

25. Paper folding A rectangular sheet of 21.6-cm-by-28-cm paper is placed on a flat surface. One of the corners is placed on the opposite longer edge, as shown in the figure, and held there as the paper is smoothed flat. The problem is to make the length of the crease as small as possible. Call the length L. Try it with paper. a. Show that L2 = 2 x 3 ( 2 x − 21.6 ) .

a. Suppose that instead of having a box with square ends, you have a box with square sides so that its dimensions are h by h by w and the girth is 2h 2 w. What dimensions will give the box its largest volume now?

b. What value of x minimizes L2 ? c. What is the minimum value of L? D

C

R

Girth

"L2 − x 2

L

h

Crease

Q (originally at A) x

w

x

h

T b. Graph the volume as a function of h and compare what you see with your answer in part (a). 22. A window is in the form of a rectangle surmounted by a semicircle. The rectangle is of clear glass, whereas the semicircle is of tinted glass that transmits only half as much light per unit area as clear glass does. The total perimeter is fixed. Find the proportions of the window that will admit the most light. Neglect the thickness of the frame.

A

P

B

26. Constructing cylinders Compare the answers to the following two construction problems. a. A rectangular sheet of perimeter 36 cm and dimensions x cm by y cm is to be rolled into a cylinder as shown in part (a) of the figure. What values of x and y give the largest volume? b. The same sheet is to be revolved about one of the sides of length y to sweep out the cylinder as shown in part (b) of the figure. What values of x and y give the largest volume?

y x Circumference = x

23. A silo (base not included) is to be constructed in the form of a cylinder surmounted by a hemisphere. The cost of construction per square unit of surface area is twice as great for the hemisphere as it is for the cylindrical sidewall. Determine the dimensions to

x y

y

(a)

(b)

4.6  Applied Optimization

27. Constructing cones A right triangle whose hypotenuse is 3 m long is revolved about one of its legs to generate a right circular cone. Find the radius, height, and volume of the cone of greatest volume that can be made this way.

285

37. What value of a makes f ( x ) = x 2 + ( a x ) have a. a local minimum at x = 2? b. a point of inflection at x = 1? 38. What values of a and b make f ( x ) = x 3 + ax 2 + bx have a. a local maximum at x = −1 and a local minimum at x = 3?

h

b. a local minimum at x = 4 and a point of inflection at x = 1?

"3 r

y x + = 1 that is closest to the origin. a b 29. Find a positive number for which the sum of it and its reciprocal is the smallest (least) possible.

28. Find the point on the line

39. A right circular cone is circumscribed by a sphere of radius 1. Determine the height h and radius r of the cone of maximum volume. 40. Determine the dimensions of the inscribed rectangle of maximum area.

30. Find a positive number for which the sum of its reciprocal and four times its square is the smallest possible. 31. A wire b m long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle. If the sum of the areas enclosed by each part is a minimum, what is the length of each part?

y

1

y = e−x

32. Answer Exercise 31 if one piece is bent into a square and the other into a circle. 33. Suppose a weight D is to be held 5 m below a horizontal line AB by a wire in the shape of a Y. If the points A and B are 4 m apart, what is the minimum total length of wire that can be used? 2m

x

41. Consider the accompanying graphs of y = 2 x + 3 and y = ln x. Determine the a. minimum vertical distance;

2m

A

b. minimum horizontal distance between these graphs.

B

y

y = 2x + 3

5m C 3

D

x

1

34. Suppose two different gauges of wire must be used to support the weight in Exercise 33: the vertical portion of the wire (the segment CD) costs $1 per meter, while the remaining wire (the segments AC and CB) must be sturdier and cost $2 per meter. What is the minimum total cost of the wire that can be used? 35. Determine the dimensions of the rectangle of largest area that can be inscribed in the right triangle shown in the accompanying figure.

42. Find the point on the graph of y = 20 x 3 + 60 x − 3 x 5 − 5 x 4 with the largest slope. 43. Among all triangles in the first quadrant formed by the x-axis, the y-axis, and tangent lines to the graph of y = 3 x − x 2 , what is the smallest possible area? y

w

5

y = ln x

3 − 2

4

h

(a, 3a − a2) 3

36. Determine the dimensions of the rectangle of largest area that can be inscribed in a semicircle of radius 3. (See the accompanying figure.)

w 0 h

3 y = 3x − x 2

r=3

x

286

Chapter 4 Applications of Derivatives

44. A cone is formed from a circular piece of material of radius 1 meter by removing a section of angle R and then joining the two straight edges. Determine the largest possible volume for the cone.

point of impact. If the projectile is fired with an initial velocity V 0 at an angle B with the horizontal, then in Chapter 12 we find that R 

υ0 2 sin 2α, g

where g is the downward acceleration due to gravity. Find the angle B for which the range R is the largest possible. 1 1

u

T 51. Strength of a beam The strength S of a rectangular wooden beam is proportional to its width times the square of its depth. (See the accompanying figure.)

1

a. Find the dimensions of the strongest beam that can be cut from a 30-cm diameter cylindrical log. Physical Applications

45. Vertical motion The height above ground of an object moving vertically is given by s = −4.9t 2 + 29.4t + 34.3, with s in meters and t in seconds. Find a. the object’s velocity when t  0;

b. Graph S as a function of the beam’s width w, assuming the proportionality constant to be k  1. Reconcile what you see with your answer in part (a). c. On the same screen, graph S as a function of the beam’s depth d, again taking k  1. Compare the graphs with one another and with your answer in part (a). What would be the effect of changing to some other value of k? Try it.

b. its maximum height and when it occurs; c. its velocity when s  0. 46. Quickest route Jane is 2 km offshore in a boat and wishes to reach a coastal village 6 km down a straight shoreline from the point nearest the boat. She can row 2 km/h and can walk 5 km/h. Where should she land her boat to reach the village in the least amount of time? 47. Shortest beam The 2-m wall shown here stands 5 m from the building. Find the length of the shortest straight beam that will reach to the side of the building from the ground outside the wall.

30 cm

d

w

T 52. Stiffness of a beam The stiffness S of a rectangular beam is proportional to its width times the cube of its depth. a. Find the dimensions of the stiffest beam that can be cut from a 30-cm-diameter cylindrical log.

Beam

Building

2 m wall 5m

48. Motion on a line The positions of two particles on the s-axis are s1 = sin t  and  s 2 = sin ( t + Q 3 ) , with s1 and s 2 in meters and t in seconds. a. At what time(s) in the interval 0 b t b 2Q do the particles meet?

b. Graph S as a function of the beam’s width w, assuming the proportionality constant to be k  1. Reconcile what you see with your answer in part (a). c. On the same screen, graph S as a function of the beam’s depth d, again taking k  1. Compare the graphs with one another and with your answer in part (a). What would be the effect of changing to some other value of k? Try it. 53. Frictionless cart A small frictionless cart, attached to the wall by a spring, is pulled 10 cm from its rest position and released at time t  0 to roll back and forth for 4 s. Its position at time t is s  10 cos Qt.

b. What is the farthest apart that the particles ever get?

a. What is the cart’s maximum speed? When is the cart moving that fast? Where is it then? What is the magnitude of the acceleration then?

c. When in the interval 0 b t b 2Q is the distance between the particles changing the fastest?

b. Where is the cart when the magnitude of the acceleration is greatest? What is the cart’s speed then?

49. The intensity of illumination at any point from a light source is proportional to the square of the reciprocal of the distance between the point and the light source. Two lights, one having an intensity eight times that of the other, are 6 m apart. How far from the stronger light is the total illumination least? 50. Projectile motion The range R of a projectile fired from the origin over horizontal ground is the distance from the origin to the

0

10

s

4.6  Applied Optimization

54. Two masses hanging side by side from springs have positions s1  2 sin t  and s 2  sin 2t , respectively. a. At what times in the interval 0  t do the masses pass each other? (Hint: sin 2t  2 sin t cos t.) b. When in the interval 0 b t b 2Q is the vertical distance between the masses the greatest? What is this distance? (Hint: cos 2t = 2 cos 2 t − 1.)

m1

s1 0 s2

287

57. Tin pest When metallic tin is kept below 13.2 °C, it slowly becomes brittle and crumbles to a gray powder. Tin objects eventually crumble to this gray powder spontaneously if kept in a cold climate for years. The Europeans who saw tin organ pipes in their churches crumble away years ago called the change tin pest because it seemed to be contagious, and indeed it was, for the gray powder is a catalyst for its own formation. A catalyst for a chemical reaction is a substance that controls the rate of reaction without undergoing any permanent change in itself. An autocatalytic reaction is one whose product is a catalyst for its own formation. Such a reaction may proceed slowly at first if the amount of catalyst present is small and slowly again at the end, when most of the original substance is used up. But in between, when both the substance and its catalyst product are abundant, the reaction proceeds at a faster pace. In some cases, it is reasonable to assume that the rate V  dx dt of the reaction is proportional both to the amount of the original substance present and to the amount of product. That is, V may be considered to be a function of x alone, and

m2

V = kx ( a − x ) = kax − kx 2 , where

s

x  the amount of product,

55. Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (nautical miles per hour; a nautical mile is 1852 m) and continued to do so all day. Ship B was sailing east at 8 knots and continued to do so all day. a. Start counting time with t  0 at noon and express the distance s between the ships as a function of t. b. How rapidly was the distance between the ships changing at noon? One hour later? c. The visibility that day was 5 nautical miles. Did the ships ever sight each other? T d. Graph s and ds dt together as functions of t for −1 ≤ t ≤ 3, using different colors if possible. Compare the graphs and reconcile what you see with your answers in parts (b) and (c). e. The graph of ds dt looks as if it might have a horizontal asymptote in the first quadrant. This in turn suggests that ds dt approaches a limiting value as t → ∞. What is this value? What is its relation to the ships’ individual speeds? 56. Fermat’s principle in optics Light from a source A is reflected by a plane mirror to a receiver at point B, as shown in the accompanying figure. Show that for the light to obey Fermat’s principle, the angle of incidence must equal the angle of reflection, both measured from the line normal to the reflecting surface. (This result can also be derived without calculus. There is a purely geometric argument, which you may prefer.) Normal

Angle of incidence u1

Angle of reflection u2

Plane mirror

k  a positive constant. At what value of x does the rate V have a maximum? What is the maximum value of V? 58. Airplane landing path An airplane is flying at altitude H when it begins its descent to an airport runway that is at horizontal ground distance L from the airplane, as shown in the accompanying figure. Assume that the landing path of the airplane is the graph of a cubic polynomial function y = ax 3 + bx 2 + cx + d, where y(−L ) = H and y(0)  0. a. What is dy dx  at x  0? b. What is dy dx  at x = −L ? c. Use the values for dy dx  at x  0 and x = −L together with y(0)  0 and y(−L ) = H to show that ⎡ x y( x ) = H ⎢ 2 ⎢⎣ L

( )

3

+3

( Lx )

2

⎤ ⎥. ⎥⎦ y

Landing path

H = Cruising altitude Airport x L

Business and Economics Light receiver

Light source A

a  the amount of substance at the beginning, and

B

59. It costs you c dollars each to manufacture and distribute backpacks. If the backpacks sell at x dollars each, the number sold is given by a + b (100 − x ) , n = x−c where a and b are positive constants. What selling price will bring a maximum profit?

288

Chapter 4 Applications of Derivatives

60. You operate a tour service that offers the following rates: $200 per person if 50 people (the minimum number to book the tour) go on the tour. For each additional person, up to a maximum of 80 people total, the rate per person is reduced by $2. It costs $6000 (a fixed cost) plus $32 per person to conduct the tour. How many people does it take to maximize your profit? 61. Wilson lot size formula One of the formulas for inventory management says that the average weekly cost of ordering, paying for, and holding merchandise is A(q) =

hq km , + cm + q 2

where q is the quantity you order when things run low (shoes, radios, brooms, or whatever the item might be), k is the cost of placing an order (the same, no matter how often you order), c is the cost of one item (a constant), m is the number of items sold each week (a constant), and h is the weekly holding cost per item (a constant that takes into account things such as space, utilities, insurance, and security). a. Your job, as the inventory manager for your store, is to find the quantity that will minimize A(q). What is it? (The formula you get for the answer is called the Wilson lot size formula.) b. Shipping costs sometimes depend on order size. When they do, it is more realistic to replace k by k + bq, the sum of k and a constant multiple of q. What is the most economical quantity to order now? 62. Production level Prove that the production level (if any) at which average cost is smallest is a level at which the average cost equals marginal cost. 63. Show that if r ( x ) = 6 x and c( x ) = − + 15 x are your revenue and cost functions, then the best you can do is break even (have revenue equal cost). x3

6x 2

64. Production level Suppose that c( x ) = x 3 − 20 x 2 + 20,000 x is the cost of manufacturing x items. Find a production level that will minimize the average cost of making x items. 65. You are to construct an open rectangular box with a square base and a volume of 6 m 3. If material for the bottom costs $60 m 2 and material for the sides costs $40 m 2, what dimensions will result in the least expensive box? What is the minimum cost? 66. The 800-room Mega Motel chain is filled to capacity when the room charge is $50 per night. For each $10 increase in room charge, 40 fewer rooms are filled each night. What charge per room will result in the maximum revenue per night?

68. How we cough a. When we cough, the trachea (windpipe) contracts to increase the velocity of the air going out. This raises the questions of how much it should contract to maximize the velocity and whether it really contracts that much when we cough. Under reasonable assumptions about the elasticity of the tracheal wall and about how the air near the wall is slowed by friction, the average flow velocity υ can be modeled by the equation r0 υ = c ( r0 − r ) r 2 cm s, ≤ r ≤ r0 , 2 where r0 is the rest radius of the trachea in centimeters and c is a positive constant whose value depends in part on the length of the trachea. Show that υ is greatest when r = ( 2 3 ) r0 , that is, when the trachea is about 33% contracted. The remarkable fact is that X-ray photographs confirm that the trachea contracts about this much during a cough. T b. Take r0 to be 0.5 and c to be 1, and graph υ over the interval 0 ≤ r ≤ 0.5. Compare what you see with the claim that υ is at a maximum when r = ( 2 3 ) r0 . Theory and Examples

69. An inequality for positive integers Show that if a, b, c, and d are positive integers, then ( a 2 + 1)( b 2 + 1)( c 2 + 1)( d 2 + 1)

abcd 70. The derivative dt dx in Example 4 a. Show that f (x) =

x a2 + x2

is an increasing function of x. b. Show that g( x ) =

d−x

b 2 + ( d − x )2 is a decreasing function of x. c. Show that dt x d−x = − 2 dx c1 a 2 + x 2 c 2 b + ( d − x )2 is an increasing function of x. 71. Let f ( x ) and g( x ) be the differentiable functions graphed here. Point c is the point where the vertical distance between the curves is the greatest. Is there anything special about the tangent lines to the two curves at c? Give reasons for your answer. y

Biology

67. Sensitivity to medicine (Continuation of Exercise 74, Section 3.3) Find the amount of medicine to which the body is most sensitive by finding the value of M that maximizes the derivative dR dM, where R = M2 and C is a constant.

≥ 16.

y = f (x) y = g(x)

( C2 − M3 ) a

c

b

x

4.7  Newton’s Method

72. You have been asked to determine whether the function f ( x ) = 3 + 4 cos x + cos 2 x is ever negative.

T b. Graph the distance function D( x ) and y = x together and reconcile what you see with your answer in part (a).

a. Explain why you need to consider values of x only in the interval [ 0, 2π ].

y

b. Is f ever negative? Explain.

(x, " x)

73. a. The function y = cot x − 2 csc x has an absolute maximum value on the interval 0 < x < π. Find it. T b. Graph the function and compare what you see with your answer in part (a). 74. a. The function y = tan x + 3 cot x has an absolute minimum value on the interval 0 < x < π 2. Find it. T b. Graph the function and compare what you see with your answer in part (a). 75. a. How close does the curve y = x come to the point ( 3 2, 0 )? (Hint: If you minimize the square of the distance, you can avoid square roots.)

289

0

y = "x

x 3 a , 0b 2

76. a. How close does the semicircle y = point (1, 3 )?

16 − x 2 come to the

T b. Graph the distance function and y = 16 − x 2 together and reconcile what you see with your answer in part (a).

4.7 Newton’s Method For thousands of years, one of the main goals of mathematics has been to find solutions to equations. For linear equations ( ax + b = 0 ), and for quadratic equations ( ax 2 + bx + c = 0 ) , we can explicitly solve for a solution. However, for most equations there is no simple formula that gives the solutions. In this section we study a numerical method called Newton’s method or the Newton– Raphson method, which is a technique to approximate the solutions to an equation f ( x ) = 0. Newton’s method estimates the solutions using tangent lines of the graph of y = f ( x ) near the points where f is zero. A value of x where f is zero is called a root of the function f and a solution of the equation f ( x ) = 0. Newton’s method is both powerful and efficient, and it has numerous applications in engineering and other fields where solutions to complicated equations are needed. y y = f (x)

Procedure for Newton’s Method

(x0, f (x0))

(x1, f (x1)) (x2, f(x2 )) Root sought 0

x3 Fourth

x2 Third

x1 Second

x0 First

APPROXIMATIONS

FIGURE 4.48 Newton’s method starts with an initial guess x 0 and (under favorable circumstances) improves the guess one step at a time.

x

The goal of Newton’s method for estimating a solution of an equation f ( x ) = 0 is to produce a sequence of approximations that approach the solution. We pick the first number x 0 of the sequence. Then, under favorable circumstances, the method moves step by step toward a point where the graph of f crosses the x-axis (Figure 4.48). At each step the method approximates a zero of f with a zero of one of its linearizations. Here is how it works. The initial estimate, x 0 , may be found by graphing or just plain guessing. The method then uses the tangent to the curve y = f ( x ) at ( x 0 , f ( x 0 ) ) to approximate the curve, calling the point x 1 where the tangent meets the x-axis (Figure 4.48). The number x 1 is usually a better approximation to the solution than is x 0 . The point x 2 where the tangent to the curve at ( x 1 , f ( x 1 ) ) crosses the x-axis is the next approximation in the sequence. We continue, using each approximation to generate the next, until we are close enough to the root to stop. We can derive a formula for generating the successive approximations in the following way. Given the approximation x n , the point-slope equation for the tangent line to the curve at ( x n , f ( x n ) ) is y = f ( x n ) + f ′( x n )( x − x n ).

290

Chapter 4 Applications of Derivatives

We can find where it crosses the x-axis by setting y = 0 (Figure 4.49):

y y = f (x)

0 = f ( x n ) + f ′( x n )( x − x n )

Point: (xn, f (xn )) Slope: f ′(xn ) Tangent line equation: y − f (xn ) = f ′(xn )(x − xn ) (xn, f (xn))

− Tangent line (graph of linearization of f at xn )

f (x n ) = x − xn f ′( x n ) x = xn −

f (x n ) f ′( x n )

If  f ′( x n ) ≠ 0

This value of x is the next approximation x n +1. Here is a summary of Newton’s method.

Root sought 0

xn xn+1 = xn −

x

f (xn ) f'(xn )

FIGURE 4.49 The geometry of the successive steps of Newton’s method. From x n we go up to the curve and follow the tangent line down to find x n +1 .

Newton’s Method

1. Guess a first approximation to a solution of the equation f ( x ) = 0. A graph of y = f ( x ) may help. 2. Use the first approximation to get a second, the second to get a third, and so on, using the formula x n +1 = x n −

f (x n ) , f ′( x n )

if  f ′( x n ) ≠ 0.

(1)

Applying Newton’s Method Applications of Newton’s method generally involve many numerical computations, making them well suited for computers or calculators. Nevertheless, even when the calculations are done by hand (which may be very tedious), they give a powerful way to find solutions of equations. In our first example, we find decimal approximations to 2 by estimating the positive root of the equation f ( x ) = x 2 − 2 = 0. EXAMPLE 1

Approximate the positive root of the equation f ( x ) = x 2 − 2 = 0.

Solution With f ( x ) = x 2 − 2 and f ′( x ) = 2 x , Equation (1) becomes

xn2 − 2 2x n x 1 = xn − n + 2 xn x 1 . = n + 2 xn

x n +1 = x n −

The equation x n +1 =

xn 1 + 2 xn

enables us to go from each approximation to the next with just a few keystrokes. With the starting value x 0 = 1, we get the results in the first column of the following table. (To five decimal places, or, equivalently, to six digits, 2 = 1.41421.)

4.7  Newton’s Method

x0  1 x 1  1.5 x 2  1.41667 x 3  1.41422

Error

Number of correct digits

0.41421 0.08579

1 1

0.00246 0.00001

3 5

291

Newton’s method is used by many software applications to calculate roots because it converges so fast (more about this later). If the arithmetic in the table in Example 1 had been carried to 13 decimal places instead of 5, then going one step further would have given 2 correctly to more than 10 decimal places. EXAMPLE 2 Find the x-coordinate of the point where the curve y = x 3 − x crosses the horizontal line y  1.

y 20

y = x3 − x − 1

Solution The curve crosses the line when x 3 − x = 1 or x 3 − x − 1 = 0. When does f ( x ) = x 3 − x − 1 equal zero? Since f is continuous on [ 1, 2 ] and f (1) = −1 while f (2)  5, we know by the Intermediate Value Theorem there is a root in the interval (1, 2 ) (Figure 4.50). We apply Newton’s method to f with the starting value x 0  1. The results are displayed in Table 4.1 and Figure 4.51. At n  5, we come to the result x 6  x 5  1.3247 17957. When x n +1 = x n , Equation (1) shows that f ( x n )  0, up to the accuracy of our computation. We have found a solution of f ( x )  0 to nine decimals places.

15 10 5

−1

1

0

2

3

x

FIGURE 4.50

The graph of f ( x ) = x 3 − x − 1 crosses the x-axis once; this is the root we want to find (Example 2).

TABLE 4.1 The Result of Applying Newton’s Method to f ( x ) = x 3 − x − 1

with x 0  1

y = x3 − x − 1 (1.5, 0.875) Root sought x0 1

x2

x1

f a( x n )

x n +1 = x n −

2

1.5

0.875

5.75

1.3478 26087

1.3478 26087

0.1006 82173

4.4499 05482

1.3252 00399

3

1.3252 00399

0.0020 58362

4.2684 68292

1.3247 18174

4

1.3247 18174

0.0000 00924

4.2646 34722

1.3247 17957

5

1.3247 17957

4.2646 32999

1.3247 17957

n

xn

0

1

1

1.5

2

f (xn ) 1

1.8672E-13

f (xn ) f ′( x n )

x 1.5 1.3478

(1, −1)

FIGURE 4.51

The first three x-values in Table 4.1 (four decimal places).

In Figure 4.52 we have indicated that the process in Example 2 might have started at the point B 0 ( 3, 23 ) on the curve, with x 0  3. Point B 0 is quite far from the x-axis, but the tangent at B 0 crosses the x-axis at about (2.12, 0), so x 1 is still an improvement over x 0 . If we use Equation (1) repeatedly as before, with f ( x ) = x 3 − x − 1 and f ′( x ) = 3 x 2 − 1, we obtain the nine-place solution x 7  x 6  1.3247 17957 in seven steps.

Convergence of the Approximations In Chapter 9 we define precisely the idea of convergence for the approximations x n in Newton’s method. Intuitively, we mean that as the number n of approximations increases

292

Chapter 4 Applications of Derivatives

y

without bound, the values x n get arbitrarily close to the desired root r. (This notion is similar to the idea of the limit of a function g(t ) as t approaches infinity, as defined in Section 2.5.) In practice, Newton’s method usually gives convergence with impressive speed, but this is not guaranteed. One way to test convergence is to begin by graphing the function to estimate a good starting value for x 0 . You can test that you are getting closer to a zero of the function by checking that f ( x n ) is approaching zero, and you can check that the approximations are converging by evaluating x n − x n +1 . Newton’s method does not always converge. For instance, if

25 B0(3, 23) 20 y = x3 − x − 1 15

10

⎧⎪− r − x , x < r f ( x ) = ⎪⎨ ⎪⎪⎩ x − r , x ≥ r,

B1(2.12, 6.35) 5 −1" 3 −1

Root sought 1" 3

x2 x1 1

0

1.6 2.12

x0 3

x

FIGURE 4.52 Any starting value x 0 to the right of x = 1 3 will lead to the root in Example 2. y

the graph will be like the one in Figure 4.53. If we begin with x 0 = r − h, we get x 1 = r + h, and successive approximations go back and forth between these two values. No amount of iteration brings us closer to the root than our first guess. If Newton’s method does converge, it converges to a root. Be careful, however. There are situations in which the method appears to converge but no root is there. Fortunately, such situations are rare. When Newton’s method converges to a root, it may not be the root you have in mind. Figure 4.54 shows two ways this can happen.

y = f (x) y = f (x) 0

x0

r

x1

x

Starting point Root sought

FIGURE 4.53 Newton’s method fails to converge. You go from x 0 to x 1 and back to x 0 , never getting any closer to r.

EXERCISES

FIGURE 4.54

x0

Root found

y = f (x) x1

x

x1

x2

Root sought

Root found

x

x0

Starting point

If you start too far away, Newton’s method may miss the root you want.

4.7

Root Finding

1. Use Newton’s method to estimate the solutions of the equation x 2 + x − 1 = 0. Start with x 0 = −1 for the left-hand solution and with x 0 = 1 for the solution on the right. Then, in each case, find x 2 . 2. Use Newton’s method to estimate the one real solution of x 3 + 3 x + 1 = 0. Start with x 0 = 0 and then find x 2 . 3. Use Newton’s method to estimate the two zeros of the function f ( x ) = x 4 + x − 3. Start with x 0 = −1 for the left-hand zero and with x 0 = 1 for the zero on the right. Then, in each case, find x 2 . 4. Use Newton’s method to estimate the two zeros of the function f ( x ) = 2 x − x 2 + 1. Start with x 0 = 0 for the left-hand zero and with x 0 = 2 for the zero on the right. Then, in each case, find x 2 .

5. Use Newton’s method to find the positive fourth root of 2 by solving the equation x 4 − 2 = 0. Start with x 0 = 1 and find x 2 . 6. Use Newton’s method to find the negative fourth root of 2 by solving the equation x 4 − 2 = 0. Start with x 0 = −1 and find x 2 . T 7. Use Newton’s method to find an approximate solution of 3 − x = ln x. Start with x 0 = 2 and find x 2 . T 8. Use Newton’s method to find an approximate solution of x − 1 = arctan x. Start with x 0 = 1 and find x 2 . T 9. Use Newton’s method to find an approximate solution of xe x = 1. Start with x 0 = 0 and find x 2 . Dependence on Initial Point

10. Using the function shown in the figure, and, for each initial estimate x 0 , determine graphically what happens to the sequence of Newton’s method approximations a. x 0 = 0

b. x 0 = 1

293

4.7  Newton’s Method

c. x 0 = 2

d. x 0 = 4

19. a. How many solutions does the equation sin 3 x = 0.99 − x 2 have?

e. x 0 = 5.5

b. Use Newton’s method to find them.

y

20. Intersection of curves 3

a. Does cos 3x ever equal x? Give reasons for your answer.

y = f (x)

2

b. Use Newton’s method to find where. 21. Find the four real zeros of the function f ( x ) = 2 x 4 − 4 x 2 + 1.

1 −4 −3 −2 −1 0

1

2

3

4

5

6

7

8

x

−1 −2

11. Guessing a root Suppose that your first guess is lucky, in the sense that x 0 is a root of f ( x ) = 0. Assuming that f ′( x 0 ) is defined and is not 0, what happens to x 1 and later approximations?

22. Estimating pi Estimate π to as many decimal places as your calculator will display by using Newton’s method to solve the equation tan x = 0 with x 0 = 3. 23. Intersection of curves At what value(s) of x does cos x = 2 x ? 24. Intersection of curves At what value(s) of x does cos x = −x ? 25. The graphs of y = x 2 ( x + 1) and y = 1 x ( x > 0 ) intersect at one point x = r. Use Newton’s method to estimate the value of r to four decimal places. y

12. Estimating pi You plan to estimate π 2 to five decimal places by using Newton’s method to solve the equation cos x = 0. Does it matter what your starting value is? Give reasons for your answer.

3 2

Theory and Examples

13. Oscillation

Show that if h > 0, applying Newton’s method to ⎧⎪ x , x ≥ 0 f ( x ) = ⎪⎨ ⎪⎪⎩ −x , x < 0

leads to x 1 = −h if x 0 = h and to x 1 = h if x 0 = −h. Draw a picture that shows what is going on.

y = x 2(x + 1)

ar, 1rb y = 1x

1 −1

0

1

2

x

26. The graphs of y = x and y = 3 − x 2 intersect at one point x = r. Use Newton’s method to estimate the value of r to four decimal places.

14. Approximations that get worse and worse Apply Newton’s method to f ( x ) = x 1 3 with x 0 = 1 and calculate x 1 , x 2 , x 3 , and  x 4 . Find a formula for x n . What happens to x n as n → ∞ ? Draw a picture that shows what is going on.

27. Intersection of curves At e −x 2 = x 2 − x + 1?

what

value(s)

of

x

does

28. Intersection of curves At ln (1 − x 2 ) = x − 1?

what

value(s)

of

x

does

15. Explain why the following four statements ask for the same information:

29. Use the Intermediate Value Theorem from Section 2.6 to show that f ( x ) = x 3 + 2 x − 4 has a root between x = 1 and x = 2. Then find the root to five decimal places.

i) Find the roots of f ( x ) = x 3 − 3 x − 1. ii) Find the x-coordinates of the intersections of the curve y = x 3 with the line y = 3 x + 1. iii) Find the x-coordinates of the points where the curve y = x 3 − 3 x crosses the horizontal line y = 1.

30. Factoring a quartic Find the approximate values of r1 through r4 in the factorization 8 x 4 − 14 x 3 − 9 x 2 + 11x − 1 = 8( x − r1 )( x − r2 )( x − r3 )( x − r4 ). y

iv) Find the values of x where the derivative of g( x ) = (1 4 ) x 4 − ( 3 2 ) x 2 − x + 5 equals zero. T When solving Exercises 16–34, you may need to use appropriate technology (such as a calculator or a computer). 16. Locating a planet To calculate a planet’s space coordinates, we have to solve equations like x = 1 + 0.5 sin x. Graphing the function f ( x ) = x − 1 − 0.5 sin x suggests that the function has a root near x = 1.5. Use one application of Newton’s method to improve this estimate. That is, start with x 0 = 1.5 and find x 1 . (The value of the root is 1.49870 to five decimal places.) Remember to use radians. 17. Intersecting curves The curve y = tan x crosses the line y = 2 x between x = 0 and x = π 2. Use Newton’s method to find where. 18. Real solutions of a quartic Use Newton’s method to find the two real solutions of the equation x 4 − 2 x 3 − x 2 − 2 x + 2 = 0.

y = 8x 4 − 14x 3 − 9x 2 + 11x − 1

2 −1

−2 −4 −6 −8 −10 −12

1

2

x

31. Converging to different zeros Use Newton’s method to find the zeros of f ( x ) = 4 x 4 − 4 x 2 using the given starting values. a. x 0 = −2 and x 0 = −0.8, lying in (−∞, − 2 2 ) b. x 0 = −0.5 and x 0 = 0.25, lying in (− 21 7, 21 7 ) c. x 0 = 0.8 and x 0 = 2, lying in ( 2 2, ∞ ) d. x 0 = − 21 7 and x 0 =

21 7

294

Chapter 4 Applications of Derivatives

with a starting value of x 0 = 2 to see how close your machine comes to the root x = 1. See the accompanying graph.

32. The sonobuoy problem In submarine location problems, it is often necessary to find a submarine’s closest point of approach (CPA) to a sonobuoy (sound detector) in the water. Suppose that the submarine travels on the parabolic path y = x 2 and that the buoy is located at the point ( 2, −1 2 ).

y

a. Show that the value of x that minimizes the distance between the submarine and the buoy is a solution of the equation x = 1 ( x 2 + 1). b. Solve the equation x = 1 ( x 2 + 1) with Newton’s method. y = (x − 1) 40 Slope = 40

Slope = −40

y

1 y=

Submarine track in two dimensions

1

Nearly flat 0

CPA 0

(2, 1)

x2

1

2

x

34. The accompanying figure shows a circle of radius r with a chord of length 2 and an arc s of length 3. Use Newton’s method to solve for r and θ (radians) to four decimal places. Assume 0 < θ < π.

x

2

1

1 Sonobuoy a2, − 2b

s=3

r

33. Curves that are nearly flat at the root Some curves are so flat that, in practice, Newton’s method stops too far from the root to give a useful estimate. Try Newton’s method on f ( x ) = ( x − 1) 40

u

2

r

4.8 Antiderivatives Many problems require that we recover a function from its derivative, or from its rate of change. For instance, the laws of physics tell us the acceleration of an object falling from an initial height, and we can use this to compute its velocity and its height at any time. More generally, starting with a function f , we want to find a function F whose derivative is f . If such a function F exists, it is called an antiderivative of f . Antiderivatives are the link connecting the two major elements of calculus: derivatives and definite integrals. Antiderivatives have an important connection to the theory of integrals that is developed in Chapter 5. For this reason the process of taking an antiderivative is also called “integration.”

Finding Antiderivatives DEFINITION A function F is an antiderivative of f on an interval I if F ′( x ) = f ( x ) for all x in I.

The process of recovering a function F ( x ) from its derivative f ( x ) is called antidifferentiation. We use capital letters such as F to represent an antiderivative of a function f , G to represent an antiderivative of g, and so forth. EXAMPLE 1 (a) f ( x ) = 2 x

Find an antiderivative for each of the following functions. 1 (b) g( x ) = cos x (c) h( x ) = + 2e 2 x x

4.8  Antiderivatives

295

Solution We need to think backward here: What function do we know has a derivative equal to the given function?

(a) F ( x ) = x 2

(b) G ( x ) = sin x

(c) H ( x ) = ln x + e 2 x

Each answer can be checked by differentiating. The derivative of F ( x ) = x 2 is 2x. The derivative of G ( x ) = sin x is cos x , and the derivative of H ( x ) = ln x + e 2 x is (1 x ) + 2e 2 x . The function F ( x ) = x 2 is not the only function whose derivative is 2x. The function x 2 + 1 has the same derivative. So does x 2 + C for any constant C. Are there others? Corollary 2 of the Mean Value Theorem in Section 4.2 gives the answer: Any two antiderivatives of a function differ by a constant. So the functions x 2 + C, where C is an arbitrary constant, form all the antiderivatives of f ( x ) = 2 x. More generally, we have the following result.

If F is an antiderivative of f on an interval I, then the most general antiderivative of f on I is

THEOREM 8

F (x) + C where C is an arbitrary constant.

y

y = x3 + C 2 1

C=2 C=1 C=0 C = −1 C = −2

EXAMPLE 2

Find an antiderivative of f ( x ) = 3 x 2 that satisfies F(1) = −1.

x

0 −1

Thus the most general antiderivative of f on I is a family of functions F ( x ) + C whose graphs are vertical translations of one another. We can select a particular antiderivative from this family by assigning a specific value to C. Here is an example showing how such an assignment might be made.

(1, −1)

−2

Solution Since the derivative of x 3 is 3 x 2, the general antiderivative

F (x) = x 3 + C gives all the antiderivatives of f ( x ). The condition F(1) = −1 determines a specific value for C. Substituting x = 1 into f ( x ) = x 3 + C gives F (1) = (1)3 + C = 1 + C. Since F(1) = −1, solving 1 + C = −1 for C gives C = −2. So

The curves y = x 3 + C fill the coordinate plane without overlapping. In Example 2, we identify the curve y = x 3 − 2 as the one that passes through the given point (1, −1).

FIGURE 4.55

F (x) = x 3 − 2 is the antiderivative satisfying F(1) = −1. Notice that this assignment for C selects the particular curve from the family of curves y = x 3 + C that passes through the point (1, −1) in the plane (Figure 4.55). By working backward from assorted differentiation rules, we can derive formulas and rules for antiderivatives. In each case there is an arbitrary constant C in the general expression representing all antiderivatives of a given function. Table 4.2 gives antiderivative formulas for a number of important functions. The rules in Table 4.2 are easily verified by differentiating the general antiderivative formula to obtain the function to its left. For example, the derivative of ( tan kx ) k + C is sec 2 kx, whatever the value of the constants C or k ≠ 0, and this shows that Formula 4 gives the general antiderivative of sec 2 kx.

296

Chapter 4 Applications of Derivatives

TABLE 4.2 Antiderivative formulas, k a nonzero constant

Function

General antiderivative

1. x n

1 x n +1 + C , n+1 1 − cos kx + C k 1 sin kx + C k 1 tan kx + C k 1 − cot kx + C k 1 sec kx + C k 1 − csc kx + C k

2. sin kx 3. cos kx 4. sec 2 kx 5. csc 2 kx 6. sec kx tan kx 7. csc kx cot kx

Function

n ≠ −1

General antiderivative 1 kx e +C k

8. e kx 9.

1 x

ln x + C ,

1 1 − k 2x 2 1 11. 1 + k 2x 2 1 12. 2 x k x2 − 1

x ≠ 0

1 arcsin kx + C k 1 arctan kx + C k

10.

arcsec kx + C , ⎛ 1 ⎞⎟ kx ⎜⎜ a + C, ⎜⎝ k ln a ⎟⎟⎠

13. a kx

kx > 1 a > 0, a ≠ 1

EXAMPLE 3

Find the general antiderivative of each of the following functions. 1 (a) f ( x ) = (b) g( x ) = (c) h( x ) = sin 2 x x x (d) i( x ) = cos (e) j( x ) = e −3 x (f) k ( x ) = 2 x 2 Solution In each case, we can use one of the formulas listed in Table 4.2. x5

(a) F ( x ) =

x6 +C 6

Formula 1 with  n = 5

(b) g( x ) = x −1 2 ,  so G(x) =

x1 2 +C = 2 x +C 12

Formula 1 with  n = −1 2

− cos 2 x +C 2 sin ( x 2 ) x (d) I ( x ) = + C = 2 sin + C 12 2 1 (e) J ( x ) = − e −3 x + C 3 ⎛ 1 ⎞⎟ x (f) K ( x ) = ⎜⎜ 2 +C ⎜⎝ ln 2 ⎟⎟⎠ (c) H ( x ) =

Formula 2 with  k = 2 Formula 3 with  k = 1 2 Formula 8 with  k = −3 Formula 13 with  a = 2, k = 1

Other derivative rules also lead to corresponding antiderivative rules. We can add and subtract antiderivatives and multiply them by constants. TABLE 4.3 Antiderivative linearity rules

1. Constant Multiple Rule: 2. Sum or Difference Rule:

Function

General antiderivative

k f (x) f ( x ) ± g( x )

k F ( x ) + C , k  a constant F (x) ± G(x) + C

The formulas in Table 4.3 are easily proved by differentiating the antiderivatives and verifying that the result agrees with the original function.

4.8  Antiderivatives

EXAMPLE 4

297

Find the general antiderivative of f (x) =

3 + sin 2 x. x

Solution We have that f ( x ) = 3g( x ) + h( x ) for the functions g and h in Example 3. Since G ( x ) = 2 x is an antiderivative of g( x ) from Example 3b, it follows from the Constant Multiple Rule for antiderivatives that 3G ( x ) = 3 ⋅ 2 x = 6 x is an antiderivative of 3g( x ) = 3 x . Likewise, from Example 3c we know that H ( x ) = ( −1 2 ) cos 2 x is an antiderivative of h( x ) = sin 2 x. From the Sum Rule for antiderivatives, we then get that

F ( x ) = 3G ( x ) + H ( x ) + C 1 = 6 x − cos 2 x + C 2 is the general antiderivative formula for f ( x ), where C is an arbitrary constant.

Initial Value Problems and Differential Equations Antiderivatives play several important roles in mathematics and its applications. Methods and techniques for finding them are a major part of calculus, and we take up that study in Chapter 8. Finding an antiderivative for a function f ( x ) is the same problem as finding a function y( x ) that satisfies the equation dy = f ( x ). dx This is called a differential equation, since it is an equation involving an unknown function y that is being differentiated. To solve it, we need a function y( x ) that satisfies the equation. This function is found by taking the antiderivative of f ( x ). We can fix the arbitrary constant arising in the antidifferentiation process by specifying an initial condition y( x 0 ) = y 0 . This condition means the function y( x ) has the value y 0 when x = x 0 . The combination of a differential equation and an initial condition is called an initial value problem. Such problems play important roles in all branches of science. The most general antiderivative F ( x ) + C of the function f ( x ) (such as x 3 + C for the function 3 x 2 in Example 2) gives the general solution y = F ( x ) + C of the differential equation dy dx = f ( x ). The general solution gives all the solutions of the equation (there are infinitely many, one for each value of C). We solve the differential equation by finding its general solution. We then solve the initial value problem by finding the particular solution that satisfies the initial condition y( x 0 ) = y 0 . In Example 2, the function y = x 3 − 2 is the particular solution of the differential equation dy dx = 3 x 2 satisfying the initial condition y(1) = −1.

Antiderivatives and Motion We have seen that the derivative of the position function of an object gives its velocity, and the derivative of its velocity function gives its acceleration. If we know an object’s acceleration, then by finding an antiderivative we can recover the velocity, and from an antiderivative of the velocity we can recover its position function. This procedure was used as an application of Corollary 2 in Section 4.2. Now that we have a terminology and conceptual framework in terms of antiderivatives, we revisit the problem from the point of view of differential equations. EXAMPLE 5 A hot-air balloon ascending at the rate of 3.6 m s is at a height 24.5 m above the ground when a package is dropped. How long does it take the package to reach the ground?

298

Chapter 4 Applications of Derivatives

Solution Let υ(t ) denote the velocity of the package at time t, and let s(t ) denote its height above the ground. The acceleration of gravity near the surface of the earth is 9.8 m s 2. Assuming no other forces act on the dropped package, we have

dυ = −9.8. dt

Negative because gravity acts in the direction of decreasing s

This leads to the following initial value problem (Figure 4.56): Differential   equation:   Initial   condition:

dυ = −9.8 dt υ(0) = 3.6.

Balloon initially rising

This is our mathematical model for the package’s motion. We solve the initial value problem to obtain the velocity of the package.

s

υ (0) = 3.6

1. Solve the differential equation: The general formula for an antiderivative of −9.8 is υ = −9.8t + C . Having found the general solution of the differential equation, we use the initial condition to find the particular solution that solves our problem. 2. Evaluate C: 3.6 = −9.8( 0 ) + C

dυ = −9.8 dt

Initial condition  υ ( 0 ) = 3.6

C = 3.6. The solution of the initial value problem is

s(t)

0

ground

FIGURE 4.56 A package dropped from a rising hot-air balloon (Example 5).

υ = −9.8t + 3.6. Since velocity is the derivative of height, and the height of the package is 24.5 m at time t = 0 when it is dropped, we now have a second initial value problem: Differential   equation: Initial   condition:

ds = −9.8t + 3.6 dt s(0) = 24.5.

Set  υ = ds dt  in the previous equation.

We solve this initial value problem to find the height as a function of t. 1. Solve the differential equation: Finding the general antiderivative of −9.8t + 3.6 gives s = −4.9t 2 + 3.6t + C . 2. Evaluate C: 24.5 = −4.9( 0 ) 2 + 3.6( 0 ) + C C

Initial condition  s ( 0 ) = 24.5

= 24.5.

The package’s height above ground at time t is s = −4.9t 2 + 3.6t + 24.5. Use the solution: To find how long it takes the package to reach the ground, we set s equal to 0 and solve for t: −4.9t 2 + 3.6t + 24.5 = 0 −3.6 ± 493.16 −9.8 t ≈ −1.90, t ≈ 2.63. t =

Quadratic formula

The package hits the ground about 2.63 s after it is dropped from the balloon. (The negative root has no physical meaning.)

4.8  Antiderivatives

299

Indefinite Integrals A special symbol is used to denote the collection of all antiderivatives of a function f .

The collection of all antiderivatives of f is called the indefinite integral of f with respect to x; it is denoted by

DEFINITION

∫

f ( x ) dx.

The symbol ∫ is an integral sign. The function f is the integrand of the integral, and x is the variable of integration.

After the integral sign in the notation we just defined, the integrand function is always followed by a differential to indicate the variable of integration. We will have more to say about why this is important in Chapter 5. Using this notation, we restate the solutions of Example 1, as follows:

∫ 2 x dx

= x 2 + C,

∫ cos x dx

= sin x + C ,

∫ ( x + 2e 2 x ) dx 1

= ln x + e 2 x + C.

This notation is related to the main application of antiderivatives, which will be explored in Chapter 5. Antiderivatives play a key role in computing limits of certain infinite sums, an unexpected and wonderfully useful role that is described in a central result of Chapter 5, the Fundamental Theorem of Calculus. EXAMPLE 6

Evaluate

∫ ( x 2 − 2 x + 5 ) dx. Solution If we recognize that ( x 3 3 ) − x 2 + 5 x is an antiderivative of x 2 − 2 x + 5, we can evaluate the integral as antiderivative    3 x 2 ∫ ( x − 2 x + 5 ) dx = 3 − x 2 + 5 x + C.

arbitrary constant

If we do not recognize the antiderivative right away, we can generate it term-by-term with the Sum, Difference, and Constant Multiple Rules:

∫ ( x 2 − 2 x + 5 ) dx = ∫ x 2 dx − ∫ 2 x dx + ∫ 5 dx =

∫ x 2 dx − 2∫ x dx + 5∫ 1 dx

=

( x3

=

x3 + C 1 − x 2 − 2C 2 + 5 x + 5C 3 . 3

3

) ( x2

+ C1 − 2

2

)

+ C 2 + 5( x + C 3 )

300

Chapter 4 Applications of Derivatives

This formula is more complicated than it needs to be. If we combine C 1 , −2C 2 , and 5C 3 into a single arbitrary constant C = C 1 − 2C 2 + 5C 3 , the formula simplifies to x3 − x 2 + 5x + C 3 and still gives all the possible antiderivatives there are. For this reason, we recommend that you go right to the final form even if you elect to integrate term-by-term. Write

∫ ( x 2 − 2 x + 5 ) dx = ∫ x 2 dx − ∫ 2 x dx + ∫ 5 dx =

x3 − x 2 + 5 x + C. 3

Find the simplest antiderivative you can for each part, and add the arbitrary constant of integration at the end. We conclude this section with a list of basic antidifferentiation formulas in Table 4.4, using the integral sign to indicate an antiderivative. TABLE 4.4 Integration formulas

EXERCISES

x n +1 +C n+1

1.

∫ x n dx

8.

∫ e x dx

2.

∫ sin x dx

= −cos x + C

9.

∫

3.

∫ cos x dx

= sin x + C

10.

∫

4.

∫ sec 2 x dx

= tan x + C

11.

∫

5.

∫ csc 2 x dx

= −cot x + C

12.

∫

6.

∫ sec x tan x dx

= sec x + C

13.

∫

7.

∫ csc x cot x dx

= − csc x + C

=

In Exercises 1–24, find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation. 1. a. 2x

b. x 2

c. x 2 − 2 x + 1

2. a. 6x

b. x 7

c. x 7 − 6 x + 8

3. a. −3 x −4

b. x −4 x −3 b. + x2 2 5 b. 2 x 1 b. 2x 3

c. x −4 + 2 x + 3

5. a.

1 x2

6. a. −

2 x3

≠ − 1)

= ex + C

dx = ln x + C ( x ≠ 0 ) x dx = arcsin x + C 1 − x2 dx = arctan x + C 1 + x2 dx = arcsec x + C ( x > 1) x x2 − 1 ax a x dx = + C (a > 0, a ≠ 1) ln a

4.8

Finding Antiderivatives

4. a. 2 x −3

(n

7. a. 8. a. 9. a.

c. −x −3 + x − 1

10. a.

5 x2 1 c. x 3 − 3 x

11. a.

c. 2 −

12. a.

3 x 2 43 x 3 2 −1 3 x 3 1 −1 2 x 2 1 x 1 3x

b. b. b. b. b. b.

1 2 x 1 33 x 1 −2 3 x 3 1 − x −3 2 2 7 x 2 5x

c. c. c. c. c. c.

1 x 1 3 x + 3 x 1 −4 3 − x 3 3 − x −5 2 2 5 1− x 4 1 1+ − 2 3x x x +

4.8  Antiderivatives

13. a. −π sin π x

17. a. csc x cot x

b. 3 sin x π πx b. cos 2 2 2 x 2 b. sec 3 3 3 3x b. − csc 2 2 2 b. − csc 5 x cot 5 x

18. a. sec x tan x

b. 4 sec 3x tan 3x

19. a. e 3 x

b. e −x

14. a. π cos πx 15. a. sec 2 x 16. a. csc 2 x

20. a.

e −2 x

c. 1 − 8 csc 2 2 x πx πx cot c. −π csc 2 2 πx πx tan c. sec 2 2 x 2 c. e c. e −x 5 5 x c. 3 c. x 2 −1 1 c. 1 + 4x 2

b. e 4 x 3

( )

b. 2 −x

21. a. 3 x

b. x π

3

22. a. x

2 1 − x2

23. a.

c. sin π x − 3 sin 3 x πx c. cos + π cos x 2 3x c. − sec 2 2

24. a. x −

( 12 )

b.

x

1 2 ( x 2 + 1)

b. x 2 + 2 x

c. π x − x −1

Finding Indefinite Integrals

In Exercises 25–70, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation. 25.

∫ ( x + 1) dx

26.

∫ ( 5 − 6 x ) dx

27.

∫(

28.

∫(

29.

∫ ( 2 x 3 − 5 x + 7 ) dx

30.

∫ (1 − x 2 − 3x 5 ) dx

31.

∫ (x2

32.

∫ (5 − x3

33.

∫ x −1 3 dx

34.

∫ x −5 4 dx

35.

∫(

37.

∫ ⎜⎜⎝ 8 y −

39.

∫ 2 x (1 − x −3 ) dx

)

t 3t 2 + dt 2

1

− x2 −

x +

⎛

t t + t2

)

1 dx 3

x ) dx

3

2 ⎞⎟ ⎟ dy y 1 4 ⎟⎠

t

)

2

40.

∫ ⎜⎜⎝

∫ x −3 ( x + 1) dx 4+ t dt t3

41.

∫

dt

42.

∫

43.

∫ (−2 cos t ) dt

44.

∫ (−5 sin t ) dt

45.

∫ 7 sin 3 d θ

46.

∫ 3 cos 5θ d θ

47.

∫ (−3 csc 2 x ) dx

θ

csc θ cot θ dθ 2

)

+ 2 x dx

⎛ x 2 ⎞⎟ + ⎟ dx 2 x ⎟⎠ ⎛1 1 ⎞ 38. ∫ ⎜⎜ − 5 4 ⎟⎟⎟ dy ⎝7 y ⎠ 36.

⎛ sec 2 x ⎞⎟ ⎟ dx 3 ⎟⎠ 2 50. ∫ sec θ tan θ d θ 5 48.

∫ ⎜⎜⎝−

49.

∫

51.

∫ ( e 3x

+ 5e −x ) dx

52.

∫ ( 2e x

− 3e −2 x ) dx

53.

∫ ( e −x

+ 4 x ) dx

54.

∫ (1.3)

x

55.

∫ ( 4 sec x tan x − 2 sec 2 x ) dx

dx

∫ 2 ( csc 2 x − csc x cot x ) dx

57.

∫ ( sin 2 x − csc 2 x ) dx

59.

∫

61.

∫(

1 + cos 4 t dt 2 1 5 − 2 dx x x +1

)

∫ 3x 3 dx 65. ∫ (1 + tan 2 θ ) d θ 63.

(Hint: 1 + tan 2 θ = sec 2 θ)

58.

1 − cos 6t dt 2 ⎛ 2 1 ⎞ 62. ∫ ⎜⎜⎜ − 1 4 ⎟⎟⎟ dy ⎝ 1 − y2 y ⎠

60.

∫

∫ x 2 −1 dx 66. ∫ ( 2 + tan 2 θ ) d θ 64.

68.

∫ (1 − cot 2 x ) dx

∫ cos θ ( tan θ + sec θ ) d θ

70.

∫ csc θ − sin θ d θ

(Hint: 1 + cot 2 x = csc 2 x) 69.

∫ ( 2 cos 2 x − 3 sin 3x ) dx

∫ cot 2 x dx

csc θ

Checking Antiderivative Formulas

Verify the formulas in Exercises 71–82 by differentiation. ( 7 x − 2 )4 +C 71. ∫ ( 7 x − 2 )3 dx = 28 ( 3 x + 5 )−1 72. ∫ ( 3 x + 5 )−2 dx = − +C 3 1 73. ∫ sec 2 ( 5 x − 1) dx = tan ( 5 x − 1) + C 5 x −1 x −1 dx = −3 cot +C 74. ∫ csc 2 3 3 1 1 dx = − +C 75. ∫ ( x + 1) 2 x +1 1 x dx = +C 76. ∫ ( x + 1) 2 x +1 1 dx = ln x + 1 + C , x ≠ −1 77. ∫ x +1

(

t2 + 4 t 3 dt 2

1

1

56.

67.

301

78.

∫ xe x dx

79.

∫ a2

80.

∫

)

(

= xe x − e x + C

( ) x = arcsin ( ) + C a

dx 1 x = arctan +C + x2 a a

dx a2 − x2 arctan x arctan x 1 dx = ln x − ln (1 + x 2 ) − +C 2 x2 x

∫ 82. ∫ ( arcsin x ) 2 dx 81.

)

= x ( arcsin x ) 2 − 2 x + 2 1 − x 2 arcsin x + C

83. Right, or wrong? Say which for each formula and give a brief reason for each answer. x2 a. ∫ x sin x dx = sin x + C 2 b. ∫ x sin x dx = −x cos x + C c.

∫ x sin x dx

= −x cos x + sin x + C

84. Right, or wrong? Say which for each formula and give a brief reason for each answer. sec 3 θ a. ∫ tan θ sec 2 θ d θ = +C 3 1 b. ∫ tan θ sec 2 θ d θ = tan 2 θ + C 2 1 2 c. ∫ tan θ sec θ d θ = sec 2 θ + C 2

302

Chapter 4 Applications of Derivatives

85. Right, or wrong? Say which for each formula and give a brief reason for each answer. ( 2 x + 1) 3 +C a. ∫ ( 2 x + 1) 2 dx = 3 b. ∫ 3( 2 x + 1) 2 dx = ( 2 x + 1)3 + C c.

∫ 6( 2 x + 1)

2

dx = ( 2 x + 1)3 + C

Solve the initial value problems in Exercises 91–112. 91. 92. 93.

86. Right, or wrong? Say which for each formula and give a brief reason for each answer.

94.

a.

∫

2 x + 1 dx =

x2 + x + C

95.

b.

∫

2 x + 1 dx =

x2 + x + C

96.

c.

∫

2 x + 1 dx =

3 1 ( 2x + 1) + C 3

97.

87. Right, or wrong? Give a brief reason why.

∫

−15( x (x

+ − 2 )4

3)2

dx =

98.

( xx −+ 23 )

3

+C

99.

88. Right, or wrong? Give a brief reason why.

∫

100.

x cos( x 2 ) − sin ( x 2 ) sin ( x 2 ) dx = +C 2 x x

89. Which of the following graphs shows the solution of the initial value problem dy = 2 x , y = 4 when  x = 1? dx y

y

4

y

4

(1, 4)

3

4

(1, 4)

3

2

3 2

1

1

1

−1 0

1

x

107.

2

−1 0

(a)

(1, 4)

1

x

1

x

(c)

90. Which of the following graphs shows the solution of the initial value problem dy = −x , y = 1 when x = −1? dx y

x

(a)

Give reasons for your answer.

y(0) = −1

=

1 + x , x > 0; x2

= 9 x 2 − 4 x + 5, = 3 x −2 3 , =

1 , 2 x

y(2) = 1 y(−1) = 0

y(−1) = −5 y(4) = 0

= 1 + cos t , s(0) = 4 = cos t + sin t , s(π ) = 1 = −π sin πθ, r (0) = 0 = cos πθ, r (0) = 1 =

d 2r 2 = 3; dt 2 t

( )

dr dt

t =1

= 1, r (1) = 1

d 2s 3t ds = ; = 3, s(4) = 4 dt 2 8 dt t = 4 d 3y 109. = 6; y′′(0) = −8, y′(0) = 0, dx 3

y(0) = 5

d 3θ 1 = 0; θ ′′(0) = −2, θ ′(0) = − , θ(0) = dt 3 2 (4) 111. y = − sin t + cos t; y′′′(0) = 7, y′′(0) = y′(0) = −1, y(0) = 0 112. y ( 4 ) = − cos x + 8 sin 2 x; y′′′(0) = 0, y′′(0) = y′(0) = 1,

y

(−1, 1) 0 (b)

x

0 (c)

2

y(0) = 3

113. Find the curve y = f ( x ) in the xy-plane that passes through the point ( 9, 4 ) and whose slope at each point is 3 x .

y

(−1, 1) 0

= 10 − x ,

110.

Give reasons for your answer.

(−1, 1)

y(2) = 0

108.

−1 0

(b)

= 2 x − 7,

1 sec t tan t , υ(0) = 1 2 π 102. = 8t + csc 2 t , υ = −7 2 3 = , t > 1, υ(2) = 0 103. t t2 − 1 8 dυ + sec 2 t , υ(0) = 1 = 104. 1 + t2 dt d 2y 105. = 2 − 6 x; y′(0) = 4, y(0) = 1 dx 2 d 2y 106. = 0; y′(0) = 2, y(0) = 0 dx 2

101.

Initial Value Problems

dy dx dy dx dy dx dy dx dy dx dy dx ds dt ds dt dr dθ dr dθ dυ dt dυ dt dυ dt

x

114. a. Find a curve y = f ( x ) with the following properties: d 2y = 6x i) dx 2 ii) Its graph passes through the point ( 0, 1) and has a horizontal tangent line there. b. How many curves like this are there? How do you know?

4.8  Antiderivatives

In Exercises 115–118, the graph of f ′ is given. Assume that f (0) = 1 and sketch a possible continuous graph of f . 116. 115. y

y y = f 9(x)

2 0

2

6

4

x

8

−2

0

2

6

4

8

118.

y

124. Liftoff from Earth A rocket lifts off the surface of Earth with a constant acceleration of 20 m s 2 . How fast will the rocket be going 1 min later?

y

y = f 9(x)

y = f 9(x)

3

2 0

2

6

4

x

8

0

2

6

4

125. Stopping a car in time You are driving along a highway at a steady 108 km/h (30 m/s) when you see an accident ahead and slam on the brakes. What constant deceleration is required to stop your car in 75 m? To find out, carry out the following steps.

x

8

−2

1. Solve the initial value problem

−3

Differential equation:

Solution (Integral) Curves

Exercises 119–122 show solution curves of differential equations. In each exercise, find an equation for the curve through the labeled point. 120. 119.

(1, 0.5)

1

2

x

−1

121.

−1

126. Stopping a motorcycle The State of Illinois Cycle Rider Safety Program requires motorcycle riders to be able to brake from 48 km/h (13.3 m/s) to 0 in 13.7 m. What constant deceleration does it take to do that?

dy = 1 + psin px dx 2" x

127. Motion along a coordinate line A particle moves on a coordinate line with acceleration a = d 2 s dt 2 = 15 t − ( 3 t ) , subject to the conditions that ds dt = 4 and s = 0 when t = 1. Find

0

1

x

2

122. dy = sin x − cos x dx y

y 6

a. the velocity υ = ds dt in terms of t.

1 0 (−p, −1)

2

Measuring time and distance from when the brakes are applied

3. Find the value of k that makes s = 75 for the value of t you found in Step 2.

1 (−1, 1) −1

x

4

2

0

d 2s ( k  constant ) = −k dt 2 ds = 30 and  s = 0 when  t = 0. dt

2. Find the value of t that makes ds dt = 0. (The answer will involve k.)

2 1

0

Initial conditions:

y dy = x − 1 dx

dy = 1 − 4 x13 3 dx

y

ii) Find the body’s displacement from t = 1 to t = 3 given that s = −2 when t = 0.

b. Suppose that the position s of a body moving along a coordinate line is a differentiable function of time t. Is it true that once you know an antiderivative of the velocity function ds dt , you can find the body’s displacement from t = a to t = b even if you do not know the body’s exact position at either of those times? Give reasons for your answer.

x

−2

117.

i) Find the body’s displacement over the time interval from t = 1 to t = 3 given that s = 5 when t = 0.

iii) Now find the body’s displacement from t = 1 to t = 3 given that s = s 0 when t = 0.

y = f 9(x)

2

303

(1, 2)

1

2

3

x

−2

Applications

123. Finding displacement from an antiderivative of velocity a. Suppose that the velocity of a body moving along the s-axis is ds = υ = 9.8t − 3. dt

b. the position s in terms of t. T 128. The hammer and the feather When Apollo 15 astronaut David Scott dropped a hammer and a feather on the moon to demonstrate that in a vacuum all bodies fall with the same (constant) acceleration, he dropped them from about 1.2 m above the ground. The television footage of the event shows the hammer and the feather falling more slowly than on Earth, where, in a vacuum, they would have taken only half a second to fall the 1.2 m. How long did it take the hammer and feather to fall 1.2 m on the moon? To find out, solve the following initial value problem for s as a function of t. Then find the value of t that makes s equal to 0. Differential equation: Initial conditions:

d 2s = − 1.6 m s 2 dt 2 ds = 0 and s = 1.2 when t = 0 dt

304

Chapter 4 Applications of Derivatives

129. Motion with constant acceleration The standard equation for the position s of a body moving with a constant acceleration a along a coordinate line is a s = t 2 + υ0t + s0 , (1) 2 where υ 0 and s 0 are the body’s velocity and position at time t = 0. Derive this equation by solving the initial value problem d 2s =a Differential equation: dt 2 ds = υ 0  and  s = s 0  when t = 0. Initial conditions: dt 130. Free fall near the surface of a planet For free fall near the surface of a planet where the acceleration due to gravity has a constant magnitude of g length-units s 2 , Equation (1) in Exercise 129 takes the form 1 s = − gt 2 + υ 0 t + s 0 , (2) 2 where s is the body’s height above the surface. The equation has a minus sign because the acceleration acts downward, in the direction of decreasing s. The velocity υ 0 is positive if the object is rising at time t = 0 and negative if the object is falling. Instead of using the result of Exercise 129, you can derive Equation (2) directly by solving an appropriate initial value problem. What initial value problem? Solve it to be sure you have the right one, explaining the solution steps as you go along.

CHAPTER 4

131. Suppose that f (x) =

d (1 − dx

x ) and g( x ) =

d( x + 2 ). dx

Find:

∫ f ( x ) dx c. ∫ [ − f ( x ) ] dx e. ∫ [ f ( x ) + g( x ) ] dx a.

∫ g( x ) dx d. ∫ [ −g( x ) ] dx f. ∫ [ f ( x ) − g( x ) ] dx b.

132. Uniqueness of solutions If differentiable functions y = F ( x ) and y = g( x ) both solve the initial value problem dy = f ( x ), y( x 0 ) = y 0 , dx on an interval I, must F ( x ) = G ( x ) for every x in I? Give reasons for your answer. COMPUTER EXPLORATIONS

Use a CAS to solve the initial value problems in Exercises 133–136. Plot the solution curves. 133. y′ = cos 2 x + sin x , y(π ) = 1 1 134. y′ = + x , y(1) = −1 x 1 135. y′ = , y(0) = 2 4 − x2 2 136. y ″ = + x , y(1) = 0, y′(1) = 0 x

Questions to Guide Your Review

1. What can be said about the extreme values of a function that is continuous on a closed interval? 2. What does it mean for a function to have a local extreme value on its domain? An absolute extreme value? How are local and absolute extreme values related, if at all? Give examples.

14. What is a cusp? Give examples. 15. List the steps you would take to graph a rational function. Illustrate with an example. 16. Outline a general strategy for solving max-min problems. Give examples.

3. How do you find the absolute extrema of a continuous function on a closed interval? Give examples.

17. Describe l’Hôpital’s Rule. How do you know when to use the rule and when to stop? Give an example.

4. What are the hypotheses and conclusion of Rolle’s Theorem? Are the hypotheses really necessary? Explain.

18. How can you sometimes handle limits that lead to indeterminate forms ∞ ∞ , ∞ ⋅ 0, and ∞ − ∞? Give examples.

5. What are the hypotheses and conclusion of the Mean Value Theorem? What physical interpretations might the theorem have?

19. How can you sometimes handle limits that lead to indeterminate forms 1∞ , 0 0 , and ∞ ∞ ? Give examples.

6. State the Mean Value Theorem’s three corollaries. 7. How can you sometimes identify a function f ( x ) by knowing f ′ and knowing the value of f at a point x = x 0 ? Give an example.

20. Describe Newton’s method for solving equations. Give an example. What is the theory behind the method? What are some of the things to watch out for when you use the method?

8. What is the First Derivative Test for Local Extreme Values? Give examples of how it is applied.

21. Can a function have more than one antiderivative? If so, how are the antiderivatives related? Explain.

9. How do you test a twice-differentiable function to determine where its graph is concave up or concave down? Give examples.

22. What is an indefinite integral? How do you evaluate one? What general formulas do you know for finding indefinite integrals?

10. What is an inflection point? Give an example. What physical significance do inflection points sometimes have?

23. How can you sometimes solve a differential equation of the form dy dx = f ( x )?

11. What is the Second Derivative Test for Local Extreme Values? Give examples of how it is applied.

24. What is an initial value problem? How do you solve one? Give an example.

12. What do the derivatives of a function tell you about the shape of its graph?

25. If you know the acceleration of a body moving along a coordinate line as a function of time, what more do you need to know to find the body’s position function? Give an example.

13. List the steps you would take to graph a polynomial function. Illustrate with an example.

Chapter 4  Practice Exercises

CHAPTER 4

Practice Exercises

Finding Extreme Values In Exercises 1–16, find the extreme values (absolute and local) of the function over its natural domain, and where they occur.

1. y = 2 x 2 − 8 x + 9

2. y = x 3 − 2 x + 4

3. y = x 3 + x 2 − 8 x + 5

4. y = x 3 ( x − 5 ) 2

5. y =

6. y = x − 4 x

x2 − 1 1 7. y = 3 1 − x2 x 9. y = 2 x +1 11. y = e x + e −x

305

8. y =

3 + 2x − x 2

not closed. Is this consistent with the Extreme Value Theorem for continuous functions? Why? T 29. A graph that is large enough to show a function’s global behavior may fail to reveal important local features. The graph of f ( x ) = ( x 8 8 ) − ( x 6 2 ) − x 5 + 5 x 3 is a case in point. a. Graph f over the interval −2.5 ≤ x ≤ 2.5. Where does the graph appear to have local extreme values or points of inflection?

x +1 x 2 + 2x + 2 12. y = e x − e −x

b. Now factor f a( x ) and show that f has a local maximum at x = 3 5 ≈ 1.70998 and local minima at x = ± 3  ≈ o1.73205.

13. y  x ln x

14. y  x 2 ln x

15. y 

16. y =

c. Zoom in on the graph to find a viewing window that shows the presence of the extreme values at x  3 5 and x  3.

 arccos ( x 2 )

10. y =

sin −1

(e x )

Extreme Values

17. Does f ( x ) = x 3 + 2 x + tan x have any local maximum or minimum values? Give reasons for your answer. 18. Does g( x ) = csc x + 2 cot x have any local maximum values? Give reasons for your answer. 19. Does f ( x ) = ( 7 + x )(11 − 3 x ) have an absolute minimum value? An absolute maximum? If so, find them or give reasons why they fail to exist. List all critical points of f . 13

20. Find values of a and b such that the function ax + b f (x) = 2 x −1 has a local extreme value of 1 at x  3. Is this extreme value a local maximum or a local minimum? Give reasons for your answer. 21. Does g( x ) = e x − x have an absolute minimum value? An absolute maximum? If so, find them or give reasons why they fail to exist. List all critical points of g. 22. Does f ( x ) = 2e x (1 + x 2 ) have an absolute minimum value? An absolute maximum? If so, find them or give reasons why they fail to exist. List all critical points of f . In Exercises 23 and 24, find the absolute maximum and absolute minimum values of f over the interval. 23. f ( x ) = x − 2 ln x , 1 ≤ x ≤ 3 24. f ( x ) = ( 4 x ) + ln x 2 , 1 ≤ x ≤ 4 25. The greatest integer function f ( x ) = ⎣ x ⎦ , defined for all values of x, assumes a local maximum value of 0 at each point of [ 0, 1). Could any of these local maximum values also be local minimum values of f ? Give reasons for your answer. 26. a. Give an example of a differentiable function f whose first derivative is zero at some point c, even though f has neither a local maximum nor a local minimum at c. b. How is this consistent with Theorem 2 in Section 4.1? Give reasons for your answer. 27. The function y  1 x does not take on either a maximum or a minimum on the interval 0  x  1 even though the function is continuous on this interval. Does this contradict the Extreme Value Theorem for continuous functions? Why? 28. What are the maximum and minimum values of the function y  x on the interval −1 ≤ x < 1? Notice that the interval is

The moral here is that without calculus, the existence of two of the three extreme values would probably have gone unnoticed. On any normal graph of the function, the values would lie close enough together to fall within the dimensions of a single pixel on the screen. (Source: Uses of Technology in the Mathematics Curriculum, by Benny Evans and Jerry Johnson, Oklahoma State University, published in 1990 under a grant from the National Science Foundation, USE-8950044.) T 30. (Continuation of Exercise 29) a. Graph f ( x ) = ( x 8 8 ) − ( 2 5 ) x 5 − 5 x − ( 5 x 2 ) + 11 over the interval −2 ≤ x ≤ 2. Where does the graph appear to have local extreme values or points of inflection? b. Show that f has a local maximum value at x = 7 5 ≈ 1.2585 and a local minimum value at x = 3 2 ≈ 1.2599. c. Zoom in to find a viewing window that shows the presence of the extreme values at x  7 5 and x  3 2. The Mean Value Theorem

31. a. Show that g(t ) = sin 2 t − 3t decreases on every interval in its domain. b. How many solutions does the equation sin 2 t − 3t = 5 have? Give reasons for your answer. 32. a. Show that y  tan R increases on every open interval in its domain. b. If the conclusion in part (a) is really correct, how do you explain the fact that tan Q  0 is less than tan ( Q 4 ) = 1? 33. a. Show that the equation x 4 + 2 x 2 − 2 = 0 has exactly one solution on [ 0, 1 ]. T b. Find the solution to as many decimal places as you can. 34. a. Show that f ( x ) = x ( x + 1) increases on every open interval in its domain. b. Show that f ( x ) = x 3 + 2 x has no local maximum or minimum values. 35. Water in a reservoir As ume of water in a reservoir in 24 hours. Show that at the reservoir’s volume was 500,000 L min.

a result of a heavy rain, the volincreased by million cubic meters some instant during that period, increasing at a rate in excess of

306

Chapter 4 Applications of Derivatives

36. The formula F ( x ) = 3 x + C gives a different function for each value of C. All of these functions, however, have the same derivative with respect to x, namely F ′( x ) = 3. Are these the only differentiable functions whose derivative is 3? Could there be any others? Give reasons for your answers. 37. Show that

(

)

(

d x d 1 = − dx x + 1 dx x + 1

)

even though x 1 ≠ − . x +1 x +1 Doesn’t this contradict Corollary 2 of the Mean Value Theorem? Give reasons for your answer. 38. Calculate the first derivatives of f ( x ) = x 2 ( x 2 + 1) and g( x ) = −1 ( x 2 + 1). What can you conclude about the graphs of these functions? Analyzing Graphs In Exercises 39 and 40, use the graph to answer the questions.

39. Identify any global extreme values of f and the values of x at which they occur. y y = f (x)

Graphs and Graphing Graph the curves in Exercises 43–58.

43. y = x 2 − ( x 3 6 )

44. y = x 3 − 3 x 2 + 3

45. y = −x 3 + 6 x 2 − 9 x + 3 46. y = (1 8 )( x 3 + 3 x 2 − 9 x − 27 ) 47. y = x 3 ( 8 − x )

48. y = x 2 ( 2 x 2 − 9 )

49. y = x − 3 x

50. y = x 1 3 ( x − 4 )

23

51. y = x 3 − x

52. y = x 4 − x 2

53. y = ( x −

54. y = xe −x 2

2

3) e x

55. y = ln ( x 2 − 4 x + 3 ) 56. y = ln ( sin x ) 1 1 57. y = arcsin 58. y = tan −1 x x Each of Exercises 59–64 gives the first derivative of a function y = f ( x ). (a) At what points, if any, does the graph of f have a local maximum, local minimum, or inflection point? (b) Sketch the general shape of the graph.

( )

( )

59. y ′ = 16 − x 2

60. y ′ = x 2 − x − 6

61. y ′ = 6 x ( x + 1)( x − 2 )

62. y ′ = x 2 ( 6 − 4 x )

63. y ′ =

64. y ′ = 4 x 2 − x 4

x4

−

2x 2

In Exercises 65–68, graph each function. Then use the function’s first derivative to explain what you see. 65. y = x 2 3 + ( x − 1)1 3

66. y = x 2 3 + ( x − 1) 2 3

67. y = x 1 3 + ( x − 1)

68. y = x 2 3 − ( x − 1)1 3

13

(1, 1) 1 a2, 2b

x

0

40. Estimate the open intervals on which the function y = f ( x ) is a. increasing. b. decreasing. c. Use the given graph of f ′ to indicate where any local extreme values of the function occur, and whether each extreme is a relative maximum or minimum. y

(−3, 1) x −1

Each of the graphs in Exercises 41 and 42 is the graph of the position function s = f (t ) of an object moving on a coordinate line (t represents time). At approximately what times (if any) is each object’s (a) velocity equal to zero? (b) Acceleration equal to zero? During approximately what time intervals does the object move (c) forward? (d) Backward? 42.

s = f (t)

s

s = f (t) 3

6

9

12 14

78. lim

83. lim − sec 7 x cos 3 x

84. lim+

85. lim ( csc x − cot x )

86. lim

x →1

x→0

x→π 2

−2

0

xa − 1 xb − 1 tan x 80. lim x → 0 x + sin x sin mx 82. lim x → 0 sin nx

77. lim

x →1

y = f ′(x)

s

Using L’Hôpital’s Rule Use l’Hôpital’s Rule to find the limits in Exercises 77–88.

x 2 + 3x − 4 x −1 tan x 79. lim x→π x sin 2 x 81. lim x → 0 tan ( x 2 )

(2, 3)

41.

Sketch the graphs of the rational functions in Exercises 69–76. x +1 2x 70. y = 69. y = x−3 x +5 x2 + 1 x2 − x + 1 71. y = 72. y = x x x3 + 2 x4 − 1 73. y = 74. y = 2x x2 x2 x2 − 4 75. y = 2 76. y = 2 x −4 x −3

t 0

2

4

6

8

t

x→0

x→0

87. lim

(

88. lim

( x x− 1 − x x+ 1)

x →∞ x →∞

x2

+ x +1−

( x1

4

−

1 x2

)

− x)

3

3

2

x2

x sec x

2

Find the limits in Exercises 89–102. 10 x − 1 3θ − 1 90. lim 89. lim x→0 θ→0 x θ 2 sin x − 1 2 −sin x − 1 91. lim x 92. lim x→0 e − 1 x→0 ex − 1 5 − 5 cos x 4 − 4e x 93. lim x 94. lim x→0 e − x − 1 x→0 xe x

307

Chapter 4  Practice Exercises

t − ln (1 + 2t ) t→0 t2 t e 1 97. lim+ − t→0 t t 95. lim+

(

(

99. lim 1 + x →∞

x → 4 e x −4

)

b x

)

kx

98. lim+ e −1 y ln y y→ 0

x →∞

cos 2 x − 1 − 1 − cos x sin 2 x

102. lim

1 + tan x − 1 + sin x x3

x→0

(

100. lim 1 +

101. lim

x→0

sin 2 (π x ) +3−x

96. lim

2 7 + 2 x x

)

T 111. Open-top box An open-top rectangular box is constructed from a 25-cm-by-40-cm piece of cardboard by cutting squares of equal side length from the corners and folding up the sides. Find analytically the dimensions of the box of largest volume and the maximum volume. Support your answers graphically. 112. The ladder problem What is the approximate length (in meters) of the longest ladder you can carry horizontally around the corner of the corridor shown here? Round your answer down to the nearest meter. y

Optimization

103. The sum of two nonnegative numbers is 36. Find the numbers if

(2.4, 1.8)

1.8

a. the difference of their square roots is to be as large as possible. 0

b. the sum of their square roots is to be as large as possible. 104. The sum of two nonnegative numbers is 20. Find the numbers a. if the product of one number and the square root of the other is to be as large as possible. b. if one number plus the square root of the other is to be as large as possible. 105. An isosceles triangle has its vertex at the origin and its base parallel to the x-axis with the vertices above the axis on the curve y = 27 − x 2 . Find the largest area the triangle can have. 106. A customer has asked you to design an open-top rectangular stainless steel vat. It is to have a square base and a volume of 1 m 3, to be welded from 6-mm-thick plate, and to weigh no more than necessary. What dimensions do you recommend? 107. Find the height and radius of the largest right circular cylinder that can be put in a sphere of radius 3. 108. The figure here shows two right circular cones, one upside down inside the other. The two bases are parallel, and the vertex of the smaller cone lies at the center of the larger cone’s base. What values of r and h will give the smaller cone the largest possible volume?

Newton’s Method

113. Let f ( x ) = 3 x − x 3 . Show that the equation f ( x ) = −4 has a solution in the interval [ 2, 3 ] and use Newton’s method to find it. 114. Let f ( x ) = x 4 − x 3 . Show that the equation f ( x ) = 75 has a solution in the interval [ 3, 4 ] and use Newton’s method to find it. Finding Indefinite Integrals Find the indefinite integrals (most general antiderivatives) in Exercises  115–138. You may need to try a solution and then adjust your guess. Check your answers by differentiation. t2 115. ∫ ( x 3 + 5 x − 7 ) dx 116. ∫ 8t 3 − + t dt 2 1 3 4 117. ∫ 3 t + 2 dt 118. ∫ − 4 dt t t 2 t 6 dr dr 120. ∫ 119. ∫ 3 ( r + 5 )2 (r − 2 ) θ 121. ∫ 3θ θ 2 + 1 d θ 122. ∫ dθ 7 + θ2

(

( (

)

−1 4

∫ x 3 (1 + x 4 ) s 125. ∫ sec 2 ds 10

123.

12′

∫ csc

129.

∫ sin 2 4 dx ( Hint: sin 2 θ

h

130.

∫ cos 2 2 dx

109. Manufacturing tires Your company can manufacture x hundred grade A tires and y hundred grade B tires a day, where 0 ≤ x ≤ 4 and 40 − 10 x y = . 5−x Your profit on a grade A tire is twice your profit on a grade B tire. What is the most profitable number of each kind to make? 110. Particle motion The positions of two particles on the s-axis are s1 = cos t and s 2 = cos ( t + π 4 ). a. What is the farthest apart the particles ever get? b. When do the particles collide?

35

124.

2θ cot 2θ d θ x

)

)

∫ ( 2 − x ) dx 126. ∫ csc 2 πs ds

dx

127. r

6′

x

2.4

128. =

θ

θ

∫ sec 3 tan 3 d θ

1 − cos 2θ 2

)

x

131.

∫ ( x − x ) dx

133.

∫ ( 2 et

135.

∫ θ 1−π d θ

137.

∫ 2x

3

)

1

− e −t dt

3 dx x2 − 1

132.

∫ (x2

+

134.

∫ (5s

+ s 5 ) ds

136.

∫ 2 π+ r dr

138.

∫

5

dθ 16 − θ 2

Initial Value Problems Solve the initial value problems in Exercises 139–142.

dy x2 + 1 = , y(1) = −1 dx x2 dy 1 2 140. = x+ , y(1) = 1 x dx 139.

(

)

)

2 dx x2 + 1

308

Chapter 4 Applications of Derivatives

d 2r 3 = 15 t + ; r ′(1) = 8, r (1) = 0 dt 2 t d 3r 142. 3 = − cos t; r ′′(0) = r ′(0) = 0, r (0) = −1 dt

In Exercises 147 and 148, find the absolute maximum and minimum values of each function on the given interval. 1 e 147. y = x ln 2 x − x , ⎡⎢ , ⎤⎥ ⎣ 2e 2 ⎦

Applications and Examples

148. y = 10 x ( 2 − ln x ) , ( 0, e 2 ]

143. Can the integrations in (a) and (b) both be correct? Explain. dx = arcsin x + C a. ∫ 1 − x2 dx dx = −∫ − = −arccos x + C b. ∫ 1 − x2 1 − x2

In Exercises 149 and 150, find the absolute maxima and minima of the functions and give the x-coordinates where they occur.

144. Can the integrations in (a) and (b) both be correct? Explain. dx dx = −∫ − = −arccos x + C a. ∫ 1 − x2 1 − x2 x = −u dx −du b. ∫ =∫ dx = −du 1 − x2 1 − (−u) 2 −du =∫ 1 − u2 = arccos u + C

T 151. Graph the following functions and use what you see to locate and estimate the extreme values, identify the coordinates of the inflection points, and identify the intervals on which the graphs are concave up and concave down. Then confirm your estimates by working with the functions’ derivatives.

141.

= arccos(−x ) + C

u = −x

145. The rectangle shown here has one side on the positive y-axis, one side on the positive x-axis, and its upper right-hand vertex on the curve y = e −x 2 . What dimensions give the rectangle its largest area, and what is that area? y 1

y = e−x

2

x 4 +1

149. f ( x ) = e x 150. g( x ) = e

3− 2 x − x 2

a. y = ( ln x ) b. y =

x

e −x 2

c. y = (1 + x ) e −x T 152. Graph f ( x ) = x ln x. Does the function appear to have an absolute minimum value? Confirm your answer with calculus. sin x over [ 0, 3π ]. Explain what you see. T 153. Graph f ( x ) = ( sin x ) 154. A round underwater transmission cable consists of a core of copper wires surrounded by nonconducting insulation. If x denotes the ratio of the radius of the core to the thickness of the insulation, it is known that the speed of the transmission signal is given by the equation υ = x 2 ln (1 x ). If the radius of the core is 1 cm, what insulation thickness h will allow the greatest transmission speed?

x

0

Insulation

146. The rectangle shown here has one side on the positive y-axis, one side on the positive x-axis, and its upper right-hand vertex on the curve y = ( ln x ) x 2 . What dimensions give the rectangle its largest area, and what is that area? y

y=

0.2

x= r h

Core

h r

ln x x2

0.1 0

CHAPTER 4

1

x

Additional and Advanced Exercises

Functions and Derivatives

1. What can you say about a function whose maximum and minimum values on an interval are equal? Give reasons for your answer. 2. Is it true that a discontinuous function cannot have both an absolute maximum value and an absolute minimum value on a closed interval? Give reasons for your answer. 3. Can you conclude anything about the extreme values of a continuous function on an open interval? On a half-open interval? Give reasons for your answer. 4. Local extrema Use the sign pattern for the derivative df = 6 ( x − 1)( x − 2 ) 2 ( x − 3 )3 ( x − 3 ) 4 dx to identify the points where f has local maximum and minimum values.

5. Local extrema a. Suppose that the first derivative of y = f ( x ) is y ′ = 6( x + 1)( x − 2 ) 2 . At what points, if any, does the graph of f have a local maximum, local minimum, or point of inflection? b. Suppose that the first derivative of y = f ( x ) is y ′ = 6 ( x + 1)( x − 2 ) . At what points, if any, does the graph of f have a local maximum, local minimum, or point of inflection? 6. If f ′( x ) ≤ 2 for all x, what is the most the values of f can increase on [ 0, 6 ]? Give reasons for your answer.

Chapter 4  Additional and Advanced Exercises

7. Bounding a function Suppose that f is continuous on [ a, b ] and that c is an interior point of the interval. Show that if f ′( x ) ≤ 0 on [ a, c ) and f ′( x ) ≥ 0 on ( c, b ], then f ( x ) is never less than f (c) on [ a, b ]. 8. An inequality a. Show that −1 2 ≤ x (1 + x 2 ) ≤ 1 2 for every value of x. b. Suppose that f is a function whose derivative is f ′( x ) = x (1 + x 2 ). Use the result in part (a) to show that 1 f (b) − f (a) ≤ b − a 2 for any a and b. 9. The derivative of f ( x )  x 2 is zero at x  0, but f is not a constant function. Doesn’t this contradict the corollary of the Mean Value Theorem that says that functions with zero derivatives are constant? Give reasons for your answer. 10. Extrema and inflection points Let h  fg be the product of two differentiable functions of x. a. If f and g are positive, with local maxima at x  a, and if f a and g a change sign at a, does h have a local maximum at a? b. If the graphs of f and g have inflection points at x  a, does the graph of h have an inflection point at a?

309

for the hole? (Hint: How long will it take an exiting droplet of water to fall from height y to the ground?) Tank kept full, top open

y

h Exit velocity = " 64(h − y) y

Ground

0

x

Range

16. Kicking a field goal An American football player wants to kick a field goal with the ball being on a right hash mark. Assume that the goal posts are b meters apart and that the hash mark line is a distance a  0 meters from the right goal post. (See the accompanying figure.) Find the distance h from the goal post line that gives the kicker his largest angle C. Assume that the football field is flat. Goal posts

In either case, if the answer is yes, give a proof. If the answer is no, give a counterexample.

b

Goal post line

a

11. Finding a function Use the following information to find the values of a, b, and c in the formula f ( x ) = ( x + a ) ( bx 2 + cx + 2 ). a. The values of a, b, and c are either 0 or 1. b. The graph of f passes through the point (−1, 0 ). h

c. The line y  1 is an asymptote of the graph of f . 12. Horizontal tangent For what value or values of the constant k will the curve y = x 3 + kx 2 + 3 x − 4 have exactly one horizontal tangent? Optimization 13. Largest inscribed triangle Points A and B lie at the ends of a diameter of a unit circle and point C lies on the circumference. Is it true that the area of triangle ABC is largest when the triangle is isosceles? How do you know? 14. Proving the second derivative test The Second Derivative Test for Local Maxima and Minima (Section 4.4) says: a. f has a local maximum value at x  c if f ′(c) = 0 and f ′′(c) < 0;

b. f has a local minimum value at x  c if f ′(c) = 0 and f ′′(c) > 0. To prove statement (a), let F = (1 2 ) f ′′(c) . Then use the fact that f ′′(c) = lim

h→ 0

b u

Football

17. A max-min problem with a variable answer Sometimes the solution of a max-min problem depends on the proportions of the shapes involved. As a case in point, suppose that a right circular cylinder of radius r and height h is inscribed in a right circular cone of radius R and height H, as shown here. Find the value of r (in terms of R and H) that maximizes the total surface area of the cylinder (including top and bottom). As you will see, the solution depends on whether H b 2 R or H  2 R.

f ′(c + h) − f ′(c) f ′(c + h) = lim h→ 0 h h

r

to conclude that for some E  0, 0 < h < δ

⇒

f ′(c + h) < f ′′(c) + ε < 0. h

Thus, f ′ ( c + h ) is positive for −E < h < 0 and negative for 0  h  E. Prove statement (b) in a similar way. 15. Hole in a water tank You want to bore a hole in the side of the tank shown here at a height that will make the stream of water coming out hit the ground as far from the tank as possible. If you drill the hole near the top, where the pressure is low, the water will exit slowly but spend a relatively long time in the air. If you drill the hole near the bottom, the water will exit at a higher velocity but have only a short time to fall. Where is the best place, if any,

H

h

R

18. Minimizing a parameter Find the smallest value of the positive constant m that will make mx − 1 + (1 x ) greater than or equal to zero for all positive values of x.

310

Chapter 4 Applications of Derivatives

19. Determine the dimensions of the rectangle of largest area that can be inscribed in the right triangle in the accompanying figure.

26. The family of straight lines y = ax + b (a, b arbitrary constants) can be characterized by the relation y ′′ = 0. Find a similar relation satisfied by the family of all circles (x

10 8

6

20. A rectangular box with a square base is inscribed in a right circular cone of height 4 and base radius 3. If the base of the box sits on the base of the cone, what is the largest possible volume of the box? Limits 21. Evaluate the following limits.

2 sin 5 x 3x c. lim x csc 2 2 x

a. lim

xl0 xl0

x − sin x e. lim x → 0 x − tan x sec x − 1 g. lim x→0 x2

b. lim sin 5 x cot 3 x xl0

d. lim ( sec x − tan x ) x→Q 2

f. lim

sin x 2 x sin x

h. lim

x3 − 8 x2 − 4

xl0

x→2

22. L’Hôpital’s Rule does not help with the following limits. Find them some other way. a. lim

x →∞

x +5 x +5

b. lim

x →∞

2x x +7 x

Theory and Examples 23. Suppose that it costs a company y = a + bx dollars to produce x units per week. It can sell x units per week at a price of P = c − ex dollars per unit. Each of a, b, c, and e represents a positive constant. (a) What production level maximizes the profit? (b) What is the corresponding price? (c) What is the weekly profit at this level of production? (d) At what price should each item be sold to maximize profits if the government imposes a tax of t dollars per item sold? Comment on the difference between this price and the price before the tax. 24. Estimating reciprocals without division You can estimate the value of the reciprocal of a number a without ever dividing by a if you apply Newton’s method to the function f ( x ) = (1 x ) − a. For example, if a  3, the function involved is f ( x ) = (1 x ) − 3. a. Graph y = (1 x ) − 3. Where does the graph cross the x-axis?

b. Show that the recursion formula in this case is x n +1 = x n ( 2 − 3 x n ) , so there is no need for division. q 25. To find x  a, we apply Newton’s method to f ( x ) = x q − a. Here we assume that a is a positive real number and q is a positive integer. Show that x 1 is a “weighted average” of x 0 and a x 0 q 1 , and find the coefficients m 0 , m1 such that ⎛ a ⎞ x 1 = m 0 x 0 + m1 ⎜⎜⎜ q −1 ⎟⎟⎟ , m 0 > 0, m1 > 0, m 0 + m1 = 1. ⎝ x0 ⎠ What conclusion would you reach if x 0 and a x 0 q 1 were equal? What would be the value of x 1 in that case?

− h )2 + ( y − h )2 = r 2 ,

where h and r are arbitrary constants. (Hint: Eliminate h and r from the set of three equations including the given one and two obtained by successive differentiation.) 27. Free fall in the fourteenth century In the middle of the fourteenth century, Albert of Saxony (1316–1390) proposed a model of free fall, which assumed that the velocity of a falling body was proportional to the distance fallen. It seemed reasonable to think that a body that had fallen 6 m might be moving twice as fast as a body that had fallen 3 m. And besides, none of the instruments in use at the time were accurate enough to prove otherwise. Today we can see just how far off Albert of Saxony’s model was by solving the initial value problem implicit in his model. Solve the problem and compare your solution graphically with the equation s  4.9t 2 . You will see that it describes a motion that starts too slowly and then becomes too fast to be realistic. 28. Group testing During World War II it was necessary to administer blood tests to large numbers of recruits. There are two standard ways to administer a blood test to N people. In method 1, each person is tested separately. In method 2, the blood samples of x people are pooled and tested as one large sample. If the test is negative, this one test is enough for all x people. If the test is positive, then each of the x people is tested separately, requiring a total of x 1 tests. Using the second method and some probability theory it can be shown that, on the average, the total number of tests y will be

(

y = N 1 − qx +

29.

30.

31.

32. 33.

)

1 . x

With q  0.99 and N  1000, find the integer value of x that minimizes y. Also find the integer value of x that maximizes y. (This second result is not important to the real-life situation.) The group testing method was used in World War II with a savings of 80% over the individual testing method, but not with the given value of q. Group testing has been implemented in diagnosing various diseases, including COVID-19. Assume that the brakes of an automobile produce a constant deceleration of k ms2. (a) Determine what k must be to bring an automobile traveling 108 kmh (30 ms) to rest in a distance of 30 m from the point where the brakes are applied. (b) With the same k, how far would a car traveling 54 kmhr go before being brought to a stop? Let f ( x ) and g( x ) be two continuously differentiable functions satisfying the relationships f ′( x ) = g( x ) and f ′′( x ) = − f ( x ). Let h( x ) = f 2 ( x ) + g 2 ( x ). If h(0)  5, find h(10). Can there be a curve satisfying the following conditions? d 2 y dx 2 is everywhere equal to zero and, when x  0, y  0 and dy dx  1. Give a reason for your answer. Find the equation for the curve in the xy-plane that passes through the point (1, −1) if its slope at x is always 3 x 2 2. A particle moves along the x-axis. Its acceleration is a = −t 2 . At t  0, the particle is at the origin. In the course of its motion, it reaches the point x  b, where b  0, but no point beyond b. Determine its velocity at t  0.

Chapter 4  Technology Application Projects

311

34. A particle moves with acceleration a = t − (1 t ). Assuming that the velocity υ = 4 3 and the position s = −4 15 when t = 0, find a. the velocity υ in terms of t.

In our model, we assume that AC = a and BC = b are fixed. Thus we have the relations d1 + d 2 cos θ = a d 2 sin θ = b,

b. the position s in terms of t. 35. Given f ( x ) = ax 2 + 2bx + c with a > 0. By considering the minimum, prove that f ( x ) ≥ 0 for all real x if and only if b 2 − ac ≤ 0. 36. The Cauchy–Schwarz inequality a. In Exercise 35, let

d 2 = b csc θ, d1 = a − d 2 cos θ = a − b cot θ.

so that

We can express the total loss L as a function of θ: L = k

f ( x ) = ( a1 x + b1 ) + ( a 2 x + b 2 ) +  + ( a n x + b n ) , 2

2

2

≤ ( a1 2 + a 2 2 +  + a n 2 )( b1 2 + b 2 2 +  + b n 2 ). b. Show that equality holds in The Cauchy–Schwarz inequality only if there exists a real number x that makes a i x equal −bi for every value of i from 1 to n. 37. The best branching angles for blood vessels and pipes When a smaller pipe branches off from a larger one in a flow system, we may want it to run off at an angle that is best from some energysaving point of view. We might require, for instance, that energy loss due to friction be minimized along the section AOB shown in the accompanying figure. In this diagram, B is a given point to be reached by the smaller pipe, A is a point in the larger pipe upstream from B, and O is the point where the branching occurs. A law formulated by Poiseuille states that the loss of energy due to friction in nonturbulent flow is proportional to the length of the path and inversely proportional to the fourth power of the radius. Thus, the loss along AO is ( kd1 ) R 4 and along OB is ( kd 2 ) r 4 , where k is a constant, d1 is the length of AO, d 2 is the length of OB, R is the radius of the larger pipe, and r is the radius of the smaller pipe. The angle θ is to be chosen to minimize the sum of these two losses: d d L = k 14 + k 42 . R r

O

d1

y (a, b)

B

y = ln x C

A x

39. Consider the unit circle centered at the origin and with a vertical tangent line passing through point A in the accompanying figure. Assume that the lengths of segments AB and AC are equal, and let point D be the intersection of the x-axis with the line passing through points B and C. Find the limit of t as B approaches A. y B

C

b = d 2 sin u

d 2 cos u

r4 . R4

b. If the ratio of the pipe radii is r R = 5 6 estimate to the nearest degree the optimal branching angle given in part (a). 38. Consider point (a, b) on the graph of y = ln x and triangle ABC formed by the tangent line at (a, b), the y-axis, and the line y = b. Show that a ( area triangle  ABC ) = . 2

u

A

4

θ c = cos −1

B d2

4

a. Show that the critical value of θ for which dL d θ equals zero is

and deduce The Cauchy–Schwarz inequality: ( a1b1 + a 2 b 2 +  + a n b n ) 2

θ ( a − Rb cot θ + b csc ). r

D A

C

1

0

1

t

x

a

CHAPTER 4

Technology Application Projects

Mathematica/Maple Projects

Projects can be found within MyLab Math. • Motion Along a Straight Line: Position → Velocity → Acceleration You will observe the shape of a graph through dramatic animated visualizations of the derivative relations among the position, velocity, and acceleration. Figures in the text can be animated. • Newton’s Method: Estimate π to How Many Places? Plot a function, observe a root, pick a starting point near the root, and use Newton’s Iteration Procedure to approximate the root to a desired accuracy. The numbers π , e, and 2 are approximated.

5

Integrals

OVERVIEW A great achievement of classical geometry was obtaining formulas for the areas and volumes of triangles, spheres, and cones. In this chapter we develop a method, called integration, to calculate the areas and volumes of more general shapes. The definite integral is the key tool in calculus for defining and calculating areas and volumes. We also use it to compute quantities such as the lengths of curved paths, probabilities, averages, energy consumption, the mass of an object, and the force against a dam’s floodgates. Like the derivative, the definite integral is defined as a limit. The definite integral is a limit of increasingly fine approximations. The idea is to approximate a quantity (such as the area of a curvy region) by dividing it into many small pieces, each of which we can approximate by something simple (such as a rectangle). Summing the contributions of each of the simple pieces gives us an approximation to the original quantity. As we divide the region into more and more pieces, the approximation given by the sum of the pieces will generally improve, converging to the quantity we are measuring. We take a limit as the number of terms increases to infinity, and when the limit exists, the result is a definite integral. We develop this idea in Section 5.3. We also show that the process of computing these definite integrals is closely connected to finding antiderivatives. This is one of the most important relationships in calculus; it gives us an efficient way to compute definite integrals, providing a simple and powerful method that eliminates the difficulty of directly computing limits of approximations. This connection is captured in the Fundamental Theorem of Calculus.

5.1 Area and Estimating with Finite Sums y

The basis for formulating definite integrals is the construction of approximations by finite sums. In this section we consider three examples of this process: finding the area under a graph, the distance traveled by a moving object, and the average value of a function. Although we have yet to define precisely what we mean by the area of a general region in the plane, or the average value of a function over a closed interval, we do have intuitive ideas of what these notions mean. We begin our approach to integration by approximating these quantities with simpler finite sums related to these intuitive ideas. We then consider what happens when we take more and more terms in the summation process. In subsequent sections we look at taking the limit of these sums as the number of terms goes to infinity, which leads to a precise definition of the definite integral.

1 y = 1 − x2

0.5 R

0

FIGURE 5.1

0.5

1

x

The area of the shaded region R cannot be found by a simple formula.

312

Area Suppose we want to find the area of the shaded region R that lies above the x-axis, below the graph of y = 1 − x 2 , and between the vertical lines x = 0 and x = 1 (see Figure 5.1).

5.1  Area and Estimating with Finite Sums

y

y

1

313

y = 1 − x2

(0, 1)

1

y = 1 − x2

(0, 1) Q1 , 15R 4 16

1 3

1 3

Q2 , 4R

Q2 , 4R

3

7

Q4 , 16R

R

R

0

1

x

0

(a)

1

x

(b)

FIGURE 5.2 (a) We get an upper sum approximation of the area of R by using two rectangles containing R. (b) Four rectangles give a better upper sum approximation. Both estimates overshoot the true value for the area by the amount shaded in light red.

Unfortunately, there is no simple geometric formula for calculating the areas of general shapes having curved boundaries like the region R. How, then, can we find the area of R? Although we do not yet have a method for determining the exact area of R, we can approximate it in a simple way. Figure 5.2a shows two rectangles that together contain the region R. Each rectangle has width 1 2 and they have heights 1 and 3 4 (left to right). The height of each rectangle is the maximum value of the function f in each subinterval. Because the function f is decreasing, the height is its value at the left endpoint of the subinterval of [ 0, 1 ] that forms the base of the rectangle. The total area of the two rectangles approximates the area A of the region R: A ≈ 1⋅

1 3 1 7 + ⋅ = = 0.875. 2 4 2 8

This estimate is larger than the true area A since the two rectangles contain R. We say that 0.875 is an upper sum because it is obtained by taking the height of the rectangle corresponding to the maximum (uppermost) value of f ( x ) over points x lying in the base of each rectangle. In Figure 5.2b, we improve our estimate by using four thinner rectangles, each of width 1 4, which taken together contain the region R. These four rectangles give the approximation A ≈ 1⋅

1 15 1 3 1 7 1 25 + ⋅ + ⋅ + ⋅ = = 0.78125, 4 16 4 4 4 16 4 32

which is still greater than A since the four rectangles contain R. Suppose instead we use four rectangles contained inside the region R to estimate the area, as in Figure 5.3a. Each rectangle has width 1 4, as before, but the rectangles are shorter and lie entirely beneath the graph of f . The function f ( x ) = 1 − x 2 is decreasing on [ 0, 1 ], so the height of each of these rectangles is given by the value of f at the right endpoint of the subinterval forming its base. The fourth rectangle has zero height and therefore contributes no area. Summing these rectangles, whose heights are the minimum value of f ( x ) over points x in the rectangle’s base, gives a lower sum approximation to the area: A ≈

15 1 3 1 7 1 1 17 ⋅ + ⋅ + ⋅ +0⋅ = = 0.53125. 16 4 4 4 16 4 4 32

This estimate is smaller than the area A since the rectangles all lie inside of the region R. The true value of A lies somewhere between these lower and upper sums: 0.53125 < A < 0.78125.

314

Chapter 5 Integrals

y

y 1

y = 1 − x2

15 Q14 , 16 R

1

63 Q18 , 64 R 55 Q38 , 64 R

y = 1 − x2

1 3 Q2 , 4R 39 Q58 , 64 R 7 Q34 , 16 R

0.5

0.5 15 Q78 , 64 R

0

0.25

0.5 (a)

0.75

1

x

0

0.25 0.5 0.75 1 0.125 0.375 0.625 0.875

x

(b)

FIGURE 5.3 (a) Rectangles contained in R give a lower sum approximation for the area. This estimate undershoots the true value by the amount shaded in light blue. (b) The midpoint rule uses rectangles whose heights are the values of y = f ( x ) at the midpoints of their bases. The estimate appears closer to the true value of the area because the light red overshoot areas roughly balance the light blue undershoot areas.

y

Considering both lower and upper sum approximations gives us estimates for the area and a bound on the size of the possible error in these estimates since the true value of the area lies somewhere between them. Here the error cannot be greater than the difference 0.78125 − 0.53125 = 0.25. Yet another estimate can be obtained by using rectangles whose heights are the values of f at the midpoints of the bases of the rectangles (Figure 5.3b). This method of estimation is called the midpoint rule for approximating the area. The midpoint rule gives an estimate that is between a lower sum and an upper sum, but it is not clear whether it overestimates or underestimates the true area. With four rectangles of width 1 4, as before, the midpoint rule estimates the area of R to be

1 y = 1 − x2

1

0

x

(a)

A ≈

63 1 55 1 39 1 15 1 172 1 ⋅ + ⋅ + ⋅ + ⋅ = ⋅ = 0.671875. 64 4 64 4 64 4 64 4 64 4

In each of the sums that we computed, the interval [ a, b ] over which the function f is defined was subdivided into n subintervals of equal width (or length) Δx = ( b − a ) n , and f was evaluated at a point in each subinterval: c1 in the first subinterval, c 2 in the second subinterval, and so on. For the upper sum we chose c k so that f (c k ) was the maximum value of f in the kth subinterval, for the lower sum we chose it so that f (c k ) was the minimum, and for the midpoint rule we chose c k to be the midpoint of the kth subinterval. In each case the finite sums have the form

y 1 y = 1 − x2

f (c1 ) Δx + f (c 2 ) Δx + f (c 3 ) Δx +  + f (c n ) Δx.

1

0

x

(b)

FIGURE 5.4

(a) A lower sum using 16 rectangles of equal width Δx = 1 16. (b) An upper sum using 16 rectangles.

As we take more and more rectangles, with each rectangle thinner than before, it appears that these finite sums give better and better approximations to the true area of the region R. Figure 5.4a shows a lower sum approximation for the area of R using 16 rectangles of equal width. The sum of their areas is 0.634765625, which appears close to the true area but is still somewhat smaller since the rectangles lie inside R. Figure 5.4b shows an upper sum approximation using 16 rectangles of equal width. The sum of their areas is 0.697265625, which is somewhat larger than the true area because the rectangles taken together contain R. The midpoint rule for 16 rectangles gives a total area approximation of 0.6669921875, but it is not immediately clear whether this estimate is larger or smaller than the true area.

5.1  Area and Estimating with Finite Sums

315

Table 5.1 shows the values of upper and lower sum approximations to the area of R, using up to 1000 rectangles. The values of these approximations appear to be approaching 2 3. In Section 5.2 we will see how to get an exact value of the area of regions such as R by taking a limit as the base width of each rectangle goes to zero and the number of rectangles goes to infinity. With the techniques developed there, we will be able to show that the area of R is exactly 2 3. TABLE 5.1 Finite approximations for the area of R

Number of subintervals

Lower sum

Midpoint sum

Upper sum

2 4 16 50 100 1000

0.375 0.5313 0.6348 0.6566 0.66165 0.6661665

0.6875 0.6719 0.6670 0.6667 0.666675 0.66666675

0.875 0.7813 0.6973 0.6766 0.67165 0.6671665

Distance Traveled Suppose we know the velocity function υ(t ) of a car that moves straight down a highway without changing direction, and we want to know how far it traveled between times t = a and t = b. The position function s(t ) of the car has derivative υ(t ). If we can find an antiderivative F (t ) of υ(t ), then we can find the car’s position function s(t ) by setting s(t ) = F (t ) + C. The distance traveled can then be found by calculating the change in position, s(b) − s(a) = F (b) − F (a). However, if the velocity is known only by the readings at various times of a speedometer on the car, then we have no formula for the velocity from which to obtain an antiderivative that gives the position function. So what do we do in this situation? If we know the velocity at a collection of times t 1 , t 2 , … t n , we can approximate the distance traveled by using finite sums in a way similar to the area estimates that we discussed before. We first subdivide the interval [ a, b ] into short time intervals and assume that the velocity on each subinterval is fairly constant. Then we approximate the distance traveled on each time subinterval with the usual distance formula distance = velocity × time and add the results across [ a, b ]. Suppose the subdivided interval looks like Δt a

t1

Δt t2

Δt t3

b

t (sec)

with the subintervals all of equal length Δt. Pick a number t 1 in the first interval. If Δt is so small that the velocity barely changes over a short time interval of duration Δt, then the distance traveled in the first time interval is about υ(t 1 ) Δt. If t 2 is a number in the second interval, the distance traveled in the second time interval is about υ(t 2 ) Δt. The sum of the distances traveled over all the time intervals is approximated by υ( t 1 ) Δt + υ( t 2 ) Δt +  + υ( t n ) Δt , where n is the total number of subintervals. This sum is only an approximation to the true distance D, but the approximation increases in accuracy as we take more and more subintervals.

316

Chapter 5 Integrals

EXAMPLE 1 The velocity function of a projectile fired straight into the air is f (t ) = 160 − 9.8t m sec. Use the summation technique just described to estimate how far the projectile rises during the first 3 sec. How close do the sums come to the exact value of 435.9 m? (We will see how to compute the exact value in Section 5.4.) Solution We explore the results for different numbers of subintervals and different choices of evaluation points. Notice that f (t ) is decreasing, so choosing left endpoints gives an upper sum estimate, and choosing right endpoints gives a lower sum estimate.

(a) Three subintervals of length 1, with  f  evaluated at left endpoints giving an upper sum: t1

t2

t3

0

1

2

t

3

Δt

With f evaluated at t = 0, 1, and 2, we have D ≈ f ( t 1 ) Δt + f ( t 2 ) Δt + f ( t 3 ) Δt = [ 160 − 9.8( 0 ) ](1) + [ 160 − 9.8(1) ](1) + [ 160 − 9.8( 2 ) ](1) = 450.6. (b) Three subintervals of length 1, with  f  evaluated at right endpoints giving a lower sum:

0

t1

t2

t3

1

2

3

t

Δt

With f evaluated at t = 1, 2, and 3, we have D ≈ f ( t 1 ) Δt + f ( t 2 ) Δt + f ( t 3 ) Δt = [ 160 − 9.8(1) ](1) + [ 160 − 9.8( 2 ) ](1) + [ 160 − 9.8( 3 ) ](1) = 421.2. (c) With six subintervals of length 1 2, we get t1 t 2 t 3 t 4 t 5 t 6 0

1

2

t1 t 2 t 3 t 4 t 5 t 6 3

Δt

t

0

1

2

3

t

Δt

These estimates give an upper sum using left endpoints: D ≈ 443.25, and a lower sum using right endpoints: D ≈ 428.55. These six-interval estimates are somewhat closer than the three-interval estimates. The results improve as the subintervals get shorter. As we can see in Table 5.2, the left-endpoint upper sums approach the true value 435.9 from above, whereas the right-endpoint lower sums approach it from below. The true value lies between these upper and lower sums. The magnitude of the error in the closest entry is 0.23, a small percentage of the true value. Error magnitude = true value − calculated value = 435.9 − 435.67 = 0.23. Error percentage =

0.23 ≈ 0.05%. 435.9

It is reasonable to conclude from the table’s last entries that the projectile rose about 436 m during its first 3 sec of flight.

5.1  Area and Estimating with Finite Sums

317

TABLE 5.2 Travel-distance estimates

Number of subintervals 3 6 12 24 48 96 192

Length of each subinterval

Upper sum

Lower sum

1

450.6

421.2

12 14 18

443.25 439.58 437.74

428.55 432.23 434.06

1 16 1 32 1 64

436.82 436.36 436.13

434.98 435.44 435.67

Displacement Versus Distance Traveled If an object with position function s(t ) moves along a coordinate line without changing direction, we can calculate the total distance it travels from t  a to t  b by summing the distance traveled over small intervals, as in Example 1. If the object reverses direction one or more times during the trip, then we need to use the object’s speed V(t ) , which is the absolute value of its velocity function, V(t ), to find the total distance traveled. Using the velocity itself, as in Example 1, gives instead an estimate of the object’s displacement, s(b)  s(a), the difference between its initial and final positions. To see the difference, think about what happens when you walk a kilometer from your home and then walk back. The total distance traveled is two kilometers, but your displacement is zero, because you end up back where you started. To see why using the velocity function in the summation process gives an estimate of the displacement, partition the time interval [ a, b ] into small enough equal subintervals %t so that the object’s velocity does not change very much from time t k 1 to t k . Then V(t k ) gives a good approximation of the velocity throughout the interval. Accordingly, the change in the object’s position coordinate, which is its displacement during the time interval, is about V ( t k ) % t. The change is positive if V(t k ) is positive and negative if V(t k ) is negative. In either case, the distance traveled by the object during the subinterval is about V ( t k ) % t.

s

The total distance traveled over the time interval is approximately the sum s(5)

Height (m)

122.5

78.4

s=0

V ( t 1 ) Δ t + V ( t 2 ) Δ t + " + V ( t n ) Δ t.

(+)

(−) 44.1

s(2)

s(8)

s(0)

FIGURE 5.5 The rock in Example 2. The height s  78.4 m is reached at t  2 s and t  8 s. The rock falls 44.1 m from its maximum height when t  8.

We will revisit these ideas in Section 5.4. EXAMPLE 2 In Example 4 in Section 3.4, we analyzed the motion of a heavy rock blown straight up by a dynamite blast. In that example, we found the velocity of the rock at time t was V(t)  49  9.8t m/s. The rock was 78.4 m above the ground 2 s after the explosion, continued upward to reach a maximum height of 122.5 m at 5 s after the explosion, and then fell back down a distance of 44.1 m to reach the height of 78.4 m again at t  8 s after the explosion. (See Figure 5.5.) The total distance traveled in these 8 seconds is 122.5 44.1  166.6 m. Solution If we follow a procedure like the one presented in Example 1, using the velocity function V(t) in the summation process from t  0 to t  8, we obtain an estimate of the rock’s height above the ground at time t  8. Starting at time t  0, the rock traveled upward a total of 78.4 44.1  122.5 m, but then it peaked and traveled downward

318

Chapter 5 Integrals

TABLE 5.3 Velocity function

t

V( t )

t

V( t )

0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

49 44.1 39.2 34.3 29.4 24.5 19.6 14.7 9.8

4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0

4.9 0 −4.9 −9.8 −14.7 −19.6 −24.5 −29.4

44.1 m ending at a height of 78.4 m at time t  8. The velocity V(t) is positive during the upward travel, but negative while the rock falls back down. When we compute the sum V(t 1 ) Δt + V(t 2 ) Δt + " + V(t n ) Δt , part of the upward positive distance change is canceled by the negative downward movement, giving in the end an approximation of the displacement from the initial position, equal to a positive change of 78.4 m. On the other hand, if we use the speed ]V(t)], which is the absolute value of the velocity function, then distances traveled while moving up and distances traveled while moving down are both counted positively. Both the total upward motion of 122.5 m and the downward motion of 44.1 m are now counted as positive distances traveled, so the sum V(t 1 ) Δt + V(t 2 ) Δt + " + V(t n ) Δt gives us an approximation of 166.6 m, the total distance that the rock traveled from time t  0 to time t  8. As an illustration of our discussion, we subdivide the interval [0, 8] into 16 subintervals of length Δt = 1 2 and take the right endpoint of each subinterval as the value of t k. Table 5.3 shows the values of the velocity function at these endpoints. Using V(t) in the summation process, we estimate the displacement at t  8: (44.1 + 39.2 + 34.3 + 29.4 + 24.5 + 19.6 + 14.7 + 9.8 + 4.9 1 + 0 − 4.9 − 9.8 − 14.7 − 19.6 − 24.5 − 29.4) ⋅ = 58.8 2 Error magnitude = 78.4 − 58.8 = 19.6 Using ]V(t)] in the summation process, we estimate the total distance traveled over the time interval [0, 8]: + 39.2 + 34.3 + 29.4 + 24.5 + 19.6 + 14.7 + 9.8 + 4.9 1 + 0 + 4.9 + 9.8 + 14.7 + 19.6 + 24.5 + 29.4 ) ⋅ = 161.7 2 Error magnitude = 166.6 − 161.7 = 4.9

( 44.1

If we take more and more subintervals of [0, 8] in our calculations, the estimates to the heights 78.4 m and 166.6 m improve, as shown in Table 5.4.

TABLE 5.4 Travel estimates for a rock blown straight up during

the time interval [ 0, 8 ]

Number of subintervals

Length of each subinterval

16 32

12

58.8

161.7

14 18 1 16 1 32

68.6 73.5 75.95 77.175

164.15 165.375 165.9875 166.29375

1 64

77.7875

166.446875

64 128 256 512

Displacement

Total distance

Average Value of a Nonnegative Continuous Function The average value of a collection of n numbers x 1 , x 2 , ! , x n is obtained by adding them together and dividing by n. But what is the average value of a continuous function f on an interval [ a, b ]? Such a function can assume infinitely many values. For example, the temperature at a certain location in a town is a continuous function that goes up and down each day. What does it mean to say that the average temperature in the town over the course of a day is 73 degrees?

5.1  Area and Estimating with Finite Sums

y

y y = g(x)

y=c

c

0

319

a

(a)

b

c

x

0

a

(b)

b

x

(a) The average value of f ( x ) = c on [ a, b ] is the area of the rectangle divided by b − a. (b) The average value of g( x ) on [ a, b ] is the area beneath its graph divided by b − a.

FIGURE 5.6

When a function is constant, this question is easy to answer. A function with constant value c on an interval [ a, b ] has average value c. When c is positive, its graph over [ a, b ] gives a rectangle of height c. The average value of the function can then be interpreted geometrically as the area of this rectangle divided by its width b − a (see Figure 5.6a). What if we want to find the average value of a nonconstant function, such as the function g in Figure 5.6b? We can think of this graph as a snapshot of the height of some water that is sloshing around in a tank between enclosing walls at x = a and x = b. As the water moves, its height over each point changes, but its average height remains the same. To get the average height of the water, we let it settle down until it is level and its height is constant. The resulting height c equals the area under the graph of g divided by b − a. We are led to define the average value of a nonnegative function on an interval [ a, b ] to be the area under its graph divided by b − a. For this definition to be valid, we need a precise understanding of what is meant by the area under a graph. This will be obtained in Section 5.3, but for now we look at an example.

y f(x) = sin x 1

0

EXAMPLE 3 [ 0, π ]. p 2

p

x

FIGURE 5.7 Approximating the area under f ( x ) = sin x between 0 and π to compute the average value of sin x over [ 0, π ] , using eight rectangles (Example 3).

Estimate the average value of the function f ( x ) = sin x on the interval

Solution Looking at the graph of sin x between 0 and π in Figure 5.7, we can see that its average height is somewhere between 0 and 1. To find the average, we need to calculate the area A under the graph and then divide this area by the length of the interval, π − 0 = π. We do not have a simple way to determine the area, so we approximate it with finite sums. To get an upper sum approximation, we add the areas of eight rectangles of equal width π 8 that together contain the region that is beneath the graph of y = sin x and above the x-axis on [ 0, π ]. We choose the heights of the rectangles to be the largest value of sin x on each subinterval. Over a particular subinterval, this largest value may occur at the left endpoint, at the right endpoint, or somewhere between them. We evaluate sin x at this point to get the height of the rectangle for an upper sum. The sum of the rectangular areas then gives an estimate of the total area (Figure 5.7):

(

) ⋅ π8

π π 3π π π 5π 3π 7π + sin + sin + sin + sin + sin + sin + sin 8 2 2 8 4 8 8 4 π ≈ ( 0.38 + 0.71 + 0.92 + 1 + 1 + 0.92 + 0.71 + 0.38 ) ⋅ = ( 6.02 ) ⋅ 8

A ≈ sin

π ≈ 2.364. 8

To estimate the average value of sin x on [ 0, π ] we divide the estimated area by the length π of the interval and obtain the approximation 2.364 π ≈ 0.753. Since we used an upper sum to approximate the area, this estimate is greater than the actual average value of sin x over [ 0, π ]. If we use more and more rectangles, with each rectangle getting thinner and thinner, we get closer and closer to the exact average

320

Chapter 5 Integrals

TABLE 5.5 Average value of sin x

on 0 ≤ x ≤ π

Number of subintervals

Upper sum estimate

8 16 32 50 100 1000

0.75342 0.69707 0.65212 0.64657 0.64161 0.63712

value, as shown in Table 5.5. We will show in Section 5.3 that the true average value is 2 π ≈ 0.63662. As before, we could just as well have used rectangles lying under the graph of y = sin x and calculated a lower sum approximation, or we could have used the midpoint rule. In each case, the approximations are close to the true area if all the rectangles are sufficiently thin.

Summary The area under the graph of a positive function, the distance traveled by a moving object that doesn’t change direction, and the average value of a nonnegative function f over an interval can all be approximated by finite sums constructed in a certain way. First we subdivide the interval into subintervals, treating f as if it were constant over each subinterval. Then we multiply the width of each subinterval by the value of f at some point within it and add these products together. If the interval [ a, b ] is subdivided into n subintervals of equal widths Δx = ( b − a ) n, and if f (c k ) is the value of f at the chosen point c k in the kth subinterval, this process gives a finite sum of the form f (c1 ) Δx + f (c 2 ) Δx + f (c 3 ) Δx +  + f (c n ) Δx. The choices for the c k could maximize or minimize the value of f in the kth subinterval, or give some value in between. The true value lies somewhere between the approximations given by upper sums and lower sums. In the examples that we looked at, the finite sum approximations improved as we took more subintervals of smaller width.

EXERCISES

5.1

Area

Distance

In Exercises 1–4, apply finite approximations to estimate the area under the graph of the function using a. a lower sum with two rectangles of equal width. b. a lower sum with four rectangles of equal width. c. an upper sum with two rectangles of equal width. d. an upper sum with four rectangles of equal width. 1. f ( x ) =

x2

between x = 0 and x = 1.

2. f ( x ) =

x3

between x = 0 and x = 1.

3. f ( x ) = 1 x between x = 1 and x = 5. 4. f ( x ) = 4 − x 2 between x = −2 and x = 2. Using rectangles, each of whose height is given by the value of the function at the midpoint of the rectangle’s base (the midpoint rule), estimate the area under the graphs of the following functions, using first two and then four rectangles. 5. f ( x ) = x 2 between x = 0 and x = 1. 6. f ( x ) = x 3 between x = 0 and x = 1. 7. f ( x ) = 1 x between x = 1 and x = 5. 8. f ( x ) = 4 − x 2 between x = −2 and x = 2.

9. Distance traveled The accompanying table shows the velocity of a model train engine moving along a track for 10 s. Estimate the distance traveled by the engine using 10 subintervals of length 1 with a. left-endpoint values. b. right-endpoint values. Time (s)

Velocity (cms)

Time (s)

Velocity (cms)

0

0

6

11

1

12

7

6

2

22

8

2

3

10

9

6

4

5

10

0

5

13

10. Distance traveled upstream You are sitting on the bank of a tidal river watching the incoming tide carry a bottle upstream. You record the velocity of the flow every 5 minutes for an hour, with the results shown in the accompanying table. About how far upstream did the bottle travel during that hour? Find an estimate using 12 subintervals of length 5 with

5.1  Area and Estimating with Finite Sums

a. left-endpoint values.

321

kmh

b. right-endpoint values.

256 224

Time (min)

Velocity (m s)

Time (min)

Velocity (m s)

192

0

1

35

1.2

160

5

1.2

40

1.0

128

10

1.7

45

1.8

15

2.0

50

1.5

20

1.8

55

1.2

25

1.6

60

0

30

1.4

96 64 32 0

11. Length of a road You and a companion are about to drive a twisty stretch of dirt road in a car whose speedometer works but whose odometer (kilometer counter) is broken. To find out how long this particular stretch of road is, you record the car’s velocity at 10-second intervals, with the results shown in the accompanying table. Estimate the length of the road using a. left-endpoint values. b. right-endpoint values.

0.002 0.004 0.006 0.008 0.01

hours

a. Use rectangles to estimate how far the car traveled during the 36 s it took to reach 228 km/h. b. Roughly how many seconds did it take the car to reach the halfway point? About how fast was the car going then? 13. Free fall with air resistance An object is dropped straight down from a helicopter. The object falls faster and faster but its acceleration (rate of change of its velocity) decreases over time because of air resistance. The acceleration is measured in m s 2 and recorded every second after the drop for 5 s, as shown: t

0

1

2

3

4

5

a

9.8

5.944

3.605

2.187

1.326

0.805

Time (s)

Velocity (converted to m s) (36 km h = 10 m s)

Time (s)

Velocity (converted to m s) (30 km h = 10 m s)

0

0

70

5

10

15

80

7

c. Find an upper estimate for the distance fallen when t = 3.

20

5

90

12

30

12

100

15

14. Distance traveled by a projectile An object is shot straight upward from sea level with an initial velocity of 122.5 m s.

40

10

110

10

50

15

120

12

60

12

a. Find an upper estimate for the speed when t = 5. b. Find a lower estimate for the speed when t = 5.

a. Assuming that gravity is the only force acting on the object, give an upper estimate for its velocity after 5 s have elapsed. Use g = 9.8 m s 2 for the gravitational acceleration. b. Find a lower estimate for the height attained after 5 s.

12. Distance from velocity data The accompanying table gives data for the velocity of a vintage sports car accelerating from 0 to 228 km/h in 36 s (10 thousandths of an hour).

Time (h)

Velocity (km h)

Time (h)

Velocity (km h)

0

0.006

187

0.001

64

0.007

201

0.002

100

0.008

212

0.003

132

0.009

220

0.004

154

0.010

228

0.005

174

0.0

Average Value of a Function

In Exercises 15–18, use a finite sum to estimate the average value of f on the given interval by partitioning the interval into four subintervals of equal length and evaluating f at the subinterval midpoints. 15. f ( x ) = x 3 on [ 0, 2 ] 16. f ( x ) = 1 x on [ 1, 9 ] 17. f (t ) = (1 2 ) + sin 2 πt on [ 0, 2 ] y 1.5

y = 1 + sin 2 pt 2

1 0.5 0

1

2

t

322

Chapter 5 Integrals

(

T 18. f (t ) = 1 − cos

πt 4

)

4

on [ 0, 4 ]

y

4 y = 1 − acos ptb 4

1

0

1

2

3

Estimations

Aug

Sep

Oct

Nov

Dec

Pollutant release rate (tons day)

0.63

0.70

0.81

0.85

0.89

0.95

b. In the best case, approximately when will a total of 125 tons of pollutants have been released into the atmosphere?

19. Water pollution Oil is leaking out of a tanker damaged at sea. The damage to the tanker is worsening as evidenced by the increased leakage each hour, recorded in the following table. Time (h)

0

1

2

3

4

Leakage (L h)

50

70

97

136

190

Time (h)

5

6

7

8

265

369

516

720

Leakage (L h)

Jul

a. Assuming a 30-day month and that new scrubbers allow only 0.05 ton day to be released, give an upper estimate of the total tonnage of pollutants released by the end of June. What is a lower estimate?

t

4

Month

T 21. Inscribe a regular n-sided polygon inside a circle of radius 1 and compute the area of the polygon for the following values of n: a. 4 (square)

b. 8 (octagon)

c. 16

d. Compare the areas in parts (a), (b), and (c) with the area inside the circle. 22. (Continuation of Exercise 21.) a. Inscribe a regular n-sided polygon inside a circle of radius 1 and compute the area of one of the n congruent triangles formed by drawing radii to the vertices of the polygon.

a. Give an upper and a lower estimate of the total quantity of oil that has escaped after 5 hours. b. Repeat part (a) for the quantity of oil that has escaped after 8 hours.

b. Compute the limit of the area of the inscribed polygon as n → ∞.

c. The tanker continues to leak 720 L h after the first 8 hours. If the tanker originally contained 25,000 L of oil, approximately how many more hours will elapse in the worst case before all the oil has spilled? In the best case?

c. Repeat the computations in parts (a) and (b) for a circle of radius r. COMPUTER EXPLORATIONS

20. Air pollution A power plant generates electricity by burning oil. Pollutants produced as a result of the burning process are removed by scrubbers in the smokestacks. Over time, the scrubbers become less efficient and eventually they must be replaced when the amount of pollution released exceeds government standards. Measurements are taken at the end of each month determining the rate at which pollutants are released into the atmosphere, recorded as follows. Month

Jan

Feb

Mar

Apr

May

Jun

Pollutant release rate (tons day)

0.20

0.25

0.27

0.34

0.45

0.52

In Exercises 23–26, use a CAS to perform the following steps. a. Plot the functions over the given interval. b. Subdivide the interval into n = 100, 200, and 1000 subintervals of equal length and evaluate the function at the midpoint of each subinterval. c. Compute the average value of the function values generated in part (b). d. Solve the equation f ( x ) = ( average value ) for x using the average value calculated in part (c) for the n = 1000 partitioning. 23. f ( x ) = sin x on [ 0, π ] π 1 25. f ( x ) = x sin on ⎡⎢ , π ⎤⎥ ⎣4 ⎦ x

24. f ( x ) = sin 2 x on [ 0, π ] π 1 26. f ( x ) = x sin 2 on ⎡⎢ , π ⎤⎥ ⎣4 ⎦ x

5.2 Sigma Notation and Limits of Finite Sums While estimating with finite sums in Section 5.1, we encountered sums that had many terms (up to 1000 terms in Table 5.1). In this section we introduce a notation for sums that have a large number of terms. After describing this notation and its properties, we consider what happens as the number of terms in a sum approaches infinity.

Finite Sums and Sigma Notation Sigma notation enables us to write a sum with many terms in the compact form n

∑ ak k =1

= a1 + a 2 + a 3 +  + a n −1 + a n .

5.2  Sigma Notation and Limits of Finite Sums

∑ is the capital Greek letter sigma

323

The Greek letter ∑ (capital sigma, corresponding to our letter S), stands for “sum.” The index of summation k tells us where the sum begins (at the number below the ∑ symbol) and where it ends (at the number above ∑ ). Any letter can be used to denote the index, but the letters i, j, k, and n are customary. The index k ends at k = n.

n The summation symbol (Greek letter sigma)

ak a k is a formula for the kth term.

k=1

The index k starts at k = 1.

Thus we can write the sum of the squares of the numbers 1 through 11 as 1 2 + 2 2 + 3 2 + 4 2 + 5 2 + 6 2 + 7 2 + 8 2 + 9 2 + 10 2 + 11 2 =

11

∑ k 2, k =1

and the sum of f (i) for integers i from 1 to 100 as f (1) + f (2) + f (3) +  + f (100) =

100

∑ f (i). i =1

The starting index does not have to be 1; it can be any integer. EXAMPLE 1 A sum in sigma notation 5

∑k

The sum written out, one term for each value of k

The value of the sum

1+ 2+3+ 4 + 5

15

( − 1)1 (1)

−1 + 2 − 3 = −2

k =1 3

∑ ( − 1) k k

+ (−1) 2 ( 2 ) + (−1)3 ( 3 )

k =1 2

1 2 + 1+1 2+1

1 2 7 + = 2 3 6

∑ k k− 1

42 52 + 4 −1 5−1

16 25 139 + = 3 4 12

EXAMPLE 2

Express the sum 1 + 3 + 5 + 7 + 9 in sigma notation.

∑ k +k 1 k =1 5

2

k=4

Solution The formula generating the terms depends on what we choose the lower limit of summation to be, but the terms generated remain the same. It is often simplest to choose the starting index to be k = 0 or k = 1, but we can start with any integer.

Starting   with   k = 0:

1+3+ 5+ 7+ 9 =

4

∑ ( 2k + 1) k=0

Starting   with   k = 1:

1+3+ 5+ 7+ 9 =

5

∑ ( 2k − 1) k =1

Starting   with   k = 2:

1+3+ 5+ 7+ 9 =

6

∑ ( 2k − 3 ) k=2

Starting   with   k = −3: 1 + 3 + 5 + 7 + 9 =

1

∑ ( 2k + 7 )

k =−3

324

Chapter 5 Integrals

When we have a sum such as 3

∑ ( k + k 2 ), k =1

we can rearrange its terms to form two sums: 3

∑ ( k + k 2 ) = (1 + 1 2 ) + ( 2 + 2 2 ) + ( 3 + 3 2 ) k =1

= (1 + 2 + 3 ) + (1 2 + 2 2 + 3 2 ) =

3

3

k =1

k =1

Regroup terms.

∑ k + ∑ k 2.

This illustrates a general rule for finite sums: n

∑ (ak k =1

+ bk ) =

n

∑ ak

+

k =1

n

∑ bk . k =1

This and three other rules are given below. Proofs of these rules can be obtained using mathematical induction (see Appendix A.3).

Algebra Rules for Finite Sums n

∑ (ak

1. Sum Rule:

k =1 n

∑ (ak

2. Difference Rule:

k =1 n

∑ ca k

3. Constant Multiple Rule:

k =1 n

∑c

4. Constant Value Rule:

n

∑ ak

+ bk ) =

+

k =1 n

∑ ak

− bk ) =

n

∑ bk k =1

−

k =1

n

∑ bk k =1

n

= c ⋅ ∑ ak

(Any number c)

k =1

= n⋅c

(Any number c)

k =1

EXAMPLE 3

We demonstrate the use of the algebra rules.

n

(a)

k =1 n

(b) ∑ (−a k ) = k =1

HISTORICAL BIOGRAPHY Carl Friedrich Gauss (1777–1855) Gauss was born in Brunswick, Germany. The list of Gauss's accomplishments in science and mathematics is astonishing, ranging from the invention of the electric telegraph (with Wilhelm Weber in 1833) to the development of a theory of planetary orbits and the development of an accurate theory of non-Euclidean geometry. To know more, visit the companion Website.

(c)

n

n

k =1

k =1

∑ ( 3k − k 2 ) = 3 ∑ k − ∑ k 2 n

∑ ( − 1) ⋅ a k k =1

3

3

3

k =1

k =1

k =1

Difference Rule and Constant Multiple Rule n

n

k =1

k =1

= −1 ⋅ ∑ a k = −∑ a k

∑ ( k + 4) = ∑ k + ∑ 4 = (1 + 2 + 3 ) + ( 3 ⋅ 4 )

Constant Multiple Rule

Sum Rule Constant Value Rule

= 6 + 12 = 18 n

(d)

∑ 1n k =1

= n⋅

1 =1 n

Constant Value Rule (1 n is constant )

Over the years, people have discovered a variety of formulas for the values of finite sums. The most famous of these are the formula for the sum of the first n positive integers (Gauss is said to have discovered it at age 8) and the formulas for the sums of the squares and cubes of the first n positive integers.

5.2  Sigma Notation and Limits of Finite Sums

325

Show that the sum of the first n positive integers is

EXAMPLE 4

n

∑k

n ( n + 1) . 2

=

k =1

Solution The formula tells us that the sum of the first 4 positive integers is ( 4 )( 5 )

2

= 10.

Addition verifies this prediction: 1 + 2 + 3 + 4 = 10. To prove the formula in general, we write out the terms in the sum twice, once forward and once backward. 1 + n +

(n

2 + − 1) +

(n

3 + " + n − 2) + " + 1

If we add the two terms in the first column we get 1 + n = n + 1. Similarly, if we add the two terms in the second column we get 2 + ( n − 1) = n + 1. The two terms in any column sum to n 1. When we add the n columns together we get n terms, each equal to n 1, for a total of n ( n + 1). Since this is twice the desired quantity, the sum of the first n integers is n ( n + 1) 2. Formulas for the sums of the squares and cubes of the first n positive integers are proved using mathematical induction (see Appendix A.3). We state them here. n

Sum of the first  n  squares:

∑ k2

=

n ( n + 1)( 2n + 1) 6

=

( n( n 2+ 1) )

k =1 n

Sum of the first  n  cubes:

∑ k3 k =1

2

Limits of Finite Sums The finite sum approximations that we considered in Section 5.1 became more accurate as the number of terms increased and the subinterval widths (lengths) narrowed. The next example shows how to calculate a limiting value as the widths of the subintervals go to zero and the number of subintervals grows to infinity. EXAMPLE 5 Find the limiting value of lower sum approximations to the area of the region R below the graph of y = 1 − x 2 and above the interval [ 0, 1 ] on the x-axis using equal-width rectangles whose widths approach zero and whose number approaches infinity. (See Figure 5.4a.) Solution We compute a lower sum approximation using n rectangles of equal width %x and then see what happens as n → ∞. We start by subdividing [ 0, 1 ] into n equal width subintervals

⎡ 0, 1 ⎤ , ⎡ 1 , 2 ⎤ , ! , ⎡ n − 1 , n ⎤ . ⎢⎣ n ⎦⎥ ⎢⎣ n n ⎥⎦ ⎢⎣ n n ⎥⎦ Each subinterval has width Δx = 1 n . The function 1  x 2 is decreasing on [ 0, 1 ], and its smallest value in a subinterval occurs at the subinterval’s right endpoint. So a lower sum is constructed with rectangles whose height over the subinterval [ ( k − 1) n, k n ] is f ( k n ) = 1 − ( k n ) 2, giving the sum f

( 1n ) ⋅ 1n + f ( n2 ) ⋅ 1n + " + f ( nk ) ⋅ 1n + " + f ( nn ) ⋅ 1n .

326

Chapter 5 Integrals

We write this in sigma notation and simplify, n

∑

k =1

f

( )

k 1 ⋅ = n n = =

⎛

∑ ⎜⎜⎝1 − ( nk ) n

k =1 n

2

⎞⎟ 1 ⎟⎟⎠ n

∑ ( n − n3 ) k =1 n

k2

1

n

1

k2

∑ n − ∑ n3 k =1

= n⋅

Difference Rule

k =1

1 1 − 3 n n

( )

n

∑ k2

Constant Value and Constant Multiple Rules

k =1

1 n ( n + 1)( 2n + 1) = 1− 3 6 n 3 2 2n + 3n + n . = 1− 6n 3

Sum of the first  n  squares Numerator expanded

We have obtained an expression for the lower sum that holds for any n. Taking the limit of this expression as n → ∞, we see that the lower sums converge as the number of subintervals increases and the subinterval widths approach zero:

(

lim 1 −

n →∞

)

2n 3 + 3n 2 + n 2 2 = 1− = . 6n 3 6 3

The lower sum approximations converge to 2 3. A similar calculation shows that the upper sum approximations also converge to 2 3. Any finite sum approximation n ∑ k =1 f (c k )(1 n ) also converges to the same value, 2 3. This is because it is possible to show that any finite sum approximation is trapped between the lower and upper sum approximations. For this reason we are led to define the area of the region R as this limiting value. In Section 5.3 we study the limits of such finite approximations in a general setting. HISTORICAL BIOGRAPHY Georg Friedrich Bernhard Riemann (1826–1866) Riemann was born in Hanover, Germany. His doctorate was obtained under the direction of Gauss in the theory of complex variables. He also worked with physicist Wilhelm Weber. He introduced the foundational ideas of differential geometry and contributed to dynamics, non-Euclidean geometry, and computational physics.

Riemann Sums The theory of limits of finite approximations was made precise by the German mathematician Bernhard Riemann. We now introduce the notion of a Riemann sum, which underlies the theory of the definite integral that will be presented in the next section. We begin with an arbitrary bounded function f defined on a closed interval [ a, b ]. Like the function pictured in Figure 5.8, f may have negative as well as positive values. We subdivide the interval [ a, b ] into subintervals, not necessarily of equal width (or length), and form sums in the same way as for the finite approximations in Section 5.1. To do so, we choose n  1 points { x 1 , x 2 , x 3 , ! , x n 1} between a and b that are in increasing order, so that

To know more, visit the companion Website.

a < x 1 < x 2 < " < x n −1 < b.

y

To make the notation consistent, we set x 0  a and x n  b , so that

y = f(x)

0 a

a = x 0 < x 1 < x 2 < " < x n −1 < x n = b. b

x

The set of all of these points, P = { x 0 , x 1 , x 2 , ! , x n −1 , x n},

FIGURE 5.8 A typical continuous function y  f ( x ) over a closed interval [ a, b ].

is called a partition of [ a, b ]. The partition P divides [ a, b ] into the n closed subintervals

[ x 0 , x 1 ], [ x 1 , x 2 ], ! , [ x n −1 , x n ].

327

5.2  Sigma Notation and Limits of Finite Sums

HISTORICAL BIOGRAPHY Richard Dedekind (1831–1916) Dedekind grew up in Germany and in 1850 entered the University of Gottingen.There he studied with Bernhard Riemann and Carl Gauss. Like Gauss, Dedekind preferred to study the theoretical aspects of number theory. His work on irrational numbers gave the subject a logical foundation.

The first of these subintervals is [ x 0 , x 1 ], the second is [ x 1 , x 2 ], and the kth subinterval is [ x k −1 , x k ] (where k is an integer between 1 and n). kth subinterval x0 = a

x1

...

x2

x k−1

x xk

...

xn = b

x n−1

The width of the first subinterval [ x 0 , x 1 ] is denoted %x 1 , the width of the second [ x 1 , x 2 ] is %x 2 , and the width of the kth subinterval is Δx k = x k − x k −1. Δ x1

To know more, visit the companion Website.

x0 = a

Δ x2 x1

Δ xn

Δ xk ...

x2

x k−1

xk

x

...

xn−1

xn = b

If all n subintervals have equal width, then their common width, which we call %x, is equal to ( b − a ) n. Using equal width subintervals is often the simplest choice when doing computations. In each subinterval we select some point. The point chosen in the kth subinterval [ x k −1 , x k ] is called c k . Then on each subinterval, we stand a vertical rectangle that stretches from the x-axis to touch the curve at (c k , f (c k )). These rectangles can be above or below the x-axis, depending on whether f (c k ) is positive or negative, or on the x-axis if f (c k )  0 (see Figure 5.9). y

y = f (x)

(cn, f (cn ))

(ck, f(ck )) kth rectangle

c1 0 x0 = a

c2 x1

x2

x k−1

ck xk

cn x n−1

xn = b

x

(c1, f (c1))

(c 2, f (c 2))

FIGURE 5.9 The rectangles approximate the region between the graph of the function y  f ( x ) and the x-axis. Figure 5.8 has been repeated and enlarged, the partition of [ a, b ] and the points c k have been added, and the corresponding rectangles with heights f (c k ) are shown.

On each subinterval we form the product f (c k ) ⋅ Δx k . This product is positive, negative, or zero, depending on the sign of f (c k ). When f (c k )  0, the product f (c k ) ⋅ Δx k is the area of a rectangle with height f (c k ) and width %x k . When f (c k )  0, the product f (c k ) ⋅ Δx k is a negative number, the negative of the area of a rectangle of width %x k that drops from the x-axis to the negative number f (c k ). Finally, we sum all these products to get SP =

n

∑ f (c k ) Δx k . k =1

The sum S P is called a Riemann sum for  f  on the interval [ a, b ]. There are many such sums, depending on the partition P we choose and on the choices of the points c k in the

328

Chapter 5 Integrals

subintervals. For instance, we could choose n subintervals all having equal width Δx = ( b − a ) n to partition [ a, b ], and then choose the point c k to be the right-hand endpoint of each subinterval when forming the Riemann sum (as we did in Example 5). This choice leads to the Riemann sum formula Sn = y

y = f(x)

0 a

x

b

Δx1 0

0 a

x

b

(b)

FIGURE 5.10 The curve of Figure 5.9 with rectangles from finer partitions of [ a, b ]. Finer partitions create collections of rectangles with thinner bases that approximate the region between the graph of f and the x-axis with increasing accuracy.

Δ x2 0.2

Δ x3 0.6

1

6k

2.

6

k −1 k k =1

∑

∑ (−2 ) k −1

4.

k =1

a.

∑ sin kπ

3

k =1

∑ (−1) k cos kπ k =1

7. Which of the following express 1 + 2 + 4 + 8 + 16 + 32 in sigma notation? 6

5

b.

∑ 2k

k=0

∑ (−1) k 2 k

3

c.

k=0

∑ (−1) k +1 2 k + 2

k =−2

( − 1) k −1

k −1

2

b.

∑ k−+1 1 (

)k

1

c.

k=0

∑

( − 1) k

k =−1 k

+2

10. Which formula is not equivalent to the other two? 4

4

6.

∑

k=2

k =1

∑ (−1) k +1 sin πk

5

b.

k =1

5

∑ cos kπ

k =1

x

2

9. Which formula is not equivalent to the other two? 4

4

∑ 2 k −1

1.5

8. Which of the following express 1 − 2 + 4 − 8 + 16 − 32 in sigma notation? a.

3

∑k +1

a.

Δ x5

Δ x4

5.2

k =1

5.

)

Any Riemann sum associated with a partition of a closed interval [ a, b ] defines rectangles that approximate the region between the graph of a continuous function f and the x-axis. Partitions with norm approaching zero lead to collections of rectangles that approximate this region with increasing accuracy, as suggested by Figure 5.10. We will see in the next section that if the function f is continuous over the closed interval [ a, b ], then no matter how we choose the partition P and the points c k in its subintervals, the Riemann sums corresponding to these choices will approach a single limiting value as the subinterval widths (which are controlled by the norm of the partition) approach zero.

Write the sums in Exercises 1–6 without sigma notation. Then evaluate them.

3.

)(

The lengths of the subintervals are Δx 1 = 0.2, Δx 2 = 0.4, Δx 3 = 0.4, Δx 4 = 0.5, and Δx 5 = 0.5. The longest subinterval length is 0.5, so the norm of the partition is P = 0.5. In this example, there are two subintervals of this length.

Sigma Notation

1.

k =1

b−a b−a ⋅ . n n

Similar formulas can be obtained if instead we choose c k to be the left-hand endpoint, or the midpoint, of each subinterval. In the cases in which the subintervals all have equal width Δx = ( b − a ) n, we can make them thinner by simply increasing their number n. When a partition has subintervals of varying widths, we can ensure they are all thin by controlling the width of a widest (longest) subinterval. We define the norm of a partition P, written P , to be the largest of all the subinterval widths. If P is a small number, then all of the subintervals in the partition P have a small width.

y = f(x)

2

n

EXAMPLE 6 The set P = { 0, 0.2, 0.6, 1, 1.5, 2 } is a partition of [ 0, 2 ]. There are five subintervals of P: [ 0, 0.2 ], [ 0.2, 0.6 ], [ 0.6, 1 ], [ 1, 1.5 ], and [ 1.5, 2 ]:

(a) y

EXERCISES

∑ f (a + k

4

c.

∑ 2 k +1

k =−1

a.

∑ ( k − 1)2 k =1

−1

3

b.

∑ ( k + 1)2

k =−1

c.

∑ k2

k =−3

Express the sums in Exercises 11–16 in sigma notation. The form of your answer will depend on your choice for the starting index. 11. 1 + 2 + 3 + 4 + 5 + 6 12. 1 + 4 + 9 + 16 1 1 1 1 14. 2 + 4 + 6 + 8 + 10 13. + + + 2 4 8 16

5.3  The Definite Integral

15. 1 −

1 1 1 1 + − + 2 3 4 5

1 2 3 4 5 + − + − 5 5 5 5 5

16. −

n

k =1

k =1

17. Suppose that ∑ a k = −5 and ∑ b k = 6. Find the values of n

n

∑

b.

k =1 n

∑ (ak

d.

k =1

k =1

− bk )

n

bk 6

c.

k =1

n

∑ ( bk

e.

∑ (ak

n

k =1

k =1

n

b.

n

c.

∑ (ak

∑ 250b k k =1 n

+ 1)

d.

k =1

∑ ( bk

− 1)

k =1

10

10

∑k

b.

k =1

20. a.

c.

k =1

13

b.

k =1

∑

c.

k =1

∑ (−2 k ) k =1

5

27.

3

⎛

k =1

26.

5

k =1

∑3

500

b.

k =1

∑k

17

b.

k=9

∑4 k =1

∑ k2 ∑c k =1

38. f ( x ) = −x 2 ,

[ 0, 1 ]

[ −π , π ] [ −π , π ]

7

3

∑ k4 k =1

46. f ( x ) = 3 x 2 over the interval [ 0, 1 ]. 47. f ( x ) = x + x 2 over the interval [ 0, 1 ].

264

48. f ( x ) = 3 x + 2 x 2 over the interval [ 0, 1 ].

k =3

49. f ( x ) = 2 x 3 over the interval [ 0, 1 ].

∑ 10 ∑ k ( k − 1)

50. f ( x ) = x 2 − x 3 over the interval [ −1, 0 ].

k =18

n

b.

[ 0, 2 ]

45. f ( x ) = x 2 + 1 over the interval [ 0, 3 ].

71

c.

k =3

n

31. a.

c.

k =1

36

30. a.

∑7

In Exercises 37–40, graph each function f ( x ) over the given interval. Partition the interval into four subintervals of equal length. Then add to your sketch the rectangles associated with the Riemann sum 4 ∑ k =1 f (c k ) Δx k , given that c k is the (a) left-hand endpoint, (b) righthand endpoint, (c) midpoint of the kth subinterval. (Make a separate sketch for each set of rectangles.)

44. f ( x ) = 2 x over the interval [ 0, 3 ].

∑ k ( 2k + 1)

⎛ 7 ⎞2 28. ⎜⎜ ∑ k ⎟⎟⎟ − ⎜⎝ k =1 ⎠

⎞3

Riemann Sums

43. f ( x ) = 1 − x 2 over the interval [ 0, 1 ].

k =1

7

29. a.

∑ ( k 2 − 5) 7

k + ⎜⎜ ∑ k ⎟⎟⎟ ∑ 225 ⎜⎝ ⎠

)

For the functions in Exercises 43–50, find a formula for the Riemann sum obtained by dividing the interval [ a, b ] into n equal subintervals and using the right-hand endpoint for each c k . Then take a limit of these sums as n → ∞ to calculate the area under the curve over [ a, b ].

k =1

5

k =1

1 1 − k k +1

Limits of Riemann Sums

πk ∑ 15 6

24.

∑ k ( 3k + 5 )

=

41. Find the norm of the partition P = { 0, 1.2, 1.5, 2.3, 2.6, 3 }.

k =1

6

25.

)

k =1

22.

∑ (3 − k 2 )

(

42. Find the norm of the partition P = { −2, −1.6, −0.5, 0, 0.8, 1 }.

k3

5

k =1

23.

∑

∑ [ sin( k − 1) − sin k ] k=2

( Hint:  k k 1+ 1

40. f ( x ) = sin x + 1,

13

k2

7

21.

∑ k3 k =1

13

∑k

∑ k ( k 1+ 1)

39. f ( x ) = sin x ,

10

∑ k2

20

34.

k − 3)

k−4−

37. f ( x ) = x 2 − 1,

Evaluate the sums in Exercises 19–36. 19. a.

∑( k=7

36.

n

k =1

35.

k

∑ n2 k =1

k =1

k =1

n

∑ 8a k

c.

k =1

∑ [ ( k + 1) 2 − k 2 ]

40

− 2a k )

k =1

33.

30

+ bk )

18. Suppose that ∑ a k = 0 and ∑ b k = 1. Find the values of a.

n

c

∑n

50

n

∑ 3a k

n

b.

k =1

Values of Finite Sums

a.

∑ ( 1n + 2n ) n

32. a.

329

n

c.

∑ ( k − 1) k =1

5.3 The Definite Integral In this section we consider the limit of general Riemann sums as the norm of the partitions of a closed interval [ a, b ] approaches zero. This limiting process leads us to the definition of the definite integral of a function over a closed interval [ a, b ].

Definition of the Definite Integral The definition of the definite integral is based on the fact that for some functions, as the norm of the partitions of [ a, b ] approaches zero, the values of the corresponding Riemann

330

Chapter 5 Integrals

sums approach a limiting value J. In particular, this is true for continuous and piecewisecontinuous functions. We again use the symbol ε to represent a small positive number, and use it to specify how close to J the Riemann sum must be. The symbol δ is used for a second small positive number that specifies how small the norm of a partition must be in order for the Riemann sum to differ from J by no more than ε. We now define this limit precisely.

DEFINITION Let f ( x ) be a function defined on a closed interval [ a, b ]. We say that a number J is the definite integral of f over [ a, b ] and that J is the limit of n the Riemann sums ∑ k =1 f (c k ) Δx k if the following condition is satisfied: Given any number ε > 0, there is a corresponding number δ > 0 such that for every partition P = { x 0 , x 1 , … , x n } of [ a, b ] with P < δ and any choice of c k in [ x k −1 , x k ], we have n

∑ f (c k ) Δx k

− J < ε.

k =1

The definition involves a limiting process in which the norm of the partition goes to zero. When this limit exists, the function f is said to be integrable over [ a, b ]. We have many choices for a partition P with norm going to zero, and many choices of points c k for each partition. The definite integral exists when we always get the same limit J, no matter what choices are made. When the limit exists we write n

∑ f (c k ) Δx k , P →0

J = lim

k =1

and we say that the definite integral exists. Leibniz introduced a notation for the definite integral that captures its construction as n a limit of Riemann sums. He envisioned the finite sums ∑ k =1 f (c k ) Δx k becoming an infinite sum of function values f ( x ) multiplied by “infinitesimal” subinterval widths dx. The sum symbol ∑ is replaced in the limit by the integral symbol ∫ , whose origin is in the letter “S” (for sum). The function values f (c k ) are replaced by a continuous selection of function values f ( x ). The subinterval widths Δx k become the differential dx. It is as if we were summing all products of the form f ( x ) ⋅ dx as x goes from a to b. While this notation captures the underlying ideas, it is Riemann’s definition that gives a precise meaning to the definite integral. If the definite integral exists, then instead of writing J, we write b

∫a

f ( x ) dx.

We read this as “the integral from a to b of f of x dee x” or sometimes as “the integral from a to b of f of x with respect to x.” The component parts in the integral symbol also have names: The function f (x) is the integrand. Upper limit of integration

b

Integral sign

La

x is the variable of integration.

f(x) dx

Lower limit of integration Integral of f from a to b

When you find the value of the integral, you have evaluated the integral.

331

5.3  The Definite Integral

When the definite integral exists, we say that the Riemann sums of f on [ a, b ] converge b to the definite integral J = ∫a f ( x ) dx and that f is integrable over [ a, b ]. If we choose all the subintervals in a partition to have equal width Δx = ( b − a ) n, the Riemann sums have the form Sn =

n

∑ f (c k ) Δx k

=

k =1

∑ f (c k ) ( b −n a ), n

Δx k = Δx = ( b − a ) n  for all  k

k =1

where c k is any point in the kth subinterval. If the definite integral exists, then these Riemann sums converge to the definite integral of f over [ a, b ], so J =

b

∫a

∑ f (c k ) ( n →∞

f ( x ) dx = lim

n

k =1

)

b−a . n

For equal-width subintervals, the condition  P → 0 is equivalent to n → ∞.

If we pick the point c k to be the right endpoint of the kth subinterval, so that c k = a + k Δx = a + k ( b − a ) n , then the formula for the definite integral becomes The Definite Integral as a Limit of Riemann Sums with Equal-Width Subintervals b

∫a

f ( x ) dx = lim

n →∞

∑ f (a + k b n

k =1

−a n

)( b −n a )

(1)

Equation (1) gives an explicit formula that can be used to compute definite integrals. When the definite integral exists, the Riemann sums coming from other choices of partitions and locations of points c k will have the same limit as n → ∞, provided that the norms of the partitions approach zero. Another choice of Riemann sums converging to the definite integral is given by the Midpoint Rule, discussed later in this section. The value of the definite integral of a function over any particular interval depends on the function, not on the letter we choose to represent its independent variable. If we decide to use t or u instead of x, we simply write the integral as b

∫a

f (t ) dt

or

b

∫a

f (u) du

instead of

b

∫a

f ( x ) dx.

No matter how we write the integral, it is still the same number, the limit of the Riemann sums as the norm of the partition approaches zero. Since it does not matter what letter we use, the variable of integration is called a dummy variable. In the three integrals given above, the dummy variables are t, u, and x.

Integrable and Nonintegrable Functions Not every function defined over a closed interval [ a, b ] is integrable even if the function is bounded. That is, the Riemann sums for some functions might not converge to the same limiting value, or to any value at all. Understanding which functions defined over [ a, b ] are integrable and which are not requires advanced mathematical analysis, but fortunately most functions that commonly occur in applications are integrable. In particular, every continuous function over [ a, b ] is integrable over this interval, and so is every function that has no more than a finite number of jump discontinuities on [ a, b ]. (See Figures 1.9 and 1.10. Such functions are called piecewise continuous functions, and they are defined in Additional Exercises 11–18 at the end of this chapter.) The following theorem, which is proved in more advanced courses, establishes these results. If a function f is continuous over the interval [ a, b ], or if f has at most finitely many jump discontinuib ties there, then the definite integral ∫a f ( x ) dx exists and f is integrable over [ a, b ].

THEOREM 1—Continuous Functions Are Integrable

332

Chapter 5 Integrals

The idea behind Theorem 1 for continuous functions is given in Exercises 86 and 87. Briefly, when f is continuous, we can choose each c k so that f (c k ) gives the maximum value of f on the subinterval [ x k −1 , x k ], resulting in an upper sum. Likewise, we can choose c k to give the minimum value of f on [ x k −1 , x k ] to obtain a lower sum. The upper and lower sums can be shown to converge to the same limiting value as the norm of the partition P tends to zero. Moreover, every Riemann sum is trapped between the values of the upper and lower sums, so every Riemann sum converges to the same limit as well. Therefore, the number J in the definition of the definite integral exists, and the continuous function f is integrable over [ a, b ]. For integrability to fail, a function needs to be sufficiently discontinuous that the region between its graph and the x-axis cannot be approximated well by increasingly thin rectangles. Our first example is a function that is not integrable over a closed interval. EXAMPLE 1

The function ⎪⎧ 1, if  x  is rational, f ( x ) = ⎪⎨ ⎪⎪ 0, if  x  is irrational, ⎩

has no Riemann integral over [ 0, 1 ]. Underlying this is the fact that between any two numbers there are both a rational number and an irrational number. Thus the function jumps up and down too erratically over [ 0, 1 ] to allow the region beneath its graph and above the x-axis to be approximated by rectangles, no matter how thin they are. In fact, we will show that upper sum approximations and lower sum approximations converge to different limiting values. If we choose a partition P of [ 0, 1 ], then the lengths of the intervals in the partition sum n to 1; that is, ∑ k =1 Δx k = 1. In each subinterval [ x k −1 , x k ] there is a rational point, say c k. Because c k is rational, f (c k ) = 1. Since 1 is the maximum value that f can take anywhere, the upper sum approximation for this choice of c k ’s is U =

n

∑ f (c k ) Δx k

=

k =1

n

∑ ( 1) Δ x k

= 1.

k =1

As the norm of the partition approaches 0, these upper sum approximations converge to 1 (because each approximation is equal to 1). On the other hand, we could pick the c k ’s differently and get a different result. Each subinterval [ x k −1 , x k ] also contains an irrational point c k , and for this choice f (c k ) = 0. Since 0 is the minimum value that f can take anywhere, this choice of c k gives us the minimum value of f on the subinterval. The corresponding lower sum approximation is L =

n

∑ f (c k ) Δx k k =1

=

n

∑ ( 0 ) Δx k

= 0.

k =1

These lower sum approximations converge to 0 as the norm of the partition converges to 0 (because they each equal 0). Thus making different choices for the points c k results in different limits for the corresponding Riemann sums. We conclude that the definite integral of f over the interval [ 0, 1 ] does not exist and that f is not integrable over [ 0, 1 ]. Theorem 1 says nothing about how to calculate definite integrals. A method of calculation will be developed in Section 5.4, through a connection of definite integrals to antiderivatives. Meanwhile, finite approximations can be used to calculate an approximation for a definite integral.

The Midpoint Rule When setting up a Riemann sum n

∑ f (c k ) Δx k k =1

333

5.3  The Definite Integral

we have many options on how to choose a partition and where to choose a point c k in the kth interval of the partition. The simplest choice for a partition with n subintervals is to take each subinterval to have equal length Δx = ( b − a ) n. In Equation (1) we obtained a formula by choosing c k at the right endpoint of the kth interval. Setting c k to be at the middle of the kth interval is often a better choice for approximating the integral using Riemann sums, as we saw in Figure 5.3. The midpoint is a good choice because on a small interval, the graph of a differentiable function can be approximated by a linear function, and the value of a linear function at an interval’s midpoint is its average value over that interval. With these choices, [ x k −1 , x k ] = [ a + ( k − 1) Δx , a + k Δx ] is the kth interval in the partition, its midpoint is located at c k = ( x k −1 + x k ) 2 , and its width is Δx = ( b − a ) n. The midpoint rule for forming Riemann sums to approximate a definite integral takes the following form. Midpoint Rule for Approximating a Definite Integral b

∫a

f ( x ) dx ≈

∑ f (c k ) ( n

k =1

)

(

b−a b−a = [ f (c1 ) + f (c 2 ) +  + f (c n ) ] n n

)

x k −1 + x k b−a . and x k = a + k 2 n

(

with c k =

EXAMPLE 2 of equal length.

Approximate ∫

0

1

)

1 dx using the midpoint rule with five intervals 1 + x2

Solution The five intervals of the partition are [ 0, 0.2 ], [ 0.2, 0.4 ], [ 0.4, 0.6 ], [ 0.6, 0.8 ], and [ 0.8, 1.0 ], each of length 1 5. Their midpoints are at 0.1, 0.3, 0.5, 0.7, and 0.9. Applying the midpoint rule gives 1

1

∫0 1 + x 2 dx

( )

1 ≈ [ f (0.1) + f (0.3) + f (0.5) + f (0.7) + f (0.9) ] 5 1 1 1 1 1 ⎤( ) ⎡ 0.2 = ⎢ + + + + ⎣ 1.01 1.09 1.25 1.49 1.81 ⎥⎦ ≈ 0.786.

The midpoint rule becomes more powerful when we can combine it with an error bound that tells us how close the approximation it gives with n intervals is to the integral. In Chapter 8 we will examine error bounds in approximations of integrals.

Properties of Definite Integrals In defining ∫a f ( x ) dx as a limit of sums ∑ k =1 f (c k ) Δx k , we moved from left to right across the interval [ a, b ]. What would happen if we instead move right to left, starting with x 0 = b and ending at x n = a? Each Δx k in the Riemann sum would change its sign, with x k − x k −1 now negative instead of positive. With the same choices of c k in each subinterval, the sign of any Riemann sum would change, as would the sign of the limit, the a integral ∫b f ( x ) dx. Since we have not previously given a meaning to integrating backward, we are led to define n

b

a

∫b

b

f ( x ) dx = −∫ f ( x ) dx. a

a  and  b  interchanged

It is convenient to have a definition for the integral over [ a, b ] when a = b, so that we are computing an integral over an interval of zero width. Since a = b gives Δx = 0, whenever f (a) exists we define a

∫a

f ( x ) dx = 0.

a is both the lower and the upper limit of integration.

334

Chapter 5 Integrals

TABLE 5.6   Rules satisfied by definite integrals a

b

1. Order of Integration: ∫ f ( x ) dx = −∫ f ( x ) dx b

A definition

a

a

2. Zero Width Interval: ∫ f ( x ) dx = 0

A definition when  f ( a ) exists

a

b

b

3. Constant Multiple: ∫ k f ( x ) dx = k ∫ f ( x ) dx a

b

b

4. Sum and Difference: ∫ ( f ( x ) ± g( x ) ) dx =

∫a

a

c

5. Additivity: ∫ f ( x ) dx + a

Any constant  k

a

∫c

b

b

∫a

f ( x ) dx =

b

∫a

f ( x ) dx ±

f ( x ) dx

g( x ) dx

c ] ∪ [ c, b ] = [ a, b ]

[ a,

6. Max-Min Inequality: If f has maximum value max f and minimum value min f on [ a, b ], then

( min f ) ⋅ ( b − a ) ≤

b

∫a

f ( x ) dx ≤ ( max f ) ⋅ ( b − a ). b

7. Domination: If f ( x ) ≥ g( x ) on [ a, b ], then ∫ f ( x ) dx ≥ a

b

∫a

g( x ) dx.

b

If f ( x ) ≥ 0 on  [ a, b ],  then  ∫ f ( x ) dx ≥ 0. a

Special case of Rule 6.

Theorem 2 states some basic properties of integrals, including the two just discussed. These properties, listed in Table 5.6, are very useful for computing integrals. We will refer to them repeatedly to simplify our calculations. Rules 2 through 7 have geometric interpretations, which are shown in Figure 5.11. The graphs in these figures show only positive functions, but the rules apply to general integrable functions, which could take both positive and negative values. When f and g are integrable over the interval [ a, b ], the definite integral satisfies the rules listed in Table 5.6.

THEOREM 2

Rules 1 and 2 are definitions, but Rules 3 to 7 of Table 5.6 must be proved. Below we give a proof of Rule 6. Similar proofs can be given to verify the other properties in Table 5.6. Proof of Rule 6 Rule 6 says that the integral of f over [ a, b ] is never smaller than the minimum value of f times the length of the interval and never larger than the maximum value of f times the length of the interval. The reason is that for every partition of [ a, b ] and for every choice of the points c k , ( min f ) ⋅ ( b

n

− a ) = ( min f ) ⋅ ∑ Δx k k =1

=

n

∑ Δx k

= b−a

k =1

n

∑ ( min f ) ⋅ Δx k

Constant Multiple Rule

k =1

≤

n

∑ f (c k ) Δx k

min f ≤ f (c k )

k =1

≤

n

∑ ( max f ) ⋅ Δx k

f (c k ) ≤ max f

k =1

n

= ( max f ) ⋅ ∑ Δx k k =1

= ( max f ) ⋅ ( b − a ).

Constant Multiple Rule

335

5.3  The Definite Integral

y

y

y

y = 2 f (x)

y = f (x) + g(x)

y = f (x)

y = g(x) y = f (x) y = f (x) x

a

0

∫a

b

b

∫a

f ( x ) dx = 0

y

L c

L a a

f (x) dx

f ( x ) dx +

∫c

f ( x ) dx =

b

∫a

g( x ) dx

y = f (x)

b

b

∫a

f ( x ) dx

x

y = g(x)

0 a

x

b

0 a

(e) Max-Min Inequality:

( min f ) · ( b − a ) ≤

b

∫a

x

b

(f) Domination: If f ( x ) ≥ g( x ) on [ a, b ], then

f ( x ) dx

b

∫a

≤ ( max f ) · ( b − a ) FIGURE 5.11

f ( x ) dx +

y

min f

c

b

b

∫a

=

y = f (x)

(d) Additivity for Definite Integrals:

∫a

∫a ( f ( x ) + g( x ) ) dx

a

max f

c

c

b

k f ( x ) dx = k ∫ f ( x ) dx

b

f (x) dx

x

b

(c) Sum: (areas add)

b

y y = f (x)

0

0 a

(b) Constant Multiple: ( k = 2)

(a) Zero Width Interval: a

a

0

x

f ( x ) dx ≥

b

∫a

g( x ) dx.

Geometric interpretations of Rules 2–7 in Table 5.6.

In short, all Riemann sums for f on [ a, b ] satisfy the inequalities n

( min

f ) ⋅ ( b − a ) ≤ ∑ f (c k ) Δx k ≤ ( max f ) ⋅ ( b − a ). k =1

Hence their limit, which is the integral, satisfies the same inequalities. EXAMPLE 3

To illustrate some of the rules, we suppose that

1

∫−1 f ( x ) dx

∫1

= 5,

4

f ( x ) dx = −2,

and

1

∫−1 h ( x ) dx

= 7.

Then 1

4

1.

∫4

f ( x ) dx = −∫ f ( x ) dx = −(−2) = 2

2.

∫−1 [ 2 f ( x ) + 3h( x ) ] dx

Rule 1

1

1

= 2∫

1

−1

1

f ( x ) dx + 3 ∫

−1

h( x ) dx

Rules 3 and 4

= 2( 5 ) + 3( 7 ) = 31 3.

4

∫−1 f ( x ) dx

EXAMPLE 4

=

1

∫−1 f ( x ) dx + ∫1

4

f ( x ) dx = 5 + ( −2 ) = 3

Show that the value of ∫

1 0

Rule 5

1 + cos x dx is less than or equal to 2.

Solution The Max-Min Inequality for definite integrals (Rule 6) says that b

( min f ) ⋅ ( b − a ) is a lower bound for the value of ∫ f ( x ) dx and that ( max f ) ⋅ ( b − a ) a

is an upper bound. The maximum value of 1 + cos x on [ 0, 1 ] is 1 + 1 = 1

∫0

1 + cos x dx ≤

2 ⋅ (1 − 0 ) =

2.

2, so

336

Chapter 5 Integrals

Area Under the Graph of a Nonnegative Function We now return to the problem that started this chapter, which is defining what we mean by the area of a region having a curved boundary. In Section 5.1 we approximated the area under the graph of a nonnegative continuous function using several types of finite sums of areas of rectangles that approximate the region—upper sums, lower sums, and sums using the midpoints of each subinterval—all of which are Riemann sums constructed in special ways. Theorem 1 guarantees that all of these Riemann sums converge to a single definite integral as the norm of the partitions approaches zero and the number of subintervals goes to infinity. As a result, we can now define the area under the graph of a nonnegative integrable function to be the value of that definite integral. DEFINITION If y = f ( x ) is nonnegative and integrable over a closed interval [ a, b ], then the area under the curve y = f ( x ) over [ a, b ] is the integral of f from a to b,

A =

b

∫a

f ( x ) dx.

For the first time, we have a rigorous definition for the area of a region whose boundary is the graph of a continuous function. We now apply this to a simple example, the area under a straight line, and we verify that our new definition agrees with our previous notion of area. EXAMPLE 5 [ 0, b ], b > 0.

y=x

P →0

b

is a triangle.

x dx and find the area A under y = x over the interval

(a) To compute the definite integral as the limit of Riemann sums, we calculate n lim ∑ k =1 f (c k ) Δx k for partitions whose norms go to zero. Theorem 1 tells us

b

FIGURE 5.12

b

∫0

Solution The region of interest is a triangle (Figure 5.12). We compute the area in two ways.

y

0

Compute

b

x

The region in Example 5

that it does not matter how we choose the partitions or the points c k as long as the norms approach zero. All choices give the exact same limit. So we consider the partition P that subdivides the interval [ 0, b ] into n subintervals of equal width Δx = ( b − 0 ) n = b n, and we choose c k to be the right endpoint in each subinterval kb b 2b 3b nb . So , , , as in formula (1). The partition is P = 0, , and c k = n n n n n

{

n

∑ f (c k ) Δx

=

k =1

=

}

n

∑ kbn ⋅ bn k =1 n

f (c k ) = c k

2

∑ kbn2 k =1

b2 = 2 n

n

∑k

Constant Multiple Rule

k =1

b 2 n ( n + 1) ⋅ n2 2 2 b 1 = 1+ . 2 n =

(

As n → ∞ and Therefore,

Sum of first  n  integers

)

P → 0, this last expression on the right has the limit b 2 2. b

b2 . 2 (b) Since the area equals the definite integral for a nonnegative function, we can quickly derive the definite integral by using the formula for the area of a triangle having base

∫0

x dx =

337

5.3  The Definite Integral

length b and height y = b. The area is A = (1 2 ) b ⋅ b = b 2 2. Again we conclude b that ∫ 0 x dx = b 2 2.

y b

Example 5 can be generalized to integrate f ( x ) = x over any closed interval [ a, b ] for which 0 < a < b . First write

y=x b a

b

∫0

a a

0

b

b−a

x

x dx =

a

∫0

x dx +

x dx

Rule 5

Then, by rearranging this equation and applying Example 5, we obtain b b a b2 a2 Example 5 ∫a x dx = ∫0 x dx − ∫0 x dx = 2 − 2 . In conclusion, we have the following rule for integrating f ( x ) = x over the interval [ a, b ]:

(a) y

b

a

b

∫a

∫a

b

b2 a2 − , 2 2

x dx =

a < b

(2)

x

0

This computation gives the area of the trapezoid in Figure 5.13a. Equation (2) remains valid when a and b are negative, but the interpretation of the definite integral changes. When a < b < 0, the definite integral value ( b 2 − a 2 ) 2 is a negative number, the negative of the area of a trapezoid dropping down to the line y = x below the x-axis (Figure 5.13b). When a < 0 and b > 0, Equation (2) is still valid and the definite integral gives the difference between two areas, the area under the graph and above [ 0, b ] minus the area below [ a, 0 ] and over the graph (Figure 5.13c). The following results can also be established by using a Riemann sum calculation similar to the one we used in Example 5 (Exercises 63 and 65).

y=x (b) y y=x a b

0

x

b

∫a c dx b

∫a

= c( b − a ),

x 2 dx =

c  any constant

(3)

a < b

(4)

b3 a3 − , 3 3

(c)

FIGURE 5.13 (a) The area of this trapezoidal region is A = ( b 2 − a 2 ) 2. (b) The definite integral in Equation (2) gives the negative of the area of this trapezoidal region. (c) The definite integral in Equation (2) gives the area of the blue triangular region added to the negative of the area of the tan triangular region.

y y = f (x) (ck, f(ck )) x1 0 x0 = a

ck

x xn = b

Average Value of a Continuous Function Revisited In Section 5.1 we informally introduced the average value of a nonnegative continuous function f over an interval [ a, b ], leading us to define this average as the area under the graph of y = f ( x ) divided by b − a. In integral notation we write this as b 1 f ( x ) dx. ∫ b−a a This formula gives us a precise definition of the average value of a continuous (or integrable) function, whether it is positive, negative, or both. Alternatively, we can justify this formula through the following reasoning. We start with the idea from arithmetic that the average of n numbers is their sum divided by n. A continuous function f on [ a, b ] may have infinitely many values, but we can still sample them in an orderly way. We divide [ a, b ] into n subintervals of equal width Δx = ( b − a ) n and evaluate f at a point c k in each (Figure 5.14). The average of the n sampled values is

Average =

f (c1 ) + f (c 2 ) +  + f (c n ) 1 = n n =

FIGURE 5.14 A sample of values of a function on an interval [ a, b ].

=

n

∑

f (c k )

Δx b−a

∑

f (c k )

Δx =

1 b−a

∑

f (c k ) Δ x.

Constant Multiple Rule

k =1

n

k =1 n

k =1

b−a 1 Δx , so = b−a n n

338

Chapter 5 Integrals

The average of the samples is obtained by dividing a Riemann sum for f on [ a, b ] by − a ). As we increase the number of samples and let the norm of the partition approach b zero, the average approaches (1 ( b − a )) ∫ a f ( x ) dx. Both points of view lead us to the following definition.

(b

If f is integrable on [ a, b ], then its average value on [ a, b ], also called its mean, is

DEFINITION

av( f ) =

EXAMPLE 6

1 b−a

b

∫a

f ( x ) dx.

Find the average value of f ( x ) = x on [ 1, 3 ]. 3

Solution From Equation (2) we see that ∫ x dx = 9 2 − 1 2 = 4. 1

So the average value of f on the interval [ 1, 3 ] is (1 2 )( 4 ) = 2. y 2 2 f(x) = "4 − x

y=p 2

1

−2

EXAMPLE 7

−1

1

2

x

4 − x 2 on [ −2, 2 ].

Solution We recognize f ( x ) = 4 − x 2 as the function whose graph is the upper semicircle of radius 2 centered at the origin (Figure 5.15). Since we know the area inside a circle, we do not need to take the limit of Riemann sums. The area between the semicircle and the x-axis from −2 to 2 can be computed using the geometry formula

FIGURE 5.15

The average value of f ( x ) = 4 − x 2 on [ −2,2 ] is π 2 (Example 7). The area of the rectangle shown here is 4 ⋅ ( π 2 ) = 2π , which is also the area of the semicircle.

Find the average value of f ( x ) =

Area =

1 1 ⋅ πr 2 = ⋅ π ( 2 ) 2 = 2π. 2 2

Because f is nonnegative, the area is also the value of the integral of f from −2 to 2, 2

∫−2

4 − x 2 dx = 2π.

Therefore, the average value of f is av( f ) =

2 1 2 − ( −2 ) ∫−2

4 − x 2 dx =

1( ) π 2π = . 4 2

Notice that the average value of f over [ −2, 2 ] is the same as the height of a rectangle over [ −2, 2 ] whose area equals the area of the upper semicircle (see Figure 5.15).

5.3

EXERCISES

Interpreting Limits of Sums as Integrals

Express the limits in Exercises 1–8 as definite integrals.

n

∑ P →0

6. lim

k =1

n

1. lim

P →0

∑ c k2 Δx k , where P is a partition of [ 0, 2 ] k =1

n

∑ ( sec c k ) Δx k , where P is a partition of [ − π 4 , 0 ] P →0

7. lim

k =1

n

2. lim

P →0

∑ 2c k3 Δx k , where P is a partition of [ −1, 0 ] k =1 n

3. lim

P →0

∑

k =1

− 3c k ) Δx k , where P is a partition of [ −7, 5 ]

⎛ ⎞ ∑ ⎜⎜⎜⎝ c1 ⎟⎟⎟⎠ Δx k , where P is a partition of [ 1, 4 ] →0 k k =1 n

n

5. lim

P →0

∑ 1 −1 c k =1

n

∑( tan c k ) Δx k , where P is a partition of [ 0, π 4 ] P →0

8. lim

k =1

( c k2

4. lim P

4 − c k2 Δx k , where P is a partition of [ 0, 1 ]

Δx k , where P is a partition of [ 2, 3 ] k

Using the Definite Integral Rules

9. Suppose that f and g are integrable and that

∫1

2

f ( x ) dx = −4,

∫1

5

f ( x ) dx = 6,

∫1

5

g( x ) dx = 8.

5.3  The Definite Integral

Use the rules in Table 5.6 to find 2

a.

∫2 g( x ) dx

c.

∫1

2

3 f ( x ) dx

5

∫1 [ f ( x ) − g( x ) ] dx

e.

27. f ( x ) = 1

b.

∫5 g( x ) dx

d.

∫2

f.

5

28. f ( x ) = 3 x +

∫1

f ( x ) dx = −1,

9

5

∫1 [ 4 f ( x ) − g( x ) ] dx

∫7

9

∫7

f ( x ) dx = 5,

h( x ) dx = 4.

Use the rules in Table 5.6 to find 9

∫1 −2 f ( x ) dx

a.

9

∫7 [ 2 f ( x ) − 3h( x ) ] dx

c.

7

∫1

e.

f ( x ) dx

11. Suppose that 2

∫1

a.

1

∫2

c.

2 ∫1

b. d. f.

b.

f (t ) dt

d.

12. Suppose that −3

f ( x ) dx

7

∫ 9 [ h( x ) −

f ( x ) ] dx

f ( x ) dx = 5. Find

f (u) du

0 ∫−3

∫7 [ f ( x ) + h( x ) ] dx ∫9

g(t ) dt =

∫1

2

∫1

2

3 f ( z ) dz [ − f ( x ) ] dx

2. Find

∫0

g(t ) dt

b.

∫−3 g(u) du

c.

∫−3 [ −g( x ) ] dx

d.

∫−3

g(r ) dr 2

0

3

13. Suppose that f is integrable and that ∫0 f ( z ) dz = 3 and 4 ∫ 0 f ( z ) dz = 7. Find 4

∫3

a.

f ( z ) dz

b.

3

∫4

∫1

3

h ( r ) dr

1

b. −∫ h ( u ) du 3

Using Known Areas to Find Integrals

In Exercises 15–22, graph the integrands and use known area formulas to evaluate the integrals. 15.

∫−2 (

17.

∫−3

19.

∫−2

21.

4

3

1

)

32

x + 3 dx 2

16.

∫1 2

9 − x 2 dx

18.

∫−4

20.

∫−1 (1 −

x dx

1

∫−1 ( 2 −

x ) dx

22.

0

(−2 x

+ 4 ) dx

16 − x 2 dx

1

1

∫−1 (1 +

2

32.

∫

5 2

35.

∫0

38.

∫a

23. 25.

∫0

b

∫a

x dx , 2

b > 0

24.

2s ds,

0 < a < b

26.

b

∫0 4 x dx , b

∫a

3t dt ,

a. [ −1, 0 ], b. [ −1, 1 ]

2 12

3

x dx

30.

2.5

∫0.5

r dr

33.

∫0

t 2 dt

36.

∫0

x dx

39.

∫0

3

7

π2

3

b

b > 0 0 < a < b

31.

∫π

x 2 dx

34.

∫0

θ 2 dθ

37.

∫a

x 2 dx

40.

∫0

1

θ dθ

0.3

2a

3b

s 2 ds x dx

x 2 dx

2

42.

∫0 5 x dx

∫0 ( 2t − 3) dt

44.

∫0

45.

∫2 (1 + 2 ) dz

46.

∫3 ( 2z − 3) dz

47.

∫1

48.

∫1 2 24u 2 du

49.

∫0 ( 3x 2 + x − 5 ) dx

50.

∫1 ( 3x 2 + x − 5 ) dx

41.

∫3 7 dx

43.

2

1

2

z

3u 2 du

2

2

( t − 2 ) dt

0

1

0

Finding Area by Definite Integrals

In Exercises 51–54, use a definite integral to find the area of the region between the given curve and the x-axis on the interval [ 0, b ]. 52. y = π x 2 x 54. y = + 1 2

51. y = 3 x 2

Finding Average Value

In Exercises 55–62, graph the function and find its average value over the given interval. x2 on [ 0, 3 ] 55. f ( x ) = x 2 − 1 on [ 0, 3 ] 56. f ( x ) = − 2 57. f ( x ) = −3 x 2 − 1 on [ 0, 1 ] 58. f ( x ) = 3 x 2 − 3 on [ 0, 1 ] 59. f (t ) = ( t − 1) 2 on [0, 3]

60. f (t ) = t 2 − t on [ −2, 1 ]

61. g( x ) = x − 1 on a. [ −1, 1 ], b. [ 1, 3 ], and c. [ −1, 3 ] 62. h( x ) = − x on a. [ −1, 0 ], b. [ 0, 1 ], and c. [ −1, 1 ] Definite Integrals as Limits of Sums

Use the method of Example 5a or Equation (1) to evaluate the definite integrals in Exercises 63–70.

x ) dx 1 − x 2 ) dx

2π

x dx

Use the rules in Table 5.6 and Equations (2)–(4) to evaluate the integrals in Exercises 41–50.

b

63.

∫a c dx

65.

∫a

67.

∫−1 ( 3x 2 − 2 x + 1) dx

69.

∫a

Use known area formulas to evaluate the integrals in Exercises 23–28. b

on

53. y = 2 x

f (t ) dt

14. Suppose that h is integrable and that ∫−11 h ( r ) dr = 0 and 3 ∫−1 h ( r ) dr = 6. Find a.

∫1

29.

0

a.

0

1 − x2

Use the results of Equations (2) and (4) to evaluate the integrals in Exercises 29–40.

9

1

a. [ −2, 2 ], b. [ 0, 2 ]

on

Evaluating Definite Integrals

f ( x ) dx

10. Suppose that f and h are integrable and that 9

4 − x2

339

b

x 2 dx ,

a < b

2

b

x 3 dx ,

a < b

2

64.

∫0 ( 2 x + 1) dx

66.

∫−1 ( x − x 2 ) dx

68.

∫−1 x 3 dx

70.

∫0 ( 3x − x 3 ) dx

0

1

1

340

Chapter 5 Integrals

Theory and Examples

71. What values of a and b, with a < b, maximize the value of b

∫a ( x − x 2 ) dx ? (Hint: Where is the integrand positive?) 72. What values of a and b, with a < b, minimize the value of

b. Suppose that instead of being equal, the lengths Δx k of the subintervals of the partition of [ a, b ] vary in size. Show that

b

∫a ( x 4 − 2 x 2 ) dx ? 73. Use the Max-Min Inequality to find upper and lower bounds for the value of 1 1 ∫0 1 + x 2 dx. 74. (Continuation of Exercise 73.) Use the Max-Min Inequality to find upper and lower bounds for 0.5

∫0

1 dx 1 + x2

figure that the difference between the upper and lower sums for f on this partition can be represented graphically as the area of a rectangle R whose dimensions are [ f (b) − f (a) ] by Δx. (Hint: The difference U − L is the sum of areas of rectangles whose diagonals Q 0 Q1 , Q1Q 2 , … , Q n−1Q n lie approximately along the curve. There is no overlapping when these rectangles are shifted horizontally onto R.)

1

U − L ≤ f (b) − f (a) Δx max , where Δx max is the norm of P, and that hence lim (U − L ) = 0. P →0

y

1

∫0.5 1 + x 2 dx.

and

y = f (x)

Add these to arrive at an improved estimate of 1

∫0 75. Show that the value of

1 ∫ 0 sin( x 2 )

76. Show that the value of and 3.

1 ∫0

on

on

[ a, b ]

Use the Max-Min Inequality

[ a, b ]

b

∫a

⇒

f ( x ) dx ≥ 0.

Show that if f is integrable, b

∫a

⇒

f ( x ) dx ≤ 0.

79. Use the inequality sin x ≤ x , which holds for x ≥ 0, to find an 1 upper bound for the value of ∫0 sin x dx. 80. The inequality sec x ≥ 1 + ( x 2 2 ) holds on (−π 2, π 2 ). Use it 1 to find a lower bound for the value of ∫ 0 sec x dx. 81. If av( f ) really is a typical value of the integrable function f ( x ) on [ a, b ], then the constant function av( f ) should have the same integral over [ a, b ] as f. Does it? That is, does b

∫a av ( f ) dx

=

b

∫a

f ( x ) dx ?

Q2 Δx

0 x 0 = a x1 x 2

82. It would be nice if average values of integrable functions obeyed the following rules on an interval [ a, b ]. a. av ( f + g ) = av( f ) + av( g) b. av( k f ) = k av( f )

(any number  k )

c. av( f ) ≤ av( g)

f ( x ) ≤ g( x )

on

x

xn = b

84. Upper and lower sums for decreasing functions of Exercise 83.)

(Continuation

a. Draw a figure like the one in Exercise 83 for a continuous function f ( x ) whose values decrease steadily as x moves from left to right across the interval [ a, b ]. Let P be a partition of [ a, b ] into subintervals of equal length. Find an expression for U − L that is analogous to the one you found for U − L in Exercise 83a. b. Suppose that instead of being equal, the lengths Δx k of the subintervals of P vary in size. Show that the inequality U − L ≤ f (b) − f (a) Δx max of Exercise 83b still holds and hence lim (U − L ) = 0. P →0

85. Use the formula sin h + sin 2h + sin 3h +  + sin kh =

Give reasons for your answer.

if

Q1

R

x + 8 dx lies between 2 2 ≈ 2.8

78. Integrals of nonpositive functions then f (x) ≤ 0

Q3

dx cannot possibly be 2.

77. Integrals of nonnegative functions to show that if f is integrable, then f (x) ≥ 0

f (b) − f(a)

1 dx. 1 + x2

cos( h 2 ) − cos(( k + (1 2 )) h ) 2 sin ( h 2 )

to find the area under the curve y = sin x from x = 0 to x = π 2 in two steps: a. Partition the interval [ 0, π 2 ] into n subintervals of equal length and calculate the corresponding upper sum U; then

[ a, b ].

Do these rules ever hold? Give reasons for your answers. 83. Upper and lower sums for increasing functions a. Suppose the graph of a continuous function f ( x ) rises steadily as x moves from left to right across an interval [ a, b ]. Let P be a partition of [ a, b ] into n subintervals of equal length Δx = ( b − a ) n. Show by referring to the accompanying

b. Find the limit of U as n → ∞ and Δx = ( b − a ) n → 0. 86. Suppose that f is continuous and nonnegative over [ a, b ], as in the accompanying figure. By inserting points x 1 , x 2 , … , x k −1 , x k , … , x n−1 as shown, divide [ a, b ] into n subintervals of lengths Δx 1 = x 1 − a, Δx 2 = x 2 − x 1 , … , Δx n = b − x n−1 , which need not be equal.

341

5.3  The Definite Integral

a. If m k = min { f ( x ) for x in the kth subinterval } , explain the connection between the lower sum L = m 1 Δx 1 + m 2 Δx 2 +  + m n Δx n

88. If you average 48 km h on a 240-km trip and then return over the same 240 km at the rate of 80 km h, what is your average speed for the trip? Give reasons for your answer. 89. Integrals of functions that are equal except at one point Suppose that f ( x ) is a continuous function over the interval [ a, b ] and that g( x ) is a function on [ a, b ] such that g( x ) = f ( x ) except at a single point c ∈ [ a, b ]. Show that g( x ) is also inteb b grable over [ a, b ] and that ∫a g( x ) dx = ∫a f ( x ) dx . (Hint: For a given n, by how much can two different Riemann sums as given in Equation (1) differ?)

and the shaded regions in the first part of the figure. b. If M k = max { f ( x ) for x in the k th subinterval } , explain the connection between the upper sum U = M 1 Δx 1 + M 2 Δx 2 +  + M n Δx n and the shaded regions in the second part of the figure. c. Explain the connection between U − L and the shaded regions along the curve in the third part of the figure.

90. Some integrable functions that are not continuous a. The floor function f ( x ) = ⎣ x ⎦ gives the greatest integer smaller than or equal to x (Example 5 of Section 1.1). This function is not continuous on the interval [ 1, 3 ]. Show that f 3 is integrable on [ 1, 3 ] and that ∫1 ⎣ x ⎦ dx = 3 . ⎧⎪ −1 if x < 0, b. The function f ( x ) = ⎪⎨ is not continuous on ⎪⎪⎩ 2 if 0 ≤ x , the interval [ −1, 1 ]. Show that f is integrable on [ −1, 1 ] and 1 that ∫−1 f ( x ) dx = 1.

y y = f (x)

a

0

x1

x2

x3

x k−1 x k

x n−1

b

x

If your CAS can draw rectangles associated with Riemann sums, use it to draw rectangles associated with Riemann sums that converge to the integrals in Exercises 91–96. Use n = 4, 10, 20, and 50 subintervals of equal length in each case.

y

a

0

xk

x k+1

b

COMPUTER EXPLORATIONS

x

1

91.

∫0 (1 − x ) dx

93.

∫−π cos x dx

95.

∫−1

96.

∫1

π

1

2

=

= 0

1 2

1

92.

∫0 ( x 2 + 1) dx

94.

∫0

π/4

=

4 3

sec 2 x dx = 1

x dx = 1 1 dx x

(The integral’s value is about 0.693.)

In Exercises 97–104, use a CAS to perform the following steps:

y

a. Plot the functions over the given interval. b. Partition the interval into n = 100, 200, and 1000 subintervals of equal length, and evaluate the function at the midpoint of each subinterval. c. Compute the average value of the function values generated in part (b).

a

0

xk

x k+1

b

e b−a

87. We say f is uniformly continuous on [ a, b ] if, given any ε > 0, there is a δ > 0 such that if x 1 , x 2 are in [ a, b ] and x 1 − x 2 < δ , then f ( x 1 ) − f ( x 2 ) < ε. It can be shown that a continuous function on [ a, b ] is uniformly continuous. Use this and the figure for Exercise 86 to show that if f is continuous and ε > 0 is given, it is possible to make U − L ≤ ε ⋅ (b − a) by making the largest of the Δx k’s sufficiently small.

x

d. Solve the equation f ( x ) = ( average value ) for x using the average value calculated in part (c) for the n = 1000 partitioning. 97. f ( x ) = sin x on [ 0, π ] 98. f ( x ) = sin 2 x 1 99. f ( x ) = x sin x 100. f ( x ) = x sin 2

1 x

on [ 0, π ] π on ⎡⎢ , π ⎤⎥ ⎣4 ⎦ π on ⎡⎢ , π ⎤⎥ ⎣4 ⎦

101. f ( x ) = x e −x

on [ 0, 1 ]

102. f ( x ) =

e −x 2

on [ 0, 1 ]

103. f ( x ) =

ln x x

on [ 2, 5 ]

104. f ( x ) =

1 1 − x2

1 on ⎢⎡ 0, ⎤⎥ ⎣ 2⎦

342

Chapter 5 Integrals

5.4 The Fundamental Theorem of Calculus HISTORICAL BIOGRAPHY Sir Isaac Newton (1642–1727) In his youth in England, Newton was interested in mechanical devices and their underlying theories. He even constructed lanterns and windmills that he designed. During the 1670s and 1680s, he built his reputation as a scientific genius. His contributions included the theory of universal gravitation, the laws of motion, methods of calculus, and the composition of white light. To know more, visit the companion Website.

y y = f(x)

f(c), average height 0

a

b

c b−a

x

FIGURE 5.16 The value f (c) in the Mean Value Theorem is, in a sense, the average (or mean) height of f on [ a, b ]. When f ≥ 0, the area of the rectangle is the area under the graph of f from a to b,

f (c)( b − a ) =

y

b

∫a

0

In the previous section we defined the average value of a continuous function over a closed b interval [ a, b ] to be the definite integral ∫a f ( x ) dx divided by the length or width b − a of the interval. The Mean Value Theorem for Definite Integrals asserts that this average value is always taken on at least once by the function f in the interval. The graph in Figure 5.16 shows a positive continuous function y = f ( x ) defined over the interval [ a, b ]. Geometrically, the Mean Value Theorem says that there is a number c in [ a, b ] such that the rectangle with height equal to the average value f (c) of the function and base width b − a has exactly the same area as the region beneath the graph of f from a to b.

If f is continuous on [ a, b ], then at some point c in [ a, b ], f (c) =

1 b−a

b

∫a

f ( x ) dx.

f ( x ) dx.

Proof If we divide all three expressions in the Max-Min Inequality (Table 5.6, Rule 6) by ( b − a ) , we obtain b 1 min f ≤ f ( x ) dx ≤ max f . b − a ∫a Since f is continuous, the Intermediate Value Theorem for Continuous Functions (Section 2.6) says that f must assume every value between min f and max f. It must therefore assume the b value (1 ( b − a )) ∫a f ( x ) dx at some point c in [ a, b ].

y = f(x) Average value 12 is not assumed by f 1

Mean Value Theorem for Definite Integrals

THEOREM 3—The Mean Value Theorem for Definite Integrals

1 1 2

In this section we present the Fundamental Theorem of Calculus, which is the central theorem of integral calculus. It connects integration and differentiation, enabling us to compute integrals by using an antiderivative of the integrand function, rather than by taking limits of Riemann sums as we did in Section 5.3. Leibniz and Newton exploited this relationship and started mathematical developments that fueled the scientific revolution for the next 200 years. Along the way, we will present an integral version of the Mean Value Theorem, which is another important theorem of integral calculus and is used to prove the Fundamental Theorem. We also find that the net change of a function over an interval is the integral of its rate of change, as suggested by Example 2 in Section 5.1.

2

x

The continuity of f is important here. It is possible for a discontinuous function to never equal its average value (Figure 5.17). EXAMPLE 1

Show that if f is continuous on [ a, b ], a ≠ b, and if b

∫a

FIGURE 5.17

A discontinuous function need not assume its average value.

f ( x ) dx = 0,

then f ( x ) = 0 at least once in [ a, b ]. Solution The average value of f on [ a, b ] is

av( f ) =

1 b−a

b

∫a

f ( x ) dx =

1 ⋅ 0 = 0. b−a

5.4  The Fundamental Theorem of Calculus

y 8

343

By the Mean Value Theorem for Definite Integrals, f assumes this value at some point c ∈ [ a, b ]. This is illustrated in Figure 5.18 for the function f ( x ) = 9 x 2 − 16 x + 4 on the interval [ 0, 2 ].

y = f(x) = 9x 2 − 16x + 4

6 4

Fundamental Theorem, Part 1

2 0 −2

c

c

1

2

x

−4

FIGURE 5.18

The function f ( x ) = 9 x 2 − 16 x + 4 satisfies 2 ∫ 0 f ( x ) dx = 0 , and there are two values of c in the interval [ 0, 2 ] where f (c) = 0.

F (x) =

area = F(x)

y

It can be very difficult to compute definite integrals by taking the limit of Riemann sums. We now develop a powerful new method for evaluating definite integrals, based on using antiderivatives. This method combines the two strands of calculus. One strand involves the idea of taking the limits of finite sums to obtain a definite integral, and the other strand contains derivatives and antiderivatives. They come together in the Fundamental Theorem of Calculus. We begin by considering how to differentiate a certain type of function that is described as an integral. If f (t ) is an integrable function over a finite interval I, then the integral from any fixed number a ∈ I to another number x ∈ I defines a new function F whose value at x is

y = f (t)

a

0

x

b

t

FIGURE 5.19 The function F ( x ) defined by Equation (1) gives the area under the graph of f from a to x when f is nonnegative and x > a.

x

∫a

f (t ) dt.

(1)

For example, if f is nonnegative and x lies to the right of a, then F ( x ) is the area under the graph from a to x (Figure 5.19). The variable x is the upper limit of integration of an integral, but F is just like any other real-valued function of a real variable. For each value of the input x, there is a single numerical output, in this case the definite integral of f from a to x. Equation (1) gives a useful way to define new functions (as we will see in Section 7.1), but its key importance is the connection that it makes between integrals and derivatives. If f is a continuous function, then the Fundamental Theorem asserts that F is a differentiable function of x whose derivative is f itself. That is, at each x in the interval [ a, b ] we have F ′( x ) = f ( x ). To gain some insight into why this holds, we look at the geometry behind it. If f ≥ 0 on [ a, b ], then to compute F ′( x ) from the definition of the derivative, we must take the limit as h → 0 of the difference quotient F ( x + h ) − F (x) . h

y

If h > 0, then F ( x + h ) is the area under the graph of f from a to x + h, while F ( x ) is the area under the graph of f from a to x. Subtracting the two gives us the area under the graph of f between x and x + h (see Figure 5.20). As shown in Figure 5.20, if h is small, the area under the graph of f from x to x + h is approximated by the area of the rectangle whose height is f ( x ) and whose base is the interval [ x , x + h ]. That is,

y = f (t)

f(x) 0

a

FIGURE 5.20

x x+h

b

In Equation (1), F ( x ) is the area to the left of x. Also, F ( x + h ) is the area to the left of x + h. The difference quotient [ F ( x + h ) − F ( x ) ] h is then approximately equal to f ( x ), the height of the rectangle shown here.

t

F ( x + h ) − F ( x ) ≈ h f ( x ). Dividing both sides by h, we see that the value of the difference quotient is very close to the value of f ( x ): F ( x + h ) − F (x) ≈ f ( x ). h This approximation improves as h approaches 0. It is reasonable to expect that F ′( x ), which is the limit of this difference quotient as h → 0, equals f ( x ), so that F ′( x ) = lim

h→0

F ( x + h ) − F (x) = f ( x ). h

This equation is true even if the function f is not positive, and it forms the first part of the Fundamental Theorem of Calculus.

344

Chapter 5 Integrals

THEOREM 4—The Fundamental Theorem of Calculus, Part 1 x

If f is continuous on [ a, b ], then F ( x ) = ∫a f (t ) dt is continuous on [ a, b ] and differentiable on ( a, b ), and its derivative is f ( x ): F ′( x ) =

d dx

x

∫a

f (t ) dt = f ( x ).

(2)

Before proving Theorem 4, we look at several examples to gain an understanding of what it says. In each of these examples, notice that the independent variable x appears in either the upper or the lower limit of integration (either as part of a formula or by itself). The independent variable on which y depends in these examples is x, while t is merely a dummy variable in the integral. Use the Fundamental Theorem to find dy dx if

EXAMPLE 2 (a) y = (c) y =

x

∫a ( t 3 + 1) dt ∫1

x2

cos t dt

5

(b) y =

∫ x 3t sin t dt

(d) y =

∫1+ 3 x

4

2

1 dt 2 + et

Solution We calculate the derivatives with respect to the independent variable x.

(a)

d dy = dx dx

∫a ( t 3 + 1) dt

(b)

d dy = dx dx

∫ x 3t sin t dt

x

5

= x3 + 1

Eq.  ( 2)  with  f ( t ) = t 3 + 1

(

x d −∫ 3t sin t dt 5 dx x d = − ∫ 3t sin t dt dx 5

=

= −3 x sin x

)

Table 5.6, Rule 1

Eq.  ( 2)  with  f ( t ) = 3t sin t

(c) The upper limit of integration is not x but x 2 . This makes y a composition of the two functions u

y = ∫ cos t dt 1

u = x 2.

and

We must therefore apply the Chain Rule to find dy dx : dy dy du = ⋅ dx du dx u d du cos t dt ⋅ = ∫ 1 du dx du = cos u ⋅ dx = (cos x 2 ) ⋅ 2 x

(

= 2 x cos x 2

)

Eq.  (2)  with  f ( t ) = cos t

5.4  The Fundamental Theorem of Calculus

(d) d dx

4

∫1+ 3 x

2

1+ 3 x 2 ⎞ d ⎛⎜ 1 1 = − dt dt ⎟⎟ ⎜ ∫ t t ⎝ ⎠ 4 2+e dx 2+e

d dx

Table 5.6, Rule 1

1+ 3 x 2

1 dt 2 + et 1 d (1 + 3 x 2 ) = − 2 + e (1+ 3 x 2 ) dx 6x = − 2 + e (1+ 3 x 2 ) = −

345

∫4

Eq. (2) and the Chain Rule

Proof of Theorem 4 We prove the Fundamental Theorem, Part 1, by applying the definition of the derivative directly to the function F ( x ), when x and x + h are in ( a, b ). This means writing out the difference quotient F ( x + h ) − F (x) h

(3)

and showing that its limit as h → 0 is the number f ( x ). Doing so, we find that F ′( x ) = lim

F ( x + h ) − F (x) h

= lim

1 ⎡ x+h f (t ) dt − ⎢ h ⎣ ∫a

= lim

1 h

h→0

h→0

h→0

∫x

x+h

x

∫a

f (t ) dt ⎤⎥ ⎦

f (t ) dt.

Table 5.6,  Rule 5

According to the Mean Value Theorem for Definite Integrals, there is some point c between x and x + h where f (c) equals the average value of f on the interval [ x , x + h ]. That is, there is some number c in [ x , x + h ] such that 1 h

∫x

x+h

f (t ) dt = f (c).

(4)

As h → 0, x + h approaches x, which forces c to approach x also (because c is trapped between x and x + h). Since f is continuous at x, f (c) therefore approaches f ( x ): lim f (c) = f ( x ).

(5)

h→0

Hence we have shown that, for any x in ( a, b ), F ′( x ) = lim

h→0

1 h

∫x

x+h

f (t ) dt

= lim f (c)

Eq. (4)

= f ( x ),

Eq. (5)

h→0

and therefore F is differentiable at x. Since differentiability implies continuity, this also shows that F is continuous on the open interval ( a, b ). To complete the proof, we just have to show that F is also continuous at x = a and x = b. To do this, we make a very similar argument, except that at x = a we need only consider the one-sided limit as h → 0 + , and similarly at x = b we need only consider h → 0 −. This shows that F has a one-sided derivative at x = a and at x = b, and therefore Theorem 1 in Section 3.2 implies that F is continuous at those two points.

346

Chapter 5 Integrals

Fundamental Theorem, Part 2 (The Evaluation Theorem) We now come to the second part of the Fundamental Theorem of Calculus. This part describes how to evaluate definite integrals without having to calculate limits of Riemann sums. Instead we find and evaluate an antiderivative at the upper and lower limits of integration.

THEOREM 4 (Continued)—The Fundamental Theorem of Calculus, Part 2

If f is continuous over [ a, b ] and F is any antiderivative of f on [ a, b ], then b

∫a

f ( x ) dx = F (b) − F (a).

Part 1 of the Fundamental Theorem tells us that an antiderivative of f exists, namely

Proof

G(x) =

x

∫a

f (t ) dt.

Thus, if F is any antiderivative of f, then F ( x ) = G ( x ) + C for some constant C for a < x < b (by Corollary 2 of the Mean Value Theorem for Derivatives, Section 4.2). Since both F and G are continuous on [ a, b ], we see that the equality F ( x ) = G ( x ) + C also holds when x = a and x = b by taking one-sided limits (as x → a + and x → b − ). Evaluating F (b) − F (a), we have F (b) − F (a) = [ G (b) + C ] − [ G (a) + C ] = G (b) − G (a) b

=

∫a

=

∫a

=

∫a

b

b

f (t ) dt −

a

∫a

f (t ) dt

f (t ) dt − 0 f (t ) dt.

The Evaluation Theorem is important because it says that to calculate the definite integral of f over an interval [ a, b ] we need do only two things: 1. Find an antiderivative F of f, and b 2. Calculate the number F (b) − F (a), which is equal to ∫a f ( x ) dx. This process is much easier than computing Riemann sums and finding their limit. The power of the theorem follows from the realization that the definite integral, which is defined by a complicated process involving all of the values of the function f over [ a, b ], can be found by knowing the values of any antiderivative F at only the two endpoints a and b. The usual notation for the difference F (b) − F (a) is b

b

⎤ ⎡ ⎤ F (x) ⎥ or ⎢ F (x) ⎥ , ⎦a ⎦a ⎣ depending on whether F has one or more terms. EXAMPLE 3 We calculate several definite integrals using the Evaluation Theorem, rather than by taking limits of Riemann sums. (a)

π

∫0

π

cos x dx = sin x ]

0

= sin π − sin 0 = 0 − 0 = 0

d dx sin x = cos x

347

5.4  The Fundamental Theorem of Calculus

(b)

⎤0 sec x tan x dx = sec x ⎥ ∫−π / 4 ⎥⎦ −π 4 0

d sec x = sec x tan x dx

( π4 ) = 1 −

= sec 0 − sec − (c)

∫1 ( 2 4

3

x −

)

4 4 ⎡x3 2 + 4⎤ dx = ⎢ ⎥ ⎣ x2 x ⎦1 4 4 = ⎡⎢ ( 4 )3 2 + ⎤⎥ − ⎡⎢ (1)3 2 + ⎤⎥ ⎣ 4⎦ ⎣ 1⎦ = [ 8 + 1] − [ 5 ] = 4

2

2 1 dx = ln x ⎤⎥ ⎦1 x = ln 2 − ln 1 = ln 2

1

⎤1 dx arctan x = ⎥ ⎥⎦ 0 x2 + 1

(d)

∫1

(e)

∫0

2

(

)

d 32 4 3 4 x + = x1 2 − 2 dx x 2 x

d 1 ln x = dx x

d 1 arctan x = = 2 dx x +1

= arctan 1 − arctan 0 =

π π −0 = . 4 4

Exercise 82 offers another proof of the Evaluation Theorem, bringing together the ideas of Riemann sums, the Mean Value Theorem, and the definition of the definite integral.

The Integral of a Rate We can interpret Part 2 of the Fundamental Theorem in another way. If F is any antiderivative of f, then F ′ = f . The equation in the theorem can then be rewritten as b

∫a

F ′( x ) dx = F (b) − F (a).

Now F ′( x ) represents the rate of change of the function F ( x ) with respect to x, so the last equation asserts that the integral of F ′ is just the net change in F as x changes from a to b. Formally, we have the following result.

THEOREM 5—The Net Change Theorem

The net change in a differentiable function F ( x ) over an interval a ≤ x ≤ b is the integral of its rate of change: F (b) − F (a) =

EXAMPLE 4

b

∫a

F ′( x ) dx.

(6)

Here are several interpretations of the Net Change Theorem.

(a) If c( x ) is the cost of producing x units of a certain commodity, then c′( x ) is the marginal cost (Section 3.4). From Theorem 5,

∫x

x2

c′( x ) dx = c( x 2 ) − c( x 1 ),

1

which is the cost of increasing production from x 1 units to x 2 units.

348

Chapter 5 Integrals

(b) If an object with position function s(t ) moves along a coordinate line, its velocity is υ(t ) = s′(t ). Theorem 5 says that

∫t

t2

υ(t ) dt = s(t 2 ) − s(t 1 ),

1

so the integral of velocity is the displacement over the time interval t 1 ≤ t ≤ t 2 . On the other hand, the integral of the speed υ(t ) is the total distance traveled over the time interval. This is consistent with our discussion in Section 5.1. If we rearrange Equation (6) as b

F (b) = F (a) + ∫ F ′( x ) dx , a

we see that the Net Change Theorem also says that the final value of a function F ( x ) over an interval [ a, b ] equals its initial value F (a) plus its net change over the interval. So if υ(t ) represents the velocity function of an object moving along a coordinate line, this means that the object’s final position s(t 2 ) over a time interval t 1 ≤ t ≤ t 2 is its initial position s(t 1 ) plus its net change in position along the line (see Example 4b). EXAMPLE 5 Consider again our analysis of a heavy rock blown straight up from the ground by a dynamite blast (Example 2, Section 5.1). The velocity of the rock at any time t during its motion was given as υ(t ) = 49 − 9.8t m s. (a) Find the displacement of the rock during the time period 0 ≤ t ≤ 8. (b) Find the total distance traveled during this time period. Solution

(a) From Example 4b, the displacement is the integral 8

∫0

υ(t ) dt = ∫ (49 − 9.8t ) dt = ⎡ 49t − 4.9t 2 8

⎣

0

= (49)(8) − (4.9)(64) = 78.4.

⎤ ⎦0 8

This means that the height of the rock is 78.4 m above the ground 8 s after the explosion, which agrees with our conclusion in Example 2, Section 5.1. (b) As we noted in Table 5.3, the velocity function υ(t ) is positive over the time interval [0, 5] and negative over the interval [5, 8]. Therefore, from Example 4b, the total distance traveled is the integral 8

∫0

υ(t ) dt = = =

5

∫0

5

υ(t ) dt +

8

∫5

υ(t ) dt 8

∫0 (49 − 9.8t ) dt − ∫5 (49 − 9.8t ) dt 5 8 ⎡ 49t − 4.9t 2 ⎤ − ⎡ 49t − 4.9t 2 ⎤ ⎣ ⎦0 ⎣ ⎦5

= [ (49)(5) − (4.9)(25) ] − [(49)(8) − (4.9)(64) − ((49)(5) − (4.9)(25))] = 122.5 − (−44.1) = 166.6 Again, this calculation agrees with our conclusion in Example 2, Section 5.1. That is, the total distance of 166.6 m traveled by the rock during the time period 0 ≤ t ≤ 8 is (i) the maximum height of 122.5 m it reached over the time interval [0, 5] plus (ii) the additional distance of 44.1 m the rock fell over the time interval [5, 8].

5.4  The Fundamental Theorem of Calculus

349

The Relationship Between Integration and Differentiation The conclusions of the Fundamental Theorem tell us several things. Equation (2) can be rewritten as d dx

x

∫a

f (t ) dt = f ( x ),

which says that if you first integrate the function f and then differentiate the result, you get the function f back again. Likewise, replacing b by x and x by t in Equation (6) gives x

∫a

F ′(t ) dt = F ( x ) − F (a),

so that if you first differentiate the function F and then integrate the result, you get the function F back (adjusted by an integration constant). In a sense, the processes of integration and differentiation are “inverses” of each other. The Fundamental Theorem also says that every continuous function f has an antiderivative F. It shows the importance of finding antiderivatives in order to evaluate definite integrals easily. Furthermore, it says that the differential equation dy dx = f ( x ) has a solution (namely, any of the functions y = F ( x ) + C ) when f is a continuous function.

Total Area Area is always a nonnegative quantity. The Riemann sum approximations contain terms such as f (c k ) Δx k that give the area of a rectangle when f (c k ) is positive. When f (c k ) is negative, then the product f (c k ) Δx k is the negative of the rectangle’s area. When we add up such terms for a negative function, we get the negative of the area between the curve and the x-axis. If we then take the absolute value, we obtain the correct positive area.

y −2

−1

0

1

2

x

EXAMPLE 6 Figure 5.21 shows the graph of f ( x ) = x 2 − 4 and its mirror image g( x ) = 4 − x 2 reflected across the x-axis. For each function, compute

−1 −2 −3

f (x) = x 2 − 4

−4

Solution

(a)

y 4

g(x) = 4 − x 2

2

FIGURE 5.21

(

) (

)

2

⎡ 32 x3 ⎤ . = ⎢ 4x − ⎥ = ⎢⎣ 3 ⎥⎦ −2 3

(b) In both cases, the area between the curve and the x-axis over [ −2, 2 ] is 32 3 square units. Although the definite integral of f ( x ) is negative, the area is still positive.

1 0

∫−2

2

⎤ ⎡ x3 8 8 32 f ( x ) dx = ⎢ − 4x ⎥ = −8 − − +8 = − , ⎢⎣ 3 ⎥⎦ −2 3 3 3

∫−2 g( x ) dx

2

−1

2

and

3

−2

(a) the definite integral over the interval [ −2, 2 ], and (b) the area between the graph and the x-axis over [ −2, 2 ].

1

2

These graphs enclose the same amount of area with the x-axis, but the definite integrals of the two functions over [ −2, 2 ] differ in sign (Example 6).

x

To compute the area of the region bounded by the graph of a function y = f ( x ) and the x-axis when the function takes on both positive and negative values, we must be careful to break up the interval [ a, b ] into subintervals on which the function doesn’t change sign. Otherwise, we might get cancelation between positive and negative signed areas, leading to an incorrect total. The correct total area is obtained by adding the absolute value of the definite integral over each subinterval where f ( x ) does not change sign. The term “area” will be taken to mean this total area.

350

Chapter 5 Integrals

EXAMPLE 7 Figure 5.22 shows the graph of the function f ( x ) = sin x between x = 0 and x = 2 π. Compute (a) the definite integral of f ( x ) over [ 0, 2π ], (b) the area between the graph of f ( x ) and the x-axis over [ 0, 2π ]. Solution

(a) The definite integral for f ( x ) = sin x is given by

y

2π

∫0

y = sin x

1

Area = 2 p

0

Area = 0−2 0 = 2

2p

x

−1

The definite integral is zero because the portions of the graph above and below the x-axis make canceling contributions. (b) The area between the graph of f ( x ) and the x-axis over [ 0, 2π ] is calculated by breaking up the domain of sin x into two pieces: the interval [ 0, π ] over which it is nonnegative and the interval [ π, 2π ] over which it is nonpositive. π

∫0

FIGURE 5.22

The total area between y = sin x and the x-axis for 0 ≤ x ≤ 2π is the sum of the absolute values of two integrals (Example 7).

2π sin x dx = − cos x ⎤⎥ = −[ cos 2π − cos 0 ] = −[ 1 − 1 ] = 0. ⎦0

2π

∫π

π sin x dx = − cos x ⎤⎥ = −[ cos π − cos 0 ] = −[ −1 − 1 ] = 2 ⎦0

2π sin x dx = − cos x ⎥⎤ = −[ cos 2π − cos π ] = −[ 1 − (−1) ] = −2 ⎦π

The second integral gives a negative value. The area between the graph and the axis is obtained by adding the absolute values, Area = 2 + −2 = 4.

Summary:

To find the area between the graph of y = f ( x ) and the x-axis over the interval [ a, b ]: 1. Subdivide [ a, b ] at the zeros of f. 2. Integrate f over each subinterval. 3. Add the absolute values of the integrals.

EXAMPLE 8 Find the area of the region between the x-axis and the graph of f ( x ) = x 3 − x 2 − 2 x , −1 ≤ x ≤ 2. Solution First find the zeros of f. Since

f ( x ) = x 3 − x 2 − 2 x = x ( x 2 − x − 2 ) = x ( x + 1)( x − 2 ) ,

y Area = 5 12

−1

y = x 3 − x 2 − 2x

0

8 Area = P – P 3 =8 3

2

FIGURE 5.23 The region between the curve y = x 3 − x 2 − 2 x and the x-axis (Example 8).

x

the zeros are x = 0, −1, and 2 (Figure 5.23). The zeros subdivide [ −1, 2 ] into two subintervals: [ −1, 0 ], on which f ≥ 0, and [ 0, 2 ], on which f ≤ 0. We integrate f over each subinterval and add the absolute values of the calculated integrals. 0

∫−1 ( x 3 − x 2 − 2 x ) dx 2

∫0 ( x 3 − x 2 − 2 x ) dx

0 x4 x3 1 1 5 = ⎡⎢ − − x 2 ⎤⎥ = 0 − ⎡⎢ + − 1⎤⎥ = ⎣4 ⎦ 12 3 3 ⎣ 4 ⎦ −1 2 x4 x3 8 8 = ⎡⎢ − − x 2 ⎤⎥ = ⎡⎢ 4 − − 4 ⎤⎥ − 0 = − ⎣ ⎦ 3 3 3 ⎣ 4 ⎦0

The total enclosed area is obtained by adding the absolute values of the calculated integrals. Total enclosed area =

5 8 37 + − = 12 3 12

351

5.4  The Fundamental Theorem of Calculus

EXERCISES

5.4 In Exercises 35–38, guess an antiderivative for the integrand function. Validate your guess by differentiation, and then evaluate the given definite integral. (Hint: Keep the Chain Rule in mind when trying to guess an antiderivative. You will learn how to find such antiderivatives in the next section.)

Evaluating Integrals

Evaluate the integrals in Exercises 1–34. 2

∫−1 ( x 2 − 2 x + 3) dx

dx

4.

∫−1 x 299 dx

x3 dx 4

)

6.

∫−2 ( x 3 − 2 x + 3) dx

x ) dx

8.

∫1

∫0

3.

∫−2 ( x + 3)

5.

∫1

7.

∫0 ( x 2 +

9.

11.

x ( x − 3 ) dx

2

( 3x

4

2

−

1

π3

∫0

2

2 sec 2 x dx

3π 4

∫π 4

csc θ cot θ dθ

1 + cos 2t dt 13. ∫ π2 2 0

∫0

17.

∫0

π8

10.

12.

1

32

x −6 5 dx

1

35.

∫0

37.

∫2

3

xe x 2 dx

x dx 1 + x2

5

19.

21.

23.

25.

27.

∫1

π3

∫0

π3

16.

∫0

sin 2 x dx

18.

∫−π 3

∫

(

1 2

2

∫1

π2

∫π 6 4

∫−4 ln 2

29.

∫0

31.

∫0

33.

∫2

12

4

)

u7 1 − 5 du 2 u s2 + s ds s2 sin 2 x dx 2 sin x

x dx

e 3x

dx

4 dx 1 − x2

x π−1 dx

20.

22.

24.

π6

sin 2

t dt

( sec x + tan x ) 2 dx

−π 4

26.

28.

∫−

3 3

−1

∫−3 ∫1

∫0

π3

sin 2 x cos x dx

b. by differentiating the integral directly.

sin u 4 du cos 2 u

tan 2 x dx

+ 1) 2 dr

38.

ln x dx x

a. by evaluating the integral and differentiating the result.

π

∫0 (1 + cos x ) dx

∫−π 3

(r

∫1

Derivatives of Integrals

( 4 sec

2

t+

)

π dt t2

39.

d dx

∫0

41.

d dt

∫0

43.

d dx

∫0

8

(t

∫0

∫1

u du

42.

d dθ

∫0

e −t dt

44.

d dt

∫0

x

∫0

tan θ

t

∫1

x

46. y =

0

sin t 2 dt

48. y = x ∫

49. y =

∫−1 t 2 + 4 dt − ∫3

50. y =

(∫

51. y =

∫0

1 − e −x dx x

)

52. y =

∫tan x 1 + t 2

3

dx 1 + 4x 2

54. y =

∫2

56. y =

∫−1

13

+ 1)( 2 − x x1 3

23

)

dx

( cos x + sec x ) dx 2

(

30.

∫1

32.

∫0

34.

∫−1 π x −1 dx

3t 2 dt

sec 2 y dy

(x

1 + t 2 dt

∫

2

0

d dx

4

+

)

3 dx 1 − x2

Find dy dx in Exercises 45–56.

y 5 − 2y dy y3

1 ⎛⎜ cos x + cos x 2 ⎜⎝

1

40.

+ 1)( t 2 + 4 ) dt

π

∫0

x3

sin x

cos t dt

47. y =

(x

π3

x

t4

45. y = −1

2

36.

Find the derivatives in Exercises 39–44.

14.

π4

15.

1

2.

1.

⎞⎟ ⎟⎟ dx ⎠

x

x

t2

x 0

10

( t 3 + 1)

sin x

dt , 1 − t2

0

dt

1

3

x

x1 π

t dt

arcsin t dt

dt

)

x

t2

1 dt , t x2 2

x > 0

sin t 3 dt

t2 dt +4

3

x
0. Express

∫1

3

sin 2 x dx x

1

∫0

a. f ( x ) = x 2 , a = 0, b = 1, c = 1 c. f ( x ) =

f ( x ) dx =

1

∫0

f (1 − x ) dx.

119. Suppose that 1

∫0 f ( x ) dx

= 3.

Find 0

∫−1 f ( x ) dx a.   f is odd,

124. For each of the following functions, graph f ( x ) over [ a, b ] and f ( x + c ) over [ a − c, b − c ] to convince yourself that Equation (1) is reasonable. b. f ( x ) = sin x , a = 0, b = π , c = π 2

in terms of F. 118. Show that if f is continuous, then

if

123. Use a substitution to verify Equation (1).

b.   f is even.

120. a. Show that if f is odd on [ −a, a ], then a

∫−a

f ( x ) dx = 0.

b. Test the result in part (a) with f ( x ) = sin x and a = π 2.

x − 4, a = 4, b = 8, c = 5

COMPUTER EXPLORATIONS

In Exercises 125–128, you will find the area between curves in the plane when you cannot find their points of intersection using simple algebra. Use a CAS to perform the following steps: a. Plot the curves together to see what they look like and how many points of intersection they have. b. Use the numerical equation solver in your CAS to find all the points of intersection. c. Integrate f ( x ) − g( x ) over consecutive pairs of intersection values. d. Sum together the integrals found in part (c). x3 x2 1 125. f ( x ) = − − 2 x + , g( x ) = x − 1 3 3 2 x4 3 126. f ( x ) = − 3 x + 10, g( x ) = 8 − 12 x 2 127. f ( x ) = x + sin ( 2 x ), g( x ) = x 3 128. f ( x ) = x 2 cos x ,

g( x ) = x 3 − x

372

Chapter 5 Integrals

CHAPTER 5

Questions to Guide Your Review

1. How can you sometimes estimate quantities like distance traveled, area, and average value with finite sums? Why might you want to do so? 2. What is sigma notation? What advantage does it offer? Give examples. 3. What is a Riemann sum? Why might you want to consider such a sum? 4. What is the norm of a partition of a closed interval?

9. What is the Fundamental Theorem of Calculus? Why is it so important? Illustrate each part of the theorem with an example. 10. What is the Net Change Theorem? What does it say about the integral of velocity? The integral of marginal cost? 11. Discuss how the processes of integration and differentiation can be considered as “inverses” of each other. 12. How does the Fundamental Theorem provide a solution to the initial value problem dy dx  f ( x ), y( x 0 )  y 0 , when f is continuous?

5. What is the definite integral of a function f over a closed interval [ a, b ]? When can you be sure it exists?

13. How is integration by substitution related to the Chain Rule?

6. What is the relation between definite integrals and area? Describe some other interpretations of definite integrals.

14. How can you sometimes evaluate indefinite integrals by substitution? Give examples.

7. What is the average value of an integrable function over a closed interval? Must the function assume its average value? Explain.

15. How does the method of substitution work for definite integrals? Give examples.

8. Describe the rules for working with definite integrals (Table 5.6). Give examples.

16. How do you define and calculate the area of the region between the graphs of two continuous functions? Give an example.

CHAPTER 5

Practice Exercises

Finite Sums and Estimates

b. Sketch a graph of s as a function of t for 0 b t b 10, assuming s(0)  0. 5 4 Velocity (m/s)

1. The accompanying figure shows the graph of the velocity ( m s ) of a model rocket for the first 8 s after launch. The rocket accelerated straight up for the first 2 s and then coasted to reach its maximum height at t  8 s.

Velocity (m/s)

200 150

3 2 1

100 0 50 0

2

4

6

8

2. a. The accompanying figure shows the velocity ( m s ) of a body moving along the s-axis during the time interval from t  0 to t  10 s. About how far did the body travel during those 10 s?

10

10

k =1

k =1

10

a.

b. Sketch a graph of the rocket’s height above ground as a function of time for 0 b t b 8.

4 6 Time (s)

8

10

3. Suppose that ∑ a k = −2 and ∑ b k = 25. Find the value of

Time after launch (s)

a. Assuming that the rocket was launched from ground level, about how high did it go? (This is the rocket in Section 3.4, Exercise 17, but you do not need to do Exercise 17 to do the exercise here.)

2

10

c.

10

ak k =1 4

∑

∑ (ak k =1

b.

∑ ( bk k =1

− 3a k )

∑ ( 52 − b k ) 10

+ b k − 1)

d.

k =1

20

20

k =1

k =1

4. Suppose that ∑ a k = 0 and ∑ b k = 7. Find the value of 20

a.

20

∑ 3a k

b.

k =1

∑ ( 12 − 20

c.

k =1

∑ (ak k =1

2b k 7

)

20

d.

∑ (ak k =1

+ bk ) − 2)

Chapter 5  Practice Exercises

Definite Integrals In Exercises 5–8, express each limit as a definite integral. Then evaluate the integral to find the value of the limit. In each case, P is a partition of the given interval, and the numbers c k are chosen from the subintervals of P. n

∑ ( 2c k P → 0

5. lim

k =1

18. x 3 +

y = 1, x = 0,

y = 0, for

x 3 + " y = 1, 0 ≤ x ≤ 1

1

− 1)−1 2 Δx k , where P is a partition of [ 1, 5 ]

n

∑ c k ( c k2 − 1)1 3 Δx k , where P is a partition of [ 1, 3 ] P → 0

0

k =1

∑ ( cos( 2k )) Δx k , where P is a partition of [ −π, 0 ] P → 0 n

c

7. lim

k =1 n

∑ (sin c k )(cos c k ) Δx k , where P is a partition of [ 0, π 2 ]

P → 0

k =1

2 ∫−2

5 ∫−2

9. If 3 f ( x ) dx = 12, f ( x ) dx = 6, and find the value of each of the following. a.

2

∫−2

f ( x ) dx

−2

c.

∫5

e.

∫−2 ( 5

b.

g( x ) dx

d.

5

∫2

5

∫−2 (−πg( x )) dx

2 ∫0

2

∫0 g( x ) dx

c.

∫2

f ( x ) dx

2

b.

∫1

d.

∫0

2

1 ∫ 0 g( x )

dx = 2, find

g( x ) dx 2 f ( x ) dx

2

∫0 ( g( x ) − 3 f ( x ) ) dx

Area In Exercises 11–14, find the total area of the region between the graph of f and the x-axis.

11. f ( x ) = x 2 − 4 x + 3, 0 ≤ x ≤ 3 13. f ( x ) = 5 − 5 x 2 3 , −1 ≤ x ≤ 8 x, 0 ≤ x ≤ 4

y = 1 x 2, x = 2

16. y = x ,

y =1

17.

x +

y = 4x − 2

22. y 2 = 4 x + 4,

y = 4 x − 16

23. y = sin x , y = x , 0 ≤ x ≤ π 4 24. y = sin x , y = 1, −π 2 ≤ x ≤ π 2 25. y = 2 sin x , y = sin 2 x , 0 ≤ x ≤ π 27. Find the area of the “triangular” region bounded on the left by x + y = 2, on the right by y = x 2 , and above by y = 2. 28. Find the area of the “triangular” region bounded on the left by y = x , on the right by y = 6 − x , and below by y = 1. 29. Find the extreme values of f ( x ) = x 3 − 3 x 2 , and find the area of the region enclosed by the graph of f and the x-axis. 30. Find the area of the region cut from the first quadrant by the curve x 1 2 + y1 2 = a1 2. 31. Find the total area of the region enclosed by the curve x = y 2 3 and the lines x = y and y = −1. 32. Find the total area of the region between the curves y = sin x and y = cos x for 0 ≤ x ≤ 3π 2. 33. Find the area between the curve y = 2 ( ln x ) x and the x-axis from x = 1 to x = e. 34. a. Show that the area between the curve y = 1 x and the x-axis from x = 10 to x = 20 is the same as the area between the curve and the x-axis from x = 1 to x = 2.

35. Show that y = x 2 +

∫1

x1

t

dt solves the initial value problem

d 2y 1 = 2 − 2; dx 2 x

x, x = 2

y = 1, x = 0, y = 0

36. Show that y = problem

y

d 2y = dx 2

1

"x + "y = 1

0

21. y 2 = 4 x ,

20. x = 4 − y 2 , x = 0

Initial Value Problems

Find the areas of the regions enclosed by the curves and lines in Exercises 15–26. 15. y = x ,

19. x = 2 y 2 , x = 0, y = 3

x

b. Show that the area between the curve y = 1 x and the x-axis from ka to kb is the same as the area between the curve and the x-axis from x = a to x = b ( 0 < a < b, k > 0 ).

12. f ( x ) = 1 − ( x 2 4 ), −2 ≤ x ≤ 3 14. f ( x ) = 1 −

1

26. y = 8 cos x , y = sec 2 x , −π 3 ≤ x ≤ π 3

f ( x ) dx

f ( x ) dx = π, 7 g( x ) dx = 7, and 10. If the value of each of the following.

e.

g( x ) dx = 2,

)

2 ∫0

0

5 ∫−2

f ( x ) + g( x ) dx 5

a.

0 ≤ x ≤1

y

6. lim

8. lim

373

1

x

x

∫ 0 (1 + 2

y′(1) = 3,

y(1) = 1.

sec t ) dt solves the initial value

sec x tan x; y′(0) = 3, y(0) = 0.

Express the solutions of the initial value problems in Exercises 37 and 38 in terms of integrals. sin x dy 37. = , y(5) = −3 dx x dy = 2 − sin 2 x , y(−1) = 2 38. dx

374

Chapter 5 Integrals

Solve the initial value problems in Exercises 39–42. dy 1 39. = , y(0) = 0 dx 1 − x2 40.

dy 1 − 1, = 2 dx x +1

dy 1 = − 1 + x2 dx

2 , 1 − x2

∫−1 ( 3x 2 − 4 x + 7 ) dx

79.

∫1

y(2) = π

∫1

x

t f (t ) dt

45.

∫ 2( cos x )

47.

∫ ( 2θ + 1 + 2 cos( 2θ + 1)) d θ

48. 49.

∫( ∫(

)(

52.

∫ ( sec θ tan θ )

53.

∫ e x sec 2 ( e x

54.

∫ e y csc( e y

55.

∫

59.

( sec 2

dx ∫−1 3x − 4

∫0

71.

dx

∫

dx (2 x − 1) ( 2 x − 1) 2 − 4 e arcsin

x

dx

2 x − x2 dy arctan y (1 + y 2 )

73.

∫

75.

∫ ( sin 2θ + cos 2θ )3 d θ

sin 2θ − cos 2θ

68. 70.

72.

∫π 4

x dx 6

92.

∫ 0 tan 2 3 dθ

94.

∫π 4

5( sin x )3 2 cos x dx

96.

∫−π 2 15 sin 4 3x cos 3x dx

3 sin x cos x dx 1 + 3 sin 2 x

98.

∫0

100.

∫1

102.

∫−ln 2 e 2 w dw

99.

∫1

101.

∫−2

ln x dx x

103.

∫0

∫

tan ( ln υ ) dυ υ

105.

1 csc 2 (1 + ln r ) dr r

) dt

∫

( csc 2

∫1

e

x)e

cot x

dx

∫

∫ 2 + ( x − 1) 2

90.

∫0

t sin ( 2t

66.

3 dr 1 − 4 ( r − 1) 2

sec 2 θ dθ

97.

∫

∫ 2 tan x sec 2 x dx

∫

∫0

∫0

64.

65.

88.

95.

32

∫

dx

∫0

∫−π 3 sec x tan x dx

62.

2

86.

93.

( ln x )−3 dx x

∫ x 3x

∫0

∫π

60.

63.

∫

58.

84.

91.

2t dt t 2 − 25

∫

69.

56.

4

∫0

sec 2 x dx

+ 1) cot ( e y + 1) dy dx

∫1

89.

∫ ( tan x )

−3 2

− 7 ) dx

tan x

82.

∫ 0 sin 2 5r dr

1 + sec θ d θ

1

61.

67.

x)e

51.

dt t t

87.

)

∫

4

∫1 8 x −1 3 (1 − x 2 3 )

f (t ) 2 ) dt

)

(t + 1) 2 − 1 dt t4

4

46.

27

85.

x

∫ 0 (1 +

1 + 2 sec 2 ( 2θ − π ) d θ 2θ − π 2 2 t− t+ dt t t

50.

57.

sin x dx

∫1

∫ 0 ( 2 x + 1) 3

Evaluating Indefinite Integrals Evaluate the integrals in Exercises 45–76. −1 2

80.

83. 44. f ( x ) =

dx

∫ 1 + ( 3x + 1)2 ∫ ( x + 3) ∫

dx ( x + 3 ) 2 − 25

arcsin x dx 1 − x2

( arctan x ) dx 1 + x2 2

74.

∫

76.

∫ cos θ ⋅ sin ( sin θ ) d θ

36 dx

1

1

32

dx

π

π3

3π

cot 2

0

π2

π2

( 8x + 21x ) dx

4

−1

e −( x +1) dx

x −4 3 dx 12

(1 + u )

1 3

12

x 3 (1 + 9 x 4 )−3 2 dx

π4

θ

3π 4

π4

8

2

0

(1 + 7 ln x )−1 3 dx

106.

∫1

107.

∫1

8

108.

∫1

109.

∫−3 4

110.

∫−1 5

111.

∫−2 4 + 3t 2

112.

∫

113.

∫1

114.

∫4

115.

∫

116.

∫−2

3 dt

2

1 3

dy y 4y2 − 1

23 23

y

dy 9y 2 − 1

sec 2 x dx (1 + 7 tan x ) 2 3

( 32x − x8 ) dx

e1

9 − 4x 2

csc z cot z dz

π2

∫1

6 dx

)

π dt 4

csc 2 x dx

π

∫0

34

(

cos 2 4 t −

3π 4

104.

log 4 θ dθ θ

dr − 5r ) 2

(7

e r ( 3e r + 1)−3 2 dr

x

du

u

ln 5

6 dr 4 − ( r + 1) 2

∫0 ( 8s 3 − 12s 2 + 5 ) ds

4 dυ υ2

∫1

y(0) = 2

1

78.

2

81.

For Exercises 43 and 44, find a function f that satisfies each equation. 43. f ( x ) = 1 +

1

77.

y(0) = 1

dy 1 41. = , x > 1; dx x x2 − 1 42.

Evaluating Definite Integrals Evaluate the integrals in Exercises 77–116.

ln 9

e θ ( e θ − 1)1 2 d θ

3 ( ln ( υ

+ 1) ) 2 dυ υ+1

e

8 ln 3 log 3 θ dθ θ 6 dx

15

3 3

4 − 25 x 2

dt 3 + t2 24 dy

8

y y 2 − 16

2

− 6 5

5

y

dy 5y 2 − 3

Average Values

117. Find the average value of f ( x ) = mx + b a. over [ −1, 1 ]

b. over [ −k , k ]

118. Find the average value of a. y =

3 x over [ 0, 3 ]

b. y =

ax over [ 0, a ]

375

Chapter 5  Additional and Advanced Exercises

119. Let f be a function that is differentiable on [ a, b ]. In Chapter 2 we defined the average rate of change of f over [ a, b ] to be f (b) − f (a) b−a and the instantaneous rate of change of f at x to be f ′( x ). In this chapter we defined the average value of a function. For the new definition of average to be consistent with the old one, we should have f (b) − f (a) = average value of  f ′  on  [ a, b ]. b−a

133. Is it true that every function y = f ( x ) that is differentiable on [ a, b ] is itself the derivative of some function on [ a, b ]? Give reasons for your answer. 134. Suppose that f ( x ) is an antiderivative of f ( x ) = Express answer.

1 ∫0

120. Is it true that the average value of an integrable function over an interval of length 2 is half the function’s integral over the interval? Give reasons for your answer. 121. a. Verify that ∫ ln x dx = x ln x − x + C. b. Find the average value of ln x over [ 1, e ]. 122. Find the average value of f ( x ) = 1 x on [ 1, 2 ]. T 123. Compute the average value of the temperature function 2π ( x − 101) − 4 f ( x ) = 20 sin 365

1 + x 4 dx in terms of F and give a reason for your 1

135. Find dy dx if y = ∫ x 1 + t 2 dt . Explain the main steps in your calculation. 136. Find dy dx if y = ∫cos x (1/(1 − t 2 ) ) dt . Explain the main steps in your calculation. 137. A new parking lot To meet the demand for parking, your town has allocated the area shown here. As the town engineer, you have been asked by the town council to find out if the lot can be built for $10,000. The cost to clear the land will be $1.00 a square meter, and the lot will cost $2.00 a square meter to pave. Can the job be done for $10,000? Use a lower sum estimate to see. (Answers may vary slightly, depending on the estimate used.)

)

0m 12 m

for a 365-day year. (See Exercise 98, Section 3.6.) This is one way to estimate the annual mean air temperature in Fairbanks, Alaska. The National Weather Service’s official figure, a numerical average of the daily normal mean air temperatures for the year, is −3.5 ° C, which is slightly higher than the average value of f ( x ).

18 m 17 m 16.5 m

T 124. Specific heat of a gas Specific heat C υ is the amount of heat required to raise the temperature of one mole (gram molecule) of a gas with constant volume by 1° C. The specific heat of oxygen depends on its temperature T and satisfies the formula

Vertical spacing = 5 m

18 m 21 m

C υ = 8.27 + 10 −5 ( 26T − 1.87T 2 ).

22 m

Find the average value of C υ for 20 ° C ≤ T ≤ 675 ° C and the temperature at which it is attained.

14 m

Differentiating Integrals In Exercises 125–132, find dy dx. x

125. y =

∫2

127. y =

∫x

129. y =

∫ln x

131. y =

1

2 + cos 3 t dt 6 dt 3 + t4

0

2

e cos t dt

sin −1 x

∫0

dt 1 − 2t 2

CHAPTER 5

7x 2

126. y =

∫2

128. y =

∫sec x t 2 + 1 dt

130. y =

∫1

132. y =

2 + cos 3 t dt

2

1

e x

ln ( t 2 + 1) dt

π4

∫tan

1 + x4.

0

Is this the case? Give reasons for your answer.

(

Theory and Examples

−1 x

e

t

dt

Ignored

138. Skydivers A and B are in a helicopter hovering at 2000 m. Skydiver A jumps and descends for 4 s before opening her parachute. The helicopter then climbs to 2200 m and hovers there. Forty-five seconds after A leaves the aircraft, B jumps and descends for 13 s before opening his parachute. Both skydivers descend at 4.9 m s with parachutes open. Assume that the skydivers fall freely (no effective air resistance) before their parachutes open. a. At what altitude does A’s parachute open? b. At what altitude does B’s parachute open? c. Which skydiver lands first?

Additional and Advanced Exercises

Theory and Examples 1

1

1. a. If ∫ 7 f ( x ) dx = 7, does ∫ f ( x ) dx = 1? 0

0

2. Suppose ∫

−2

0

1

∫0

f ( x ) dx =

Give reasons for your answers.

4 = 2?

f ( x ) dx = 4,

5

∫2

f ( x ) dx = 3,

5

∫−2 g( x ) dx

Which, if any, of the following statements are true?

1

b. If ∫ f ( x ) dx = 4 and f ( x ) ≥ 0,  does

2

a.

2

∫5

f ( x ) dx = −3

b.

5

∫−2 ( f ( x ) + g( x ) ) dx

c. f ( x ) ≤ g( x ) on the interval −2 ≤ x ≤ 5

= 9

= 2.

376

Chapter 5 Integrals

3. Initial value problem Show that 1 a

y =

x

∫0

function, we integrate the individual extensions and add the results. The integral of

f (t ) sin a ( x − t ) dt

⎪⎧⎪ 1 − x , ⎪ f ( x ) = ⎪⎨ x 2 , ⎪⎪ ⎪⎪ −1, ⎩

solves the initial value problem d 2y + a 2 y = f ( x ), dx 2

dy = 0 and  y = 0 when x = 0. dx

y

0

x2

∫0

f (t ) dt = x cos π x

b.

∫0

f (x)

a2 a π + sin a + cos a. 2 2 2 7. The area of the region in the xy-plane enclosed by the x-axis, the curve y = f ( x ), f ( x ) ≥ 0, and the lines x = 1 and x = b is

∫ 0 ( ∫0 x

u

2 for all b > 1. Find f ( x ).

)

f (t ) dt du =

x

∫0

f (u) ( x − u ) du.

Piecewise Continuous Functions Although we are mainly interested in continuous functions, many functions in applications are piecewise continuous. A function f ( x ) is piecewise continuous on a closed interval I if f has only finitely many discontinuities in I, the limits

lim f ( x )

and

3

∫2 (−1) dx

4 y = x2

3 2 y=1−x

−1

1

0 −1

1

2 3 y = −1

x

FIGURE 5.33 Piecewise continuous functions like this are integrated piece by piece.

(Hint: Express the integral on the right-hand side as the difference of two integrals. Then show that both sides of the equation have the same derivative with respect to x.) 9. Finding a curve Find the equation for the curve in the xy-plane that passes through the point (1, −1) if its slope at x is always 3 x 2 + 2. 10. Shoveling dirt You sling a shovelful of dirt up from the bottom of a hole with an initial velocity of 9.8 m s. The dirt must rise 5.2 m above the release point to clear the edge of the hole. Is that enough speed to get the dirt out, or had you better duck?

x → c−

x 2 dx +

y

t 2 dt = x cos π x.

6. Find f ( π 2 ) from the following information. i) f is positive and continuous. ii) The area under the curve y = f ( x ) from x = 0 to x = a is

equal to b 2 + 1 − 8. Prove that

2

3 x2 ⎤0 x3 2 = ⎡⎢ x − + ⎡⎢ ⎤⎥ + ⎡⎢ −x ⎤⎥ ⎥ ⎣ ⎦2 2 ⎦ −1 ⎣ ⎣ 3 ⎦0 3 8 19 . = + −1 = 2 3 6

Show that d 2 y dx 2 is proportional to y, and find the constant of proportionality. 5. Find f (4) if a.

2 ≤ x ≤ 3

∫−1 f ( x ) dx = ∫−1 (1 − x ) dx + ∫0

1 dt. 1 + 4t 2

∫0

x =

0 ≤ x < 2

(Figure 5.33) over [ −1, 3 ] is 3

(Hint: sin ( ax − at ) = sin ax cos at − cos ax sin at. ) 4. Proportionality Suppose that x and y are related by the equation

−1 ≤ x < 0

lim f ( x )

x →c+

exist and are finite at every interior point of I, and the appropriate onesided limits exist and are finite at the endpoints of I. All piecewise continuous functions are integrable. The points of discontinuity subdivide I into open and half-open subintervals on which f is continuous, and the limit criteria above guarantee that f has a continuous extension to the closure of each subinterval. To integrate a piecewise continuous

The Fundamental Theorem applies to piecewise continuous funcx tions with the restriction that ( d dx ) ∫ a f (t ) dt is expected to equal f ( x ) only at values of x at which f is continuous. There is a similar restriction on Leibniz’s Rule (see Exercises 25–32). Graph the functions in Exercises 11–16 and integrate them over their domains. ⎪⎧⎪ x 2 3 , −8 ≤ x < 0 11. f ( x ) = ⎨ ⎪⎪ −4, 0 ≤ x ≤ 3 ⎪⎩ ⎧⎪ −x , −4 ≤ x < 0 12. f ( x ) = ⎪⎨ 2 ⎪⎪ x − 4, 0 ≤ x ≤ 3 ⎪⎩ ⎧⎪ t , 0 ≤ t a ) to generate a solid shaped like a doughnut and called a torus. Find its volume.

y (cm) x2 + y2 = 162 = 256

a

(Hint: ∫−a a 2 − y 2 dy = πa 2 2, since it is the area of a semicircle of radius a.) 58. Volume of a bowl A bowl has a shape that can be generated by revolving the graph of y = x 2 2 between y = 0 and y = 5 about the y-axis.

0 −7

x (cm)

a. Find the volume of the bowl. b. Related rates If we fill the bowl with water at a constant rate of 3 cubic units per second, how fast will the water level in the bowl be rising when the water is 4 units deep? 59. Volume of a bowl a. A hemispherical bowl of radius a contains water to a depth h. Find the volume of water in the bowl. b. Related rates Water runs into a sunken concrete hemispherical bowl of radius 5 m at the rate of 0.2 m 3 s. How fast is the water level in the bowl rising when the water is 4 m deep?

9 cm deep −16

64. Max-min The arch y = sin x ,  0 ≤ x ≤ π, is revolved about the line y = c, 0 ≤ c ≤ 1, to generate the solid in the accompanying figure. a. Find the value of c that minimizes the volume of the solid. What is the minimum volume? b. What value of c in [ 0, 1 ] maximizes the volume of the solid?

391

6.2  Volumes Using Cylindrical Shells

T c. Graph the solid’s volume as a function of c, first for 0 ≤ c ≤ 1 and then on a larger domain. What happens to the volume of the solid as c moves away from [ 0, 1 ]? Does this make sense physically? Give reasons for your answers. y

65. Consider the region R bounded by the graphs of y = f ( x ) > 0, x = a > 0, x = b > a, and y = 0 (see accompanying figure). If the volume of the solid formed by revolving R about the x-axis is 4 π , and the volume of the solid formed by revolving R about the line y = −1 is 8π , find the area of R. y y = f (x)

y = sin x

R 0 0 y=c p

x

a

b

x

66. Consider the region R given in Exercise 65. If the volume of the solid formed by revolving R around the x-axis is 6π , and the volume of the solid formed by revolving R around the line y = − 2 is 10 π , find the area of R.

6.2 Volumes Using Cylindrical Shells b

In Section 6.1 we defined the volume of a solid to be the definite integral V = ∫a A( x ) dx , where A( x ) is an integrable cross-sectional area of the solid from x = a to x = b. The area A( x ) was obtained by slicing through the solid with a plane perpendicular to the x-axis. However, this method of slicing is sometimes awkward to apply, as we will illustrate in our first example. To overcome this difficulty, we use the same integral definition for volume, but obtain the area by slicing through the solid in a different way.

Slicing with Cylinders Suppose we slice through the solid using circular cylinders of increasing radii, like cookie cutters. We slice straight down through the solid so that the axis of each cylinder is parallel to the y-axis. The vertical axis of each cylinder is always the same line, but the radii of the cylinders increase with each slice. In this way the solid is sliced up into thin cylindrical shells of constant thickness that grow outward from their common axis, like circular tree rings. Unrolling a cylindrical shell shows that its volume is approximately that of a rectangular slab with area A( x ) and thickness Δx. This slab interpretation allows us to apply the same integral definition for volume as before. The following example provides some insight. EXAMPLE 1 The region enclosed by the x-axis and the parabola y = f ( x ) = 3 x − x 2 is revolved about the vertical line x = −1 to generate a solid (see Figure 6.16). Find the volume of the solid. Solution Using the washer method from Section 6.1 would be awkward here because we would need to express the x-values of the left and right sides of the parabola in Figure 6.16a in terms of y. This is because these x-values, which describe the inner and outer radii of a typical washer, are solutions to the equation y = 3 x − x 2 , and this gives a complicated formula for x. Therefore, instead of rotating a horizontal strip of thickness Δy, we rotate a vertical strip of thickness Δx. This rotation produces a cylindrical shell of height y k above a point x k within the base of the vertical strip and of thickness Δx. An example of a cylindrical shell is shown as the orange-shaded region in Figure 6.17. We can think of the cylindrical shell shown in the figure as approximating a slice of the solid obtained by cutting straight down through it, parallel to the axis of revolution, all the way around. We

392

Chapter 6 Applications of Definite Integrals

y

y y = 3x −

2

x2

1 −2 −1

0

1

2

3

x

0

3

x

−1 Axis of revolution x = −1

Axis of revolution x = −1

−2 (a)

(b)

FIGURE 6.16 (a) The graph of the region in Example 1, before revolution. (b) The solid formed when the region in part (a) is revolved about the axis of revolution x = −1.

y

yk −5

0

xk

3

x = −1

FIGURE 6.17 A cylindrical shell of height y k obtained by rotating a vertical strip of thickness Δx k about the line x = −1. The outer radius of the cylinder occurs at x k , where the height of the parabola is y k = 3 x k − x k2 (Example 1).

x

start by cutting close to the inside hole and then cut another cylindrical slice around the enlarged hole, then another, and so on, obtaining n cylinders. The radii of the cylinders gradually increase, and the heights of the cylinders follow the contour of the parabola: shorter to taller, then back to shorter (Figure 6.16a). The sum of the volumes of the shells is a Riemann sum that approximates the volume of the entire solid. Each shell sits over a subinterval [ x k −1 , x k ] in the x-axis. The thickness of the shell is Δx k = x k − x k −1. Because the parabola is rotated around the line x = −1, the outer radius of the shell is 1 + x k . The height of the shell is the height of the parabola at some point in the interval [ x k −1 , x k ], or approximately y k = f ( x k ) = 3 x k − x k2 . If we unroll this cylinder and flatten it out, it becomes (approximately) a rectangular slab with thickness Δx k (see Figure 6.18). The height of the rectangular slab is approximately y k = 3 x k − x k2 , and its length is the circumference of the shell, which is approximately 2π ⋅ radius = 2π (1 + x k ). Hence the volume of the shell is approximately the volume of the rectangular slab, which is ΔVk = circumference × height × thickness = 2π (1 + x k ) ⋅ ( 3 x k − x k2 ) ⋅ Δx k . Δ xk

Outer circumference = 2p • radius = 2p(1 + xk) Radius = 1 + xk

(3xk − xk2)

h = (3xk − xk2)

Δ xk = thickness l = 2p(1 + xk)

FIGURE 6.18

Cutting and unrolling a cylindrical shell gives a nearly rectangular solid (Example 1).

6.2  Volumes Using Cylindrical Shells

393

Summing together the volumes ΔVk of the individual cylindrical shells over the interval [ 0, 3 ] gives the Riemann sum n

∑ ΔV k

n

∑ 2π ( x k

=

k =1

k =1

+ 1)( 3 x k − x k2 ) Δx k .

Taking the limit as the thickness Δx k → 0 and n → ∞ gives the volume integral n

∑ 2π ( x k n →∞

V = lim

k =1

+ 1)( 3 x k − x k2 ) Δx k

3

=

∫0 2π ( x + 1)( 3x − x 2 ) dx

=

∫0 2π ( 3x 2 + 3x − x 3 − x 2 ) dx

3

3

= 2π ∫ ( 2 x 2 + 3 x − x 3 ) dx 0

3 2 3 1 45π . = 2π ⎡⎢ x 3 + x 2 − x 4 ⎤⎥ = ⎣3 2 4 ⎦0 2

We now generalize this procedure to a broader class of solids.

The Shell Method Suppose that the region bounded by the graph of a nonnegative continuous function y = f ( x ) and the x-axis over the finite closed interval [ a, b ] lies to the right of the vertical line x = L (see Figure 6.19a). We assume a ≥ L, so the vertical line may touch the region but cannot pass through it. We generate a solid S by rotating this region about the vertical line L. Let P be a partition of the interval [ a, b ] by the points a = x 0 < x 1 <  < x n = b. As usual, we choose a point c k in each subinterval [ x k −1 , x k ]. In Example 1 we chose c k to be the endpoint x k , but now it will be more convenient to let c k be the midpoint of the subinterval [ x k −1 , x k ]. We approximate the region in Figure 6.19a with rectangles based on this partition of [ a, b ]. A typical approximating rectangle has height f (c k ) and width Vertical axis of revolution Vertical axis of revolution

y = f(x)

y = f (x) Δ xk

ck

a a x=L

xk−1

ck

(a)

FIGURE 6.19

xk

b

x

xk−1

b Δ xk (b)

xk

Rectangle height = f(ck) x

When the region shown in (a) is revolved about the vertical line x = L, a solid is produced which can be sliced into cylindrical shells. A typical shell is shown in (b).

394

Chapter 6 Applications of Definite Integrals

Δx = x − x

. If this rectangle is rotated about the vertical line x = L, then a shell is

k k k −1 The volume of a cylindrical shell of swept out, as in Figure 6.19b. A formula from geometry tells us that the volume of the shell height h with inner radius r and outer swept out by the rectangle is radius R is R+r ΔVk = 2π × average shell radius × shell height × thickness π R 2 h − πr 2 h = 2π (h)( R − r ). 2 R = x k − L  and  r = x k −1 − L = 2π ⋅ ( c k − L ) ⋅ f (c k ) ⋅ Δx k .

(

)

We approximate the volume of the solid S by summing the volumes of the shells swept out by the n rectangles: V ≈

n

∑ ΔV k . k =1

The limit of this Riemann sum as each Δx k → 0 and n → ∞ gives the volume of the solid as a definite integral: n

∑ ΔV k n →∞

V = lim

b

=

∫a 2π ( shell radius )( shell height ) dx

=

∫a 2π ( x − L ) f ( x ) dx.

k =1

b

We refer to the variable of integration, here x, as the thickness variable. To emphasize the process of the shell method, we state the general formula in terms of the shell radius and shell height. This will allow for rotations about a horizontal line y = L as well.

Shell Formula for Revolution About a Vertical Line

The volume of the solid generated by revolving the region between the x-axis and the graph of a continuous function y = f ( x ) ≥ 0,  L ≤ a ≤ x ≤ b, about a vertical line x = L is V =

b

⎛ shell ⎞ ⎛ shell ⎞

∫a 2π ⎜⎜⎜⎝ radius ⎟⎟⎟⎠ ⎜⎜⎜⎝ height ⎟⎟⎟⎟⎠ dx.

EXAMPLE 2 The region bounded by the curve y = x, the x-axis, and the line x = 4 is revolved about the y-axis to generate a solid. Find the volume of the solid. Solution Sketch the region and draw a line segment across it parallel to the axis of revolution (Figure 6.20a). Label the segment’s height (shell height) and distance from the axis of revolution (shell radius). (We drew the shell in Figure 6.20b, but you need not do that.) Shell radius y

y

x

Shell radius

2

y = "x

4

Interval of integration (a)

FIGURE 6.20

4 x

f(x) = "x x

" x = Shell height x

2

Shell height

x

0

y = "x (4, 2)

Interval of integration

0

x

–4

(b)

(a) The region, shell dimensions, and interval of integration in Example 2. (b) The shell swept out by the vertical segment in part (a) with a width Δx.

395

6.2  Volumes Using Cylindrical Shells

The shell thickness variable is x, so the limits of integration for the shell formula are a  0 and b  4 (Figure 6.20). The volume is V= =

b

∫a

⎛ shell ⎞⎟ ⎛⎜ shell ⎞⎟ ⎜ ⎟⎟ dx ⎟⎟ ⎜⎜ 2Q ⎜⎜ ⎜⎝ radius ⎟⎟⎠ ⎜⎜⎝ height ⎟⎟⎟⎠

4

∫ 0 2Q ( x ) (

= 2Q

4

∫0

x ) dx

4 2 128Q x 3 2 dx = 2Q ⎡⎢ x 5 2 ⎤⎥ = . ⎣5 ⎦0 5

So far, we have used vertical axes of revolution. For horizontal axes, we replace the x’s with y’s. EXAMPLE 3 The region bounded by the curve y  x, the x-axis, and the line x  4 is revolved about the x-axis to generate a solid. Find the volume of the solid by the shell method. Solution This is the solid whose volume was found by the disk method in Example 3 of Section 6.1. Now we find its volume by the shell method. First, sketch the region and draw a line segment across it parallel to the axis of revolution (Figure 6.21a). Label the segment’s length (shell height) and distance from the axis of revolution (shell radius). (We drew the shell in Figure 6.21b, but you need not do that.) In this case, the shell thickness variable is y, so the limits of integration for the shell formula method are a  0 and b  2 (along the y-axis in Figure 6.21). The volume of the solid is

⎛ shell ⎞⎟ ⎛⎜ shell ⎞⎟ b ⎜ ⎟⎟ dy ⎟⎟ ⎜⎜ V = ∫ 2Q ⎜⎜ ⎜⎜⎝ radius ⎟⎟⎠ ⎜⎜ height ⎟⎟⎟ a ⎠ ⎝ 2

= ∫ 2Q ( y)( 4 − y 2 ) dy 0

= 2Q

2

∫0 ( 4 y − y 3 ) dy

y4 ⎤ 2 ⎡ = 2Q ⎢ 2 y 2 − ⎥ = 8Q. 4 ⎦0 ⎣ y Shell height 2 y

y

4 − y2 (4, 2)

y = "x

4 − y2 Shell height 0

Interval of integration

2

(4, 2)

x = y2

y 4

y y Shell radius 0

4 (a)

Shell radius

x (b)

FIGURE 6.21 (a) The region, shell dimensions, and interval of integration in Example 3. (b) The shell swept out by the horizontal segment in part (a) with a width %y.

x

396

Chapter 6 Applications of Definite Integrals

Summary of the Shell Method

Regardless of the position of the axis of revolution (horizontal or vertical), the steps for implementing the shell method are these. 1. Draw the region and sketch a line segment across it parallel to the axis of revolution. Label the segment’s height or length (shell height) and distance from the axis of revolution (shell radius). 2. Find the limits of integration for the thickness variable. 3. Integrate the product 2Q (shell radius) (shell height) with respect to the thickness variable (x or y) to find the volume.

The shell method gives the same answer as the washer method when both are used to calculate the volume of a region. We do not prove that result here, but it is illustrated in Exercises 37 and 38. (Exercise 45 outlines a proof.) Both volume formulas are actually special cases of a general volume formula we will look at when studying double and triple integrals in Chapter 14. That general formula also allows for computing volumes of solids other than those swept out by regions of revolution.

6.2

EXERCISES

5. The y-axis

Revolution About the Axes

In Exercises 1–6, use the shell method to find the volumes of the solids generated by revolving the shaded region about the indicated axis. 1.

2.

y

y 2 y=1+ x 4

2 y=2− x 4

y

y

5 2

2

6. The y-axis

x = "3

1

0

x

2

0

3.

0

x

x=

"3

y2

2

x

0

7. y = x , y = − x 2, x = 2 8. y  2 x , y  x 2, x  1

y = "3

9. y = x 2 , y = 2 − x , x = 0, for  x ≥ 0 10. y = 2 − x 2 , y = x 2 , x = 0

x = 3 − y2 0

x 3

x

Use the shell method to find the volumes of the solids generated by revolving the regions bounded by the curves and lines in Exercises 7–12 about the y-axis. y

y = "2

"3

Revolution About the y-Axis

4. y

"2

2

9x "x 3 + 9

y = "x 2 + 1

1

0

y=

3

x

11. y = 2 x − 1, y =

x, x = 0

12. y = 3 ( 2 x ), y = 0, x = 1, x = 4

6.2  Volumes Using Cylindrical Shells

⎪⎧ ( sin x ) x , 13. Let f ( x ) = ⎪⎨ ⎪⎪ 1, ⎪⎩

25. y = x + 2, y = x 2

0 < x ≤ π

a. The line x = 2

x = 0.

b. Find the volume of the solid generated by revolving the shaded region about the y-axis in the accompanying figure. y

p

⎧⎪ ( tan x ) 2 x , ⎪ 14. Let g( x ) = ⎨ ⎪⎪ 0, ⎪⎩

26. y =

x 4,

d. The line y = 4 3x 2

y = 4−

a. The line x = 1

b. The x-axis

In Exercises 27 and 28, use the shell method to find the volumes of the solids generated by revolving the shaded regions about the indicated axes.

sin x , 0 < x ≤ p x y= x=0 1,

0

b. The line x = −1

c. The x-axis

a. Show that x f ( x ) = sin x ,  0 ≤ x ≤ π.

1

397

27. a. The x-axis x

b. The line y = 1

c. The line y = 8 5

d. The line y = − 2 5

y

0 < x ≤ π4 x = 0.

x = 12(y 2 − y 3)

1

a. Show that x g( x ) = ( tan x ) ,  0 ≤ x ≤ π 4. 2

b. Find the volume of the solid generated by revolving the shaded region about the y-axis in the accompanying figure. y 4 p

y=

p tan2 x , 0 0, and suppose that f has a differentiable inverse, f −1. Revolve about the y-axis the region bounded by the graph of f and the lines x = a and y = f (b) to generate a solid. Then the values of the integrals given by the washer and shell methods for the volume are identical. f (b)

b

2 ∫ f (a) π (( f −1 ( y) ) − a 2 ) dy = ∫a 2π x ( f (b) −

y x = 3y2 − 2 −2

1

(1, 1) x = y2

0

1

x

f ( x ) ) dx.

To prove this equality, define

Theory and Examples

39. The region shown here is to be revolved about the x-axis to generate a solid. Which of the methods (disk, washer, shell) could you use to find the volume of the solid? How many integrals would be required in each case? Explain.

x

1

0

W (t ) = S (t ) =

f (t )

2 ∫ f (a) π (( f −1 ( y) ) − a 2 ) dy t

∫ a 2π x ( f (t ) −

f ( x ) ) dx.

Then show that the functions W and S agree at a point of [ a, b ] and have identical derivatives on [ a, b ]. As you saw in Section 4.8, Exercise 132, this will guarantee W (t ) = S (t ) for all t in [ a, b ]. In particular, W (b) = S (b). (Source: “Disks and Shells Revisited” by Walter Carlip, in American Mathematical Monthly, Feb. 1991, vol. 98, no. 2, pp. 154–156.)

6.3  Arc Length

46. The region between the curve y = sec −1 x and the x-axis from x = 1 to x = 2 (shown here) is revolved about the y-axis to generate a solid. Find the volume of the solid. y

p 3

399

49. Consider the region R bounded by the graphs of y = f ( x ) > 0,   x = a > 0, and x = b > a. If the volume of the solid formed by revolving R about the y-axis is 2π , and the volume formed by revolving R about the line x = −2 is 10 π , find the area of R.

y = sec−1 x

y y = f (x)

0

1

2

x

47. Find the volume of the solid generated by revolving the region enclosed by the graphs of y = e −x 2 , y = 0, x = 0 , and x = 1 about the y-axis. 48. Find the volume of the solid generated by revolving the region enclosed by the graphs of y = e x 2 , y = 1, and x = ln 3 about the x-axis.

R 0

a

b

x

50. Consider the region R given in Exercise 49. If the area of region R is 1, and the volume of the solid formed by revolving R about the line x = −3 is 10 π , find the volume of the solid formed by revolving R about the y-axis.

6.3 Arc Length We know what is meant by the length of a straight-line segment, but without calculus, we have no precise definition of the length of a general winding curve. If the curve is the graph of a continuous function defined over an interval, then we can find the length of the curve using a procedure similar to that we used for defining the area between the curve and the x-axis. We divide the curve into many pieces, and we approximate each piece by a straightline segment. The sum of the lengths of these segments is an approximation to the total curve length that we seek. The total length of the curve is the limiting value of these approximations as the number of segments goes to infinity.

Length of a Curve y = f ( x) Suppose the curve whose length we want to find is the graph of the function y = f ( x ) from x = a  to x = b. In order to derive an integral formula for the length of the curve, we assume that f has a continuous derivative at every point of [ a, b ]. Such a function is called smooth, and its graph is a smooth curve because it does not have any breaks, corners, or cusps. We partition the interval [ a, b ] into n subintervals with a = x 0 < x 1 < x 2 <  < x n = b. If y k = f ( x k ), then the corresponding point Pk ( x k , y k ) lies on the curve. Next we connect successive points Pk −1 and Pk with straight-line segments that, taken together, form a polygonal path whose length approximates the length of the curve (Figure 6.22). If we set Δx k = x k − x k −1 and Δy k = y k − y k −1, then a representative line segment in the path has length Lk =

( Δx k ) 2 + ( Δy k ) 2

(see Figure 6.23), so the length of the curve is approximated by the sum n

∑ Lk = k =1

n

∑

k =1

( Δx k ) 2 + ( Δy k ) 2 .

(1)

We expect the approximation to improve as the partition of [ a, b ] becomes finer. In order to evaluate this limit, we use the Mean Value Theorem, which tells us that there is a point c k , with x k −1 < c k < x k , such that Δy k = f ′(c k ) Δx k .

400

Chapter 6 Applications of Definite Integrals

FIGURE 6.22

The length of the polygonal path P0 P1P2  Pn approximates the length of the curve y = f ( x ) from point A to point B.

y

Pk−1 Pk B = Pn y = f (x)

x0 = a

x1

x2

xk−1

xk

b = xn

x

P0 = A P1

y

When this is substituted for Δy k , the sums in Equation (1) take the form

Pk−1 Δ yk

n

y = f (x) Lk Δx k

xk−1

0

P2

∑ Lk = k =1

Pk

x

xk

FIGURE 6.23 The arc Pk −1 Pk of the curve y = f ( x ) is approximated by the straight-line segment shown here, which has length L k = ( Δx k ) 2 + ( Δy k ) 2 .

n

∑

( Δx k ) 2 + ( f ′(c k )Δx k )

k =1

n

lim

n →∞

n

∑ L k = lim

∑ n →∞

k =1

k =1

y = 4" 2 x 3/2 − 1 3

k =1

1 + [ f ′(c k ) ]2 Δx k =

b

∫a

1 + [ f ′( x ) ]2 dx =

b

∫a

4 2 32 x − 1, 3

b

∫a

1 + [ f ′( x ) ]2 dx.

1+

( dxdy )

2

dx.

0 ≤ x ≤ 1.

4 2 32 x −1 3 dy 4 2 3 12 = ⋅ x = 2 2x 1 2 3 2 dx 2 2 dy = ( 2 2 x 1 2 ) = 8 x. dx y =

x

A

The length of the curve is slightly larger than the length of the line segment joining points A and B (Example 1).

(2)

If  x = 1, then y ≈ 0.89.

( )

The length of the curve over x = 0 to x = 1 is L= =

(3)

Find the length of the curve shown in Figure 6.24, which is the graph

B

FIGURE 6.24

1 + [ f ′(c k ) ]2 Δx k .

Solution We use Equation (3) with a = 0, b = 1, and

y

−1

∑

DEFINITION If f ′ is continuous on [ a, b ], then the length (arc length) of the curve y = f ( x ) from the point A = ( a, f (a) ) to the point B = ( b, f (b) ) is the value of the integral

y =

1

n

We define the length of the curve to be this integral.

EXAMPLE 1 of the function

0

=

This is a Riemann sum whose limit we can evaluate. Because 1 + [ f ′( x ) ]2 is continuous on [ a, b ], the limit of the Riemann sum on the right-hand side of Equation (2) exists and has the value

L =

1

2

1

∫0

1+

( dxdy )

2

dx =

1

∫0

1 + 8 x dx

1 2 1( 13 ⋅ 1 + 8 x )3 2 ⎤⎥ = ≈ 2.17. ⎦0 3 8 6

Eq. (3) with  a = 0,  b = 1 Let u = 1 + 8 x , integrate, and replace u by 1 + 8 x.

401

6.3  Arc Length

Notice that the length of the curve is slightly larger than the length of the straight-line segment joining the points A = ( 0, −1) and B = (1, 4 2 3 − 1) on the curve (see Figure 6.24): 2.17 > y

EXAMPLE 2

1 2 + (1.89 ) 2 ≈ 2.14.

1

x3 1 + , x 12

f (x) =

1 ≤ x ≤ 4.

Solution A graph of the function is shown in Figure 6.25. To use Equation (3), we find

x2 1 − 2 x 4

f ′( x ) =

A 0

Decimal approximations

Find the length of the graph of

3 y = x + 1x 12

B

 

4

x

so

( x4

2

1 + [ f ′( x ) ]2 = 1 +

FIGURE 6.25

The curve in Example 2, where A = (1, 13 12 ) and B = ( 4, 67 12 ).

=

x4

1 x2

)

( 16x − 12 + x1 ) x 1 =( + 4 x ) 2

4

= 1+

4

2

2

1 1 + 4 2 x

+

16

−

2

and

( x4

2

1 + [ f ′( x ) ]2 =

1 x2

+

)

2

=

x2 1 x2 1 + 2 = + 2. 4 x 4 x

 

x2 + 1 > 0 4 x2

The length of the graph over [ 1, 4 ] is

∫1

L=

4

1 + [ f ′( x ) ]2 dx =

(

∫1

4

( x4

2

) (

+

)

1 dx x2

)

x3 1 4 64 1 1 72 = ⎡⎢ − ⎤⎥ = − − −1 = = 6. x ⎦1 12 4 12 12 ⎣ 12 EXAMPLE 3

Find the length of the curve 1( x e + e − x ), 2

y =

0 ≤ x ≤ 2.

Solution We use Equation (3) with a = 0, b = 2, and

1 y = ( e x + e −x ) 2 dy 1 = ( e x − e −x ) dx 2

( dxdy ) 1+ 1+

( dxdy )

2

=

2

( dxdy )

2

=

1 ( 2x e − 2 + e −2 x ) 4

=

2 1 ( 2x 1 e + 2 + e −2 x ) = ⎡⎢ ( e x + e − x )⎤⎥ ⎣2 ⎦ 4

⎡1 x −x ⎤ ⎢⎣ 2 ( e + e )⎥⎦

2

=

1( x 1 e + e − x ) = ( e x + e − x ). 2 2

1 (e x 2

+ e −x ) > 0

The length of the curve from x = 0 to x = 2 is L= =

2

∫0

1+

( ) dy dx

2

2

dx =

2

∫0

1( x e + e − x ) dx 2

1⎡ x 1 ⎤ ⎢ e − e − x ⎥ = (( e 2 − e −2 ) − (1 − 1)) ≈ 3.63. 2⎣ 2 ⎦0

Eq. (3) with a = 0,  b = 2

402

Chapter 6 Applications of Definite Integrals

Dealing with Discontinuities in dy dx Even if the derivative dy dx does not exist at some point on a curve, it is possible that dx dy could exist. This can happen, for example, when a curve has a vertical tangent. In this case, we may be able to find the curve’s length by expressing x as a function of y and applying the following analogue of Equation (3).

Formula for the Length of x = g ( y ), c ≤ y ≤ d

If g′ is continuous on [ c, d ], the length of the curve x = g( y) from A = ( g(c), c ) to B = ( g(d ), d ) is L =

∫c

⎛ dx ⎞ 2 1 + ⎜⎜ ⎟⎟⎟ dy = ⎝ dy ⎠

d

∫c

d

1 + [ g′( y) ]2 dy.

(4)

Find the length of the curve y = ( x 2 ) 2 3 from x = 0 to x = 2.

EXAMPLE 4

Solution The derivative

( ) ( 12 ) = 13( 2x ) −1 3

dy 2 x = dx 3 2

is not defined at x = 0, so we cannot find the curve’s length with Equation (3). We therefore rewrite the equation to express x in terms of y:

y

1

x 23 y= Q R 2

y =

(2, 1)

y3 2 = 0

FIGURE 6.26

13

1

2

The graph of y = ( x 2 ) 2 3 from x = 0 to x = 2 is also the graph of x = 2 y 3 2 from y = 0 to  y = 1 (Example 4).

x

( 2x )

23

x 2

Raise both sides to the power 3 2.

x = 2 y 3 2.

Solve for x.

From this we see that the curve whose length we want is also the graph of x = 2 y 3 2 from y = 0 to y = 1 (see Figure 6.26). The derivative

( )

dx 3 12 = 2 y = 3y1 2 dy 2 is continuous on [ 0, 1 ]. We may therefore use Equation (4) to find the curve’s length: L=

∫c

d

⎛ dx ⎞ 2 1 + ⎜⎜ ⎟⎟⎟ dy = ⎝ dy ⎠

1

∫0

=

1 1 2( ⋅ 1 + 9 y )3 2 ⎤⎥ ⎦0 9 3

=

2 (10 10 − 1) ≈ 2.27. 27

1 + 9 y dy

Eq. (4) with c = 0,  d = 1 Let u = 1 + 9 y,  du 9 = dy, integrate, and substitute back.

The Differential Formula for Arc Length If y = f ( x ) and if f ′ is continuous on [ a, b ], then by the Fundamental Theorem of Calculus, we can define a new function s( x ) =   ∫

x a

1 + [ f ′(t ) ]2 dt.

(5)

6.3  Arc Length

y

ds

403

From Equation (3) and Figure 6.22, we see that this function s( x ) is continuous and measures the length along the curve y = f ( x ) from the initial point P0 ( a, f (a) ) to the point Q ( x , f ( x ) ) for each x ∈ [ a, b ]. The function s is called the arc length function for y = f ( x ). From the Fundamental Theorem, the function s is differentiable on (a, b) and dy

ds = dx

dx x

0

1 + [ f ′( x ) ]2 =

1+

( dxdy ) . 2

Then the differential of arc length is

(a)

ds =

y

1+

( dxdy )

2

dx.

(6)

A useful way to remember Equation (6) is to write ds = Δ s ≈ ds

Δy ≈ dy

dx x

0

dx 2 + dy 2 ,

(7)

which can be integrated between appropriate limits to give the total length of a curve. From this point of view, all the arc length formulas are simply different expressions for the equation L = ∫ ds. Figure 6.27a, which corresponds to Equation (7), can be thought of as a simplified approximation of Figure 6.27b. That is, ds is approximately equal to the exact arc length Δs.

(b)

FIGURE 6.27 Diagrams for remembering the equation ds = dx 2 + dy 2 .

EXAMPLE 5 Find the arc length function for the curve in Example 2, taking A = (1, 13 12 ) as the starting point (see Figure 6.25). Solution In the solution to Example 2, we found that

1 + [ f ′( x ) ]2 =

x2 1 + 2. x 4

Therefore the arc length function is given by s( x ) =

∫1

x

1 + [ f ′(t ) ]2 dt =

∫1

x

( t4 + t1 ) dt 2

2

x

t3 1 x3 1 11 = ⎡⎢ − ⎤⎥ = − + . t ⎦1 12 x 12 ⎣ 12 To compute the arc length along the curve from A = (1, 13 12 )  to B = ( 4, 67 12 ), for instance, we simply calculate s(4) =

43 1 11 − + = 6. 12 4 12

This is the same result we obtained in Example 2.

EXERCISES

6.3

Finding Lengths of Curves

Find the lengths of the curves in Exercises 1–16. If you have graphing software, you may want to graph these curves to see what they look like. 1. y = (1 3 )( x 2 + 2 ) 2. y = x

32

32

from x = 0 to x = 3

from x = 0 to x = 4

3. x = ( y 3 3 ) + 1 (4 y) from y = 1 to y = 3

4. x = ( y 3 2 3 ) − y 1 2 from y = 1 to y = 9 5. x = ( y 4 4 ) + 1 ( 8 y 2 ) from y = 1 to y = 2 6. x = ( y 3 6 ) + 1 ( 2 y ) from y = 2 to y = 3 7. y = ( 3 4 ) x 4 3 − ( 3 8 ) x 2 3 + 5, 1 ≤ x ≤ 8 8. y = ( x 3 3 ) + x 2 + x + 1 ( 4 x + 4 ), 0 ≤ x ≤ 2

404

Chapter 6 Applications of Definite Integrals

9. y = ln x − 10. y =

x2

−

2

x2 from x = 1 to x = 2 8

ln x from x = 1 to x = 3 4

x3 1 11. y = + , 1≤ x ≤ 3 3 4x x5 1 1 + , ≤ x ≤1 5 12 x 3 2

13. y =

3 23 1 x + 1, ≤ x ≤1 2 8

14. y =

1( x e + e −x ), −1 ≤ x ≤ 1 2

15. x =

∫0

16. y =

∫−2

x

y =

x

∫0

cos 2t dt

from x = 0 to x = π 4. 28. The length of an astroid The graph of the equation x 2 3 + y 2 3 = 1 is one of a family of curves called astroids (not “asteroids”) because of their starlike appearance (see the accompanying figure). Find the length of this particular astroid by finding the length of half the first-quadrant portion, 32 y = (1 − x 2 3 ) ,   2 4 ≤ x ≤ 1, and multiplying by 8.

12. y =

y

27. Find the length of the curve

y 1 x 23 + y 23 = 1

sec 4 t − 1 dt , −π 4 ≤ y ≤ π 4 3t 4 − 1 dt , −2 ≤ x ≤ −1 −1

0

1

x

T Finding Integrals for Lengths of Curves In Exercises 17–24, do the following. a. Set up an integral for the length of the curve.

−1

b. Graph the curve to see what it looks like. c. Use your grapher’s or computer’s integral evaluator to find the curve’s length numerically. 17. y = x 2 , −1 ≤ x ≤ 2 18. y = tan x , −π 3 ≤ x ≤ 0

30. Circumference of a circle Set up an integral to find the circumference of a circle of radius r centered at the origin. You will learn how to evaluate the integral in Section 8.3.

19. x = sin y, 0 ≤ y ≤ π 20. x =

1 − y 2 , −1 2 ≤ y ≤ 1 2

(−1, −1) to ( 7, 3 )

21. y 2 + 2 y = 2 x + 1 from

29. Length of a line segment Use the arc length formula (Equation 3) to find the length of the line segment y = 3 − 2 x ,  0 ≤ x ≤ 2. Check your answer by finding the length of the segment as the hypotenuse of a right triangle.

31. If 9 x 2 = y ( y − 3 ) 2, show that

22. y = sin x − x cos x , 0 ≤ x ≤ π 23. y = 24. x =

x

∫0

∫0

y

ds 2 =

tan t dt , 0 ≤ x ≤ π 6

(y

+ 1) 2 2 dy . 4y

32. If 4 x 2 − y 2 = 64, show that

sec 2 t − 1 dt , −π 3 ≤ y ≤ π 4

ds 2 =

4( 2 5 x − 16 ) dx 2 . y2

Theory and Examples

25. a. Find a curve with a positive derivative through the point (1, 1) whose length integral (Equation 3) is L =

∫1

4

1+

1 dx. 4x

b. How many such curves are there? Give reasons for your answer. 26. a. Find a curve with a positive derivative through the point ( 0, 1) whose length integral (Equation 4) is L =

∫1

2

1+

1 dy. y4

b. How many such curves are there? Give reasons for your answer.

33. Is there a smooth (continuously differentiable) curve y = f ( x ) whose length over the interval 0 ≤ x ≤ a is always 2a ? Give reasons for your answer. 34. Using tangent fins to derive the length formula for curves Assume that f is smooth on [ a, b ] and partition the interval [ a, b ] in the usual way. In each subinterval [ x k −1 , x k ], construct the tangent fin at the point ( x k −1 , f ( x k −1 )), as shown in the accompanying figure. a. Show that the length of the kth tangent fin over the interval

[ x k −1 , x k ] equals ( Δx k ) 2 + ( f ′( x k −1 ) Δx k ) 2 . b. Show that n

b

∑ ( length of kth tangent fin ) = ∫a n →∞ k =1 lim

1 + ( f ′( x ) ) 2 dx ,

6.4  Areas of Surfaces of Revolution

which is the length L of the curve y  f ( x ) from a to b.

y = f (x)

37. Find the arc length function for the graph of f ( x )  2 x 3 2 using ( 0, 0 ) as the starting point. What is the length of the curve from ( 0, 0 ) to (1, 2 )? 38. Find the arc length function for the curve in Exercise 8, using ( 0, 1 4 ) as the starting point. What is the length of the curve from ( 0, 1 4 ) to (1, 59 24 )?

Tangent fin with slope f ′(xk−1)

(xk−1, f (xk−1)) Δ xk xk−1

405

xk

COMPUTER EXPLORATIONS

In Exercises 39–44, use a CAS to perform the following steps for the given graph of the function over the closed interval. a. Plot the curve together with the polygonal path approximations for n  2, 4, 8 partition points over the interval. (See Figure 6.22.)

x

35. Approximate the arc length of one-quarter of the unit circle (which is Q 2 by computing the length of the polygonal approximation with n  4 segments (see accompanying figure).

b. Find the corresponding approximation to the length of the curve by summing the lengths of the line segments. c. Evaluate the length of the curve using an integral. Compare your approximations for n  2, 4, 8 with the actual length given by the integral. How does the actual length compare with the approximations as n increases? Explain your answer.

y

39. f ( x ) =

1 − x 2 , −1 ≤ x ≤ 1

40. f ( x ) = x 1 3 + x 2 3 , 0 ≤ x ≤ 2 0

0.25 0.5 0.75 1

x

36. Distance between two points Assume that the two points ( x 1 , y1 ) and ( x 2 , y 2 ) lie on the graph of the straight line y = mx + b. Use the arc length formula (Equation 3) to find the distance between the two points.

41. f ( x ) = sin ( Q x 2 ), 0 ≤ x ≤

2

42. f ( x ) = x 2 cos x , 0 ≤ x ≤ Q x −1 1 43. f ( x ) = , − ≤ x ≤1 4x 2 + 1 2 44. f ( x ) = x 3 − x 2 , −1 ≤ x ≤ 1

6.4 Areas of Surfaces of Revolution When you jump rope, the rope sweeps out a surface in the space around you similar to what is called a surface of revolution. The surface surrounds a volume of revolution, and many applications require that we know the area of the surface rather than the volume it encloses. In this section we define areas of surfaces of revolution. More general surfaces are treated in Chapter 15.

Defining Surface Area Δx y A

Δx

B

2py

y x

0

x NOT TO SCALE

(a)

(b)

FIGURE 6.28 (a) A cylindrical surface generated by rotating the horizontal line segment AB of length %x about the x-axis has area 2Q y %x. (b) The cut and rolledout cylindrical surface as a rectangle.

If you revolve a region in the plane that is bounded by the graph of a function over an interval, it sweeps out a solid of revolution, as we saw earlier in the chapter. However, if you revolve only the bounding curve itself, it does not sweep out any interior volume but rather a surface that surrounds the solid and forms part of its boundary. Just as we were interested in defining and finding the length of a curve in the last section, we are now interested in defining and finding the area of a surface generated by revolving a curve about an axis. Before considering general curves, we begin by rotating horizontal and slanted line segments about the x-axis. If we rotate the horizontal line segment AB having length %x about the x-axis (Figure 6.28a), we generate a cylinder with surface area 2Q y %x. This area is the same as that of a rectangle with side lengths %x and 2Qy (Figure 6.28b). The length 2Qy is the circumference of the circle of radius y generated by rotating the point (x, y) on the line AB about the x-axis. Suppose the line segment AB has length L and is slanted rather than horizontal. Now when AB is rotated about the x-axis, it generates a frustum of a cone (Figure 6.29a). From

406

Chapter 6 Applications of Definite Integrals

L y

y L

A y*

y1

B

2py*

y = f(x)

0

y2

a x

0

PQ

xk−1 x k

b

x

NOT TO SCALE

(a)

(b)

FIGURE 6.29 (a) The frustum of a cone generated by rotating the slanted line segment AB of length L about the x-axis has area y + y2 2πy* L. (b) The area of the rectangle for y* = 1 , the average 2 height of AB above the x-axis.

P Q

xk−1

xk

x

FIGURE 6.31 The line segment joining P and Q sweeps out a frustum of a cone.

P

Segment length: L = "(Δxk )2 + (Δyk )2 y = f (x)

Δ yk

Q

f(xk − 1)

f (xk ) xk – 1

FIGURE 6.32

Δxk

classical geometry, the surface area of this frustum is 2πy* L , where y* = ( y1 + y 2 ) 2 is the average height of the slanted segment AB above the x-axis. This surface area is the same as that of a rectangle with side lengths L and 2πy* (Figure 6.29b). Let’s build on these geometric principles to define the area of a surface swept out by revolving more general curves about the x-axis. Suppose we want to find the area of the surface swept out by revolving the graph of a nonnegative continuous function y = f ( x ),  a ≤ x ≤ b, about the x-axis. We partition the closed interval [ a, b ] in the usual way and use the points in the partition to subdivide the graph into short arcs. Figure 6.30 shows a typical arc PQ and the band it sweeps out as part of the graph of f . As the arc PQ revolves about the x-axis, the line segment joining P and Q sweeps out a frustum of a cone whose axis lies along the x-axis (Figure 6.31). The surface area of this frustum approximates the surface area of the band swept out by the arc PQ. The surface area of the frustum of the cone shown in Figure 6.31 is 2πy* L , where y* is the average height of the line segment joining P and Q, and L is its length (just as before). Since f ≥ 0, from Figure 6.32 we see that the average height of the line segment is y* = ( f ( x k −1 ) + f ( x k ) ) 2, and the slant length is L = ( Δx k ) 2 + ( Δy k ) 2 . Therefore, Frustum surface area = 2π ⋅

f ( x k −1 ) + f ( x k ) ⋅ ( Δx k ) 2 + ( Δy k ) 2 2

= π ( f ( x k −1 ) + f ( x k ) ) ( Δx k ) 2 + ( Δy k ) 2 .

xk

Dimensions associated with the arc and line segment PQ.

FIGURE 6.30 The surface generated by revolving the graph of a nonnegative function y = f ( x ),   a ≤ x ≤ b, about the x-axis. The surface is a union of bands like the one swept out by the arc PQ.

The area of the original surface, being the sum of the areas of the bands swept out by arcs like arc PQ, is approximated by the frustum area sum n

∑ π ( f ( x k −1 ) + k =1

f ( x k ) ) ( Δx k ) 2 + ( Δy k ) 2 .

(1)

We expect the approximation to improve as the partition of [ a, b ] becomes finer. To find the limit, we first need to find an appropriate substitution for Δy k . If the function f is differentiable, then by the Mean Value Theorem, there is a point ( c k , f (c k ) ) on the curve between P and Q where the tangent is parallel to the segment PQ (Figure 6.33). At this point, f ′(c k ) =

Δy k , Δx k

Δy k = f ′(c k ) Δx k .

407

6.4  Areas of Surfaces of Revolution

(ck , f (ck )) Δyk

Tangent parallel to chord

P

With this substitution for Δy k , the sums in Equation (1) take the form n

∑ π ( f ( x k −1 ) + k =1

Q y = f (x) xk−1

ck Δ xk

xk

FIGURE 6.33 If f is smooth, the Mean Value Theorem guarantees the existence of a point c k where the tangent is parallel to segment PQ.

f ( x k ) ) ( Δx k ) 2 + ( f ′(c k ) Δx k ) 2

n

∑ π ( f ( x k −1 ) +

=

f ( x k ) ) 1 + ( f ′(c k ) ) 2 Δx k .

k =1

(2)

These sums are not the Riemann sums of any function because the points x k −1 , x k , and c k are not the same. However, the points x k −1 , x k , and c k are very close to each other, and so we expect (and it can be proved) that as the norm of the partition of [ a, b ] goes to zero, the sums in Equation (2) converge to the integral b

∫ a 2π f ( x )

1 + ( f ′( x ) ) 2 dx.

We therefore define this integral to be the area of the surface swept out by the graph of f from a to b.

If the function f ( x ) ≥ 0 is continuously differentiable on [ a, b ], the area of the surface generated by revolving the graph of y = f ( x ) about the x-axis is

DEFINITION

S =

b

∫a

2π y 1 +

( dxdy )

2

dx =

b

∫ a 2π f ( x )

1 + ( f ′( x ) ) 2 dx.

(3)

Note that the square root in Equation (3) is similar to the one that appears in the formula for the arc length of the generating curve in Equation (6) of Section 6.3.

y y = 2"x (2, 2" 2) (1, 2)

EXAMPLE 1 Find the area of the surface generated by revolving the curve y = 2 x , 1 ≤ x ≤ 2, about the x-axis (Figure 6.34). Solution We evaluate the formula

0 1

2

x

S =

b

∫ a 2π y

1+

( dxdy )

2

dx

  

Eq. (3)

with a = 1,

FIGURE 6.34

In Example 1 we calculate the area of this surface.

b = 2,

y = 2 x,

dy 1 = . dx x

First, we perform some algebraic manipulation on the radical in the integrand to transform it into an expression that is easier to integrate. 1+

( dxdy )

2

=

1+

( 1x )

2

=

1+

1 = x

x +1 = x

With these substitutions, we have 2 x +1 x + 1 dx dx = 4 π ∫ 1 x 2 2 8π = 4 π ⋅ ( x + 1)3 2 ⎤⎥ = ( 3 3 − 2 2 ). ⎦1 3 3

S =

∫1

2

2π ⋅ 2 x

x +1 x

408

Chapter 6 Applications of Definite Integrals

Revolution About the y-Axis For revolution about the y-axis, we interchange x and y in Equation (3). Surface Area for Revolution About the y-Axis

If x = g( y) ≥ 0 is continuously differentiable on [ c, d ], the area of the surface generated by revolving the graph of x = g( y) about the y-axis is

∫c

S =

y

d

⎛ dx ⎞ 2 2π x 1 + ⎜⎜ ⎟⎟⎟ dy = ⎝ dy ⎠

∫c

d

2π g( y) 1 + ( g′( y) ) 2 dy.

(4)

EXAMPLE 2 The line segment x = 1 − y, 0 ≤ y ≤ 1, is revolved about the y-axis to generate the cone in Figure 6.35. Find its lateral surface area (which excludes the base area).

A(0, 1) x+y=1

Solution Here we have a calculation we can check with a formula from geometry:

Lateral surface area = 0

base circumference × slant height = π 2. 2

To see how Equation (4) gives the same result, we take

B(1, 0) x

c = 0,

x = 1 − y,

⎛ dx ⎞ 2 1 + ⎜⎜ ⎟⎟⎟ = ⎝ dy ⎠

FIGURE 6.35

Revolving line segment AB about the y-axis generates a cone whose lateral surface area we can now calculate in two different ways (Example 2).

d = 1,

dx = −1, dy

1 + (−1) 2 =

2

and calculate S =

∫c

d

⎛ dx ⎞ 2 2π x 1 + ⎜⎜ ⎟⎟⎟ dy = ⎝ dy ⎠

1

∫ 0 2 π (1 − y )

(

2 dy

)

y2 ⎤1 1 ⎡ = 2π 2 ⎢ y − = π 2. ⎥ = 2π 2 1 − 2 ⎦0 2 ⎣

EXERCISES

6.4

Finding Integrals for Surface Area

In Exercises 1–8: a. Set up an integral for the area of the surface generated by revolving the given curve about the indicated axis. T b. Graph the curve to see what it looks like. If you can, graph the surface too. T c. Use your utility’s integral evaluator to find the surface’s area numerically. 1. y = tan x , 0 ≤ x ≤ π 4; x -axis 2. y = x 2 , 0 ≤ x ≤ 2; x -axis 3. xy = 1, 1 ≤ y ≤ 2; y-axis 4. x = sin y, 0 ≤ y ≤ π; 5. x

12

+ y

12

y-axis

= 3 from ( 4, 1) to (1, 4 ); x-axis

6. y + 2 y = x , 1 ≤ y ≤ 2; y-axis

y

7. x =

∫0 tan t dt ,

8. y =

∫1

x

0 ≤ y ≤ π 3;

t 2 − 1 dt , 1 ≤ x ≤

y-axis 5; x -axis

Finding Surface Area

9. Find the lateral (side) surface area of the cone generated by revolving the line segment y = x 2,  0 ≤ x ≤ 4, about the x-axis. Check your answer with the geometry formula Lateral surface area =

1 × base circumference × slant height. 2

10. Find the lateral surface area of the cone generated by revolving the line segment y = x 2,  0 ≤ x ≤ 4, about the y-axis. Check your answer with the geometry formula given in Exercise 9.

6.4  Areas of Surfaces of Revolution

11. Find the surface area of the cone frustum generated by revolving the line segment y = ( x 2 ) + (1 2 ),  1 ≤ x ≤ 3, about the x-axis. Check your result with the geometry formula Frustum surface area = π ( y1 + y 2 ) × slant height. 12. Find the surface area of the cone frustum generated by revolving the line segment y = ( x 2 ) + (1 2 ),  1 ≤ x ≤ 3, about the y-axis. Check your result with the geometry formula given in Exercise 11. Find the areas of the surfaces generated by revolving the curves in Exercises 13–23 about the indicated axes. If you have a grapher, you may want to graph these curves to see what they look like. 13. y = x 3 9, 0 ≤ x ≤ 2; x -axis 14. y =

x , 3 4 ≤ x ≤ 15 4; x -axis

15. y =

2 x − x 2 , 0.5 ≤ x ≤ 1.5; x -axis

16. y = 17. x =

x + 1, 1 ≤ x ≤ 5; x -axis y3

3, 0 ≤ y ≤ 1; y-axis

18. x = (1 3 ) y 3 2 − y 1 2 , 1 ≤ y ≤ 3; y-axis 19. x = 2 4 − y , 0 ≤ y ≤ 15 4; y-axis y

15 a1, b 4

15 4

x = 2" 4 − y

23. x = ( y 4 4 ) + 1 ( 8 y 2 ), 1 ≤ y ≤ 2; x -axis (Hint: Express ds =

dx 2 + dy 2 in terms of dy, and evaluate the integral

S = ∫ 2π y ds with appropriate limits.) 24. Write an integral for the area of the surface generated by revolving the curve y = cos x , −π 2 ≤ x ≤ π 2, about the x-axis. In Section 8.3 we will see how to evaluate such integrals. 25. Testing the new definition Show that the surface area of a sphere of radius a is still 4 πa 2 by using Equation (3) to find the area of the surface generated by revolving the curve y = a 2 − x 2 , −a ≤ x ≤ a, about the x-axis. 26. Testing the new definition The lateral (side) surface area of a cone of height h and base radius r should be πr r 2 + h 2 , the semiperimeter of the base times the slant height. Show that this is still the case by finding the area of the surface generated by revolving the line segment y = ( r h ) x ,  0 ≤ x ≤ h, about the x-axis. T 27. Enameling woks Your company decided to put out a deluxe version of a wok you designed. The plan is to coat it inside with white enamel and outside with blue enamel. Each enamel will be sprayed on 0.5 mm thick before baking. (See accompanying figure.) Your manufacturing department wants to know how much enamel to have on hand for a production run of 5000 woks. What do you tell them? (Neglect waste and unused material and give your answer in liters. Remember that 1 cm 3 = 1 mL, so 1 L = 1000 cm 3 .) y (cm)

0 x

4

20. x =

409

x2 + y2 = 162 = 256

2 y − 1, 5 8 ≤ y ≤ 1; y-axis 0 −7

y

x (cm) 9 cm deep

1

x = " 2y − 1

1 5 a2 , 8b 0

1 2

1

21. x = ( e y + e − y ) 2, 0 ≤ y ≤ ln 2; y

x=

28. Here is a schematic drawing of the 30-m dome used by the U.S. National Weather Service to house radar in Bozeman, Montana. a. How much outside surface is there to paint (not counting the bottom)?

x

y-axis

T b. Express the answer to the nearest square meter.

ey + e−y 2

Axis

5 8

−16

(1, 1)

ln 2

0

15 m 1

Radius 15 m

x 7.5 m

22. y = (1 3 )( x 2 + 2 )3 2, 0 ≤ x ≤ 2; y-axis (Hint: Express ds = dx 2 + dy 2 in terms of dx, and evaluate the integral S = ∫ 2π x ds with appropriate limits.)

Center

410

Chapter 6 Applications of Definite Integrals

29. The shaded band shown here is cut from a sphere of radius R by parallel planes h units apart. Show that the surface area of the band is 2πRh.

b. Show that the length L k of the tangent line segment in the kth subinterval is L k =

( Δx k ) 2 + ( f ′( m k ) Δx k ) . 2

y = f (x)

h r2

R

r1 xk−1

c. Show that the lateral surface area of the frustum of the cone swept out by the tangent line segment as it revolves about the x-axis is 2π f ( m k ) 1 + ( f ′( m k ) ) 2 Δx k . d. Show that the area of the surface generated by revolving y = f ( x ) about the x-axis over [ a, b ] is n

−r

lim

k =1

B

r 0

a

h

⎛ lateral surface area ⎞⎟ ⎟= of  kth frustum ⎟⎠

∑ ⎜⎜⎜⎝ n →∞

y A

a+h

b

∫a 2π f ( x )

1 + ( f ′( x ) ) 2 dx.

32. The surface of an astroid Find the area of the surface generated by revolving about the x-axis the portion of the astroid x 2 3 + y 2 3 = 1 shown in the accompanying figure. 32

(Hint: Revolve the first-quadrant portion y = (1 − x 2 3 ) , 0 ≤ x ≤ 1, about the x-axis and double your result.)

x

y

31. An alternative derivation of the surface area formula Assume f is smooth on [ a, b ] and partition [ a, b ] in the usual way. In the kth subinterval [ x k −1 , x k ], construct the tangent line to the curve at the midpoint m k = ( x k −1 + x k ) 2, as in the accompanying figure.

1 x 23 + y 23 = 1

a. Show that r1 = f ( m k ) − f ′( m k )

x

xk

Δxk

30. Slicing bread Did you know that if you cut a spherical loaf of bread into slices of equal width, each slice will have the same amount of crust? To see why, suppose the semicircle y = r 2 − x 2 shown here is revolved about the x-axis to generate a sphere. Let AB be an arc of the semicircle that lies above an interval of length h on the x-axis. Show that the area swept out by AB does not depend on the location of the interval. (It does depend on the length of the interval.) y = "r 2 − x 2

mk

Δx k Δx k . and r2 = f ( m k ) + f ′( m k ) 2 2

−1

0

1

x

6.5 Work and Fluid Forces In everyday life, work means an activity that requires muscular or mental effort. In science, the term refers specifically to a force acting on an object and the object’s subsequent displacement. This section shows how to calculate work. The applications run from compressing railroad car springs and emptying subterranean tanks to forcing subatomic particles to collide and lifting satellites into orbit.

Work Done by a Constant Force When an object moves a distance d along a straight line as a result of being acted on by a force of constant magnitude F in the direction of motion, we define the work W done by the force on the object with the formula W = Fd

( Constant-force formula for work ).

(1)

From Equation (1) we see that the unit of work in any system is the unit of force multiplied by the unit of distance. In SI units (SI stands for Système International, or International System), the unit of force is a newton (N), the unit of distance is a meter (m), and the unit of work is a newton-meter ( N ⋅ m ). This combination appears so often, it has

411

6.5  Work and Fluid Forces

Joules The joule, abbreviated J, is named after the English physicist James Prescott Joule (1818–1889). The defining equation is 1 joule = (1 newton )(1 meter ). In symbols, 1 J = 1 N ⋅ m.

a special name, the joule (J). Taking gravitational acceleration at sea level to be 9.8 m s 2, to lift one kilogram one meter requires work of 9.8 joules. This is seen by multiplying the force of 9.8 newtons exerted on one kilogram by the one-meter distance moved. EXAMPLE 1 Suppose you jack up the side of a 1000-kg car 35 cm to change a tire. The jack applies a constant vertical force of about 5000 N in lifting the side of the car (but because of the mechanical advantage of the jack, the force you apply to the jack itself is only about 150 N). The total work performed by the jack on the car is 5000 × 0.35 = 1750 J.

Work Done by a Variable Force Along a Line If the force you apply varies along the way, as it will if you are stretching or compressing a spring, the formula W = Fd has to be replaced by an integral formula that takes the variation in F into account. Suppose that the force performing the work acts on an object moving along a straight line, which we take to be the x-axis. We assume that the magnitude of the force is a continuous function F of the object’s position x. We want to find the work done over the interval from x = a to x = b. We partition [ a, b ] in the usual way and choose an arbitrary point c k in each subinterval [ x k −1 , x k ]. If the subinterval is short enough, the continuous function F will not vary much from x k −1 to x k . The amount of work done across the interval will be about F (c k ) times the distance Δx k , the same as it would be if F were constant and we could apply Equation (1). The total work done from a to b is therefore approximated by the Riemann sum Work ≈

n

∑ F (c k ) Δx k . k =1

We expect the approximation to improve as the norm of the partition goes to zero, so we define the work done by the force from a to b to be the integral of F from a to b: n

∑ F (c k ) Δx k n →∞ lim

=

k =1

b

∫a

F ( x ) dx.

DEFINITION The work done by a variable force F ( x ) in moving an object along the x-axis from x = a to x = b is

W =

b

∫a

F ( x ) dx.

(2)

The units of the integral are joules if F is in newtons and x is in meters. So the work done by a force of F ( x ) = 1 x 2 newtons in moving an object along the x-axis from x = 1 m to x = 10 m is W =

10

∫1

1 1 10 1 dx = − ⎥⎤ = − + 1 = 0.9 J. 2 x ⎦1 10 x

Hooke’s Law for Springs: F = kx One calculation for work arises in finding the work required to stretch or compress a spring. Hooke’s Law says that the force required to hold a stretched or compressed spring x units from its natural (unstressed) length is proportional to x. In symbols, F = kx.

(3)

412

Chapter 6 Applications of Definite Integrals

The constant k, measured in force units per unit length, is a characteristic of the spring, called the force constant (or spring constant) of the spring. Hooke’s Law, Equation (3), gives good results as long as the force doesn’t distort the metal in the spring. We assume that the forces in this section are too small to do that.

Compressed F x 0

Uncompressed

0.3

x

(a)

Force (N)

F

EXAMPLE 2 Find the work required to compress a spring from its natural length of 30 cm to a length of 20 cm if the force constant is k 240 Nm. Solution We picture the uncompressed spring laid out along the x-axis with its movable end at the origin and its fixed end at x 0.3 m (Figure 6.36). This enables us to describe the force required to compress the spring from 0 to x with the formula F 240x. To compress the spring from 0 to 0.1 m, the force must increase from

F = 240x Work done by F from x = 0 to x = 0.1

24

0

0.1 Amount compressed

x (m)

F (0) = 240 ⋅ 0 = 0 N The work done by F over this interval is W =

(b)

FIGURE 6.36 The force F needed to hold a spring under compression increases linearly as the spring is compressed (Example 2).

F (0.1) = 240 ⋅ 0.1 = 24 N.

to

0.1

∫0

0.1

240 x dx = 120 x 2 ⎤⎥ ⎦0

= 1.2 J.

Eq. (2) with a = 0,   b = 0.1, F ( x ) = 240 x

EXAMPLE 3 A spring has a natural length of 1 m. A force of 24 N holds the spring stretched to a total length of 1.8 m. (a) Find the force constant k. (b) How much work will it take to stretch the spring from its natural length to a length of 3 m? (c) How far will a 45-N force stretch the spring? Solution

(a) The force constant. We find the force constant from Equation (3). A force of 24 N maintains the spring at a position where it is stretched 0.8 m from its natural length, so

x=0

0.8 1

24 = k (0.8) k = 24 0.8 = 30 N m. 24 N

Eq. (3) with F = 24,  x = 0.8

(b) The work to stretch the spring 2 m. We imagine the unstressed spring hanging along the x-axis with its free end at x = 0 (Figure 6.37). The force required to stretch the spring x meters beyond its natural length is the force required to hold the free end of the spring x units from the origin. Hooke’s Law with k = 30 says that this force is F ( x ) = 30 x.

x (m)

FIGURE 6.37

A 24-N weight stretches this spring 0.8 m beyond its unstressed length (Example 3).

The work done by F on the spring from x = 0 m to x = 2 m is W =

2

∫0

2

30 x dx = 15 x 2 ⎤⎥ = 60 J. ⎦0

(c) How far will a 45-N force stretch the spring? We substitute F = 45 in the equation F = 30 x to find

x

45 = 30 x ,

6

or

x = 1.5 m.

A 45-N force will keep the spring stretched 1.5 m beyond its natural length.

Lifting Objects and Pumping Liquids from Containers The work integral is useful for calculating the work done in lifting objects whose weights vary with their elevation. 0

FIGURE 6.38

Example 4.

Lifting the bucket in

EXAMPLE 4 A 2-kg bucket is lifted from the ground into the air by pulling in 6 m of rope at a constant speed (Figure 6.38). The rope weighs 0.1 kgm. How much work was spent lifting the bucket and rope?

6.5  Work and Fluid Forces

413

Solution The bucket has constant weight, so the work done lifting it alone is

weight × distance = 2 ⋅ 9.8 ⋅ 6 = 117.6 J. The weight of the rope varies with the bucket’s elevation, because less of it is freely hanging. When the bucket is x m off the ground, the remaining proportion of the rope still being lifted weighs ( 0.1 ⋅ 9.8 ) ⋅ ( 6 − x ) N. So the work in lifting the rope is Work on rope =

6

∫0

( 0.98 )( 6

− x ) dx =

6

∫0

( 5.88

− 0.98 x ) dx

6

⎡ ⎤ = ⎢ 5.88 x − 0.49 x 2 ⎥ = 35.28 − 17.64 = 17.64 J. ⎣ ⎦0 The total work for the bucket and rope combined is 117.6 + 17.64 = 135.24 J. How much work does it take to pump all or part of the liquid from a container? Engineers often need to know the answer in order to design or choose the right pump, or to compute the cost to transport water or some other liquid from one place to another. To find out how much work is required to pump the liquid, we imagine lifting the liquid out one thin horizontal slab at a time and applying the equation W  Fd to each slab. We then evaluate the integral that this leads to as the slabs become thinner and more numerous. y

10 − y

y = 2x or x = 1 y 2

10 8

(5, 10) 1y 2

y

0 5

Example 5.

Solution We imagine the oil divided into thin slabs by planes perpendicular to the y-axis at the points of a partition of the interval [ 0, 8 ]. The typical slab between the planes at y and y + Δy has a volume of about

ΔV = Q ( radius ) 2 ( thickness ) = Q

Δy

FIGURE 6.39

EXAMPLE 5 The conical tank in Figure 6.39 is filled to within 2 m of the top with olive oil weighing 0.9 g cm 3 or 8820 N m 3 . How much work does it take to pump the oil to the rim of the tank?

x

The olive oil and tank in

( 12 y )

2

Δy =

Q 2 y Δy m 3 . 4

The force F ( y) required to lift this slab is equal to its weight, F ( y ) = 8820 ΔV =

8820 Q 2 y Δy N. 4

Weight = ( weight per unit volume ) × volume

The distance through which F ( y) must act to lift this slab to the level of the rim of the cone is about (10 − y ) m, so the work done lifting the slab is about ΔW =

8820 Q ( 10 − y ) y 2 Δy J. 4

Assuming there are n slabs associated with the partition of [ 0, 8 ], and that y  y k denotes the plane associated with the kth slab of thickness %y k , we can approximate the work done lifting all of the slabs with the Riemann sum W ≈

n

8820 Q (10 − y k ) y k2 Δy k J. 4 k =1

∑

The work of pumping the oil to the rim is the limit of these sums as the norm of the partition goes to zero, and the number of slabs tends to infinity: n

8820 Q (10 − y k ) y k2 Δy k = 4 k =1

∑ n →∞

W = lim

8

∫0

8820 Q ( 10 − y ) y 2 dy 4

8820 Q 8 (10 y 2 − y 3 ) dy 4 ∫0 y4 ⎤8 8820 Q ⎡ 10 y 3 = − ⎢ ⎥ ≈ 4,728, 977 J. 4 ⎣ 3 4 ⎦0 =

414

Chapter 6 Applications of Definite Integrals

Fluid Pressure and Forces

FIGURE 6.40 To withstand the increasing pressure, dams are built thicker as they go down.

Weight-density A fluid’s weight-density w is its weight per unit volume. Typical values ( N m 3 ) are listed below. Gasoline 6600 Mercury 133,000 Milk 10,100 Molasses 15,700 Olive oil 8820 Seawater 10,050 Freshwater 9800

Dams are built thicker at the bottom than at the top (Figure 6.40) because the pressure against them increases with depth. The pressure at any point on a dam depends only on how far below the surface the point is and not on how much the surface of the dam happens to be tilted at that point. The pressure, in newtons per square meter at a point h meters below the surface, is always 9800h. The number 9800 is the weight-density of freshwater in newtons per cubic meter. The pressure h meters below the surface of any fluid is the fluid’s weight-density times h.

The Pressure-Depth Equation

In a fluid that is standing still, the pressure p at depth h is the fluid’s weightdensity w times h: p  wh.

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In a container of fluid with a flat horizontal base, the total force exerted by the fluid against the base can be calculated by multiplying the area of the base by the pressure at the base. We can do this because total force equals force per unit area (pressure) times area. (See Figure 6.41.) If F, p, and A are the total force, pressure, and area, then F = total force = force per unit area × area = pressure × area = pA = whA.

p  wh from Eq. (4)

h

Fluid Force on a Constant-Depth Surface

F  pA  whA

(5)

FIGURE 6.41

These containers are filled with water to the same depth and have the same base area. The total force is therefore the same on the bottom of each container. The containers’ shapes do not matter here. y Surface of fluid

b

Submerged vertical plate Strip depth Δy

y a L(y) Strip length at level y

FIGURE 6.42 The force exerted by a fluid against one side of a thin, flat horizontal strip is about ΔF = pressure × area = w × ( strip depth ) × L ( y) Δy.

For example, the weight-density of freshwater is 9800 Nm3, so the fluid force at the bottom of a 3 m q 6 m rectangular swimming pool 1 m deep is F = whA = ( 9800 N m 3 )(1 m )( 3 ⋅ 6 m 2 ) = 176,400 N. For a flat plate submerged horizontally, like the bottom of the swimming pool just discussed, the downward force acting on its upper face due to liquid pressure is given by Equation (5). If the plate is submerged vertically, however, then the pressure against it will be different at different depths and Equation (5) no longer is usable in that form (because h varies). Suppose we want to know the force exerted by a fluid against one side of a vertical plate submerged in a fluid of weight-density w. To find it, we model the plate as a region extending from y  a to y  b in the xy-plane (Figure 6.42). We partition [ a, b ] in the usual way and imagine the region to be cut into thin horizontal strips by planes perpendicular to the y-axis at the partition points. The typical strip from y to y + Δy is %y units wide by L ( y) units long. We assume L ( y) to be a continuous function of y. The pressure varies across the strip from top to bottom. If the strip is narrow enough, however, the pressure will remain close to its bottom-edge value of w × ( strip depth ). The force exerted by the fluid against one side of the strip will be about ΔF = ( pressure along bottom edge ) × ( area ) = w ⋅ ( strip depth ) ⋅ L ( y) Δy.

415

6.5  Work and Fluid Forces

Assume there are n strips associated with the partition of a b y b b and that y k is the bottom edge of the kth strip having length L ( y k ) and width %y k . The force against the entire plate is approximated by summing the forces against each strip, giving the Riemann sum F ≈

n

∑ w ⋅ ( strip depth ) k k =1

⋅ L ( y k ) Δy k .

(6)

The sum in Equation (6) is a Riemann sum for a continuous function on [ a, b ], and we expect the approximations to improve as the norm of the partition goes to zero. The force against the plate is the limit of these sums: n

∑ w ⋅ ( strip depth ) k n →∞

⋅ L ( y k ) Δy k =

lim

k =1

b

∫a

w ⋅ ( strip depth ) ⋅ L ( y) dy.

The Integral for Fluid Force Against a Vertical Flat Plate

Suppose that a plate submerged vertically in fluid of weight-density w runs from y  a to y  b on the y-axis. Let L ( y) be the length of the horizontal strip measured from left to right along the surface of the plate at level y. Then the force exerted by the fluid against one side of the plate is F =

y (m) y = x or x = y

Pool surface at Depth: 1.6 − y

x=y y

y = 1.6

y=1

(1, 1) (x, x) = (y, y)

Δy

x (m)

0

FIGURE 6.43

To find the force on one side of the submerged plate in Example 6, we can use a coordinate system like the one here.

b

∫a

w ⋅ ( strip depth ) ⋅ L ( y) dy.

EXAMPLE 6 A flat isosceles right-triangular plate with base 2 m and height 1 m is submerged vertically, base up, 0.6 m below the surface of a swimming pool. Find the force exerted by the water against one side of the plate. Solution We establish a coordinate system to work in by placing the origin at the plate’s bottom vertex and running the y-axis upward along the plate’s axis of symmetry (Figure 6.43). The surface of the pool lies along the line y  1.6, and the plate’s top edge along the line y  1. The plate’s right-hand edge lies along the line y  x, with the upper-right vertex at (1, 1). The length of a thin strip at level y is

L ( y)  2 x  2 y. The depth of the strip beneath the surface is (1.6 − y ). The force exerted by the water against one side of the plate is therefore F = =

b

∫a

⎛ strip ⎞⎟ ⎟⎟ ⋅ L ( y ) dy w ⋅ ⎜⎜⎜ ⎜⎝ depth ⎟⎠

Eq. (7)

1

∫0 9800 (1.6 − y ) 2 y dy 1

= 19,600 ∫ (1.6 y − y 2 ) dy 0

y3 ⎤1 ⎡ = 19,600 ⎢ 0.8 y 2 − ⎥ = 9147 N. 3 ⎦0 ⎣

EXERCISES

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6.5

For some exercises, a calculator may be helpful when expressing answers in decimal form.

1.

F (N) 24

Springs

The graphs of force functions (in newtons) are given in Exercises 1 and 2. How much work is done by each force in moving an object 10 m?

16 8 0

2

4

6

8

10

x (m)

416 2.

Chapter 6 Applications of Definite Integrals

F (N)

12. Force of attraction When a particle of mass m is at ( x, 0 ), it is attracted toward the origin with a force whose magnitude is k x 2 . If the particle starts from rest at x  b and is acted on by no other forces, find the work done on it by the time it reaches x = a,  0 < a < b.

Quarter-circle 6 4 2 0

2

4

6

8

10

x (m)

3. Spring constant It took 1800 J of work to stretch a spring from its natural length of 2 m to a length of 5 m. Find the spring’s force constant.

13. Leaky bucket Assume the bucket in Example 4 is leaking. It starts with 8 L of water (78 N) and leaks at a constant rate. It finishes draining just as it reaches the top. How much work was spent lifting the water alone? (Hint: Do not include the rope and bucket, and find the proportion of water left at elevation x m.)

b. How much work is done in stretching the spring from 10 cm to 12 cm?

14. (Continuation of Exercise 11) The workers in Example 4 and Exercise 11 changed to a larger bucket that held 20 L (195 N) of water, but the new bucket had an even larger leak so that it, too, was empty by the time it reached the top. Assuming that the water leaked out at a steady rate, how much work was done lifting the water alone? (Do not include the rope and bucket.)

c. How far beyond its natural length will a 1600-N force stretch the spring?

Pumping Liquids from Containers

4. Stretching a spring A spring has a natural length of 10 cm. An 800-N force stretches the spring to 14 cm. a. Find the force constant.

5. Stretching a rubber band A force of 2 N will stretch a rubber band 2 cm (0.02 m). Assuming that Hooke’s Law applies, how far will a 4-N force stretch the rubber band? How much work does it take to stretch the rubber band this far? 6. Stretching a spring If a force of 90 N stretches a spring 1 m beyond its natural length, how much work does it take to stretch the spring 5 m beyond its natural length? 7. Subway car springs It takes a force of 96,000 N to compress a coil spring assembly on a New York City Transit Authority subway car from its free height of 20 cm to its fully compressed height of 12 cm. a. What is the assembly’s force constant? b. How much work does it take to compress the assembly the first centimeter? the second centimeter? Answer to the nearest joule. 8. Bathroom scale A bathroom scale is compressed 1.5 mm when a 70-kg person stands on it. Assuming that the scale behaves like a spring that obeys Hooke’s Law, how much does someone who compresses the scale 3 mm weigh? How much work is done compressing the scale 3 mm? Work Done by a Variable Force

15. Pumping water The rectangular tank shown here, with its top at ground level, is used to catch runoff water. Assume that the water weighs 9800 N m 3. a. How much work does it take to empty the tank by pumping the water back to ground level once the tank is full? b. If the water is pumped to ground level with a 5-horsepower (hp) motor (work output 3678 W), how long will it take to empty the full tank (to the nearest minute)? c. Show that the pump in part (b) will lower the water level 10 m (halfway) during the first 266 min of pumping. d. The weight of water What are the answers to parts (a) and (b) in a location where water weighs 9780 N m 3?  9820 N m 3?

10 m Ground level 0

12 m

y Δy 20

9. Lifting a rope A mountain climber is about to haul up a 50-m length of hanging rope. How much work will it take if the rope weighs 0.624 N m?

y

10. Leaky sandbag A bag of sand originally weighing 600 N was lifted at a constant rate. As it rose, sand also leaked out at a constant rate. The sand was half gone by the time the bag had been lifted to 6 m. How much work was done lifting the sand this far? (Neglect the weight of the bag and lifting equipment.)

16. Emptying a cistern The rectangular cistern (storage tank for rainwater) shown has its top 3 m below ground level. The cistern, currently full, is to be emptied for inspection by pumping its contents to ground level.

11. Lifting an elevator cable An electric elevator with a motor at the top has a multistrand cable weighing 60 N m . When the car is at the first floor, 60 m of cable are paid out, and effectively 0 m are out when the car is at the top floor. How much work does the motor do just lifting the cable when it takes the car from the first floor to the top?

a. How much work will it take to empty the cistern? b. How long will it take a 1 2-hp pump, rated at 370 W, to pump the tank dry? c. How long will it take the pump in part (b) to empty the tank halfway? (It will be less than half the time required to empty the tank completely.)

6.5  Work and Fluid Forces

d. The weight of water What are the answers to parts (a) through (c) in a location where water weighs 9780 N m 3? 9820 N m 3? Ground level

0 3

24. You are in charge of the evacuation and repair of the storage tank shown here. The tank is a hemisphere of radius 3 m and is full of benzene weighing 8800 N m 3 . A firm you contacted says it can empty the tank for 0.4¢ per joule of work. Find the work required to empty the tank by pumping the benzene to an outlet 0.6 m above the top of the tank. If you have $5000 budgeted for the job, can you afford to hire the firm?

3m

y

6 4m

6m

y

18. Pumping a half-full tank Suppose that, instead of being full, the tank in Example 5 is only half full. How much work does it take to pump the remaining oil to a level 1 m above the top of the tank? 19. Emptying a tank A vertical right-circular cylindrical tank measures 9 m high and 6 m in diameter. It is full of kerosene weighing 7840 N m 3 . How much work does it take to pump the kerosene to the level of the top of the tank? 20. a. Pumping milk Suppose that the conical container in Example 5 contains milk (weighing 10,100 N m 3) instead of olive oil. How much work will it take to pump the contents to the rim? b. Pumping oil How much work will it take to pump the oil in Example 5 to a level 1 m above the cone’s rim? 21. The graph of y  x 2 on 0 b x b 2 is revolved about the y-axis to form a tank that is then filled with salt water from the Dead Sea (weighing approximately 11, 500 N m 3). How much work does it take to pump all of the water to the top of the tank? 22. A right-circular cylindrical tank of height 3 m and radius 1.5 m is lying horizontally and is full of diesel fuel weighing 8300 N m 3. How much work is required to pump all of the fuel to a point 4.5 m above the top of the tank? 23. Emptying a water reservoir We model pumping from spherical containers the way we do from other containers, with the axis of integration along the vertical axis of the sphere. Use the figure here to find how much work it takes to empty a full hemispherical water reservoir of radius 5 m by pumping the water to a height of 4 m above the top of the reservoir. Water weighs 9800 N m 3 . y

5

Outlet pipe

x2 + y2 + z2 = 9 3

0.6 m 0

17. Pumping oil How much work would it take to pump oil from the tank in Example 5 to the level of the top of the tank if the tank were completely full?

0

417

3

x

z

Work and Kinetic Energy

25. Kinetic energy If a variable force of magnitude F ( x ) moves an object of mass m along the x-axis from x 1 to x 2 , the object’s velocity V can be written as dx dt (where t represents time). Use Newton’s second law of motion F = m ( d V dt ) and the Chain Rule dV d V dx dV   V dt dx dt dx to show that the net work done by the force in moving the object from x 1 to x 2 is W =

∫x

x2 1

F ( x ) dx =

1 1 m V 2 2 − m V1 2 , 2 2

where V1 and V 2 are the object’s velocities at x 1 and x 2 . In physics, the expression (1 2 ) mV 2 is called the kinetic energy of an object of mass m moving with velocity V. Therefore, the work done by the force equals the change in the object’s kinetic energy, and we can find the work by calculating this change. In Exercises 26–30, use the result of Exercise 25. 26. Tennis A 57 g tennis ball was served at 50 m s (180 kmh). How much work was done on the ball to make it go this fast? 27. Baseball How many joules of work does it take to throw a baseball 144 kmh? A baseball’s mass is 150 g. 28. Golf A 50 g golf ball is driven off the tee at a speed of 84 m s (302.4 kmh). How many joules of work are done on the ball getting it into the air? 29. Tennis At the 2012 Busan Open Challenger Tennis Tournament in Busan, South Korea, the Australian Samuel Groth hit a serve measured at 263 kmph. How much work was required by Groth to serve a 0.0567-kg tennis ball at that speed? 30. Softball How much work has to be performed on a 200 g softball to pitch it 40 m s (144 kmh)?

4m

y

x Δy "25 − y2

0 y 0 = −y

31. Drinking a milkshake The truncated conical container shown here is full of strawberry milkshake, which has a density of 0.8 g cm 3. As you can see, the container is 18 cm deep, 6 cm across at the base, and 9 cm across at the top (a standard size at Brigham’s in Boston). The straw sticks up 3 cm above the top. About how much work does it take to suck up the milkshake through the straw (neglecting friction)?

418

Chapter 6 Applications of Definite Integrals

y y = 12x − 36

21 18

21 − y

34. Forcing electrons together each other with a force of F =

(4.5, 18) y + 36 12

y Δy 0

x

3 Dimensions in centimeters

32. Water tower Your town has decided to drill a well to increase its water supply. As the town engineer, you have determined that a water tower will be necessary to provide the pressure needed for distribution, and you have designed the system shown here. The water is to be pumped from a 90 m well through a vertical 10 cm pipe into the base of a cylindrical tank 6 m in diameter and 7.5 m high. The base of the tank will be 18 m above ground. The pump is a 3-hp pump, rated at 2200 W ( J s ) . To the nearest hour, how long will it take to fill the tank the first time? (Include the time it takes to fill the pipe.) Assume that water weighs 9800 N m 3 . 3m

Two electrons r meters apart repel

23 × 10 −29 newtons. r2

a. Suppose one electron is held fixed at the point (1, 0 ) on the x-axis (units in meters). How much work does it take to move a second electron along the x-axis from the point (−1, 0 ) to the origin? b. Suppose an electron is held fixed at each of the points (−1, 0 ) and (1, 0 ). How much work does it take to move a third electron along the x-axis from ( 5, 0 ) to ( 3, 0 )? Finding Fluid Forces

35. Triangular plate Calculate the fluid force on one side of the plate in Example 6 using the coordinate system shown here. y (m)

Surface of pool 1.6

0 Depth 0 y 0

x (m)

y = −0.6 y

7.5 m

x

(x, y)

−1.6 Ground

18 m

36. Triangular plate Calculate the fluid force on one side of the plate in Example 6 using the coordinate system shown here.

0.1 m 90 m Water surface

y (m) Pool surface at y = 0.6 1

Submersible pump NOT TO SCALE

33. Putting a satellite in orbit The strength of Earth’s gravitational field varies with the distance r from Earth’s center, and the magnitude of the gravitational force experienced by a satellite of mass m during and after launch is mMG F (r ) = . r2 Here, M = 5.975 × 10 24 kg is Earth’s mass, G = 6.6720 × 10 −11 N ⋅ m 2 kg −2 is the universal gravitational constant, and r is measured in meters. The work it takes to lift a 1000-kg satellite from Earth’s surface to a circular orbit 35,780 km above Earth’s center is therefore given by the integral Work =

35,780,000

∫6,370,000

1000 MG dr joules. r2

Evaluate the integral. The lower limit of integration is Earth’s radius in meters at the launch site. (This calculation does not take into account energy spent lifting the launch vehicle or energy spent bringing the satellite to orbit velocity.)

−1

0

1

x (m)

−1

37. Rectangular plate In a pool filled with water to a depth of 3 m, calculate the fluid force on one side of a 0.9 m by 1.2 m rectangular plate if the plate rests vertically at the bottom of the pool a. on its 1.2-m edge. b. on its 0.9-m edge. 38. Semicircular plate Calculate the fluid force on one side of a semicircular plate of radius 5 m that rests vertically on its diameter at the bottom of a pool filled with water to a depth of 6 m. y Surface of water

6 5

x

419

6.5  Work and Fluid Forces

39. Triangular plate The isosceles triangular plate shown here is submerged vertically 1 m below the surface of a freshwater lake.

4m

y (m) (−1, 1)

a. Find the fluid force against one face of the plate. b. What would be the fluid force on one side of the plate if the water were seawater instead of freshwater?

A

Surface level 4m

4m

y = x2 −1

4m 1m

B

4m

40. Rotated triangular plate The plate in Exercise 37 is revolved 180° about line AB so that part of the plate sticks out of the lake, as shown here. What force does the water exert on one face of the plate now?

(1, 1)

0

x (m)

1

Enlarged view of parabolic gate

Parabolic gate

46. The end plates of the trough shown here were designed to withstand a fluid force of 25,000 N. How many cubic meters of water can the tank hold without exceeding this limitation? Round down to the nearest cubic meter. What is the value of h? y (m)

2m

(1, 3)

(−1, 3) (0, h)

y = 3x 3m

3m

A

Surface level 4m

B

1m

41. New England Aquarium The viewing portion of the rectangular glass window in a typical fish tank at the New England Aquarium in Boston is 1.6 m wide and runs from 0.01 m below the water’s surface to 0.85 m below the surface. Find the fluid force against this portion of the window. The weight-density of seawater is 10, 050 N m 3 . (In case you were wondering, the glass is 2 cm thick and the tank walls extend 10 cm above the water to keep the fish from jumping out.) 42. Semicircular plate A semicircular plate 2 m in diameter sticks straight down into freshwater with the diameter along the surface. Find the force exerted by the water on one side of the plate. 43. Tilted plate Calculate the fluid force on one side of a 1 m by 1 m square plate if the plate is at the bottom of a pool filled with water to a depth of 2 m and

x (m)

0

10 m Dimensional view of trough

End view of trough

47. A vertical rectangular plate a units long by b units wide is submerged in a fluid of weight-density w with its long edges parallel to the fluid’s surface. Find the average value of the pressure along the vertical dimension of the plate. Explain your answer. 48. (Continuation of Exercise 47.) Show that the force exerted by the fluid on one side of the plate is the average value of the pressure (found in Exercise 47) times the area of the plate. 49. Water pours into the tank shown here at the rate of 0.5 m 3 min . The tank’s cross-sections are 2-m-diameter semicircles. One end of the tank is movable, but moving it to increase the volume compresses a spring. The spring constant is k  3000 N m. If the end of the tank moves 2.5 m against the spring, the water will drain out of a safety hole in the bottom at the rate of 0.6 m 3 min . Will the movable end reach the hole before the tank overflows? Movable end

Water in

a. lying flat on its 1 m by 1 m face.

Water in

b. resting vertically on a 1 m edge. c. resting on a 1 m edge and tilted at 45 n to the bottom of the pool. 44. Tilted plate Calculate the fluid force on one side of a right-triangular plate with edges 3 m, 4 m, and 5 m if the plate sits at the bottom of a pool filled with water to a depth of 6 m on its 3-m edge and tilted at 60 n to the bottom of the pool. 45. The cubical metal tank shown here has a parabolic gate held in place by bolts and designed to withstand a fluid force of 25,000 N without rupturing. The liquid you plan to store has a weight-density of  8000 N m 3. a. What is the fluid force on the gate when the liquid is 2 m deep? b. What is the maximum height to which the container can be filled without exceeding the gate’s design limitation?

y

Drain hole

2.5 m Side view

2m

Movable end

x x 2 + y2 = 1 Drain hole

1m

50. Watering trough The vertical ends of a watering trough are squares 1 m on a side. a. Find the fluid force against the ends when the trough is full. b. How many centimeters do you have to lower the water level in the trough to reduce the fluid force by 25%?

420

Chapter 6 Applications of Definite Integrals

6.6 Moments and Centers of Mass Many structures and mechanical systems behave as if their masses were concentrated at a single point, called the center of mass (Figure 6.44). It is important to know how to locate this point, and doing so is basically a mathematical enterprise. Here we consider masses distributed along a line or region in the plane. Masses distributed across a region or curve in three-dimensional space are treated in Chapters 14 and 15.

Masses Along a Line We develop our mathematical model in stages. The first stage is to imagine masses m1 ,  m 2 , and m 3 on a rigid x-axis supported by a fulcrum at the origin. x1

x2

x3

m2

m3

0

m1

x

Fulcrum at origin

The resulting system might balance, or it might not, depending on how large the masses are and how they are arranged along the x-axis. Each mass m k exerts a downward force m k g (the weight of m k ) equal to the magnitude of the mass times the acceleration due to gravity. Note that gravitational acceleration is downward, hence negative. Each of these forces has a tendency to turn the x-axis about the origin, the way a child turns a seesaw. This turning effect, called a torque, is measured by multiplying the force m k g by the signed distance x k from the point of application to the origin. By convention, a positive torque induces a counterclockwise turn. Masses to the left of the origin exert positive (counterclockwise) torque. Masses to the right of the origin exert negative (clockwise) torque. The sum of the torques measures the tendency of a system to rotate about the origin. This sum is called the system torque. System torque = m1gx 1 + m 2 gx 2 + m 3 gx 3

(1)

The system will balance if and only if its torque is zero. If we factor out the g in Equation (1), we see that the system torque is g ⋅ ( m1 x 1 + m 2 x 2 + m 3 x 3 ).   a feature of the environment

a feature of the system

Thus, the torque is the product of the gravitational acceleration g, which is a feature of the environment in which the system happens to reside, and the number ( m1 x 1 + m 2 x 2 + m 3 x 3 ), which is a feature of the system itself. The number ( m1 x 1 + m 2 x 2 + m 3 x 3 ) is called the moment of the system about the origin. It is the sum of the moments m1 x 1 , m 2 x 2 , m 3 x 3 of the individual masses. M0 = Moment of system about origin =

∑ mk x k

(We shift to sigma notation here to allow for sums with more terms.) We usually want to know where to place the fulcrum to make the system balance; that is, we want to know at what point x to place the fulcrum to make the torques add to zero. FIGURE 6.44

A wrench gliding on ice turning about its center of mass as the center glides in a vertical line. (Source: Berenice Abbott/ScienceSource)

x1 m1

0

x2

x

m2

x3 m3

Special location for balance

x

6.6  Moments and Centers of Mass

421

The torque of each mass about the fulcrum in this special location is ⎛ signed distance ⎞⎟ ⎛ downward ⎞ ⎟⎟ = ( x − x ) m g. ⎟⎟ ⎜⎜ Torque of  m k about  x = ⎜⎜⎜ k k ⎜⎝ of  m k from  x ⎟⎟⎠ ⎜⎝ force ⎟⎠ When we write the equation that says that the sum of these torques is zero, we get an equation we can solve for x:

∑ (xk

− x ) mkg = 0 x =

Sum of the torques equals zero.

∑ mk x k . ∑ mk

Solved for x

This last equation tells us to find x by dividing the system’s moment about the origin by the system’s total mass: x =

∑ mk x k ∑ mk

=

system moment about origin . system mass

(2)

The point x is called the system’s center of mass.

Thin Wires Instead of a discrete set of masses arranged in a line, suppose that we have a straight wire or rod located on interval [ a, b ] on the x-axis. Suppose further that this wire is not homogeneous, but rather the density varies continuously from point to point. If a short segment of a rod containing the point x with length Δx has mass Δm, then the density at x is given by δ ( x ) = lim Δm Δx .  Δx → 0

Mass Δm k a

Δx k x k−1 xk

b

x

FIGURE 6.45 A rod of varying density can be modeled by a finite number of point masses of mass Δm k = δ ( x k ) Δx k located at points x k along the rod.

We often write this formula in one of the alternative forms δ = dm dx and dm = δ dx. Partition the interval [ a, b ] into finitely many subintervals [ x k −1 , x k ]. If we take n subintervals and replace the portion of a wire along a subinterval of length Δx k containing x k by a point mass located at x k with mass Δm k = δ ( x k ) Δx k , then we obtain a collection of point masses that have approximately the same total mass and same moment as the wire as indicated in Figure 6.45. The mass M of the wire and the moment M0 are approximated by the Riemann sums M ≈

n

∑ Δm k

=

k =1

n

∑ δ ( x k ) Δx k , k =1

n

∑ x k Δm k

M0 ≈

k =1

=

n

∑ x k δ ( x k ) Δx k . k =1

By taking a limit of these Riemann sums as the length of the intervals in the partition approaches zero, we get integral formulas for the mass and the moment of the wire about the origin. The mass M, moment about the origin M0 , and center of mass x are b

M =

b

∫a δ ( x ) dx ,

M0 =

b

∫a

x δ ( x ) dx ,

x =

M0 = M

∫a x δ ( x ) dx . b ∫a δ ( x ) dx

EXAMPLE 1 Find the mass M and the center of mass x of a rod lying on the x-axis over the interval [ 1, 2 ] whose density is given by δ ( x ) = 2 + 3 x 2 . Solution The mass of the rod is obtained by integrating the density,

M =

2

∫1 ( 2 + 3x 2 ) dx

= [ 2 x + x 3 ]12 = ( 4 + 8 ) − ( 2 + 1) = 9,

and the center of mass is x =

M0 = M

∫1

2

x ( 2 + 3 x 2 ) dx 9

2 ⎡ x 2 + 3x 4 ⎤ ⎢⎣ ⎥ 4 ⎦ 1 = 19 . = 9 12

422

Chapter 6 Applications of Definite Integrals

y

Masses Distributed over a Plane Region

yk

xk

0

mk

Suppose that we have a finite collection of masses located in the plane, with mass m k at the point ( x k , y k ) (see Figure 6.46). The mass of the system is

(xk, yk )

System mass:

yk xk

x

Moment about  x -axis:

Each mass m k has a moment about each axis.

The x-coordinate of the system’s center of mass is defined to be Ba x = la nc x el in e

x =

y c.m. y=y Balan ce line

0 x

x

FIGURE 6.47 A two-dimensional array of masses balances on its center of mass.

~ y

Strip of mass Δm

~~ (x, y) ~ y

0

~ x

∑ mk yk , = ∑ mk x k.

Mx =

Moment about  y-axis:   My

y

Strip c.m. ~ x

∑ mk.

Each mass m k has a moment about each axis. Its moment about the x-axis is m k y k , and its moment about the y-axis is m k x k . The moments of the entire system about the two axes are

FIGURE 6.46

y

M =

x

FIGURE 6.48 A plate cut into thin strips parallel to the y-axis. The moment exerted by a typical strip about each axis is the moment its mass Δm would exert if concentrated at the strip’s center of mass ( x, y ).

My = M

∑ mk x k . ∑ mk

(3)

With this choice of x , as in the one-dimensional case, the system balances about the line x = x (Figure 6.47). The y-coordinate of the system’s center of mass is defined to be M ∑ mk yk . y = x = (4) M ∑ mk

With this choice of y, the system balances about the line y = y as well. The torques exerted by the masses about the line y = y cancel out. Thus, as far as balance is concerned, the system behaves as if all its mass were at the single point ( x ,  y ). We call this point the system’s center of mass (c.m.).

Thin, Flat Plates In many applications, we need to find the center of mass of a thin, flat plate: a disk of aluminum, say, or a triangular sheet of steel. In such cases, we assume the distribution of mass to be continuous, and the formulas we use to calculate x and y contain integrals instead of finite sums. The integrals arise in the following way. Imagine that the plate occupying a region in the xy-plane is cut into thin strips parallel to one of the axes (in Figure 6.48, the y-axis). The center of mass of a typical strip is ( x, y ). We treat the strip’s mass Δm as if it were concentrated at ( x, y ). The moment of the strip about the y-axis is then x Δm. The moment of the strip about the x-axis is y Δm. Equations (3) and (4) then become My ∑ x Δm , y = Mx = ∑ y Δm . x = = M M ∑ Δm ∑ Δm These sums are Riemann sums for integrals, and they approach these integrals in the limit as the strips become narrower and narrower. We write these integrals symbolically as x =

∫ x dm ∫ dm

and

y =

∫ y dm . ∫ dm

Moments, Mass, and Center of Mass of a Thin Plate Covering a Region in the xy-Plane

Density of a plate A material’s density is its mass per unit area. For wires, rods, and narrow strips, the density is given in terms of mass per unit length.

Moment about the  x -axis:

Mx =

∫ y dm

Moment about the  y-axis:

My =

∫ x dm

Mass:

M =

∫ dm

Center of mass:

x =

My M , y = x M M

(5)

6.6  Moments and Centers of Mass

y (cm)

Although we do not indicate the limits of integration, the integrals that appear in Equation (5) are definite integrals. The differential dm in these integrals is the mass of the strip. For this section, we assume the density δ of the plate is a constant or a continuous function of x or of y. Then dm = δ dA, which is the mass per unit area δ times the area dA of the strip. To evaluate the integrals in Equations (5), we picture the plate in the coordinate plane and sketch a strip of mass parallel to one of the coordinate axes. We then express the strip’s mass dm and the coordinates ( x, y ) of the strip’s center of mass in terms of x or y. Finally, we integrate y dm, x dm, and dm between limits of integration determined by the plate’s location in the plane.

(1, 2)

2

y = 2x x=1

y=0

0

x (cm)

1

FIGURE 6.49

423

The plate in Example 2.

EXAMPLE 2 δ = 3 g cm 2 .

The triangular plate shown in Figure 6.49 has a constant density of

(a) Find the plate’s moment My   about the y-axis. (b) Find the plate’s mass M. (c) Find the x-coordinate of the plate’s center of mass (c.m.). (Figure 6.50)

Solution Method 1: Vertical Strips

(a) The moment My : The typical vertical strip has the following relevant data.

( x, y ) = ( x , x )

center of mass  ( c.m. ): y (cm)

length:

2x

width:

dx

(1, 2)

2

dA = 2 x dx

area:

y = 2x

dm = δ dA = 3 ⋅ 2 x dx = 6 x dx

mass:

(x, 2x)

x = x

distance of c.m. from  y-axis: Strip c.m. is halfway. ~~ (x, y) = (x, x)

x

The moment of the strip about the y-axis is x dm = x ⋅ 6 x dx = 6 x 2 dx.

2x

The moment of the plate about the y-axis is therefore dx

0

x (cm)

1

FIGURE 6.50 Modeling the plate in Example 2 with vertical strips.

∫

x dm =

1

∫0

1

6 x 2 dx = 2 x 3 ⎥⎤ = 2 g ⋅ cm. ⎦0

(b) The plate’s mass: M =

∫

dm =

1

∫0

1

6 x dx = 3 x 2 ⎥⎤ = 3 g. ⎦0

(c) The x-coordinate of the plate’s center of mass:

y (cm) 2 x= y a , yb 2

y 1+

x =

(1, 2)

y 2

1−

Strip c.m. is halfway. y+2 ~~ , yb (x, y) = a 4

y 2

dy

y 2 1

My 2 g ⋅ cm 2 = = cm. M 3g 3

By a similar computation, we could find Mx and y = Mx M . Method 2: Horizontal Strips

(Figure 6.51)

(a) The moment My : The y-coordinate of the center of mass of a typical horizontal strip is y (see the figure), so y = y.

(1, y)

2 0

My =

x (cm)

FIGURE 6.51 Modeling the plate in Example 2 with horizontal strips.

The x-coordinate is the x-coordinate of the point halfway across the triangle. This makes it the average of y 2 (the strip’s left-hand x-value) and 1 (the strip’s right-hand x-value): x =

( y 2) + 1 y y+2 1 = + = . 2 4 2 4

424

Chapter 6 Applications of Definite Integrals

We also have length:

1−

width:

dy

y 2−y = 2 2 2−y dy 2

area:

dA =

mass:

dm = δ dA = 3 ⋅

distance of c.m. to y-axis:

x =

2−y dy 2

y+2 . 4

The moment of the strip about the y-axis is x dm =

y+2 2−y 3 ⋅3⋅ dy = ( 4 − y 2 ) dy. 4 2 8

The moment of the plate about the y-axis is My =

∫ x dm

( )

2

y3 ⎤ 2 3( 3⎡ 3 16 4 − y 2 ) dy = ⎢ 4 y − = 2 g ⋅ cm. ⎥ = 8 8⎣ 3 ⎦0 8 3

2

y2 ⎤ 2 3 3( 3⎡ 2 − y ) dy = ⎢ 2 y − ⎥ = ( 4 − 2 ) = 3 g. 2 2⎣ 2 ⎦0 2

=

∫0

=

∫0

(b) The plate’s mass: M =

∫ dm

(c) The x-coordinate of the plate’s center of mass: x =

My 2 g ⋅ cm 2 = = cm. M 3g 3

By a similar computation, we could find Mx and y. If the distribution of mass in a thin, flat plate has an axis of symmetry, the center of mass will lie on this axis. If there are two axes of symmetry, the center of mass will lie at their intersection. These facts often help to simplify our work. y 4 y = 4 − x2 Center of mass 4 − x2 ~~ b (x, y) = ax, 2

4 − x2

−2

0

FIGURE 6.52

y 2 x dx

2

x

Modeling the plate in Example 3 with vertical strips.

EXAMPLE 3 Find the center of mass of a thin plate covering the region bounded above by the parabola y = 4 − x 2 and below by the x-axis (Figure 6.52). Assume the density of the plate at the point (x, y) is δ = 2 x 2 , which is twice the square of the distance from the point to the y-axis. Solution The mass distribution is symmetric about the y-axis, so x = 0. We model the distribution of mass with vertical strips, since the density is given as a function of the variable x. The typical vertical strip (see Figure 6.52) has the following relevant data.

center of mass  ( c.m. ): length: width: area: mass: distance from c.m. to x -axis:

(

( x, y ) = x ,

4 − x2 2

)

4 − x2 dx dA = ( 4 − x 2 ) dx dm = δ dA = δ ( 4 − x 2 ) dx 2 y = 4 − x . 2

The moment of the strip about the x-axis is 2 y dm = 4 − x ⋅ δ ( 4 − x 2 ) dx = δ ( 4 − x 2 ) 2 dx. 2 2

425

6.6  Moments and Centers of Mass

The moment of the plate about the x-axis is Mx = =

∫ y dm

δ

2

∫−2 2 ( 4 − x 2 )

=

2

2

∫−2 (16 x 2 − 8 x 4 + x 6 ) dx

dx =

2

∫−2 x 2 ( 4 − x 2 )

2

dx

2048 . 105

=

The mass of the plate is M = =

∫ dm

=

2

∫−2 δ ( 4 − x 2 ) dx

2

∫−2 ( 8 x 2 − 2 x 4 ) dx

=

=

2

∫−2 2 x 2 ( 4 − x 2 ) dx

256 . 15

Therefore, Mx 2048 15 8 = ⋅ = . M 105 256 7

y = The plate’s center of mass is

( x ,  y )

y

)

8 . 7

Plates Bounded by Two Curves y = f (x)

0

(

= 0, 

Suppose a plate covers a region that lies between two curves y = g( x ) and y = f ( x ), where f ( x ) ≥ g( x ) and a ≤ x ≤ b. The typical vertical strip (see Figure 6.53) has

(

center of mass  ( c.m. ):

y = g(x)

length:

f ( x ) − g( x )

width:

dx

a

dx

b

x

Modeling the plate bounded by two curves with vertical strips. The strip c.m. is halfway, so y = 1 [ f ( x ) + g( x ) ]. 2

dA = [ f ( x ) − g( x ) ] dx

area:

FIGURE 6.53

)

1 ( x, y ) = x , [ f ( x ) + g( x ) ] 2

~~ (x, y)

dm = δ dA = δ [ f ( x ) − g( x ) ] dx.

mass:

The moment of the plate about the y-axis is My =

∫ x dm

=

b

∫a

xδ [ f ( x ) − g( x ) ] dx ,

and the moment about the x-axis is Mx = =

∫ y dm b

∫a

=

b

∫a

1 [ f ( x ) + g( x ) ] ⋅ δ [ f ( x ) − g( x ) ] dx 2

δ 2 [ f ( x ) − g 2 ( x ) ] dx. 2

These moments give us the following formulas. b

x =

1 M

∫a

y =

1 M

∫a

b

δ x [ f ( x ) − g( x ) ] dx δ 2 [ f ( x ) − g 2 ( x ) ] dx 2

(6) (7)

426

Chapter 6 Applications of Definite Integrals

y

EXAMPLE 4 Find the center of mass for the thin plate bounded by the curves g( x ) = x 2 and f ( x ) = x , 0 ≤ x ≤ 1 (Figure 6.54), using Equations (6) and (7) with the density function δ ( x ) = x 2 .

1 f (x) = "x

Solution We first compute the mass of the plate, using dm = δ [ f ( x ) − g( x ) ] dx : c.m.

M = g(x) = x 2

FIGURE 6.54

1

x −

)

x dx = 2

∫0 ( x 5 2 − 1

)

1 2 1 9 x3 . dx = ⎢⎡ x 7 2 − x 4 ⎤⎥ = ⎣7 2 8 ⎦0 56

Then, from Equations (6) and (7) we get

x

1

0

∫0 x 2 (

x =

The region in Example 4.

56 1 2 x ⋅x 9 ∫0

(

x −

(

)

x dx 2

)

=

x4 56 1 7 2 x − dx ∫ 9 0 2

=

56 ⎡ 2 9 2 1 5 ⎤1 308 x − x ⎥ = , ⎢ 9 ⎣9 10 ⎦ 0 405

and y =

(

)

56 1 x 2 x2 x− dx ∫ 9 0 2 4

(

)

=

28 1 3 x4 x − dx ∫ 9 0 4

=

28 ⎡ 1 4 1 5 ⎤1 28 x − x = . ⎢ 9 ⎣4 20 ⎥⎦ 0 45

The center of mass is shown in Figure 6.54. y

Centroids

A typical small segment of wire has dm = d ds = da d u.

y = "a2 − x 2

~~ (x, y) = (a cos u, a sin u)

du

u −a

a

0

x

(a)

The center of mass in Example 4 is not located at the geometric center of the region. This is due to the region’s nonuniform density. When the density function is constant, it cancels out of the numerator and denominator of the formulas for x and y . Thus, when the density is constant, the location of the center of mass is a feature of the geometry of the object and not of the material from which it is made. In such cases, engineers may call the center of mass the centroid of the shape, as in “Find the centroid of a triangle or a solid cone.” To do so, just set δ equal to 1 and proceed to find x and y as before, by dividing moments by masses.

y

EXAMPLE 5 Find the center of mass (centroid) of a thin wire of constant density δ shaped like a semicircle of radius a.

a c.m.

−a

2 a0, p ab

0

a

(b)

FIGURE 6.55 The semicircular wire in Example 5. (a) The dimensions and variables used in finding the center of mass. (b) The center of mass does not lie on the wire.

x

Solution We model the wire with the semicircle y = a 2 − x 2 (Figure 6.55). The distribution of mass is symmetric about the y-axis, so x = 0. To find y, we imagine the wire divided into short subarc segments. If ( x, y ) is the center of mass of a subarc and θ is the angle between the x-axis and the radial line joining the origin to ( x, y ), then y = a sin θ is a function of the angle θ measured in radians (see Figure 6.55a). The length ds of the subarc containing ( x, y ) subtends an angle of dθ radians, so ds = a dθ. Thus a typical subarc segment has these relevant data for calculating y:

length: mass: distance of c.m. to x -axis:

ds = a dθ dm = δ ds = δ a dθ y = a sin θ.

Mass per unit length times length

427

6.6  Moments and Centers of Mass

Hence, y =

∫ y dm ∫ dm

π

=

∫0

a sin θ ⋅ δ a dθ π

∫0

δ a dθ

=

δ a 2 [ − cos θ ]π0 2 = a. δ aπ π

The center of mass lies on the axis of symmetry at the point ( 0, 2a Q ), about two-thirds of the way up from the origin (Figure 6.55b). Notice how E cancels in the equation for y , so we could have set E  1 everywhere and obtained the same value for y. In Example 5 we found the center of mass of a thin wire lying along the graph of a differentiable function in the xy-plane. In Chapter 15 we will learn how to find the center of mass of a wire lying along a more general smooth curve in the plane or in space.

Fluid Forces and Centroids Surface level of fluid h = centroid depth

If we know the location of the centroid of a submerged flat vertical plate (Figure 6.56), we can take a shortcut to find the force against one side of the plate. From Equation (7) in Section 6.5, and the definition of the moment about the x-axis, we have F =

Plate centroid

b

∫a

w × ( strip depth ) × L ( y) dy b

= w ∫ ( strip depth ) × L ( y) dy a

= w × ( moment about surface level line of region occupied by plate ) = w × ( depth of plate’s centroid ) × ( area of plate ).

FIGURE 6.56 The force against one side of the plate is w ¸ h ¸ plate area.

Fluid Forces and Centroids

The force of a fluid of weight-density w against one side of a submerged flat vertical plate is the product of w, the distance h from the plate’s centroid to the fluid surface, and the plate’s area: F  whA.

(8)

EXAMPLE 6 A flat isosceles triangular plate with base 2 m and height 1 m is submerged vertically, base up with its vertex at the origin, so that the base is 0.6 m below the surface of a swimming pool. (This is Example 6, Section 6.5.) Use Equation (8) to find the force exerted by the water against one side of the plate. Solution The centroid of the triangle (Figure 6.43) lies on the y-axis, one-third of the way from the base to the vertex, so h = 2.8/3 (where y = 2/3), since the pool’s surface is y  1.6. The triangle’s area is 1 1 A = ( base )( height ) = ( 2 )(1) = 1. 2 2

Hence, F = whA = ( 9800 )( 2.8/3 )(1) = 9147 N.

The Theorems of Pappus In the fourth century, an Alexandrian Greek named Pappus discovered two formulas that relate centroids to surfaces and solids of revolution. The formulas provide shortcuts to a number of otherwise lengthy calculations.

428

Chapter 6 Applications of Definite Integrals

y

THEOREM 1—Pappus’s Theorem for Volumes

If a plane region is revolved once about a line in the plane that does not cut through the region’s interior, then the volume of the solid it generates is equal to the region’s area times the distance traveled by the region’s centroid during the revolution. If ρ is the distance from the axis of revolution to the centroid, then

d L(y)

r=y

V = 2πρ A.

R c

(9)

Centroid x

0

FIGURE 6.57 The region R is to be revolved (once) about the x-axis to generate a solid. A 1700-year-old theorem says that the solid’s volume can be calculated by multiplying the region’s area by the distance traveled by its centroid during the revolution.

Proof We draw the axis of revolution as the x-axis with the region R in the first quadrant (Figure 6.57). We let L ( y) denote the length of the cross-section of R perpendicular to the y-axis at y. We assume L ( y) to be continuous. By the method of cylindrical shells, the volume of the solid generated by revolving the region about the x-axis is V =

∫c

d

d

2π ( shell radius )( shell height ) dy = 2π ∫ y L ( y) dy. c

(10)

The y-coordinate of R’s centroid is y =

∫c

d

y dA A

=

∫c

d

y L ( y) dy A

,

y = y,   dA = L ( y) dy

so that

z

∫c

Distance from axis of revolution to centroid b

d

y L ( y) dy = Ay .

Substituting Ay for the last integral in Equation (10) gives V = 2π yA. With ρ equal to y, we have V = 2πρ A.

a

EXAMPLE 7 Find the volume of the torus (doughnut) generated by revolving a circular disk of radius a about an axis in its plane at a distance b ≥ a from its center (Figure 6.58). y x

Area: pa2 Circumference: 2pa

Solution We apply Pappus’s Theorem for volumes. The centroid of a disk is located at its center, the area is A = πa 2 , and ρ = b is the distance from the centroid to the axis of revolution (see Figure 6.58). Substituting these values into Equation (9), we find the volume of the torus to be

FIGURE 6.58

With Pappus’s first theorem, we can find the volume of a torus without having to integrate (Example 7).

y

V = 2π(b)(πa 2 ) = 2π 2 ba 2. The next example shows how we can use Equation (9) in Pappus’s Theorem to find one of the coordinates of the centroid of a plane region of known area A when we also know the volume V of the solid generated by revolving the region about the other coordinate axis. That is, if y is the coordinate we want to find, we revolve the region around the x-axis so that y = ρ is the distance from the centroid to the axis of revolution. The idea is that the rotation generates a solid of revolution whose volume V is an already known quantity. Then we can solve Equation (9) for ρ, which is the value of the centroid’s coordinate y.

a

EXAMPLE 8

4 a Centroid 3p −a

0

a

x

FIGURE 6.59 With Pappus’s first theorem, we can locate the centroid of a semicircular region without having to integrate (Example 8).

Locate the centroid of a semicircular region of radius a.

Solution We consider the region between the semicircle y = a 2 − x 2 (Figure 6.59) and the x-axis and imagine revolving the region about the x-axis to generate a solid sphere. By symmetry, the x-coordinate of the centroid is x = 0. With y = ρ in Equation (9), we have

y =

( 4 3 ) πa 3 V 4 = = a. 2π A 2 π (1 2 ) π a 2 3π

429

6.6  Moments and Centers of Mass

THEOREM 2—Pappus’s Theorem for Surface Areas

If an arc of a smooth plane curve is revolved once about a line in the plane that does not cut through the arc’s interior, then the area of the surface generated by the arc equals the length L of the arc times the distance traveled by the arc’s centroid during the revolution. If ρ is the distance from the axis of revolution to the centroid, then S = 2πρ L.

The proof we give assumes that we can model the axis of revolution as the x-axis and the arc as the graph of a continuously differentiable function of x.

y ds

Proof We draw the axis of revolution as the x-axis with the arc extending from x = a to x = b in the first quadrant (Figure 6.60). The area of the surface generated by the arc is

Arc ~ y 0

(11)

S =

a b

x

x=b

∫x =a

2π y ds = 2π ∫

x=b x=a

y ds.

(12)

The y-coordinate of the arc’s centroid is x=b

FIGURE 6.60

Figure for proving Pappus’s Theorem for surface area. The arc length differential ds is given by Equation (6) in Section 6.3.

y =

∫ x = a y ds x=b ∫ x = a ds

x=b

=

∫x =a

L

y ds

.

  

L = ∫ ds is the arc’s length and y = y.

Hence x=b

∫x =a

y ds = yL.

Substituting y L for the last integral in Equation (12) gives S = 2π y L. With ρ equal to y, we have S = 2πρ L. EXAMPLE 9 Example 7.

Use Pappus’s area theorem to find the surface area of the torus in

Solution From Figure 6.58, the surface of the torus is generated by revolving a circle of radius a about the z-axis, and b ≥ a is the distance from the centroid to the axis of revolution. The arc length of the smooth curve generating this surface of revolution is the circumference of the circle, so L = 2πa. Substituting these values into Equation (11), we find the surface area of the torus to be

S = 2π (b)(2πa) = 4 π 2 ba.

EXERCISES

6.6

Mass of a wire

In Exercises 1–6, find the mass M and center of mass x of the linear wire covering the given interval and having the given density δ ( x ). 1. 1 ≤ x ≤ 4, δ ( x ) =

x

2. −3 ≤ x ≤ 3, δ ( x ) = 1 + 3 x 2 1 x +1 8 4. 1 ≤ x ≤ 2, δ ( x ) = 3 x

3. 0 ≤ x ≤ 3, δ ( x ) =

⎧⎪ 4, 0 ≤ x ≤ 2 5. δ ( x ) = ⎪⎨ ⎪⎪⎩ 5, 2 < x ≤ 3 ⎧⎪ 2 − x , 0 ≤ x < 1 6. δ ( x ) = ⎪⎨ ⎪⎪⎩ x , 1≤ x ≤ 2 Thin Plates with Constant Density

In Exercises 7–20, find the center of mass of a thin plate of constant density δ covering the given region. 7. The region bounded by the parabola y = x 2 and the line y = 4

430

Chapter 6 Applications of Definite Integrals

8. The region bounded by the parabola y = 25 − x 2 and the x-axis 9. The region bounded by the parabola y = x − x 2 and the line y = −x 10. The region enclosed by the parabolas y =

x2

− 3 and y =

−2 x 2

11. The region bounded by the y-axis and the curve x = y − y 3 , 0 ≤ y ≤1

b. Find the center of mass of a thin plate covering the region if the plate’s density at the point (x, y) is δ ( x ) = 1 x . c. Sketch the plate and show the center of mass in your sketch. 26. The region between the curve y = 2 x and the x-axis from x = 1 to x = 4 is revolved about the x-axis to generate a solid. a. Find the volume of the solid.

12. The region bounded by the parabola x = y 2 − y and the line y = x

b. Find the center of mass of a thin plate covering the region if the plate’s density at the point (x, y) is δ ( x ) = x .

13. The region bounded by the x-axis and the curve y = cos x , −π 2 ≤ x ≤ π 2

c. Sketch the plate and show the center of mass in your sketch.

14. The region between the curve y = sec 2 x ,   −π 4 ≤ x ≤ π 4 and the x-axis T 15. The region between the curve y = 1 x and the x-axis from x = 1 to x = 2. Give the coordinates to two decimal places. 16. a. The region cut from the first quadrant by the circle x 2 + y2 = 9 b. The region bounded by the x-axis and the semicircle y = 9 − x2 Compare your answer in part (b) with the answer in part (a). 17. The region in the first and fourth quadrants enclosed by the curves y = 1 (1 + x 2 ) and y = −1 (1 + x 2 ) and by the lines x = 0 and x = 1 18. The region bounded by the parabolas y = 2 x 2 − 4 x and y = 2x − x 2

Centroids of Triangles

27. The centroid of a triangle lies at the intersection of the triangle’s medians You may recall that the point inside a triangle that lies one-third of the way from each side toward the opposite vertex is the point where the triangle’s three medians intersect. Show that the centroid lies at the intersection of the medians by showing that it too lies one-third of the way from each side toward the opposite vertex. To do so, take the following steps. i) Stand one side of the triangle on the x-axis as in part (b) of the accompanying figure. Express dm in terms of L and dy. ii) Use similar triangles to show that L = ( b h )( h − y ). Substitute this expression for L in your formula for dm. iii) Show that y = h 3. iv) Extend the argument to the other sides. y

19. The region between the curve y = 1 x and the x-axis from x = 1 to x = 16 20. The region bounded above by the curve y = 1 x 3 , below by the curve y = −1 x 3, and on the left and right by the lines x = 1 and x = a > 1. Also, find lim x. h

21. Consider a region bounded by the graphs of y = x 4 and y = x 5 . Show that the center of mass lies outside the region.

0

a. moment about the x-axis. (a)

b. moment about the y-axis. d. moment about the line x = −1. e. moment about the line y = 2.

L

Centroid

22. Consider a thin plate of constant density δ lies in the region bounded by the graphs of y = x and x = 2 y. Find the plate’s

c. moment about the line x = 5.

h−y

dy

a →∞

y x b (b)

Use the result in Exercise 27 to find the centroids of the triangles whose vertices appear in Exercises 28–32. Assume a, b > 0. 28. ( −1, 0 ) , (1, 0 ) , ( 0, 3 )

29. ( 0, 0 ) , (1, 0 ) , ( 0, 1)

f. moment about the line y = −3.

30. ( 0, 0 ) , ( a, 0 ) , ( 0, a )

31. ( 0, 0 ) , ( a, 0 ) , ( 0, b )

g. mass.

32. ( 0, 0 ) , ( a, 0 ) , ( a 2, b )

h. center of mass. Thin Wires Thin Plates with Varying Density

23. Find the center of mass of a thin plate covering the region between the x-axis and the curve y = 2 x 2,  1 ≤ x ≤ 2, if the plate’s density at the point (x, y) is δ ( x ) = x 2 . 24. Find the center of mass of a thin plate covering the region bounded below by the parabola y = x 2 and above by the line y = x if the plate’s density at the point (x, y) is δ ( x ) = 12 x. 25. The region bounded by the curves y = ±4 x and the lines x = 1 and x = 4 is revolved about the y-axis to generate a solid. a. Find the volume of the solid.

33. Constant density Find the moment about the x-axis of a wire of constant density that lies along the curve y = x from x = 0 to x = 2. 34. Constant density Find the moment about the x-axis of a wire of constant density that lies along the curve y = x 3 from x = 0 to x = 1. 35. Variable density Suppose that the density of the wire in Example 5 is δ = k sin θ (k constant). Find the center of mass. 36. Variable density Suppose that the density of the wire in Example 5 is δ = 1 + k cos θ (k constant). Find the center of mass.

Chapter 6  Questions to Guide Your Review

Plates Bounded by Two Curves

In Exercises 37–40, find the centroid of the thin plate bounded by the graphs of the given functions. Use Equations (6) and (7) with δ = 1 and M = area of the region covered by the plate. 37. g( x ) = x 2

f (x) = x + 6

and

38. g( x ) = x 2 ( x + 1),

f ( x ) = 2,

39. g( x ) = x 2 ( x − 1) and 40. g( x ) = 0,

x = 0

and

x = 2π

and

(Hint: ∫ x sin x dx = sin x − x cos x + C.) Theory and Examples

Verify the statements and formulas in Exercises 41 and 42. 41. The coordinates of the centroid of a differentiable plane curve are

∫

x =

x ds

length

,

y =

∫

y ds

length

.

y

47. Use Pappus’s Theorem for surface area and the fact that the surface area of a sphere of radius a is 4 πa 2 to find the centroid of the semicircle y = a 2 − x 2 . 48. As found in Exercise 47, the centroid of the semicircle y = a 2 − x 2 lies at the point ( 0, 2a π ). Find the area of the surface swept out by revolving the semicircle about the line y = a. 49. The area of the region R enclosed by the semiellipse y = ( b a ) a 2 − x 2 and the x-axis is (1 2 ) πab, and the volume of the ellipsoid generated by revolving R about the x-axis is ( 4 3 ) πab 2 . Find the centroid of R. Notice that the location is independent of a.

51. The region of Exercise 50 is revolved about the line y = x − a to generate a solid. Find the volume of the solid.

y x

0

42. Whatever the value of p > 0 in the equation y = x 2 (4 p), the y-coordinate of the centroid of the parabolic segment shown here is y = ( 3 5 ) a. y

52. As found in Exercise 47, the centroid of the semicircle y = a 2 − x 2 lies at the point ( 0, 2a π ). Find the area of the surface generated by revolving the semicircle about the line y = x − a. In Exercises 53 and 54, use a theorem of Pappus to find the centroid of the given triangle. Use the fact that the volume of a cone of radius r and height h is V = 13 πr 2 h. y

53.

2 y= x 4p

a

45. Find the volume of the torus generated by revolving the circle ( x − 2 ) 2 + y 2 = 1 about the y-axis.

50. As found in Example 8, the centroid of the region enclosed by the x-axis and the semicircle y = a 2 − x 2 lies at the point ( 0, 4 a 3π ). Find the volume of the solid generated by revolving this region about the line y = −a.

ds

x

44. Use a theorem of Pappus to find the volume generated by revolving about the line x = 5 the triangular region bounded by the coordinate axes and the line 2 x + y = 6 (see Exercise 27).

46. Use the theorems of Pappus to find the lateral surface area and the volume of a right-circular cone.

f (x) = x 2

f ( x ) = 2 + sin x , x = 0,

(0, b)

y=3a 5 (0, 0) 0

431

x

(a, 0) y

54.

x

(a, c)

The Theorems of Pappus

43. The square region with vertices ( 0, 2 ), ( 2, 0 ), ( 4, 2 ), and ( 2, 4 ) is revolved about the x-axis to generate a solid. Find the volume and surface area of the solid.

CHAPTER 6

(a, b) (0, 0)

x

Questions to Guide Your Review

1. How do you define and calculate the volumes of solids by the method of slicing? Give an example. 2. How are the disk and washer methods for calculating volumes derived from the method of slicing? Give examples of volume calculations by these methods. 3. Describe the method of cylindrical shells. Give an example.

4. How do you find the length of the graph of a smooth function over a closed interval? Give an example. What about functions that do not have continuous first derivatives? 5. How do you define and calculate the area of the surface swept out by revolving the graph of a smooth function y = f ( x ),   a ≤ x ≤ b, about the x-axis? Give an example.

432

Chapter 6 Applications of Definite Integrals

6. How do you define and calculate the work done by a variable force directed along a portion of the x-axis? How do you calculate the work it takes to pump a liquid from a tank? Give examples. 7. What is a center of mass? What is a centroid?

CHAPTER 6

8. How do you locate the center of mass of a thin flat plate of material? Give an example. 9. How do you locate the center of mass of a thin plate bounded by two curves y  f ( x ) and y = g( x ) over  a ≤ x ≤ b ?

Practice Exercises

Volumes Find the volumes of the solids in Exercises 1–18.

1. The solid lies between planes perpendicular to the x-axis at x  0 and x  1. The cross-sections perpendicular to the x-axis between these planes are circular disks whose diameters run from the parabola y  x 2 to the parabola y  x. 2. The base of the solid is the region in the first quadrant between the line y  x and the parabola y  2 x . The cross-sections of the solid perpendicular to the x-axis are equilateral triangles whose bases stretch from the line to the curve. 3. The solid lies between planes perpendicular to the x-axis at x  Q 4 and x  5Q 4. The cross-sections between these planes are circular disks whose diameters run from the curve y  2 cos x to the curve y  2 sin x. 4. The solid lies between planes perpendicular to the x-axis at x  0 and x  6. The cross-sections between these planes are squares whose bases run from the x-axis up to the curve x 1 2 + y 1 2 = 6. y 6

9. Find the volume of the solid generated by revolving the region bounded on the left by the parabola x = y 2 + 1 and on the right by the line x  5 about (a) the x-axis; (b) the y-axis; (c) the line x  5. 10. Find the volume of the solid generated by revolving the region bounded by the parabola y 2  4 x and the line y  x about (a) the x-axis; (b) the y-axis; (c) the line x  4; (d) the line y  4. 11. Find the volume of the solid generated by revolving the “triangular” region bounded by the x-axis, the line x  Q 3, and the curve y  tan x in the first quadrant about the x-axis. 12. Find the volume of the solid generated by revolving the region bounded by the curve y  sin x and the lines x  0,   x  Q , and y  2 about the line y  2. 13. Find the volume of the solid generated by revolving the region bounded by the curve x  e y 2 and the lines y  0,   x  0, and y  1 about the x-axis. 14. Find the volume of the solid generated by revolving about the x-axis the region bounded by y = 2 tan x ,   y = 0,   x = −Q 4 , and x  Q 4. (The region lies in the first and third quadrants and resembles a skewed bowtie.) 15. Volume of a solid sphere hole A round hole of radius 3 m is bored through the center of a solid sphere of radius 2 m. Find the volume of material removed from the sphere.

x12 + y12 = " 6

16. Volume of a football A football resembles the surface of revolution obtained by revolving the ellipse shown here around the x-axis. Find the football’s volume to the nearest cubic centimeter. 6

x

5. The solid lies between planes perpendicular to the x-axis at x  0 and x  4. The cross-sections of the solid perpendicular to the x-axis between these planes are circular disks whose diameters run from the curve x 2  4 y to the curve y 2  4 x. 6. The base of the solid is the region bounded by the parabola y 2  4 x and the line x  1 in the xy-plane. Each cross-section perpendicular to the x-axis is an equilateral triangle with one edge in the plane. (The triangles all lie on the same side of the plane.) 7. Find the volume of the solid generated by revolving the region bounded by the x-axis, the curve y  3 x 4 , and the lines x  1 and x = −1 about (a) the x-axis; (b) the y-axis; (c) the line x  1; (d) the line y  3. 8. Find the volume of the solid generated by revolving the “triangular” region bounded by the curve y  4 x 3 and the lines x  1 and y  1 2 about (a) the x-axis; (b) the y-axis; (c) the line x  2; (d) the line y  4.

y y2 x2 + =1 196 75 −14

14

0

x

17. Set up and evaluate an integral to find the volume of the given circular frustum of height h and radii a and b. a h b

Chapter 6  Practice Exercises

18. The graph of x 2 3 + y 2 3 = 1 is called an astroid and is given below. Find the volume of the solid formed by revolving the region enclosed by the astroid about the x-axis. y 1

−1

1

x

−1

Lengths of Curves Find the lengths of the curves in Exercises 19–24.

19. y = x 1 2 − (1 3 ) x 3 2 , 1 ≤ x ≤ 4 20. x = y 2 3 , 1 ≤ y ≤ 8 21. y = x 2 − ( ln x ) 8, 1 ≤ x ≤ 2 22. x = ( y 3 12 ) + (1 y ), 1 ≤ y ≤ 2 23. y =   arcsin x − 1 − x 2 , 0 ≤ x ≤ 24. y =

3 4

2 32 x − 1, 0 ≤ x ≤ 1 3

Areas of Surfaces of Revolution In Exercises 25–28, find the areas of the surfaces generated by revolving the curves about the given axes.

25. y =

2 x + 1, 0 ≤ x ≤ 3; x -axis

26. y = x 3 3, 0 ≤ x ≤ 1; x -axis 27. x =

4 y − y 2 , 1 ≤ y ≤ 2; y-axis

28. x =

y , 2 ≤ y ≤ 6; y-axis

force stretch the spring? How much work does it take to stretch the spring this far from its unstressed length? 33. Pumping a reservoir A reservoir shaped like a right-circular cone, point down, 20 m across the top and 8 m deep, is full of water. How much work does it take to pump the water to a level 6 m above the top? 34. Pumping a reservoir (Continuation of Exercise 33.) The reservoir is filled to a depth of 5 m, and the water is to be pumped to the same level as the top. How much work does it take?

x 2/3 + y 2/3 = 1

0

433

Work

29. Lifting equipment A rock climber is about to haul up 100 N of equipment that has been hanging beneath her on 40 m of rope that weighs 0.8 N/m. How much work will it take? (Hint: Solve for the rope and equipment separately, then add.) 30. Leaky tank truck You drove an 4000-L tank truck of water from the base of Mt. Washington to the summit and discovered on arrival that the tank was only half full. You started with a full tank, climbed at a steady rate, and accomplished the 1500-m elevation change in 50 min. Assuming that the water leaked out at a steady rate, how much work was spent in carrying water to the top? Do not count the work done in getting yourself and the truck there. Water weighs 9.8 NL. 31. Earth’s attraction The force of attraction on an object below Earth’s surface is directly proportional to its distance from Earth’s center. Find the work done in moving a weight of w N located a km below Earth’s surface up to the surface itself. Assume Earth’s radius is a constant r km. 32. Garage door spring A force of 200 N will stretch a garage door spring 0.8 m beyond its unstressed length. How far will a 300-N

35. Pumping a conical tank A right-circular conical tank, point down, with top radius 5 m and height 10 m, is filled with a liquid whose weight-density is 9000 Nm3. How much work does it take to pump the liquid to a point 2 m above the tank? If the pump is driven by a motor rated at 41,250 Js, how long will it take to empty the tank? 36. Pumping a cylindrical tank A storage tank is a right-circular cylinder 6 m long and 2.5 m in diameter with its axis horizontal. If the tank is half full of olive oil weighing 8950 Nm3, find the work done in emptying it through a pipe that runs from the bottom of the tank to an outlet that is 2 m above the top of the tank. 37. Assume that a spring does not follow Hooke’s Law. Instead, the force required to stretch the spring x m from its natural length is F ( x )  5 x 3 2 N. How much work does it take to a. stretch the spring 2 m from its natural length? b. stretch the spring from an initial 1 m past its natural length to 3 m past its natural length? 38. Assume that a spring does not follow Hooke’s Law. Instead, the force required to stretch the spring x m from its natural length is F ( x ) = k 5 + x 2  N. a. If a 3-N force stretches the spring 2 m, find the value of k. b. How much work is required to stretch the spring 1 m from its natural length? Centers of Mass and Centroids

39. Find the centroid of a thin, flat plate covering the region enclosed by the parabolas y  2 x 2 and y = 3 − x 2 . 40. Find the centroid of a thin, flat plate covering the region enclosed by the x-axis, the lines x  2 and x = −2, and the parabola y  x 2. 41. Find the centroid of a thin, flat plate covering the “triangular” region in the first quadrant bounded by the y-axis, the parabola y  x 2 4 , and the line y  4. 42. Find the centroid of a thin, flat plate covering the region enclosed by the parabola y 2  x and the line x  2 y. 43. Find the center of mass of a thin, flat plate covering the region enclosed by the parabola y 2  x and the line x  2 y if the density function is E ( y) = 1 + y. (Use horizontal strips.) 44. a. Find the center of mass of a thin plate of constant density covering the region between the curve y  3 x 3 2 and the x-axis from x  1 to x  9. b. Find the plate’s center of mass if, instead of being constant, the density is E ( x )  x. (Use vertical strips.)

434

Chapter 6 Applications of Definite Integrals

y

Fluid Force

45. Trough of water The vertical triangular plate shown here is the end plate of a trough full of water ( w = 9800 ) . What is the fluid force against the plate?

y=x−2

1 −2

0

x

2

y y= x 2

2 −4

0

4

UNITS IN METERS

x

UNITS IN METERS

46. Trough of maple syrup The vertical trapezoidal plate shown here is the end plate of a trough full of maple syrup weighing 11,000 Nm3. What is the force exerted by the syrup against the end plate of the trough when the syrup is 0.5 m deep?

CHAPTER 6

α

T 48. You plan to store mercury ( w = 133,350 N/m 3 ) in a vertical rectangular tank with a 0.3 m square base side whose interior side wall can withstand a total fluid force of 150,000 N. About how many cubic meters of mercury can you store in the tank at any one time?

Additional and Advanced Exercises

Volume and Length 1. A solid is generated by revolving about the x-axis the region bounded by the graph of the positive continuous function y  f ( x ), the x-axis, the fixed line x  a, and the variable line x = b,   b > a. Its volume, for all b, is b 2  ab. Find f ( x ). 2. A solid is generated by revolving about the x-axis the region bounded by the graph of the positive continuous function y  f ( x ), the x-axis, and the lines x  0 and x  a. Its volume, for all a  0, is a 2 a. Find f ( x ). 3. Suppose that the increasing function f ( x ) is smooth for x p 0 and that f (0)  a. Let s( x ) denote the length of the graph of f from (0, a) to ( x ,   f ( x ) ) ,   x > 0. Find f ( x ) if s( x )  Cx for some constant C. What are the allowable values for C? 4. a. Show that for 0 < α ≤ π 2,

∫0

47. Force on a parabolic gate A flat vertical gate in the face of a dam is shaped like the parabolic region between the curve y  4 x 2 and the line y  4, with measurements in meters. The top of the gate lies 5 m below the surface of the water. Find the force exerted by the water against the gate ( w = 9800 ) .

1 + cos 2 θ dθ >

Surface Area 7. At points on the curve y  2 x , line segments of length h  y are drawn perpendicular to the xy-plane. (See accompanying figure.) Find the area of the surface formed by these perpendiculars from ( 0, 0 ) to ( 3, 2 3 ). y

2"3 2" x

y = 2" x

α 2 + sin 2 α.

b. Generalize the result in part (a). 5. Find the volume of the solid formed by revolving the region bounded by the graphs of y  x and y  x 2 about the line y  x. 6. Consider a right-circular cylinder of diameter 1. Form a wedge by making one slice parallel to the base of the cylinder completely through the cylinder, and another slice at an angle of 45 n to the first slice and intersecting the first slice at the opposite edge of the cylinder (see accompanying diagram). Find the volume of the wedge.

(3, 2"3)

0

x

3

x

8. At points on a circle of radius a, line segments are drawn perpendicular to the plane of the circle, the perpendicular at each point P being of length ks, where s is the length of the arc of the circle measured counterclockwise from (a, 0) to P, and k is a positive constant, as shown here. Find the area of the surface formed by the perpendiculars along the arc beginning at (a, 0) and extending once around the circle.

45° wedge

0 r=

1 2

a x

a

y

Chapter 6  Technology Application Projects

Work 9. A particle of mass m starts from rest at time t  0 and is moved along the x-axis with constant acceleration a from x  0 to x  h against a variable force of magnitude F (t )  t 2 . Find the work done. 10. Work and kinetic energy Suppose a 50-g golf ball is placed on a vertical spring with force constant k  2 N cm . The spring is compressed 15 cm and released. About how high does the ball go (measured from the spring’s rest position)? Centers of Mass 11. Find the centroid of the region bounded below by the x-axis and above by the curve y = 1 − x n , n an even positive integer. What is the limiting position of the centroid as n → ∞ ? 12. If you haul a telephone pole on a two-wheeled carriage behind a truck, you want the wheels to be 1 m or so behind the pole’s center of mass to provide an adequate “tongue” weight. The 12-m wooden telephone poles used by Verizon have a 66-cm circumference at the top and a 104-cm circumference at the base. About how far from the top is the center of mass? 13. Suppose that a thin metal plate of area A and constant density E occupies a region R in the xy-plane, and let My be the plate’s moment about the y-axis. Show that the plate’s moment about the line x  b is a. My  bE A if the plate lies to the right of the line, and

435

14. Find the center of mass of a thin plate covering the region bounded by the curve y 2  4 ax and the line x  a,   a  positive constant, if the density at (x, y) is directly proportional to (a) x, (b) y . 15. a. Find the centroid of the region in the first quadrant bounded by two concentric circles and the coordinate axes, if the circles have radii a and b, 0  a  b, and their centers are at the origin. b. Find the limits of the coordinates of the centroid as a l b and discuss the meaning of the result. 16. A triangular corner is cut from a square 40 cm on a side. The area of the triangle removed is 400 cm2. If the centroid of the remaining region is 22 cm from one side of the original square, how far is it from the remaining sides? Fluid Force 17. A triangular plate ABC is submerged in water with its plane vertical. The side AB, 4 m long, is 6 m below the surface of the water, while the vertex C is 2 m below the surface. Find the force exerted by the water on one side of the plate. 18. A vertical rectangular plate is submerged in a fluid with its top edge parallel to the fluid’s surface. Show that the force exerted by the fluid on one side of the plate equals the average value of the pressure up and down the plate times the area of the plate.

b. bE A  M y if the plate lies to the left of the line.

CHAPTER 6

Technology Application Projects

Mathematica/Maple Projects

Projects can be found within MyLab Math. • Using Riemann Sums to Estimate Areas, Volumes, and Lengths of Curves Visualize and approximate areas and volumes in Part I and Part II: Volumes of Revolution; and Part III: Lengths of Curves. • Modeling a Bungee Cord Jump Collect data (or use data previously collected) to build and refine a model for the force exerted by a jumper’s bungee cord. Use the work-energy theorem to compute the distance fallen for a given jumper and a given length of bungee cord.

7

Integrals and Transcendental Functions OVERVIEW Our treatment of the logarithmic and exponential functions has been rather informal. In this chapter, we give a rigorous analytic approach to the definitions and properties of these functions. We also introduce the hyperbolic functions and their inverses. Like the trigonometric functions, these functions belong to the class of transcendental functions.

7.1 The Logarithm Defined as an Integral In Chapter 1, we introduced the natural logarithm function ln x as the inverse of the exponential function e x . The function e x was chosen as that function in the family of general exponential functions a x, a > 0, whose graph has slope 1 as it crosses the y-axis. The function a x was presented intuitively, however, based on its graph at rational values of x. In this section we recreate the theory of logarithmic and exponential functions from an entirely different point of view. Here we define these functions analytically and derive their behaviors. To begin, we use the Fundamental Theorem of Calculus to define the natural logarithm function ln x as an integral. We quickly develop its properties, including the algebraic, geometric, and analytic properties with which we are already familiar. Next we introduce the function e x as the inverse function of ln x , and establish its properties. Defining ln x as an integral and e x as its inverse is an indirect approach that gives an elegant and powerful way to obtain and validate the key properties of logarithmic and exponential functions.

Definition of the Natural Logarithm Function The natural logarithm of a positive number x, written as ln x, is the value of an integral. The appropriate integral is suggested by our earlier results in Chapter 5.

DEFINITION

The natural logarithm is the function given by ln x =

∫1

x

1 dt , x > 0. t

From the Fundamental Theorem of Calculus, we know that ln x is a continuous function. Geometrically, if x > 1, then ln x is the area under the curve y = 1 t from t = 1 to t = x (Figure 7.1). For 0 < x < 1, ln x gives the negative of the area under the curve from x to 1,

436

7.1  The Logarithm Defined as an Integral

437

and the function is not defined for x ≤ 0. From the Zero Width Interval Rule for definite integrals, we also have ln 1 =

1

∫1

1 dt = 0. t

y

x

If 0 < x < 1, then ln x =

L 1

1

1 dt = − t

x L

1 dt t

gives the negative of this area. x

If x > 1, then ln x =

1 L

gives this area.

1 dt t y = ln x

1 y = 1x 0

x

x

x

1

1

If x = 1, then ln x =

L 1

1 dt = 0. t

y = ln x

The graph of y = ln x and its relation to the function y = 1 x, x > 0. The graph of the logarithm rises above the x-axis as x moves from 1 to the right, and it falls below the axis as x moves from 1 to the left. FIGURE 7.1

Notice that we show the graph of y = 1 x in Figure 7.1 but use y = 1 t in the integral. Using x for everything would have us writing TABLE 7.1 Typical 2-place

values of ln x

x

ln x

  0   0.05   0.5   1   2   3   4 10

undefined −3.00 −0.69 0 0.69 1.10 1.39 2.30

ln x =

∫1

x

1 dx , x

with x meaning two different things. So we change the variable of integration to t. By using rectangles to obtain finite approximations of the area under the graph of y = 1 t and over the interval between t = 1 and t = x, as in Section 5.1, we can approximate the values of the function ln x. Several values are given in Table 7.1. There is an important number between x = 2 and x = 3 whose natural logarithm equals 1. This number, which we now define, exists because ln x is a continuous function and therefore satisfies the Intermediate Value Theorem on the interval [ 2, 3 ].

DEFINITION The number e is the number in the domain of the natural logarithm that satisfies

ln (e) =

∫1

e

1 dt = 1. t

Interpreted geometrically, the number e corresponds to the point on the x-axis for which the area under the graph of y = 1 t and above the interval [ 1, e ] equals the area of the unit square. That is, the area of the region shaded blue in Figure 7.1 is 1 square unit when x = e. We will see further on that this is the same number e ≈ 2.718281828 we have encountered before.

438

Chapter 7 Integrals and Transcendental Functions

The Derivative of y  ln x By the first part of the Fundamental Theorem of Calculus (Section 5.4), d d ln x = dx dx

∫1

x

1 1 dt = , t x

so we have

d 1 ln x = , x > 0. dx x

(1)

Therefore, the function y  ln x is a solution to the initial value problem dy dx  1 x , x  0, with y (1)  0. Notice that the derivative is always positive. If u is a differentiable function of x whose values are positive so that ln u is defined, then by applying the Chain Rule, we obtain d 1 du ln u = , u > 0. dx u dx

(2)

The derivative of ln x can be found just as in Example 3(c) of Section 3.8, giving

d 1 ln x = , x ≠ 0. dx x

y

Moreover, if b is any constant with bx  0, Equation (2) gives y = ln x

0

(3)

d 1 1 1 d ln bx = ⋅ (bx ) = (b) = . dx bx dx bx x x

(1, 0)

The Graph and Range of ln x The derivative d (ln x ) dx  1 x is positive for x  0, so ln x is an increasing function of x. The second derivative, 1 x 2, is negative, so the graph of ln x is concave down. (See Figure 7.2a.) The function ln x has the following familiar algebraic properties, which we stated in Section 1.5. In Section 4.2 we showed these properties are a consequence of Corollary 2 of the Mean Value Theorem, and those derivations still apply.

(a) y y = 1x 1 1 2

1. ln bx = ln b + ln x 0

1 (b)

2

x

FIGURE 7.2 (a) The graph of the natural logarithm. (b) The rectangle of height y  1 2 fits beneath the graph of y  1 x for the interval 1 b x b 2.

3. ln

1 = − ln x x

2. ln

b = ln b − ln x x

4. ln x r  r ln x ,  r rational

We can estimate the value of ln 2 by considering the area under the graph of y  1 x and above the interval [ 1, 2 ]. In Figure 7.2(b) a rectangle of height 1 2 over the interval

439

7.1  The Logarithm Defined as an Integral

[ 1, 2 ] fits under the graph. Therefore, the area under the graph, which is ln 2, is greater than the area of the rectangle, which is 1 2. So ln 2 > 1 2. Knowing this we have ln 2 n = n ln 2 > n

( 12 ) = n2 .

This result shows that ln (2 n ) → ∞ as n → ∞. Since ln x is an increasing function, it follows that lim ln x = ∞.

  

x →∞

ln x is increasing and not bounded above.

We also have lim ln x = lim ln

x →0+

t →∞

1 = lim ( − ln t ) = −∞. t →∞ t

  

x = 1t

We defined ln x for x > 0, so the domain of ln x is the set of positive real numbers. The above discussion and the Intermediate Value Theorem show that its range is the entire real line, giving the familiar graph of y = ln x shown in Figure 7.2(a).

The Integral

∫1

x dx

Equation (3) leads to the following integral formula:

∫

1 dx = ln x + C. x

(4)

If u is a differentiable function that is never zero, then 1 (5) du = ln u + C. u Equation (5) applies anywhere on the domain of 1 u , which is the set of points where du u ≠ 0. It says that integrals that have the form ∫ lead to logarithms. Whenever u u = f ( x ) is a differentiable function that is never zero, we have that du = f ′( x ) dx and

∫

∫ EXAMPLE 1 π 2

f ′( x ) dx = ln f ( x ) + C. f (x)

We rewrite an integral so that it has the form ∫

4 cos θ

∫−π 2 3 + 2 sin θ d θ = ∫1

5

2 du u

du . u

u = 3 + 2 sin θ, du = 2 cos θ dθ, u (−π 2 ) = 1, u ( π 2 ) = 5

⎤5 = 2 ln u ⎥ ⎥⎦ 1 = 2 ln 5 − 2 ln 1 = 2 ln 5

     

Note that u = 3 + 2 sin θ is always positive on [ − π 2, π 2 ], so Equation (5) applies.

The Inverse of ln x and the Number e The function ln x , being an increasing function of x with domain ( 0, ∞ ) and range (−∞, ∞ ) , has an inverse ln −1 x with domain (−∞, ∞ ) and range ( 0, ∞ ). The graph of ln −1 x is the graph of ln x reflected across the line y = x. As you can see in Figure 7.3, lim ln −1 x = ∞

x →∞

and

lim ln −1 x = 0.

x →−∞

440

Chapter 7 Integrals and Transcendental Functions

y

The inverse function ln −1 x is also denoted by exp x. We have not yet established that exp x is an exponential function, only that exp x is the inverse of the function ln x. We will now show that ln −1 x = exp x is, in fact, the exponential function with base e. The number e was defined to satisfy the equation ln(e) = 1, so e = exp(1). We can raise the number e to a rational power r using algebra:

y = ln−1x or x = ln y

8 7 6

e 2 = e ⋅ e,

5 4 e

(1, e)

e1 2 =

e2 3 =

e,

3

e2 ,

ln e r = r ln e = r ⋅ 1 = r.

y = ln x

Applying the function ln −1 to both sides of the equation ln e r = r , we find that

1 0

1 , e2

and so on. Since e is positive, e r is positive too. Therefore, e r has a logarithm. When we take the logarithm, we find that if r is rational, then

2

−2 −1

e −2 =

1

2

e

4

x

FIGURE 7.3 The graphs of y = ln x and y = ln −1 x = exp x. The number e is ln −1 1 = exp(1).

e r = exp r for  r  rational.

exp is ln −1.

(6)

Thus exp r coincides with the exponential function e r for all rational values of r. We have not yet found a way to give an exact meaning to e x for x irrational, but we can use Equation (6) to do so. The function exp x = ln −1 x has domain (−∞, ∞ ) , so it is defined for every x. We have exp r = e r for r rational by Equation (6), and we now define e x to equal exp x for all x. For every real number x, we define the natural exponential function to be e x = exp x.

DEFINITION

The notations ln −1 x , exp x, and e x all refer to the natural exponential function.

For the first time we have a precise meaning for a number raised to an irrational power. Usually the exponential function is denoted by e x rather than exp x. Since ln x and e x are inverses of one another, we have the following relations. Inverse Equations for e x and ln x

e ln x = x ln(e x ) Typical values of e x

x

e x (rounded)

−1 0 1 2 10

0.37 1 2.72 7.39 22026

100

2.6881 × 10 43

= x

(all  x > 0) (all  x )

The Derivative and Integral of e x The exponential function is differentiable because it is the inverse of a differentiable function whose derivative is never zero. We calculate its derivative by using Theorem 3 of Section 3.8 and our knowledge of the derivative of ln x. Let f ( x ) = ln x

and

y = e x = ln −1 x = f −1 ( x ).

Then dy d x d −1 = e = ln x dx dx dx d −1 = f (x) dx 1 = f ′( f −1 ( x )) 1 = f ′(e x ) 1 = 1 ex = e x.

( )

Theorem 3, Section 3.8 f

−1

(x) = e x

f ′( z ) =

1  with  z = e x z

441

7.1  The Logarithm Defined as an Integral

That is, for y  e x , we find that dy dx  e x, so the natural exponential function e x is its own derivative, just as we claimed in Section 3.3.

d x e  e x. dx

(7)

We will see in the next section that the only functions that behave this way are constant multiples of e x . The Chain Rule extends the derivative result in the usual way to a more general form: If u is any differentiable function of x, then d u du e  eu . dx dx

(8)

Since e x  0, its derivative is also everywhere positive, so it is an increasing and continuous function for all x, having limits lim e x = 0

and

x →−∞

lim e x = ∞.

x →∞

It follows that the x-axis ( y = 0 ) is a horizontal asymptote of the graph y  e x (see Figure 7.3). Equation (7) also tells us the indefinite integral of e x.

∫ e x dx

= ex + C

If f ( x )  e x , then we see from Equation (7) that f ′(0) = e 0 = 1. That is, the exponential function e x has slope 1 as it crosses the y-axis at x  0. This agrees with our assertion for the natural exponential in Section 3.3.

Logarithms and Laws of Exponents The familiar algebraic properties of logarithms and exponential functions were stated in Section 1.5. We now show that these follow from the definition of the logarithm as an integral that we have used. The properties of logarithms are stated in Theorem 1.

THEOREM 1—Algebraic Properties of the Natural Logarithm

For any numbers b  0 and x  0, the natural logarithm satisfies the following rules: 1. Product Rule:

ln bx = ln b + ln x

2. Quotient Rule:

ln

b = ln b − ln x x

3. Reciprocal Rule:

ln

1 = − ln x x

4. Power Rule:

ln x r  r ln x

Rule 2 with b  1

The proof uses the Mean Value Theorem, which was established in Section 4.2.

442

Chapter 7 Integrals and Transcendental Functions

Proof that ln bx = ln b + ln x ln x have the same derivative:

The argument starts by observing that ln bx and

d b 1 d ln(bx ) = = = ln x. dx bx x dx According to Corollary 2 of the Mean Value Theorem, the functions must differ by a constant, which means that ln bx = ln x + C for some C. Since this last equation holds for all positive values of x, it must hold for x = 1. Hence, ln ( b ⋅ 1) = ln 1 + C ln b = 0 + C

   ln 1 = 0

C = ln b. By substituting, we conclude ln bx = ln b + ln x.

Proof that ln x r = r ln x tive values of x,

We use the same-derivative argument again. For all posi-

d 1 d r ln x r = r (x ) dx x dx =

1 r −1 rx xr

=r⋅

Chain Rule

Derivative Power Rule

1 d = (r ln x ). x dx

Since ln x r and r ln x have the same derivative, ln x r = r ln x + C for some constant C. Taking x to be 1 identifies C as zero, and we’re done.

You are asked to prove the Quotient Rule for logarithms, ln

( bx ) = ln b − ln x,

in Exercises 63. The Reciprocal Rule, ln (1 x ) = − ln x , is a special case of the Quotient Rule, obtained by taking b = 1 and noting that ln 1 = 0 . We now state the algebraic properties of exponential functions.

THEOREM 2—Laws of Exponents for e x

For all numbers x and y, the natural exponential function e x obeys the following laws. 1. e x e y = e x + y 3.

ex = e x−y ey

2. e − x =

1 ex

4. (e x ) y = e xy = (e y ) x

7.1  The Logarithm Defined as an Integral

443

Proof that e x e y = e x + y e x e y = e ln(e x e y ) = e ln(e x ) + ln(e y ) =

e x + y.

u = e ln u . Product formula for ln ln e u = u.

Theorem 1 and the inverse relationship between the logarithmic and exponential functions also imply the other algebraic properties of the exponential function (see Exercise 67).

The General Exponential Function a x Since a  e ln a for any positive number a, we can express a x as (e ln a ) x  e x ln a. We therefore make the following definition, consistent with what we stated in Section 1.5.

DEFINITION

For any numbers a  0 and x, the exponential function with base

a is given by a x  e x ln a.

The General Power Function

x r is the function e r ln x.

When a  e, the definition gives a x = e x ln a = e x ln e = e x ⋅ 1 = e x. Similarly, the power function f ( x )  x r is defined for positive x by the formula x r  e r ln x, which holds for any real number r, rational or irrational. Theorem 2 is also valid for a x , the exponential function with base a. For example, ax ⋅ ay = = = =

e x ln a ⋅ e y ln a e x ln a + y ln a e ( x + y ) ln a a x+ y.

Definition of a x Law 1 Common factor ln a Definition of a x

Starting with the definition a x = e x ln a , a > 0, we get the derivative d x d x ln a a  e  (ln a) e x ln a  (ln a) a x , dx dx so

d x a  a x ln a. dx

Alternatively, we get the same derivative rule by applying logarithmic differentiation: y  ax ln y  x ln a 1 dy  ln a y dx dy  y ln a  a x ln a. dx

Take logarithms. Differentiate with respect to x.

With the Chain Rule, we get a more general form, as in Section 3.8: If a  0 and u is a differentiable function of x, then a u is a differentiable function of x and d u du a  a u ln a . dx dx

444

Chapter 7 Integrals and Transcendental Functions

y

The integral equivalent of this derivative rule is

y = 2x y=x

1 2

=

ax +C ln a

Logarithms with Base a

y = log 2 x

2 1 0

∫ a x dx

x

If a is any positive number other than 1, the function a x is one-to-one and has a nonzero derivative at every point. It therefore has a differentiable inverse. For any positive number a v 1, the logarithm of x with base a, denoted by log a x , is the inverse function of a x .

DEFINITION FIGURE 7.4

The graph of 2 x and its

inverse, log 2 x.

The graph of y  log a x can be obtained by reflecting the graph of y  a x across the 45 n line y  x (Figure 7.4). When a  e, we have log e x  inverse of e x  ln x. Since log a x and a x are inverses of one another, composing them in either order gives the identity function. Inverse Equations for a x and log a x

a log a x = x log a (a x ) = x

( x > 0) (all x )

As stated in Section 1.5, the function log a x is just a numerical multiple of ln x . We see this from the following derivation. y  log a x TABLE 7.2 Rules for base a

logarithms

For any numbers x  0 and y  0,

2. Quotient Rule: x log a = log a x − log a y y 3. Reciprocal Rule: 1 log a = − log a y y

ay  x ln a y  ln x

Equivalent equation

y ln a  ln x

Algebra Rule 4 for natural log

Natural log of both sides

ln x y  ln a

1. Product Rule: log a xy = log a x + log a y

Defining equation for y

log a x 

Solve for y.

ln x ln a

Substitute for y.

It then follows easily that the arithmetic rules satisfied by log a x are the same as the ones for ln x. These rules, given in Table 7.2, can be proved by dividing the corresponding rules for the natural logarithm function by ln a. For example,

4. Power Rule: log a x y  y log a x

ln xy = ln x + ln y

Rule 1 for natural logarithms ... 

ln xy ln x ln y = + ln a ln a ln a

... divided by ln a ... 

log a xy = log a x + log a y.

... gives Rule 1 for base a logarithms.

Derivatives and Integrals Involving log a x To find derivatives or integrals involving base a logarithms, we can convert them to natural logarithms. In particular, differentiating log a x gives d d ⎛⎜ ln x ⎞⎟ 1 d 1 1 ( log a x ) = ⋅ , ( ln x ) = ⎟= ⎜ dx dx ⎜⎝ ln a ⎟⎠ ln a dx ln a x so d 1 ( log a x ) = . dx x ln a

7.1  The Logarithm Defined as an Integral

445

If u is a positive differentiable function of x, then d 1 1 du ( log a u ) = ⋅ . dx ln a u dx EXAMPLE 2 We illustrate the derivative and integral results. d 1 1 d( 3 (a) ⋅ log 10 ( 3 x + 1) = 3 x + 1) = dx ln 10 3 x + 1 dx ( ln 10 )( 3 x + 1)

Transcendental Numbers and Transcendental Functions Numbers that are solutions of polynomial equations with rational coefficients are called algebraic: −2 is algebraic because it satisfies the equation x + 2 = 0, and 3 is algebraic because it satisfies the equation x 2 − 3 = 0. Numbers such as e and π that are not algebraic are called transcendental. We call a function y = f ( x ) algebraic if it satisfies an equation of the form

(b)

∫

ln x log 2 x 1 dx = dx ln 2 ∫ x x

log 2 x =

ln x ln 2

=

1 u du ln 2 ∫

=

(ln x ) 2 1 u2 1 (ln x ) 2 +C = +C = +C ln 2 2 ln 2 2 2 ln 2

u = ln x , du =

1 dx x

Summary

In this section we used calculus to give precise definitions of the logarithmic and exponential functions. This approach is somewhat different from our earlier treatments of the polyin which the P’s are polynomials in x nomial, rational, and trigonometric functions. There we first defined the function and then with rational coefficients. The function we studied its derivatives and integrals. Here we started with an integral from which the y = 1 x + 1 is algebraic because it functions of interest were obtained. The motivation behind this approach was to address satisfies the equation ( x + 1) y 2 − 1 = 0. mathematical difficulties that arise when we attempt to define functions such as a x for any Here the polynomials are P2 = x + 1, real number x, rational or irrational. Defining ln x as the integral of the function 1 t from P1 = 0, and P0 = −1. Functions that are t = 1 to t = x enabled us to define all of the exponential and logarithmic functions and not algebraic are called transcendental. then to derive their key algebraic and analytic properties. Pn y n +  + P1 y + P0 = 0

EXERCISES

7.1

Integration

Evaluate the integrals in Exercises 1–46. −2

dx x

3 dx

0

22.

∫ e csc π + t (

ln( π 2 )

)

csc ( π + t ) cot ( π + t ) dt

2.

∫−1 3x − 2

23.

∫ln( π 6)

4.

∫

8r dr 4r 2 − 5

25.

∫ 1 + e r dr

3 sec 2 t dt 6 + 3 tan t

6.

∫

sec y tan y dy 2 + sec y

27.

∫0

2 −θ dθ

dx 2 x + 2x

∫

29.

∫1

2

8. 10.

∫

8e ( x +1) dx

31.

∫0

12.

∫

ln(ln x ) dx x ln x

33.

∫2

14.

∫

tan x ln (cos x ) dx

35.

∫

e−

dr

1.

∫−3

3.

∫

y2

5.

∫

7.

∫

9.

∫ln 2

2 y dy − 25

ln 3

4

e x dx

(ln x ) 3 dx 2x

11.

∫1

13.

∫ln 4

ln 9

e x 2 dx r

15.

∫

e

17.

∫

2t e −t 2 dt

18.

19.

∫

e1 x dx x2

20.

21.

∫ e sec πt sec πt tan πt dt

r

dr

16.

ln π

24.

∫0

26.

∫

28.

∫−2 5 −θ dθ

x 2 ( x 2 ) dx

30.

∫1

7 cos t sin t dt

32.

∫0

x 2 x (1 + ln x ) dx

34.

∫1

2

∫0

( 2 + 1) x

36.

∫1

e

37.

∫

log 10 x dx x

38.

∫1

4

log 2 x dx x

∫

ln x dx x ln 2 x + 1

39.

∫1

e

2 ln 10 log 10 x dx x

41.

∫0

10

∫

e −1 x dx x3

log 10 (10 x ) dx x

43.

∫0

3

2 log 2 ( x − 1) dx x −1

sec x dx ln ( sec x + tan x )

r

r

2

2e υ cos e υ d υ

er

1

π2

4

3

2

dx

2 x e x 2 cos(e x 2 ) dx

dx 1 + ex 0

4

2

x

x π4

dx

( 13 )

tan t

sec 2 t dt

2 ln x dx x x (ln 2)−1 dx

4

ln 2 log 2 x dx x

40.

∫1

2

log 2 ( x + 2 ) dx x+2

42.

∫1 10

9

2 log 10 ( x + 1) dx x +1

44.

∫2

446

45.

∫

Chapter 7 Integrals and Transcendental Functions

dx x log 10 x

46.

∫

dx x (log 8 x ) 2

Initial Value Problems

Solve the initial value problems in Exercises 47–52. 47.

dy = e t sin ( e t − 2 ), y (ln 2) = 0 dt

48.

dy = e −t sec 2 (πe −t ), y (ln 4) = 2 π dt

49.

d 2y = 2e −x , y (0) = 1 and dx 2

50.

d 2y dt 2

y′(1) = 0

dy 1 = 1 + , y (1) = 3 51. dx x 52.

d 2y = sec 2 x , y(0) = 0 and dx 2

y

y′(0) = 1

y = ln x

ln a

0

y′(0) = 0

= 1 − e 2 t , y (1) = −1 and

as suggested by the accompanying figure.

60. The geometric, logarithmic, and arithmetic mean inequality a. Show that the graph of e x is concave up over every interval of x-values. b. Show, by reference to the accompanying figure, that if 0 < a < b, then ln b e ln a + e ln b e ( ln a + ln b ) 2 ⋅ ( ln b − ln a ) < ∫ e x dx < ⋅ ( ln b − ln a ). ln a 2 y = ex

Theory and Applications

F

53. The region between the curve y = 1 x 2 and the x-axis from x = 1 2 to x = 2 is revolved about the y-axis to generate a solid. Find the volume of the solid. 54. In Section 6.2, Exercise 6, we revolved about the y-axis the region between the curve y = 9 x x 3 + 9 and the x-axis from x = 0 to x = 3 to generate a solid of volume 36π. What volume do you get if you revolve the region about the x-axis instead? (See Section 6.2, Exercise 6, for a graph.)

C E B A ln a

M

ln a + ln b 2

56. x = ( y 4 ) − 2 ln ( y 4 ), 4 ≤ y ≤ 12 2

T 57. The linearization of ln ( 1 + x ) at x = 0 Instead of approximating ln x near x = 1, we approximate ln (1 + x ) near x = 0. We get a simpler formula this way. a. Derive the linearization ln (1 + x ) ≈ x at x = 0. b. Estimate to five decimal places the error involved in replacing ln (1 + x ) by x on the interval [ 0, 0.1 ]. c. Graph ln (1 + x ) and x together for 0 ≤ x ≤ 0.5. Use different colors, if available. At what points does the approximation of ln (1 + x ) seem best? Least good? By reading coordinates from the graphs, find as good an upper bound for the error as your grapher will allow. 58. The linearization of e x at x = 0 a. Derive the linear approximation e x ≈ 1 + x at x = 0. T b. Estimate to five decimal places the magnitude of the error involved in replacing e x by 1 + x on the interval [ 0, 0.2 ]. T c. Graph e x and 1 + x together for −2 ≤ x ≤ 2. Use different colors, if available. On what intervals does the approximation appear to overestimate e x ? Underestimate e x ? 59. Show that for any number a > 1

∫1

ln x dx +

∫0

x

c. Use the inequality in part (b) to conclude that

55. y = ( x 2 8 ) − ln x , 4 ≤ x ≤ 8

ln a

D ln b

NOT TO SCALE

Find the lengths of the curves in Exercises 55 and 56.

a

x

a

1

e y dy = a ln a,

ab
0 be given, and use Figure 7.1 to show that 1 < x +1

∫x

x +1

1 1 dt < . t x

[−3, 6] by [−3, 3]

b. Conclude from part (a) that 1 1 1 < ln 1 + < . x +1 x x

(

b. Give an argument based on the graphs of y = ln x and the tangent line to explain why ln x < x e for all positive x ≠ e.

)

c. Show that ln ( x e ) < x for all positive x ≠ e.

c. Conclude from part (b) that e

x x +1

(

1 < 1+ x

)

x

d. Conclude that x e < e x for all positive x ≠ e.

< e.

e. So which is bigger, π e or e π ? 76. A decimal representation of e Find e to as many decimal places as you can by solving the equation ln x = 1 using Newton’s method in Section 4.7.

d. Conclude from part (c) that

(

lim 1 +

x →∞

1 x

)

x

= e.

Calculations with Other Bases

Grapher Explorations

T When solving Exercises 69–76, you may need to use appropriate technology (such as a graphing calculator or a computer). 69. Graph ln x, ln 2x, ln 4x, ln 8x, and ln 16x (as many as you can) together for 0 < x ≤ 10. What is going on? Explain.

T 77. Most scientific calculators have keys for log 10 x and ln x. To find logarithms to other bases, we use the equation log a x = (ln x ) (ln a).

70. Graph y = ln sin x in the window 0 ≤ x ≤ 22, −2 ≤ y ≤ 0. Explain what you see. How could you change the formula to turn the arches upside down? 71. a. Graph y = sin x and the curves y = ln ( a + sin x ) for a = 2, 4, 8, 20, and 50 together for 0 ≤ x ≤ 23. b. Why do the curves flatten as a increases? (Hint: Find an a-dependent upper bound for y′ .) 72. Does the graph of y = x − ln x , x > 0, have an inflection point? Try to answer the question (a) by graphing, (b) by using calculus. 73. The equation x 2 = 2 x has three solutions: x = 2, x = 4, and one other. Estimate the third solution as accurately as you can by graphing. 74. Could x ln 2 possibly be the same as 2 ln x for some x > 0? Graph the two functions and explain what you see. 75. Which is bigger, S e or e S ? Calculators have taken some of the mystery out of this once-challenging question. (Go ahead and check; you will see that it is a surprisingly close call.) You can answer the question without a calculator, though.

Find the following logarithms to five decimal places. a. log 3 8 b. log 7 0.5 c. log 20 17 d. log 0.5 7 e. ln x, given that log 10 x = 2.3 f. ln x, given that log 2 x = 1.4 g. ln x, given that log 2 x = −1.5 h. ln x, given that log 10 x = −0.7 78. Conversion factors a. Show that the equation for converting base 10 logarithms to base 2 logarithms is ln 10 log 2 x = log 10 x. ln 2 b. Show that the equation for converting base a logarithms to base b logarithms is ln a log b x = log a x. ln b

7.2 Exponential Change and Separable Differential Equations Exponential functions increase or decrease very rapidly with changes in the independent variable. They describe growth or decay in many natural and industrial situations. The variety of models based on these functions partly accounts for their importance.

Exponential Change In modeling many real-world situations, a quantity y increases or decreases at a rate proportional to its size at a given time t. Examples of such quantities include the size of a

448

Chapter 7 Integrals and Transcendental Functions

population, the amount of a decaying radioactive material, and the temperature difference between a hot object and its surrounding medium. Such quantities are said to undergo exponential change. If the amount present at time t = 0 is called y 0 , then we can find y as a function of t by solving the following initial value problem: dy = ky dt Initial condition: y = y0

Differential equation:

(1a) when t = 0. 

(1b)

If y is positive and increasing, then k is positive, and we use Equation (1a) to say that the rate of growth is proportional to what has already been accumulated. If y is positive and decreasing, then k is negative, and we use Equation (1a) to say that the rate of decay is proportional to the amount still left. The constant function y = 0 is a solution of Equation (1a) if y 0 = 0. To find the solutions when y 0 is not zero, we divide Equation (1a) by y:

∫

k = 1.3

k=1

t (a)

k = −0.5

k = −1 t (b)

FIGURE 7.5

y = e kt +C

Exponentiate.

y = e C ⋅ e kt

e a+b = e a ⋅ e b

y = ±e C e kt

If y = r ,  then  y = ±r.

y = Ae kt.

A is a shorter name for ±e C.

dy = ky, dt

y (0) = y 0

is (2)

Quantities changing in this way are said to undergo exponential growth if k > 0 and exponential decay if k < 0. The number k is called the rate constant of the change. (See Figure 7.5.) The derivation of Equation (2) shows also that the only functions that are their own derivatives ( k = 1) are constant multiples of the exponential function. Before presenting several examples of exponential change, let us consider the process we used to derive it.

y = y0 ekt

k = −1.3

∫ (1 u ) du = ln u + C

y = y 0 e kt.

y y0

Integrate with respect to t.

The solution of the initial value problem

k = 0.6 y0

y ≠ 0

By allowing A to take on the value 0 in addition to all possible values ±e C , we can include the solution y = 0 (which occurs when y 0 = 0) in the formula. We find the value of A for the initial value problem by solving for A when y = y 0 and t = 0: y 0 = Ae k ⋅ 0 = A.

y

y = y0 ekt

1 dy ⋅ =k y dt 1 dy dt = ∫ k dt y dt ln y = kt + C

Graphs of (a) exponential growth and (b) exponential decay. As k increases, the growth ( k > 0 ) or decay ( k < 0 ) intensifies.

Separable Differential Equations Exponential change is modeled by a differential equation of the form dy dx = ky, where k is a nonzero constant. More generally, suppose we have a differential equation of the form dy = f ( x, y ), dx

(3)

7.2  Exponential Change and Separable Differential Equations

449

where f is a function of both the independent and dependent variables. A solution of the equation is a differentiable function y = y ( x ) defined on an interval of x-values (perhaps infinite) such that d y (x) = f ( x, y (x)) dx on that interval. That is, when y ( x ) and its derivative y′( x ) are substituted into the differential equation, the resulting equation is true for all x in the solution interval. The general solution is a solution y ( x ) that contains all possible solutions, and it always contains an arbitrary constant. Equation (3) is separable if f can be expressed as a product of a function of x and a function of y. The differential equation then has the form dy = g( x ) H ( y). dx

  

g is a function of x; H is a function of y.

Then collect all of the y factors so that they are together with dy, and likewise collect all of the x factors together with dx: 1 dy = g( x ) dx. H ( y) Now we integrate both sides of this equation:

∫

1 dy = H ( y)

∫ g( x ) dx.

(4)

After completing the integrations, we obtain the solution y, defined implicitly as a function of x. The justification that we can integrate both sides in Equation (4) in this way is based on the Substitution Rule (Section 5.5):

∫

EXAMPLE 1

1 dy = H ( y)

∫

dy 1 dx H ( y ( x )) dx

=

∫

1 H ( y ( x )) g( x ) dx H ( y ( x ))

=

∫ g( x ) dx.

  

dy = H ( y( x )) g( x ) dx

Solve the differential equation dy = (1 + y )e x, y > −1. dx

Solution Since 1 + y is never zero for y > −1, we can solve the equation by separating the variables.

dy = (1 + y ) e x dx dy = (1 + y )e x dx dy = e x dx 1+ y dy

∫1+ y

=

∫ e x dx

ln (1 + y ) = e x + C

Treat dy dx as a quotient of differentials and multiply both sides by dx. Divide by (1 + y ).

Integrate both sides. C represents the combined constants of integration.

The last equation gives y as an implicit function of x.

450

Chapter 7 Integrals and Transcendental Functions

EXAMPLE 2

Solve the equation y ( x + 1)

dy = x ( y 2 + 1). dx

Solution We change to differential form, separate the variables, and integrate:

y ( x + 1) dy = x ( y 2 + 1) dx y dy x dx = y2 + 1 x +1 y dy

∫ 1 + y2

=

x ≠ −1

∫ (1 − x + 1 ) dx 1

Divide  x  by  x + 1.

1 ( ln 1 + y 2 ) = x − ln x + 1 + C. 2 The last equation gives the solution y as an implicit function of x. The initial value problem dy = ky, dt y

involves a separable differential equation, and the solution y = y 0 e kt expresses exponential change. We now present several examples of such change.

500 Yeast biomass (mg)

y (0) = y 0

400 300

Unlimited Population Growth

200 100 0 0

5 Time (hr)

10

FIGURE 7.6 Graph of the growth of a yeast population over a 10-hour period, based on the data in Table 7.3.

TABLE 7.3 Population of yeast

Time (hr)

Yeast biomass (mg)

  0   1   2   3   4   5   6   7   8   9 10

    9.6   18.3   29.0   47.2   71.1 119.1 174.6 257.3 350.7 441.0 513.3

t

Strictly speaking, the number of individuals in a population (of people, plants, animals, or bacteria, for example) is a discontinuous function of time because it takes on discrete values. However, when the number of individuals becomes large enough, the population can be approximated by a continuous function. Differentiability of the approximating function is another reasonable hypothesis in many settings, allowing for the use of calculus to model and predict population sizes. If we assume that the proportion of reproducing individuals remains constant and assume a constant fertility, then at any instant t the birth rate is proportional to the number y (t ) of individuals present. We likewise assume that the death rate of the population is stable and proportional to y (t ). If, further, we neglect departures and arrivals, the growth rate dy dt is the birth rate minus the death rate, which is the difference of the two proportionalities under our assumptions. In other words, dy dt = ky so that y = y 0 e kt , where y 0 is the size of the population at time t = 0. As with all kinds of growth, there may be limitations imposed by the surrounding environment, but we will not go into these here. (We treat one model imposing such limitations in Section 16.4.) When k is positive, the proportionality dy dt = ky models unlimited population growth. (See Figure 7.6.)

EXAMPLE 3 The biomass of a yeast culture in an experiment is initially 29 grams. After 30 minutes the mass is 37 grams. Assuming that the equation for unlimited population growth gives a good model for the growth of the yeast when the mass is below 100 grams, how long will it take for the mass to double from its initial value? Solution Let y (t ) be the yeast biomass after t minutes. We use the exponential growth model dy dt = ky for unlimited population growth, with solution y = y 0 e kt . We have y 0 = y (0) = 29. We are also told that

y (30) = 29 e k(30) = 37.

7.2  Exponential Change and Separable Differential Equations

451

Solving this equation for k, we find e k(30) =

37 29

30 k = ln k =

( 37 29 ) ( )

1 37 ≈ 0.008118. ln 30 29

Then the mass of the yeast in grams after t minutes is given by the equation y = 29 e ( 0.008118 ) t. To solve the problem, we find the time t for which y (t ) = 58, which is twice the initial amount. 29 e ( 0.008118 )t = 58 ( 0.008118 ) t

= ln

t =

( 58 29 )

ln 2 ≈ 85.38 0.008118

It takes about 85 minutes for the yeast population to double. In the next example we model the number of people within a given population who are infected by a disease that is being eradicated from the population. Here the constant of proportionality k is negative, and the model describes an exponentially decaying number of infected individuals.

EXAMPLE 4 One model for the way diseases die out when properly treated assumes that the rate dy dt at which the number of infected people changes is proportional to the number y. The number of people cured is proportional to the number y that are infected with the disease. Suppose that in the course of any given year, the number of cases of a disease is reduced by 20%. If there are 10,000 cases today, how many years will it take to reduce the number to 1000? Solution We use the equation y = y 0 e kt. There are three things to find: the value of y 0, the value of k, and the time t when y = 1000. The value of y 0. We are free to count time beginning anywhere we want. If we count from today, then y = 10, 000 when t = 0, so y 0 = 10, 000. Our equation is now

y = 10, 000 e kt.

(5)

The value of k. When t = 1 year, the number of cases will be 80% of its present value, or 8000. Hence, 8000 = 10,000 e k (1) ek ln(e k )

Eq. (5) with t = 1 and  y = 8000

= 0.8 = ln 0.8

Take logs of both sides

k = ln 0.8 < 0.

ln 0.8 ≈ −0.223

At any given time t, y = 10,000 e ( ln 0.8 )t.

(6)

452

Chapter 7 Integrals and Transcendental Functions

y

The value of t that makes y  1000. We set y equal to 1000 in Equation (6) and solve for t:

10,000

1000 = 10,000 e ( ln 0.8 )t 5,000

e ( ln 0.8 )t = 0.1

y = 10,000 e(ln 0.8)t

1,000 0

5

10

t

FIGURE 7.7

A graph of the number of people infected by a disease exhibits exponential decay (Example 4).

For radon-222 gas, t is measured in days and k  0.18. For radium-226, which used to be painted on watch dials to make them glow at night (a dangerous practice), t is measured in years and k = 4.3 × 10 −4.

( ln 0.8 ) t = ln 0.1 ln 0.1 t = ≈ 10.32 years. ln 0.8

  

Take logs of both sides

It will take a little more than 10 years to reduce the number of cases to 1000. (See Figure 7.7.)

Radioactivity Some atoms are unstable and can spontaneously emit mass or radiation. This process is called radioactive decay, and an element whose atoms go spontaneously through this process is called radioactive. Sometimes when an atom emits some of its mass through this process of radioactivity, the remainder of the atom re-forms to make an atom of some new element. For example, radioactive carbon-14 decays into nitrogen; radium, through a number of intermediate radioactive steps, decays into lead. Experiments have shown that at any given time, the rate at which a radioactive element decays (as measured by the number of nuclei that change per unit time) is approximately proportional to the number of radioactive nuclei present. Thus, the decay of a radioactive element is described by the equation dy dt = −ky, k > 0. It is conventional to use k, with k  0, to emphasize that y is decreasing. If y 0 is the number of radioactive nuclei present at time zero, the number still present at any later time t will be y = y 0 e −kt , k > 0. In Section 1.5, we defined the half-life of a radioactive element to be the time required for half of the radioactive nuclei present in a sample to decay. It is an interesting fact that the half-life is a constant that does not depend on the number of radioactive nuclei initially present in the sample, but only on the radioactive substance. We found that the half-life is given by the following equation.

Half-life 

ln 2 k

(7)

For example, the half-life for radon-222 is half-life =

Carbon-14 dating uses the half-life of 5730 years.

ln 2 ≈ 3.9 days. 0.18

EXAMPLE 5 The decay of radioactive elements can sometimes be used to date events from Earth’s past. In a living organism, the ratio of radioactive carbon, carbon-14, to ordinary carbon stays fairly constant during the lifetime of the organism, being approximately equal to the ratio in the organism’s atmosphere at the time. After the organism’s death, however, no new carbon is ingested, and the proportion of carbon-14 in the organism’s remains decreases as the carbon-14 decays. Scientists who do carbon-14 dating often use a figure of 5730 years for its half-life. Find the age of a sample in which 10% of the radioactive nuclei originally present have decayed. Solution We use the decay equation y = y 0 e −kt. There are two things to find: the value of k and the value of t when y is 0.9 y 0 (90% of the radioactive nuclei are still present). That is, find t when y 0 e −kt = 0.9 y 0 , or e −kt = 0.9.

7.2  Exponential Change and Separable Differential Equations

453

The value of k. We use the half-life Equation (7): k =

ln 2 ln 2 ( about 1.2 × 10 −4 ). = half-life 5730

The value of t that makes e −kt = 0.9. e −kt = 0.9 e −( ln 2 5730)t = 0.9 −

ln 2 t = ln 0.9 5730 5730 ln 0.9 t = − ≈ 871 years ln 2

  

Take logs of both sides

The sample is about 871 years old.

Heat Transfer: Newton’s Law of Cooling Hot soup left in a tin cup cools to the temperature of the surrounding air. A hot silver bar immersed in a large tub of water cools to the temperature of the surrounding water. In situations like these, the rate at which an object’s temperature is changing at any given time is roughly proportional to the difference between its temperature and the temperature of the surrounding medium. This observation is called Newton’s Law of Cooling, although it applies to warming as well. If H is the temperature of the object at time t, and H S is the constant surrounding temperature, then the differential equation is dH = −k ( H − H S ). dt

(8)

If we substitute y for ( H − H S ), then dy d dH d = (H − HS ) = − ( HS ) dt dt dt dt dH = −0 dt dH = dt = −k ( H − H S ) = −ky.

H S is a constant.

Eq. (8) H − HS = y

We know that the solution of the equation dy dt = −ky is y = y 0 e −kt , where y (0) = y 0 . Substituting ( H − H S ) for y, this says that H − H S = ( H 0 − H S ) e −kt ,

(9)

where H 0 is the temperature at t = 0. This equation is the solution to Newton’s Law of Cooling. EXAMPLE 6 A hard-boiled egg at 98 °C is put in a sink of 18 °C water. After 5 min, the egg’s temperature is 38 °C. Assuming that the water has not warmed appreciably, how much longer will it take the egg to reach 20 °C? Solution We find how long it would take the egg to cool from 98 °C to 20 °C and subtract the 5 min that have already elapsed. Using Equation (9) with H S = 18 and H 0 = 98, the egg’s temperature t min after it is put in the sink is

H = 18 + ( 98 − 18 ) e −kt = 18 + 80 e −kt.

454

Chapter 7 Integrals and Transcendental Functions

To find k, we use the information that H = 38 when t = 5: 38 = 18 + 80 e −5 k e −5 k =

1 4

−5 k = ln k =

1 = − ln 4 4

1 ln 4 = 0.2 ln 4 5

(about 0.28).

The egg’s temperature at time t is H = 18 + 80 e −( 0.2 ln 4 )t. Now find the time t when H = 20: 20 = 18 + 80 e −( 0.2 ln 4 )t 80 e −( 0.2 ln 4 )t = 2 e −( 0.2 ln 4 )t =

1 40

1 = − ln 40 40 ln 40 ≈ 13 min. t = 0.2 ln 4

−( 0.2 ln 4 ) t = ln

The egg’s temperature will reach 20 °C about 13 min after it is put in the water to cool. Since it took 5 min to reach 38 °C, it will take about 8 min more to reach 20 °C.

7.2

EXERCISES Verifying Solutions

Separable Differential Equations

In Exercises 1–4, show that each function y = f ( x ) is a solution of the accompanying differential equation.

Solve the differential equation in Exercises 9–22.

1. 2 y′ + 3 y = e −x b. y = e −x + e −( 3 2 )x

a. y = e −x c. y =

e −x

+ Ce

−( 3 2 )x

2. y′ = y 2 a. y = − 3. y = 4. y =

1 x

∫1

1 x

x

b. y = −

1 x+3

c. y = −

1 x+C

et dt , x 2 y′ + xy = e x t

1 1 + x4

∫1

x

1 + t 4 dt ,

y′ +

2x 3 y =1 1 + x4

Initial Value Problems

In Exercises 5–8, show that each function is a solution of the given initial value problem. Differential equation

Initial equation

Solution candidate

2 5. y′ + y = 1 + 4e 2 x

π y (− ln 2 ) = 2

y =

6. y′ = e −x 2 − 2 xy

y (2) = 0

y = ( x − 2 )e −x 2

7. xy′ + y = − sin x , x > 0

π y = 0 2

cos x y = x

8. x 2 y′ = xy − y 2, x >1

y (e) = e

y =

( )

e −x

tan −1 (2e x )

x ln x

9. 2 xy

dy = 1, x , y > 0 dx

10.

dy = x 2 y, dx dy = 3x 2 e −y dx

11.

dy = e x−y dx

12.

13.

dy = dx

14.

15.

x

y cos 2 y

dy = e y+ dx

x,

x > 0

2 xy

dy = 2 x 1 − y 2 , −1 < y < 1 dx

18.

dy e 2x−y = x+ y dx e

dy = e y + sin x dx

dy = 3x 2 y 3 − 6 x 2 dx

20.

dy = xy + 3 x − 2 y − 6 dx

1 dy = ye x 2 + 2 y   e x 2 x dx

22.

dy = e x− y + e x + e −y + 1 dx

19. y 2 21.

dy =1 dx

16. (sec x )

17.

y > 0

Applications and Examples

The answers to most of the following exercises are in terms of logarithms and exponentials. A calculator can be helpful, enabling you to express the answers in decimal form. 23. Human evolution continues The analysis of tooth shrinkage by C. Loring Brace and colleagues at the University of Michigan’s

7.2  Exponential Change and Separable Differential Equations

Museum of Anthropology indicates that human tooth size is continuing to decrease and that the evolutionary process has not yet come to a halt. In northern Europeans, for example, tooth size reduction now has a rate of 1% per 1000 years. a. If t represents time in years and y represents tooth size, use the condition that y = 0.99 y 0 when t = 1000 to find the value of k in the equation y = y 0 e kt. Then use this value of k to answer the following questions. b. In about how many years will human teeth be 90% of their present size? c. What will be our descendants’ tooth size 20,000 years from now (as a percentage of our present tooth size)? 24. Atmospheric pressure The earth’s atmospheric pressure p is often modeled by assuming that the rate dp dh at which p changes with the altitude h above sea level is proportional to p. Suppose that the pressure at sea level is 1013 hectopascals and that the pressure at an altitude of 20 km is 90 hectopascals. a. Solve the initial value problem Differential equation: dp dh = kp (k a constant) Initial condition: p = p 0 when h = 0 to express p in terms of h. Determine the values of p 0 and k from the given altitude-pressure data. b. What is the atmospheric pressure at h = 50 km ? c. At what altitude does the pressure equal 900 hectopascals? 25. First-order chemical reactions In some chemical reactions, the rate at which the amount of a substance changes with time is proportional to the amount present. For the change of δ -gluconolactone into gluconic acid, for example, dy = −0.6 y dt when t is measured in hours. If there are 100 grams of δ -gluconolactone present when t = 0, how many grams will be left after the first hour? 26. The inversion of sugar The processing of raw sugar has a step called “inversion” that changes the sugar’s molecular structure. Once the process has begun, the rate of change of the amount of raw sugar is proportional to the amount of raw sugar remaining. If 1000 kg of raw sugar reduces to 800 kg of raw sugar during the first 10 hours, how much raw sugar will remain after another 14 hours? 27. Working underwater The intensity L ( x ) of light x meters beneath the surface of the ocean satisfies the differential equation dL = −kL. dx As a diver, you know from experience that diving to 6 m in the Caribbean Sea cuts the intensity in half. You cannot work without artificial light when the intensity falls below one-tenth of the surface value. About how deep can you expect to work without artificial light? 28. Voltage in a discharging capacitor Suppose that electricity is draining from a capacitor at a rate that is proportional to the voltage V across its terminals and that, if t is measured in seconds, dV 1 = − V. dt 40

455

Solve this equation for V, using V0 to denote the value of V when t = 0. How long will it take the voltage to drop to 10% of its original value? 29. Cholera bacteria Suppose that the bacteria in a colony can grow unchecked, by the law of exponential change. The colony starts with 1 bacterium and doubles every half-hour. How many bacteria will the colony contain at the end of 24 hours? (Under favorable laboratory conditions, the number of cholera bacteria can double every 30 min. In an infected person, many bacteria are destroyed, but this example helps explain why a person who feels well in the morning may be dangerously ill by evening.) 30. Growth of bacteria A colony of bacteria is grown under ideal conditions in a laboratory so that the population increases exponentially with time. At the end of 3 hours there are 10,000 bacteria. At the end of 5 hours there are 40,000. How many bacteria were present initially? 31. The incidence of a disease (Continuation of Example 4.) Suppose that in any given year the number of cases can be reduced by 25% instead of 20%. a. How long will it take to reduce the number of cases to 1000? b. How long will it take to eradicate the disease—that is, reduce the number of cases to less than 1? 32. Drug concentration An antibiotic is administered intravenously into the bloodstream at a constant rate r. As the drug flows through the patient’s system and acts on the infection that is present, it is removed from the bloodstream at a rate proportional to the amount in the bloodstream at that time. Since the amount of blood in the patient is constant, this means that the concentration y = y (t ) of the antibiotic in the bloodstream can be modeled by the differential equation dy = r − ky, k > 0 and constant. dt a. If y (0) = y 0 , find the concentration y (t ) at any time t. b. Assume that y 0 < ( r k ) and find lim y (t ). Sketch the soluy→∞ tion curve for the concentration. 33. Endangered species Biologists consider a species of animal or plant to be endangered if it is expected to become extinct within 20 years. If a certain species of wildlife is counted to have 1147 members at the present time, and the population has been steadily declining exponentially at an annual rate averaging 39% over the past 7 years, do you think the species is endangered? Explain your answer. 34. The U.S. population The U.S. Census Bureau keeps a running clock totaling the U.S. population. On April 19, 2021, the total was increasing at the rate of 1 person every 40 s. The population figure for 1:32 p.m. EST on that day was 330,215,841. a. Assuming exponential growth at a constant rate, find the rate constant for the population’s growth (people per 365-day year). b. At this rate, what will the U.S. population be at 1:32 p.m. EST on April 19, 2028? 35. Oil depletion Suppose the amount of oil pumped from one of the canyon wells in Whittier, California, decreases at the continuous rate of 10% per year. When will the well’s output fall to onefifth of its present value?

456

Chapter 7 Integrals and Transcendental Functions

36. Continuous price discounting To encourage buyers to place 100-unit orders, your firm’s sales department applies a continuous discount that makes the unit price a function p ( x ) of the number of units x ordered. The discount decreases the price at the rate of $0.01 per unit ordered. The price per unit for a 100-unit order is p(100 ) = $20.09. a. Find p ( x ) by solving the following initial value problem. Differential equation: Initial condition:

dp 1 p = − 100 dx p (100) = 20.09

b. Find the unit price p(10) for a 10-unit order and the unit price p(90) for a 90-unit order. c. The sales department has asked you to find out if it is discounting so much that the firm’s revenue, r ( x ) = x ⋅ p ( x ), will actually be less for a 100-unit order than, say, for a 90-unit order. Reassure them by showing that r has its maximum value at x  100. d. Graph the revenue function r ( x )  xp ( x ) for 0 b x b 200. 37. Plutonium-239 The half-life of the plutonium isotope is 24,360 years. If 10 g of plutonium is released into the atmosphere by a nuclear accident, how many years will it take for 80% of the isotope to decay? 38. Polonium-210 The half-life of polonium is 139 days, but your sample will not be useful to you after 95% of the radioactive nuclei present on the day the sample arrives has disintegrated. For about how many days after the sample arrives will you be able to use the polonium? 39. The mean life of a radioactive nucleus Physicists using the radioactivity equation y = y 0 e −kt call the number 1 k the mean life of a radioactive nucleus. The mean life of a radon nucleus is about 1 0.18  5.6 days. The mean life of a carbon-14 nucleus is more than 8000 years. Show that 95% of the radioactive nuclei originally present in a sample will disintegrate within three mean lifetimes, i.e., by time t  3 k . Thus, the mean life of a nucleus gives a quick way to estimate how long the radioactivity of a sample will last. 40. Californium-252 What costs $27 million per gram and can be used to treat brain cancer, analyze coal for its sulfur content, and detect explosives in luggage? The answer is californium-252, a radioactive isotope discovered by Glenn Seaborg in 1950. Much less than 1g of it is produced each year. The half-life of the isotope is 2.645 years—long enough for a useful service life and short enough to have a high radioactivity per unit mass. One microgram of the isotope releases 170 million neutrons per minute. a. What is the value of k in the decay equation for this isotope? b. What is the isotope’s mean life? (See Exercise 39.) c. How long will it take 95% of a sample’s radioactive nuclei to disintegrate? 41. Cooling soup Suppose that a cup of soup cooled from 90 nC to 60 nC after 10 min in a room where the temperature was 20 nC. Use Newton’s Law of Cooling to answer the following questions. a. How much longer would it take the soup to cool to 35 nC? b. Instead of being left to stand in the room, the cup of 90 nC soup is put in a freezer where the temperature is −15 °C. How long will it take the soup to cool from 90 nC to 35 nC?

42. A beam of unknown temperature An aluminum beam was brought from the outside cold into a machine shop where the temperature was held at 18 nC. After 10 min, the beam warmed to 2 nC and after another 10 min, it was 10 nC. Use Newton’s Law of Cooling to estimate the beam’s initial temperature. 43. Surrounding medium of unknown temperature A pan of warm water (46 nC) was put in a refrigerator. Ten minutes later, the water’s temperature was 39 nC; 10 min after that, it was 33 nC. Use Newton’s Law of Cooling to estimate how cold the refrigerator was. 44. Silver cooling in air The temperature of an ingot of silver is 60 nC above room temperature right now. Twenty minutes ago, it was 70 nC above room temperature. How far above room temperature will the silver be a. 15 min from now? b. 2 hours from now? c. When will the silver be 10 nC above room temperature? 45. The age of Crater Lake The charcoal from a tree killed in the volcanic eruption that formed Crater Lake in Oregon contained 44.5% of the carbon-14 found in living matter. About how old is Crater Lake? 46. The sensitivity of carbon-14 dating to measurement To see the effect of a relatively small error in the estimate of the amount of carbon-14 in a sample being dated, consider this hypothetical situation: a. A bone fragment found in central Illinois in the year 2000 contains 17% of its original carbon-14 content. Estimate the year the animal died. b. Repeat part (a), assuming 18% instead of 17%. c. Repeat part (a), assuming 16% instead of 17%. 47. Carbon-14 The oldest known frozen human mummy, discovered in the Schnalstal glacier of the Italian Alps in 1991 and called Otzi, was found wearing straw shoes and a leather coat with goat fur, and holding a copper ax and stone dagger. It was estimated that Otzi died 5000 years before he was discovered in the melting glacier. How much of the original carbon-14 remained in Otzi at the time of his discovery? 48. Art forgery A painting attributed to Vermeer (1632–1675), which should contain no more than 96.2% of its original carbon-14, contains 99.5% instead. About how old is the forgery? 49. Lascaux Cave paintings Prehistoric cave paintings of animals were found in the Lascaux Cave in France in 1940. Scientific analysis revealed that only 15% of the original carbon-14 in the paintings remained. What is an estimate of the age of the paintings? 50. Incan mummy The frozen remains of a young Incan woman were discovered by archeologist Johan Reinhard on Mt. Ampato in Peru during an expedition in 1995. a. How much of the original carbon-14 was present if the estimated age of the “Ice Maiden” was 500 years? b. If a 1% error can occur in the carbon-14 measurement, what is the oldest possible age for the Ice Maiden?

7.3  Hyperbolic Functions

457

7.3 Hyperbolic Functions The hyperbolic functions are formed by taking combinations of the two exponential functions e x and e − x . The hyperbolic functions simplify many mathematical expressions and occur frequently in mathematical and engineering applications.

Definitions and Identities The hyperbolic sine and hyperbolic cosine functions are defined by the equations sinh x =

e x − e −x 2

cosh x =

and

e x + e −x . 2

We pronounce sinh x as “cinch x,” rhyming with “pinch x ,” and cosh x as “kosh x ,” rhyming with “gosh x.” From this basic pair, we define the hyperbolic tangent, cotangent, secant, and cosecant functions. The defining equations and graphs of these functions are shown in Table 7.4. We will see that the hyperbolic functions bear many similarities to the trigonometric functions after which they are named.

TABLE 7.4 The six basic hyperbolic functions y y = cosh x

y

y

3

x y= e 2

3 y = sinh x

2

y=

e−x 2

1 −3 −2 −1

y = coth x

x

y=e 2

2

2 y=1

1 1

2

−1 y=−

−2

x

3

−3 −2 −1

1

2

x

3

−2

e−x 2

y = tanh x x 1 2

−1

−2

−3

y = −1

y = coth x (b)

(a)

(c)

Hyperbolic sine:

Hyperbolic cosine:

Hyperbolic tangent:

sinh x =

cosh x =

tanh x =

e x − e −x 2

e x + e −x 2

y

Hyperbolic cotangent:

y

coth x = 2

2 y=1 1

−2

−1

1

0

2

x

−2

−1

sech x =

1 2 = x e + e −x cosh x

2

x

y = csch x

(d)

Hyperbolic secant:

1 −1

y = sech x

(e)

Hyperbolic cosecant:

csch x =

sinh x e x − e −x = x e + e −x cosh x

1 2 = x e − e −x sinh x

cosh x e x + e −x = x e − e −x sinh x

458

Chapter 7 Integrals and Transcendental Functions

TABLE 7.5 Identities for

hyperbolic functions

cosh 2 x − sinh 2 x = 1 sinh 2 x = 2 sinh x cosh x cosh 2 x = cosh 2 x + sinh 2 x cosh 2 x + 1 2 cosh 2 x − 1 sinh 2 x = 2 cosh 2 x =

tanh 2 x = 1 − sech 2 x coth 2 x = 1 + csch 2 x

Hyperbolic functions satisfy the identities in Table 7.5. Except for differences in sign, these resemble identities we know for the trigonometric functions. The identities are proved directly from the definitions, as we show here for the second one:

(e

− e −x 2 e 2 x − e −2 x = 2 = sinh 2 x.

2 sinh x cosh x = 2

x

)( e

x

+ e −x 2

) Simplify. Definition of sinh

The other identities are obtained similarly, by substituting in the definitions of the hyperbolic functions and using algebra. For any real number u, we know the point with coordinates (cos u, sin u) lies on the unit circle x 2 + y 2 = 1. So the trigonometric functions are sometimes called the circular functions. Because of the first identity cosh 2 u − sinh 2 u = 1,

TABLE 7.6 Derivatives of

hyperbolic functions

d ( sinh x ) = cosh x dx d ( cosh x ) = sinh x dx d ( tanh x ) = sech 2 x dx d ( coth x ) = − csch 2 x dx d ( sech x ) = − sech x tanh x dx d ( csch x ) = − csch x coth x dx

with u substituted for x in Table 7.5, the point having coordinates (cosh u, sinh u) lies on the right-hand branch of the hyperbola x 2 − y 2 = 1. This is where the hyperbolic functions get their names (see Exercise 86). Hyperbolic functions are useful in finding integrals, which we will see in Chapter 8. They play an important role in science and engineering as well. The hyperbolic cosine describes the shape of a hanging cable or wire that is strung between two points at the same height and hanging freely (see Exercise 83). The shape of the St. Louis Arch is an inverted hyperbolic cosine. The hyperbolic tangent occurs in the formula for the velocity of an ocean wave moving over water having a constant depth, and the inverse hyperbolic tangent describes how relative velocities sum according to Einstein’s Law in the Special Theory of Relativity.

Derivatives and Integrals of Hyperbolic Functions The six hyperbolic functions, being rational combinations of the differentiable functions e x and e − x , have derivatives at every point at which they are defined (Table 7.6). Again, there are similarities to trigonometric functions. The derivative formulas are obtained from the derivative of e x :

(

d d e x − e −x ( sinh x ) = dx dx 2 =

TABLE 7.7 Integral formulas for

hyperbolic functions

∫ sinh x dx ∫

= cosh x + C

cosh x dx = sinh x + C

∫ sech 2 x dx

= tanh x + C

∫ csch 2 x dx

= − coth x + C

∫ sech x tanh x dx

= −sech x + C

∫ csch x coth x dx

= − csch x + C

e x + e −x 2

= cosh x.

)

Definition of sinh x

Derivative of e x Definition of cosh x

This gives the first derivative formula. From the definition, we can calculate the derivative of the hyperbolic cosecant function, as follows: d d ⎛⎜ 1 ⎞⎟ ( csch x ) = ⎟ ⎜ dx dx ⎜⎝ sinh x ⎟⎠ cosh x sinh 2 x 1 cosh x =− sinh x sinh x =−

= − csch x coth x

Definition of csch x

Quotient Rule for derivatives Rearrange factors. Definitions of csch x and coth x

The other formulas in Table 7.6 are obtained similarly. The derivative formulas lead to the integral formulas in Table 7.7.

7.3  Hyperbolic Functions

EXAMPLE 1

We illustrate the derivative and integral formulas.

d ( tanh 1 + t 2 ) = sech 2 1 + t 2 ⋅ dtd ( 1 + t 2 ) dt t sech 2 1 + t 2 = 1 + t2

(a)

(b)

459

∫ coth 5 x dx = ∫

cosh 5 x 1 du dx = ∫ sinh 5 x 5 u

u = sinh 5 x , du = 5 cosh 5 x dx

1 1 ln u + C = ln sinh 5 x + C 5 5 1 1 cosh 2 x − 1 dx (c) ∫ sinh 2 x dx = ∫ 0 0 2 1 ⎤ 1 1 1 ⎡ sinh 2 x = ∫ ( cosh 2 x − 1) dx = ⎢ − x⎥ 2 0 2⎣ 2 ⎦0 sinh 2 1 = − ≈ 0.40672 4 2 =

(d)

ln 2

∫0

4e x sinh x dx =

ln 2

∫0

4e x

e x − e −x dx = 2

ln 2

∫0

Table 7.5

Evaluate with a calculator.

( 2e 2 x − 2 ) dx

ln 2 = ⎡⎢ e 2 x − 2 x ⎤⎥ = ( e 2 ln 2 − 2 ln 2 ) − (1 − 0 ) ⎣ ⎦0 = 4 − 2 ln 2 − 1 ≈ 1.6137

Inverse Hyperbolic Functions The inverses of the six basic hyperbolic functions are very useful in integration (see Chapter 8). Since d ( sinh x ) dx = cosh x > 0, the hyperbolic sine is an increasing function of x. We denote its inverse by y = sinh −1 x. For every value of x in the interval −∞ < x < ∞, the value of y = sinh −1 x is the number whose hyperbolic sine is x. The graphs of y = sinh x and y = sinh −1 x are shown in Figure 7.8a. Other notations used for the inverse hyperbolic sine function include arcsinh x , arsinh x , and argsinh x . The function y = cosh x is not one-to-one because its graph in Table 7.4 does not pass the horizontal line test. The restricted function y = cosh x , x ≥ 0, however, is oneto-one and therefore has an inverse, denoted by y = cosh −1 x. y y

y = sinh x y = x y = sinh−1 x (x = sinh y)

2 1 −6 −4 −2

2

4

6

x

8 7 6 5 4 3 2 1 0

y = cosh x, y=x x≥0

y

3

y = sech−1 x (x = sech y, y ≥ 0)

y=x

2 y = cosh−1 x (x = cosh y, y ≥ 0) x 1 2 3 4 5 6 7 8 (b)

1 0

1

y = sech x x≥0 x 2 3 (c)

(a)

FIGURE 7.8 The graphs of the inverse hyperbolic sine, cosine, and secant of x. Notice the symmetries about the line y = x .

460

Chapter 7 Integrals and Transcendental Functions

For every value of x ≥ 1, y = cosh −1 x is the number in the interval 0 ≤ y < ∞ whose hyperbolic cosine is x. The graphs of y = cosh x , x ≥ 0, and y = cosh −1 x are shown in Figure 7.8b. Like y = cosh x , the function y = sech x = 1 cosh x fails to be one-to-one, but its restriction to nonnegative values of x does have an inverse, denoted by y = sech −1 x. For every value of x in the interval ( 0, 1 ], y = sech −1 x is the nonnegative number whose hyperbolic secant is x. The graphs of y = sech x , x ≥ 0, and y = sech −1 x are shown in Figure 7.8c. The hyperbolic tangent, cotangent, and cosecant are one-to-one on their domains and therefore have inverses, denoted by y = tanh −1 x ,

y = coth −1 x ,

y = csch −1 x.

These functions are graphed in Figure 7.9. y

y x = tanh y y = tanh−1 x

−1

x = coth y y = coth−1 x

0

x

1

−1

(a)

FIGURE 7.9

y x = csch y y = csch−1 x

0

x

1

(b)

0

x

(c)

The graphs of the inverse hyperbolic tangent, cotangent, and cosecant of x.

Useful Identities TABLE 7.8 Identities for

inverse hyperbolic functions

1 x 1 csch −1 x = sinh −1 x 1 coth −1 x = tanh −1 x

sech −1 x = cosh −1

We can use the identities in Table 7.8 to express sech −1 x , csch −1 x , and coth −1 x in terms of cosh −1 x , sinh −1 x , and tanh −1 x. These identities are direct consequences of the definitions. For example, if 0 < x ≤ 1, then

(

sech cosh −1

1 1 = = x. ( 1x )) = cosh cosh 1 1 ( ( x )) ( x ) −1

We also know that sech ( sech −1 x ) = x , so because the hyperbolic secant is one-to-one on ( 0, 1 ], we have cosh −1

( 1x ) = sech

−1

x.

Derivatives of Inverse Hyperbolic Functions An important use of inverse hyperbolic functions lies in antiderivatives that reverse the derivative formulas in Table 7.9. The restrictions x < 1 and x > 1 on the derivative formulas for tanh −1 x and coth −1 x come from the natural restrictions on the values of these functions. (See Figure 7.9a and b.) The distinction between x < 1 and x > 1 becomes important when we convert the derivative formulas into integral formulas.

7.3  Hyperbolic Functions

461

We illustrate how the derivatives of the inverse hyperbolic functions are found in Example 2, where we calculate d ( cosh −1 x ) dx . The other derivatives are obtained by similar calculations. TABLE 7.9 Derivatives of inverse hyperbolic functions

d ( sinh −1 x ) = dx

1 1 + x2

d ( cosh −1 x ) = dx

EXAMPLE 2

1 x2

−1

x >1

,

d ( tanh −1 x ) 1 = , dx 1 − x2

x 1

d ( sech −1 x ) 1 = − , dx x 1 − x2

0 < x 0

( )

x > a > 0

1.

∫

dx x = sinh −1 + C, a a2 + x 2

2.

∫

x2

dx x = cosh −1 + C, 2 a −a

∫

dx a2 − x2

4.

∫

x

a2

5.

∫x

a2

3.

⎧⎪ ⎪⎪ ⎪ = ⎨ ⎪⎪ ⎪⎪ ⎪⎩

1 tanh −1 a 1 coth −1 a

( ax ) + C , ( ax ) + C , ( )

x2 < a2 x2 > a2

dx 1 x = − sech −1 + C, 2 a a −x

0 < x < a

dx 1 x = − csch −1 + C, 2 a a + x

x ≠ 0 and a > 0

Solution The indefinite integral is

2 dx

∫

3 + 4x 2

=

∫

du 3 + u2

= sinh −1 = sinh −1

u = 2 x , du = 2 dx

( u3 ) + C ( 2x3 ) + C.

Formula 1 from Table 7.10 with a 2 = 3

Therefore, 1

∫0

EXERCISES

2 dx 3 + 4x 2

Values and Identities

3. cosh x =

( ) ( )

7.3

Each of Exercises 1–4 gives a value of sinh x or cosh x. Use the definitions and the identity cosh 2 x − sinh 2 x = 1 to find the values of the remaining five hyperbolic functions. 1. sinh x = −

( )

2x ⎤ 1 2 −1 = ⎡⎢ sinh −1 − sinh −1 (0) ⎥ = sinh ⎣ 3 ⎦0 3 2 = sinh −1 − 0 ≈ 0.98665. 3

3 4

17 , x > 0 15

2. sinh x =

4 3

4. cosh x =

13 , x > 0 5

Rewrite the expressions in Exercises 5–10 in terms of exponentials and simplify the results as much as you can. 5. 2 cosh (ln x )

6. sinh (2 ln x )

7. cosh 5 x + sinh 5 x

8. cosh 3 x − sinh 3 x

9. ( sinh x + cosh x ) 4 10. ln ( cosh x + sinh x ) + ln ( cosh x − sinh x ) 11. Prove the identities sinh ( x + y ) = sinh x cosh y + cosh x sinh y, cosh ( x + y ) = cosh x cosh y + sinh x sinh y. Then use them to show that a. sinh 2 x = 2 sinh x cosh x. b. cosh 2 x = cosh 2 x + sinh 2 x. 12. Use the definitions of cosh x and sinh x to show that cosh 2 x − sinh 2 x = 1.

7.3  Hyperbolic Functions

Finding Derivatives

sech t tanh t dt t

In Exercises 13–24, find the derivative of y with respect to the appropriate variable.

49.

∫

x 13. y = 6 sinh 3

1 14. y = sinh ( 2 x + 1) 2 1 16. y = t 2 tanh t

51.

∫ln 2

53.

∫−ln 4

18. y = ln ( cosh z )

15. y = 2 t tanh t 17. y = ln ( sinh z )

∫

52.

∫0

2e θ cosh θ dθ

54.

∫0

55.

∫−π 4 cosh ( tan θ ) sec 2 θ dθ

56.

∫0

57.

∫1

58.

∫1

59.

∫−ln 2 cosh 2 ( 2 ) dx

60.

∫0

19. y = ( sech θ )(1 − ln sech θ ) 20. y = ( csch θ )(1 − ln csch θ ) 21. y = ln cosh υ −

1 1 tanh 2 υ 22. y = ln sinh υ − coth 2 υ 2 2

23. y = ( x 2 + 1) sech ( ln x ) (Hint: Before differentiating, express in terms of exponentials and simplify.) 24. y = ( 4 x 2 − 1) csch ( ln 2 x )

csch ( ln t ) coth ( ln t ) dt t

50.

ln 4

coth x dx

− ln 2

π4

2

cosh ( ln t ) dt t

0

x

463

ln 2

ln 2

π2

4

tanh 2 x dx 4e −θ sinh θ dθ 2 sinh ( sin θ ) cos θ dθ

8 cosh x dx x

ln 10

4 sinh 2

( 2x ) dx

Inverse Hyperbolic Functions and Integrals

Since the hyperbolic functions can be expressed in terms of exponential functions, it is possible to express the inverse hyperbolic functions in terms of logarithms, as shown in the following table.

In Exercises 25–36, find the derivative of y with respect to the appropriate variable. 25. y = sinh −1 x

26. y = cosh −1 2 x + 1

sinh −1 x = ln ( x +

x 2 + 1 ),

−∞ < x < ∞

27. y = (1 − θ ) tanh −1 θ

28. y = ( θ 2 + 2θ ) tanh −1 ( θ + 1)

cosh −1 x = ln ( x +

x 2 − 1 ),

x ≥1

29. y = (1 − t ) coth −1 t

30. y = (1 − t 2 ) coth −1 t

31. y =

32. y = ln x +

1 1+ ln 2 1− ⎛1 + sech −1 x = ln ⎜⎜⎜ ⎝

33. y =

cos −1

x−x

csch −1

( ) 1 2

sech −1

θ

x

34. y =

1−

x2

tanh −1 x =

sech −1

x

csch −1 2 θ

x , x

x 1

Verify the integration formulas in Exercises 37–40. 37. a.

∫ sech x dx

= tan −1 ( sinh x ) + C

b.

∫ sech x dx

= sin −1 ( tanh x ) + C x2 1 sech −1 x − 1 − x2 + C 2 2

38.

∫ x sech −1 x dx

39.

∫

x coth −1 x dx =

40.

∫

tanh −1 x dx = x tanh −1 x +

=

Use these formulas to express the numbers in Exercises 61–66 in terms of natural logarithms.

x2 − 1 x coth −1 x + + C 2 2 1 ( ln 1 − x 2 ) + C 2

61. sinh −1 (− 5 12 )

62. cosh −1 ( 5 3 )

63. tanh −1 (−1 2 )

64. coth −1 ( 5 4 )

65. sech −1 ( 3 5 )

66. csch −1 (−1

Evaluate the integrals in Exercises 67–74 in terms of a. inverse hyperbolic functions. b. natural logarithms.

Evaluating Integrals

Evaluate the integrals in Exercises 41–60. 41. 43.

∫ sinh 2 x dx ∫

45.

∫

47.

∫

42.

(

)

x 6 cosh − ln 3 dx 2 x tanh dx 7 1 sech 2 x − dx 2

(

)

44.

∫ 4 cosh( 3x − ln 2 ) dx

46.

∫

θ coth dθ 3

48.

∫

csch 2 ( 5 − x ) dx

2 3

dx 4 + x2

67.

∫0

69.

∫5 4 1 − x 2

71.

∫1 5

73.

∫0

x

∫ sinh 5 dx

3)

2

dx

13

68.

∫0

70.

∫0

12

3 13

dx x 1 − 16 x 2

72.

∫1

2

π

cos x dx 1 + sin 2 x

74.

∫1

e

6 dx 1 + 9x 2 dx 1 − x2

dx x 4 + x2 dx x 1 + ( ln x ) 2

464

Chapter 7 Integrals and Transcendental Functions

Applications and Examples

75. Show that if a function f is defined on an interval symmetric about the origin (so that f is defined at x whenever it is defined at x), then f ( x ) + f (−x ) f ( x ) − f (−x ) f (x) = + . 2 2 Then show that ( f ( x ) + f (−x )) 2 is even and that ( f ( x )  f (x )) 2 is odd. 76. Derive the formula sinh −1 x = ln ( x + x 2 + 1 ) for all real x. Explain in your derivation why the plus sign is used with the square root instead of the minus sign. 77. Skydiving If a body of mass m falling from rest under the action of gravity encounters an air resistance proportional to the square of the velocity, then the body’s velocity t s into the fall satisfies the differential equation dV m = mg − k V 2, dt where k is a constant that depends on the body’s aerodynamic properties and the density of the air. (We assume that the fall is short enough so that the variation in the air’s density will not affect the outcome significantly.) a. Show that V =

⎛ gk ⎞⎟ mg tanh ⎜⎜⎜ t⎟ ⎝ m ⎟⎠ k

satisfies the differential equation and the initial condition that V  0 when t  0. b. Find the body’s limiting velocity, lim V.

83. Hanging cables Imagine a cable, like a telephone line or TV cable, strung from one support to another and hanging freely. The cable’s weight per unit length is a constant w, and the horizontal tension at its lowest point is a vector of length H. If we choose a coordinate system for the plane of the cable in which the x-axis is horizontal, the force of gravity is straight down, the positive y-axis points straight up, and the lowest point of the cable lies at the point y  H w on the y-axis (see accompanying figure), then it can be shown that the cable lies along the graph of the hyperbolic cosine H w cosh x. w H

y  y

y = H cosh w x w H

Hanging cable

H w

H 0

x

Such a curve is sometimes called a chain curve or a catenary, the latter deriving from the Latin catena, meaning “chain.” a. Let P ( x , y ) denote an arbitrary point on the cable. The next accompanying figure displays the tension at P as a vector of length (magnitude) T, as well as the tension H at the lowest point A. Show that the cable’s slope at P is

t →∞

c. For a 75-kg skydiver ( mg = 735 N ) , with time in seconds and distance in meters, a typical value for k is 0.235. What is the diver’s limiting velocity?

tan G  y

78. Accelerations whose magnitudes are proportional to displacement Suppose that the position of a body moving along a coordinate line at time t is a. s = a cos kt + b sin kt.

79. Volume A region in the first quadrant is bounded above by the curve y  cosh x , below by the curve y  sinh x , and on the left and right by the y-axis and the line x  2, respectively. Find the volume of the solid generated by revolving the region about the x-axis. 80. Volume The region enclosed by the curve y  sech x , the x-axis, and the lines x = ± ln 3 is revolved about the x-axis to generate a solid. Find the volume of the solid. 81. Arc length Find the length of the graph of y = (1 2 ) cosh 2 x from x  0 to x  ln 5. 82. Use the definitions of the hyperbolic functions to find each of the following limits. a. lim tanh x

b. lim tanh x

c. lim sinh x

d. lim sinh x

e. lim sech x

f . lim coth x

g. lim+ coth x

h. lim− coth x

x →∞ x →∞ x →∞

x→0

i. lim csch x x →−∞

y = H cosh w x w H

b. s = a cosh kt + b sinh kt.

Show in both cases that the acceleration d 2 s dt 2 is proportional to s but that in the first case it is directed toward the origin, whereas in the second case it is directed away from the origin.

x →−∞

dy w  sinh x. dx H

T A Q0, H wR

P(x, y)

f T cos f

H x

0

b. Using the result from part (a) and the fact that the horizontal tension at P must equal H (the cable is not moving), show that T  wy. Hence, the magnitude of the tension at P ( x , y ) is exactly equal to the weight of y units of cable. 84. (Continuation of Exercise 83.) The length of arc AP in the Exercise 83 figure is s = (1 a ) sinh ax , where a  w H . Show that the coordinates of P may be expressed in terms of s as x =

1 sinh −1 as, a

y =

s2 +

1 . a2

x →−∞ x →∞

x→0

85. Area Show that the area of the region in the first quadrant enclosed by the curve y = (1 a ) cosh ax , the coordinate axes, and the line x  b is the same as the area of a rectangle of height 1/a and length s, where s is the length of the curve from x  0 to x  b. Draw a figure illustrating this result.

465

7.4  Relative Rates of Growth

86. The hyperbolic in hyperbolic functions Just as x = cos u and y = sin u are identified with points ( x , y ) on the unit circle, the functions x = cosh u and y = sinh u are identified with points ( x , y ) on the right-hand branch of the unit hyperbola, x 2 − y 2 = 1.

P(cosh u, sinh u)

u=0

0

1 . 2

c. Solve this last equation for A(u). What is the value of A(0)? What is the value of the constant of integration C in your solution? With C determined, what does your solution say about the relationship of u to A(u)?

x

1

y

u:

x2 − y 2 = 1

−

−1

A′(u) =

sy m pt ot e

∞

Since cosh 2 u − sinh 2 u = 1, the point ( cosh u, sinh u ) lies on the right-hand branch of the hyperbola x 2 − y 2 = 1 for every value of u.

A

1

u:

∞

y

b. Differentiate both sides of the equation in part (a) with respect to u to show that

O

P(cosh u, sinh u) u is twice the area of sector AOP. x u=0

y x 2 + y2 = 1

P(cos u, sin u)

O u is twice the area of sector AOP.

A

x u=0

e

ot

pt

m

sy

A

Another analogy between hyperbolic and circular functions is that the variable u in the coordinates (cosh u, sinh u) for the points of the right-hand branch of the hyperbola x 2 − y 2 = 1 is twice the area of the sector AOP pictured in the figure following part (c). To see why this is so, carry out the following steps.

A

x 2 − y2 = 1

a. Show that the area A(u) of sector AOP is 1 cosh u sinh u − 2

A(u) =

∫1

cosh u

x 2 − 1 dx.

One of the analogies between hyperbolic and circular functions is revealed by these two diagrams (Exercise 86).

7.4 Relative Rates of Growth y y = ex 160

y=

It is often important in mathematics, computer science, and engineering to compare the rates at which functions of x grow as x becomes large. Exponential functions are important in these comparisons because of their very fast growth, and logarithmic functions because of their very slow growth. In this section we introduce the little-oh and big-oh notation used to describe the results of these comparisons. We restrict our attention to functions whose values eventually become and remain positive as x → ∞.

2x

140 120 100 80

y = x2

60 40 20 0

1

2

FIGURE 7.10

and

x 2.

3

4

5

6

7

x

The graphs of e x , 2 x ,

Growth Rates of Functions You may have noticed that exponential functions like 2 x and e x seem to grow more rapidly as x gets large than do polynomials and rational functions. These exponentials certainly grow more rapidly than x itself, and you can see 2 x outgrowing x 2 as x increases in Figure 7.10. In fact, as x → ∞, the functions 2 x and e x grow faster than any power of x, even x 1,000,000 (Exercise 19). In contrast, logarithmic functions like y = log 2 x and y = ln x grow more slowly as x → ∞ than any positive power of x (Exercise 21). To get a feeling for how rapidly the values of y = e x grow with increasing x, think of graphing the function on a large blackboard, with the axes scaled in centimeters. At x = 1 cm, the graph is e 1 ≈ 3 cm above the x-axis. At x = 6 cm, the graph is e 6 ≈ 403 cm ≈ 4 m high (it is about to go through the ceiling if it hasn’t done so already). At x = 10 cm, the graph is e 10 ≈ 22,026 cm ≈ 220 m high, higher than most buildings. At x = 24 cm, the graph is more than halfway to the moon, and at x = 43 cm

466

Chapter 7 Integrals and Transcendental Functions

y

from the origin, the graph is high enough to reach past the sun’s closest stellar neighbor, the red dwarf star Proxima Centauri. By contrast, with axes scaled in centimeters, you have to go nearly 5 light-years out on the x-axis to find a point where the graph of y = ln x is even y = 43 cm high. See Figure 7.11. These important comparisons of exponential, polynomial, and logarithmic functions can be made precise by defining what it means for a function f ( x ) to grow faster than another function g( x ) as x → ∞.

y = ex 70 60 50 40 30

Let f ( x ) and g( x ) be positive for x sufficiently large.

DEFINITION 20

1. f grows faster than g as x → ∞ if

10 0

y = ln x 10

20

FIGURE 7.11

30

40

50

60

Scale drawings of the graphs of e x and ln x.

x

lim

f (x) = ∞ g( x )

lim

g( x ) = 0. f (x)

x →∞

or, equivalently, if x →∞

We also say that g grows slower than f as x → ∞. 2. f and g grow at the same rate as x → ∞ if lim

x →∞

f (x) = L g( x )

where L is finite and positive.

According to these definitions, y = 2 x does not grow faster than y = x. The two functions grow at the same rate because lim

x →∞

2x = lim 2 = 2, x →∞ x

which is a finite, positive limit. The reason for this departure from more colloquial usage is that we want “ f grows faster than g” to mean that for large x-values g is negligible when compared with f . EXAMPLE 1

We compare the growth rates of several common functions.

(a) e x grows faster than x 2 as x → ∞ because ex ex ex = lim = ∞. lim 2 = lim →∞ →∞ x →∞ 2 2 x x x  x  ∞∞

  

Using l’Hôpital’s Rule twice

∞∞

(b) 3 x grows faster than 2 x as x → ∞ because

( )

3x 3 = lim x →∞ 2 x →∞ 2 x lim

x

= ∞.

(c) x 2 grows faster than ln x as x → ∞ because x2 2x = lim = lim 2 x 2 = ∞. x →∞ ln x x →∞ 1 x x →∞ lim

  

l’Hôpital’s Rule

7.4  Relative Rates of Growth

467

(d) ln x grows slower than x 1 n as x → ∞ for any positive integer n because lim

x →∞

ln x 1x = lim x →∞ (1 n ) x (1 n )−1 x1 n = lim

x →∞

l’Hôpital’s Rule

n = 0. x1 n

n is constant.

(e) As part (b) suggests, exponential functions with different bases never grow at the same rate as x → ∞. If a > b > 0, then a x grows faster than b x . Since ( a b ) > 1,

( )

ax a = lim x →∞ b x →∞ b x lim

x

= ∞.

(f ) In contrast to exponential functions, logarithmic functions with different bases a > 1 and b > 1 always grow at the same rate as x → ∞: lim

x →∞

ln x ln a ln b log a x = lim = . x →∞ log b x ln x ln b ln a

The limiting ratio is always finite and never zero. If f grows at the same rate as g as x → ∞, and g grows at the same rate as h as x → ∞, then f grows at the same rate as h as x → ∞. The reason is that lim

x →∞

f = L1 g

and

lim

x →∞

g = L2 h

together imply lim

x →∞

f f g = lim ⋅ = L1 L 2 . x →∞ g h h

If L1 and L 2 are finite and nonzero, then so is L1 L 2 . EXAMPLE 2

Show that

2

x 2 + 5 and ( 2 x − 1) grow at the same rate as x → ∞.

Solution We show that the functions grow at the same rate by showing that they both grow at the same rate as the function g( x ) = x :

lim

x →∞

lim

x →∞

x2 + 5 5 = lim 1 + 2 = 1, x →∞ x x

( 2 x − 1)

2

x

2

(

⎛ 2 x − 1 ⎞⎟ 1 = lim ⎜⎜ 2− ⎟ = xlim x →∞ ⎝ →∞ x ⎠ x

)

2

= 4.

Order and Oh-Notation The “little-oh” and “big-oh” notation was invented by number theorists over a hundred years ago and is now commonplace in mathematical analysis and computer science. According to this definition, saying f = o ( g) as x → ∞ is another way to say that f grows slower than g as x → ∞. DEFINITION A function f is of smaller order than g as x → ∞ if

lim

x →∞

f (x) = 0. We indicate this by writing f = o( g ) (“ f is little-oh of g”). g( x )

468

Chapter 7 Integrals and Transcendental Functions

EXAMPLE 3

Here we use little-oh notation.

(a) ln x = o( x ) as x → ∞ because

lim

x →∞

(b) x 2 = o( x 3 + 1) as x → ∞ because

ln x = 0 x lim

x →∞

x2 = 0 +1

x3

Let f ( x ) and g( x ) be positive for x sufficiently large. Then f is of at most the order of g as x → ∞ if there is a positive integer M for which

DEFINITION

f (x) ≤ M, g( x ) for x sufficiently large. We indicate this by writing f = O( g ) (“ f is big-oh of g”).

EXAMPLE 4

Here we use big-oh notation.

x + sin x ≤ 2 for x sufficiently large. x ex + x2 (b) e x + x 2 = O (e x ) as x → ∞ because → 1 as x → ∞. ex x (c) x = O (e x ) as x → ∞ because → 0 as x → ∞. ex (a) x + sin x = O( x ) as x → ∞ because

If you look at the definitions again, you will see that f = o ( g) implies f = O ( g) for functions that are positive for all sufficiently large x. Also, if f and g grow at the same rate, then f = O ( g) and g = O ( f ) (Exercise 11).

Sequential vs. Binary Search Computer scientists often measure the efficiency of an algorithm by counting the number of steps a computer must take to execute the algorithm. There can be significant differences in how efficiently algorithms perform, even if they are designed to accomplish the same task. These differences are often described using big-oh notation. Here is an example. One edition of Webster’s International Dictionary lists about 26,000 words that begin with the letter a. One way to look up a word, or to learn if it is not there, is to read through the list one word at a time until you either find the word or determine that it is not there. This method, called sequential search, makes no particular use of the words’ alphabetical arrangement in the list. You are sure to get an answer, but it might take 26,000 steps. Another way to find the word or to learn it is not there is to go straight to the middle of the list (give or take a few words). If you do not find the word, then go to the middle of the half that contains it and forget about the half that does not. (You know which half contains it because you know the list is ordered alphabetically.) This method, called a binary search, eliminates roughly 13,000 words in a single step. If you do not find the word on the second try, then jump to the middle of the half that contains it. Continue this way until you have either found the word or divided the list in half so many times there are no words left. How many times do you have to divide the list to find the word or learn that it is not there? At most 15, because

( 26,000 2 15 ) < 1. That certainly beats a possible 26,000 steps. For a list of length n, a sequential search algorithm takes on the order of n steps to find a word or determine that it is not in the list. A binary search, as the second algorithm is called, takes on the order of log 2 n steps. The reason is that if 2 m −1 < n ≤ 2 m , then

7.4  Relative Rates of Growth

469

m − 1 < log 2 n ≤ m, and the number of bisections required to narrow the list to one word will be at most m = ⎡ log 2 n ⎤ , the integer ceiling of the number log 2 n. Big-oh notation provides a compact way to say all this. The number of steps in a sequential search of an ordered list is O(n); the number of steps in a binary search is O ( log 2 n ). In our example, there is a big difference between the two (26,000 versus 15), and the difference can only increase with n because n grows faster than log 2 n as n → ∞.

EXERCISES

7.4 Ordering Functions by Growth Rates

Comparisons with the Exponential e x

1. Which of the following functions grow faster as x → ∞ ? Which grow at the same rate as e x ? Which grow slower? than e x

b. x 3 + sin 2 x

a. x − 3 c.

e. ( 3 2 )

x

f. e

g. e x 2

x 2

2. Which of the following functions grow faster than e x as x → ∞ ? Which grow at the same rate as e x ? Which grow slower? c.

1+

x4

b. x ln x − x d. ( 5 2 ) x

e. e −x

f. xe x

g. e cos x

h. e x −1

3. Which of the following functions grow faster than x 2 as x → ∞ ? Which grow at the same rate as x 2 ? Which grow slower? c.

x4 + x3

b. x 5 − x 2 d. ( x + 3 )

e. x ln x

f. 2 x

g. x 3e −x

h. 8 x 2

2

4. Which of the following functions grow faster than x 2 as x → ∞ ? Which grow at the same rate as x 2 ? Which grow slower? a. x 2 +

x

b. 10 x 2

c. x 2e −x

d. log 10 ( x 2 )

e. x 3 − x 2

f. (1 10 ) x

g. (1.1) x

h. x 2 + 100 x

Comparisons with the Logarithm ln x

5. Which of the following functions grow faster than ln x as x → ∞ ? Which grow at the same rate as ln x? Which grow slower? a. log 3 x

b. ln 2x

c. ln x

d.

e. x

f. 5 ln x

6. Which of the following functions grow faster than ln x as x → ∞ ? Which grow at the same rate as ln x? Which grow slower? a. log 2 ( x 2 ) c. 1

x

b. log 10 10 x d. 1

a. 2 x c. ( ln 2 )

b. x 2 x

d. e x

Big-oh and Little-oh; Order

a. x = o ( x )

b. x = o( x + 5 )

c. x = O ( x + 5 )

d. x = O (2 x )

e. e x = o (e 2 x )

f. x + ln x = O ( x )

g. ln x = o (ln 2 x )

h.

10. True, or false? As x → ∞, 1 1 = O a. x+3 x

( )

c.

( )

1 1 1 − 2 = o x x x

b.

x 2 + 5 = O (x)

( )

1 1 1 + 2 = O x x x

d. 2 + cos x = O (2)

e. e x + x = O (e x )

f. x ln x = o( x 2 )

g. ln(ln x ) = O(ln x )

h. ln( x ) = o( ln ( x 2 + 1))

11. Show that if positive functions f ( x ) and g( x ) grow at the same rate as x → ∞, then f = O ( g) and g = O ( f ). 12. When is a polynomial f ( x ) of smaller order than a polynomial g( x ) as x → ∞ ? Give reasons for your answer. 13. When is a polynomial f ( x ) of at most the order of a polynomial g( x ) as x → ∞ ? Give reasons for your answer. 14. What do the conclusions we drew in Section 2.8 about the limits of rational functions tell us about the relative growth of polynomials as x → ∞ ?

x

h. e x

g. 1 x

d. e x 2

9. True, or false? As x → ∞,

Comparisons with the Power x 2

a. x 2 + 4 x

b. x x x

8. Order the following functions from slowest growing to fastest growing as x → ∞.

h. log 10 x

a. 10 x 4 + 30 x + 1

a. e x c. ( ln x )

d. 4 x

x

7. Order the following functions from slowest growing to fastest growing as x → ∞.

x2

e. x − 2 ln x

f. e −x

g. ln(ln x )

h. ln ( 2 x + 5 )

Other Comparisons

T 15. Investigate lim

x →∞

ln ( x + 1) and ln x

lim

x →∞

ln ( x + 999 ) . ln x

Then use l’Hôpital’s Rule to explain what you find.

470

Chapter 7 Integrals and Transcendental Functions

16. (Continuation of Exercise 15.) Show that the value of lim

x →∞

+ ln x

ln ( x

a)

is the same no matter what value you assign to the constant a. What does this say about the relative rates at which the functions f ( x ) = ln ( x + a ) and g( x ) = ln x grow? 17. Show that 10 x + 1 and  x + 1 grow at the same rate as x → ∞ by showing that they both grow at the same rate as x as x → ∞. 18. Show that x 4 + x  and  x 4 − x 3 grow at the same rate as x → ∞ by showing that they both grow at the same rate as x 2 as x → ∞.

T d. (Continuation of part (c).) The value of x at which ln x = 10 ln(ln x ) is too far out for some graphers and root finders to identify. Try it on the equipment available to you and see what happens. 22. The function ln x grows slower than any polynomial Show that ln x grows slower as x → ∞ than any nonconstant polynomial. Algorithms and Searches

23. a. Suppose you have three different algorithms for solving the same problem and each algorithm takes a number of steps that is of the order of one of the functions listed here: n log 2 n, n 3 2, n ( log 2 n ) 2.

19. Show that e x grows faster as x → ∞ than x n for any positive integer n, even x 1,000,000. (Hint: What is the nth derivative of x n ?)

Which of the algorithms is the most efficient in the long run? Give reasons for your answer.

Show that e x grows

T b. Graph the functions in part (a) together to get a sense of how rapidly each one grows.

20. The function e x outgrows any polynomial faster as x → ∞ than any polynomial

a n x n + a n−1 x n−1 +  + a1 x + a 0 . 21. a. Show that ln x grows slower as x → ∞ than x 1 n for any positive integer n, even x 1 1,000,000. T b. Although the values of x 1 1,000,000 eventually overtake the values of ln x , you have to go way out on the x-axis before this happens. Find a value of x greater than 1 for which x 1 1,000,000 > ln x. You might start by observing that when x > 1 the equation ln x = x 1 1,000,000 is equivalent to the equation ln(ln x ) = (ln x ) 1,000,000.

24. Repeat Exercise 23 for the functions n,

n log 2 n, ( log 2 n ) 2.

T 25. Suppose you are looking for an item in an ordered list one million items long. How many steps might it take to find that item with a sequential search? A binary search? T 26. You are looking for an item in an ordered list 450,000 items long (the length of Webster’s Third New International Dictionary). How many steps might it take to find the item with a sequential search? A binary search?

T c. Even x 1 10 takes a long time to overtake ln x. Experiment with a calculator to find the value of x > 10 at which the graphs of x 1 10 and ln x cross, or, equivalently, at which ln x = 10 ln(ln x ). Bracket the crossing point between powers of 10 and then close in by successive halving.

CHAPTER 7

Questions to Guide Your Review

1. How is the natural logarithm function defined as an integral? What are its domain, range, and derivative? What arithmetic properties does it have? Comment on its graph. 2. What integrals lead to logarithms? Give examples. 3. What are the integrals of tan x and cot x ? Of sec x and csc x ? 4. How is the exponential function e x defined? What are its domain, range, and derivative? What laws of exponents does it obey? Comment on its graph.

9. What are the derivatives of the six basic hyperbolic functions? What are the corresponding integral formulas? What similarities do you see here to the six basic trigonometric functions? 10. How are the inverse hyperbolic functions defined? Comment on their domains, ranges, and graphs. How can you find values of sech −1 x , csch −1 x , and coth −1 x using a calculator’s keys for cosh −1 x , sinh −1 x , and tanh −1 x ? 11. What integrals lead naturally to inverse hyperbolic functions?

5. How are the functions and log a x defined? Are there any restrictions on a? How is the graph of log a x related to the graph of ln x ? What truth is there in the statement that there are really only one exponential function and one logarithmic function?

12. How do you compare the growth rates of positive functions as x → ∞?

6. How do you solve separable first-order differential equations?

15. Which is more efficient—a sequential search or a binary search? Explain.

ax

7. What is the law of exponential change? How can it be derived from an initial value problem? What are some of the applications of the law? 8. What are the six basic hyperbolic functions? Comment on their domains, ranges, and graphs. What are some of the identities relating them?

13. What roles do the functions e x and ln x play in growth comparisons? 14. Describe big-oh and little-oh notation. Give examples.

Chapter 7  Practice Exercises

CHAPTER 7

Practice Exercises

Integration Evaluate the integrals in Exercises 1–12.

1.

∫ e x sin (e x ) dx

3.

∫0

5.

∫−π 2 1 − sin t dt

π

x tan dx 3

π6

7.

∫

9.

∫1

11.

∫e

cos t

ln ( x − 5 ) dx x−5

Theory and Applications

2.

∫ e t cos( 3e t

4.

∫1 6

6.

∫

e x sec e x dx

8.

∫

cos(1 − ln υ ) dυ υ

14

7

3 dx x

10.

∫1

e2

1 dx x ln x

12.

∫2

32

4

− 2 ) dt

23. The function f ( x ) = e x + x , being differentiable and one-toone, has a differentiable inverse f −1 ( x ). Find the value of df −1 dx at the point f (ln 2). 24. Find the inverse of the function f ( x ) = 1 + (1 x ) , x ≠ 0. Then show that f −1 ( f ( x )) = f ( f −1 ( x )) = x and that

2 cot π x dx

1 dx 5x

(1 + ln t ) t ln t dt

df −1 dx

13. 3 y = 2 y +1

14. 4 − y = 3 y + 2

15. 9e 2 y = x 2, x > 0

16. 3 y = 3 ln x

17. ln ( y − 1) = x + ln y

18. ln (10 ln y ) = ln 5 x

Comparing Growth Rates of Functions

c. f ( x ) = x 100 ,

g( x ) = log 3 x 1 g( x ) = x + x g( x ) = xe −x

d. f ( x ) = x ,

g( x ) = arctan x

b. f ( x ) = x ,

e. f ( x ) = arccsc x ,

g( x ) = 1 x

f. f ( x ) = sinh x ,

g( x ) = e x

dy dt = (−1 4 ) 3 − y m s. At approximately what rate is her x-coordinate changing when she reaches the bottom of the slide at x = 3 m ? 27. The functions f ( x ) = ln 5 x and g( x ) = ln 3 x differ by a constant. What constant? Give reasons for your answer. 28. a. If ( ln x ) x = ( ln 2 ) 2, must x = 2? b. If ( ln x ) x = −2 ln 2, must x = 1 2?

g( x ) =

b. f ( x ) = ln 2 x ,

Give reasons for your answers. 29. The quotient ( log 4 x ) ( log 2 x ) has a constant value. What value? Give reasons for your answer. T 30. log x (2)   vs. log 2 ( x ) How does f ( x ) = log x (2) compare with g( x ) = log 2 ( x )? Here is one way to find out. a. Use the equation log a b = ( ln b ) ( ln a ) to express f ( x ) and g( x ) in terms of natural logarithms.

20. Does f grow faster, slower, or at the same rate as g as x → ∞ ? Give reasons for your answers. 3−x ,

1 . f ′( x )

26. A girl is sliding down a slide shaped like the curve y = 3e −x 3. Her y-coordinate is changing at the rate

19. Does f grow faster, slower, or at the same rate as g as x → ∞ ? Give reasons for your answers. a. f ( x ) = log 2 x ,

= f (x)

25. A particle is traveling upward and to the right along the curve y = ln x. Its x-coordinate is increasing at the rate ( dx dt ) = x m s. At what rate is the y-coordinate changing at the point ( e 2 , 2 )?

Solving Equations with Logarithmic or Exponential Terms In Exercises 13–18, solve for y.

a. f ( x ) =

471

2 −x

b. Graph f and g together. Comment on the behavior of f in relation to the signs and values of g. In Exercises 31–34, solve the differential equation. 31.

g( x ) = ln x 2

dy = dx

y cos 2

y

32. y′ =

3 y ( x + 1) 2 y −1

g( x ) = e x

33. yy′ = sec y 2 sec 2 x

d. f ( x ) = arctan (1 x ) ,

g( x ) = 1 x

e. f ( x ) = arcsin (1 x ),

g( x ) = 1 x 2

f. f ( x ) = sech x ,

g( x ) = e − x

In Exercises 35–38, solve the initial value problem. y ln y dy dy 35. , y (0) = e 2 = e −x − y−2, y (0) = −2 36. = dx dx 1 + x2

c. f ( x ) =

10 x 3

+

2 x 2,

21. True, or false? Give reasons for your answers. 1 1 1 1 1 1 a. 2 + 4 = O 2 b. 2 + 4 = O 4 x x x x x x

( )

( )

c. x = o( x + ln x )

d. ln (ln x ) = o (ln x )

e. arctan x = O (1)

f. cosh x = O (e x )

22. True, or false? Give reasons for your answers. 1 1 1 1 1 1 b. 4 = o 2 + 4 a. 4 = O 2 + 4 x x x x x x

(

c. ln x = o( x + 1) e.

sec −1

x = O (1)

)

(

d. ln 2 x = O (ln x ) f. sinh x = O (e x )

37. x dy − ( y + 38. y −2

34. y cos 2 x dy + sin x dx = 0

y ) dx = 0, y (1) = 1 ex

dx = 2x , y (0) = 1 dy e +1

39. What is the age of a sample of charcoal in which 90% of the carbon-14 originally present has decayed?

)

40. Cooling a pie A deep-dish apple pie, whose internal temperature was 104 °C when removed from the oven, was set out on a breezy 5 °C porch to cool. Fifteen minutes later, the pie’s internal temperature was 82 °C. How long did it take the pie to cool from there to 21 °C?

472

Chapter 7 Integrals and Transcendental Functions

41. Find the length of the curve y = ln ( e x − 1) − ln ( e x + 1) , ln 2 b x b ln 3.

Assume that L (t ) is the length in meters of a shark of age t years. In addition, assume A  3, L (0)  0.5 m, and L (5)  1.75 m.

42. In 1934, the Austrian biologist Ludwig von Bertalanffy derived and published the von Bertalanffy growth equation, which continues to be widely used and is especially important in fisheries studies. Let L (t ) denote the length of a fish at time t and assume L (0)  L 0 . The von Bertalanffy equation is

a. Solve the von Bertalanffy differential equation. b. What is L(10)? c. When does L  2.5 m ?

dL = k ( A − L ), dt where A = lim L (t ) is the asymptotic length of the fish and k is t →∞

a proportionality constant.

CHAPTER 7

Additional and Advanced Exercises

1. Let A(t ) be the area of the region in the first quadrant enclosed by the coordinate axes, the curve y = e −x , and the vertical line x = t , t > 0. Let V (t ) be the volume of the solid generated by revolving the region about the x-axis. Find the following limits. a. lim A(t ) t →∞

b. lim V (t ) A(t ) t →∞

c. lim+ V (t ) A(t ) t→0

2. Varying a logarithm’s base a. Find lim log a 2 as a → 0 +, 1− , 1+, and d. T b. Graph y  log a 2 as a function of a over the interval 0 < a ≤ 4. T 3. Graph f ( x ) = tan −1 x + tan −1 (1 x ) for −5 ≤ x ≤ 5. Then use calculus to explain what you see. How would you expect f to behave beyond the interval [ −5, 5 ]? Give reasons for your answer. sin x over [ 0, 3Q ]. Explain what you see. T 4. Graph f ( x ) = ( sin x )

5. Even-odd decompositions a. Suppose that g is an even function of x and h is an odd function of x. Show that if g( x ) + h( x ) = 0 for all x, then g( x )  0 for all x and h( x )  0 for all x. b. Use the result in part (a) to show that if f ( x ) = f E ( x ) + f O ( x ) is the sum of an even function f E ( x ) and an odd function f O ( x ), then f E ( x ) = ( f ( x ) + f (−x ) ) 2

and

f O ( x ) = ( f ( x ) − f (−x ) ) 2.

c. What is the significance of the result in part (b)?

7. Center of mass Find the center of mass of a thin plate of constant density covering the region in the first and fourth quadrants enclosed by the curves y = 1 (1 + x 2 ) and y = −1 (1 + x 2 ) and by the lines x  0 and x  1. 8. Solid of revolution The region between the curve y = 1 ( 2 x ) and the x-axis from x  1 4 to x  4 is revolved about the x-axis to generate a solid. a. Find the volume of the solid. b. Find the centroid of the region. 9. The Rule of 70 If you use the approximation ln 2 x 0.70 (in place of 0.69314 ... ), you can derive a rule of thumb that says, “To estimate how many years it will take an amount of money to double when invested at r percent compounded continuously, divide r into 70.” For instance, an amount of money invested at 5% will double in about 70 5  14 years. If you want it to double in 10 years instead, you have to invest it at 70 10  7%. Show how the Rule of 70 is derived. (A similar “Rule of 72” uses 72 instead of 70, because 72 has more integer factors.) T 10. Urban gardening A vegetable garden 15 m wide is to be grown between two buildings, which are 150 m apart along an east–west line. If the buildings are 60 m and 105 m tall, where should the garden be placed in order to receive the maximum number of hours of sunlight exposure? (Hint: Determine the value of x in the accompanying figure that maximizes sunlight exposure for the garden.)

6. Let g be a function that is differentiable throughout an open interval containing the origin. Suppose g has the following properties. g ( x ) + g ( y) for all real numbers x, y, and 1 − g ( x ) g ( y) x y in the domain of g. ii. lim g(h) = 0 i. g ( x + y ) =

h→ 0

g(h) =1 h a. Show that g(0)  0. iii. lim

h→ 0

b. Show that g′( x ) = 1 + [ g( x ) ]2 . c. Find g( x ) by solving the differential equation in part (b).

105 m tall 60 m tall

u2

West x

15

u1 135 − x

East

8

Techniques of Integration OVERVIEW The Fundamental Theorem tells us how to evaluate a definite integral once we have an antiderivative for the integrand function. However, finding antiderivatives (or indefinite integrals) is not as straightforward as finding derivatives. In this chapter we study a number of important techniques that apply to finding integrals for specialized classes of functions such as trigonometric functions, products of certain functions, and rational functions. Since we cannot always find an antiderivative, we develop numerical methods for calculating definite integrals. We also study integrals for which the domain or range is infinite, called improper integrals.

8.1 Using Basic Integration Formulas Table 8.1 summarizes the indefinite integrals of many of the functions we have studied so far, and the substitution method helps us use the table to evaluate more complicated functions involving these basic ones. In this section we combine the Substitution Rules (studied in Chapter 5) with algebraic methods and trigonometric identities to help us use Table 8.1. A more extensive Table of Integrals is given at the back of the chapter, and we discuss its use in Section 8.6. Sometimes we have to rewrite an integral to match it to a standard form of the type displayed in Table 8.1. We start with an example of this procedure. EXAMPLE 1

Evaluate the integral 5

∫3

2x − 3 dx. x 2 − 3x + 1

Solution We rewrite the integral and apply the Substitution Rule for Definite Integrals presented in Section 5.6, to find 5

∫3

2x − 3 dx = − 3x + 1

∫1

=

∫1

x2

11

11

du u

u = x 2 − 3 x + 1,  du = ( 2 x − 3 ) dx; u = 1 when  x = 3,  u = 11 when x = 5

u −1 2 du 11

⎤ = 2 u ⎥ = 2( 11 − 1) ≈ 4.63. ⎥⎦ 1

Table 8.1, Formula 2

473

474

Chapter 8 Techniques of Integration

TABLE 8.1 Basic integration formulas

1. ∫ k dx = kx + C (any number k )

12. ∫ tan x dx = ln sec x + C

n +1 2. ∫ x n dx = x + C ( n ≠ −1)

13. ∫ cot x dx = ln sin x + C

3. ∫ dx = ln x + C

14. ∫ sec x dx = ln sec x + tan x + C

4. ∫ e x dx = e x + C

15. ∫ csc x dx = − ln csc x + cot x + C

n+1

x

x 5. ∫ a x dx = a + C

ln a

(a

16. ∫ sinh x dx = cosh x + C

> 0,   a ≠ 1)

6. ∫ sin x dx = − cos x + C

17. ∫ cosh x dx = sinh x + C

7. ∫ cos x dx = sin x + C

18. ∫

8. ∫ sec 2 x dx = tan x + C

19. ∫

dx 1 x = arctan +C a2 + x2 a a

9. ∫ csc 2 x dx = − cot x + C

20. ∫

dx 1 x = arcsec +C a a x x2 − a2

10. ∫ sec x tan x dx = sec x + C

21. ∫

a2

11. ∫ csc x cot x dx = − csc x + C

22. ∫

dx x = cosh −1 +C a x2 − a2

EXAMPLE 2

( )

dx x = arcsin +C 2 a −x

a2

( )

( )

(a

> 0)

( )

(x

> a > 0)

dx x = sinh −1 +C 2 a + x

Complete the square to evaluate

∫

dx . 8x − x 2

Solution We complete the square to simplify the denominator:

8 x − x 2 = −( x 2 − 8 x ) = −( x 2 − 8 x + 16 − 16 ) = −( x 2 − 8 x + 16 ) + 16 = 16 − ( x − 4 ) 2 . Then

∫

dx = 8x − x 2

∫

dx 16 − ( x − 4 ) 2

=

∫

du a2 − u2

= arcsin

( ua ) + C

= arcsin

( x −4 4 ) + C.

a = 4, u = ( x − 4 ), du = dx Table 8.1, Formula 18

8.1  Using Basic Integration Formulas

EXAMPLE 3

475

Evaluate the integral

∫ ( cos x sin 2 x + sin x cos 2 x ) dx. Solution We can replace the integrand with an equivalent trigonometric expression using the Sine Addition Formula to obtain a simple substitution:

∫ ( cos x sin 2 x + sin x cos 2 x ) dx = ∫ sin( x + 2 x ) dx =

∫ sin 3x dx

=

∫

1 sin u du 3

u = 3 x ,  du = 3 dx

1 = − cos 3 x + C. 3

Table 8.1, Formula 6

In Section 5.5 we found the indefinite integral of the secant function by multiplying it by a fractional form equal to one, and then integrating the equivalent result. We can use that same procedure in other instances as well, as we illustrate next.

EXAMPLE 4

Find ∫

π 4 0

dx . 1 − sin x

Solution We multiply the numerator and denominator of the integrand by 1 + sin x . This procedure transforms the integral into one we can evaluate: π 4

∫0

π 4

1 + sin x 1 dx ⋅ 1 − sin x 1 + sin x

Multiply and divide by conjugate.

π 4

1 + sin x dx 1 − sin 2 x

Simplify.

1 + sin x dx cos 2 x

1 − sin 2 x = cos 2 x

( sec 2 x + sec x tan x ) dx

Use Table 8.1, Formulas 8 and 10

dx = 1 − sin x

∫0

=

∫0

=

∫0

=

∫0

π 4

π 4

⎡ ⎤π 4 = ⎢ tan x + sec x ⎥ = (1 + ⎢⎣ ⎦⎥ 0 EXAMPLE 5

)

x−3 3x 2 + 2 x −9 x   − 9 x − 6 +6

2.

Evaluate

∫ 3x + 2 3x 2 − 7 x

2 − ( 0 + 1) ) =

3x 2 − 7 x dx. 3x + 2

Solution The integrand is an improper fraction since the degree of the numerator is greater than the degree of the denominator. To integrate it, we perform long division to obtain a quotient plus a remainder that is a proper fraction:

6 3x 2 − 7 x = x−3+ . 3x + 2 3x + 2

476

Chapter 8 Techniques of Integration

Therefore,

∫

3x 2 − 7 x dx = 3x + 2

∫ ( x − 3 + 3x + 2 ) dx 6

x2 − 3 x + 2 ln 3 x + 2 + C. 2

=

Reducing an improper fraction by long division (Example 5) does not always lead to an expression we can integrate directly. We see what to do about that in Section 8.5. EXAMPLE 6

Evaluate

∫

3x + 2 dx. 1 − x2

Solution We first separate the integrand to get

∫

3x + 2 dx = 3∫ 1 − x2

x dx 1−

x2

+ 2∫

dx . 1 − x2

In the first of these new integrals, we substitute u = 1 − x 2,

du = −2 x dx ,

1 x dx = − du. 2

so

Then we obtain 3∫

x dx 1 − x2

= 3∫

(−1 2 ) du 3 = − 2 u

∫ u −1 2 du

3 u1 2 =− ⋅ + C 1 = − 3 1 − x 2 + C 1. 2 12 The second of the new integrals is a standard form, 2∫

dx = 2 arcsin x + C 2 . 1 − x2

Table 8.1, Formula 18

Combining these results and renaming C 1 + C 2 as C gives

∫

3x + 2 dx = −3 1 − x 2 + 2 arcsin x + C. 1 − x2

The question of what to substitute for in an integrand is not always quite so clear. Sometimes we simply proceed by trial-and-error, and if nothing works out, we then try another method altogether. The next several sections of the text present some of these new methods, but substitution works in the following example. EXAMPLE 7

Evaluate dx

∫ (1 +

x)

3.

Solution We might try substituting for the term x, but the derivative factor 1 x is missing from the integrand, so this substitution will not help. The other possibility is to substitute for (1 + x ), and it turns out this works:

∫

dx 3 = (1 + x ) =

∫

2( u − 1) du   u3

∫ (u2 2

−

)

2 du u3

1 dx; 2 x dx = 2 x du = 2( u − 1) du

u = 1+

x , du =

477

8.1  Using Basic Integration Formulas

2 1 =− + 2 +C u u 1 − 2u +C u2

=

1 − 2 (1 +

=

(1 + x )

=C−

x) 2

+C

1+ 2 x 2. (1 + x )

When evaluating definite integrals, a property of the integrand may help us in calculating the result. EXAMPLE 8

Evaluate π 2

∫−π 2 x 3 cos x dx. Solution No substitution or algebraic manipulation is clearly helpful here. But we observe that the interval of integration is the symmetric interval [ −π 2,   π 2 ]. Moreover, the factor x 3 is an odd function, and cos x is an even function, so their product is odd. Therefore, π 2

∫−π 2 x 3 cos x dx EXERCISES

The integrals in Exercises 1–44 are in no particular order. Evaluate each integral using any algebraic method, trigonometric identity, or substitution you think is appropriate. 1. 3. 5. 7. 9.

∫0

16 x dx 8x 2 + 2

∫ ( sec x − tan x )

2. 2

1− x dx 1 − x2

∫ ∫

4. 6.

x2 + 1 π3

∫π 4

∫

19.

∫ sec θ + tan θ

21.

∫

23.

∫0

25.

∫

27.

∫

29.

∫ ( csc x − sec x )( sin x + cos x ) dx

dθ

dx

dx cos 2 x tan x

∫

dx x− x

∫

2 ln z dz 16 z

4 t 3 − t 2 + 16t dt t2 + 4 π2

1 − cos θ d θ

dy e 2y − 1

3

8.

dz + e −z

10.

∫1

12.

∫−1

4x 2 − 7 dx 2x + 3

30.

∫ 3 sinh( 2 + ln 5 ) dx

14.

∫ csc t sin 3t dt

31.

∫

dθ 2θ − θ 2

33.

∫−1

∫ ez

4 dx

0

∫−1 1 + ( 2 x + 1)2

13.

∫ 1 − sec t

dt

π4

∫0

dx

∫

x2

ln y dy y + 4 y  ln 2 y

17.

e − cot z dz sin 2 z

11.

15.

Theorem 8, Section 5.6

8.1

Assorted Integrations

1

= 0.

1 + sin θ dθ cos 2 θ

16.

2

3

∫

8 dx x 2 − 2x + 2

2 dx x 1 − 4 ln 2 x

2 y dy 2 y

18.

∫

20.

∫t

22.

∫

24.

∫ ( sec t + cot t )

26.

∫

28.

∫ ( x − 2)

dt 3 + t2

x + 2 x −1 dx 2x x − 1 2

dt

6 dy y (1 + y ) dx x 2 − 4x + 3

x

3 2 0

1

2x 3 dx −1

32.

∫−1

1+ y dy 1− y

34.

∫ e z +e

x2

1 + x 2 sin x dx z

dz

478

35. 37.

Chapter 8 Techniques of Integration

∫ ( x − 1) ∫

7 dx x 2 − 2 x − 48

2θ 3 − 7θ 2 + 7θ dθ 2θ − 5

dx 39. ∫ 1 + ex Hint: Use long division. 41.

∫

43.

∫

e 3x dx ex + 1 1 dx x (1 + x )

dx 4x + 4x 2

36.

∫ ( 2 x + 1)

38.

dθ cos θ − 1

∫

51. The functions y = e x 3 and y = x 3e x 3 do not have elementary antiderivatives, but y = (1 + 3 x 3 )e x 3 does. Evaluate

∫ (1 + 3 x 3 ) e x

∫

2x − 1 dx 3x

44.

∫

tan θ + 3 dθ sin θ

Theory and Examples

45. Area Find the area of the region bounded above by y = 2 cos x and below by y = sec x ,   −π 4 ≤ x ≤ π 4.

dx.

52. Use the substitution u = tan x to evaluate the integral

x dx 40. ∫ 1 + x3 Hint: Let u = x 3 2 . 42.

3

dx

∫ 1 + sin 2 x . 53. Use the substitution u = x 4 + 1 to evaluate the integral

∫ x7

x 4 + 1 dx.

54. Using different substitutions Show that the integral −2 3

∫ (( x 2 − 1)( x + 1))

dx

can be evaluated with any of the following substitutions. a. u = 1 ( x + 1)

46. Volume Find the volume of the solid generated by revolving the region in Exercise 45 about the x-axis.

b. u = (( x − 1) ( x + 1)) k for k = 1,  1 2,  1 3,   −1 3,   −2 3, and −1

47. Arc length Find the length of the curve y = ln ( cos x ), 0 ≤ x ≤ π 3.

c. u = arctan x d. u = tan −1 x

48. Arc length Find the length of the curve y = ln ( sec x ), 0 ≤ x ≤ π 4.

e. u = tan −1 (( x − 1) 2 )

49. Centroid Find the centroid of the region bounded by the x-axis, the curve y = sec x, and the lines x = −π 4 ,   x = π 4.

g. u = cosh −1 x

50. Centroid Find the centroid of the region bounded by the x-axis, the curve y = csc x , and the lines x = π 6,   x = 5π 6.

f. u = arccos x What is the value of the integral?

8.2 Integration by Parts Integration by parts is a technique for simplifying integrals of the form

∫ u ( x ) υ′( x ) dx. It is useful when u can be differentiated repeatedly and υ′ can be integrated repeatedly without difficulty. The integrals

∫ x cos x dx

and

∫ x 2 e x dx

are such integrals because u ( x ) = x or u ( x ) = x 2 can be differentiated repeatedly, and υ′( x ) = cos x or υ′( x ) = e x can be integrated repeatedly without difficulty. Integration by parts also applies to integrals like

∫ ln x dx

and

∫ e x cos x dx.

In the first case, the integrand ln x can be rewritten as ( ln x )(1), and u ( x ) = ln x is easy to differentiate while υ′( x ) = 1 easily integrates to x. In the second case, each part of the integrand appears again after repeated differentiation or integration.

Product Rule in Integral Form If u and υ are differentiable functions of x, the Product Rule says that d [ u ( x ) υ ( x ) ] = u′( x ) υ ( x ) + u ( x ) υ′ ( x ). dx

8.2  Integration by Parts

479

In terms of indefinite integrals, this equation becomes d

∫ dx [ u ( x ) υ ( x ) ] dx

=

∫ [ u′( x ) υ ( x ) + u ( x ) υ′( x ) ] dx

or d

∫ dx [ u ( x ) υ ( x ) ] dx

=

∫ u′( x ) υ ( x ) dx + ∫ u ( x ) υ′( x ) dx.

Rearranging the terms of this last equation, we get

∫ u ( x ) υ′( x ) dx

=

∫

d [ u ( x ) υ ( x ) ] dx − dx

∫ υ ( x ) u′( x ) dx ,

leading to the following integration by parts formula.

Integration by Parts Formula

∫ u ( x ) υ′ ( x ) dx

= u (x) υ (x) −

∫ υ ( x ) u′( x ) dx  

(1)

This formula allows us to exchange the problem of computing the integral ∫ u ( x ) υ′( x ) dx for the problem of computing a different integral, ∫ υ ( x ) u′( x ) dx . In many cases, we can choose the functions u and υ so that the second integral is easier to compute than the first. There can be many choices for u and υ, and it is not always clear which choice works best, so sometimes we need to try several. The formula is often given in differential form. With υ′( x ) dx = d υ and u′( x ) dx = du , the integration by parts formula becomes

Integration by Parts Formula—Differential Version

∫ u dυ

= uυ −

∫ υ du

(2)

The next examples illustrate the technique. EXAMPLE 1

Find

∫ x cos x dx. Solution There is no obvious antiderivative of x cos x , so we use the integration by parts formula

∫ u ( x ) υ′( x ) dx

= u (x) υ (x) −

∫ υ ( x ) u′( x ) dx

to change this expression to one that is easier to integrate. We first decide how to choose the functions u ( x ) and υ ( x ). There is more than one way to do this, but here we choose to factor the expression x cos x into u ( x ) = x and υ′( x ) = cos x. Next we differentiate u ( x ) and find an antiderivative of υ′( x ), u′( x ) = 1 and υ ( x ) = sin x.

480

Chapter 8 Techniques of Integration

When finding an antiderivative for υ′( x ), we have a choice of how to pick a constant of integration C. We choose the constant C = 0, since that makes this antiderivative as simple as possible. We now apply the integration by parts formula:

∫

x cos x dx = x sin x − |

|

|

u ( x ) υ′( x )

u( x ) υ( x )

∫ sin| x (1| ) dx

Integration by parts formula

υ ( x ) u′ ( x )

= x sin x + cos x + C

Integrate and simplify.

and we have found the integral of the original function. There are at least four apparent choices available for u ( x ) and υ′( x ) in Example 1: 1. Let u ( x ) = 1 and υ′( x ) = x cos x . 3. Let u ( x ) = x cos x and υ′( x ) = 1.

2. Let u ( x ) = x and υ′( x ) = cos x. 4. Let u ( x ) = cos x and υ′( x ) = x .

We used choice 2 in Example 1. The other three choices lead to integrals that we do not know how to evaluate. For instance, Choice 3, with u′( x ) = cos x − x sin x , leads to the integral

∫ ( x cos x − x 2 sin x ) dx. The goal of integration by parts is to go from an integral ∫ u ( x )υ′( x ) dx that we don’t see how to evaluate to an integral ∫ υ ( x )u′( x ) dx that we can evaluate. Generally, we choose υ′( x ) first to be as much of the integrand as we can readily integrate; then we let u ( x ) be the leftover part. When finding υ ( x ) from υ′( x ), any antiderivative will work, and we usually pick the simplest one. In particular, no arbitrary constant of integration is needed in υ ( x ) because it would simply cancel out of the right-hand side of Equation (2). Find ∫ ln x dx.

EXAMPLE 2

Solution We have not yet seen how to find an antiderivative for ln x. If we set u ( x ) = ln x , then u′( x ) is the simpler function 1 x . It may not appear that a second function υ′( x ) is multiplying u ( x ) = ln x , but we can choose υ′( x ) to be the constant function υ′( x ) = 1. We use the integration by parts formula given in Equation (1), with

u ( x ) = ln x and υ′( x ) = 1. We differentiate u ( x ) and find an antiderivative of υ′( x ), u′( x ) =

1 and υ ( x ) = x. x

Then

∫

ln  x ⋅ 1 dx = ( ln x ) x − |

|

u ( x ) υ′( x )

|

|

u( x ) υ( x )

= x ln x −

∫

1 x x dx

Integration by parts formula

|

υ ( x ) u′ ( x )

∫ 1 dx

= x ln x − x + C

Simplify and integrate.

In the following examples we use the differential form to indicate the process of integration by parts. The computations are the same, with du and d υ providing shorter expressions for u′( x ) dx and υ′( x ) dx. Sometimes we have to use integration by parts more than once, as in the next example. EXAMPLE 3

Evaluate

∫ x 2 e x dx.

8.2  Integration by Parts

481

Solution We use the integration by parts formula given in Equation (1), with

u (x) = x 2

and υ′( x ) = e x .

We differentiate u ( x ) and find an antiderivative of υ′( x ), u′( x ) = 2 x and υ ( x ) = e x . We summarize this choice by setting du = u′( x ) dx and d υ = υ′( x ) dx , so du = 2 x dx and d υ = e x dx. We then have  = ∫ x| 2 ex|dx u

dυ

x2 ex − | |

u υ

 = ∫ e| x 2x| dx υ

x 2 e x − 2 ∫ x e x dx

Integration by parts formula

du

The new integral is less complicated than the original because the exponent on x is reduced by one. To evaluate the integral on the right, we integrate by parts again with u = x ,  d υ = e x dx. Then du = dx ,  υ = e x , and  = ∫ x| ex|dx u

dv

xex − | |

uυ

∫ e| x dx  υ

= x e x − e x + C.

|

du

Integration by parts Equation (2) u = x ,  dυ = e x dx v = e x,  du = dx

Using this last evaluation, we then obtain

∫ x 2 e x dx = x 2 e x

− 2 ∫ x e x dx

= x 2 e x − 2 x e x + 2e x + C , where the constant of integration is renamed after substituting for the integral on the right. The technique of Example 3 works for any integral ∫ x n e x dx in which n is a positive integer, because differentiating x n will eventually lead to a constant, and repeatedly integrating e x is easy. Integrals like the one in the next example occur in electrical engineering. Their evaluation requires two integrations by parts, followed by solving for the unknown integral.

EXAMPLE 4

Evaluate

∫ e x cos x dx. Solution Let u = e x and d υ = cos x dx. Then du = e x dx ,  υ = sin x , and

∫ e x cos x dx

= e x sin x −

∫ e x sin x dx.

The second integral is like the first except that it has sin x in place of cos x. To evaluate it, we use integration by parts with u = e x,

d υ = sin x dx , υ = − cos x , du = e x dx.

Then

∫ e x cos x dx = e x sin x − (−e x cos x − ∫ (− cos x )( e x dx ))                    = e x sin x + e x cos x −

∫ e x cos x dx.

The unknown integral now appears on both sides of the equation, but with opposite signs. Adding the integral to both sides and adding the constant of integration gives 2 ∫ e x cos x dx = e x sin x + e x cos x + C 1.

482

Chapter 8 Techniques of Integration

Dividing by 2 and renaming the constant of integration then gives

∫ e x cos x dx EXAMPLE 5

=

e x sin x + e x cos x + C. 2

Obtain a formula that expresses the integral

∫ cos n x dx in terms of an integral of a lower power of cos x. Solution We may think of cos n x as cos n−1 x ⋅ cos x. Then we let

u = cos n−1 x

and

d υ = cos x dx ,

so that du = ( n − 1)( cos n − 2 x )(− sin x dx )

and

υ = sin x.

Integration by parts then gives

∫ cos n x dx = cos n−1 x sin x + ( n − 1) ∫ sin 2 x cos n−2 x dx = cos n−1 x sin x + ( n − 1) ∫ (1 − cos 2 x ) cos n−2 x dx = cos n−1 x sin x + ( n − 1) ∫ cos n−2 x dx − ( n − 1) ∫ cos n x dx. If we add (n

− 1) ∫ cos n x dx

to both sides of this equation, we obtain n ∫ cos n x dx = cos n−1 x sin x + ( n − 1) ∫ cos n−2 x dx. We then divide through by n, and the final result is

∫ cos n x dx

=

cos n−1 x sin x n−1 + cos n−2 x dx. n n ∫

The formula found in Example 5 is called a reduction formula because it replaces an integral containing some power of a function with an integral of the same form having the power reduced. When n is a positive integer, we may apply the formula repeatedly until the remaining integral is easy to evaluate. For example, the result in Example 5 tells us that

∫

cos 2 x sin x 2 + ∫ cos x dx 3 3 1 2 2 = cos x sin x + sin x + C. 3 3

cos 3 x dx =

Evaluating Definite Integrals by Parts The integration by parts formula in Equation (1) can be combined with Part 2 of the Fundamental Theorem in order to evaluate definite integrals by parts. Assuming that both u′ and υ′ are continuous over the interval [ a,  b ], Part 2 of the Fundamental Theorem gives

Integration by Parts Formula for Definite Integrals b

∫a

⎤b u ( x ) υ′( x ) dx = u ( x ) υ( x ) ⎥ − ⎥⎦ a

b

∫a

υ( x ) u′( x ) dx

(3)

8.2  Integration by Parts

y

EXAMPLE 6 Find the area of the region bounded by the curve y = x e − x and the x-axis from x = 0 to x = 4.

1

Solution The region is shaded in Figure 8.1. Its area is

y = xe-x

0.5

-1

1

0

2

4

3

∫0

x

4

x e − x dx.

Let u = x ,  d υ = e − x dx ,  υ = −e − x , and du = dx. Then

-0.5

4

∫0

-1

⎤4 x e − x dx = −x e − x ⎥ − ⎥⎦ 0

4

∫0 (−e −x ) dx

= [ − 4e −4 − ( − 0e −0 ) ] + FIGURE 8.1

The region in Example 6.

EXERCISES

Integration by parts Formula (3) 4

∫0 e −x dx

4

⎤ = − 4e −4 − e − x ⎥ ⎦0 = −4e −4 − ( e −4 − e −0 ) = 1 − 5e −4 ≈ 0.91.

8.2

Integration by Parts

Evaluate the integrals in Exercises 1–24 using integration by parts. x 2. ∫ θ  cos πθ d θ 1. ∫ x sin dx 2 3.

∫ t 2 cos t dt

4.

∫ x 2 sin x dx

29.

∫ sin( ln x ) dx

30.

∫ z ( ln z )

2

dz

Evaluating Integrals

Evaluate the integrals in Exercises 31–56. Some integrals do not require integration by parts. 31.

∫ x sec x 2 dx

32.

∫

cos x dx x

33.

∫ x ( ln x )

34.

∫

1 dx x ( ln x ) 2

36.

∫

( ln x )3 dx x

38.

∫ x5 ex

40.

∫ x 2 sin x 3 dx

6.

∫1

e

x e x dx

8.

∫

x e 3 x dx

9.

∫ x 2 e −x dx

10.

∫ ( x 2 − 2 x + 1)e 2 x dx

35.

∫

11.

∫ tan −1 y dy

12.

∫ arcsin y dy

37.

∫ x3ex

13.

∫ x sec 2 x dx

14.

∫ 4 x sec 2 2 x dx

39.

∫ x3

15.

∫ x 3 e x dx

16.

∫

41.

∫ sin 3x  cos 2 x dx

42.

∫ sin 2 x  cos 4 x dx

17.

∫ ( x 2 − 5 x )e x dx

18.

∫ ( r 2 + r + 1)e r dr

43.

∫

44.

∫

19.

∫ x 5 e x dx

20.

∫ t 2 e 4 t dt

45.

∫ cos

x dx

46.

∫

21.

∫

e θ sin θ d θ

22.

∫

e − y cos y dy

47.

∫0

θ 2 sin 2θ d θ

48.

∫0

23.

∫

e 2 x cos 3 x dx

24.

∫

e −2 x sin 2 x dx

49.

∫2

t sec −1 t dt

50.

∫0

51.

∫ x arctan x dx

52.

∫ x 2 tan −1 2 dx

53.

∫ (1 + 2 x 2 ) e x

54.

∫ ( x + 1)2 dx

55.

∫

56.

∫

5.

∫1

2

7.

∫

x ln x dx

x 3 ln x dx

p 4 e − p dp

Using Substitution

Evaluate the integrals in Exercises 25–30 by using a substitution prior to integration by parts. 25.

∫e

27.

∫0

π3

483

1

ds

26.

∫0

x 1 − x dx

x tan 2 x dx

28.

∫

ln ( x + x 2 ) dx

3s + 9

2

dx

ln x dx x2 4

dx

x 2 + 1 dx

x ln x dx

π2

2 3

2

dx

x ( arcsin x ) dx

x

e

xe

1

dx

dx

x

π2

3

x

dx

x 3 cos 2 x dx 2

2 x arcsin ( x 2 ) dx x

xex

( sin −1 x )

1−

2

x2

dx

484

Chapter 8 Techniques of Integration

Theory and Examples

57. Finding area Find the area of the region enclosed by the curve y = x sin x and the x-axis (see the accompanying figure) for

62. Finding volume Find the volume of the solid generated by revolving the region bounded by the x-axis and the curve y = x sin x ,  0 ≤ x ≤ π , about a. the y-axis.

a. 0 ≤ x ≤ π.

b. the line x = π.

b. π ≤ x ≤ 2π .

(See Exercise 57 for a graph.)

c. 2π ≤ x ≤ 3π. d. What pattern do you see here? What is the area between the curve and the x-axis for nπ ≤ x ≤ ( n + 1) π,   n an arbitrary nonnegative integer? Give reasons for your answer.

63. Consider the region bounded by the graphs of y = ln x ,   y = 0, and x = e. a. Find the area of the region. b. Find the volume of the solid formed by revolving this region about the x-axis.

y

c. Find the volume of the solid formed by revolving this region about the line x = −2.

y = x sin x

10 5

d. Find the centroid of the region. p

0

2p

3p

x

64. Consider the region bounded by the graphs of y = arctan x ,   y = 0, and x = 1.

-5

a. Find the area of the region.

58. Finding area Find the area of the region enclosed by the curve y = x cos x and the x-axis (see the accompanying figure) for a. π 2 ≤ x ≤ 3π 2. b. 3π 2 ≤ x ≤ 5π 2.

b. Find the volume of the solid formed by revolving this region about the y-axis. 65. Average value A retarding force, symbolized by the dashpot in the accompanying figure, slows the motion of the weighted spring so that the mass’s position at time t is y = 2e −t cos t , t ≥ 0.

c. 5π 2 ≤ x ≤ 7π 2. d. What pattern do you see? What is the area between the curve and the x-axis for

Find the average value of y over the interval 0 ≤ t ≤ 2π.

( 2n 2− 1 )π ≤ x ≤ ( 2n 2+ 1 )π,

y

n an arbitrary positive integer? Give reasons for your answer. y 10

y = x cos x

0 y

0

p 2

3p 5p 2 2

7p 2

Mass

x Dashpot

-10

59. Finding volume Find the volume of the solid generated by revolving the region in the first quadrant bounded by the coordinate axes, the curve y = e x , and the line x = ln 2 about the line x = ln 2.

66. Average value In a mass-spring-dashpot system like the one in Exercise 65, the mass’s position at time t is

60. Finding volume Find the volume of the solid generated by revolving the region in the first quadrant bounded by the coordinate axes, the curve y = e −x , and the line x = 1 a. about the y-axis. b. about the line x = 1. 61. Finding volume Find the volume of the solid generated by revolving the region in the first quadrant bounded by the coordinate axes and the curve y = cos x ,  0 ≤ x ≤ π 2, about a. the y-axis. b. the line x = π 2.

y = 4e −t ( sin t − cos t ),

t ≥ 0.

Find the average value of y over the interval 0 ≤ t ≤ 2π. Reduction Formulas

In Exercises 67–73, use integration by parts to establish the reduction formula. 67.

∫ x n cos x dx

= x n sin x − n ∫ x n−1 sin x dx

68.

∫ x n sin x dx

= −x n cos x + n ∫ x n−1 cos x dx

485

8.2  Integration by Parts

69.

∫ x n e ax dx

70.

∫ ( ln x )

n

x n e ax n −   ∫ x n−1 e ax dx , a ≠ 0 a a

=

dx = x ( ln x ) n − n ∫ ( ln x ) n−1 dx

For the integral of arccos x , we get

∫ arccos x dx = x arccos x − ∫ cos y dy = x arccos x − sin y + C

71. ∫ x m ( ln x ) n dx =

= x arccos x − sin ( arccos x ) + C.

x m +1 n ( ln x ) n − m+1 m+1

∫ x m ( ln x )

n −1

dx ,   m ≠ −1

Use the formula

∫

72. ∫ x n x + 1 dx = 73.

∫

2x n ( 2n x + 1) 3 2 −   x n−1 x + 1 dx 2n + 3 2n + 3 ∫

xn dx x +1 2n 2x n = x +1− 2n + 1 2n + 1

x n−1 dx x +1

∫

π2

sin n x dx =

π2

∫0

cos n x dx

⎧⎪ π 1 ⋅ 3 ⋅ 5( n − 1) ⎪⎪ ,  n  even 2⋅4⋅6n ⎪ 2 = ⎨ ⎪⎪ 2 ⋅ 4 ⋅ 6  ( n − 1) ,  n  odd ⎪⎪ 1⋅3⋅5 n ⎪⎩

( )

75. Show that

∫a ( ∫ x b

b

)

f (t ) dt dx =

b

∫a ( x − a )  f ( x ) dx.

76. Use integration by parts to obtain the formula

∫

1 − x 2 dx =

1 1 x 1 − x2 + 2 2

∫

∫

f −1 ( x ) dx =

y = f x = f ( y) dx = f ′( y) dy

∫ y f ′( y) dy

= y f ( y) − = xf

−1 ( x )

∫

f ( y) dy

−

∫

∫ arctan x dx

79.

∫ sec −1 x dx

80.

∫

= x ln x − x + C.

(4)

log 2 x dx

Another way to integrate f −1 ( x ) (when f −1 is integrable) is to use integration by parts with u = f −1 ( x ) and d υ = dx to rewrite the integral of f −1 as

∫

f −1 ( x ) dx = x f −1 ( x ) −

∫ x ( dx f −1 ( x ) ) dx. d

(5)

Exercises 81 and 82 compare the results of using Equations (4) and (5). 81. Equations (4) and (5) give different formulas for the integral of arccos x : a.

∫ arccos x dx

= x arccos x − sin ( arccos x ) + C

Eq. (4)

b.

∫ arccos x dx

= x arccos x − 1 − x 2 + C

Eq. (5)

Can both integrations be correct? Explain. 82. Equations (4) and (5) lead to different formulas for the integral of arctan x :

Integration by parts with u = y,   d υ = f ′( y) dy

a.

∫ arctan x dx

= x arctan x − ln sec ( arctan x ) + C

Eq. (4)

b.

∫ arctan x dx

= x arctan x − ln 1 + x 2 + C

Eq. (5)

Can both integrations be correct? Explain.

The idea is to take the most complicated part of the integral, in this case f −1 ( x ), and simplify it first. For the integral of ln x, we get

= ye y − e y + C

y = f −1 ( x )

78.

f ( y) dy

∫ ln x dx = ∫ ye y dy

f ( y) dy

∫ arcsec x dx

Integrating Inverses of Functions

−1 ( x ),

∫

77.

1 dx. 1 − x2

Integration by parts leads to a rule for integrating inverses that usually gives good results:

f −1 ( x ) dx = x f −1 ( x ) −

to evaluate the integrals in Exercises 77–80. Express your answers in terms of x.

74. Use Example 5 to show that

∫0

y = arccos x

y = ln x , x = e y dx = e y dy

Evaluate the integrals in Exercises 83 and 84 with (a) Eq. (4) and (b) Eq. (5). In each case, check your work by differentiating your answer with respect to x. 83.

∫ sinh −1 x dx

84.

∫ tanh −1 x dx

486

Chapter 8 Techniques of Integration

8.3 Trigonometric Integrals Trigonometric integrals involve algebraic combinations of the six basic trigonometric functions. In principle, we can always express such integrals in terms of sines and cosines, but it is often simpler to work with other functions, as in the integral

∫ sec 2 x dx

= tan x + C.

The general idea is to use identities to transform the integrals we must find into integrals that are easier to work with.

Products of Powers of Sines and Cosines We begin with integrals of the form

∫ sin m x cos n x dx , where m and n are nonnegative integers (positive or zero). We can divide the appropriate substitution into three cases according to m and n being odd or even. Case 1 If m is odd in ∫ sin m x cos n x dx , we write m as 2 k + 1 and use the identity

sin 2 x = 1 − cos 2 x to obtain

sin m x = sin 2 k +1 x = ( sin 2 x ) k sin x = (1 − cos 2 x ) k sin x.

(1)

Then we substitute u = cos x  and  du = − sin x dx . Case 2 If n is odd in ∫ sin m x cos n x dx , we write n as 2 k + 1 and use the identity

cos 2 x = 1 − sin 2 x to obtain

cos n x = cos 2 k +1 x = ( cos 2 x ) k cos x = (1 − sin 2 x ) k cos x. We then substitute u = sin x  and  du = cos x dx . Case 3 If both m and n are even in ∫ sin m x cos n x dx , we substitute

sin 2 x =

1 − cos 2 x 1 + cos 2 x , cos 2 x = 2 2

(2)

to reduce the integrand to one in lower powers of cos 2 x . Here are some examples illustrating each case. EXAMPLE 1

Evaluate ∫ sin 3 x cos 2 x dx.

Solution This is an example of Case 1.

∫ sin 3 x cos 2 x dx = ∫ sin 2 x cos 2 x sin x dx

m is odd.

=

∫ (1 − cos 2 x )( cos 2 x )sin x dx

sin 2 x = 1 − cos 2 x

=

∫ (1 − u 2 )(u 2 )(−du )

u = cos x ,  du = −sin x dx

=

∫ ( u 4 − u 2 ) du

Distribute.

=

u5 u3 cos 5 x cos 3 x − +C = − +C 5 3 5 3

8.3  Trigonometric Integrals

EXAMPLE 2

487

Evaluate

∫ cos 5 x dx. Solution This is an example of Case 2, where m = 0 is even and n = 5 is odd.

∫ cos 5 x dx = ∫ cos 4 x cos x dx =

∫ (1 − sin 2 x )

=

∫ (1 − u 2 )

=

∫ (1 − 2u 2 + u 4 ) du

=u−

EXAMPLE 3

2

2

cos x dx u = sin x ,  du = cos x dx

du

Square 1 − u 2 .

2 3 1 2 1 u + u 5 + C = sin x − sin 3 x + sin 5 x + C 3 5 3 5

Evaluate

∫ sin 2 x cos 4 x dx. Solution This is an example of Case 3.

∫

sin 2 x cos 4 x dx =

∫(

)(

1 − cos 2 x 1 + cos 2 x 2 2

)

2

dx

m  and  n  both even

=

1 8

∫ (1 − cos 2 x )(1 + 2 cos 2 x + cos 2 2 x ) dx

=

1 8

∫ (1 + cos 2 x − cos 2 2 x − cos 3 2 x ) dx

=

1⎡ 1 x + sin 2 x − 8 ⎣⎢ 2

⎤

∫ ( cos 2 2 x + cos 3 2 x ) dx ⎦⎥

For the term involving cos 2 2 x , we use Use the identity 

∫

cos 2

1 2 x dx =   ∫ (1 + cos 4 x ) dx 2 =

(

cos 2 θ = (1 + cos 2θ ) 2,  with θ = 2 x.

)

1 1 x + sin 4 x + C 1. 2 4

For the cos 3 2 x term, we have

∫ cos 3 2 x dx = ∫ (1 − sin 2 2 x ) cos 2 x dx =

(

u = sin 2 x ,  du = 2 cos 2 x dx

)

1 ( 1 1   1 − u 2 ) du = sin 2 x − sin 3 2 x + C 2 . 2 ∫ 2 3

Combining everything and simplifying, we get

∫ sin 2 x cos 4 x dx

=

(

)

1 1 1 x − sin 4 x + sin 3 2 x + C. 16 4 3

488

Chapter 8 Techniques of Integration

Eliminating Square Roots In the next example, we use the identity cos 2 θ = (1 + cos 2θ ) 2 to eliminate a square root. EXAMPLE 4

Evaluate π 4

∫0

1 + cos 4 x dx.

Solution To eliminate the square root, we use the identity

1 + cos 2θ 2

cos 2 θ =

1 + cos 2θ = 2 cos 2 θ.

or

With θ = 2 x , this becomes 1 + cos 4 x = 2 cos 2 2 x. Therefore, π 4

∫0

π 4

1 + cos 4 x dx =

∫0

=

2

=

⎡ sin 2 x ⎤ 2⎢ ⎥ ⎣ 2 ⎦0

2 cos 2 2 x dx = π 4

∫0

cos 2 x dx = π 4

π 4

∫0

2

2 cos 2 2 x dx π 4

∫0

cos 2 x dx

cos 2 x ≥ 0  on [ 0,  π 4 ]

2[ 2 1 − 0] = . 2 2

=

Integrals of Powers of tan x and sec x We know how to integrate the tangent and secant functions and their squares. To integrate higher powers, we use the identities tan 2 x = sec 2 x − 1 and sec 2 x = tan 2 x + 1, and integrate by parts when necessary to reduce the higher powers to lower powers. EXAMPLE 5

Evaluate

∫ tan 4 x dx. Solution

∫ tan 4 x dx = ∫ tan 2 x · tan 2 x dx =

∫ tan 2 x · ( sec 2 x − 1) dx

=

∫ tan 2 x sec 2 x dx − ∫ tan 2 x dx

=

∫ tan 2 x sec 2 x dx − ∫ ( sec 2 x − 1) dx

=

∫ tan 2 x sec 2 x dx − ∫ sec 2 x dx + ∫

In the first integral, we let u = tan x ,

du = sec 2 x dx

tan 2 x = sec 2 x − 1

tan 2 x = sec 2 x − 1

dx  

8.3  Trigonometric Integrals

489

and have

∫ u 2 du

=

1 3 u + C 1. 3

The remaining integrals are standard forms, so

∫ tan 4 x dx EXAMPLE 6

1 tan 3 x − tan x + x + C. 3

=

Evaluate

∫ sec 3 x dx. Solution We integrate by parts using

u = sec x ,

d υ = sec 2 x dx ,

υ = tan x ,

du = sec x tan x dx.

Then

∫ sec 3 x dx = sec x tan x − ∫ ( tan x )( sec x tan x ) dx = sec x tan x −

∫ ( sec 2 x − 1)sec x dx

= sec x tan x −

∫ sec 3 x dx + ∫ sec x dx

Integrate by parts. tan 2 x = sec 2 x − 1

Combining the two secant-cubed integrals gives 2 ∫ sec 3 x dx = sec x tan x +

∫ sec x dx

and therefore

∫ sec 3 x dx EXAMPLE 7

=

1 1 sec x tan x + ln sec x + tan x + C. 2 2

Evaluate

∫ tan 4 x sec 4 x dx. Solution

∫ ( tan 4 x )( sec 4 x ) dx = ∫ ( tan 4 x )(1 + tan 2 x )( sec 2 x ) dx =

∫ ( tan 4 x + tan 6 x )( sec 2 x ) dx

=

∫ ( u 4 + u 6 ) du

=

tan 5 x tan 7 x + +C 5 7

=

u5 u7 + +C 5 7

sec 2 x = 1 + tan 2 x

Distribute. u = tan x , du = sec 2 x dx

490

Chapter 8 Techniques of Integration

Products of Sines and Cosines The integrals

∫ sin mx sin nx dx ,

∫ sin mx cos nx dx ,

∫ cos mx cos nx dx

and

arise in many applications involving periodic functions. We can evaluate these integrals through integration by parts, but two such integrations are required in each case. It is simpler to use the following identities. sin mx sin nx =

1[ cos( m − n ) x − cos( m + n ) x ] 2

(3)

sin mx cos nx =

1[ ( sin m − n ) x + sin ( m + n ) x ] 2

(4)

cos mx cos nx =

1[ cos( m − n ) x + cos( m + n ) x ] 2

(5)

These identities come from the angle sum formulas for the sine and cosine functions (Section 1.3). They give functions whose antiderivatives are easily found.

EXAMPLE 8

Evaluate

∫ sin 3x cos 5 x dx.  Solution From Equation (4) with m = 3 and n = 5, we get

1

∫ sin 3x cos 5 x dx = 2 ∫ [ sin(−2 x ) + sin 8 x ] dx =

1 ( sin 8 x − sin 2 x ) dx 2∫

=−

EXERCISES

cos 8 x cos 2 x + + C. 16 4

8.3

Powers of Sines and Cosines

Evaluate the integrals in Exercises 1–22. π

x 3 sin dx 3

11.

∫ sin 3 x cos 3 x dx

12.

∫ cos 3 2 x sin 5 2 x dx

13.

∫ cos 2 x dx

14.

∫0

sin 7 y dy

16.

∫

8 sin 4 x dx

18.

∫ 8 cos 4 2π x dx

π2

1.

∫ cos 2 x dx

2.

∫0

3.

∫ cos 3 x sin x dx

4.

∫ sin 4 2 x cos 2 x dx

15.

∫0

5.

∫ sin 3 x dx

6.

∫ cos 3 4 x dx

17.

∫0

7.

∫ sin 5 x dx

8.

∫0

x dx 2

19.

∫ 16 sin 2 x cos 2 x dx

20.

∫0

9.

∫ cos 3 x dx

10.

∫0

3 cos 5 3 x dx

21.

∫ 8 cos 3 2θ sin 2θ d θ

22.

∫0

π

sin 5

π6

π2

π

sin 2 x dx

7 cos 7 t dt

π

8 sin 4 y cos 2 y dy  

π2

sin 2 2θ cos 3 2θ d θ

8.3  Trigonometric Integrals

Integrating Square Roots

25.

∫0

27.

∫π 3

29.

31.

1 − cos x dx 2

2π

∫0

π

1 − sin 2 t dt

π2

sin 2 x dx 1 − cos x

π

π2

∫0

∫0

26.

∫0

28.

∫0

∫−π sin 3x sin 3x dx

56.

∫0

1 − cos 2 x dx

57.

∫ cos 3x cos 4 x dx

58.

∫−π 2 cos x cos 7 x dx

π

1 − cos 2 θ d θ

Exercises 59–64 require the use of various trigonometric identities before you evaluate the integrals.

π6

∫ sin 2 θ cos 3θ d θ

60.

∫ cos 2 2θ sin θ d θ

⎛ ⎞ ⎜⎜ Hint: Multiply by  1 − sin x . ⎟⎟ ⎜⎝ 1 − sin x ⎟⎠

61.

∫ cos 3 θ sin 2θ d θ

62.

∫ sin 3 θ cos 2θ d θ

63.

∫ sin θ cos θ cos 3θ d θ

64.

∫ sin θ sin 2θ sin 3θ d θ

θ 1 − cos 2θ d θ

32.

3π 4

∫π 2 π

∫−π

1 − sin 2 x dx

Assorted Integrations (1 − cos 2 t )

32

dt

Use any method to evaluate the integrals in Exercises 65–70. 65.

∫

sec 3 x dx tan x

66.

∫ cos 4 x dx

67.

∫

tan 2 x dx csc x

68.

∫ cos 2 x dx

69.

∫ x sin 2 x dx

70.

∫ x cos 3 x dx

Powers of Tangents and Secants

Evaluate the integrals in Exercises 33–52.

∫ sec 2 x tan x dx

35.

∫

sec 3

34.

∫ sec x tan 2 x dx

36.

∫

sec 3

37.

∫

sec 2

38.

∫

sec 4

40.

∫

ex

dθ

42.

∫

tan 4

csc 4 θ d θ

44.

∫ sec 6 x dx

x tan x dx

π2

59.

30.

33.

sin x cos x dx

1 + sin x dx

cos 4 x dx 1 − sin x

∫5 π 6

π

24.

π2

55.

Evaluate the integrals in Exercises 23–32. 23.

π

491

x

tan 3

x

tan 2

x dx

sin 3 x cot x

Applications

0

39.

∫−π 3

41.

∫

43.

∫π 4

45.

∫4

x

2 sec 3

sec 4 θ π2

tan 2

tan 3

∫ tan 5 x dx

49.

∫π 6

51.

π3

∫π 4

x dx

x dx

47.

π3

x dx

cot 3 x dx tan 5 θ sec 4 θ d θ

46.

π4

sec 3 e x

∫−π 4 6

x

dx

x dx

x dx

∫ cot 6 2 x dx

50.

∫ 8 cot 4 t dt ∫ cot 3 t csc 4 t dt

Products of Sines and Cosines

Evaluate the integrals in Exercises 53–58. 53.

∫ sin 3x cos 2 x dx

54.

π π ≤ x ≤ 6 2

72. Center of gravity Find the center of gravity of the region bounded by the x-axis, the curve y = sec x , and the lines x = −π 4 ,   x = π 4. 73. Volume Find the volume generated by revolving one arch of the curve y = sin x about the x-axis.

48.

52.

71. Arc length Find the length of the curve y = ln ( sin x ),

sec 3

tan 4

x dx

∫ sin 2 x cos 3x dx

74. Area Find the area between the x-axis and the curve y = 1 + cos 4 x ,  0 ≤ x ≤ π. 75. Centroid Find the centroid of the region bounded by the graphs of y = x + cos x and y = 0 for 0 ≤ x ≤ 2π. 76. Volume Find the volume of the solid formed by revolving the region bounded by the graphs of y = sin x + sec x ,   y = 0, x = 0, and x = π 3 about the x-axis. 77. Volume Find the volume of the solid formed by revolving the region bounded by the graphs of y = arctan x , x = 0, and y = π 4 about the y-axis. 78. Average Value Find the average value of the function 1 f (x) = on [ 0, π 6 ]. 1 − sin θ

492

Chapter 8 Techniques of Integration

8.4 Trigonometric Substitutions Trigonometric substitutions occur when we replace the variable of integration by a trigonometric function. The most common substitutions are x  a tan R,  x  a sin R, and x  a sec R. These substitutions are effective in transforming integrals involving a 2 x 2 , a 2  x 2 , and x 2  a 2 into integrals with respect to R, since they come from the reference right triangles in Figure 8.2.

"a 2 + x 2 u

a

x

x

x

u

"x 2 - a2

u

a x = a tan u

"a 2 - x 2 x = a sin u

a x = a sec u

"a 2 + x 2 = a 0 sec u 0

"a 2 - x 2 = a 0 cos u 0

"x 2 - a2 = a 0 tan u 0

FIGURE 8.2 Reference triangles for the three basic substitutions, identifying the sides labeled x and a for each substitution.

With x  a tan R, a 2 + x 2 = a 2 + a 2 tan 2 R = a 2 (1 + tan 2 R ) = a 2 sec 2 R.

u

With x  a sin R,

p 2

a 2 − x 2 = a 2 − a 2 sin 2 R = a 2 (1 − sin 2 R ) = a 2 cos 2 R.

u = arctan ax

x a

0 -

x 2 − a 2 = a 2 sec 2 R − a 2 = a 2 ( sec 2 R − 1) = a 2 tan 2 R.

p 2

We want any substitution we use in an integration to be reversible so that we can change back to the original variable afterward. For example, if x  a tan R, we want to be able to set R = arctan ( x a ) after the integration takes place. If x  a sin R, we want to be able to set R = arcsin( x a ) when we’re done, and similarly for x  a sec R. As we know from Section 1.5, the functions in these substitutions have inverses only for selected values of R (Figure 8.3). For reversibility,

u x u = arcsin a

p 2

-1

0

-

With x  a sec R,

1

x a

( ax )

with −

( ax )

with −

( ax )

⎧⎪ ⎪⎪ 0 ≤ θ < π if x ≥ 1, ⎪⎪ 2 a with ⎪⎨ ⎪⎪ π ⎪⎪ < θ ≤ π if x ≤ −1. a ⎪⎪ 2 ⎩

x = a tan θ requires θ = arctan

p 2

x = a sin θ requires θ = arcsin u

p

-1

u = arcsec ax

0 1

p 2 x a

FIGURE 8.3 The arctangent, arcsine, and arcsecant of x a, graphed as functions of x a.

x = a sec θ requires θ = arcsec

π π < θ < , 2 2

π π ≤ θ ≤ , 2 2

To simplify calculations with the substitution x  a sec R, we will restrict its use to integrals in which x a p 1. This will place R in [ 0,  Q 2 ) and make tan R p 0. We will then have x 2 − a 2 = a 2 tan 2 R = a tan R = a tan R, free of absolute values, provided a  0.

8.4  Trigonometric Substitutions

493

Procedure for a Trigonometric Substitution

1. Write down the substitution for x, calculate the differential dx, and specify the selected values of θ for the substitution. 2. Substitute the trigonometric expression and the calculated differential into the integrand, and then simplify the results algebraically. 3. Evaluate the trigonometric integral, keeping in mind the restrictions on the angle θ for reversibility. 4. Draw an appropriate reference triangle to reverse the substitution in the integration result and convert it back to the original variable x.

EXAMPLE 1

Evaluate

∫

dx . 4 + x2

Solution We set

π π < θ < , 2 2 = 4 + 4 tan 2 θ = 4 (1 + tan 2 θ ) = 4 sec 2 θ.

x = 2 tan θ, 4 + x2

−

Then

"4 + x 2

x

u

dx = 4 + x2

∫

2

=

FIGURE 8.4

Reference triangle for x = 2 tan θ (Example 1): tan θ =

2 sec 2 θ dθ

∫

4 sec 2 θ

∫

sec 2 θ dθ sec θ

∫ sec θ d θ

= ln 4+ 2

=

sec 2 θ = sec θ

sec θ > 0 for −

π π < θ < 2 2

= ln sec θ + tan θ + C

x 2

and sec θ =

dx = 2 sec 2 θ dθ,

x2

.

4 + x2 x + + C. 2 2

From Fig. 8.4

Notice how we expressed ln sec θ + tan θ in terms of x: We drew a reference triangle for the original substitution x = 2 tan θ (Figure 8.4) and read the ratios from the triangle. EXAMPLE 2 Here we find an expression for the inverse hyperbolic sine function in terms of the natural logarithm. Following the same procedure as in Example 1, we find that

∫

dx = a2 + x 2

∫ sec θ d θ

x = a tan θ, dx = a sec 2 θ dθ

= ln sec θ + tan θ + C = ln

a2 + x 2 x + +C a a

Fig. 8.2

From Table 7.9, sinh −1 ( x a ) is also an antiderivative of 1 derivatives differ by a constant, giving sinh −1

x = ln a

a 2 + x 2 , so the two anti-

a2 + x2 x + + C. a a

494

Chapter 8 Techniques of Integration

Setting x = 0 in this last equation, we find that 0 = ln 1 + C , so C = 0. Since a2 + x2 x + > 0, and therefore a a

a 2 + x 2 > x , we conclude that

sinh −1

⎛ a2 + x2 x x⎞ = ln ⎜⎜⎜ + ⎟⎟⎟ ⎝ a a a⎠

(See also Exercise 76 in Section 7.3.) EXAMPLE 3

Evaluate x 2 dx

∫

9 − x2

.

Solution We set

π π < θ < 2 2 = 9 − 9 sin 2 θ = 9(1 − sin 2 θ ) = 9 cos 2 θ.

x = 3 sin θ, 9 − x2

dx = 3 cos θ dθ,

−

Then 3

x

u

∫

x 2 dx 9 − x2

"9 - x 2

=

9 sin 2 θ ⋅ 3 cos θ dθ 3 cos θ

= 9 ∫ sin 2 θ d θ

FIGURE 8.5

Reference triangle for x = 3 sin θ (Example 3):

= 9∫

x sin θ = 3

=

and cos θ =

∫

9 − x2 . 3

cos θ > 0 for −

1 − cos 2θ dθ 2

(

sin 2 θ =

)

x x 9⎛ = ⎜⎜⎜ arcsin − ⋅ 2⎝ 3 3

EXAMPLE 4

1 − cos 2θ 2

sin 2θ 9 θ− +C 2 2

9 = ( θ − sin θ cos θ ) + C 2

=

π π < θ < 2 2

sin 2θ = 2 sin θ cos θ

9 − x 2 ⎞⎟ ⎟⎟ + C ⎠ 3

From Fig. 8.5

9 x x arcsin − 9 − x 2 + C. 2 3 2

Evaluate

∫

dx 2 , x > . 2 5 25 x − 4

Solution We first rewrite the radical as

25 x 2 − 4 =

(

25 x 2 −

= 5 x2 −

4 25

( 25 )

)

2

x 2 − a 2  with  a =

2 5

8.4  Trigonometric Substitutions

to put the radicand in the form x 2 − a 2 . We then substitute x =

2 2 π sec θ, dx = sec θ tan θ dθ, 0 < θ < . 5 5 2

We then get x2 −

( 25 )

2

=

4 4 4( 4 sec 2 θ − = sec 2 θ − 1) = tan 2 θ 25 25 25 25

and x2 − 5x

"25x 2 - 4

( 52 )

∫

2

Evaluate the integrals in Exercises 1–14.

3.

dx 9 + x2 2

2.

dx

∫−2 4 + x 2 32

4.

dx 9 − x2

5.

∫0

7.

∫

25 −

9.

∫5

x2

dx = − ( 4 25 )

( 2 5 )sec θ tan θ dθ 5 ⋅ ( 2 5 ) tan θ

1 5

=

1 5x ln + 5 2

25 x 2 − 4 + C. 2

∫ sec θ d θ

=

1 ln sec θ + tan θ + C 5 From Fig. 8.6

3 dx 1 + 9x 2

∫ 2

∫0

17.

∫

x dx x2 − 1

18.

∫

dx 1 − x2

19.

∫

x dx 9 − x2

20.

∫

x2 dx 4 + x2

21.

∫

x 3 dx x2 + 4

22.

∫

dx x2 x2 + 1

23.

∫

24.

∫

26.

∫x

28.

∫0

30.

∫ ( x 2 − 1)5 2 ,

dx 8 + 2x 2

12 2

2 dx 1 − 4x 2

∫0

dt

8.

∫

1−

∫

dx 7 , x > 2 4 x 2 − 49

10.

∫

5 dx 3 , x > 5 25 x 2 − 9

25.

∫

11.

∫

y 2 − 49 dy, y > 7 y

12.

∫

y 2 − 25 dy, y > 5 y3

27.

∫0

13.

∫

dx , x >1 x2 − 1

14.

∫

29.

∫ ( x 2 − 1)3 2 ,

x2

∫

=

6.

t2

tan θ > 0 for 0 < θ < π 2

8.4

Using Trigonometric Substitutions

∫

2 2 tan θ = tan θ. 5 5

dx = 25 x 2 − 4

FIGURE 8.6 If x = ( 2 5 ) sec θ, 0 < θ < π 2, then θ = arcsec( 5 x 2 ), and we can read the values of the other trigonometric functions of θ from this right triangle (Example 4).

1.

=

With these substitutions, we have

u

EXERCISES

2

x3

9t 2

dt

2 dx , x >1 x2 − 1

w2

8 dw 4 − w2

x +1 dx 1− x 32

4 x 2 dx 32 (1 − x 2 ) dx

x >1

9 − w2 dw w2

1

x 2 − 4 dx

dx (4 − x 2 )

32

x 2 dx

Assorted Integrations

Use any method to evaluate the integrals in Exercises 15–38. Most will require trigonometric substitutions, but some can be evaluated by other methods. 15.

∫

dx x x2 − 1

16.

dx

∫ 1 + x2

(1 − x 2 )

32

31.

∫

33.

∫ ( 4 x 2 + 1) 2

x6 8 dx

dx

12

(1 − x 2 )

32.

∫

34.

∫ ( 9t 2 + 1)2

x4 6 dt

dx

x >1

495

496

35.

37.

Chapter 8 Techniques of Integration

∫

x 3 dx x2 − 1

36.

∫

V 2 dV 52 (1 − V 2 )

38.

∫

x dx 25 + 4 x 2

∫

(1 − r 2 )

58. Area Find the area enclosed by the ellipse y2 x2 + 2 = 1. 2 a b

52

r8

dr

In Exercises 39–48, use an appropriate substitution and then a trigonometric substitution to evaluate the integrals. ln( 4 3 )

ln 4

e t dt e 2t + 9

40.

∫ln( 3 4 )

14

2 dt t + 4t t

42.

∫1

44.

∫

39.

∫0

41.

∫1 12

43.

∫

45.

4−x dx x ( Hint : Let  x = u 2 . )

46.

47.

∫

48.

x dx 1 + x4

∫

x 1 − x dx

e

e t dt 32 (1 + e 2 t ) dy

59. Consider the region bounded by the graphs of y = sin −1 x ,   y = 0, and x  1 2. a. Find the area of the region. b. Find the centroid of the region. 60. Consider the region bounded by the graphs of y  x arctan x and y  0 for 0 b x b 1. Find the volume of the solid formed by revolving this region about the x-axis (see accompanying figure).

y 1 + ( ln y ) 2

y y = "x arctan x

1 − ( ln x ) 2 dx x ln x

x dx 1 − x3 ( Hint : Let  u = x 3 2 . )

∫

x−2 dx x −1

∫

1

0

x

61. Evaluate ∫ x 3 1 − x 2 dx using a. integration by parts. b. a u-substitution.

Complete the Square Before Using Trigonometric Substitutions

For Exercises 49–52, complete the square before using an appropriate trigonometric substitution. 49.

∫

8 − 2 x − x 2 dx

50.

∫

1 dx x 2 − 2x + 5

51.

∫

x 2 + 4x + 3 dx x+2

52.

∫

x 2 + 2x + 2 dx x 2 + 2x + 1

Initial Value Problems

Solve the initial value problems in Exercises 53–56 for y as a function of x. 53. x

54.

55.

dy = dx

x 2 − 4, x ≥ 2, y (2) = 0

x2 − 9

dy = 1, x > 3, y (5) = ln 3 dx

(x2

dy + 4) = 3, y (2) = 0 dx

56. ( x 2 + 1) 2

dy = dx

x 2 + 1, y (0) = 1

Applications and Examples

57. Area Find the area of the region in the first quadrant that is enclosed by the coordinate axes and the curve y = 9 − x 2 3.

c. a trigonometric substitution. 62. Path of a water skier Suppose that a boat is positioned at the origin with a water skier tethered to the boat at the point (10,  0 ) on a rope 10 m long. As the boat travels along the positive y-axis, the skier is pulled behind the boat along an unknown path y  f ( x ), as shown in the accompanying figure. − 100 − x 2 . x (Hint: Assume that the skier is always pointed directly at the boat and the rope is on a line tangent to the path y  f ( x ).) a. Show that f ′( x ) =

b. Solve the equation in part (a) for f ( x ), using f (10)  0. y y = f (x) path of skier boat

10 m rope (x, f (x)) skier

f (x)

0

x

(10, 0)

x

NOT TO SCALE

x +1 on the interval [ 1,  3 ]. x 64. Find the length of the curve y = 1 − e −x ,  0 ≤ x ≤ 1. 63. Find the average value of f ( x ) =

8.5  Integration of Rational Functions by Partial Fractions

497

8.5 Integration of Rational Functions by Partial Fractions This section shows how to express a rational function (a quotient of polynomials) as a sum of simpler fractions, called partial fractions, which are more easily integrated. For instance, the rational function ( 5 x − 3 ) ( x 2 − 2 x − 3 ) can be rewritten as 5x − 3 2 3 = + . x 2 − 2x − 3 x +1 x−3 You can verify this equation algebraically by placing the fractions on the right side over a common denominator ( x + 1)( x − 3 ). The skill acquired in writing rational functions as such a sum is useful in other settings as well (for instance, when using certain transform methods to solve differential equations). To integrate the rational function ( 5 x − 3 ) ( x 2 − 2 x − 3 ) on the left side of the expression we are considering, we simply sum the integrals of the fractions on the right side: 5x − 3

2

3

∫ ( x + 1)( x − 3) dx = ∫ x + 1 dx + ∫ x − 3 dx = 2 ln x + 1 + 3 ln x − 3 + C. The method for rewriting rational functions as a sum of simpler fractions is called the method of partial fractions. In the case of our example, it consists of finding constants A and B such that 5x − 3 A B = + . x 2 − 2x − 3 x +1 x−3

(1)

(Pretend for a moment that we do not know that A = 2 and B = 3 will work.) We call the fractions A ( x + 1) and B ( x − 3 ) partial fractions because their denominators are only part of the original denominator x 2 − 2 x − 3. We call A and B undetermined coefficients until suitable values for them have been found. To find A and B, we first clear Equation (1) of fractions and regroup in powers of x, obtaining 5 x − 3 = A( x − 3 ) + B( x + 1) = ( A + B ) x − 3 A + B. This will be an identity in x if and only if the coefficients of like powers of x on the two sides are equal: A + B = 5,

−3 A + B = −3.

Solving these equations simultaneously gives A = 2 and B = 3.

General Description of the Method Success in writing a rational function f ( x ) g( x ) as a sum of partial fractions depends on three things:

•

The degree of f ( x ) must be less than the degree of g( x ). That is, the fraction must be proper. If it isn’t, divide f ( x ) by g( x ) and work with the remainder term. Example 3 of this section illustrates such a case.

•

We must know the factors of g( x ). In theory, any polynomial with real coefficients can be written as a product of real linear factors and real quadratic factors. In practice, the factors may be hard to find.

•

The values of the undetermined coefficients form a system of n linear equations in n unknowns. For large n, solving such systems may require linear algebra methods (such as Gaussian Elimination).

498

Chapter 8 Techniques of Integration

Here is how we find the partial fractions of a proper fraction f ( x ) g( x ) when the factors of g are known. A quadratic polynomial (or factor) is irreducible if it cannot be written as the product of two linear factors with real coefficients. That is, the polynomial has no real roots.

Method of Partial Fractions When f ( x ) g( x ) Is Proper

1. Let x − r be a linear factor of g( x ). Suppose that ( x − r ) m is the highest power of x − r that divides g( x ). Then, to this factor, assign the sum of the m partial fractions: (x

Am A1 A2 + + … +  . ( x − r )m − r ) ( x − r )2

Do this for each distinct linear factor of g( x ). 2. Let x 2 + px + q be an irreducible quadratic factor of g( x ). In this case, x 2 + px + q has no real roots. Suppose that ( x 2 + px + q ) n is the highest power of this factor that divides g( x ). Then, to this factor, assign the sum of the n partial fractions: B1 x + C 1

( x 2 + px + q )

+

B2 x + C 2 ( x 2 + px + q )

2

++

Bn x + C n ( x 2 + px + q )

n

.

Do this for each distinct quadratic factor of g( x ). 3. Set the original fraction f ( x ) g( x ) equal to the sum of all these partial fractions. Clear the resulting equation of fractions. 4. Find the values of the undetermined coefficients.

There are often multiple ways to find the values of the undetermined coefficients in Step 4. To find the values of the coefficients that satisfy Equation (1), we equated coefficients of like powers of x. In the next example, we instead will assign convenient values of x, leading to simple equations that we can solve for the undetermined coefficients.

EXAMPLE 1

Use partial fractions to evaluate x 2 + 4x + 1

∫ ( x − 1)( x + 1)( x + 3) dx. Solution Note that each of the factors ( x − 1),  ( x + 1), and ( x + 3 ) is raised only to the first power. Therefore, the partial fraction decomposition has the form

x 2 + 4x + 1 A B C = + + . ( x − 1)( x + 1)( x + 3 ) x −1 x +1 x+3 To find the values of the undetermined coefficients A, B, and C, we clear fractions and get x 2 + 4 x + 1 = A( x + 1)( x + 3 ) + B ( x − 1)( x + 3 ) + C ( x − 1)( x + 1). On the right side, we notice that a factor ( x − 1) is present in all terms except for the one containing A. Therefore, letting x = 1 allows us to solve for A. x = 1:

1 2 + 4(1) + 1 = A (2)(4) + B (0) + C (0) 6 = 8A A=

3 4

8.5  Integration of Rational Functions by Partial Fractions

499

In a similar manner, we can let x equal −1 to find B or −3 to find C. x = −1:

( − 1) 2

+ 4 (−1) + 1 = A (0) + B(−2 )(2) + C (0) −2 = −4 B B =

x = −3:

(−3 ) 2

1 2

+ 4 ( −3 ) + 1 = A (0) + B (0) + C ( −4 )(−2 ) −2 = 8C C =−

1 4

Hence we have x 2 + 4x + 1

∫ ( x − 1)( x + 1)( x + 3) dx = ∫ =

⎡ 3 1 + 1 1 − 1 1 ⎤ dx ⎢ ⎥ 2 x + 1 4 x + 3⎦ ⎣4 x − 1

3 1 1 ln x − 1 + ln x + 1 − ln x + 3 + K ,  4 2 4

where K is the arbitrary constant of integration (we call it K here to avoid confusion with the undetermined coefficient we labeled as C). You can solve for the undetermined coefficients (A, B, etc.) by equating coefficients of like powers of x or by assigning convenient values to x. You should choose the method that is most convenient for the problem at hand.

EXAMPLE 2

Use partial fractions to evaluate 6x + 7

∫ ( x + 2 )2 dx. Solution First we express the integrand as a sum of partial fractions with undetermined coefficients.

6x + 7 A B = + x + 2 ( x + 2 )2 + 2 )2

(x

6 x + 7 = A( x + 2 ) + B

Two terms because ( x + 2 ) is squared 2

Multiply both sides by ( x + 2 ) .

= Ax + ( 2 A + B ) Equating coefficients of corresponding powers of x gives A = 6

and

2 A + B = 12 + B = 7,

A = 6

or

and

B = −5.

Therefore,

∫ ( x + 2 )2 dx = ∫ ( x + 2 − ( x + 2 )2 ) dx 6x + 7

6

= 6∫

5

dx − 5 ∫ ( x + 2 )−2 dx x+2

= 6 ln x + 2 + 5( x + 2 )−1 + C. The next example shows how to handle the case when f ( x ) g( x ) is an improper fraction. It is a case where the degree of f is larger than the degree of g.

500

Chapter 8 Techniques of Integration

EXAMPLE 3

Use partial fractions to evaluate

∫

2x 3 − 4 x 2 − x − 3 dx. x 2 − 2x − 3

Solution First we divide the denominator into the numerator to get a polynomial plus a proper fraction.

2x x 2 − 2x − 3 2x 3 − 4 x 2 − x − 3 2x 3 − 4 x 2 − 6x 5x − 3

)

Then we write the improper fraction as a polynomial plus a proper fraction. 2x 3 − 4 x 2 − x − 3 5x − 3 = 2x + 2 x 2 − 2x − 3 x − 2x − 3 We found the partial fraction decomposition of the fraction on the right in the opening example, so

∫

2x 3 − 4 x 2 − x − 3 dx = x 2 − 2x − 3

5x − 3

∫ 2 x dx + ∫ x 2 − 2 x − 3 dx 2

3

∫ 2 x dx + ∫ x + 1 dx + ∫ x − 3 dx

=

= x 2 + 2 ln x + 1 + 3 ln x − 3 + C.

EXAMPLE 4

Use partial fractions to evaluate −2 x + 4

∫ ( x 2 + 1)( x − 1) 2 dx. Solution The denominator has an irreducible quadratic factor x 2 + 1 as well as a repeated linear factor ( x − 1) 2, so we write

−2 x + 4

( x 2 + 1)( x − 1) 2

=

C D Ax + B + + . x2 + 1 x − 1 ( x − 1) 2

Clearing the equation of fractions gives −2 x + 4 = ( Ax + B )( x − 1) 2 + C ( x − 1)( x 2 + 1) + D ( x 2 + 1) = ( A + C ) x 3 + (−2 A + B − C + D ) x 2 +( A − 2 B + C ) x + ( B − C + D ). Equating coefficients of like terms gives Coefficients of  x 3 :

0 = A+C

Coefficients of  x 2 :

0 = −2 A + B − C + D

Coefficients of  x 1 :

−2 = A − 2 B + C

Coefficients of  x 0 :

4 = B−C + D

We solve these equations simultaneously to find the values of A, B, C, and D. −4 = −2 A,

A = 2

Subtract fourth equation from second.

C = − A = −2

From the first equation

B = ( A + C + 2) 2 = 1

From the third equation and C = − A

D = 4 − B + C = 1.

From the fourth equation

(2)

8.5  Integration of Rational Functions by Partial Fractions

501

We substitute these values into Equation (2), obtaining −2 x + 4

=

( x 2 + 1)( x − 1) 2

2x + 1 2 1 . − + x 2 + 1 x − 1 ( x − 1) 2

Finally, using the expansion above, we can integrate:

∫ ( x 2 + 1)( x − 1)2 dx = ∫ ( x 2 + 1 − x − 1 + ( x − 1)2 ) dx −2 x + 4

2x + 1

=

2

∫ (x2 + 1 + 2x

1

)

1 2 1 − + dx x 2 + 1 x − 1 ( x − 1) 2

= ln ( x 2 + 1) + tan −1 x − 2 ln x − 1 −

1 + K. x −1

We use the letter K instead of C to represent an arbitrary constant here because we have already used C to represent a variable in the partial fraction representation.

EXAMPLE 5

Use partial fractions to evaluate dx

∫ x ( x 2 + 1) 2 . Solution The form of the partial fraction decomposition is

1 A Bx + C Dx + E = + 2 + 2 2. x x +1 ( x + 1) x ( x 2 + 1) 2 Multiplying by x ( x 2 + 1) 2, we have 1 = A( x 2 + 1) 2 + ( Bx + C ) x ( x 2 + 1) + ( Dx + E ) x = A( x 4 + 2 x 2 + 1) + B ( x 4 + x 2 ) + C ( x 3 + x ) + Dx 2 + Ex = ( A + B ) x 4 + Cx 3 + ( 2 A + B + D ) x 2 + ( C + E ) x + A. If we equate coefficients, we get the system A + B = 0,

C = 0,

2 A + B + D = 0,

C + E = 0,

A = 1.

Solving this system gives A = 1,  B = −1,  C = 0,  D = −1, and E  0. Thus, dx

∫ x ( x 2 + 1) 2

HISTORICAL BIOGRAPHY Oliver Heaviside (1850–1925) Heaviside studied electricity and languages on his own. He was able to simplify Maxwell’s 20 equations into the two we now call Maxwell’s equations. Heaviside’s contributions in mathematics are in the areas of vector algebra and vector calculus. To know more, visit the companion Website.

=

∫

⎡1 ⎤ −x −x + ⎢ + 2 ⎥ dx x + 1 ( x 2 + 1) 2 ⎥⎦ ⎢⎣ x

=

∫

dx − x

=

∫

dx 1 du 1 du − ∫ − ∫ 2 x 2 u 2 u

x dx

x dx

∫ x 2 + 1 − ∫ ( x 2 + 1) 2

= ln x −

1 1 +K ln u + 2 2u

= ln x −

1 ( 2 1 +K ln x + 1) + 2( x 2 + 1) 2

= ln

1 x + + K. 2( x 2 + 1) x2 + 1

u = x 2 + 1, du = 2 x dx

502

Chapter 8 Techniques of Integration

Determining Coefficients by Differentiating Another way to determine the constants that appear in partial fractions is to differentiate, as in the next example. EXAMPLE 6

Find A, B, and C in the equation x −1 A B C = + + . ( x + 1) 3 x + 1 ( x + 1) 2 + 1)3

(x

Solution We first clear fractions:

x − 1 = A ( x + 1) 2 + B ( x + 1 ) + C . Substituting x = −1 shows C = −2. We then differentiate both sides with respect to x, obtaining 1 = 2 A( x + 1) + B. Substituting x = −1 shows B = 1. We differentiate again to get 0 = 2 A, which shows A = 0. Hence, x −1 1 2 . 3 = 2 − ) ( ) ( +1 x +1 x + 1)3

(x

EXERCISES

8.5

Expanding Quotients into Partial Fractions

Expand the quotients in Exercises 1–8 by partial fractions. 1. 3.

5 x − 13 − 3 )( x − 2 )

(x

x+4 + 1) 2

(x

z +1 5. 2 z ( z − 1) 7.

t2 + 8 t 2 − 5t + 6

∫

11.

∫

x2

13.

∫4

15.

∫

8

x+4 dx + 5x − 6

y dy y 2 − 2y − 3

dt t 3 + t 2 − 2t

∫ ( x 2 − 1)2

5x − 7 x 2 − 3x + 2

19.

4.

2x + 2 x 2 − 2x + 1

Irreducible Quadratic Factors

z 6. 3 z − z 2 − 6z 8.

t4 + 9 t 4 + 9t 2

In Exercises 9–16, express the integrand as a sum of partial fractions and evaluate the integrals. dx 1 − x2

x 3 dx x 2 + 2x + 1

∫0

2.

Nonrepeated Linear Factors

9.

1

17.

10.

∫

dx x 2 + 2x

12.

∫

x2

14.

∫1 2 y 2 + y dy

16.

∫ 2 x 3 − 8 x dx

1

2x + 1 dx − 7 x + 12 y+4

dx

∫−1 x 2 − 2 x + 1

20.

∫ ( x − 1)( x 2 + 2 x + 1)

x 2 dx

In Exercises 21–32, express the integrand as a sum of partial fractions and evaluate the integrals. 1

dx + 1)( x 2 + 1)

21.

∫0

(x

23.

∫

y 2 + 2y + 1 dy 2 ( y 2 + 1)

25.

∫ ( s 2 + 1)( s − 1)3 ds

27.

∫

29.

3

3t 2 + t + 4 dt t3 + t

22.

∫1

24.

∫

8x 2 + 8x + 2 dx 2 ( 4 x 2 + 1)

26.

∫

s 4 + 81 ds s( s 2 + 9 )2

x2 − x + 2 dx x3 − 1

28.

∫

1 dx x4 + x

∫

x2 dx x4 − 1

30.

∫

x2 + x dx x 4 − 3x 2 − 4

31.

∫

2θ 3 + 5θ 2 + 8θ + 4 dθ 2 ( θ 2 + 2θ + 2 )

32.

∫

θ 4 − 4θ 3 + 2θ 2 − 3θ + 1 dθ 3 ( θ 2 + 1)

x+3

2s + 2

Repeated Linear Factors

In Exercises 17–20, express the integrand as a sum of partial fractions and evaluate the integrals.

x 3 dx

0

18.

503

8.5  Integration of Rational Functions by Partial Fractions

Improper Fractions

In Exercises 33–38, perform long division on the integrand, write the proper fraction as a sum of partial fractions, and then evaluate the integral. 33.

35.

37.

∫

2x 3 − 2x 2 + 1 dx x2 − x

34.

∫

9 x 3 − 3x + 1 dx x3 − x2

36.

∫

y4 + y2 − 1 dy y3 + y

38.

∫

x4 dx x2 − 1 16 x 3

∫ 4 x 2 − 4 x + 1 dx ∫

y3

2 dx x ( ln x − 2 )3

62.

∫

63.

∫

65.

∫ x5

1 dx x2 − 1 x 3 + 1 dx

2y 4 dy − y2 + y − 1

Evaluate the integrals in Exercises 39–54. 39.

∫ e 2t

dt + 3e t + 2

40.

cos y dy

41.

∫ sin 2 y + sin y − 6

43.

∫

(x

44.

∫

(x

45.

∫

1 x3 2 −

42.

∫ ∫

+ − e 2t + 1

et

dt 69. ( t 2 + 2 t )

sin θ dθ cos 2 θ + cos θ − 2

− 2 ) 2 tan −1 ( 2 x ) − 12 x 3 − 3 x dx ( 4 x 2 + 1)( x − 2 ) 2

dx

46.

∫ x2

1 − x 2 dx

70. ( t + 1)

(t

> 2 ), x (3) = 0

dx = 2 3, x (1) = −π 3 4 dt

dx = 2x + 2 dt

( t ,  x

> 0 ), x (1) = 1

dx = x 2 + 1 ( t > −1), x (0) = 0 dt

Applications and Examples

In Exercises 71 and 72, find the volume of the solid generated by revolving the shaded region about the indicated axis.

+ 1) 2 tan −1 ( 3 x ) + 9 x 3 + x dx ( 9 x 2 + 1)( x + 1) 2 x

dx =1 dt

68. ( 3 t 4 + 4 t 2 + 1) 2e 2 t

66.

x dx x2 + 2

x+

Solve the initial value problems in Exercises 67–70 for x as a function of t.

Evaluating Integrals

e 4t

∫

Initial Value Problems

67. ( t 2 − 3 t + 2 )

et

64.

71. The x-axis 1

∫ ( x 1 3 − 1)

x

y=

y

dx

3 "3x - x 2

( Hint: Let  x = u 6 . )

2

x +1 47. ∫ dx x ( Hint: Let  x + 1 = u 2 . ) 49.

∫

(

1 dx x ( x 4 + 1)

x3 Hint : Multiply by 3 . x

51.

∫

53.

∫

x

∫

0

50.

∫

)

1 dθ cos 2θ   sin θ 1+ x

48.

dx

1 dx x x+9 1 dx x 6 ( x 5 + 4)

y

52.

∫

54.

∫

x x +

55.

∫

x 3 − 2 x 2 − 3x dx x+2

56.

∫

57.

∫

2 x − 2 −x dx 2 x + 2 −x

58.

∫ 2 2x

x

∫

1 dx x4 − 1

61.

∫

ln x + 2 dx x ( ln x + 1)( ln x + 3 )

60.

∫

2x dx + 2x − 2

−1 dx x 5 − 5x + 1

y=

2 (x + 1)(2 - x)

dx

x+2 dx x 3 − 2 x 2 − 3x

x4

x

1

1 dθ cos θ + sin 2θ 2−

2.5

72. The y-axis

0

Use any method to evaluate the integrals in Exercises 55–66.

59.

0.5

1

x

73. Find the length of the curve y = ln (1 − x 2 ),  0 ≤ x ≤

1 . 2

74. Evaluate ∫ sec θ d θ by sec θ + tan θ and then using a u-substitution. sec θ + tan θ cos θ 1 d θ. Then multiply by , b. writing the integral as ∫ cos θ cos θ use a trigonometric identity and a u-substitution, and finally integrate using partial fractions. a. multiplying by

504

Chapter 8 Techniques of Integration

T 75. Find, to two decimal places, the x-coordinate of the centroid of the region in the first quadrant bounded by the x-axis, the curve y = arctan x , and the line x = 3.

Suppose t is in days, k = 1 250 , and two people start a rumor at time t = 0 in a population of N = 1000 people.

T 76. Find the x-coordinate of the centroid of this region to two decimal places.

b. When will half the population have heard the rumor? (This is when the rumor will be spreading the fastest.)

y

T 78. Second-order chemical reactions Many chemical reactions are the result of the interaction of two molecules that undergo a change to produce a new product. The rate of the reaction typically depends on the concentrations of the two kinds of molecules. If a is the amount of substance A and b is the amount of substance B at time t = 0, and if x is the amount of product at time t, then the rate of formation of x may be given by the differential equation

-9 y = 4x3 + 13x x + 2x 2 - 3x 2

1

0

3

x

5

a. Find x as a function of t.

T 77. Social diffusion Sociologists sometimes use the phrase “social diffusion” to describe the way information spreads through a population. The information might be a rumor, a cultural fad, or news about a technical innovation. In a sufficiently large population, the number of people x who have the information is treated as a differentiable function of time t, and the rate of diffusion, dx dt, is assumed to be proportional to the number of people who have the information times the number of people who do not. This leads to the equation dx = kx ( N − x ), dt

dx = k ( a − x )( b − x ), dt or (a

1 dx = k, − x )( b − x ) dt

where k is a constant for the reaction. Integrate both sides of this equation to obtain a relation between x and t (a) if a = b, and (b) if a ≠ b. Assume in each case that x = 0 when t = 0.

where N is the number of people in the population.

8.6 Integral Tables and Computer Algebra Systems In this section we discuss how to use tables and computer algebra systems (CAS) to evaluate integrals.

Integral Tables A Brief Table of Integrals is provided at the back of the text, after the index. (More extensive tables appear in compilations such as CRC Mathematical Tables, which contain thousands of integrals.) The integration formulas are stated in terms of constants a, b, c, m, n, and so on. These constants can usually assume any real value and need not be integers. Occasional limitations on their values are stated with the formulas. Formula 21 requires n ≠ −1, for example, and Formula 27 requires n ≠ −2. The formulas also assume that the constants do not take on values that require dividing by zero or taking even roots of negative numbers. For example, Formula 24 assumes that a ≠ 0, and Formulas 29a and 29b cannot be used unless b is positive. EXAMPLE 1

Find

∫ x ( 2x + 5)

−1

dx.

Solution We use Formula 24 at the back of the text (not 22, which requires n ≠ −1):

∫ x ( ax + b )

−1

dx =

x b − 2 ln ax + b + C. a a

dx =

x 5 − ln 2 x + 5 + C. 2 4

With a = 2 and b = 5, we have

∫ x ( 2x + 5)

−1

505

8.6  Integral Tables and Computer Algebra Systems

EXAMPLE 2

Find

∫x

dx . 2x − 4

Solution We use Formula 29b:

dx 2 = arctan ax − b b

∫x

ax − b + C. b

With a = 2 and b = 4, we have

∫x

2x − 4 + C = arctan 4

dx 2 = arctan 2x − 4 4

EXAMPLE 3

x−2 + C. 2

Find

∫ x arcsin x dx. Solution We begin by using Formula 106:

∫ x n arcsin ax dx

=

x n +1 a arcsin ax − n+1 n + 1∫

x n +1 dx 1 − a 2x 2

, n ≠ −1.

With n = 1 and a = 1, we have

∫

x2 1 arcsin x − ∫ 2 2

x arcsin x dx =

x 2 dx 1 − x2

.

Next we use Formula 49 to find the integral on the right:

∫

( )

x2 a2 x 1 dx = arcsin − x a 2 − x 2 + C. 2 a 2 a2 − x2

With a = 1,

∫

x 2 dx 1−

=

x2

1 1 arcsin x − x 1 − x 2 + C. 2 2

The combined result is

∫ x arcsin x dx = =

(

x2 1 1 1 arcsin x − arcsin x − x 1 − x 2 + C 2 2 2 2

( x2

2

−

)

)

1 1 arcsin x + x 1 − x 2 + C ′. 4 4

Reduction Formulas The time required for repeated integrations by parts can sometimes be shortened by applying reduction formulas like the following.

∫ tan n x dx ∫ ( ln x ) ∫ sin n x cos m x dx

= −

n

1 tan n −1 x − n −1

∫ tan n−2 x dx

(1)

dx = x ( ln x ) n − n   ∫ ( ln x ) n −1 dx

(2)

=

sin n −1 x cos m +1 x n −1 +   sin n −2 x cos m x dx m+n m+n ∫

(n

≠ −m ). (3)

506

Chapter 8 Techniques of Integration

By applying such a formula repeatedly, we can eventually express the original integral in terms of a power low enough to be evaluated directly. The next example illustrates this procedure. EXAMPLE 4

Find

∫ tan 5 x dx. Solution We apply Equation (1) with n = 5 to get

∫ tan 5 x dx

=

1 tan 4 x − 4

∫ tan 3 x dx.

We then apply Equation (1) again, with n = 3, to evaluate the remaining integral:

∫ tan 3 x dx

=

1 tan 2 x − 2

∫ tan x dx

=

1 tan 2 x + ln cos x + C 1. 2

The combined result is

∫ tan 5 x dx

=

1 1 tan 4 x − tan 2 x − ln cos x + C. 4 2

As their form suggests, reduction formulas are derived using integration by parts. (See Example 5 in Section 8.3.)

Integration with a CAS A powerful capability of computer algebra systems is their ability to integrate symbolically. This is performed with the integrate command specified by the particular system (for example, int in Maple, Integrate in Mathematica). EXAMPLE 5

Suppose that you want to evaluate the indefinite integral of the function f (x) = x 2 a 2 + x 2 .

Using Maple, you first define or name the function: > f := x ^2 * sqrt ( a^2 + x ^2 ); Then you use the integrate command on f , identifying the variable of integration: > int ( f , x ); Maple returns the answer a 4 ln ( x + a 2 + x 2 ) x ( a 2 + x 2 )3 2 a 2x a 2 + x 2 − − 4 8 8 If you want to see whether the answer can be simplified, enter > simplify ( % ); Maple returns −

a 4 ln ( x +

a2 + x2 ) 8

+

x a 2 + x 2 ( a 2 + 2x 2 ) 8

If you want the definite integral for 0 ≤ x ≤ π 2, you can use the format > int ( f , x = 0..Pi 2 );

8.6  Integral Tables and Computer Algebra Systems

507

Maple will return the expression π ( π 2 + 4 a 2 )3 2 a 4 ln(a 2 ) a 4 ln(2) a 2π π 2 + 4a 2 + − + 16 64 32 8 −

a 4 ln ( π +

π 2 + 4a 2 ) 8

You can also find the definite integral for a particular value of the constant a: > a:= 1; > int ( f , x = 0..1); Maple returns the numerical answer 3 1 2 + ln ( 2 − 1). 8 8 EXAMPLE 6

Use a CAS to find

∫ sin 2 x cos 3 x dx. Solution With Maple, we have the entry

> int (( sin^2 )( x ) * ( cos^3 )( x ), x ); with the immediate return 1 1 2 − sin( x ) cos( x ) 4 + cos( x ) 2 sin( x ) + sin( x ). 5 15 15 Computer algebra systems vary in how they process integrations. We used Maple in Examples 5 and 6. Mathematica would have returned somewhat different results: 1. In Example 5, given In[ 1] := Integrate[ x ^2 * Sqrt[ a^2 + x ^2 ], x ] Mathematica returns ⎛ x ⎜⎜ a 3 sinh −1 1 2 a Out[ 1] = a + x 2 ⎜⎜⎜ a 2 x + 2 x 3 − 8 x2 ⎜⎜ 1+ 2 ⎜⎝ a

( ) ⎞⎟⎟⎟⎟ ⎟⎟ ⎟⎟ ⎟⎠

without having to simplify an intermediate result. The answer is different from, but equivalent to, Formula 36 in the integral tables. 2. The Mathematica answer to the integral In[ 2 ] := Integrate [ Sin[ x ]^2 * Cos[ x ]^3, x ] in Example 6 is Out[ 2 ] =

Sin[ x ] 1 1 − Sin[ 3 x ] − Sin[ 5 x ] 8 48 80

differing from the Maple answer. Both answers are correct. Although a CAS is very powerful and can aid us in solving difficult problems, each CAS has its own limitations. There are even situations where a CAS may further complicate a problem (in the sense of producing an answer that is extremely difficult to use or interpret). Note, too, that neither Maple nor Mathematica returns an arbitrary constant +C. On the other hand, a little mathematical thinking on your part may reduce the problem to one that is quite easy to handle. We provide an example in Exercise 67.

508

Chapter 8 Techniques of Integration

Nonelementary Integrals Many functions have antiderivatives that cannot be expressed using the standard functions that we have encountered, such as polynomials, trigonometric functions, and exponential functions. Integrals of functions that do not have elementary antiderivatives are called nonelementary integrals. These integrals can sometimes be expressed with infinite series (Chapter 9) or approximated using numerical methods (Section 8.7). Examples of nonelementary integrals include the error function (which measures the probability of random errors) 2 Q

erf( x ) =

x

∫0

e −t 2 dt

and integrals such as

∫ sin x 2 dx

and

∫

1 + x 4 dx

that arise in engineering and physics. These and a number of others, such as

∫

ex dx , x

∫ e(e

x)

∫

1

∫ ln x dx ,

dx ,

1 − k 2 sin 2 x dx ,

∫ ln( ln x ) dx ,

∫

sin x dx , x

0 < k < 1,

look so easy they tempt us to try them just to see how they turn out. It can be proved, however, that there is no way to express any of these integrals as finite combinations of elementary functions. The same applies to integrals that can be changed into these by substitution. The functions in these integrals all have antiderivatives, as a consequence of the Fundamental Theorem of Calculus, Part 1, because they are continuous. However, none of the antiderivatives are elementary. The integrals you are asked to evaluate in this chapter have elementary antiderivatives.

EXERCISES

8.6

Using Integral Tables

Use the table of integrals at the back of the text to evaluate the integrals in Exercises 1–26. 1.

∫

dx x x−3

2.

∫

dx x x+4

3.

∫

x dx x−2

4.

∫ ( 2 x + 3 )3 2

5.

∫x

2 x − 3 dx

6.

∫ x ( 7x + 5)

7.

∫

9 − 4x dx x2

8.

∫

x dx

32

dx x 2 4x − 9

dx

9.

∫x

4 x − x 2 dx

dx x 7 + x2

10.

∫

12.

∫

14.

∫

x − x2 dx x dx x 7 − x2

11.

∫

13.

∫

15.

¨ e 2t cos 3t dt

16.

∫ e −3t sin 4t dt

17.

¨

18.

¨

4 − x2 dx x

x arccos x dx

x2 − 4 dx x

x arctan x dx

8.6  Integral Tables and Computer Algebra Systems

19.

∫ x 2 arctan x dx

21.

∫ sin 3x cos 2 x dx

23.

t 8 sin 4 t sin dt 2

25.

∫ ∫

θ θ cos cos d θ 3 4

20.

tan −1 x dx x2

∫

Evaluate the integrals in Exercises 51–56 by making a substitution (possibly trigonometric) and then applying a reduction formula.

∫ e t sec 3 ( e t

22.

∫ sin 2 x cos 3x dx

51.

53.

24.

t t sin sin dt 3 6

∫0

55.

∫1

26.

∫ ∫

509

θ cos cos 7θ d θ 2

1

2

− 1) dt

2 x 2 + 1 dx ( r 2 − 1)

∫

54.

∫0

56.

∫0

32

r

dr

csc 3 θ dθ θ

52.

1

32

dy (1 − y 2 )

3

52

dt ( t 2 + 1)

72

Applications Substitution and Integral Tables

In Exercises 27–40, use a substitution to change the integral into one you can find in the table. Then evaluate the integral. 27.

∫

x3 + x + 1 2 dx ( x 2 + 1)

∫

59. Centroid Find the centroid of the region cut from the first quadrant by the curve y = 1 x + 1 and the line x = 3.

30.

∫

cos −1 x dx x

60. Moment about y-axis A thin plate of constant density δ = 1 occupies the region enclosed by the curve y = 36 ( 2 x + 3 ) and the line x = 3 in the first quadrant. Find the moment of the plate about the y-axis.

32.

∫

2−x dx x

28.

∫ arcsin

31.

∫

33.

∫ cot t

1 − sin 2 t dt , 0 < t < π 2

34.

∫ tan t

dt 4 − sin 2 t

35.

∫

dy

37.

39.

x dx 1− x

y 3 + ( ln y ) 2

1 dx x 2 + 2x + 5 (Hint: Complete the square.)

∫ ∫

5 − 4 x − x 2 dx

58. Arc length Find the length of the curve y = x 2 , 0 ≤ x ≤ 3 2.

x 2 + 6x 2 dx ( x 2 + 3)

29.

x dx

57. Surface area Find the area of the surface generated by revolving the curve y = x 2 + 2,  0 ≤ x ≤ 2, about the x-axis.

36.

38.

∫ tan −1 ∫

T 61. Use the integral table and a calculator to find, to two decimal places, the area of the surface generated by revolving the curve y = x 2 , −1 ≤ x ≤ 1, about the x-axis. 62. Volume The head of your firm’s accounting department has asked you to find a formula she can use in a computer program to calculate the year-end inventory of gasoline in the company’s tanks. A typical tank is shaped like a right circular cylinder of radius r and length L, mounted horizontally, as shown in the accompanying figure. The data come to the accounting office as depth measurements taken with a vertical measuring stick marked in centimeters.

y dy

a. Show, in the notation of the figure, that the volume of gasoline that fills the tank to a depth d is

x2 x 2 − 4x + 5

dx V = 2L ∫

−r + d

−r

40.

∫ x2

2 x − x 2 dx

r 2 − y 2 dy.

b. Evaluate the integral. y

Using Reduction Formulas

Use reduction formulas to evaluate the integrals in Exercises 41–50.

Measuring stick r

41.

∫ sin 5 2 x dx

42.

∫ 8 cos 4 2πt dt

43.

∫ sin 2 2θ cos 3 2θ d θ

44.

∫ 2 sin 2 t sec 4 t dt −r

45.

∫ 4 tan 3 2 x dx

46.

∫ 8 cot 4 t dt

47.

∫ 2 sec 3 π x dx

48.

∫ 3 sec 4 3x dx

∫ csc 5 x dx

50.

49.

∫ 16 x 3 ( ln x )

2

d = Depth of gasoline L

63. What is the largest value that b

∫a dx

x − x 2 dx

can have for any a and b? Give reasons for your answer.

510

Chapter 8 Techniques of Integration

64. What is the largest value that b

∫a

e. What is the formula for

x 2 x − x 2 dx

can have for any a and b? Give reasons for your answer.

Check your answer using a CAS.

COMPUTER EXPLORATIONS

67. a. Use a CAS to evaluate

In Exercises 65 and 66, use a CAS to perform the integrations.

π2

∫0

65. Evaluate the integrals a.

∫ x ln x dx

b.

ln x dx , n ≥ 2? xn

∫

∫ x 2 ln x dx

c.

∫ x 3 ln x dx.

sin n x dx , x + cos n x

where n is an arbitrary positive integer. Does your CAS find the result?

d. What pattern do you see? Predict the formula for ∫ x 4 ln x dx and then see if you are correct by evaluating it with a CAS.

b. In succession, find the integral when n = 1,  2,  3,  5,  and 7. Comment on the complexity of the results.

e. What is the formula for ∫ x n ln x dx ,   n ≥ 1? Check your answer using a CAS.

c. Now substitute x = ( π 2 ) − u and add the new and old integrals. What is the value of

66. Evaluate the integrals ln x ln x ln x b. ∫ 3 dx c. ∫ 4 dx. a. ∫ 2 dx x x x d. What pattern do you see? Predict the formula for

∫

sin n

π2

∫0

sin n

sin n x dx ? x + cos n x

This exercise illustrates how a little mathematical ingenuity can sometimes solve a problem not immediately amenable to solution by a CAS.

ln x dx x5

and then see if you are correct by evaluating it with a CAS.

8.7 Numerical Integration The antiderivatives of some functions, like sin( x 2 ), 1 ln x, and 1 + x 4 , have no elementary formulas. When we cannot find a workable antiderivative for a function f that we have to integrate, we can partition the interval of integration, replace f by a closely fitting polynomial on each subinterval, integrate the polynomials, and add the results to approximate the definite integral of f . This procedure is an example of numerical integration. In this section we start by revisiting the Midpoint Rule, which we studied in Section 5.2. We then study two new methods, the Trapezoidal Rule and Simpson’s Rule. A key goal in our analysis is to control the possible error that is introduced when computing an approximation to an integral.

Approximating Integrals with the Midpoint Rule In Section 5.2 we introduced the Midpoint Rule to approximate a definite integral over an interval [ a,  b ]. The rule is based on subdividing [ a,  b ] into n equal subintervals, [ x 0 ,  x 1 ],  [ x 1 ,  x 2 ],  …,  [ x n −1 ,  x n ]

each of width Δx =

b−a . n

We then approximate the integral using n rectangles, where the height of the kth rectangle is the value of f at the midpoint c k = ( x k −1 + x k ) 2 of the kth subinterval [ x k −1 , x k ]. Midpoint Rule for Approximating a Definite Integral b

∫a

f ( x ) dx ≈

with c k =

∑ f (c k )( b −n a ) = [ f (c1 ) + n

k =1

x k −1 + x k b−a and x k = a + k . 2 n

(

)

( b −n a )

f (c 2 ) +  + f (c n ) ]

8.7  Numerical Integration

511

Trapezoidal Approximations The Trapezoidal Rule for the value of a definite integral is based on approximating the region between a curve and the x-axis with trapezoids instead of rectangles, as in Figure 8.7. It is not necessary for the subdivision points x 0 ,  x 1 ,  x 2 , …,  x n in the figure to be evenly spaced, but the resulting formula is simpler if they are. We therefore assume that the length of each subinterval is Δx =

b−a . n

The length Δx = ( b − a ) n is called the step size or mesh size. The area of the trapezoid that lies above the ith subinterval is Δx

(y

i −1

)

+ yi Δx ( y i −1 + y i ), = 2 2 y = f (x)

Trapezoid area 1 ( y1 + y 2 )¢ x 2

y1

x0 = a

x1

y2

x2

yn-1

¢x

x n-1

yn

x

xn = b

FIGURE 8.7 The Trapezoidal Rule approximates short stretches of the curve y = f ( x ) with line segments. To approximate the integral of f from a to b, we add the areas of the trapezoids made by vertically joining the ends of the segments to the x-axis.

where y i−1 = f ( x i−1 ) and y i = f ( x i ). (See Figure 8.7.) The area below the curve y = f ( x ) and above the x-axis is then approximated by adding the areas of all the trapezoids: 1 1 T = ( y 0 + y1 ) Δ x + ( y1 + y 2 ) Δ x +  2 2 +

1 1 (y + y n −1 ) Δx + ( y n −1 + y n ) Δx 2 n−2 2

= Δx =

( 12 y

0

+ y1 + y 2 +  + y n −1 +

1 y 2 n

)

Δx ( y 0 + 2 y1 + 2 y 2 +  + 2 y n −1 + y n ), 2

where y 0 = f (a), y1 = f ( x 1 ), …, y n −1 = f ( x n −1 ), y n = f (b). The Trapezoidal Rule says: Use T to estimate the integral of f from a to b.

512

Chapter 8 Techniques of Integration

y

4

The Trapezoidal Rule b

To approximate ∫a f ( x ) dx , use 49 16

T =

Δx ( y 0 + 2 y1 + 2 y 2 +  + 2 y n −1 + y n ). 2

The y’s are the values of f at the partition points

36 16

x 0 = a,  x 1 = a + Δx ,  x 2 = a + 2Δx ,  …,  x n −1 = a + ( n − 1) Δx ,   x n = b, where Δx = ( b − a ) n.

25 16

1 2

EXAMPLE 1 Use the Trapezoidal Rule with n = 4 to estimate ∫1 x 2 dx. Compare the estimate with the exact value.

y = x2 0

1

5 4

6 4

x

2

7 4

FIGURE 8.8 The trapezoidal approximation of the area under the graph of y = x 2 from x = 1 to x = 2 is a slight overestimate (Example 1). TABLE 8.2

x

y = x2

1

1

5 4 6 4 7 4

25 16 36 16 49 16

2

4

Solution Partition [ 1, 2 ] into four subintervals of equal length (Figure 8.8). Then evaluate y = x 2 at each partition point (Table 8.2). Using these y-values, n = 4, and Δx = ( 2 − 1) 4 = 1 4 in the Trapezoidal Rule, we have

T =

Δx ( y 0 + 2 y1 + 2 y 2 + 2 y 3 + y 4 ) 2

(

( )

( )

( )

=

1 25 36 49 1+ 2 +2 +2 +4 8 16 16 16

=

75 = 2.34375. 32

)

Since the parabola is concave up, the approximating segments lie above the curve, giving each trapezoid slightly more area than the corresponding strip under the curve. The exact value of the integral is

∫1

2

x 2 dx =

x3 ⎤2 8 1 7 = − = . 3 ⎥⎦ 1 3 3 3

The T approximation overestimates the integral by about half a percent of its true value of 7 3. The percentage error is ( 2.34375 − 7 3 ) ( 7 3 ) ≈ 0.00446, or 0.446%.

Simpson’s Rule: Approximations Using Parabolas y

Parabola yn-1 yn

y0

y1 y2 h h 0 a = x0 x1 x2

y = f (x)

h

xn-1 xn= b

FIGURE 8.9 Simpson’s Rule approximates short stretches of the curve with parabolas.

x

Another rule for approximating the definite integral of a continuous function results from using parabolas instead of the straight-line segments that produced trapezoids. As before, we partition the interval [ a,  b ] into n subintervals of equal length h = Δx = ( b − a ) n , but this time we require that n be an even number. On each consecutive pair of intervals we approximate the curve y = f ( x ) ≥ 0 by a parabola, as shown in Figure 8.9. A typical parabola passes through three consecutive points ( x i−1 ,  y i−1 ),  ( x i ,  y i ), and ( x i +1 ,  y i +1 ) on the curve. Let’s calculate the shaded area beneath a parabola passing through three consecutive points. To simplify our calculations, we first take the case where x 0 = −h,  x 1 = 0, and x 2 = h (Figure 8.10), where h = Δx = ( b − a ) n. The area under the parabola will be the same if we shift the y-axis to the left or right. The parabola has an equation of the form y = Ax 2 + Bx + C,

8.7  Numerical Integration

y

y0

so the area under it from x = −h to x  h is (0, y1)

(-h, y0)

y1

Ap =

(h, y2 )

y2

0

h

x

By integrating from h to h, we find the shaded area to be h( y + 4 y1 + y 2 ). 3 0

h

∫−h ( Ax 2 + Bx + C ) dx

h Ax 3 Bx 2 = ⎡⎢ + + Cx ⎤⎥ 2 ⎣ 3 ⎦ −h

y = Ax 2 + Bx + C

= -h

513

2 Ah 3 h + 2Ch = ( 2 Ah 2 + 6C ). 3 3

Since the curve passes through the three points (−h, y 0 ), ( 0, y1 ), and ( h,  y 2 ), we also have y 0 = Ah 2 − Bh + C ,

FIGURE 8.10

y1 = C ,

y 2 = Ah 2 + Bh + C.

After some algebraic manipulation, we find that Ap =

h ( y + 4 y1 + y 2 ). 3 0

Now shifting the parabola horizontally to its shaded position in Figure 8.9 does not change the area under it. Thus the area under the parabola through ( x 0 ,  y 0 ),  ( x 1 ,  y1 ), and ( x 2 ,  y 2 ) in Figure 8.9 is still h ( y + 4 y1 + y 2 ). 3 0 Similarly, the area under the parabola through the points ( x 2 ,  y 2 ),  ( x 3 ,  y 3 ), and ( x 4 ,  y 4 ) is h ( y + 4 y 3 + y 4 ). 3 2 Computing the areas under all the parabolas and adding the results give the approximation b

∫a

f ( x ) dx ≈

= HISTORICAL BIOGRAPHY Thomas Simpson (1720–1761) Simpson was a successful text writer and did most of his research on probability. Simpson’s rule to approximate definite integrals was developed before he was born. It is another of history’s beautiful quirks that one of the ablest mathematicians of the 18th century is remembered not for his own work but for a rule that was never his, that he never claimed, and that bears his name only because he happened to mention it in one of his books. To know more, visit the companion Website.

h h ( y + 4 y1 + y 2 ) + ( y 2 + 4 y 3 + y 4 ) + " 3 0 3 h + ( y n −2 + 4 y n −1 + y n ) 3

h ( y + 4 y1 + 2 y 2 + 4 y 3 + 2 y 4 + " + 2 y n − 2 + 4 y n −1 + y n ). 3 0

The result is known as Simpson’s Rule. The function need not be positive, as in our derivation, but the number n of subintervals must be even for us to apply the rule because each parabolic arc uses two subintervals.

Simpson’s Rule b

To approximate ¨a f ( x ) dx , use S =

Δx ( y 0 + 4 y1 + 2 y 2 + 4 y 3 + " + 2 y n − 2 + 4 y n −1 + y n ). 3

The y’s are the values of f at the partition points x 0 = a,  x 1 = a + Δx ,  x 2 = a + 2Δx ,  !,  x n −1 = a + ( n − 1) Δx ,  x n = b. The number n is even, and Δx = ( b − a ) n.

Note the pattern of the coefficients in the above rule: 1, 4, 2, 4, 2, 4, 2, . . . , 4, 1. EXAMPLE 2

2

Use Simpson’s Rule with n  4 to approximate ¨ 0 5x 4 dx.

514

Chapter 8 Techniques of Integration

TABLE 8.3

Solution Partition [ 0, 2 ] into four subintervals and evaluate y  5 x 4 at the partition points (Table 8.3). Then apply Simpson’s Rule with n  4 and Δx = 1 2:

x

y  5x4

0

0

S =

1 2 1

5 16 5

=

3 2 2

405 16 80

Δx ( y 0 + 4 y1 + 2 y 2 + 4 y 3 + y 4 ) 3

(

( )

( )

1 5 405 0+4 + 2( 5 ) + 4 + 80 6 16 16

= 32

)

1 . 12

This estimate differs from the exact value (32) by only 1 12, a percentage error of less than three-tenths of one percent, and this was with just four subintervals.

Error Analysis Whenever we use an approximation technique, we must consider how accurate the approximation might be. The following theorem gives formulas for estimating the errors when using the Midpoint Rule, the Trapezoidal Rule, and Simpson’s Rule. The error is the difference between the approximation obtained by using the rule and the actual value of the b definite integral ¨a f ( x ) dx . THEOREM 1—Error Estimates in the Midpoint, Trapezoidal, and Simpson’s Rules

If f aa is continuous and M is any upper bound for the values of f aa on [ a,  b ], then the error E M in the Midpoint Rule approximation of the integral of f from a to b for n steps satisfies the inequality EM ≤

M ( b − a )3 . 24 n 2

Midpoint Rule

If f aa is continuous and M is any upper bound for the values of f aa on [ a,  b ], then the error E T in the Trapezoidal Rule approximation of the integral of f from a to b for n steps satisfies the inequality ET ≤

M ( b − a )3 . 12n 2

Trapezoidal Rule

If f ( 4 ) is continuous and M is any upper bound for the values of f ( 4 ) on [ a,  b ], then the error E S in the Simpson’s Rule approximation of the integral of f from a to b for n steps satisfies the inequality ES ≤

M ( b − a )5 . 180 n 4

Simpson’s Rule

To give an idea of why Theorem 1 is true in the case of the Trapezoidal Rule, we begin with a result which says that if f aa is continuous on the interval [ a,  b ], then b

∫a

f ( x ) dx = T −

b−a ⋅ f ′′(c)( Δx ) 2 12

for some number c between a and b. This result follows from Taylor’s Remainder Theorem, which we discuss in Theorem 24 of Section 9.9. It follows that as %x approaches zero, the error defined by ET = −

b−a ⋅ f ′′(c) ( Δx ) 2 12

approaches zero at the rate of the square of %x.

8.7  Numerical Integration

515

The inequality ET ≤

b−a max f ′′( x ) ( Δx ) 2 , 12

where “max” refers to the maximum of f ′′( x ) over the interval [ a,  b ], gives an upper bound for the magnitude of the error. In practice, we usually cannot find the exact value of max f ′′( x ) and have to estimate an upper bound or “worst case” value for it instead. If M is any upper bound for the values of f ′′( x ) on [ a,  b ], so that f ′′( x ) ≤ M for every x in [ a,  b ], then ET ≤

b − a ( )2 M Δx . 12

If we substitute ( b − a ) n for Δx, we get ET ≤

M ( b − a )3 . 12n 2

The upper bound for the error in the Midpoint Rule, EM ≤

M ( b − a )3 , 24 n 2

is based on a similar argument. Taylor’s Remainder Theorem implies that the Midpoint Rule approximation and the definite integral differ by at most f ′′(c) ( b − a )3 24 n 2 for some c ∈ [ a,  b ]. Taking M to be at least as large as the maximum of f ′′(c) on [ a,  b ], we obtain the error bound in Theorem 1. Note that this is half as large as the error bound for the Trapezoidal Rule. This does not mean that the Midpoint Rule is always more accurate than the Trapezoidal Rule, but rather that the largest possible error that might occur is only half as large as the largest possible error for the Trapezoidal Rule. To estimate the error in Simpson’s Rule, we start with a result, again following from Taylor’s Remainder Theorem, that says that if the fourth derivative f (4) is continuous, then b

∫a

f ( x ) dx = S −

b−a ⋅ f (4) (c)( Δx ) 4 180

for some point c between a and b. Thus, as Δx approaches zero, the error, ES = −

b−a ⋅ f (4) (c)( Δx ) 4 , 180

approaches zero as the fourth power of Δx . (This helps to explain why Simpson’s Rule is likely to give better results than the Trapezoidal Rule.) The inequality ES ≤

b−a max f (4) ( x ) ( Δx ) 4 , 180

where “max” refers to the maximum of f (4) ( x ) over the interval [ a,  b ], gives an upper bound for the magnitude of the error. As with max f ′′ in the error formula for the Trapezoidal Rule, we usually cannot find the exact value of max f (4) ( x ) and have to replace it with an upper bound. If M is any upper bound for the values of f (4) ( x ) on [ a,  b ], then ES ≤

b − a ( )4 M Δx . 180

Substituting ( b − a ) n for Δx in this last expression gives ES ≤

M ( b − a )5 . 180 n 4

516

Chapter 8 Techniques of Integration

You might wonder why we don’t just take M to be the maximum value of f ′′( x ) on [ a,  b ] for the first two rules, or the maximum value of f (4) ( x ) for Simpson’s Rule. The reason is that sometimes this maximum is hard to compute, while a less accurate upper bound is easily found. We can certainly set M to the maximum value if we can compute it. 2

EXAMPLE 3 Find an upper bound for the error in estimating ∫ 0 5x 4 dx using Simpson’s Rule with n = 4 (Example 2). Solution To estimate the error, we first find an upper bound M for the magnitude of the fourth derivative of f ( x ) = 5 x 4 on the interval 0 ≤ x ≤ 2. Since the fourth derivative has the constant value f (4) ( x ) = 120, we take M = 120. With b − a = 2 and n = 4, the error estimate for Simpson’s Rule gives

ES ≤

M ( b − a )5 120 ( 2 ) 5 1 = = . 180 n 4 180 ⋅ 4 4 12

This estimate is consistent with the result of Example 2. Theorem 1 can also be used to estimate the number of subintervals required when using the Trapezoidal or Simpson’s Rule if we specify a certain tolerance for the error. EXAMPLE 4 Estimate the minimum number of subintervals needed to approximate the integral in Example 3 using Simpson’s Rule with an error of magnitude less than 10 −4. Solution Using the inequality in Theorem 1, if we choose the number of subintervals n to satisfy

M ( b − a )5 < 10 −4 , 180 n 4 then the error E S in Simpson’s Rule satisfies E S < 10 −4, as required. From the solution in Example 3, we have M = 120 and b − a = 2, so we want n to satisfy 120 ( 2 ) 5 1 < , 180 n 4 10 4 or, equivalently, n4 >

64 ⋅ 10 4 . 3

It follows that n > 10

( 643 )

14

≈ 21.5.

Since n must be even in Simpson’s Rule, we estimate the minimum number of subintervals required for the error tolerance to be n = 22. EXAMPLE 5 integral

As we saw in Chapter 7, the value of ln 2 can be calculated from the ln 2 =

∫1

2

1 dx. x 2

Table 8.4 shows values of T and S for approximations of ∫1 (1 x ) dx using various values of n. Notice how Simpson’s Rule dramatically improves over the Trapezoidal Rule.

8.7  Numerical Integration

517

TABLE 8.4 Trapezoidal Rule approximations (Tn ) and Simpson’s Rule 2

approximations ( S n ) of ln 2 = ∫1 ( 1 x ) dx

n 10 20 30 40 50 100

Tn

Error less than . . .

Sn

Error less than . . .

0.6937714032 0.6933033818 0.6932166154 0.6931862400 0.6931721793 0.6931534305

0.0006242227 0.0001562013 0.0000694349 0.0000390595 0.0000249988 0.0000062500

0.6931502307 0.6931473747 0.6931472190 0.6931471927 0.6931471856 0.6931471809

0.0000030502 0.0000001942 0.0000000385 0.0000000122 0.0000000050 0.0000000004

In particular, notice that when we double the value of n (thereby halving the value of h = Δx), the T error is divided by 2 squared, whereas the S error is divided by 2 to the fourth. This has a dramatic effect as Δx = ( 2 − 1) n gets very small. The Simpson approximation for n  50 rounds accurately to seven places and for n  100 is accurate to nine decimal places (billionths)! If f ( x ) is a polynomial of degree less than 4, then its fourth derivative is zero, and ES = −

b − a (4) ( ) 4 b − a ( )( ) 4 f (c) Δx = − 0 Δx = 0. 180 180

Thus, there will be no error in the Simpson approximation of any integral of f . In other words, if f is a constant, a linear function, or a quadratic or cubic polynomial, Simpson’s Rule will give the value of any integral of f exactly, whatever the number of subdivisions. Similarly, if f is a constant or a linear function, then its second derivative is zero, and ET = −

b−a b − a ( )( ) 2 f ′′(c)( Δx ) 2 = − 0 Δx = 0. 12 12

The Trapezoidal Rule will therefore give the exact value of any integral of f . This is no surprise, for the trapezoids fit the graph perfectly. Although decreasing the step size %x reduces the error in the Simpson and Trapezoidal approximations in theory, it may fail to do so in practice. When %x is very small, say Δx = 10 −8 , computer or calculator round-off errors in the arithmetic required to evaluate S and T may accumulate to such an extent that the error formulas no longer describe what is going on. Shrinking %x below a certain size can actually make things worse. You should consult a text on numerical analysis for more sophisticated methods if you are having problems with round-off error using the rules discussed in this section. 44 m

EXAMPLE 6 A town wants to drain and fill a polluted swamp (Figure 8.11). The swamp averages 1.5 m deep. About how many cubic meters of dirt will it take to fill the area after the swamp is drained?

37 m 23 m 16 m 12 m

Vertical spacing = 6 m

9m 4m Ignored

FIGURE 8.11 The dimensions of the swamp in Example 6.

Solution To calculate the volume of the swamp, we estimate the surface area and multiply by 1.5. To estimate the area, we use Simpson’s Rule with Δx = 6 m, and the y’s equal to the distances measured across the swamp, as shown in Figure 8.11.

Δx ( y 0 + 4 y1 + 2 y 2 + 4 y 3 + 2 y 4 + 4 y 5 + y 6 ) 3 6 = ( 44 + 148 + 46 + 64 + 24 + 36 + 4 ) = 732 3

S =

The volume is about ( 732 )(1.5 ) = 1098 m 3.

518

Chapter 8 Techniques of Integration

EXERCISES

8.7

For some exercises, a calculator may be helpful for expressing answers in decimal form.

17.

¨1

Estimating Definite Integrals

19.

∫0

21.

∫0

The instructions for the integrals in Exercises 1–10 have three parts, one for the Midpoint Rule, one for the Trapezoidal Rule, and one for Simpson’s Rule. I. Using the Midpoint Rule a. Estimate the integral with n  4 steps and find an upper bound for E M . b. Evaluate the integral directly and find E M . c. Use the formula ( E M ( true value )) × 100 to express E M as a percentage of the integral’s true value.

2

1 ds s2

3

2

x + 1 dx

sin ( x + 1) dx

22.

∫−1 cos( x + Q ) dx

3

1 dx x +1

1

23. Volume of water in a swimming pool A rectangular swimming pool is 5 m wide and 10 m long. The accompanying table shows the depth h ( x ) of the water at 1-m intervals from one end of the pool to the other. Estimate the volume of water in the pool using the Trapezoidal Rule with n  10 applied to the integral V =

10

∫0

5 ⋅ h ( x ) dx. Position (m) x

Depth (m) h ( x)

0

1.20

6

2.30

1

1.64

7

2.38

a. Estimate the integral with n  4 steps and find an upper bound for E S .

2

1.82

8

2.46

3

1.98

9

2.54

b. Evaluate the integral directly and find E S .

4

2.10

10

2.60

c. Use the formula ( E S ( true value )) × 100 to express E S as a percentage of the integral’s true value.

5

2.20

III. Using Simpson’s Rule

2

1.

¨1

3.

∫−1 ( x 2 + 1) dx

5.

∫0

7.

¨1

9.

¨0

∫1

4.

∫−2 ( x 2 − 1) dx

( t 3 + t ) dt

6.

∫−1 ( t 3 + 1) dt

1 ds s2

8.

∫2

sin t dt

10.

1

2

2

Q

3

2.

x dx

( 2x

− 1) dx

0

4

(s

1 ds − 1)2

1

¨0 sin Qt dt

In Exercises 11–22, estimate the minimum number of subintervals needed to approximate the integrals with an error of magnitude less than 10 4 by (a) the Trapezoidal Rule and (b) Simpson’s Rule. (The integrals in Exercises 11–18 are the integrals from Exercises 1–8.)

¨1

2

1

∫−1 2

∫0

x dx

12.

( x 2 + 1) dx

14.

( t 3 + t ) dt

16.

24. Distance traveled The accompanying table shows time-tospeed data for a car accelerating from rest to 130 kmh. How far had the car traveled by the time it reached this speed? (Use trapezoids to estimate the area under the velocity curve, but be careful: The time intervals vary in length.)

1

Estimating the Number of Subintervals

15.

∫0

Depth (m) h ( x)

c. Use the formula ( E T ( true value )) × 100 to express E T as a percentage of the integral’s true value.

13.

20.

(s

Position (m) x

b. Evaluate the integral directly and find E T .

11.

1 ds − 1) 2

∫2

Estimates with Numerical Data

II. Using the Trapezoidal Rule a. Estimate the integral with n  4 steps and find an upper bound for E T .

4

18.

∫1

3

( 2x

0

∫−2 1

∫−1

− 1) dx

Speed change

Time (s)

Zero to 30 kmh

2.2

40 kmh

3.2

50 kmh

4.5

60 kmh

5.9

70 kmh

7.8

80 kmh

10.2

90 kmh

12.7

100 kmh

16.0

110 kmh

20.6

120 kmh

26.2

130 kmh

37.1

( x 2 − 1) dx

( t 3 + 1) dt

25. Wing design The design of a new airplane requires a gasoline tank of constant cross-sectional area in each wing. A scale drawing of a cross-section is shown here. The tank must hold 2000 kg

519

8.7  Numerical Integration

of gasoline, which has a density of 673 kg m 3 . Estimate the length of the tank by Simpson’s Rule.

y2

y1

y0

y3

y4

y5

a. Use the fact that f ( 4 ) ≤ 1 on [ 0, Q 2 ] to give an upper bound for the error that will occur if Si

y6

( Q2 ) = ∫

Q2 0

sin t dt t

is estimated by Simpson’s Rule with n  4. y0 = 0.5 m, y4 = 0.7 m,

b. Estimate Si ( Q 2 ) by Simpson’s Rule with n  4.

y1 = 0.55 m, y2 = 0.6 m, y3 = 0.65 m, y5 = y6 = 0.75 m Horizontal spacing = 0.3 m

26. Oil consumption on Pathfinder Island A diesel generator runs continuously, consuming oil at a gradually increasing rate until it must be temporarily shut down to have the filters replaced. Use the Trapezoidal Rule to estimate the amount of oil consumed by the generator during that week.

c. Express the error bound you found in part (a) as a percentage of the value you found in part (b). 28. The error function

The error function,

erf( x ) =

x 2   ∫ e −t 2 dt , 0 Q

which is important in probability and in the theories of heat flow and signal transmission, must be evaluated numerically because there is no elementary expression for the antiderivative of e t 2 .

Day

Oil consumption rate (liters/hour)

Sun

0.019

a. Use Simpson’s Rule with n  10 to estimate erf (1).

Mon

0.020

b. In [ 0,  1 ],

Tue

0.021

Wed

0.023

Thu

0.025

Fri

0.028

Sat

0.031

Sun

0.035

d 4 −t 2 ( e ) ≤ 12. dt 4 Give an upper bound for the magnitude of the error of the estimate in part (a). b

29. Prove that the sum T in the Trapezoidal Rule for ¨a f ( x ) dx is a Riemann sum for f continuous on [ a,   b ]. (Hint: Use the Intermediate Value Theorem to show the existence of c k in the subinterval [ x k −1 ,   x k ] satisfying f (c k ) = ( f ( x k −1 ) + f ( x k ) ) 2.)

Theory and Examples

b

27. Usable values of the sine-integral function The sine-integral function, Si ( x ) =

x

∫0

sin t dt , t

y

-p

1

0

31. Elliptic integrals The length of the ellipse

“Sine integral of  x ”

is one of the many functions in engineering whose formulas cannot be simplified. There is no elementary formula for the antiderivative of (sin t ) t . The values of Si(x), however, are readily estimated by numerical integration. Although the notation does not show it explicitly, the function being integrated is ⎧⎪ sin t ⎪⎪ , t ≠ 0 f (t ) = ⎨ t ⎪⎪ t = 0, ⎪⎪⎩  1, the continuous extension of (sin t ) t to the interval [0, x]. The function has derivatives of all orders at every point of its domain. Its graph is smooth, and you can expect good results from Simpson’s Rule.

sin t y= t

30. Prove that the sum S in Simpson’s Rule for ¨a f ( x ) dx is a Riemann sum for f continuous on [ a,   b ]. (See Exercise 29.)

y2 x2 + 2 =1 2 a b turns out to be Length = 4 a   ∫

Q2 0

1 − e 2 cos 2 t dt ,

where e = a 2 − b 2 a is the ellipse’s eccentricity. The integral in this formula, called an elliptic integral, is nonelementary except when e  0 or 1. a. Use the Trapezoidal Rule with n  10 to estimate the length of the ellipse when a  1 and e  1 2. b. Use the fact that the absolute value of the second derivative of f (t ) = 1 − e 2 cos 2 t is less than 1 to find an upper bound for the error in the estimate you obtained in part (a). Applications

Si(x) =

x

x

L0 p

32. The length of one arch of the curve y  sin x is given by

sin t dt t

L = 2p

t

Q

∫0

1 + cos 2 x dx.

Estimate L by Simpson’s Rule with n  8.

520

Chapter 8 Techniques of Integration

T When solving Exercises 33-40, you may need to use a calculator or a computer. 33. Your metal fabrication company is bidding for a contract to make sheets of corrugated iron roofing like the one shown here. The cross-sections of the corrugated sheets are to conform to the curve 3Q y = sin x, 0 ≤ x ≤ 20 cm. 20

y

35. y = sin x , 0 ≤ x ≤ Q 36. y = x 2 4 , 0 ≤ x ≤ 2 37. Use numerical integration to estimate the value of 0.6

∫0

arcsin 0.6 =

If the roofing is to be stamped from flat sheets by a process that does not stretch the material, how wide should the original material be? To find out, use numerical integration to approximate the length of the sine curve to two decimal places. Original sheet

Find, to two decimal places, the areas of the surfaces generated by revolving the curves in Exercises 35 and 36 about the x-axis.

Corrugated sheet

dx . 1 − x2

For reference, arcsin 0.6  0.64350 to five decimal places. 38. Use numerical integration to estimate the value of Q = 4∫

1 0

1 dx. 1 + x2

39. Drug assimilation An average adult under age 60 years assimilates a 12-hour cold medicine into his or her system at a rate modeled by 20 cm 0 y = sin 3p x 20

20

x (cm)

34. Your engineering firm is bidding for the contract to construct the tunnel shown here. The tunnel is 90 m long and 15 m wide at the base. The cross-section is shaped like one arch of the curve y = 7.5 cos ( Q x 15 ) . Upon completion, the tunnel’s inside surface (excluding the roadway) will be treated with a waterproof sealer that costs $26.11 per square meter to apply. How much will it cost to apply the sealer? (Hint: Use numerical integration to find the length of the cosine curve.) y

dy = 6 − ln ( 2t 2 − 3t + 3 ), dt where y is measured in milligrams and t is the time in hours since the medication was taken. What amount of medicine is absorbed into a person’s system over a 12-hour period? 40. Effects of an antihistamine The concentration of an antihistamine in the bloodstream of a healthy adult is modeled by C = 12.5 − 4 ln ( t 2 − 3t + 4 ), where C is measured in grams per liter and t is the time in hours since the medication was taken. What is the average level of concentration in the bloodstream over a 6-hour period?

y = 7.5 cos (p x15)

-7.5 0

90 m 7.5 x (m)

NOT TO SCALE

8.8 Improper Integrals Up to now, we have required definite integrals to satisfy two properties. First, the domain of integration [ a,  b ] must be finite. Second, the range of the integrand must be finite on this domain. In practice, we may encounter problems that fail to meet one or both of these conditions. The integral for the area under the curve y = ( ln x ) x 2 from x  1 to x = ∞ is an example for which the domain is infinite (Figure 8.12a). The integral for the area under the curve of y  1 x between x  0 and x  1 is an example for which the range of the integrand is infinite (Figure 8.12b). In either case, the integrals are said to be improper and are calculated as limits. We will see in Chapter 9 that improper integrals are useful for investigating the convergence of certain infinite series.

8.8  Improper Integrals

y

y y = ln2x x

0.2

y= 1 "x

0.1

0

521

1

1

2

3

4

5

x

6

1

0

x

(b)

(a)

FIGURE 8.12 Are the areas under these infinite curves finite? We will see that the answer is yes for both curves.

y

Infinite Limits of Integration

Area = 2

x

Consider the infinite region (unbounded on the right) that lies under the curve y = e − x 2 in the first quadrant (Figure 8.13a). You might think this region has infinite area, but we will see that the value is finite. We assign a value to the area in the following way. First find the area A(b) of the portion of the region that is bounded on the right by x = b (Figure 8.13b). A(b) =

(a) y

b

∫0

b

e − x 2 dx = −2e − x 2 ⎤⎥ = −2e −b 2 + 2 = 2 − 2e −b 2 , ⎦0

which is a little less than 2. Then find the limit of A(b) as b → ∞. lim A (b) = lim ( 2 − 2e −b 2 ) = 2

Area = 2 - 2e-b2

b →∞

b →∞

Therefore, the value we assign to the area under the curve from 0 to ∞ is b

x

∞

∫0

(b)

FIGURE 8.13 (a) The area in the first quadrant under the curve y = e −x 2 . (b) The area is an improper integral of the first type.

DEFINITION

b

b →∞ ∫ 0

e − x 2 dx = lim

e − x 2 dx = 2.

Integrals with infinite limits of integration are improper integrals

of Type I. 1. If f ( x ) is continuous on [ a, ∞ ), then ∞

∫a

f ( x ) dx = lim

b →∞

b

∫a

f ( x ) dx.

2. If f ( x ) is continuous on (−∞, b ], then b

∫−∞ f ( x ) dx

= lim

a →−∞

b

∫a

f ( x ) dx.

3. If f ( x ) is continuous on (−∞, ∞ ), then ∞

∫−∞ f ( x ) dx

=

c

∞

∫−∞ f ( x ) dx + ∫c

f ( x ) dx ,

where c is any real number. In each case, if the limit exists and is finite, we say that the improper integral converges and that the limit is the value of the improper integral. If the limit fails to exist, the improper integral diverges.

522

Chapter 8 Techniques of Integration

The choice of c in Part 3 of the definition is unimportant. We can evaluate or deter∞ mine the convergence or divergence of ∫−∞ f ( x ) dx with any convenient choice. Any of the integrals in the above definition can be interpreted as an area if f p 0 on the interval of integration. For instance, we interpreted the improper integral in Figure 8.13 as an area. In that case, the area has the finite value 2. If f p 0 and the improper integral diverges, we say the area under the curve is infinite.

EXAMPLE 1 Is the area under the curve y = ( ln x ) x 2 from x  1 to x = ∞ finite? If so, what is its value?

y y = ln2x x

0.2

Solution We find the area under the curve from x  1 to x  b and examine the limit as b → ∞. If the limit is finite, we take it to be the area under the curve (Figure 8.14). The area from 1 to b is

0.1

0

1

b

x

FIGURE 8.14

The area under this curve is an improper integral (Example 1).

∫1

b

( )

ln x 1 b dx = ⎡⎢ ( ln x ) − ⎤⎥ − 2 ⎣ x x ⎦1 =−

ln b 1 b − ⎡⎢ ⎤⎥ ⎣ x ⎦1 b

=−

ln b 1 − + 1. b b

∫1

b

Integration by parts with u = ln x ,  dV = dx x 2, du = dx x ,   V = −1 x

( )( ) −

1 x

1 dx x

The limit of the area as b → ∞ is ∞

∫1

b ln x ln x dx = lim ∫ dx 2 b →∞ 1 x x2

⎡ ln b ⎤ 1 = lim ⎢ − − + 1⎥ b →∞ ⎢⎣ ⎥⎦ b b ln b ⎤ ⎡ = − ⎢ lim ⎥−0+1 ⎢⎣ b →∞ b ⎥⎦ 1 b⎤ ⎡ = − ⎢ lim ⎥ + 1 = 0 + 1 = 1. ⎢⎣ b →∞ 1 ⎥⎦

L’Hôpital’s Rule

Thus, the improper integral converges and the area has finite value 1.

EXAMPLE 2

Evaluate ∞

dx

∫−∞ 1 + x 2 . HISTORICAL BIOGRAPHY Lejeune Dirichlet (1805–1859) Dirichlet, a German mathematician, investigated the solution and equilibrium of systems of differential equations and discovered many results on the convergence of series. In 1855, Dirichlet succeeded Gauss as the professor of mathematics at Göttingen. To know more, visit the companion Website.

Solution According to Part 3 of the definition, we can choose c  0 and write ∞

dx

∫−∞ 1 + x 2

0

dx

∫−∞ 1 + x 2

=

+

∞

∫0

dx . 1 + x2

Next we evaluate each improper integral on the right side of the equation above. 0

dx

∫−∞ 1 + x 2

0

a →−∞ ∫ a

= lim

dx 1 + x2

0 = lim tan −1 x ⎤⎥ ⎦ a →−∞ a

( )

Q Q = = lim ( tan −1 0 − tan −1 a ) = 0 − − a →−∞ 2 2

8.8  Improper Integrals

∞

y

∫0

523

b dx dx = lim ∫ 2 b →∞ 0 1 + x 2 1+ x b

1 y= 1 + x2

= lim tan −1 x ⎤⎥ ⎦0 b →∞

Area = p

0

π π = lim ( tan −1 b − tan −1 0 ) = − 0 = b →∞ 2 2

x

Thus,

NOT TO SCALE

∞

dx

∫−∞ 1 + x 2

FIGURE 8.15

The area under this curve is finite (Example 2).

π π + = π. 2 2

=

Since 1 (1 + x 2 ) > 0, the improper integral can be interpreted as the (finite) area beneath the curve and above the x-axis (Figure 8.15).

The Integral ∫

∞

1

dx xp

The function y = 1 x is the boundary between the convergent and divergent improper integrals with integrands of the form y = 1 x p . As the next example shows, the improper integral converges if p > 1 and diverges if p ≤ 1. ∞

EXAMPLE 3 For what values of p does the integral ∫1 dx x p converge? When the integral does converge, what is its value? Solution If p ≠ 1, then

∫1

b

b ⎛ 1 ⎞⎟ − p +1 ⎛ 1 ⎞⎟ 1 dx x − p +1 ⎤ (b = = ⎜⎜ − 1) = ⎜⎜ −1 . ⎥ ⎝ 1 − p ⎟⎟⎠ ⎝ 1 − p ⎟⎟⎠ b p−1 xp − p + 1 ⎥⎦ 1

(

)

Thus,

∫1

∞

b dx dx = lim ∫ p b →∞ 1 x p x

⎧⎪ 1 ⎪⎪ , ⎡ ⎛ 1 ⎞⎟ 1 ⎤ = lim ⎢ ⎜⎜ − 1 ⎥ = ⎪⎨ p − 1 ⎟ ⎟ − p 1 b →∞ ⎢⎣ ⎝ 1 − p ⎠ b ⎪⎪ ⎥⎦ ⎪⎪⎩ ∞,

(

)

p >1 p 1 p < 1.

Therefore, the integral converges to the value 1 ( p − 1) if p > 1, and it diverges if p < 1. If p = 1, the integral also diverges: ∞

∫1

∫1

∞

dx x

= lim

∫1

dx = xp

b →∞

b

dx x

⎡ ⎤b = lim ⎢ ln x ⎥ b →∞ ⎢⎣ ⎦⎥ 1 = lim ( ln b − ln1) = ∞. b →∞

524

Chapter 8 Techniques of Integration

y

Integrands with Vertical Asymptotes y= 1 "x

Another type of improper integral arises when the integrand has a vertical asymptote—an infinite discontinuity—at a limit of integration or at some point between the limits of integration. If the integrand f is positive over the interval of integration, we can again interpret the improper integral as the area under the graph of f and above the x-axis between the limits of integration. Consider the region in the first quadrant that lies under the curve y = 1 x from x = 0 to x = 1 (Figure 8.12b). First we find the area of the portion from a to 1 (Figure 8.16):

Area = 2 - 2"a

1

1 dx = 2 x ⎥⎤ = 2 − 2 a. ⎦a x

1

0

a

1

∫a

x

Then we find the limit of this area as a → 0 + : FIGURE 8.16 The area under this curve is an example of an improper integral of the second kind.

lim

a→ 0+

1

dx = lim+ ( 2 − 2 a ) = 2. a→ 0 x

∫a

Therefore, the area under the curve from 0 to 1 is finite and is defined to be 1

1 dx dx = lim+ ∫ = 2. a→ 0 a x x

∫0

Integrals of functions that become infinite at a point within the interval of integration are improper integrals of Type II.

DEFINITION

1. If f ( x ) is continuous on ( a, b ] and discontinuous at a, then b

∫a

b

f ( x ) dx = lim+ ∫ f ( x ) dx. c→ a

c

2. If f ( x ) is continuous on [ a, b ) and discontinuous at b, then b

∫a

c

f ( x ) dx = lim− ∫ f ( x ) dx. c→ b

a

3. If f ( x ) is discontinuous at c, where a < c < b, and continuous on [ a, c) ∪ (c, b ], then b

∫a

c

∫a

f ( x ) dx =

f ( x ) dx +

∫c

b

f ( x ) dx.

In each case, if the limit exists and is finite, we say that the improper integral converges and that the limit is the value of the improper integral. If the limit does not exist, the integral diverges.

In Part 3 of the definition, the integral on the left side of the equation converges if both integrals on the right side converge; otherwise, it diverges. EXAMPLE 4

Investigate the convergence of 1

∫0

1 dx. 1− x

8.8  Improper Integrals

Solution The integrand f ( x ) = 1 (1 − x ) is continuous on [0, 1) but is discontinuous at x = 1 and becomes infinite as x → 1− (Figure 8.17). We evaluate the integral as

y y=

525

1 1-x

lim− ∫

b →1

b 0

⎡ ⎤b 1 dx = lim− ⎢ − ln 1 − x ⎥ b →1 ⎢⎣ 1− x ⎥⎦ 0 = lim− [ − ln (1 − b ) + 0 ] = ∞. b →1

The limit is infinite, so the integral diverges.

1

EXAMPLE 5 b

0

Evaluate 3

∫0

x

1

FIGURE 8.17

The area beneath the curve and above the x-axis for [0, 1) is not a real number (Example 4). y

(x

dx . − 1) 2 3

Solution The integrand has a vertical asymptote at x = 1 and is continuous on [0, 1) and (1, 3] (Figure 8.18). Thus, by Part 3 of the definition above, 3

∫0

1

dx = ( x − 1) 2 3

dx

∫0 ( x − 1)2 3

+

∫1

3

(x

dx . − 1) 2 3

Next, we evaluate each improper integral on the right-hand side of this equation. 1

dx

∫0 ( x − 1)2 3

1 y= (x - 1)23

= lim− ∫ b →1

b 0 (x

dx − 1) 2 3 b

= lim− 3( x − 1)1 3 ⎤⎥ ⎦0 b →1 = lim− [ 3( b − 1)1 3 + 3 ] = 3 b →1

∫1

1

3

3 dx dx 2 3 = lim+ ∫ 23 c →1 c ( x − 1) ( x − 1) 3

0

c

b

3

= lim+ 3( x − 1)1 3 ⎤⎥ ⎦c c →1

x

= lim+ [ 3( 3 − 1)1 3 − 3( c − 1)1 3 ] = 3 3 2

1

c →1

FIGURE 8.18

Example 5 shows that the area under the curve exists (so it is a real number).

We conclude that 3

∫0

(x

dx = 3 + 3 3 2. − 1) 2 3

Improper Integrals with a CAS Computer algebra systems can evaluate many convergent improper integrals. To evaluate the integral ∞

∫2

(x

x+3 dx − 1)( x 2 + 1)

(which converges) using Maple, enter > f := ( x + 3 ) (( x − 1) * ( x ^2 + 1)); Then use the integration command > int ( f , x = 2..infinity ); Maple returns the answer ln (5) + arctan(2) −

π 2

526

Chapter 8 Techniques of Integration

To obtain a numerical result, use the evaluation command evalf and specify the number of digits as follows: > evalf ( %, 6 ); The symbol % instructs the computer to evaluate the last expression on the screen, in this case (−1 2 ) Q + ln(5) + arctan(2). Maple returns 1.14579. If you are using Mathematica, entering In[ 1] := Integrate[ ( x + 3 ) (( x − 1)( x ^2 + 1)),  { x , 2,Infinity } ] returns Q + ArcTan[ 2 ] + Log[ 5 ]. 2 To obtain a numerical result with six digits, use the command “N[%, 6]”; it also yields 1.14579. Out[ 1] = −

Tests for Convergence and Divergence When we cannot evaluate an improper integral directly, we try to determine whether it converges or diverges. If the integral diverges, that’s the end of the story. If it converges, we can use numerical methods to approximate its value. The principal tests for convergence or divergence are the Direct Comparison Test and the Limit Comparison Test. THEOREM 2—Direct Comparison Test

Let f and g be continuous on [ a,  ∞ ) with 0 b f ( x ) b g( x ) for all x p a. Then 1. if ∫ 2. if ∫

∞ a ∞

g( x ) dx converges, then ∫

a

f ( x ) dx diverges, then ∫

∞

a ∞

a

f ( x ) dx also converges.

g( x ) dx also diverges.

∞

HISTORICAL BIOGRAPHY Karl Weierstrass (1815–1897) Weierstrass attended the University of Bonn to learn public administration, but he found that his passion was for mathematics. In his Berlin lectures in the 1860s, he also proved several theorems for continuous and complex functions. The standards of rigor that he set greatly affected the future of mathematics. To know more, visit the companion Website.

I (b) =

f ( x ) dx ≤

b

∫a g( x ) dx

≤

∞

∫a

g( x ) dx.

∞

Thus, the finite number M = ∫a g( x ) dx is an upper bound to the values of I (b). Therefore I (b) cannot diverge to infinity, so it must must converge to a finite value as b → ∞ . Hence ∞ ∫a f ( x ) dx converges. This establishes that statement 1 holds. Since statement 2 is the contrapositive form of statement 1, it must hold as well.

y = e -x

EXAMPLE 6

2

(a) (1,

∞ ∫1 e − x 2

1

These examples illustrate how we use Theorem 2.

dx converges because 0 < e − x 2 < e − x for every x p 1 (Figure 8.19) and

∫1

e -1)

x

b

∞

b →∞ ∫1

e − x dx = lim

b →∞

FIGURE 8.19

e x 2

The graph of lies below the graph of e x for x  1 (Example 6a).

b

e − x dx

⎡ ⎤b = lim ⎢ −e − x dx ⎥ b →∞ ⎢⎣ ⎥⎦ 1 = lim ( −e −b + e −1 )

y = e -x 0

b

∫a

Although the theorem is stated for Type I improper integrals, a similar result is true for integrals of Type II as well.

y 1

Outline of a Proof Assume that ∫a g( x ) dx converges. For each number b p a, let b I (b) = ∫a f ( x ) dx. Since f ( x ) is nonnegative, we know that I (b) is an increasing function of b. Using the completeness property of the real numbers (Appendix A.7), it can be shown that since this integral increases with b, it either has a limit or it diverges to infinity as b → ∞ . We will not give a proof of this fact here. Since f ( x ) b g( x ) for every x,

= converges.

1 e

8.8  Improper Integrals

(b)

∫1

∞

sin 2 x dx x2

0 ≤ (c)

∞

∫1

x2 (d)

π 2

∫0

converges because

sin 2 x 1 ≤ 2 2 x x

on [ 1, ∞ ) and

1 dx x 2 − 0.1

∫1

∞

1 dx x2

converges.

Example 3

diverges because ∞

1 1 ≥ on [ 1, ∞ ) and x − 0.1 cos x dx x

527

∫1

1 dx x

diverges.

Example 3

converges because 0 ≤

cos x 1 ≤ x x

on

⎡ π⎤ ⎢⎣ 0, 2 ⎥⎦ ,

π 0 ≤ cos x ≤ 1 on  ⎡⎢ 0, ⎤⎥ ⎣ 2⎦

and π 2

∫0

π 2 dx dx = lim+ ∫ a→ 0 a x x

⎤π 2 4 x ⎥⎥ ⎥⎦ a

= lim+ a→ 0

= lim+ ( 2π − a→ 0

4a ) =

2 x =

4x

2π

converges.

Although we have shown that the integrals in parts (a), (b), and (d) of Example 6 converge, we do not know the exact values of these integrals. For example, our computations in part (d) imply that π 2 cos x π 2 dx 0 ≤ ∫ dx ≤ ∫ = 2π , 0 0 x x but unless we do further calculations the most we can say is that the integral is some real number between 0 and 2π . THEOREM 3—Limit Comparison Test

If the positive functions f and g are continuous on [ a, ∞ ), and if lim

x →∞

f (x) = L , 0 < L < ∞, g( x )

then ∞

∫a

f ( x ) dx and

∞

∫a

g( x ) dx

either both converge or both diverge. We omit the proof of Theorem 3, which is similar to that of Theorem 2. If two functions f ( x ) and g( x ) satisfy the hypotheses of Theorem 3 and their improper integrals both converge, it need not be the case that these two integrals have the same value. This is illustrated in the next example. EXAMPLE 7

Show that ∞

∫1 ∞

dx 1 + x2

converges by comparison with ∫1 (1 x 2 ) dx. Find and compare the values of the two integrals.

528

Chapter 8 Techniques of Integration

Solution The functions f ( x ) = 1 x 2 and g( x ) = 1 (1 + x 2 ) are positive and continuous on [ 1, ∞ ). Also,

y y = 12 x

1

lim

x →∞

1 x2 f (x) 1 + x2 = lim = lim 2 x →∞ 1 (1 + x ) x →∞ g( x ) x2 = lim

x →∞

1 y= 1 + x2

0

1

FIGURE 8.20

2

2

)

+ 1 = 0 + 1 = 1,

∞ ∞ dx dx which is a positive finite limit (Figure 8.20). Therefore, ∫ converges because ∫ 1 1 + x2 1 x2 converges. The integrals converge to different values, however:

x

3

( x1

∫1

The functions in

∞

Example 7.

dx 1 = =1 x2 2−1

Example 3

and

∫1

∞

b dx dx π π π = lim ∫ = lim [ tan −1 b − tan −1 1 ] = − = . 2 b →∞ 1 1 + x 2 b →∞ 1+ x 2 4 4

TABLE 8.5

∫1

b 2 5 10 100 1000 10000 100000

b

1− x

e−x

EXAMPLE 8

lim

x →∞

∞

dx +1

3.

∫0

5.

∫−1 x 2 3

7.

∫0

9.

∫−∞ x 2 − 1

11.

1

x2 dx x

1

1

−2

∞

∫2

dx

dx 1 − x2 2 dx

2 dυ υ2 − υ

(

f (x) 1 − e −x = lim x →∞ g( x ) x

)( 1x ) =

lim (1 − e − x ) = 1,

x →∞

∞ dx 1 − e −x dx diverges because ∫ diverges. 1 x x Approximations to the improper integral are given in Table 8.5. Note that the values of these approximations do not appear to approach a fixed finite limit as b → ∞.

which is a positive finite limit. Therefore, ∫

∞

1

8.8

∞

15.

∫0

dx 4−x

17.

∫0

dx

19.

∫0

dr

21.

∫−∞ θe θ d θ

23.

∫−∞ e − x

25.

∫0

dx

∫1

4.

∫0

6.

∫−8 x 1 3

8.

∫0

x 1.001

1

1

r 0.999

2

2 dx

10.

∫−∞ x 2 + 4

12.

∫2

∞

2 x dx

∫−∞ ( x 2 + 1)2

2.

4

∞

13.

The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.

∫0

1 − e −x dx. x

Solution The integrand suggests a comparison of f ( x ) = (1 − e − x ) x with g( x ) = 1 x . However, we cannot use the Direct Comparison Test because f ( x ) ≤ g( x ) and the integral of g( x ) diverges. On the other hand, using the Limit Comparison Test, we find that

Evaluating Improper Integrals

1.

∞

1

dx

0.5226637569 1.3912002736 2.0832053156 4.3857862516 6.6883713446 8.9909564376 11.2935415306

EXERCISES

Investigate the convergence of ∫

2 dt t2 − 1

2

s+1 ds 4 − s2

∞

1 dx x x2 − 1

∞

16 tan −1 x dx 1 + x2

16.

∫0

18.

∫1

20.

∫0

22.

∫0

dx

24.

∫−∞ 2 x e −x

x ln x dx

26.

∫0 (− ln x ) dx

θ+1 dθ θ 2 + 2θ

∞

dx (1 + x ) x dυ + tan −1 υ )

∞

(1 +

υ 2 )(1

0

1

x dx

∫−∞ ( x 2 + 4 )3 2

1

0

∞

14.

∞

∞

1

2e −θ sin θ d θ 2

dx

8.8  Improper Integrals

2

27.

∫0

29.

∫1

31.

∫−1

33.

2

ds 4 − s2

28.

∫0

ds s s2 − 1

30.

∫2

32.

∫0

4

dx x

∞

∫−1

dθ θ 2 + 5θ + 6

34.

4r dr 1 − r4

1

4

dt −4

t2

t

2

dx x −1

∞

∫0

dx ( x + 1)( x 2 + 1)

Testing for Convergence

In Exercises 35–68, use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. 2

dx

35.

∫1 2 x ln x

37.

∫1 2

39.

∫0

41.

∫0

43.

45.

47.

49.

51.

53.

55.

∞

π2

1

ln 2

∫0

π

2

∫0

tan θ d θ

40.

∫0

42.

∫1

x −2 e −1 x dx

dx 1 − x2

∫−1 ln ∫1

∞

x dx

2

1

∫0

50.

2

∫0

∞

x +1 dx x2

58.

∫2

∞

2 + cos x dx x

60.

∫π

∞

2 dt t3 2 − 1

∞

ex x

dx

56.

62.

64.

∫4

∞

∫0

∞

∫2

∞

dx x −1 dθ 1 + eθ dx x2 − 1 x dx x4 − 1

∞

1 dx ln x

∫e

e

1

1 dx x x

69.

∫0

68.

∫−∞ e x

∞

dx + e −x

71.

∫0

73.

∫3

75.

∫0

72.

∫1

1 dx x4

74.

∫−2 x 4 dx

x 2 e x 3 dx

76.

∫−∞ x 2 e x

77.

∫−3 x 2 + 3x dx

78.

∫1

79.

∫−∞ ( x 2 + 9 )5 2 dx

80.

∫−∞ ( x 2 + 9 )2 5 dx

1 dx x

5

∞

1 dx x x

∫2

32

∞

∞

70.

0

1

4

x

a. 82.

∫1

2

dx x ( ln x ) p

∞

1 dx x

5

1

1

0

∞

3

dx

1 dx x 2 + 3x

4

b.

x

∞

∞

∫2

dx x ( ln x ) p

b

  f ( x) dx. ∫−∞ f ( x) dx may not equal blim → ∞ ∫− b ∞

2 x dx x2 + 1

∞

2 x dx

∫0

Show that

∫−∞ x 2 + 1

x dx

1 + sin x dx x2

∞

dx x4 + 1

1 dx ex − 2x

∫1

diverges and hence that

∞

∫2

∞

∞

66.

In Exercises 69–80, determine whether the improper integral converges or diverges. If it converges, evaluate the integral.

dx

dx 1− x

∞

dx x6 + 1

∫4

x

∫−1 −x ln

∞

∫π

e−

1

54.

∫−∞

1 dx −x

81. Find the values of p for which each integral converges.

dt 46. ∫ 0 t − sin t (Hint: t ≥ sin t for t ≥ 0) 48.

67.

ex

Theory and Examples

dx x ln x

1

dυ υ −1

59.

cot θ d θ

x

∞

∫0

dθ θ2 − 1

π2

52.

∫2

∫1

44.

∞

dx x3 + 1

∫1

63.

38.

1

57.

61.

dx x ( ln x )3

dt t + sin t

∫0

∫0

∫−1 θ 2 − 2θ

ln x dx x2

∫1

dθ

1

36.

∞

65.

529

ln ( ln x ) dx

diverges. Then show that lim

b →∞

b

2 x dx

∫−b x 2 + 1 =

0.

Exercises 83–86 are about the infinite region in the first quadrant between the curve y = e −x and the x-axis. 83. Find the area of the region. 84. Find the centroid of the region. 85. Find the volume of the solid generated by revolving the region about the y-axis. 86. Find the volume of the solid generated by revolving the region about the x-axis. 87. Find the area of the region that lies between the curves y = sec x and y = tan x from x = 0 to x = π 2. 88. The region in Exercise 87 is revolved about the x-axis to generate a solid. a. Find the volume of the solid. b. Show that the inner and outer surfaces of the solid have infinite area.

530

Chapter 8 Techniques of Integration

89. Consider the infinite region in the first quadrant bounded by the 1 graphs of y = 2 ,   y = 0, and x = 1. x a. Find the area of the region. b. Find the volume of the solid formed by revolving the region (i) about the x-axis; (ii) about the y-axis. 90. Consider the infinite region in the first quadrant bounded by the 1 ,   y = 0,   x = 0, and x = 1. graphs of y = x a. Find the area of the region. b. Find the volume of the solid formed by revolving the region (i) about the x-axis; (ii) about the y-axis. 91. Evaluate the integrals. 1 dt a. ∫ 0 t (1 + t )

b.

∫0

∞

a. Show that e −3 x dx =

1 −9 e < 0.000042, 3

∞

and hence that ∫ 3 e −x 2 dx < 0.000042. Explain why this ∞ 3 means that ∫ 0 e −x 2 dx can be replaced by ∫ 0 e −x 2 dx without introducing an error of magnitude greater than 0.000042. 3

2 T b. Evaluate ∫ 0 e −x dx numerically.

94. The infinite paint can or Gabriel’s horn As Example 3 shows, ∞ the integral ∫1 ( dx x ) diverges. This means that the integral ∞

1 1 1 + 4 dx , x x which measures the surface area of the solid of revolution traced out by revolving the curve y = 1 x ,  1 ≤ x , about the x-axis, diverges also. By comparing the two integrals, we see that, for every finite value b > 1,

∫1

∫1 y

b

2π

x

∫0

T a. Plot the integrand ( sin t ) t for t > 0. Is the sine-integral function everywhere increasing or decreasing? Do you think Si( x ) = 0 for x ≥ 0? Check your answers by graphing the function Si( x ) for 0 ≤ x ≤ 25.

∫0

3

∞

sin t dt , t called the sine-integral function, has important applications in optics. Si ( x ) =

dt t (1 + t )

∞

∫3

95. Sine-integral function The integral

b. Explore the convergence of ∞

dx . x x2 − 9 93. Estimating the value of a convergent improper integral whose domain is infinite 92. Evaluate ∫

amount of paint cannot cover an infinite surface. But if we fill the horn with paint (a finite amount), then we will have covered an infinite surface. Explain the apparent contradiction.

If it converges, what is its value? 96. Error function The function 2e −t 2 dt , π called the error function, has important applications in probability and statistics. erf( x ) =

x

∫0

T a. Plot the error function for 0 ≤ x ≤ 25. b. Explore the convergence of ∞

2e −t 2 dt. π If it converges, what appears to be its value? You will see how to confirm your estimate in Section 14.4, Exercise 41.

∫0

97. Normal probability distribution The function

2π

b 1 1 1 1 + 4 dx > 2π ∫ dx. 1 x x x

sin t dt. t

f (x) =

1 x −μ − 1 e 2( σ ) σ 2π

2

is called the normal probability density function with mean μ and standard deviation σ. The number μ tells where the distribution is centered, and σ measures the “scatter” around the mean. From the theory of probability, it is known that ∞

∫−∞ f ( x ) dx

y = 1x

= 1.

In what follows, let μ = 0 and σ = 1. T a. Draw the graph of f . Find the intervals on which f is increasing, the intervals on which f is decreasing, and any local extreme values and where they occur.

0 1 b

b. Evaluate n

∫−n

x

However, the integral

∫1

∞

π

( ) 1 x

f ( x ) dx

for n = 1,  2, and 3.

2

dx

c. Give a convincing argument that ∞

∫−∞ f ( x ) dx

for the volume of the solid converges. a. Calculate it. b. This solid of revolution is sometimes described as a can that does not hold enough paint to cover its own interior. Think about that for a moment. It is common sense that a finite

= 1.

(Hint: Show that 0 < f ( x ) < e −x 2 for x > 1, and for b > 1, ∞

∫b

e −x 2 dx → 0 as b → ∞.)

Chapter 8  Questions to Guide Your Review

98. Show that if f ( x ) is integrable on every interval of real numbers, and if a and b are real numbers with a < b, then a

∞

b

∞

a. ∫−∞ f ( x ) dx and ∫ a f ( x ) dx both converge if and only if ∫−∞ f ( x ) dx and ∫ b f ( x ) dx both converge. b.

a ∫−∞

∞ ∫a

b ∫−∞

f ( x ) dx + f ( x ) dx = f ( x ) dx + when the integrals involved converge.

∞ ∫b

integral converge? What is the value of the integral when it does converge? Plot the integrand for various values of p. 99. 101.

f ( x ) dx

COMPUTER EXPLORATIONS

e

∫0 x p ln x dx ∞

∫0

x p ln x dx

∞

100.

∫e

102.

∫−∞ x p ln

x p ln x dx

∞

x dx

Use a CAS to evaluate the integrals. 103.

In Exercises 99–102, use a CAS to explore the integrals for various values of p (include noninteger values). For what values of p does the

CHAPTER 8

531

2π

∫0

1 sin dx x

104.

2π

∫0

1 x sin dx x

Questions to Guide Your Review

1. What is the formula for integration by parts? Where does it come from? Why might you want to use it? 2. When applying the formula for integration by parts, how do you choose the u and d υ ? How can you apply integration by parts to an integral of the form ∫ f ( x ) dx ? 3. If an integrand is a product of the form sin n x cos m x , where m and n are nonnegative integers, how do you evaluate the integral? Give a specific example of each case.

12. What is an improper integral of Type I? Type II? How are the values of various types of improper integrals defined? Give examples. 13. What tests are available for determining the convergence and divergence of improper integrals that cannot be evaluated directly? Give examples of their use. 14. What is a random variable? What is a continuous random variable? Give some specific examples.

4. What substitutions are made to evaluate integrals of sin mx sin nx, sin mx cos nx, and cos mx cos nx? Give an example of each case.

15. What is a probability density function? What is the probability that a continuous random variable has a value in the interval [ c, d ]?

5. What substitutions are sometimes used to transform integrals involving a 2 − x 2 ,   a 2 + x 2 , and x 2 − a 2 into integrals that can be evaluated directly? Give an example of each case.

16. What is an exponentially decreasing probability density function? What are some typical events that might be modeled by this distribution? What do we mean when we say such distributions are memoryless?

6. What restrictions can you place on the variables involved in the three basic trigonometric substitutions to make sure the substitutions are reversible (have inverses)? 7. What is the goal of the method of partial fractions? 8. When the degree of a polynomial f ( x ) is less than the degree of a polynomial g( x ), how do you write f ( x ) g( x ) as a sum of partial fractions if g( x ) a. is a product of distinct linear factors? b. consists of a repeated linear factor? c. contains an irreducible quadratic factor? What do you do if the degree of f is not less than the degree of g? 9. How are integral tables typically used? What do you do if a particular integral you want to evaluate is not listed in the table? 10. What is a reduction formula? How are reduction formulas used? Give an example. 11. How would you compare the relative merits of the Midpoint Rule, the Trapezoidal Rule, and Simpson’s Rule?

17. What is the expected value of a continuous random variable? What is the expected value of an exponentially distributed random variable? 18. What is the median of a continuous random variable? What is the median of an exponential distribution? 19. What does the variance of a random variable measure? What is the standard deviation of a continuous random variable X? 20. What probability density function describes the normal distribution? What are some examples typically modeled by a normal distribution? How do we usually calculate probabilities for a normal distribution? 21. In a normal distribution, what percentage of the population lies within 1 standard deviation of the mean? Within 2 standard deviations?

532

Chapter 8 Techniques of Integration

CHAPTER 8

Practice Exercises

Integration by Parts Evaluate the integrals in Exercises 1–8 using integration by parts.

Trigonometric Integrals Evaluate the integrals in Exercises 37–44.

1.

∫ ln( x + 1) dx

2.

∫ x 2 ln x dx

37.

∫ sin 3 x cos 4 x dx

38.

∫ cos 5 x sin 5 x dx

3.

∫ arctan 3x dx

4.

∫ cos −1 ( 2 ) dx x

39.

∫ tan 4 x sec 2 x dx

40.

∫ tan 3 x sec 3 x dx

5.

∫ ( x + 1)

6.

∫ x 2 sin(1 − x ) dx

41.

∫ sin 5θ cos 6θ d θ

42.

∫ sec 2 θ sin 3 θ d θ

∫

∫

44.

8.

∫

43.

7.

∫ et

2

e x dx

e x cos 2 x dx

x sin x cos x dx

∫

x +1 dx x 2 ( x − 1)

14.

∫

cos θ dθ sin 2 θ + sin θ − 6

16.

∫

4 x dx x 3 + 4x

∫

dx x ( x + 1) 2

12.

13.

∫

sin θ dθ cos 2 θ + cos θ − 2

15.

∫

3x 2 + 4 x + 4 dx x3 + x

17.

∫

υ+3 dυ 2υ 3 − 8υ

19.

∫ t 4 + 4t 2 + 3

21.

∫

x3 + x2 dx x2 + x − 2

23.

∫

x2

25.

∫

dx x (3 x + 1)

27.

∫ es − 1

dt

x 3 + 4x 2 dx + 4x + 3

ds

( 3υ

18.

∫

− 7 ) dυ ( υ − 1)( υ − 2 )( υ − 3 )

20.

∫

t dt t4 − t2 − 2

22.

∫

x3 + 1 dx x3 − x

24.

∫

2 x 3 + x 2 − 21x + 24 dx x 2 + 2x − 8

26.

∫

dx x (1 + 3 x )

28.

∫

ds es + 1

Trigonometric Substitutions Evaluate the integrals in Exercises 29–32 (a) without using a trigonometric substitution, (b) using a trigonometric substitution. y dy x dx 29. ∫ 30. ∫ 16 − y 2 4 + x2

31.

x dx

∫ 4 − x2

32.

∫

t dt 4t 2 − 1

Evaluate the integrals in Exercises 33–36. x dx dx 33. ∫ 34. ∫ x(9 − x 2 ) 9 − x2 35.

dx

∫ 9 − x2

36.

∫

tan 2 e t + 1 dt

Numerical Integration

Partial Fractions Evaluate the integrals in Exercises 9–28. It may be necessary to use a substitution first. x dx x dx 9. ∫ 2 10. ∫ 2 x − 3x + 2 x + 4x + 3

11.

1 + cos( t 2 ) dt

dx 9 − x2

45. According to the error-bound formula for Simpson’s Rule, how many subintervals should you use to be sure of estimating the value of ln 3 =

∫1

3

1 dx x

by Simpson’s Rule with an error of no more than 10 −4 in absolute value? (Remember that for Simpson’s Rule, the number of subintervals has to be even.) 46. A brief calculation shows that if 0 ≤ x ≤ 1, then the second derivative of f ( x ) = 1 + x 4 lies between 0 and 8. Based on this, about how many subdivisions would you need to estimate the integral of f from 0 to 1 with an error no greater than 10 −3 in absolute value using the Trapezoidal Rule? 47. A direct calculation shows that π

∫0

2 sin 2 x dx = π.

How close do you come to this value by using the Trapezoidal Rule with n = 6? The Midpoint Rule with n = 6? Simpson’s Rule with n = 6? Try them and find out. 48. You are planning to use Simpson’s Rule to estimate the value of the integral

∫1

2

f ( x ) dx

with an error magnitude less than 10 −5. You have determined that f (4) ( x ) ≤ 3 throughout the interval of integration. How many subintervals should you use to ensure the required accuracy? (Remember that for Simpson’s Rule, the number has to be even.) T 49. Mean temperature Compute the average value of the temperature function 2π ( f ( x ) = 20 sin x − 101) − 4 365

(

)

for a 365-day year. This is one way to estimate the annual mean air temperature in Fairbanks, Alaska. The National Weather Service’s official figure, a numerical average of the daily normal mean air temperatures for the year, is −3.5  °C, which is slightly higher than the average value of f ( x ). 50. Heat capacity of a gas Heat capacity C υ is the amount of heat required to raise the temperature of a given mass of gas with constant volume by 1  °C, measured in units of cal deg-mol (calories per degree gram molecular weight). The heat capacity of oxygen depends on its temperature T and satisfies the formula C υ = 8.27 + 10 −5 ( 26T − 1.87T 2 ).

Chapter 8  Practice Exercises

Use Simpson’s Rule to find the average value of C υ and the temperature at which it is attained for 20  °C ≤ T ≤ 675  °C. 51. Fuel efficiency An automobile computer gives a digital readout of fuel consumption in liters per hour. During a trip, a passenger recorded the fuel consumption every 5 min for a full hour of travel. Time

L/h

Time

L/h

0

2.5

35

2.5

5

2.4

40

2.4

10

2.3

45

2.3

15

2.4

50

2.4

20

2.4

55

2.4

25

2.5

60

2.3

30

2.6

533

Which of the improper integrals in Exercises 63–68 converge and which diverge? dθ θ2 + 1

∞

63.

∫6

65.

∫1

67.

∫−∞ e x

∞

ln z dz z

∞

2 dx + e −x

∞

64.

∫0

66.

∫1

68.

∫−∞ x 2 (1 + e x )

e −u cos u du

∞

e −t dt t

∞

dx

Assorted Integrations Evaluate the integrals in Exercises 69–134. The integrals are listed in random order so you need to decide which integration technique to use. 1

69.

∫ x e 2 x dx

70.

∫0

71.

∫ ( tan 2 x + sec 2 x ) dx

72.

∫0

b. If the automobile covered 60 km in the hour, what was its fuel efficiency (in kilometers per liter) for that portion of the trip?

73.

∫ x sec 2 x dx

74.

∫ x sec 2 ( x 2 ) dx

52. A new parking lot To meet the demand for parking, your town has allocated the area shown here. As the town engineer, you have been asked by the town council to find out if the lot can be built for $11,000. The cost to clear the land will be $1.00 a square meter, and the lot will cost $20.00 a square meter to pave. Use Simpson’s Rule to find out if the job can be done for $11,000.

75.

∫ sin x cos 2 x dx

76.

∫ sin 2 x sin( cos 2 x ) dx

77.

∫−1 e x

78.

∫ (e 2x

79.

∫

x +1 dx x4 − x3

80.

∫ e x (e 2x

81.

∫

e x +  e 3 x dx e 2x

82.

∫ ( e x − e −x )( e x + e −x )

83.

∫0

84.

∫ tan 4 x sec 4 x dx

85.

∫0 ( x + 2 )

86.

∫ ( x + 1)

87.

∫ cot x csc 3 x dx

88.

∫ sin x ( tan x − cot x )

89.

∫1+

90.

∫

91.

∫

92.

∫

93.

∫

94.

∫ sin 2 θ cos 5 θ d θ

95.

∫ 81 − υ 4

96.

∫2

3υ − 1 dυ 4υ 3 − υ 2

97.

∫ θ cos( 2θ + 1) d θ

98.

∫

x e 3 x dx

99.

∫ (1 + cos 2θ )2

100.

∫π 4

x dx 2−x

102.

∫

a. Use the Trapezoidal Rule to approximate the total fuel consumption during the hour.

0m 12 m 18 m 17 m

0

π3

16.5 m Vertical spacing = 5 m

18 m 21 m

ex dx + e −x

tan 3 x sec 2 x dx

3

x + 1 dx

x 2 e x 3 dx

π4

cos 2 2 x dx

+ e −x ) 2 dx

ex + 1 dx − 4)

Ignored

Improper Integrals Evaluate the improper integrals in Exercises 53–62.

53. 55. 57.

3

dx 9 − x2

54.

∫0

2

dy ( y − 1) 2 3

56.

dθ ∫−2 ( θ + 1)3 5

∫0

∫0

∞

∫3

∞

ln x dx

0

∞

2 du u 2 − 2u

58.

x 2 e −x dx

60.

∫−∞

62.

∫−∞ x 2 + 16

59.

∫0

61.

∫−∞ 4 x 2 + 9

∞

1

dx

∫1

0

∞

4 dx

101.

x dx x

2 x − x 2 dx 2 − cos x + sin x dx sin 2 x 9 dυ

sin 2θ dθ

∫

x3 + 2 dx 4 − x2 dx −2 x − x 2

∞

(x

x2 π2

dx

x 2 + 2 x dx

22 m 14 m

3

dx − 1) 2

x 3 dx − 2x + 1 1 + cos 4 x dx

1 − υ2 dυ υ2

2

dx

534

Chapter 8 Techniques of Integration

dy y 2 − 2y + 2

103.

∫

105. 107.

∫

109.

∫ e 2t

111.

∫1

113. 115. 117.

x dx 8 − 2x 2 − x 4

104.

∫

∫ z 2 ( z 2 + 4 ) dz

106.

∫ x 2 ( x − 1)

t dt 9 − 4t 2

108.

∫

110.

∫ tan 3 t dt

112.

∫ y 3 2 ( ln y )

z +1

∞

e t dt + 3e t + 2

ln y dy y3

∫ e ln

x

dx

114.

sin 5t dt

∫ 1 + ( cos 5t )2

116.

dr

∫1+

118.

r

x3

119.

∫ 1 + x 2 dx

121.

1 + x2 dx 1 + x3

∫

120. 122.

CHAPTER 8

13

dx

arctan x dx x2

∫ eθ

2

− 20 x dx x 4 − 10 x 2 + 9 x2

∫ 1 + x 3 dx ∫

1 + x2 dx (1 + x ) 3

∫ ( arcsin x )

2.

∫

3.

∫ x arcsin x dx

4.

∫ sin −1

5.

∫t−

dt 1 − t2

6.

∫

dx

y dy

dx x4 + 4

Evaluate the limits in Exercise 7 and 8. x

∫−x sin t dt

8. lim+ x ∫ x→0

1 x

cos t dt t2

Evaluate the limits in Exercise 9 and 10 by identifying them with definite integrals and evaluating the integrals. n −1

n

∑ ln n 1 + nk n →∞

∑ n →∞

9. lim

10. lim

k =1

k=0

Applications 11. Finding arc length

y =

1 dx x ⋅ 1+ x

127.

∫

129.

126.

∫0

ln x dx x + x ln x

128.

∫

∫

x ln x ln x dx x

130.

∫ ( ln x )

131.

∫

1 dx x 1 − x4

132.

∫

1− x dx x

133.

∫ 1 + sin 2 x dx

134.

∫

1 − cos x dx 1 + cos x

sin 2 x

135. Evaluate ∫

π2 0

12

1 + x dx

1+

1 − x 2 dx

1 dx x ⋅ ln x ⋅ ln ( ln x ) ln x

⎡ 1 ln ( ln x ) ⎤ ⎢ + ⎥ dx ⎢⎣ x x ⎦⎥

sin x dx in two ways: sin x + cos x

a. By evaluating ∫ Theorem.

1+

sin x dx , then using the Evaluation sin x + cos x a

b. By showing that ∫ f ( x ) dx = 0 this result.

a

∫0

f ( a − x ) dx , then using

14. Finding volume The region in the first quadrant that is enclosed by the x-axis, the curve y = 5 ( x 5 − x ), and the lines x = 1 and x = 4 is revolved about the x-axis to generate a solid. Find the volume of the solid.

dx x ( x + 1)( x + 2 )( x + m )

x →∞

∫

∫

13. Finding volume The region in the first quadrant that is enclosed by the x-axis and the curve y = 3 x 1 − x is revolved about the y-axis to generate a solid. Find the volume of the solid.

1.

7. lim

125.

124.

x dx

Additional and Advanced Exercises

Evaluating Integrals Evaluate the integrals in Exercises 1–6. 2

x ⋅ 1+

3 + 4e θ d θ

4x 3

∫

∫

dy

dυ e 2υ − 1

∫

123.

1 n2 − k2

Find the length of the curve x

∫0

cos 2t dt , 0 ≤ x ≤ π 4.

12. Finding arc length Find the length of the graph of the function y = ln (1 − x 2 ),  0 ≤ x ≤ 1 2.

15. Finding volume The region in the first quadrant enclosed by the coordinate axes, the curve y = e x , and the line x = 1 is revolved about the y-axis to generate a solid. Find the volume of the solid. 16. Finding volume The region in the first quadrant that is bounded above by the curve y = e x − 1, below by the x-axis, and on the right by the line x = ln 2 is revolved about the line x = ln 2 to generate a solid. Find the volume of the solid. 17. Finding volume Let R be the “triangular” region in the first quadrant that is bounded above by the line y = 1, below by the curve y = ln x , and on the left by the line x = 1. Find the volume of the solid generated by revolving R about a. the x-axis.

b. the line y = 1.

18. Finding volume (Continuation of Exercise 17.) Find the volume of the solid generated by revolving the region R about a. the y-axis.

b. the line x = 1.

19. Finding volume The region between the x-axis and the curve ⎪⎧ 0, x = 0 y = f ( x ) = ⎪⎨ ⎪⎪ x ln x , 0 < x ≤ 2 ⎩

535

Chapter 8  Additional and Advanced Exercises

is revolved about the x-axis to generate the solid shown here.

solid generated by revolving the region about one of the coordinate axes is finite.

a. Show that f is continuous at x = 0.

31. Integrating the square of the derivative If f is continuously differentiable on [ 0, 1 ], and f (1) = f (0) = −1 6, prove that

b. Find the volume of the solid. y

y = x ln x

0

1

1

∫0 x

2

( f ′( x )) 2 dx ≥ 2 ∫

1 0

f ( x ) dx +

1 . 4

∫0 ( f ′( x ) + x − 2 ) 1

Hint: Consider the inequality 0 ≤

1

2

dx.

Source: Mathematics Magazine, vol. 84, no. 4, Oct. 2011. 20. Finding volume The infinite region bounded by the coordinate axes and the curve y = − ln x in the first quadrant is revolved about the x-axis to generate a solid. Find the volume of the solid.

32. (Continuation of Exercise 31.) If f is continuously differentiable on [ 0, a ] for a > 0, and f (a) = f (0) = b, prove that a

21. Centroid of a region Find the centroid of the region in the first quadrant that is bounded below by the x-axis, above by the curve y = ln x , and on the right by the line x = e. 22. Centroid of a region Find the centroid of the region in the plane enclosed by the curves y = ±(1 − x 2 )−1 2 and the lines x = 0 and x = 1.

≥ 2 ∫ f ( x ) dx − 2ab + 0

25. The surface generated by an astroid The graph of the equation x 2 3 + y 2 3 = 1 is an astroid (see accompanying figure). Find the area of the surface generated by revolving the curve about the x-axis.

a

∫0

Hint: Consider the inequality 0 ≤

)

a3 . 12

( f ′( x) + x − a2 )

2

dx.

Source: Mathematics Magazine, vol. 84, no. 4, Oct. 2011. The Substitution z = tan ( x 2 ) The substitution

23. Length of a curve Find the length of the curve y = ln x from x = 1 to x = e. 24. Finding surface area Find the area of the surface generated by revolving the curve in Exercise 23 about the y-axis.

(

a

∫0 ( f ′( x )) 2 dx

z = tan

x 2

(1)

reduces the problem of integrating a rational expression in sin x and cos x to a problem of integrating a rational function of z. This in turn can be integrated by partial fractions. From the accompanying figure P(cos x, sin x)

y 1

1 x 23

+

y 23

x 2

=1 A

-1

0

1

x

t − 1 dt ,

tan

1 ≤ x ≤ 16.

∫1

(

cos x = 2 cos 2

)

ax 1 − dx x 2 + 1 2x

=

converge? Evaluate the corresponding integral(s). ∞

28. For each x > 0, let G ( x ) = ∫ 0 e −xt dt.  Prove that xG ( x ) = 1 for each x > 0. 29. Infinite area and finite volume What values of p have the following property? The area of the region between the curve y = x − p ,  1 ≤ x < ∞, and the x-axis is infinite but the volume of the solid generated by revolving the region about the x-axis is finite. 30. Infinite area and finite volume What values of p have the following property? The area of the region in the first quadrant enclosed by the curve y = x − p , the y-axis, the line x = 1, and the interval [ 0, 1 ] on the x-axis is infinite, but the volume of the

sin x x = . 2 1 + cos x

To see the effect of the substitution, we calculate

27. For what value or values of a does ∞

0 cos x

we can read the relation

26. Length of a curve Find the length of the curve

∫1

sin x

x

-1

y =

1

x

cos x =

( 2x ) − 1 = sec 2( x 2 ) − 1 2

2 2 −1 = −1 1 + tan 2 ( x 2 ) 1 + z2 1 − z2 , 1 + z2

(2)

and sin ( x 2 ) x x x sin x = 2 sin cos = 2 ⋅ cos 2 2 2 cos( x 2 ) 2

( )

= 2 tan sin x =

2 tan ( x 2 ) 1 x ⋅ = 2 sec 2 ( x 2 ) 1 + tan 2 ( x 2 )

2z . 1 + z2

(3)

536

Chapter 8 Techniques of Integration

y

Finally, x  2 arctan z , so 2 dz . 1 + z2

dx =

(4)

3

1

∫ 1 + cos x dx = ∫

a.

=

1 + z 2 2  dz 2 1 + z2

∫ dz

= tan

∫

b.

1 dx = 2 + sin x =

∫

1 -3

= z+C

( )

dz

dx 1 + sin x

36.

∫Q 3

π2

dθ 2 + cos θ

38.

∫π 2

37.

∫0

39.

∫

∫

dx 1 + sin x + cos x

Q2

∫0

dt sin t − cos t

40.

Q2

¨ sec R d R

42.

2π 3

∫

cos θ dθ sin θ cos θ + sin θ

cos t dt 1 − cos t

¨ csc R d R

The Gamma Function and Stirling’s Formula Euler’s gamma function (( x ) (“gamma of x”; ( is a Greek capital g) uses an integral to extend the factorial function from the nonnegative integers to other real values. The formula is

Γ( x ) =

∞

∫0

3

x

FIGURE 8.21 Euler’s gamma function (( x ) is a continuous function of x whose value at each positive integer n 1 is n!. The defining integral formula for ( is valid only for x  0, but we can extend ( to negative noninteger values of x with the formula Γ( x ) = ( Γ( x + 1)) x , which is the subject of Exercise 43.

43. If n is a nonnegative integer, then Γ( n + 1 ) = n! a. Show that Γ(1) = 1. b. Then apply integration by parts to the integral for Γ( x + 1) to show that Γ( x + 1) = xΓ( x ). This gives Γ(2) = 1Γ(1) = 1 Γ(3) = 2Γ(2) = 2 Γ(4) = 3Γ(3) = 6 # Γ ( n + 1) = n Γ ( n ) = n !

dx 1 − cos x

Use the substitution z = tan ( R 2 ) to evaluate the integrals in Exercises 41 and 42. 41.

2

dz

Use the substitutions in Equations (1)–(4) to evaluate the integrals in Exercises 33–40. Integrals like these arise in calculating the average angular velocity of the output shaft of a universal joint when the input and output shafts are not aligned. 34.

1

-3

∫ z 2 + z + 1 = ∫ ( z + (1 2 )) 2 + 3 4

dx 1 − sin x

0

-2

2 dz 1 + z2 2 + 2z + 2z 2 1 + z 2

( )

35.

-1

-1

du ∫ u2 + a2 u 1 = arctan +C a a 2 2z + 1 = +C arctan 3 3 1 + 2 tan ( x 2 ) 2 = +C arctan 3 3

∫

-2

x +C 2

=

33.

y = ≠(x)

2

Examples

t x −1e −t dt , x > 0.

For each positive x, the number (( x ) is the integral of t x 1e t with respect to t from 0 to d. Figure 8.21 shows the graph of ( near the origin. You will see how to calculate Γ(1 2 ) if you do Additional Exercise 23 in Chapter 14.

(1)

c. Use mathematical induction to verify Equation (1) for every nonnegative integer n. 44. Stirling’s formula Scottish mathematician James Stirling (1692–1770) showed that lim

x →∞

( ex )

x

x Γ( x ) = 1, 2Q

so, for large x, Γ( x ) =

( ex )

x

2π ( 1 + ε( x ) ) x

ε( x ) → 0 as  x → ∞.

(2)

Dropping F( x ) leads to the approximation Γ( x ) ≈

( ex )

x

2Q x

( Stirling’s   formula ).

(3)

a. Stirling’s approximation for n! Use Equation (3) and the fact that n! = nΓ(n) to show that n! ≈

( ne )

n

2 nQ

( Stirling’s   approximation ).

(4)

Chapter 8  Technology Application Projects

As you will see if you do Exercise 114 in Section 9.1, Equation (4) leads to the approximation n

n n! x . e

T c. A refinement of Equation (2) gives Γ( x ) =

CHAPTER 8

( ex )

x

or Γ( x ) ≈

(5)

T b. Compare your calculator’s value for n! with the value given by Stirling’s approximation for n  10, 20, 30, !, as far as your calculator can go.

537

( ex )

x

2Q 1 (12 x ) e , x

which tells us that n! ≈

( ne )

n

2nQ e 1 (12 n ).

(6)

Compare the values given for 10! by your calculator, Stirling’s approximation, and Equation (6).

2π 1 (12 x ) (  e 1 + ε( x ) ) x

Technology Application Projects

Mathematica/Maple Projects

Projects can be found within MyLab Math. • Riemann, Trapezoidal, and Simpson Approximations Part I: Visualize the error involved in using Riemann sums to approximate the area under a curve. Part II: Build a table of values and compute the relative magnitude of the error as a function of the step size %x. Part III: Investigate the effect of the derivative function on the error. Parts IV and V: Trapezoidal Rule approximations. Part VI: Simpson’s Rule approximations. • Games of Chance: Exploring the Monte Carlo Probabilistic Technique for Numerical Integration Graphically explore the Monte Carlo method for approximating definite integrals. • Computing Probabilities with Improper Integrals More explorations of the Monte Carlo method for approximating definite integrals.

9

Infinite Sequences and Series OVERVIEW In this chapter we introduce the topic of infinite series. Such series give us precise ways to express many numbers and functions, both familiar and new, as arithmetic sums with infinitely many terms. For example, we will learn that

Q 1 1 1 1 = 1− + − + −" 3 5 7 9 4 and cos x = 1 −

x2 x4 x6 x8 + − + − ". 2 24 720 40,320

We need to develop a method to make sense of such expressions. Everyone knows how to add two numbers together, or even several. But how do you add together infinitely many numbers? Or, when adding together functions, how do you add infinitely many powers of x? In this chapter we answer these questions, which are part of the theory of infinite sequences and series. As with the differential and integral calculus, limits play a major role in the development of infinite series. One common and important application of series occurs in making computations with complicated functions. A hard-to-compute function is replaced by an expression that looks like an “infinite degree polynomial,” an infinite series in powers of x, as we see with the cosine function given above. Using the first few terms of this infinite series can allow for highly accurate approximations of functions by polynomials, enabling us to work with more general functions than those we have encountered before. These new functions are commonly obtained as solutions to differential equations arising in important applications of mathematics to science and engineering. The terms “sequence” and “series” are sometimes used interchangeably in spoken language. In mathematics, however, each has a distinct meaning. A sequence is a type of infinite list, whereas a series is an infinite sum. To understand the infinite sums described by series, we first must understand infinite sequences.

9.1 Sequences HISTORICAL ESSAY Sequences and Series To read this essay, visit the companion Website.

538

Sequences are fundamental to the study of infinite series and to many aspects of mathematics. We saw one example of a sequence when we studied Newton’s Method in Section 4.7. Newton’s Method produces a sequence of approximations x n that become closer and closer to the root of a differentiable function. Now we will explore general sequences of numbers and the conditions under which they converge to a finite number.

9.1  Sequences

539

Representing Sequences A sequence is a list of numbers a 1 , a 2 , a 3 , …, a n , … in a given order. Each of a1 , a 2 , a 3 , and so on represents a number. These are the terms of the sequence. For example, the sequence 2, 4, 6, 8, 10, 12, …, 2n, … has first term a1 = 2, second term a 2 = 4, and nth term a n = 2n. The integer n is called the index of a n and indicates where a n occurs in the list. Order is important. The sequence 2, 4, 6, 8 …  is not the same as the sequence 4, 2, 6, 8 …. We can think of the sequence a 1 , a 2 , a 3 , …, a n , … as a function that sends 1 to a1 , 2 to a 2 , 3 to a 3 , and in general sends the positive integer n to the nth term a n . More precisely, an infinite sequence of numbers is a function whose domain is the set of positive integers. For example, the function associated with the sequence 2, 4, 6, 8, 10, 12, …, 2n, … sends 1 to a1 = 2, 2 to a 2 = 4, and so on. The general behavior of this sequence is described by the formula a n = 2n. We can change the index to start at any given number n. For example, the sequence 12, 14, 16, 18, 20, 22 … is described by the formula a n = 10 + 2n, if we start with n = 1. It can also be described by the simpler formula b n = 2n, where the index n starts at 6 and increases. To allow such simpler formulas, we let the first index of the sequence be any appropriate integer. In the sequence above, { a n } starts with a1 while { b n } starts with b 6 . Sequences can be described by writing rules that specify their terms, such as an =

n,

1 b n = (−1) n +1 , n

cn =

n −1 , n

d n = (−1) n +1,

or by listing terms: {an } =

{ 1, 2, 3, …, n , …}

{

1 1 2 3

1 4

1 n

}

{ b n } = 1, − , , − , …, ( −1) n +1 , …

{

1 2 3 4 2 3 4 5

{ c n } = 0, , , , , …,

}

n −1 ,… n

{ d n } = {1, −1, 1, −1, 1, −1, …, ( −1) n +1, …}.

We also sometimes write a sequence using its rule, as with {an } =

∞

{ n } n =1

and

{

{ b n } = ( − 1) n + 1

1 n

}

∞

.

n =1

Figure 9.1 shows two ways to represent sequences graphically. The first marks the first few points from a1 , a 2 , a 3 , …, a n , …  on the real axis. The second method shows the graph of the function defining the sequence. The function is defined only on integer inputs, and the graph consists of some points in the xy-plane located at (1, a1 ) , ( 2, a 2 ) , …, ( n, a n ) , ….

540

Chapter 9 Infinite Sequences and Series

an a1 1

0

3 2 1

a2 a3 a4 a5 2 an = " n

a3 a 2

0 an

a1

0

n

1 2 3 4 5

1

1 1 an = n

0

1

2

3

4

5

n

an a2 a4

a5 a3 0

a1

1

1 an = (−1) n+1 1n

n

0

FIGURE 9.1 Sequences can be represented as points on the real line or as points in the plane where the horizontal axis n is the index number of the term and the vertical axis a n is its value. L−e 0

a2 a3

aN

a1

L L+e

an

Sometimes the numbers in a sequence approach a single value as the index n increases. This happens in the sequence

an L+e (n, an)

L

(N, aN)

0

1 2 3

N

n

Convergence and Divergence

{1, 12 , 13 , 14 , …, 1n , …}, whose terms approach 0 as n gets large, and in the sequence

L−e

n

FIGURE 9.2

In the representation of a sequence as points in the plane, a n → L if y = L is a horizontal asymptote of the sequence of points {( n, a n )}. In this figure, all the a n’s after a N lie within ε of L.

HISTORICAL BIOGRAPHY Nicole Oresme (ca. 1320–1382) Frenchman Oresme went to the University of Paris in the 1340s, studying theology and liberal arts. Later he was a faculty member and administrator at the same university. His work entitled De configurationibus (1350s) contained results in geometry and was the first to present graphs of velocities. The argument we use to show the divergence of the harmonic series was devised by Oresme in this publication. To know more, visit the companion Website.

{ 0, 12 , 23 , 43 , 45 , …, 1 − 1n , …}, whose terms approach 1. On the other hand, sequences like

{ 1, 2, 3, …, n , …} have terms that get larger than any number as n increases, and sequences like

{1, −1, 1, −1, 1, −1, …, (−1) n +1, …} bounce back and forth between 1 and −1, never converging to a single value. The following definition captures the meaning of having a sequence converge to a limiting value. It says that if we specify any number ε > 0, then by going far enough out in the sequence, taking the index n to be larger than some value N, the difference between a n and the limit of the sequence becomes less than ε. DEFINITIONS The sequence { a n } converges to the number L if for every positive number ε there corresponds an integer N such that

an − L < ε

whenever

n > N.

If no such number L exists, we say that { a n } diverges. If { a n } converges to L, we write lim a n = L , or simply a n → L, and call n →∞

L the limit of the sequence (Figure 9.2).

The definition is very similar to the definition of lim f ( x ), the limit of a function f ( x ) x →∞

as x tends to ∞, discussed in Section 2.5. We will exploit this connection to calculate limits of sequences.

9.1  Sequences

EXAMPLE 1 (a) lim

n →∞

1 = 0 n

541

Show that (b) lim k = k

( where  k  is a constant )

n →∞

Solution

(a) Let ε > 0 be given. We must show that there exists an integer N such that 1 −0 < ε n

whenever

n > N.

The inequality 1 n − 0 < ε will hold if 1 n < ε or n > 1 ε. If N is any integer greater than 1 ε , the inequality will hold for all n > N. This proves that lim 1 n = 0. n →∞

(b) Let k be a constant, and let ε > 0 be given. We must show that there exists an integer N such that k−k < ε

whenever

n > N.

Since k − k = 0, we can use any positive integer for N and the inequality k − k < ε will hold. This proves that lim k = k. n →∞

EXAMPLE 2

Show that the sequence {1, −1, 1, −1, 1, −1, …, (−1) n +1, …} diverges.

Solution Since the terms in this sequence alternate between 1 and −1, our intuition is that they cannot approach arbitrarily close to some single limit L. To prove this, suppose that there did exist a number L that is the limit of this sequence. The definition of convergence then tells us that for every number ε > 0, there is an integer N such that a n − L < ε for all n > N. No matter what positive number ε we consider, such an integer N must exist. In particular, for the number ε = 1 2 , there must exist some integer N such that a n − L < 1 2 for all n > N. If n > N is odd, then a n = 1, so we have 1 − L < 1 2, or

an

M

1 3 < L < . 2 2 N

0 12 3

n

−

(a) an

0 12 3

N

n

m

3 1 < L N is even, then a n = −1, and so we have −1 − L < 1 2, or

(a) The sequence diverges to ∞ because no matter what number M is chosen, the terms of the sequence after some index N all lie in the yellow band above M. (b) The sequence diverges to −∞ because all terms after some index N lie below any chosen number m.

lim

n →∞

n = ∞.

In writing infinity as the limit of a sequence, we are not saying that the differences between the terms a n and ∞ become small as n increases, nor are we asserting that there is some number infinity that the sequence approaches. We are merely using a notation that captures the idea that a n eventually gets and stays larger than any fixed number as n gets large (see Figure 9.3a). The terms of a sequence might also decrease to negative infinity, as in Figure 9.3b.

542

Chapter 9 Infinite Sequences and Series

The sequence { a n } diverges to infinity if for every number M there is an integer N such that for all n larger than N, a n > M . If this condition holds, then we write

DEFINITION

lim a n = ∞

n →∞

or

a n → ∞.

Similarly, if for every number m there is an integer N such that for all n > N, we have a n < m, then we say { a n } diverges to negative infinity and write lim a n = −∞

n →∞

or

a n → −∞.

A sequence may diverge without diverging to infinity or negative infinity, as we saw in Example 2. The sequences {1, − 2, 3, − 4, 5, − 6, 7, −8, …} and {1, 0, 2, 0, 3, 0, …} are also sequences that diverge but do not diverge to infinity or negative infinity. The convergence or divergence of a sequence is not affected by the values of any number of its initial terms (whether we omit or change the first 10, the first 1000, or even the first million terms does not matter). From Figure 9.2, we can see that only the part of the sequence that remains after discarding some initial number of terms determines whether the sequence has a limit and the value of that limit when it does exist.

Calculating Limits of Sequences Since sequences are functions with domain restricted to the positive integers, it is not surprising that the theorems on limits of functions given in Chapter 2 have versions for sequences. THEOREM 1 Let { a n } and { b n } be sequences of real numbers, and let A and B be real numbers. The following rules hold if lim a n = A and lim b n = B. n →∞

n →∞

lim ( a n + b n ) = A + B

1. Sum Rule:

n →∞

2. Difference Rule: 3. Constant Multiple Rule:

lim ( a n − b n ) = A − B

n →∞

lim ( k ⋅ b n ) = k ⋅ B ( any number k )

n →∞

lim ( a n ⋅ b n ) = A ⋅ B

4. Product Rule:

n →∞

5. Quotient Rule:

lim

n →∞

an A = bn B

if B ≠ 0

The proof is similar to that of Theorem 1 in Section 2.2 and is omitted. EXAMPLE 3

By combining Theorem 1 with the limits of Example 1, we have:

Constant Multiple Rule ( 1n ) = −1 ⋅ lim 1n = −1 ⋅ 0 = 0 and Example 1a Rule n −1 (b) lim ( ) = lim (1 − n1 ) = lim 1 − lim n1 = 1 − 0 = 1 Difference and Example 1a n

(a) lim − n →∞

n →∞

n →∞

n →∞

n →∞

n →∞

5 1 1 (c) lim 2 = 5 ⋅ lim ⋅ lim = 5 ⋅ 0 ⋅ 0 = 0 n →∞ n n →∞ n n →∞ n (4 n6 ) − 7 4 − 7n 6 0−7 = lim = = −7. (d) lim 6 n →∞ n + 3 n →∞ 1 + ( 3 n 6 ) 1+ 0

Product Rule Divide numerator and denominator by n 6 and use the Sum and Quotient Rules.

9.1  Sequences

543

Be cautious in applying Theorem 1. It does not say, for example, that both of the sequences { a n } and { b n } have limits if their sum { a n + b n } has a limit. For instance, { a n } = {1, 2, 3, …} and { b n } = {−1, − 2, − 3, …} both diverge, but their sum { a n + b n } = { 0, 0, 0, …} clearly converges to 0. One consequence of Theorem 1 is that every nonzero multiple of a divergent sequence { a n } diverges. Suppose, to the contrary, that { ca n } converges for some number c ≠ 0. Then, by taking k = 1 c in the Constant Multiple Rule in Theorem 1, we see that the sequence

cn L bn an

{ 1c ⋅ ca } = { a

n

0

n

FIGURE 9.4 The terms of sequence { b n } are sandwiched between those of { a n } and { c n } , forcing them to the same

common limit L.

n}

converges. Thus, { ca n } cannot converge unless { a n } also converges. If { a n } does not converge, then { ca n } does not converge. The next theorem is the sequence version of the Sandwich Theorem in Section 2.2. You are asked to prove the theorem in Exercise 119. (See Figure 9.4.) THEOREM 2—The Sandwich Theorem for Sequences Let { a n } , { b n } , and { c n } be sequences of real numbers. If a n ≤ b n ≤ c n

holds for all n beyond some index N, and if lim a n = lim c n = L , then n →∞ n →∞ lim b n = L also.

n →∞

An immediate consequence of Theorem 2 is that if b n ≤ c n and c n → 0, then b n → 0 because −c n ≤ b n ≤ c n . We use this fact in the next example. EXAMPLE 4

Since 1 n → 0, we know that

cos n → 0 n 1 (b) n → 0 2 1 (c) ( −1) n → 0 n (a)

y=

y 2

2x

(1, 2)

because because

(d) If a n → 0, then a n → 0

1 1/2 a ,2 b 2

because

cos n 1 1 ≤ ≤ ; n n n 1 1 0 ≤ n ≤ ; 2 n 1 1 1 − ≤ ( − 1) n ≤ . n n n −

− an ≤ an ≤ an .

The application of Theorems 1 and 2 is broadened by a theorem stating that applying a continuous function to a convergent sequence produces a convergent sequence. We state the theorem, leaving the proof as an exercise (Exercise 120).

1 1/3 a ,2 b 3

1

because

THEOREM 3—The Continuous Function Theorem for Sequences Let { a n } be a sequence of real numbers. If a n → L and if f is a function that is

continuous at L and defined at all a n , then f (a n ) → f ( L ).

0

1 3

1 2

1

x

As n → ∞, 1 n → 0 and 2 1 n → 2 0 (Example 6). The terms of {1 n } are shown on the x-axis; the terms of { 2 1 n } are shown as the y-values on the graph of f ( x ) = 2 x. FIGURE 9.5

EXAMPLE 5

Show that

(n

+ 1) n → 1.

Solution We know that ( n + 1) n → 1. Taking f ( x ) = gives ( n + 1) n → 1 = 1.

x and L = 1 in Theorem 3

EXAMPLE 6 The sequence {1 n } converges to 0. By taking a n = 1 n , f ( x ) = 2 x , and L = 0 in Theorem 3, we see that 2 1 n = f (1 n ) → f ( L ) = 2 0 = 1. The sequence { 2 1 n } converges to 1 (Figure 9.5).

Using L’Hôpital’s Rule The next theorem formalizes the connection between lim a n and lim f ( x ). It enables us n →∞

to use l’Hôpital’s Rule to find the limits of some sequences.

x →∞

544

Chapter 9 Infinite Sequences and Series

THEOREM 4 Suppose that f ( x ) is a function defined for all x ≥ n 0 and that { a n } is a sequence of real numbers such that a n = f (n)  for  n ≥ n 0 . Then

lim a n = L

lim f ( x ) = L.

whenever

n →∞

x →∞

Proof Suppose that lim f ( x ) = L. Then for each positive number ε, there is a x →∞ number M such that f (x) − L < ε

x > M.

whenever

Let N be an integer that is both greater than M and greater than or equal to n 0 . Since a n = f (n), it follows that for all n > N, we have a n − L = f (n) − L < ε. EXAMPLE 7

Show that lim

n →∞

ln n = 0. n

Solution The function ( ln x ) x is defined for all x ≥ 1 and agrees with the given sequence at positive integers. Therefore, by Theorem 4, lim ( ln n ) n will equal n →∞

lim ( ln x ) x if the latter exists. A single application of l’Hôpital’s Rule shows that

x →∞

ln x 1x 0 = lim = = 0. x →∞ x x →∞ 1 1 lim

We conclude that lim ( ln n ) n = 0. n →∞

Does the sequence whose nth term is

EXAMPLE 8

an =

( nn +− 11) , n

n ≥ 2

converge? If so, find lim a n . n →∞

Solution The function f ( x ) =

( xx +− 11)

x

is defined for all real numbers x ≥ 2 and x +1 x = L , then by agrees with a n at all integers n ≥ 2. If we can show that lim x →∞ x − 1 n+1 n Theorem 4, lim = L. n →∞ n − 1 The limit lim f ( x ) leads to the indeterminate form 1∞. To evaluate this limit, we

(

(

)

)

x →∞

apply l’Hôpital’s Rule after taking the natural logarithm of f ( x ): ln f ( x ) = ln Then

( xx +− 11)

x

= x ln

( xx +− 11) x +1 ln ( x − 1) lim

lim ln f ( x ) = lim x ln

x →∞

x →∞

=

x →∞

1x

− 2 ( x 2 − 1) x →∞ −1 x 2

= lim = lim

x →∞

Therefore, lim

x →∞

( xx +− 11)

x

2x 2 = 2. x2 − 1

( xx +− 11). ∞ ⋅ 0 form

0 form 0 Apply I’Hôpital’s Rule.

Simplify and evaluate.

= lim f ( x ) = lim e ln f ( x ) = e 2. Applying Theorem 4, we x →∞

x →∞

conclude that the sequence { a n } also converges to e 2.

9.1  Sequences

545

Commonly Occurring Limits The next theorem gives some limits that arise frequently. Factorial Notation The notation n! (“n factorial”) means the product 1 ⋅ 2 ⋅ 3  n of the integers from 1 to n. Notice that ( n + 1)! = ( n + 1) ⋅ n!. Thus, 4! = 1 ⋅ 2 ⋅ 3 ⋅ 4 = 24 and 5! = 1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5 = 5 ⋅ 4! = 120. We define 0! to be 1. Factorials grow even faster than exponentials, as the table suggests. The values in the table are rounded. n   1

en

The following six sequences converge to the limits listed below.

THEOREM 5

ln n = 0 n →∞ n

1. lim

2. lim

n →∞

3. lim x 1 n = 1

(x

(

= ex

n →∞

5. lim 1 + n →∞

x n

)

n

> 0)

( any x )

n

n =1

4. lim x n = 0

(

n →∞

6. lim

n →∞

xn = 0 n!

x < 1)

( any x )

In Formulas (3) through (6),  x remains fixed as n → ∞.

n! 3

1

  5

148

120

10

22,026

3,628,800

20

4.9 × 10 8

2.4 × 10 18

Proof The first limit was computed in Example 7. The next two can be proved by taking logarithms and applying Theorem 4 (Exercises 117 and 118). The remaining proofs are given in Appendix A.6. EXAMPLE 9 (a)

These are examples involving the limits in Theorem 5.

ln (n 2 ) 2 ln n = → 2⋅0 = 0 n n 2

Formula 1

(b)

n

n 2 = n 2 n = ( n 1 n ) → ( 1) 2 = 1

Formula 2

(c)

n

3n = 31 n ( n 1 n ) → 1 ⋅ 1 = 1

Formula 3 with x = 3 and Formula 2

( 12 ) → 0 n−2 (e) ( ) = (1 + −n2 ) n (d) −

n

n

(f )

100 n → 0 n!

Formula 4 with x = − n

→ e −2

1 2

Formula 5 with x = −2 Formula 6 with x = 100

Recursive Definitions So far, we have calculated each a n directly from the value of n. But sequences are often defined recursively by giving 1. The value(s) of the initial term or terms, and 2. a rule called a recursion formula for calculating any later term from terms that precede it. EXAMPLE 10 (a) The statements a1 = 1 and a n = a n −1 + 1 for n > 1 define the sequence 1, 2, 3, …, n, … of positive integers. With a1 = 1, we have a 2 = a1 + 1 = 2, a 3 = a 2 + 1 = 3, and so on. (b) The statements a1 = 1 and a n = n ⋅ a n −1 for n > 1 define the sequence 1, 2, 6, 24, …, n!, … of factorials. With a1 = 1, we have a 2 = 2 ⋅ a1 = 2, a 3 = 3 ⋅ a 2 = 6, a 4 = 4 ⋅ a 3 = 24, and so on. (c) The statements a1 = 1, a 2 = 1, and a n +1 = a n + a n −1 for n > 2 define the sequence 1, 1, 2, 3, 5, … of Fibonacci numbers. With a1 = 1 and a 2 = 1, we have a 3 = 1 + 1 = 2, a 4 = 2 + 1 = 3, a 5 = 3 + 2 = 5, and so on.

546

Chapter 9 Infinite Sequences and Series

(d) As we can see by applying Newton’s method (see Exercise 145), the statements x 0 = 1 and x n +1 = x n − [ ( sin x n − x n 2 ) ( cos x n − 2 x n ) ] for n > 0 define a sequence that, when it converges, gives a solution to the equation sin x − x 2 = 0.

Bounded Monotonic Sequences Two concepts that play a key role in determining the convergence of a sequence are those of a bounded sequence and a monotonic sequence. First we define bounded sequences.

DEFINITION A sequence { a n } is bounded from above if there exists a number M such that a n ≤ M for all n. The number M is an upper bound for { a n }. If M is an upper bound for { a n } but no number less than M is an upper bound for { a n } , then M is the least upper bound for { a n }. A sequence { a n } is bounded from below if there exists a number m such that a n ≥ m for all n. The number m is a lower bound for { a n }. If m is a lower bound for { a n } but no number greater than m is a lower bound for { a n } , then m is the greatest lower bound for { a n }. If { a n } is bounded from above and below, then { a n } is bounded. If { a n } is not bounded, then we say that { a n } is an unbounded sequence.

EXAMPLE 11 (a) The sequence 1, 2, 3, …, n, …  has no upper bound because it eventually surpasses every number M. However, it is bounded below by every real number less than or equal to 1. The number m = 1 is the greatest lower bound of the sequence. 1 2 3 n , , , …, , …  is bounded above by every real number greater 2 3 4 n+1 than or equal to 1. The upper bound M = 1 is the least upper bound (Exercise 137). 1 The sequence is also bounded below by every number less than or equal to , which is 2 its greatest lower bound.

(b) The sequence

Convergent sequences are bounded.

Suppose that a sequence { a n } converges to a number L. Applying the definition of convergence with the particular value ε = 1, there must exist a number N such that a n − L < 1 if n > N. That is, L − 1 < an < L + 1

an M

0 123

n

for n > N .

If M is a number larger than both L + 1 and all of the finitely many numbers a1 , a 2 , …, a N , then for every index n we have a n ≤ M, and therefore { a n } is bounded from above. Similarly, if m is a number smaller than both L − 1 and all of the numbers a1 , a 2 , …, a N , then m is a lower bound for the sequence. Therefore, all convergent sequences are bounded. Although it is true that every convergent sequence is bounded, there are bounded sequences that fail to converge. One example is the bounded sequence {(−1) n +1 } discussed in Example 2. The problem here is that some bounded sequences bounce around in the band determined by any lower bound m and any upper bound M but do not converge (Figure 9.6). An important type of sequence that does not behave that way is one for which each term is at least as large, or at least as small, as its predecessor.

m

FIGURE 9.6

Some bounded sequences bounce around between their bounds and fail to converge to any limiting value.

DEFINITIONS A sequence { a n } is nondecreasing if a n ≤ a n +1 for all n. That

is, a1 ≤ a 2 ≤ a 3 ≤ …. The sequence is nonincreasing if a n ≥ a n +1 for all n. The sequence { a n } is monotonic if it is either nondecreasing or nonincreasing.

9.1  Sequences

547

EXAMPLE 12 (a) The sequence 1, 2, 3, …, n, …  is nondecreasing. 1 2 3 n , …  is nondecreasing. (b) The sequence , , , …, 2 3 4 n+1 1 1 1 1 (c) The sequence 1, , , , …, n , …  is nonincreasing. 2 2 4 8 (d) The constant sequence 3, 3, 3, …, 3, …  is both nondecreasing and nonincreasing. (e) The sequence 1, −1, 1, −1, 1, −1, …  is not monotonic. A sequence that is bounded from above always has a least upper bound. Likewise, a sequence bounded from below always has a greatest lower bound. These results are based on the completeness property of the real numbers, discussed in Appendix A.9. We now prove that if L is the least upper bound of a nondecreasing sequence, then the sequence converges to L, and that if L is the greatest lower bound of a nonincreasing sequence, then the sequence converges to L.

THEOREM 6—The Monotonic Sequence Theorem If a sequence { a n } is both bounded and monotonic, then the sequence converges.

Proof Suppose { a n } is nondecreasing, L is its least upper bound, and we plot the points 1, a , 2, a ( 1) ( 2 ) , …, ( n, a n ) , …  in the xy-plane. If M is an upper bound of the sequence, all these points will lie on or below the line y = M (Figure 9.7). The line y = L is the lowest such line. None of the points ( n, a n ) lies above y = L, but some do lie above any lower line y = L − ε, if ε is a positive number (because L − ε is not an upper bound). That is,

y y=M

M

y=L

L

a. a n ≤ L for all values of n, and b. given any ε > 0, there exists at least one integer N for which a N > L − ε.

L−e

The fact that { a n } is nondecreasing tells us further that an ≥ a N > L − ε

x

N

0

FIGURE 9.7

If the terms of a nondecreasing sequence have an upper bound M, then they have a limit L ≤ M.

for all  n ≥ N .

Thus, all the numbers a n beyond the Nth number lie within ε of L. This is precisely the condition for L to be the limit of the sequence { a n }. The proof for nonincreasing sequences bounded from below is similar.

It is important to realize that Theorem 6 does not say that convergent sequences are monotonic. The sequence {(−1) n +1 n } converges and is bounded, but it is not monotonic since it alternates between positive and negative values as it tends toward zero. What the theorem does say is that a nondecreasing sequence converges when it is bounded from above, but it diverges to infinity otherwise.

EXERCISES

9.1

Finding Terms of a Sequence

Each of Exercises 1–6 gives a formula for the nth term a n of a sequence { a n }. Find the values of a1 , a 2 , a 3 , and a 4 . 1− n 1. a n = n2 3. a n =

( − 1) n +1

2n − 1

1 2. a n = n! 4. a n = 2 + (−1) n

5. a n =

2n 2 n +1

6. a n =

2n − 1 2n

Each of Exercises 7–12 gives the first term or two of a sequence along with a recursion formula for the remaining terms. Write out the first ten terms of the sequence. 7. a1 = 1, a n +1 = a n + (1 2 n ) 8. a1 = 1, a n +1 = a n ( n + 1)

548

Chapter 9 Infinite Sequences and Series

9. a1 = 2, a n +1 = (−1) n +1 a n 2 10. a1 = −2, a n +1 =

na n ( n

+

41. a n =

1)

11. a1 = a 2 = 1, a n + 2 = a n +1 + a n

43. a n =

12. a1 = 2, a 2 = −1, a n + 2 = a n +1 a n

( n 2+n 1 )(1 − n1 ) ( − 1)

13. 1, −1, 1, −1, 1, …

1’s with alternating signs

14. −1, 1, −1, 1, −1, …

1’s with alternating signs

15. 1, − 4, 9, −16, 25, …

Squares of the positive integers, with alternating signs Reciprocals of squares of the positive integers, with alternating signs Powers of 2 divided by multiples of 3

1 1 1 1 16. 1, − , , − , , … 4 9 16 25 17.

1 2 22 23 24 , , , , ,… 9 12 15 18 21

Integers differing by 2 divided by products of consecutive integers Squares of the positive integers diminished by 1 Integers, beginning with −3

3 1 1 3 5 18. − , − , , , , … 2 6 12 20 30 19. 0, 3, 8, 15, 24, … 20. −3, − 2, −1, 0, 1, … 21. 1, 5, 9, 13, 17, …

23.

5 8 11 14 17 , , , , ,… 1 2 6 24 120

Every other odd positive integer Every other even positive integer Integers differing by 3 divided by factorials

24.

1 8 27 64 125 , , , , ,… 25 125 625 3125 15,625

Cubes of positive integers divided by powers of 5

22. 2, 6, 10, 14, 18, …

25. 1, 0, 1, 0, 1, …

Alternating 1’s and 0’s

26. 0, 1, 1, 2, 2, 3, 3, 4, … 27. 28.

1 1 1 1 1 1 1 1 − , − , − , − ,… 2 3 3 4 4 5 5 6 5−

4, 6 −

5, 7 −

Each positive integer repeated

6, 8 −

( 12 )

n

46. a n =

( π2 + 1n )

)( 3 + 21 ) n

1 ( 0.9 ) n

48. a n = nπ cos ( nπ ) sin 2 n 2n

49. a n =

sin n n

50. a n =

51. a n =

n 2n

53. a n =

ln ( n + 1) n

3n n3 ln n 54. a n = ln 2n 52. a n =

55. a n = 8 1 n

56. a n = ( 0.03 )1 n

(

)

7 n

57. a n = 1 +

(

n

58. a n = 1 − 60. a n =

1 n

)

n

59. a n =

n

61. a n =

( n3 )

62. a n = ( n + 4 )1 ( n + 4 )

63. a n =

ln n n1 n

64. a n = ln n − ln ( n + 1)

68. a n =

10 n 1n

65. a n =

n

67. a n =

n! (Hint: Compare with 1 n .) nn

70. a n = 72. a n = 74. a n =

7, …

⎛ 2 ⎞⎟ ⎛ 3 ⎞⎟ ⎛ ⎛ 4 ⎞⎟ 5 ⎞⎟ 29. sin ⎜⎜ , sin ⎜⎜ , sin ⎜⎜ , sin ⎜⎜ ,… ⎝ 1 + 4 ⎟⎟⎠ ⎝ 1 + 9 ⎟⎟⎠ ⎝ 1 + 16 ⎟⎟⎠ ⎝ 1 + 25 ⎟⎟⎠ 30.

47. a n = sin

1 2n

44. a n = −

2n n+1

Finding a Sequence’s Formula

In Exercises 13–30, find a formula for the nth term of the sequence.

n +1

2n − 1

45. a n =

(

42. a n = 2 −

66. a n =

4nn

(−4 ) n

n

n

n2

3 2 n +1

69. a n =

n! 10 6 n

n! ⋅ 3n

71. a n =

( 1n )

+ 1)! + 3 )!

73. a n =

3e n + e −n e n + 3e −n

75. a n =

n! 2n (n (n

(

76. a n = 1 −

5 7 9 11 , , ,… , 8 11 14 17

) ( (

1 ( ln n )

( 2n ( 2n

+ 2 )! − 1)!

e −2 n − 2e −3n e − 2 n − e −n

) ( 13 − 14 ) +  1 1 1 + − − 1) ( n − 1 n )

1 1 1 + − + 2 2 3 1 + − n−2 n

77. a n = ( ln 3 − ln 2 ) + ( ln 4 − ln 3 ) + ( ln 5 − ln 4 ) +  Convergence and Divergence Which of the sequences { a n } in Exercises 31–100 converge, and which

1 − 2n 1 + 2n 1 − 5n 4 35. a n = 4 n + 8n 3

n + (−1) n n 2n + 1 34. a n = 1−3 n n+3 36. a n = 2 n + 5n + 6

n 2 − 2n + 1 ,n ≥ 2 37. a n = n −1

1 − n3 38. a n = 70 − 4 n 2

39. a n = 1 + (−1) n

40. a n = (−1) n 1 −

33. a n =

(

78. a n = ln 1 +

diverge? Find the limit of each convergent sequence. 31. a n = 2 + ( 0.1) n

+ ( ln ( n − 1) − ln ( n − 2 )) + ( ln n − ln ( n − 1))

32. a n =

(

80. a n =

1 n

)

( n +n 1)

n

(

)

82. a n = 1 − 84. a n =

1 n

1 n2

)

n

n

(10 11) n ( 9 10 ) n + (11 12 ) n

86. a n = sinh ( ln n )

79. a n =

( 33nn +− 11 )

81. a n =

( 2nx+ 1)

83. a n =

3n ⋅ 6n 2 −n ⋅ n !

n

n

1n

, x > 0

85. a n = tanh n 87. a n =

1 n2 sin 2n − 1 n

9.1  Sequences

(

88. a n = n 1 − cos

1 n

)

89. a n =

90. a n = ( 3 n + 5 n )1 n 92. a n =

( 13 )

n2 + n

95. a n =

( ln n ) 200 n

96. a n =

( ln n ) 5 n

99. a n =

91. a n = tan −1 n 93. a n =

n

n

97. a n = n −

1 −1−

n2

b. The fractions rn = x n y n approach a limit as n increases. What is that limit? (Hint: Use part (a) to show that rn2 − 2 = ±(1 y n ) 2 and that y n is not less than n.)

1 n

1 arctan n n

94. a n =

98. a n =

n sin

n2

+

111. Newton’s method The following sequences come from the recursion formula for Newton’s method,

1 2n

x n +1 = x n −

n2 − n

+n

n

1 1 dx n ∫1 x

100. a n =

∫1

n

1 dx , xp

p >1

Recursively Defined Sequences

In Exercises 101–108, assume that each sequence converges and find its limit. 72 101. a1 = 2, a n +1 = 1 + an 102. a1 = −1, a n +1 = 103. a1 = −4, a n +1 =

an + 6 an + 2 8 + 2a n

104. a1 = 0, a n +1 =

8 + 2a n

105. a1 = 5, a n +1 =

5a n

106. a1 = 3, a n +1 = 12 − 107. 2, 2 +

108.

1+

an

1 1 1 ,2 + ,… ,2 + 1 1 2 2+ 2+ 1 2 2+ 2

1, 1 +

1, 1 +

1+

1+

1+

549

f (x n ) . f ′( x n )

Do the sequences converge? If so, to what value? In each case, begin by identifying the function f that generates the sequence. a. x 0 = 1, x n +1 = x n −

x n2 − 2 x 1 = n + 2x n 2 xn

b. x 0 = 1, x n +1 = x n −

tan x n − 1 sec 2 x n

c. x 0 = 1, x n +1 = x n − 1 112. a. Suppose that f ( x ) is differentiable for all x in [ 0, 1 ] and that f (0) = 0. Define sequence { a n } by the rule a n = n f (1 n ). Show that lim a n = f ′(0). Use the result in part (a) to find n →∞

the limits of the following sequences { a n }. 1 b. a n = n tan −1 c. a n = n ( e 1 n − 1) n

(

d. a n = n ln 1 +

2 n

)

113. Pythagorean triples A triple of positive integers a, b, and c is called a Pythagorean triple if a 2 + b 2 = c 2. Let a be an odd positive integer and let a2 a2 b = ⎢⎢ ⎥⎥ and c = ⎡⎢ ⎤⎥ ⎣ 2 ⎦ ⎢ 2 ⎥ be, respectively, the integer floor and ceiling for a 2 2.

1,

1 ,…

Theory and Examples

109. The first term of a sequence is x 1 = 1. Each succeeding term is the sum of all those that come before it:

2 la m 2

2 ja k 2

x n +1 = x 1 + x 2 +  + x n . Write out enough early terms of the sequence to deduce a general formula for x n that holds for n ≥ 2. 110. A sequence of rational numbers is described as follows: 1 3 7 17 a a + 2b , …. , , , , …, , 1 2 5 12 b a+b Here the numerators form one sequence, the denominators form a second sequence, and their ratios form a third sequence. Let x n and y n be, respectively, the numerator and the denominator of the nth fraction rn = x n y n. a. Verify that x 12 − 2 y12 = −1, that x 22 − 2 y 22 = +1, and, more generally, that if a 2 − 2b 2 = −1  or  +1, then (a

+ 2b ) 2 − 2( a + b ) 2 = +1 or −1,

respectively.

u a

a. Show that a 2 + b 2 = c 2 . (Hint: Let a = 2n + 1 and express b and c in terms of n.) b. By direct calculation, or by appealing to the accompanying figure, find ⎢ a2 ⎥ ⎢ ⎥ lim ⎣ 2 ⎦ . a →∞ ⎡ a 2 ⎤ ⎢⎢ 2 ⎥⎥

550

Chapter 9 Infinite Sequences and Series

114. The nth root of n! a. Show that lim ( 2nπ )1 ( 2 n ) = 1 and hence, using Stirling’s n →∞

approximation (Chapter 8, Additional Exercise 44a), that n! ≈

n

n e

for large values of  n.

T b. Test the approximation in part (a) for n = 40, 50, 60, …,  as far as your calculator will allow. 115. a. Assuming that lim (1 n c ) = 0 if c is any positive constant, n →∞ show that lim

n →∞

ln n = 0 nc

b. Prove that lim (1 n c ) = 0 if c is any positive constant. n →∞

(Hint: If ε = 0.001 and c = 0.04, how large should N be to ensure that 1 n c − 0 < ε if n > N ?) 116. The zipper theorem Prove the “zipper theorem” for sequences: If { a n } and { b n } both converge to L, then the sequence a1 , b1 , a 2 , b 2 , … , a n , b n , … converges to L. n

n →∞

n = 1.

118. Prove that lim x 1 n = 1, ( x > 0 ). n →∞

119. Prove Theorem 2.

120. Prove Theorem 3.

In Exercises 121–124, determine whether the sequence is monotonic and whether it is bounded. ( 2n + 3 )! 3n + 1 121. a n = 122. a n = ( n + 1)! n+1 123. a n =

2 n3 n n!

124. a n = 2 −

2 1 − n 2 n

In Exercises 125–134, determine whether the sequence is monotonic, whether it is bounded, and whether it converges. 1 1 125. a n = 1 − 126. a n = n − n n 127. a n =

2n − 1 2n

128. a n =

2n − 1 3n

( n +n 1 )

129. a n = ((−1) n + 1)

130. The first term of a sequence is x 1 = cos (1). The next terms are x 2 = x 1 or cos (2), whichever is larger; and x 3 = x 2 or cos (3), whichever is larger (farther to the right). In general, x n +1 = max { x n , cos ( n + 1)}. 131. a n =

1+

2n n

4 n +1 + 3 n 133. a n = 4n

132. a n =

139. Is it true that a sequence { a n } of positive numbers must converge if it is bounded from above? Give reasons for your answer.

a m − a n < ε whenever

m > N

and n > N .

141. Uniqueness of limits Prove that limits of sequences are unique. That is, show that if L1 and L 2 are numbers such that a n → L1 and a n → L 2 , then L1 = L 2 . 142. Limits and subsequences If the terms of one sequence appear in another sequence in their given order, we call the first sequence a subsequence of the second. Prove that if two subsequences of a sequence { a n } have different limits L1 ≠ L 2 , then { a n } diverges. 143. For a sequence { a n } the terms of even index are denoted by a 2 k and the terms of odd index by a 2 k +1. Prove that if a 2 k → L and a 2 k +1 → L, then a n → L. 144. Prove that a sequence { a n } converges to 0 if and only if the sequence of absolute values { a n } converges to 0. 145. Sequences generated by Newton’s method Newton’s method, applied to a differentiable function f ( x ), begins with a starting value x 0 and constructs from it a sequence of numbers { x n } that under favorable circumstances converges to a zero of f . The recursion formula for the sequence is x n +1 = x n −

f (x n ) . f ′( x n )

a. Show that the recursion formula for f ( x ) = x 2 − a, a > 0, can be written as x n +1 = ( x n + a x n ) 2. T b. Starting with x 0 = 1 and a = 3, calculate successive terms of the sequence until the display begins to repeat. What number is being approximated? Explain. T 146. A recursive definition of π 2 If you start with x 1 = 1 and if you define the subsequent terms of { x n } by the rule x n = x n−1 + cos x n−1 , you generate a sequence that converges rapidly to π 2. (a) Try it. (b) Use the accompanying figure to explain why the convergence is so rapid. y

n+1 n

134. a1 = 1, a n +1 = 2a n − 3

In Exercises 135–136, use the definition of convergence to prove the given limit. sin n 1 135. lim = 0 136. lim 1 − 2 = 1 n →∞ n n →∞ n

(

138. Uniqueness of least upper bounds Show that if M 1 and M 2 are least upper bounds for the sequence { a n } , then M 1 = M 2 . That is, a sequence cannot have two different least upper bounds.

140. Prove that if { a n } is a convergent sequence, then to every positive number ε there corresponds an integer N such that

if c is any positive constant.

117. Prove that lim

137. The sequence { n ( n + 1)} has a least upper bound of 1 Show that if M is a number less than 1, then the terms of { n ( n + 1)} eventually exceed M. That is, if M < 1, there is an integer N such that n ( n + 1) > M whenever n > N. Since n ( n + 1) < 1 for every n, this proves that 1 is a least upper bound for { n ( n + 1)}.

)

1

cos xn − 1 xn − 1 xn − 1

0

1

x

9.2  Infinite Series

COMPUTER EXPLORATIONS

Use a CAS to perform the following steps for the sequences in Exercises 147–158. a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit L?

1 5n

150. a1 = 1, a n +1 = a n + (−2 ) n 151. a n = sin n 153. a n =

b. If the sequence converges, find an integer N such that a n − L ≤ 0.01 for n ≥ N. How far in the sequence do you have to get for the terms to lie within 0.0001 of L? 0.5 n 147. a n = n n 148. a n = 1 + n

(

149. a1 = 1, a n +1 = a n +

152. a n = n sin

sin n n

154. a n =

155. a n = ( 0.9999 ) n 157. a n =

)

551

1 n

ln n n

156. a n = (123456 )1 n

8n n!

158. a n =

n 41 19 n

9.2 Infinite Series An infinite series is the sum of an infinite sequence of numbers a1 + a 2 + a 3 +  + a n +  . The goal of this section is to understand the meaning of such an infinite sum and to develop methods to calculate it. Since there are infinitely many terms to add in an infinite series, we cannot just keep adding to see what comes out. Instead we look at the result of summing just the first n terms of the sequence. The sum of the first n terms s n = a1 + a 2 + a 3 +  + a n is an ordinary finite sum and can be calculated by normal addition. It is called the nth partial sum. As n gets larger, we expect the partial sums to get closer and closer to a limiting value in the same sense that the terms of a sequence approach a limit, as discussed in Section 9.1. For example, to assign meaning to an expression like 1+

1 1 1 1 + + + + , 2 4 8 16

we add the terms one at a time from the beginning and look for a pattern in how these partial sums grow.

Partial sum

Value

Suggestive expression for partial sum

First:

s1 = 1

1

2−1

Second:

1 s2 = 1 + 2

3 2

2−

1 2

Third:

s3 = 1 +

7 4

2−

1 4





 nth:

1 1 + 2 4 

sn = 1 +

1 1 1 + +  + n −1 2 4 2

−1 2 n −1

2n

2−

Indeed there is a pattern. The partial sums form a sequence whose nth term is sn = 2 −

1 . 2 n −1

1 2 n −1

552

Chapter 9 Infinite Sequences and Series

This sequence of partial sums converges to 2 because lim (1 2 n −1 ) = 0. We say n →∞

1 1 1 “the sum of the infinite series 1 + + + " + n −1 + " is 2.” 2 4 2 Is the sum of any finite number of terms in this series equal to 2? No. Can we actually add an infinite number of terms one by one? No. But we can still define their sum by defining it to be the limit of the sequence of partial sums as n → ∞, in this case 2 (Figure 9.8). Our knowledge of sequences and limits enables us to break away from the confines of finite sums. 1 HISTORICAL BIOGRAPHY Blaise Pascal (1623–1662) Pascal was born in France and was encouraged by his father to study science. He met Fermat and was inspired to work on applied science problems. As early as 1640, he wrote an essay on conic sections and earned praise for his work from Descartes. Despite his poor health, Pascal designed an “arithmetic machine” to perform computations for tax collecting. Pascal also contributed to the development of differential calculus.

14 12

0

FIGURE 9.8

18

2

As the lengths 1, 1 2, 1 4 , 1 8, !  are added one by one, the sum

approaches 2.

DEFINITIONS

Given a sequence of numbers { a n } , an expression of the form a1 a 2 a 3 " a n "

is an infinite series. The number a n is the nth term of the series. The sequence

{ s n } defined by

s 1 = a1

To know more, visit the companion Website.

s 2 = a1 + a 2  s n = a1 + a 2 +  + a n =

y

is the sequence of partial sums of the series, the number s n being the nth partial sum. If the sequence of partial sums converges to a limit L, we say that the series converges and that its sum is L. In this case, we also write

a1

1

a3

a2 1

2

a4 3

a5 4

a6 5

6 ...

x

a1 + a 2 + " + a n + " =

(a)

∞

∑ an

= L.

n =1

If the sequence of partial sums of the series does not converge, we say that the series diverges.

y an = 1 + 1 n

2 1

k =1



an = 22 n

2

n

∑ ak

a1

a2 1

a3 2

a4 3

a5 4

a6 5

6 ...

x

(b)

FIGURE 9.9 The sum of a series with positive terms can be interpreted as a total area of an infinite collection of rectangles. The series converges when the total area of the rectangles is finite (a) and diverges when the total area is unbounded (b). Note that the total area can be infinite even if the area of the rectangles is decreasing.

If all the terms a n in an infinite series are positive, then we can represent each term in the series by the area of a rectangle. The series converges if the total area is finite, and diverges otherwise. Figure 9.9a shows an example where the series converges, and Figure 9.9b shows an example where it diverges. The convergence of the total area is related to the convergence or divergence of improper integrals, as we found in Section 8.8. We make this connection explicit in the next section, where we develop an important test for convergence of series, the Integral Test. When we begin to study a given series a1 a 2 " a n ", we might not know whether it converges or diverges. In either case, it is convenient to use sigma notation to write the series as ∞

∑ an , n =1

∞

∑ ak , k =1

or

∑ an

A useful shorthand when summation from 1 to d is understood

9.2  Infinite Series

553

Geometric Series Geometric series are series of the form a + ar + ar 2 +  + ar n −1 +  =

∞

∑ ar n−1 , n =1

in which a and r are fixed real numbers and a ≠ 0. The series can also be written ∞ as ∑ n = 0 ar n. The ratio r can be positive, as in 1+

( )

1 1 1 + ++ 2 4 2

n −1

+ ,

r = 1 2, a = 1

+ .

r = −1 3, a = 1

or negative, as in 1−

( )

1 1 1 + −+ − 3 9 3

n −1

If r = 1, the nth partial sum of the geometric series is s n = a + a ⋅ 1 + a ⋅ 1 2 +  + a ⋅ 1 n −1 = na, and the series diverges because lim s n = ±∞, depending on the sign of a. If r = −1, n →∞

the series diverges because the nth partial sums alternate between a and 0 and never approach a single limit. If r ≠ 1, we can determine the convergence or divergence of the series in the following way: s n = a + ar + ar 2 +  + ar n −1 rs n = ar + ar 2 +  + ar n −1 + ar n s n − rs n = a − ar n

Multiply s n by r. Subtract rs n from s n . Most of the terms on the right cancel.

s n (1 − r ) = a (1 − r n ) a (1 − r n ) sn = , 1−r

Write the nth partial sum.

Factor. (r

≠ 1).

We can solve for s n if r ≠ 1.

If r < 1, then r n → 0 as n → ∞ (as in Section 9.1), so s n → a (1 − r ) in this case. On the other hand, if r > 1, then r n → ∞ and the series diverges. We summarize these results as follows.

Geometric Series

If r < 1, the geometric series a + ar + ar 2 +  + ar n −1 +  converges to a (1 − r ): ∞

∑ ar n−1 n =1

=

a , 1−r

r < 1.

If r ≥ 1, the series diverges.

The formula a (1 − r ) for the sum of a geometric series applies only when the sum∞ mation index begins with n = 1 in the expression ∑ n =1 ar n −1 (or with the index n = 0 if ∞ we write the series as ∑ n = 0 ar n ). EXAMPLE 1

The geometric series with a = 1 9 and r = 1 3 is 1 1 1 + + += 9 27 81

∞

∑ 19 ( 13 ) n =1

n −1

.

554

Chapter 9 Infinite Sequences and Series

Since r = 1 3 < 1, the series converges to 19 1 = . 1 − (1 3 ) 6

EXAMPLE 2

The series ∞

∑

( − 1) n 5

n=0

4n

= 5−

5 5 5 + − + 4 16 64

is a geometric series with a = 5 and r = −1 4. Since r = 1 4 < 1, the series converges to a 5 = = 4. 1−r 1 + (1 4 ) a

EXAMPLE 3 You drop a ball from a meters above a flat surface. Each time the ball hits the surface after falling a distance h, it rebounds a distance rh, where r is positive but less than 1. Find the total distance the ball travels up and down (Figure 9.10).

ar

Solution The total distance is ar

2

2ar 1+r s = a + 2ar + 2ar 2 + 2ar 3 + ⋅⋅⋅ = a + = a . 1 r 1−r − 

ar 3

This sum is 2 ar (1 − r ).

If a = 6 m and r = 2 3, for instance, the distance is s = 6⋅ (a)

EXAMPLE 4

⎛ 5 3 ⎞⎟ 1 + ( 2 3) = 6⎜⎜ = 30 m. ⎜⎝ 1 3 ⎟⎟⎠ 1 − ( 2 3)

Express the repeating decimal 5.232323 …  as the ratio of two integers.

Solution From the definition of a decimal number, we get a geometric series.

23 23 23 + + + (100 ) 2 (100 ) 3 100 ⎞ 23 ⎛⎜ 1 1 2 1+ = 5+ + + ⎟⎟ ⎜ 100 ⎝ 100 100 ⎠

5.232323 … = 5 +

( )

1 (1 − 1 100 )

= 5+

(

a = 1,    r = 1 100 r = 1 100 < 1

)

1 518 23 23 = 5+ = 100 0.99 99 99

(b)

FIGURE 9.10 (a) Example 3 shows how to use a geometric series to calculate the total vertical distance traveled by a bouncing ball if the height of each rebound is reduced by the factor r. (b) A stroboscopic photo of a bouncing ball. (Source: Berenice Abbott/Science Source)

Unfortunately, formulas like the one for the sum of a convergent geometric series are rare, and we usually have to settle for an estimate of a series’ sum (more about this later). The next example, however, is another case in which we can find the sum exactly.

∞

EXAMPLE 5

1 . ( n n + 1) n =1

Find the sum of the “telescoping” series  ∑

9.2  Infinite Series

555

Solution We look for a pattern in the sequence of partial sums that might lead to a formula for s k . The key observation is the partial fraction decomposition

1 1 1 , = − n ( n + 1) n n+1 so

∑ n ( n 1+ 1) = ∑ ( 1n − n +1 1 ) k

k

n =1

n =1

and sk =

(11 − 12 ) + ⎛⎜⎜⎝ 12 − 13 ⎞⎟⎟⎠ + ⎛⎜⎜⎝ 13 − 14 ⎞⎟⎟⎠ +  + ⎜⎜⎝⎛ 1k − k +1 1⎞⎟⎟⎠.

Removing parentheses and canceling adjacent terms of opposite sign collapse the sum to 1 . k +1 We now see that s k → 1 as k → ∞. The series converges, and its sum is 1: sk = 1 −

∞

∑ n ( n 1+ 1) = 1. n =1

The nth-Term Test for a Divergent Series One reason why a series may fail to converge is that its terms don’t become small. EXAMPLE 6

The series ∞

∑ n +n 1 = n =1

2 3 4 n+1 + + ++ + 1 2 3 n

diverges because the partial sums eventually outgrow every preassigned number. Each term is greater than 1, so the sum of n terms is greater than n. ∞

We now show that lim a n must equal zero if the series ∑ n =1 a n converges. To see n →∞ why, let S represent the series’ sum, and s n = a1 + a 2 +  + a n the nth partial sum. When n is large, both s n and s n −1 are close to S, so their difference, a n , is close to zero. More formally, s n and s n −1 both converge to S as n increases, so lim a n = lim ( s n − s n −1 ) = lim s n − lim s n −1 = S − S = 0.

n →∞

n →∞

n →∞

n →∞

Difference Rule for sequences

This establishes the following theorem. ∞

Caution ∞ Theorem 7 does not say that ∑ n =1 a n converges if a n → 0. It is possible for a series to diverge when a n → 0. (See Example 8.)

THEOREM 7

If  ∑ a n converges, then a n → 0. n =1

Theorem 7 leads to a test for detecting the kind of divergence that occurred in Example 6. The nth-Term Test for Divergence ∞

a n fails to exist or is different from zero. ∑ a n diverges if nlim →∞ n =1

EXAMPLE 7 ∞

(a)

The following are all examples of divergent series.

∑ n 2 diverges because n 2 n =1

→ ∞.

lim a n fails to exist.

n →∞

556

Chapter 9 Infinite Sequences and Series

∞

(b)

∑ n +n 1 diverges because n +n 1 → 1. n =1

lim a n ≠ 0

n →∞

∞

(c)

( − 1) n +1 does not exist. ∑ (−1)n +1  diverges because nlim →∞ n =1 ∞

(d)

−n ∑ 2n−+n 5   diverges because nlim →∞ 2n + 5 n =1

EXAMPLE 8

= −

1 ≠ 0. 2

The series

1 1 1 1 1 1 1 1 1 1+ + + + + + + + n + n + + n +  2 2 4 4 4 4 2 2 2    2 terms

2 n terms

4 terms

diverges because the terms can be grouped into infinitely many clusters each of which adds to 1, so the partial sums increase without bound. However, the terms of the series form a sequence that converges to 0. Example 1 of Section 9.3 shows that the harmonic series ∑ 1 n also behaves in this manner.

Combining Series Whenever we have two convergent series, we can add them term by term, subtract them term by term, or multiply them by constants to make new convergent series. THEOREM 8

If ∑ a n = A and ∑ b n = B are convergent series, then

1. Sum Rule: 2. Difference Rule: 3. Constant Multiple Rule:

∑ ( a n + bn ) = ∑ a n + ∑ bn = A + B ∑ ( a n − bn ) = ∑ a n − ∑ bn = A − B ∑ ka n = k ∑ a n = kA ( any number k ).

Proof The three rules for series follow from the analogous rules for sequences in Theorem 1, Section 9.1. To prove the Sum Rule for series, let An = a1 + a 2 +  + a n ,

B n = b1 + b 2 +  + b n .

Then the partial sums of ∑ ( a n + b n ) are s n = ( a1 + b1 ) + ( a 2 + b 2 ) +  + ( a n + b n ) = ( a1 +  + a n ) + ( b1 +  + b n ) = An + B n . Since An → A and B n → B, we have s n → A + B by the Sum Rule for sequences. The proof of the Difference Rule is similar. To prove the Constant Multiple Rule for series, observe that the partial sums of  ∑ ka n form the sequence s n = ka1 + ka 2 +  + ka n = k ( a1 + a 2 +  + a n ) = kAn , which converges to kA by the Constant Multiple Rule for sequences. As corollaries of Theorem 8, we have the following results. We omit the proofs. 1. Every nonzero constant multiple of a divergent series diverges. 2. If ∑ a n converges and ∑ b n diverges, then ∑ ( a n + b n ) and ∑ ( a n − b n ) both diverge.

9.2  Infinite Series

557

∑ ( a n + bn ) can converge even if both œ a n and œ bn diverge. For = 1 + 1 + 1 + "  and  ∑ b n = ( −1) + ( −1) + ( −1) + "  diverge,

Caution Remember that example, both whereas

∑

∑ an

( a n + b n ) = 0 + 0 + 0 + "  converges to 0.

EXAMPLE 9 (a)

∞

Find the sums of the following series. ∞

∑ 3 6 n−−1 1 = ∑ ( 2 n1−1 − 6 n1−1 ) n −1

n =1

n =1

= =

∞

∞

n =1

n =1

∑ 2 n1−1 − ∑ 6 n1−1

Difference Rule

1 1 − 1 − (1 2 ) 1 − (1 6 )

Geometric series with  a = 1 and r = 1 2,  1 6 Both series converge since  1 2 < 1 and  1 6 < 1.

=2− ∞

(b)

∑ 24n

n=0

∞

= 4∑

n=0

6 4 = 5 5

1 2n

Constant Multiple Rule

⎛ ⎞⎟ 1 = 4 ⎜⎜ ⎜⎝ 1 − (1 2 ) ⎟⎟⎠

Geometric series with  a = 1 and r = 1 2 Converges since  r = 1 2 < 1.

=8

Adding or Deleting Terms We can add a finite number of terms to a series or delete a finite number of terms without altering the series’ convergence or divergence, although in the case of convergence, this ∞ ∞ will usually change the sum. If ∑ n =1 a n converges, then ∑ n = k a n converges for any k  1 and ∞

∑ an n =1

= a1 + a 2 + " + a k −1 +

∞

∞

∑ an.

n= k

∞

Conversely, if  ∑ n = k a n converges for any k  1, then  ∑ n =1 a n converges. Thus, ∞

∑ 51n

=

n =1

1 1 1 + + + 5 25 125

∞

∑ 51n

n=4

and ∞

∑ 51n

n=4

HISTORICAL BIOGRAPHY Richard Dedekind (1831–1916) Dedekind grew up in Germany and in 1850 entered the University of Gottingen.There he studied with Bernhard Riemann and Carl Gauss. Like Gauss, Dedekind preferred to study the theoretical aspects of number theory. His work on irrational numbers gave the subject a logical foundation. To know more, visit the companion Website.

⎛∞ 1⎞ 1 1 1 = ⎜⎜ ∑ n ⎟⎟⎟ − − . − ⎜⎝ 5 25 125 n =1 5 ⎠

The convergence or divergence of a series is not affected by its first few terms. Only the “tail” of the series, the part that remains when we sum beyond some finite number of initial terms, influences whether it converges or diverges.

Reindexing As long as we preserve the order of its terms, we can reindex any series without altering its convergence. To raise the starting value of the index h units, replace the n in the formula for a n by n  h: ∞

∑ an = n =1

∞

∑

n = 1+ h

a n − h = a1 + a 2 + a 3 + ".

To lower the starting value of the index h units, replace the n in the formula for a n by n h: ∞

∑ an n =1

=

∞

∑

n =1− h

a n + h = a1 + a 2 + a 3 + ".

558

Chapter 9 Infinite Sequences and Series

We saw this reindexing in starting a geometric series with the index n = 0 instead of the index n = 1, but we can use any other starting index value as well. We usually give preference to indexings that lead to simple expressions. EXAMPLE 10

We can write the geometric series ∞

∑ 2 n1−1

1 1 + + 2 4

= 1+

n =1

as ∞

∑

n=0

∞

1 , 2n

∑2

n=5

∞

1

, n−5

1 . 2 n+4

∑

or even

n =−4

The partial sums remain the same no matter what indexing we choose to use.

9.2

EXERCISES

Finding nth Partial Sums

In Exercises 1–6, find a formula for the nth partial sum of each series and use it to find the series’ sum if the series converges. 2 2 2 2 1. 2 + + + +  + n−1 +  3 9 27 3 2.

( 81 ) + ( 81 )

18.

( −32 )

19. 1 −

9 9 9 9 + + ++ + 100 100 2 100 3 100 n 1 1 1 1 + − +  + (−1) n−1 n−1 +  2 2 4 8

3. 1 −

17.

20.

( 13 )

21. 1 +

1 1 1 1 + + ++ + 5. ( n + 1)( n + 2 ) 2⋅3 3⋅4 4⋅5

22.

5 5 5 5 + + ++ + 1⋅2 2⋅3 3⋅4 n ( n + 1)

∞

n=0

8.

4n

10.

∞

∑ ( 25n

+

n=0

∑ ( 21n ∞

13.

n=0

+

∞

)

12.

∑ ( 25n

−

n=0

( − 1) n

5n

)

∞

14.

∑ ( 25 n n=0

( 25 ) + ( 25 )

2

+

( 25 ) + ( 25 ) 3

4

+

2

−

( 13 )

−1

3

+1−

5

( −32 ) 4

+

5

( 13 ) + ( 13 )

4

( −32 )

6

+

− 2

−

( 109 ) + ( 109 ) + ( 109 ) + ( 109 ) 2

+

6

8

+

9 27 81 243 729 − + − + − 4 8 16 32 64

23. 0.23 = 0.23 23 23 … 24. 0.234 = 0.234 234 234 …

28. 1.414 = 1.414 414 414 …

n +1

1 3n

)

)

In Exercises 15–22, determine whether the geometric series converges or diverges. If a series converges, find its sum. 15. 1 +

4

( 81 )

27. 0.06 = 0.06666 …

∑ (−1)n 45n n=0

1 3n

3

+

26. 0.d = 0.dddd …, where d is a digit

∞

∑ (1 − 47n ) n =1

11.

∑ 41n n=2

∞

9.

( −32 ) + ( −32 )

+

4

25. 0.7 = 0.7777 …

∞

( − 1) n

3

Express each of the numbers in Exercises 23–30 as the ratio of two integers.

In Exercises 7–14, write out the first eight terms of each series to show how the series starts. Then find the sum of the series or show that it diverges.

∑

( 81 ) + ( 81 )

Repeating Decimals

Series with Geometric Terms

7.

+

( 2e ) + ( 2e ) − ( 2e ) + ( 2e )

−2

4. 1 − 2 + 4 − 8 +  + (−1) n−1 2 n−1 + 

6.

2

2

29. 1.24123 = 1.24 123 123 123 … 30. 3.142857 = 3.142857 142857 … Using the nth-Term Test

In Exercises 31–38, use the nth-Term Test for divergence to show that the series is divergent, or state that the test is inconclusive. ∞

31.

∞

32.

n =1

+

16. 1 + (−3 ) + (−3 ) 2 + (−3 )3 + (−3 ) 4 + 

∑ n +n 10 ∑ n +1 4 n=0

(

n =1

∞

33.

∑ ( n +n 2n)(+n 1+ 3) ∞

34.

∑ n 2 n+ 3 n =1

)

9.2  Infinite Series

∞

35.

∞

∑ cos 1n

36.

n =1

69.

n=0

∞

37.

∞

n

∑ e n e+ n

∑ ln 1n

38.

n =1

∞

∑ ln( n +n 1 )

∑ ln( 2n n+ 1 )

70.

n =1

∞

∞

∑ cos nπ

71.

n=0

∑ ( πe )

559

n =1 ∞

n

nπ

∑ πe ne

72.

n=0

n=0 ∞

73.

Telescoping Series

In Exercises 39–44, find a formula for the nth partial sum of the series and use it to determine whether the series converges or diverges. If a series converges, find its sum. ∞

39.

∑( n =1

1 1 − n n+1

∞

41.

∑ ( ln

)

40.

n =1

n + 1 − ln n )

n =1

∑( ∞

3 3 − ( n + 1) 2 n2

)

n =1 ∞

74.

∑ ( tan (n) − tan ( n − 1)) n =1

∞

75.

∞

76.

∑(

n+4−

n =1

In each of the geometric series in Exercises 77–80, write out the first few terms of the series to find a and r, and find the sum of the series. Then express the inequality r < 1 in terms of x and find the values of x for which the inequality holds and the series converges.

n + 3)

Find the sum of each series in Exercises 45–52. ∞

45.

4

∑ ( 4n − 3)( 4n + 1)

46.

n ∑ ( 2n − 1)40 2 ( 2n + 1) 2 ∞

∑( n =1

1 − n

1 n+1

)

∑ ( 2n − 1)( 2n + 1) ∞

48.

n =1

49.

6

77.

∞

∑ ( 211 n

−

n =1

1 2 1 ( n +1)

)

∑ ( ln ( n1+ 2 ) − ln ( n1+ 1) )

83.

n =1

∞

n=0 ∞

55.

∑

1 2

)

∞

n

( − 1) n +1

n =1

54.

∞

56.

∑ cos( n2π )

∞

n =1 ∞

63.

∑ 2 3−n 1 n

∞

n=0

n! ∑ 1000 n n=0

67.

∑2

∞

n =1

n

+ 3n 4n

n=0

Theory and Examples

∑ ( n + 1)(1 n + 2 )

n

∞

66.

n ∑ nn! n =1

68.

∑ 23 n n =1

n

∞

∑ ( n + 3)(1 n + 4 ).

and

n =1

n =1

x >1

∑ (1 − 1n )

∞

n=0

∞

n =1

∞

65.

∑ ( ln x )n

86.

∑ n ( n 5+ 1)

n=0

64.

∞

∑ sin n x

n =1

∑ x1n ,

− 3)n

n =−1

88. The series in Exercise 6 can also be written as

∞

62.

(x

Write this series as a sum beginning with (a) n = −2, (b) n = 0, (c) n = 5.

cos nπ 5n n=0

∑ ln 31n

n

n=0

∞

60.

∞

∑ 102 n

∑ (− 12 )

84.

87. The series in Exercise 5 can also be written as

∞

n=0

61.

∑

n

85.

( − 1) n +1 n

∑

∞

∑ e −2 n

2)

58.

n=0

59.

∞

∑ (−1)n ( x + 1)n

n =1

∞

57.

∑( n=0

3 2n

n=0

∞

Which series in Exercises 53–76 converge, and which diverge? Give reasons for your answers. If a series converges, find its sum.

∑(

∑ (−1)n x −2n

82.

n=0

Convergence or Divergence

2

⎛ ⎞⎟ n 1 ⎜⎜ ⎜⎝ 3 + sin x ⎟⎟⎠

∞

∑ 2n x n ∞

∑ ( tan −1 (n) − tan −1 ( n + 1))

( − 1) n

n=0

n=0

∞

53.

∞

∑

80.

∞

81.

n =1

52.

∑ 3( x −2 1 )

n=0

n

In Exercises 81–86, find the values of x for which the given geometric series converges. Also, find the sum of the series (as a function of x) for those values of x.

∞

51.

∑ (−1)n x 2n

78.

n=0

n =1

50.

∞

79.

∑ n 22(nn ++ 11)2

∞

∑ (−1)n x n n=0

n =1

∞

47.

∞

∞

n =1

+ 1))

Geometric Series with a Variable x

n =1

44.

∑ ( ln( 4e n − 1) − ln( 2e n n=0

∑ ( arccos( n +1 1 ) − arccos( n +1 2 )) ∞

∑ ( cos( πn ) + sin( πn )) n =1

∞

43.

π ∑ ( sin ( πn ) − sin ( n − 1 )) n=2

∞

42.

2 ∑ ( n +n 1 − nn + + 3)

∞

and

∑ ( n + 1)(5 n + 2 ). n=0

Write this series as a sum beginning with (a) n = −1, (b) n = 3, (c) n = 20. 89. Make up an infinite series of nonzero terms whose sum is a. 1

b. −3

c. 0.

90. (Continuation of Exercise 89.) Can you make an infinite series of nonzero terms that converges to any number you want? Explain. + 4n + 4n

91. Show by example that ∑ ( a n b n ) may diverge even though ∑ a n and ∑ b n converge and no b n equals 0.

560

Chapter 9 Infinite Sequences and Series

92. Find convergent geometric series A = ∑ a n and B = ∑ b n that illustrate the fact that ∑ a n b n may converge without being equal to AB. 93. Show by example that ∑ ( a n b n ) may converge to something other than A B even when A = ∑ a n , B = ∑ b n ≠ 0, and no b n equals 0.

∑ a n converges and a n > 0 for all n, can anything be said about ∑ (1 a n )? Give reasons for your answer.

94. If

95. What happens if you add a finite number of terms to a divergent series or delete a finite number of terms from a divergent series? Give reasons for your answer. 96. If ∑ a n converges and ∑ b n diverges, can anything be said about their term-by-term sum ∑ ( a n + b n )? Give reasons for your answer. 97. Make up a geometric series ∑ ar n−1 that converges to the number 5 if a. a = 2

b. a = 13 2.

98. Find the value of b for which 1 + e b + e 2 b + e 3b +  = 9. 99. For what values of r does the infinite series 1 + 2r + r 2 + 2r 3 + r 4 + 2r 5 + r 6 +  converge? Find the sum of the series when it converges. 100. The accompanying figure shows the first five of a sequence of squares. The outermost square has an area of 4 m 2. Each of the other squares is obtained by joining the midpoints of the sides of the squares before it. Find the sum of the areas of all the squares.

103. The Cantor set To construct this set, we begin with the closed interval [ 0, 1 ]. From that interval, we remove the middle open interval (1 3, 2 3 ), leaving the two closed intervals [ 0, 1 3 ] and [ 2 3, 1 ]. At the second step we remove the open middle third interval from each of those remaining. From [ 0, 1 3 ] we remove the open interval (1 9, 2 9 ), and from [ 2 3, 1 ] we remove ( 7 9, 8 9 ), leaving behind the four closed intervals [ 0, 1 9 ], [ 2 9, 1 3 ], [ 2 3, 7 9 ], and [ 8 9, 1 ]. At the next step, we remove the open middle third interval from each closed interval left behind, so (1 27, 2 27 ) is removed from [ 0, 1 9 ], leaving the closed intervals [ 0, 1 27 ] and [ 2 27,1 9 ]; ( 7 27, 8 27 ) is removed from [ 2 9, 1 3 ], leaving behind [ 2 9, 7 27 ] and [ 8 27,1 3 ], and so forth. We continue this process repeatedly without stopping, at each step removing the open third interval from every closed interval remaining behind from the preceding step. The numbers remaining in the interval [ 0, 1 ], after all open middle third intervals have been removed, are the points in the Cantor set (named after Georg Cantor, 1845–1918). The set has some interesting properties. a. The Cantor set contains infinitely many numbers in [ 0, 1 ]. List 12 numbers that belong to the Cantor set. b. Show, by summing an appropriate geometric series, that the total length of all the open middle third intervals that have been removed from [ 0, 1 ] is equal to 1. 104. Helge von Koch’s snowflake curve Helge von Koch’s snowflake is a curve of infinite length that encloses a region of finite area. To see why this is so, suppose the curve is generated by starting with an equilateral triangle whose sides have length 1. a. Find the length L n of the nth curve C n and show that lim L n = ∞. n →∞

b. Find the area An of the region enclosed by C n and show that lim An = ( 8 5 ) A1. n →∞

C1

101. Drug dosage A patient takes a 300 mg tablet for the control of high blood pressure every morning at the same time. The concentration of the drug in the patient’s system decays exponentially at a constant hourly rate of k = 0.12. a. How many milligrams of the drug are in the patient’s system just before the second tablet is taken? Just before the third tablet is taken? b. After the patient has taken the medication for at least six months, what quantity of drug is in the patient’s body just before the next regularly scheduled morning tablet is taken? 102. Show that the error ( L − s n ) obtained by replacing a convergent geometric series with one of its partial sums s n is ar n (1 − r ).

C2

C3

C4

105. The largest circle in the accompanying figure has radius 1. Consider the sequence of circles of maximum area inscribed in semicircles of diminishing size. What is the sum of the areas of all of the circles?

9.3  The Integral Test

561

9.3 The Integral Test The most basic question we can ask about a series is whether it converges. In this section we begin to study this question, starting with series that have nonnegative terms. Such a series converges if its sequence of partial sums is bounded. If we establish that a given series does converge, we generally do not have a formula available for its sum. So to get an estimate for the sum of a convergent series, we investigate the error involved when using a partial sum to approximate the total sum.

Nondecreasing Partial Sums ∞

Suppose that ∑ n =1 a n is an infinite series with a n ≥ 0 for all n. Then each partial sum is greater than or equal to its predecessor because s n +1 = s n + a n , so s1 ≤ s 2 ≤ s 3 ≤  ≤ s n ≤ s n +1 ≤ . Since the partial sums form a nondecreasing sequence, the Monotonic Sequence Theorem (Theorem 6, Section 9.1) gives the following result.

Corollary of Theorem 6 ∞

A series ∑ n =1 a n of nonnegative terms converges if and only if its partial sums are bounded from above.

EXAMPLE 1

As an application of the above corollary, consider the harmonic series ∞

∑ 1n

= 1+

n =1

1 1 1 + +  + + . 2 3 n

Although the nth term 1 n does go to zero, the series diverges because there is no upper bound for its partial sums. To see why, group the terms of the series in the following way: 1+

(

) (

) (

)

1 1 1 1 1 1 1 1 1 1 + + + + + + + + ++ + . 2 3 4 5 6 7 8 9 10 16    >2=1 4 2

>4=1 8 2

> 8 =1 16 2

The sum of the first two terms is 1.5. The sum of the next two terms is 1 3 + 1 4, which is greater than 1 4 + 1 4 = 1 2. The sum of the next four terms is 1 5 + 1 6 + 1 7 + 1 8, which is greater than 1 8 + 1 8 + 1 8 + 1 8 = 1 2. The sum of the next eight terms is 1 9 + 1 10 + 1 11 + 1 12 + 1 13 + 1 14 + 1 15 + 1 16, which is greater than 8 16 = 1 2. The sum of the next 16 terms is greater than 16 32 = 1 2, and so on. In general, the sum of 2 m terms ending with 1 2 m +1 is greater than 2 m 2 m +1 = 1 2. Therefore, if n = 2 k , then the partial sum s n is greater than k 2, so the sequence of partial sums is not bounded from above. The harmonic series diverges.

The Integral Test We introduce the Integral Test with a series that is related to the harmonic series, but whose nth term is 1 n 2 instead of 1 n. EXAMPLE 2

Does the following series converge? ∞

∑ n12 n =1

= 1+

1 1 1 1 + + ++ 2 + 4 9 16 n

562

Chapter 9 Infinite Sequences and Series

Caution The series and integral need not have the same value in the convergent case. You will see in Example 6 that ∞

∑ n=1 (1 n 2 ) ≠ ∫1

∞

(1 x 2 ) dx = 1.

Solution We determine the convergence of

∞

∑ n =1 (1 n 2 )

by comparing it with carry out the comparison, we think of the terms of the series as values of the function f ( x ) = 1 x 2 and interpret these values as the areas of rectangles under the curve y = 1 x 2. As Figure 9.11 shows,

∞ ∫ 1 (1

x 2 ) dx. To

1 1 1 1 + 2 + 2 ++ 2 n 12 2 3 = f (1) + f (2) + f (3) +  + f (n)

sn = y

< f (1) +

(1, f (1))

(2, f (2)) 1 32 (3, f (3))

1 42

1 22 0

1

3

2

4 …

n

1 dx x2

1 n2

(n, f (n))

n−1 n…

∫1

∞

Thus the partial sums of converges.

∞

n

∫1 (1 x 2 ) dx < ∫1 (1 x 2 ) dx As in Section 8.8, ∞ Example 3, ∫1 (1 x 2 ) dx = 1.

∞

∑ n =1 (1 n 2 ) are bounded from above (by 2), and the series

x

THEOREM 9—The Integral Test Let { a n } be a sequence of positive terms. Suppose that a n = f (n), where f

FIGURE 9.11 The sum of the areas of the rectangles under the graph of f ( x ) = 1 x 2 is less than the area under the graph (Example 2).

is a continuous, positive, decreasing function of x for all x ≥ N (N a positive ∞

∞

integer). Then the series ∑ n = N a n and the integral ∫ f ( x ) dx both converge or N both diverge.

Proof We establish the test for the case N = 1. The proof for general N is similar. We start with the assumption that f is a decreasing function with f (n) = a n for every n. This leads us to observe that the rectangles in Figure 9.12a, which have areas a1 , a 2 , …, a n , collectively enclose more area than that under the curve y = f ( x ) from x = 1 to x = n + 1. That is,

y

∫1 y = f (x)

a1

1

n

3

2

n+1

n +1

f ( x ) dx ≤ a1 + a 2 +  + a n .

In Figure 9.12b the rectangles have been faced to the left instead of to the right. If we momentarily disregard the first rectangle of area a1 , we see that

a2 an

0

x

a2 + a3 +  + an ≤

(a) y

∫1

n

f ( x ) dx.

If we include a1 , we have a1 + a 2 +  + a n ≤ a1 +

y = f (x)

a1

0

Rectangle areas sum to less than area under graph.

1 dx x2 = 1 + 1 = 2.

1, diverges if p ≤ 1.

=

n =1

1 1 1 1 + p + p ++ p + p 1 2 3 n

( p a real constant) converges if p > 1 and diverges if p ≤ 1. Solution If p > 1, then f ( x ) = 1 x p is a positive, continuous, and decreasing function of x. Since

∫1

∞

⎡ x − p +1 ⎤ b x − p dx = lim ⎢ ⎥ b →∞ ⎢⎣ − p + 1 ⎥⎦ 1 1 1 lim = −1 1 − p b →∞ b p−1 1 ( 1 0 − 1) = , = 1− p p−1

1 dx = xp

∫1

∞

(

Evaluate the improper integral

)

b p −1 → ∞ as b → ∞ because p − 1 > 0.

the series converges by the Integral Test. We emphasize that the sum of the p-series is not 1 ( p − 1). The series converges, but we don’t know the value it converges to. If p ≤ 0, the series diverges by the nth-term test. If 0 < p < 1, then 1 − p > 0 and ∞

∫1

1 1 dx = lim ( b 1− p − 1) = ∞. xp 1 − p b →∞

Therefore, the series diverges by the Integral Test. If p = 1, we have the (divergent) harmonic series 1+

1 1 1 + +  + + . 2 3 n

In summary, we have convergence for p > 1 but divergence for all other values of p. The p-series with p = 1 is the harmonic series (Example 1). The p-Series Test shows that the harmonic series is just barely divergent; if we increase p to 1.000000001, for instance, the series converges! The slowness with which the partial sums of the harmonic series approach infinity is impressive. For instance, it takes more than 178 million terms of the harmonic series to move the partial sums beyond 20. (See also Exercise 49b.) ∞

EXAMPLE 4 The series ∑ n =1 (1 ( n 2 + 1)) is not a p-series, but we will show that it converges by the Integral Test. The function f ( x ) = 1 ( x 2 + 1) is positive, continuous, and decreasing for x ≥ 1 since it is the reciprocal of a function that is positive, continuous, and increasing on this interval, and

∫1

∞

b 1 ⎡ arctan x ⎤ lim dx = ⎢ ⎥ b →∞ ⎣ ⎦1 x2 + 1 = lim [ arctan b − arctan 1 ] b →∞

=

π π π − = . 2 4 4

The Integral Test tells us that the series converges, but it does not say that π 4 is the sum of the series. EXAMPLE 5 ∞

(a)

∑ ne −n n =1

2

Determine the convergence or divergence of the series. ∞

(b)

∑ 2 1ln n n =1

564

Chapter 9 Infinite Sequences and Series

Solutions

(a) The function f ( x ) = xe − x 2 is continuous and positive for x ≥ 1. The first derivative f ′( x ) = e − x 2 − 2 x 2e − x 2 = (1 − 2 x 2 ) e − x 2 is negative on that interval (since x ≥ 1 implies 1 − 2 x 2 ≤ 1 − 2 < 0), so f is decreasing and we can apply the Integral Test:

∫1

∞

x e x2

1 ∞ du 2 ∫1 e u b 1 = lim ⎡⎢ − e −u ⎤⎥ b →∞ ⎣ 2 ⎦1 1 1 1 = . = lim − b + b →∞ 2e 2e 2e

dx =

(

u = x 2, du = 2 x dx

)

Since the integral converges, the series also converges. (b) The integrand 1 ( 2 ln x ) is a positive, continuous, and decreasing function on (1, ∞ ) , so we can apply the Integral Test. ∞

∫1

∞

e u du 2u ∞ e u du = ∫ 0 2 1 e = lim e b →∞ 2 ln 2

dx = 2 ln x

∫0

u = ln x , x = e u, dx = e u du

( )

(( ) ( )

b

)

−1 = ∞

(e 2) > 1

The improper integral diverges, so the series diverges also.

Error Estimation

y Remainder terms an+1 an+2

···

0

an+3

an+4

n + 1 n + 2 n +3 n+ 4 ···

n

x

(a)

R n = S − s n = a n +1 + a n + 2 + a n + 3 + .

y Remainder terms

an+1

0

For some convergent series, such as the geometric series or the telescoping series in Example 5 of Section 9.2, we can actually find the total sum of the series. That is, we can find the limiting value S of the sequence of partial sums. For most convergent series, however, we cannot easily find the total sum. Nevertheless, we can estimate the sum by adding the first n terms to get s n , but we need to know how far off s n is from the total sum S. An approximation to a function or to a number is more useful when it is accompanied by a bound on the size of the worst possible error that could occur. With such an error bound we can try to make an estimate or approximation that is close enough for the problem at hand. Without a bound on the error size, we are just guessing and hoping that we are close to the actual answer. We now show a way to bound the error size using integrals. Suppose that a series ∑ a n with positive terms is shown to be convergent by the Integral Test, and we want to estimate the size of the remainder R n measuring the difference between the total sum S of the series and its nth partial sum s n . That is, we wish to estimate

···

n

an+2 a n+3 an+4

n + 1 n + 2 n +3 n+ 4 ···

(b)

FIGURE 9.13

A geometric interpretation of Remainder Formula (1).

To get a lower bound for the remainder, we compare the sum of the areas of the rectangles with the area under the curve y = f ( x ) for x ≥ n (see Figure 9.13a). We see that x

R n = a n +1 + a n + 2 + a n + 3 +  ≥

∞

∫n +1 f ( x ) dx.

Similarly, from Figure 9.13b, we find an upper bound with R n = a n +1 + a n + 2 + a n + 3 +  ≤

∞

∫n

f ( x ) dx.

565

9.3  The Integral Test

These comparisons prove the following result, giving bounds on the size of the remainder. Bounds for the Remainder in the Integral Test Suppose { a k } is a sequence of positive terms with a k = f ( k ), where f is a con-

tinuous positive decreasing function of x for all x ≥ n, and suppose that ∑ a n converges to S. Then the remainder R n = S − s n satisfies the inequalities ∞

∫n +1 f ( x ) dx

≤ Rn ≤

∞

∫n

f ( x ) dx.

(1)

Since s n + R n = S, if we add the partial sum s n to each side of the inequalities in (1), we get sn +

∞

∫n +1 f ( x ) dx

≤ S ≤ sn +

∞

∫n

f ( x ) dx.

(2)

The inequalities in (2) are useful for estimating the error in approximating the sum of a series known to converge by the Integral Test. The error can be no larger than the length of the interval containing S, with endpoints given by (2). EXAMPLE 6 and n = 10.

Estimate the sum of the series

∑ (1 n 2 ) using the inequalities in (2)

Solution We have that ∞

∫n

(

)

b 1 ⎡ − 1 ⎤ = lim − 1 + 1 = 1 . = dx lim ⎢ ⎥ b →∞ ⎣ b →∞ x2 x⎦n b n n

Using this result with the inequalities in (2) gives 1 1 ≤ S ≤ s10 + . 11 10

s10 +

Since s10 = 1 + (1 4 ) + (1 9 ) + (1 16 ) +  + (1 100 ) ≈ 1.54977, these last inequalities give 1.64068 ≤ S ≤ 1.64977. If we approximate the sum S by the midpoint of this interval, we find that ∞

∑ n12 The p-series for p = 2

The error in this approximation is then less than half the length of the interval, so the error is less than 0.005. Using a trigonometric Fourier series, we prove in Section 19.4 that S is equal to π 2 6 ≈ 1.64493.

∞

2 ∑ n12 = π6 ≈ 1.64493 n =1

EXERCISES

9.3 ∞

Applying the Integral Test

Use the Integral Test to determine whether the series in Exercises 1–12 converge or diverge. Be sure to check that the conditions of the Integral Test are satisfied. ∞

1.

∑ n12

∞

2.

n =1

∑ n +1 4 n =1

∑ n10.2

∞

3.

n =1

∞

4.

≈ 1.6452.

n =1

n =1

∞

5.

∑ e −2 n n =1

∑ n 2 1+ 4 ∞

6.

∑ n ( ln1 n )2 n=2

7.

∑ n 2 n+ 4

∞

8.

n =1 ∞

9.

2

∑ enn 3

∞

10.

n =1 ∞

11.

∑

n =1

7 n+4

ln (n 2 ) n n=2

∑

∑ n 2 −n −2n4+ 1 n=2 ∞

12.

∑ 5n +110 n=2

n

566

Chapter 9 Infinite Sequences and Series

Determining Convergence or Divergence

The partial sums just grow too slowly. To see what we mean, suppose you had started with s1 = 1 the day the universe was formed, 13 billion years ago, and added a new term every second. About how large would the partial sum s n be today, assuming a 365-day year?

Which of the series in Exercises 13–46 converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series’ convergence or divergence.) ∞

13.

∞

∑ 101 n n =1

∞

5 16. ∑ n + 1 n =1

∑

17.

n =1

∞

19.

∞

ln n n n=2

∑

24.

∑ (1 + 1n ) n =1

30.

n=2

n =1

∞

31.

∑ ( ln12 )n

32.

∞

∑ ( ln n ) n=3

34.

39.

∑ 10 e+ e n

f (x) =

∑ 1 +2 e n n =1

40.

∑ (10 +e e n )2

∞

n

0

∞

42.

∑

n=3

∞

sn =

7 n + 1 ln n + 1

∞

8 tan −1 n 43. ∑ 2 n =1 1 + n

44.

∑ n 2 n+ 1 n =1

∞

∞

∑ sech n

46.

n =1

54.

∑ i=1 (1 n

∑( n =1

i + 1 ) satisfy s n > 1000?

∑ n=1 ( 1 n 4 ) converges ∞

a. Use the accompanying graph to find an upper bound for the 30 error if s 30 = ∑ n =1 (1 n 4 ) is used to estimate the value of ∞ ∑ n=1 (1 n 4 ). y

n =1

f (x) = 14 x

2×10−6

For what values of a, if any, do the series in Exercises 47 and 48 converge? 47.

x

1 2 3 4 5 ··· 48 49 50 51

∑ sech 2 n

Theory and Examples

∞

"x + 1

b. What should n be in order that the partial sum

n

n =1

n+2− n+1 n+1 n+2

1

···

∞

38.

n =1

45.

∑ n (1 +1ln 2 n )

1 36. ∑ n tan n n =1

n ∑ 1 +e e 2n n =1

n=2

y 1

n =1

∞

∞

∑ ( ln13)n

∞

37.

∑

51

∞

(1 n ) ln 2 n − 1

1 35. ∑ n sin n n =1

41.

n + 1 ) diverges

50 1 1 dx < s 50 < ∫ dx. 0 x +1 x +1 Conclude that 11.5 < s 50 < 12.3.

n =1

∞

∞

∞

∫1

1 n ( n + 1)

∞

n =1

33.

∑

∑ n=1 ( 1

a. Use the accompanying graph to show that the partial sum 50 s 50 = ∑ n =1 (1 n + 1 ) satisfies

n

∑ n 2+ 1 ∞

∑ ln nn

29.

53.

n =1

∞

n

n

∑ 4 n5+ 3 ∞

27.

n =1

∞

52. (Continuation of Exercise 51.) Is there a “largest” convergent series of positive numbers? Explain.

n =1

∞

n=0

28.

n

∑ 2n 1− 1

26.

ln n n=2 n

∑ ∞

n =1

∑ n−+2 1

∞

then there is also a divergent series ∑ n =1 b n of positive numbers with b n < a n for every n? Is there a “smallest” divergent series of positive numbers? Give reasons for your answers.

∞

21.

∑ 23 n

23.

∞

25.

−2 18. ∑ n n n =1

n =1

∞

∞

51. Is it true that if ∑ n =1 a n is a divergent series of positive numbers,

∞

3 n

∑ −n8

20.

∑ n +n 1 n =1

∞

∑ −81n n =1

22.

15.

n =1

∞

∞

50. Are there any values of x for which ∑ n =1 (1 nx ) converges? Give reasons for your answer.

∞

∑ e −n

14.

a 1 − n+2 n+4

∞

)

48.

∑( n=3

1 2a − n −1 n +1

)

49. a. Draw illustrations like those in Figures 9.12a and 9.12b to show that the partial sums of the harmonic series satisfy the inequalities n +1

1 1 1 dx ≤ 1 + +  + x 2 n n 1 dx = 1 + ln n. ≤1+ ∫ 1 x

ln ( n + 1) =

∫1

T b. There is absolutely no empirical evidence for the divergence of the harmonic series even though we know it diverges.

29

30

31

32

··· 33

x

∞

b. Find n so that the partial sum s n = ∑ i =1 (1 i 4 ) estimates ∞ the value of ∑ n =1 (1 n 4 ) with an error of at most 0.000001. (The exact value of this series is computed in Exercise 20 of Section 19.4.) ∞

55. Estimate the value of ∑ n =1 (1 n 3 ) to within 0.01 of its exact value. 56. Estimate the value of exact value.

∞

∑ n= 2 (1 ( n 2 + 4 )) to within 0.1 of its ∞

57. How many terms of the convergent series ∑ n =1 (1 n 1.1 ) should be used to estimate its value with error at most 0.00001?

9.4  Comparison Tests

∞

58. How many terms of the convergent series ∑ n = 4 1 ( n ( ln n )3 ) should be used to estimate its value with error at most 0.01? 59. The Cauchy condensation test The Cauchy condensation test says: Let { a n } be a nonincreasing sequence (a n ≥ a n +1 for all n) of positive terms that converges to 0. Then ∑ a n converges if and only if  ∑ 2 n a 2 n converges. For example,

∑ (1 n ) diverges because

∑ 2 n ⋅ (1 2 n ) = ∑ 1 diverges. Show why the test works.

567

a. By taking f ( x ) = 1 x in the proof of Theorem 9, show that ln ( n + 1) ≤ 1 +

1 1 +  + ≤ 1 + ln n 2 n

or 0 < ln ( n + 1) − ln n ≤ 1 +

1 1 +  + − ln n ≤ 1. 2 n

Thus, the sequence

60. Use the Cauchy condensation test from Exercise 59 to show that an = 1 +

∞

a.

∑ n ln1 n diverges; n=2

is bounded from below and from above.

∞

1 b. ∑ p converges if p > 1 and diverges if p ≤ 1. n =1 n

b. Show that 1 < n+1

61. Logarithmic p-series a. Show that the improper integral ∞

∫2

dx x ( ln x ) p

( p  a positive constant )

converges if and only if p > 1.

n=2

1+

∞

62. (Continuation of Exercise 61.) Use the result in Exercise 61 to determine which of the following series converge and which diverge. Support your answer in each case.

d.

n=2

2

n=0

converges. 65. a. For the series ∑ (1 n 3 ), use the inequalities in Equation (2) with n = 10 to find an interval containing the sum S.

∞

b. As in Example 5, use the midpoint of the interval found in part (a) to approximate the sum of the series. What is the maximum error for your approximation?

∑ n ( ln1 n )3 n=2

63. Euler’s constant Graphs like those in Figure 9.12 suggest that as n increases there is little change in the difference between the sum 1 1 1+ ++ 2 n and the integral ln n =

∑ e −n

∞

1 b. ∑ n ( ln n )1.01 n=2

∞

∑ n ln1(n 3 )

1 1 +  + − ln n → γ. n 2

64. Use the Integral Test to show that the series

Give reasons for your answer.

c.

1 dx = ln ( n + 1) − ln n, x

The number γ , whose value is 0.5772 … , is called Euler’s constant.

∞

∑ n ( ln1n ) p ?

1 a. ∑ ln n ( n) n=2

n +1

∫n

and use this result to show that the sequence { a n } in part (a) is decreasing. Since a decreasing sequence that is bounded from below converges, the numbers a n defined in part (a) converge:

b. What implications does the fact in part (a) have for the convergence of the series

∞

1 1 +  + − ln n 2 n

∫1

n

1 dx. x

To explore this idea, carry out the following steps.

66. Repeat Exercise 65 using the series ∑ (1 n 4 ).

67. Area Consider the sequence {1 n }∞ . On each subinterval n =1 (1 ( n + 1) , 1 n ) within the interval [ 0, 1 ], erect the rectangle with area a n having height 1 n and width equal to the length of the subinterval. Find the total area ∑ a n of all the rectangles. (Hint: Use the result of Example 5 in Section 9.2.) 68. Area Repeat Exercise 67, using trapezoids instead of rectangles. That is, on the subinterval (1 ( n + 1) , 1 n ) , let a n denote the area of the trapezoid having heights y = 1 ( n + 1) at x = 1 ( n + 1) and y = 1 n at x = 1 n .

9.4 Comparison Tests We have seen how to determine the convergence of geometric series, p-series, and a few others. We can test the convergence of many more series by comparing their terms to those of a series whose convergence is already known.

568

Chapter 9 Infinite Sequences and Series

THEOREM 10—Direct Comparison Test

Let œ a n  and  œ b n be two series with 0 b a n b b n for all n. Then 1. If œ b n converges, then œ a n also converges. 2. If œ a n diverges, then œ b n also diverges. Proof The series œ a n and œ b n have nonnegative terms. The Corollary of Theorem 6 stated in Section 9.3 tells us that the series œ a n and œ b n converge if and only if their partial sums are bounded from above. In Part (1) we assume that œ b n converges to some number M. The partial sums N ∑ n =1 a n are all bounded from above by M = ∑ bn because

y b1 b2 b3 a1

a 2 a3 1

2

b4 b 5

bn−1 b n a a4 a5 n 3 4 5 ··· n−1 n

n

If the total area œ b n of the taller b n rectangles is finite, then so is the total area œ a n of the shorter a n rectangles. FIGURE 9.14

s N = a1 + a 2 + " + a N ≤ b1 + b 2 + " + b N ≤

∞

∑ bn

= M.

n =1

Since the partial sums of œ a n are bounded from above, the Corollary of Theorem 6 implies that œ a n converges. We conclude that if œ b n converges, then so does œ a n . Figure 9.14 illustrates this result, with each term of each series interpreted as the area of a rectangle. ∞ In Part (2), where we assume that œ a n diverges, the partial sums of ∑ n =1 b n are not bounded from above. If they were, the partial sums for œ a n would also be bounded from above, since a1 + a 2 + " + a N ≤ b1 + b 2 + " + b N , and this would mean that œ a n converges. We conclude that if œ a n diverges, then so does œ b n . EXAMPLE 1

We apply Theorem 10 to several series.

(a) The series

∞

∑ 5n 5− 1 n =1

diverges because its nth term

HISTORICAL BIOGRAPHY Albert of Saxony (ca. 1316–1390) Albert attended the University of Paris and gained recogintion as a teacher at its faculty of arts. He wrote on squaring the circle and other geometric problems. He also published books on physics and mechanics, Tractatus proportionum being the most popular one. To know more, visit the companion Website.

∞ 1 5 1 1 = > diverges and 1 n > 0. ∑ 1 5n − 1 n n =1 n n− 5 is positive and is greater than the nth term of the (positive) divergent harmonic series. (b) The series ∞ ∑ n1! = 1 + 1!1 + 2!1 + 3!1 + " n=0

converges because its terms are all positive and less than or equal to the corresponding terms of ∞ 1 1 1 1 + ∑ n = 1 + 1 + + 2 + ". 2 2 2 n=0 The geometric series on the left converges (since r = 1 2 < 1) and we have 1+

∞

∑ 21n

n=0

= 1+

1 = 3. 1 − (1 2 ) ∞

The fact that 3 is an upper bound for the partial sums of ∑ n = 0 (1 n!) does not mean that the series converges to 3. As we will see in Section 9.9, the series converges to e.

9.4  Comparison Tests

569

(c) The series 5+

1 1 1 1 2 1 + +1+ + + ++ n 3 7 2 + 2+ 1 4+ 2 8+ 3

n

+

converges. To see this, we ignore the first three terms and compare the remaining terms ∞ with those of the convergent geometric series ∑ n = 0 (1 2 n ). The term 1 ( 2 n + n ) of the truncated sequence is positive and is less than the corresponding term 1 2 n of the geometric series. We see that term-by-term we have the comparison of positive terms 1+

1 1 1 1 1 1 + + +  ≤ 1 + + + + . 2 4 8 2+ 1 4+ 2 8+ 3

So the truncated series and the original series converge by an application of the Direct Comparison Test.

The Limit Comparison Test We now introduce a comparison test that is particularly useful for series in which a n is a rational function of n.

THEOREM 11—Limit Comparison Test

Suppose that a n > 0 and b n > 0 for all n ≥ N (N an integer). a 1. If lim n = c  and c > 0, then ∑ a n and ∑ b n both converge or both diverge. n →∞ b n a 2. If lim n = 0 and ∑ b n converges, then ∑ a n converges. n →∞ b n a 3. If lim n = ∞ and ∑ b n diverges, then ∑ a n diverges. n →∞ b n Proof

We will prove Part 1. Parts 2 and 3 are left as Exercises 57a and b. a We assume that lim n = c where c > 0. Then ε = c 2 is a positive number, so by n →∞ b n the definition of convergence there exists an integer N such that an c −c < bn 2

whenever

n > N.

Limit definition with ε = c 2, L = c, and a n replaced by a n b n

Thus, for n > N, a c c < n −c < , 2 bn 2 a c 3c < n < , bn 2 2 c 3c b < an < b . 2 n 2 n −

( )

( )

If ∑ b n converges, then ∑ ( 3c 2 ) b n converges and ∑ a n converges by the Direct Comparison Test. If ∑ b n diverges, then ∑ ( c 2 ) b n diverges and ∑ a n diverges by the Direct Comparison Test. EXAMPLE 2 (a)

Which of the following series converge, and which diverge?

3 5 7 9 + + + += 4 9 16 25

∞

∑ ( n2n++1)12 n =1

=

∞

∑ n 2 2+n 2+n 1+ 1 n =1

570

Chapter 9 Infinite Sequences and Series

(b) (c)

1 1 1 1 + + + += 1 3 7 15

∞

∑ 2 n 1− 1 n =1

1 + 2 ln 2 1 + 3 ln 3 1 + 4 ln 4 + + += 9 14 21

∞

1 + n ln n 2 n=2 n + 5

∑

Solution We apply the Limit Comparison Test to each series.

(a) Let a n = ( 2n + 1) ( n 2 + 2n + 1). For large n, we expect a n to behave like 2n n 2 = 2 n since the leading terms dominate for large n, so we let b n = 1 n. The numbers a n and b n are positive for each n, the series ∞

∑ bn

=

n =1

∞

∑ 1n diverges, n =1

and an 2n 2 + n = lim 2 = 2, n →∞ n + 2n + 1 n →∞ b n lim

so  ∑ a n diverges by Part 1 of the Limit Comparison Test. We could just as well have taken b n = 2 n , but 1 n is simpler. (b) Let a n = 1 ( 2 n − 1). For large n, we expect a n to behave like 1 2 n , so we let b n = 1 2 n. The numbers a n and b n are positive for each n, the series ∞

∑ bn = n =1

∞

∑ 21n converges, n =1

and lim

n →∞

an 2n 1 = lim n = lim = 1, n →∞ →∞ n bn 2 −1 1 − (1 2 n )

so  ∑ a n converges by Part 1 of the Limit Comparison Test.

(c) Let a n = (1 + n ln n ) ( n 2 + 5 ). For large n, we expect a n to behave like ( n ln n ) n 2 = ( ln n ) n , which is greater than 1 n for n ≥ 3, so we let b n = 1 n . The numbers a n and b n are positive for each n, the series ∞

∑ bn =

n=2

∞

∑ 1n diverges,

n=2

and n + n 2 ln n an = ∞, = lim n →∞ b n n →∞ n2 + 5 lim

so  ∑ a n diverges by Part 3 of the Limit Comparison Test. ∞

EXAMPLE 3

ln n   converge? 32 n n =1

Does ∑

Solution First note that both ln n and n 3 2 are positive for n ≥ 3, so the Limit Comparison Theorem can be applied. Because ln n grows more slowly than n c for any positive constant c (Section 9.1, Exercise 115), we can compare the series to a convergent p-series. To get the p-series, we see that ln n n1 4 1 < 32 = 54 32 n n n

571

9.4  Comparison Tests

for n sufficiently large. Then, taking a n = ( ln n ) n 3 2 and b n = 1 n 5 4 , we have lim

n →∞

ln n an = lim 1 4 n →∞ n bn 1n n →∞ (1 4 ) n −3 4

= lim

= lim

n →∞

  

L’Hôpital’s Rule

4 = 0. n1 4

Since ∑ b n = ∑ (1 n 5 4 ) is a p-series with p > 1, it converges. Therefore, ∑ a n converges by Part 2 of the Limit Comparison Test.

9.4

EXERCISES

∞

Direct Comparison Test

In Exercises 1–8, use the Direct Comparison Test to determine whether each series converges or diverges. ∞

1.

∞

∑ n 2 +1 30

2.

n =1

∞

∑ nn2 +− 2n

5.

n=2 ∞

7.

∑

n =1

3.

n =1

∞

4.

∞

∑ nn4 −+ 12 ∑

cos 2

n =1 ∞

n+4 n4 + 4

8.

∑

n =1

∑

n=2

1 n −1

20.

n

6.

∑ n13 n n =1

n +1 n2 + 3

∞

10.

∑

n =1

∞

∑ ( n 2 n+n1)+( n1− 1) (

)

12.

n=2 ∞

13.

∑

n =1

40.

n

∑ 3 +2 4 n ∞

14.

43.

∑ ( 52nn ++ 43 )

(Hint: Limit Comparison with ∑ n = 2 (1 n ))

)

50.

Determining Convergence or Divergence

n =1

∞

n

18.

∑ n +3 n =1

53.

∑3

n −1

19.

∞

36.

3n

n =1 ∞

41.

n=2

1 n2 − 1

∞

n

∑ nn+2 22n n =1

+1

∞

39.

∑ n n2 ++ 13n ⋅ 51n n =1

∑ 2 n2−n n n

∞

42.

n =1

∑ ( nn +− 12 )!! (

)

ln n n en

∑

n =1

∞

45.

tan −1 n n 1.1 n =1

∑

tanh n n2 n =1

∑

sin 2 n n n =1 2

∑

arcsec n n 1.3 n =1

∑

∞

49.

∞

∑ n n1 n

52.

n =1

n

∑ n n2

∑ 1 + 2 2 + 312 +  + n 2 n =1 ∞

56.

coth n n2

n =1

∞

54.

∑

n =1

∞

51.

∑ tan 1n n =1

∞

48.

∑ 1 + 2 + 31+  + n ∑ ( lnnn )2 n=2

46.

n =1

∞

55.

∞

∑ sin 1n

n =1

∞

n

)

∞

Which of the series in Exercises 17–56 converge, and which diverge? Use any method, and give reasons for your answers. 3

∑n

(

∑ 1n−2 nn ∞

38.

∞

∞

1 n +

33.

∑ ln nn++11 n =1

∞

47.

(Hint: Limit Comparison with ∑ n =1 (1 n 2 ))

∑2

( ln n ) 2 ∑ n3 2 n =1

∞

∞

35.

n =1

∞

17.

30.

n=3

∑ n1! ∞

44.

(

1 n ln n

1 ∑ ln( ln n)

(Hint: First show that (1 n!) ≤ (1 n ( n − 1)) for n ≥ 2.)

n =1

1 16. ∑ ln 1 + 2 n n =1

27.

n=2

n

n=2

∞

+ 3n + 4n

∞

1 n3 + 2

∞

n =1

5n n 4n

n

∑ 32 n n =1

∑ ln1n ∞

∑ 3 n−11 + 1 ∞

n   ))

∞

15.

∑ n 2 +n 1

3

n=2

n =1

∞

∞

∞

32.

∞

37.

n+1 n2 + 2

(Hint: Limit Comparison with ∑ n =1 (1 11.

∑ 1 +1ln n n =1

∞

∑

n=2

∞

(Hint: Limit Comparison with ∑ n =1 (1 n 2 ))

∑ ∞

29.

n =1

34.

n =1

∞

( ln n ) ∑ n3 n =1

3n ∑ n 2 ( n 5−n 2 − )( n 2 + 5 ) n=3

n =1

2

∞

31.

∞

∑ n 3 −n −n 2 2+ 3

26.

∑ nn 2+ n1 n =1

∞

24. ∞

n

n =1

Limit Comparison Test

9.

∑ ( 3n n+ 1 )

22.

n =1

n+1 ∑ n( n 10 + 1)( n + 2 ) ∞

25.

∞

∑ 3n2−n 1

n =1

28.

In Exercises 9–16, use the Limit Comparison Test to determine whether each series converges or diverges.

∞

21.

∞

23.

∞

n3 2

1 + cos n n2 n =1

∑

( ln n ) 2 ∑ n n=2

572

Chapter 9 Infinite Sequences and Series

Theory and Examples

COMPUTER EXPLORATIONS

57. Prove (a) Part 2 and (b) Part 3 of the Limit Comparison Test.

73. It is not yet known whether the series

58. If

∞

∑ n=1 a n

is a convergent series of nonnegative numbers, can

∞

1 ∑ n 3 sin 2n

∞

anything be said about ∑ n =1 ( a n n )? Explain.

n =1

59. Suppose that a n > 0 and b n > 0 for n ≥ N (N an integer). If lim ( a n b n ) = ∞ and

n →∞

∑ an

converges or diverges. Use a CAS to explore the behavior of the series by performing the following steps.

converges, can anything be said

about ∑ b n? Give reasons for your answer.

60. Prove that if

∑ an

a. Define the sequence of partial sums

is a convergent series of nonnegative terms,

then ∑ a n2 converges.

61. Suppose that a n > 0 and lim a n = ∞. Prove that n →∞ diverges.

sk =

∑ an

b. Plot the first 100 points ( k , s k ) for the sequence of partial sums. Do they appear to converge? What would you estimate the limit to be?

∞

63. Show that ∑ n = 2 (( ln n ) q n p ) converges for −∞ < q < ∞ and

c. Next plot the first 200 points ( k , s k ). Discuss the behavior in your own words.

∞

(Hint: Limit Comparison with ∑ n = 2 1 n r for 1 < r < p.) 64. (Continuation of Exercise 63.) Show that

∞

∑ n= 2 (( ln n )q

np)

d. Plot the first 400 points ( k , s k ). What happens when k = 355? Calculate the number 355 113. Explain from your calculation what happened at k = 355. For what values of k would you guess this behavior might occur again?

diverges for −∞ < q < ∞ and 0 < p < 1.

(Hint: Limit Comparison with an appropriate p-series.) 65. Decimal numbers Any real number in the interval [ 0, 1 ] can be represented by a decimal (not necessarily unique) as 0.d1d 2 d 3 d 4  =

74. a. Use Theorem 8 to show that

d d1 d d + 22 + 33 + 44 + , 10 10 10 10

S =

where d i is one of the integers 0, 1, 2, 3, … , 9. Prove that the series on the right-hand side always converges.

∑ a n is a convergent ∑ sin (a n ) converges.

66. If

where S =

67.

( ln n ) n4 n=2

∑

( ln n )1000 69. ∑ n 1.001 n=2

∞

68.

∑ n 1.1 ( 1ln n )3 n=2

∑

n=2

∞

n =1

n =1

72.

∑

n=2

)

1 , n ( n + 1)

∞

∑ n=1 (1 n 2 ), the sum of a convergent p-series. ∞

∑ n 2 ( n1+ 1).

c. Explain why taking the first M terms in the series in part (b) gives a better approximation to S than taking the first M terms ∞ in the original series ∑ n =1 (1 n 2 ).

ln n n

d. We know the exact value of S is π 2 6. Which of these sums,

( ln n )1 5 70. ∑ n 0.99 n=2 ∞

−

n =1

∞

∞

71.

∞

S = 1+

3

∞

∑ n ( n 1+ 1) + ∑ ( n12

b. From Example 5, Section 9.2, show that

series of positive terms, prove that

In Exercises 67–72, use the results of Exercises 63 and 64 to determine whether each series converges or diverges. ∞

n =1

What happens when you try to find the limit of s k as k → ∞ ? Does your CAS find a closed form answer for this limit?

62. Suppose that a n > 0 and lim n 2 a n = 0. Prove that ∑ a n conn →∞ verges. p > 1.

k

1 . ∑ n 3 sin 2n

1000000

∑

n =1

1 n ⋅ ln n

1 n2

or 1 +

1000

∑ n 2 ( n1+ 1) , n =1

gives a better approximation to S?

9.5 Absolute Convergence; The Ratio and Root Tests When some of the terms of a series are positive and others are negative, the series may or may not converge. For example, the geometric series 5− converges ( since r =

5 5 5 + − += 4 16 64

∞

∑ 5( −41 )

n

(1)

n=0

1 < 1) , whereas the different geometric series 4 1−

5 25 125 + − += 4 16 64

∞

∑ ( −45 )

n=0

n

(2)

9.5  Absolute Convergence; The Ratio and Root Tests

573

diverges (since r = 5 4 > 1). In series (1), there is some cancelation in the partial sums, which may be assisting the convergence property of the series. However, if we make all of the terms positive in series (1) to form the new series 5+

5 5 5 + + += 4 16 64

∞

∑

n=0

5

( −41)

n

=

∞

∑ 5( 4 ) , 1

n

n=0

we see that it still converges. For a general series with both positive and negative terms, we can apply the tests for convergence that we studied before to the series of absolute values of its terms. In doing so, we are led naturally to the following concept. DEFINITION A series ∑ a n converges absolutely (is absolutely convergent)

if the corresponding series of absolute values, ∑ a n , converges.

So the geometric series (1) is absolutely convergent. We observed, too, that it is also convergent. This situation is always true: An absolutely convergent series is convergent as well, which we now prove. Caution Be careful when using Theorem 12. A convergent series need not converge absolutely, as you will see in the next section.

THEOREM 12—The Absolute Convergence Test ∞

∞

n =1

n =1

If ∑ a n converges, then ∑ a n converges.

For each n,

Proof

− an ≤ an ≤ an , ∞

0 ≤ an + an ≤ 2 an .

so

∞

If  ∑ n =1 a n converges, then  ∑ n =1 2 a n converges and, by the Direct Comparison Test, ∞

the nonnegative series ∑ n =1 ( a n + a n )  converges. The equality  a n = ( a n + a n ) − a n ∞

now lets us express  ∑ n =1 a n   as the difference of two convergent series: ∞

∑ an = n =1

∞

∑

n =1

( an + an − an ) =

∞

∑

n =1

( an + an ) −

∞

∑

n =1

an .

∞

Therefore,  ∑ n =1 a n converges. EXAMPLE 1 ∞

This example gives two series that converge absolutely.

1 1 1 + − + , the corresponding series of absolute 4 9 16 n =1 values is the convergent series

(a) For

∑ (−1)n +1 n12

= 1− ∞

∑ n12 n =1

= 1+

1 1 1 + + + . 4 9 16

The original series converges because it converges absolutely. ∞ sin n sin 1 sin 2 sin 3 (b) For  ∑ 2 = + + + ,  which contains both positive and negative n 1 4 9 n =1 terms, the corresponding series of absolute values is ∞

∑

n =1

sin n sin 1 sin 2 = + + , n2 1 4 ∞

which converges by comparison with ∑ n =1 (1 n 2 ) because sin n ≤ 1 for every n. The original series converges absolutely; therefore, it converges.

574

Chapter 9 Infinite Sequences and Series

The Ratio Test The Ratio Test measures the rate of growth (or decline) of a series by examining the ratio a n +1 a n. For a geometric series ∑ ar n , this rate is a constant (( ar n +1 ) ( ar n ) = r ) , and the series converges if and only if its ratio is less than 1 in absolute value. The Ratio Test is a powerful rule extending that result.

THEOREM 13—The Ratio Test

Let ∑ a n be any series and suppose that ρ is the Greek lowercase letter rho,

lim

which is pronounced “row.”

n →∞

a n +1 = ρ. an

Then (a) the series converges absolutely if ρ < 1, (b) the series diverges if ρ > 1 or ρ is infinite, and (c) the test is inconclusive if ρ = 1.

Proof (a) ρ < 1. Let r be a number between ρ and 1. Then the number ε = r − ρ is positive. Since a n +1 → ρ, an a n +1 a n must lie within ε of ρ when n is large enough, say, for all n ≥ N. In particular, a n +1 < ρ + ε = r, an

when n ≥ N .

Hence a N +1 < r a N , a N + 2 < r a N +1 < r 2 a N , a N +3 < r a N +2 < r 3 a N ,  a N + m < r a N + m −1 < r m a N . Therefore, ∞

∑

m= N

am =

∞

∑

m=0

aN +m ≤

∞

∑

m=0

aN r m = aN

∞

∑ r m.

m=0

The geometric series on the right-hand side converges because 0 < r < 1, so the ∞ series of absolute values ∑ m = N a m converges by the Direct Comparison Test. Because adding or deleting finitely many terms in a series does not affect its conver∞ gence or divergence property, the series ∑ n =1 a n also converges. That is, the series ∑ a n is absolutely convergent. (b) 1 < ρ ≤ ∞. From some index M on, a n +1 >1 an

and

a M < a M +1 < a M + 2 < .

The terms of the series do not approach zero as n becomes infinite, so the series diverges by the nth-Term Test.

9.5  Absolute Convergence; The Ratio and Root Tests

575

(c) ρ = 1. The two series ∞

∞

1 ∑n n =1

1

∑ n2

and

n =1

show that some other test for convergence must be used when ρ = 1. ∞

For

1 ( n + 1) a n +1 n = = → 1. an 1n n+1

∑ 1n : n =1

∞

For

(

1 ( n + 1) 2 a n +1 n = = an n+1 1 n2

∑ n12 : n =1

)

2

→ 1 2 = 1.

In both cases, ρ = 1, yet the first series diverges, whereas the second converges. The Ratio Test is often effective when the terms of a series contain factorials of expressions involving n or expressions raised to a power involving n. EXAMPLE 2 ∞

(a)

∑2

n=0

n

Investigate the convergence of the following series. ∞

+5 3n

(b)

∑

∞

( 2n )!

(c)

n! n!

n =1

n

∑ 4( 2nn!)n!! n =1

Solution We apply the Ratio Test to each series. ∞

(a) For the series ∑ n = 0

2n + 5 , 3n

(

( 2 n +1 + 5 ) 3 n +1 a n +1 1 2 n +1 + 5 1 2 + 5 ⋅ 2 −n = ⋅ = ⋅ n = ( 2 n + 5) 3n an 3 2 +5 3 1 + 5 ⋅ 2 −n

) → 13 ⋅ 12 = 23 .

The series converges absolutely (and thus converges) because ρ = 2 3 is less than 1. This does not mean that 2 3 is the sum of the series. In fact, ∞

∑2

n=0

(b) If a n =

n

+5 = 3n

( 2n )!

n! n!

∞

∑ ( 23 )

n

+

n=0

, then a n +1 =

∞

∑ 35n

n=0

=

1 5 21 + = . 1 − ( 2 3 ) 1 − (1 3 ) 2

+ 2 )!   and ( n + 1)!( n + 1)! ( 2n

a n +1 n! n!( 2n + 2 )( 2n + 1)( 2n )! = ( n + 1)!( n + 1)!( 2n )! an =

( 2n (n

+ 2 )( 2n + 1) 4n + 2 = → 4. + 1)( n + 1) n+1

The series diverges because ρ = 4 is greater than 1. (c) If a n =

4 n n! n! ,  then ( 2n )! a n +1 ( 2n )! 4 n +1 ( n + 1)!( n + 1)! = ⋅ n ( )( )( ) an 2n + 2 2n + 1 2n ! 4 n ! n ! =

4 ( n + 1)( n + 1) 2 ( n + 1) = → 1. + 2 )( 2n + 1) 2n + 1

( 2n

Because the limit is ρ = 1, we cannot decide from the Ratio Test whether the series converges. However, when we notice that a n +1 a n = ( 2n + 2 ) ( 2n + 1), we conclude that a n +1 is always greater than a n because ( 2n + 2 ) ( 2n + 1) is always greater than 1. Therefore, all terms are greater than or equal to a1 = 2, and the nth term does not approach zero as n → ∞. The series diverges.

576

Chapter 9 Infinite Sequences and Series

The Root Test The convergence tests for ∑ a n that we have studied so far work best when the formula for an is relatively simple. However, consider the series with the terms n ⎪⎧ n 2 , n odd a n = ⎪⎨ n ⎪⎪⎩1 2 , n even.

To investigate convergence we write out several terms of the series: ∞

∑ an

=

1 1 3 1 5 1 7 + 2 + 3 + 4 + 5 + 6 + 7 + 21 2 2 2 2 2 2

=

1 3 1 5 1 7 1 + + + + + + + . 2 4 8 16 32 64 128

n =1

Clearly, this is not a geometric series. The nth term approaches zero as n → ∞, so the nthTerm Test does not tell us whether the series diverges. The Integral Test does not look promising. The Ratio Test produces ⎪⎧⎪ 1 , n odd a n +1 ⎪ 2n = ⎪⎨ ⎪⎪ n + 1 an ⎪⎪⎩ 2 , n even. As n → ∞, the ratio is alternately small and large and therefore has no limit. However, we will see that the following test establishes that the series converges. THEOREM 14—The Root Test

Let ∑ a n be any series and suppose that lim

n

n →∞

a n = ρ.

Then (a) the series converges absolutely if ρ < 1, (b) the series diverges if ρ > 1 or ρ is infinite, and (c) the test is inconclusive if ρ = 1. Proof (a) ρ < 1. Choose an ε > 0 so small that ρ + ε < 1. Since n a n → ρ , the terms n a n eventually get to within ε of ρ. So there exists an index M such that n

an < ρ + ε

when n ≥ M .

Then it is also true that a n < ( ρ + ε )n ∞

for n ≥ M .

Now, ∑ n = M ( ρ + ε ) is a geometric series with ratio 0 < ( ρ + ε ) < 1 and there∞ fore converges. By the Direct Comparison Test, ∑ n = M a n converges. Adding finitely many terms to a series does not affect its convergence or divergence, so the series n

∞

∑

n =1

a n = a1 +  + a M −1 +

∞

∑

n= M

an

also converges. Therefore, ∑ a n converges absolutely.

(b) 1 < ρ ≤ ∞. For all indices beyond some integer M, we have n a n > 1, and therefore a n > 1 for n > M. The terms of the series do not converge to zero. The series diverges by the nth-Term Test. ∞

∞

(c) ρ = 1. The series ∑ n =1 (1 n ) and ∑ n =1 (1 n 2 ) show that the test is not conclusive when ρ = 1. The first series diverges and the second converges, but in both cases n a n → 1.

9.5  Absolute Convergence; The Ratio and Root Tests

⎧⎪ n 2 n, EXAMPLE 3 Consider again the series with terms a n = ⎪⎨ n ⎪⎪⎩ 1 2 , Does ∑ a n converge?

577

n odd n even.

Solution We apply the Root Test, finding that n ⎪⎧ n 2, a n = ⎪⎨ ⎪⎪⎩ 1 2,

n

n  odd n  even.

Therefore, 1 ≤ 2 n

Since

n

an ≤

n

n . 2

n → 1 (Section 9.1, Theorem 5), we have lim

n

n →∞

a n = 1 2 by the Sandwich

Theorem. The limit is less than 1, so the series converges absolutely by the Root Test. EXAMPLE 4 ∞

(a)

Which of the following series converge, and which diverge? ∞

2

∑ n2 n

(b)

n =1

∞

n

∑ 2n 3

(c)

n =1

∑ ( 1 +1 n )

n

n =1

Solution We apply the Root Test to each series, noting that each series has positive terms. ∞

2 ∑ n2 n converges because n =1

(b)

∑ 2n 3 diverges because n 2n 3

∞

n

n

n

∞

∑ ( 1 +1 n )

n

n

converges because

n

n =1

EXERCISES

2.

n =1

3.

)

4.

n =1

+ 2 )! n! 3 2 n

∞

8.

∞

16.

n

∑ ( 2n + 3n)5ln ( n + 1) n =1

∞

10.

n =1 ∞

∑ ( 43nn −+ 53 ) n =1

∑ (− ln( e 2 + 1n )) n =1

n2

( − 1) n

n 1+ n

17.

2

∑ n2 n

∞

18.

n =1

19.

∑ n!(−e )−n

∞

20.

n =1

n +1

∞

21.

10

n ∑ 10 n n =1

∑ (−1)n n 2e −n n =1

∞

n

∑ ( 34n )n ∞

12.

)

In Exercises 17–46, use any method to determine whether the series converges or diverges. Give reasons for your answer.

n =1

n

1 n

Determining Convergence or Divergence

∞

∞

11.

∑

n=2

In Exercises 9–16, use the Root Test to determine whether each series converges absolutely or diverges.

∑ ( 2n +7 5 )n

n =1

n →∞

Using the Root Test

9.

∑ (−1)n (1 − 1n )

∑ sin n (

(Hint: lim (1 + x n ) n = e x )

n +1

∑ n23 n−1

3 n+2 6. ∑ n = 2 ln n 2(n

14.

n =1

∞

n4 5. ∑ ( 4 )n − n =1 ∞

15.

n =1

∞

∑ ( − 1) n n

1 → 0 < 1. 1+ n

∞

∑ ( 3 + −(18n ))2n ∞

∑ (−1)n n 3+n 2 ∞

n =1

7.

=

n =1

n =1

∑ ( nn +− 11)!2 (

13.

∞

n

∑ 2n! ∞

n

∞

In Exercises 1–8, use the Ratio Test to determine whether each series converges absolutely or diverges. ∞

(1 +1 n )

9.5

Using the Ratio Test

1.

n

2

( n n) n2 12 = → < 1. n 2 2 2

2 2 3 → 3 > 1. 1 ( n)

=

n =1

(c)

n

n2 = 2n

(a)

∑ 10n!n n =1 ∞

22.

∑ ( n −n 2 ) n =1

n

578

Chapter 9 Infinite Sequences and Series

∞

23.

∞

∑ 2 +1.25−n1 (

)n

24.

n =1 ∞

25.

∑ (−1)n (1 − n3 )

26.

∑( n =1

1 1 − 2 n n

28.

)

∞

(n

n =1 ∞

35.

∑

34.

38.

n! ln n

42.

(

n =1 ∞

)2

∑ ( n2!n )!

44.

n =1

∑

∞

)

∑ [ 2 ⋅14⋅ ⋅3⋅ ⋅ (⋅2n2)n]( 3−n 1+ 1) (

n a n+1 n

51. a1 = 2, a n +1 =

2 a n n

52. a1 = 5, a n +1 = 53. a1 = 1, a n +1 =

n

n a 2 n

1 + ln n an n

n + ln n 1 , a n +1 = a 2 n + 10 n

)

∞

∑ (b n ) 1 n

b.

∞

c.

+ 3 )( 2 n + 3 ) 3n + 2

n

(b n )

∞

∑ (b n ) n

d.

n =1

n32n

∑ ( 45 ) n =1

n

∑ n1000 ! + bn n =1

66. Assume that b n is a sequence of positive numbers converging to 1 3. Determine whether the following series converge or diverge. ∞

a.

∞

b n +1b n n n =1 n 4

∑

b.

n

∑ n! b12 bn22  bn2 n =1

Theory and Examples

67. Neither the Ratio Test nor the Root Test helps with p-series. Try them on ∞ ∑ n1p n =1 and show that both tests fail to provide information about convergence. 68. Show that neither the Ratio Test nor the Root Test provides information about the convergence of ∞

1 3n − 1 , a n +1 = a 3 2n + 5 n

50. a1 = 3, a n +1 =

54. a1 =

(

n =1

Recursively Defined Terms Which of the series ∑ n =1 a n defined by the formulas in Exercises 47–56 converge, and which diverge? Give reasons for your answers. 1 + sin n 47. a1 = 2, a n +1 = an n

49. a1 =

⋅ 2n − 1 ∑1 ⋅ 3 ⋅ 4 n 2 n n!

∞

∞

1 + tan −1 n an n

n =1

65. Assume that b n is a sequence of positive numbers converging to 4 5. Determine whether the following series converge or diverge.

2 n2 46. ∑ 2 n n=3 n

48. a1 = 1, a n +1 =

n

∑ ( 2nn )2

n =1

∞

2n 45. ∑ 2 n=3 n

62.

2

(−3 ) n

n =1

)n 2

∞

n

∑ 2n(n )

a.

( 2n

(

∑ (−1)n nn(!n ) n =1

∞

64.

∑ (−nn!) n n =1

∑

)

n =1

n 40. ∑ ( )( n 2 ) ln n n=2

n =1 ∞

63.

)n (

(

∞

60.

∞

∞

∑ n ( n + 2 )!

)

n =1

∑ e −n ( n 3 )

∞

(

∑ n!( n −+11)!(3nn+! 2 )! n =1

n

∑ (nnn! ) 2 ∞

61.

∞

∑ ( 2n n+! 1)! n =1

58.

n =1

n 2 n ( n + 1)! 36. ∑ 3 n n! n =1

−n 39. ∑ ln n )n ( n=2

43.

59.

∞

∞

41.

n

n =1

+ 3 )! 3! n!3 n

∞

∞

∞

+ 1)( n + 2 ) n!

∞

37.

)

1 , a n +1 = (a n ) n +1 2

∞

n

∑ 2( 2nn!)n!! n =1

n ln n 32. ∑ ( )n n =1 − 2

(n

n =1

57.

∞

en 31. ∑ e n =1 n

∑

∑( n =1

∞

33.

∞

(− ln n ) n nn n =1 1 1 − 2 n n

56. a1 =

an

Which of the series in Exercises 57–64 converge, and which diverge? Give reasons for your answers.

∑ ∞

30.

n

Convergence or Divergence

n

∞

ln n 3 n =1 n ∞

∑ (1 − 31n ) n =1

∑

1 , a n +1 = 3

55. a1 =

3n

∞

n

∞

29.

(−2 ) n

n =1

n =1

27.

∑

∑ ( ln1n ) p

(p

constant ).

n=2

n ⎪⎧ n 2 , if  n  is a prime number 69. Let a n = ⎨ n ⎪⎪⎩1 2 , otherwise.

Does ∑ a n converge? Give reasons for your answer. 70. Show that

∞

∑ n=1 2 (n )

Exponents that

2 (n 2 )

2

n! diverges. Recall from the Laws of

= ( 2 n ) n. ∞

71. Determine whether the series ∑ n =1 c n converges, where ⎪⎧−1 n , if  n  is a perfect square, cn = ⎨ 2 ⎪⎪⎩1 n , if n  is not a perfect square.

.

9.6  Alternating Series and Conditional Convergence

579

9.6 Alternating Series and Conditional Convergence A series in which the terms are alternately positive and negative is an alternating series. Here are three examples: ( − 1) n + 1 1 1 1 1 + − + −+ + 2 3 4 5 n ( − 1) n 4 1 1 1 −2 + 1 − + − +  + + 2 4 8 2n

1−

1 − 2 + 3 − 4 + 5 − 6 +  + (−1) n +1 n + 

(1) (2) (3)

We see from these examples that the nth term of an alternating series is of the form a n = ( −1) n +1 u n

a n = ( −1) n u n ,

or

where u n = a n is a positive number. Series (1), called the alternating harmonic series, converges, as we will see in a moment. Series (2), which is a geometric series with ratio r = −1 2, converges to −2 [ 1 + (1 2 ) ] = −4 3. Series (3) diverges because the nth term does not approach zero. We prove the convergence of the alternating harmonic series by applying the Alternating Series Test. This test is for convergence of an alternating series and cannot be used to conclude that such a series diverges. If we multiply ( u1 − u 2 + u 3 − u 4 + ) by −1, we see that the test is also valid for the alternating series −u1 + u 2 − u 3 + u 4 − , as with the one in Series (2) given above. +u1

THEOREM 15—The Alternating Series Test

−u2

The series

+u3

∞

∑ ( − 1) n + 1 u n

−u4

n =1

0

s2

s4

L

s3

FIGURE 9.15 The partial sums of an alternating series that satisfies the hypotheses of Theorem 15 for N = 1 straddle the limit from the beginning.

s1

x

= u1 − u 2 + u 3 − u 4 + 

converges if the following conditions are satisfied: 1. The u n’s are all positive. 2. The u n’s are eventually nonincreasing: u n ≥ u n +1 for all n ≥ N, for some integer N. 3. u n → 0. Proof We look at the case N = 1, where we have nonincreasing terms u1 ≥ u 2 ≥ u 3 ≥ . If n is an even integer, say n = 2 m, then the sum of the first n terms is s 2 m = ( u1 − u 2 ) + ( u 3 − u 4 ) +  + ( u 2 m −1 − u 2 m ). Written this way, we see that s 2 m is the sum of n nonnegative terms, since each term in parentheses is positive or zero. In particular, s 2 m + 2 equals s 2 m plus the nonnegative term ( u 2 m +1 − u 2 m + 2 ) , so s 2 m + 2 ≥ s 2 m . This shows that the sequence { s 2 m } is nondecreasing. On the other hand, we can write s 2 m = u1 − ( u 2 − u 3 ) − ( u 4 − u 5 ) −  − ( u 2 m − 2 − u 2 m −1 ) − u 2 m . This shows that s 2 m ≤ u1 for every m. Hence the sequence { s 2 m } is both nondecreasing and bounded from above, so it has a limit, which we call L, lim s 2 m = L.

m →∞

Theorem 6

(4)

If n is an odd integer, say n = 2 m + 1, then the sum of the first n terms is s 2 m +1 = s 2 m + u 2 m +1. Since u n → 0, lim u 2 m +1 = 0.

m →∞

580

Chapter 9 Infinite Sequences and Series

Therefore, lim s 2 m +1 = lim s 2 m + lim u 2 m +1 = L + 0 = L.

m →∞

m →∞

(5)

m →∞

Combining the results of Equations (4) and (5) gives lim s n = L (Section 9.1, n →∞ Exercise 143). EXAMPLE 1

The alternating harmonic series ∞

∑ (−1)n +1 1n

= 1−

n =1

1 1 1 + − + 2 3 4

clearly satisfies the three requirements of Theorem 15 with N = 1; it therefore converges by the Alternating Series Test. Notice that the test gives no information about what the sum of the series might be. Figure 9.16 shows histograms of the partial sums of the divergent harmonic series and those of the convergent alternating harmonic series. It turns out that the alternating harmonic series converges to ln 2 (Exercise 61 in Section 9.7). sn increases, eventually becoming larger than any constant M

y

y

M 1

1

1.5 1.0

1

1

1

1

1+− 2 +− 3 +− 4 +− 5 +− 6

2.0 1 1+− 2

1

1

1

1.00

0.50 0.25

2

3

4

5

6

7

1

1

1

1−− 2 +− 3 −− 4 +− 5

0.75

1 1 1 1+− 2 +− 3 +− 4 1 1 1+− 2 +− 3

0.5 1

1

1−− 2 +− 3

1

1+− 2 +− 3 +− 4 +− 5

1

1

1

x

(a)

ln 2 1

1−− 2

1

2

1 1 1 1−− 2 +− 3 −− 4 1 1 1 1 1 1−− 2 +− 3 −− 4 +− 5 −− 6

3

4

5

6

7

x

(b)

FIGURE 9.16 (a) The harmonic series diverges, with partial sums that eventually exceed any constant. (b) The alternating harmonic series converges to ln 2 ≈ .693.

Rather than directly verifying the definition u n ≥ u n +1 , a second way to show that the sequence { u n } is nonincreasing is to define a differentiable function f ( x ) satisfying f (n) = u n . That is, the values of f match the values of the sequence at every positive integer n. If f ′( x ) ≤ 0 for all x greater than or equal to some positive integer N, then f ( x ) is nonincreasing for x ≥ N. It follows that f (n) ≥ f ( n + 1) , or u n ≥ u n +1 , for n ≥ N. EXAMPLE 2 We show that the sequence u n = 10 n ( n 2 + 16 ) is eventually nonincreasing. Define f ( x ) = 10 x ( x 2 + 16 ). Then, from the Derivative Quotient Rule, f ′( x ) =

10 (16 − x 2 ) 2 ≤ 0 ( x 2 + 16 )

whenever  x ≥ 4.

It follows that u n ≥ u n +1 for n ≥ 4. That is, the sequence { u n } is nonincreasing for n ≥ 4. A graphical interpretation of the partial sums (Figure 9.15) shows how an alternating series converges to its limit L when the three conditions of Theorem 15 are satisfied with N = 1. Starting from the origin of the x-axis, we lay off the positive distance s1 = u1. To find the point corresponding to s 2 = u1 − u 2 , we back up a distance equal to u 2 . Since u 2 ≤ u1 , we do not back up any farther than the origin. We continue in this seesaw fashion, backing up or going forward as the signs in the series demand. Each forward or backward step is shorter than (or at most the same size as) the preceding step because u n +1 ≤ u n . And since the nth term approaches zero as n increases, the size of step we take forward or backward gets smaller and smaller. We oscillate back and forth across the limit L, and the amplitude of oscillation approaches zero. The limit L lies between any two successive

9.6  Alternating Series and Conditional Convergence

581

sums s n and s n +1 and hence differs from s n by an amount less than u n +1. If N > 1, then the first few forward and backward steps need not get smaller, but from N onward they will decrease. Therefore L − s n < u n +1

for n ≥ N ,

and we can use this fact to make useful estimates of the sums of convergent alternating series. THEOREM 16—The Alternating Series Estimation Theorem ∞

If the alternating series ∑ n =1 (−1) n +1 u n satisfies the three conditions of Theorem 15, then for n ≥ N, s n = u 1 − u 2 +  + ( − 1) n + 1 u n approximates the sum L of the series with an error whose absolute value is less than u n +1 , the absolute value of the first unused term. Furthermore, the sum L lies between any two successive partial sums s n and s n +1 , and the remainder, L − s n , has the same sign as the first unused term. We leave the verification of the sign of the remainder for Exercise 87. EXAMPLE 3 ∞

We try Theorem 16 on a series whose sum we know:

∑ (−1)n 21n

n=0

= 1−

1 1 1 1 1 1 1 1 + − + − + − + − . 2 4 8 16 32 64 128 256

The theorem says that if we truncate the series after the eighth term, we throw away a total that is positive and less than 1 256. The sum of the first eight terms is s 8 = 0.6640625 and the sum of the first nine terms is s 9 = 0.66796875. The sum of the geometric series is 1 1 2 = = , 1 − (−1 2 ) 32 3 and we note that 0.6640625 < ( 2 3 ) < 0.66796875. The difference,

( 2 3 ) − 0.6640625 = 0.0026041666 …, is positive and is less than (1 256 ) = 0.00390625.

Conditional Convergence If we replace all the negative terms in the alternating series in Example 3, changing them to positive terms instead, we obtain the geometric series ∑ 1 2 n. The original series and the new series of absolute values both converge (although to different sums). For an absolutely convergent series, changing infinitely many of the negative terms in the series to positive values does not change its property of still being a convergent series. Other convergent series may behave differently. The convergent alternating harmonic series has infinitely many negative terms, but if we change its negative terms to positive values, the resulting series is the divergent harmonic series. So the presence of infinitely many negative terms is essential to the convergence of this alternating harmonic series. The following terminology distinguishes these two types of convergent series. DEFINITION A series that is convergent but not absolutely convergent is called

conditionally convergent. The alternating harmonic series is conditionally convergent, or converges conditionally. The next example extends that result to the alternating p-series.

582

Chapter 9 Infinite Sequences and Series

EXAMPLE 4 If p is a positive constant, the sequence {1 n p } is a decreasing sequence with limit zero. Therefore, the alternating p-series ∞

∑

( − 1) n −1

n =1

np

= 1−

1 1 1 + p − p + , 2p 3 4

p > 0

converges. If p > 1, the series converges absolutely as an ordinary p-series. If 0 < p ≤ 1, the series converges conditionally: It converges by the alternating series test, but the corresponding series of absolute values is a divergent p-series. For instance, 1 + 23 2 1 Conditional convergence  ( p = 1 2 ): 1 − + 2 Absolute convergence  ( p = 3 2 ): 1 −

1 1 − 32 + 33 2 4 1 1 − + 3 4

We need to be careful when using a conditionally convergent series. We have seen with the alternating harmonic series that altering the signs of infinitely many terms of a conditionally convergent series can change its convergence status. Even more, simply changing the order of occurrence of infinitely many of its terms can also have a significant effect, as we now discuss.

Rearranging Series We can always rearrange the terms of a finite collection of numbers without changing their sum. The same result is true for an infinite series that is absolutely convergent (see Exercise 96 for an outline of the proof ).

THEOREM 17—The Rearrangement Theorem for Absolutely Convergent Series ∞

If  ∑ n =1 a n  converges absolutely, and  b1 , b 2 , …, b n , …  is any arrangement of the ∞ sequence { a n } , then ∑ n =1 b n converges absolutely and ∞

∑ bn n =1

=

∞

∑ an. n =1

On the other hand, if we rearrange the terms of a conditionally convergent series, we can get different results. In fact, for any real number r, a given conditionally convergent series can be rearranged so that its sum is equal to r. (We omit the proof of this.) Here’s an example where summing the terms of a conditionally convergent series with different orderings gives different values for the sum. ∞

EXAMPLE 5 We know that the alternating harmonic series ∑ n =1 (−1) n +1 n converges to some number L. Moreover, by Theorem 16, L lies between the successive partial sums s 2 = 1 2 and s 3 = 5 6, so L ≠ 0. If we multiply the series by 2, we obtain ∞

2L = 2 ∑

n =1

( − 1) n + 1

n

(

= 2 1−

)

1 1 1 1 1 1 1 1 1 1 + − + − + − + − + − 2 3 4 5 6 7 8 9 10 11

= 2−1+

2 1 2 1 2 1 2 1 2 − + − + − + − + − . 3 2 5 3 7 4 9 5 11

Now we change the order of this last sum by grouping each pair of terms with the same odd denominator, but leaving the negative terms with the even denominators as they are

583

9.6  Alternating Series and Conditional Convergence

placed (so that the denominators are the positive integers in their natural order). This rearrangement gives (2

− 1) −

( (

)

(

)

(

)

1 2 1 1 2 1 1 2 1 1 + − − + − − + − − + 2 3 3 4 5 5 6 7 7 8 1 1 1 1 1 1 1 1 1 1 = 1− + − + − + − + − + − 2 3 4 5 6 7 8 9 10 11 =

∞

∑

n =1

)

( − 1) n + 1

n

= L. ∞

So when we rearrange the terms of the conditionally convergent series ∑ n =1 2(−1) n +1 n , ∞ the series becomes ∑ n =1 (−1) n +1 n , which is the alternating harmonic series itself. If the two series are the same, it would imply that 2 L = L , which is clearly false since L ≠ 0. Example 5 shows that we cannot rearrange the terms of a conditionally convergent series and expect the new series to be the same as the original one. When we use a conditionally convergent series, we must add the terms together in the order in which they are given to obtain a correct result. In contrast, Theorem 17 guarantees that the terms of an absolutely convergent series can be summed in any order without affecting the result.

Summary of Tests to Determine Convergence or Divergence We have developed a variety of tests to determine convergence or divergence for an infinite series of constants. Other tests that we have not presented are sometimes given in more advanced courses. Here is a summary of the tests we have considered.

1. The nth-Term Test for Divergence: Unless a n → 0, the series diverges. 2. Geometric series: ∑ ar n converges if r < 1; otherwise, it diverges. p 3. p-series: ∑ 1 n converges if p > 1; otherwise, it diverges. 4. Series with nonnegative terms: Try the Integral Test or try comparing to a known series with the Direct Comparison Test or the Limit Comparison Test. Try the Ratio or Root Test. 5. Series with some negative terms: Does ∑ a n converge by the Ratio or Root Test, or by another of the tests listed above? Remember, absolute convergence implies convergence. 6. Alternating series: ∑ a n converges if the series satisfies the conditions of the Alternating Series Test.

EXERCISES

9.6 ∞

Convergence of Alternating Series

In Exercises 1–14, determine whether the alternating series converges or diverges. Some of the series do not satisfy the conditions of the Alternating Series Test. ∞

1.

∑ (−1)n+1 n =1

1 n

∞

2.

∑ (−1)n+1 n13 n n =1

∞

7.

∑ (−1)n+1 n13 2 ∞

4.

∑ (−1)n ( ln4n )2 n=2

∑ (−1)n n 2 n+ 1

∞

6.

n =1

∞

n

∑ (−1)n+1 n2 2 ∞

9.

∑ (−1)n+1 ( 10n ) n =1

+5 ∑ (−1)n+1 nn 2 + 4 2

n =1

8.

n =1

n =1

∞

3.

5.

n

∑ (−1)n ( n10+ 1)! n =1 ∞

n

10.

∑ (−1)n+1 ln1n n=2

584

Chapter 9 Infinite Sequences and Series

∞

11.

∑ (−1)n+1 n =1 ∞

13.

∞

ln n n

12.

n =1 ∞

n +1 n+1

∑ (−1)n+1 n =1

∑ (−1)n ln (1 + 1n )

14.

n+1 n +1

∑ (−1)n+1 3 n =1

Error Estimation

In Exercises 49–52, estimate the magnitude of the error involved in using the sum of the first four terms to approximate the sum of the entire series. ∞

49.

Absolute and Conditional Convergence

∞

∞

∑ (−1)n+1 ( 0.1)n

16.

n =1 ∞

17.

n =1

18. 20. 22.

n =1

∞

∞

n ∑ (−1)n+1 35 + +n

24.

n =1

n =1

∞

26.

n =1

27.

∑

( − 1) n n 2 ( 2

3)

∞

29.

∑ ( − 1)

n

n =1 ∞

31.

∑

( − 1) n

n =1 ∞

33.

∑

30.

n =1 ∞

n n+1

32.

∞

34.

∞

n =1

39.

38.

( 2n ) n

∞

(

∑ (−1)n (

)

40.

n+1−

43.

∑ (−1)n (

n+

n =1 ∞

44.

∑

n =1

( − 1)

n +

n −

n)

n =1

n =1

47.

∑ (−1)n

(n

∞

+ 2 )!

62.

( 2n )!

( 2n )! )

n

n 2 + n − n)

n)

∞

n2

66.

2 2 − ∑ nn2 − + 3n ( 3 )

∞

68.

∑ ( nn ++ 21)!( 32 ) n=0

70. 1 −

1 1 1 1 + − + − 8 64 512 4096

∞

∑ sin (

1 n

∞

)

72.

n+1

∞

∑ lnnn

74. 76.

∑ 1 + 2 + 2 21 +  + 2 n n=2 ∞

78.

∑ 1 +1 +3 +2 +3 3++++3n n=2

∑(

)

n=2

∞

77.

1 ln n

∞

∑ ln( nn ++ 12 ) n=2

n =1

∑n n=2

∞

75.

∑ tan( n 1 n ) n =1

n=2

1 1 1 1 1 1 1 − − + + − − + 4 9 16 25 36 49 64

n

1 1 1 1 1 1 69. − + − + − +  2 2 2 2 2 2

73.

∑ (−1)n csch n

∑ ( nn ++ 12 ) n=0

n

∞

46.

∑ 10 +3 n 4 3 n=2

∑ (1 − n2 )

n=3

∞

)

∞

64.

n =1

2

(

∑ ( 3nn!)3! n=2

5

∞

67.

∑ ( 2n 1+ 1 − 2n 1+ 2 ) n =1

n =1

1 1 1 1 1 1 − + − + − + 4 6 8 10 12 14

48. 1 +

60.

n =1

65.

n

∑ (−1)n sech n

∞

∑ ( n +1 2 − n +1 3 )

∑ n −2

n

∑ 3n 3 n =1

∞

71.

∞

45.

58.

n =1

63.

( − 1) n +1 ( n !) 2

∑ (−1)n (

∞

n

∞

(

∑ (−1)n ln( ln( n1 + 2 )) n =1

∞

( − 1) n −1

∑ (−1)n ( 2nn! +31)! ∞

42.

56.

∑ n3 n

n=0

n =1

n =1 ∞

59. 61.

( − 5 )−n

∞

∑ (−1)n 2 2n nn! n! ∞

∑

n =1

n =1

41.

ln n n − ln n

∑ n 2 + 2n + 1

∞

+ 1) n

∞

1 3 (n + 3 n )

∞

cos nπ 36. ∑ n n =1

( − 1) n ( n

∑ (−1)n+1

∑ (−1)n+1 n 2 n+ 1 n =1

n =1

∞

cos nπ 35. ∑ n =1 n n

∑

57.

n =1

∞

37.

∑

∞

54.

In Exercises 57–82, use any method to determine whether the series converges or diverges. Give reasons for your answer.

n =1

n

n!

n =1

∑ (−1)

0 < t 1. At x = 1 the series becomes 1 − 1 3 + 1 5 − 1 7 + , which converges by the Alternating Series Theorem. It also converges at x = −1 because it is again an alternating series that satisfies the conditions for convergence. The value at x = −1 is the negative of the value at x = 1. Series (b) converges for −1 ≤ x ≤ 1 and diverges elsewhere. The convergence is absolute for −1 < x < 1, but conditional at the points x = −1 and x = 1. −1

0

1

x

9.7  Power Series

(c) 

u n +1 = un

x n +1 n! x ⋅ n = → 0 for every  x. ) +1! x n+1

(n

(n

589

1⋅2 ⋅3n n! = + 1)! 1 ⋅ 2 ⋅ 3  n ⋅ ( n + 1)

The series converges absolutely for all x. 0

(d) 

x

if x = 0 + 1)! x n +1 ⎪⎧ 0 ( n + 1) x → ⎨ = ⎪⎪⎩∞ if x ≠ 0. n! x n The series diverges for all values of x except x = 0. u n +1 = un

(n

0

x

The previous example illustrated how a power series might converge. The next result shows that if a power series converges at a nonzero value, then it converges over an entire interval of values. The interval might be finite or infinite and might contain one, both, or none of its endpoints. We will see that each endpoint of a finite interval must be tested independently for convergence or divergence. THEOREM 18—The Convergence Theorem for Power Series

If the power series ∞

∑ a n x n = a 0 + a1 x + a 2 x 2 +  converges at x = c ≠ 0,  then it converges

n=0

absolutely for all x with x < c . If the series diverges at x = d, then it diverges for all x with x > d . Proof The proof uses the Direct Comparison Test, with the given series compared to a converging geometric series. ∞ Suppose the series ∑ n = 0 a n c n converges. Then lim a n c n = 0 by the nth-Term Test. n →∞

Hence, there is an integer N such that a n c n < 1 for all n > N, so an
d . The corollary to Theorem 18 asserts the existence of a radius of convergence R ≥ 0. For x < R the series converges absolutely, and for x > R it diverges. FIGURE 9.20

1 cn

for n > N .

(5)

Now take any x such that x < c , so that x c < 1. Multiplying both sides of Equation (5) by x n gives xn an x n < for n > N . cn ∞

Since x c < 1, it follows that the geometric series ∑ n = 0 x c converges. By the Direct n

∞

Comparison Test (Theorem 10), the series ∑ n = 0 a n x n converges, so the original power ∞

series ∑ n = 0 a n x n converges absolutely for − c < x < c , as claimed by the theorem.

(See Figure 9.20.) ∞ Now suppose that the series ∑ n = 0 a n x n diverges at x = d. If x is a number with x > d and the series converges at x, then the first half of the theorem shows that the series also converges at d, contrary to our assumption. So the series diverges for all x with x > d. To simplify the notation, Theorem 18 deals with the convergence of series of the form a ∑ n x n. For series of the form ∑ a n ( x − a )n, we can replace x − a by t and apply the results to the series ∑ a n t n.

The Radius of Convergence of a Power Series The theorem we have just proved and the examples we have studied lead to the conclusion that a power series ∑ c n ( x − a ) n behaves in one of three possible ways. It might

590

Chapter 9 Infinite Sequences and Series

Diverges

Diverges

a

a

a −R

a +R

Converges on [a − R, a + R]

a −R

a +R

Converges on [a − R, a + R)

(a)

(b)

Diverges

Diverges

a a −R

a a +R

Converges on (a − R, a + R] (c)

a −R

a +R

Converges on (a − R, a + R) (d) Diverges

a

a

Converges everywhere

Converges only at x = a

(e)

(f)

FIGURE 9.21

The six possibilities for an interval of convergence.

converge only at x = a, or converge everywhere, or converge on some interval of radius R centered at x = a. We prove this as a corollary to Theorem 18. When we also consider the convergence at the endpoints of an interval, we see that there are six different possibilities, shown in Figure 9.21.

Corollary to Theorem 18

The convergence of the series ∑ c n ( x − a ) n is described by one of the following three cases: 1. There is a positive number R such that the series diverges for x with x − a > R but converges absolutely for x with x − a < R. The series may or may not converge at either of the endpoints x = a − R and x = a + R. 2. The series converges absolutely for every x ( R = ∞ ). 3. The series converges at x = a and diverges elsewhere ( R = 0 ).

We first consider the case where a = 0, so that we have a power series ∑ n = 0 c n centered at 0. If the series converges everywhere we are in Case 2. If it converges only at x = 0 then we are in Case 3. Otherwise there is a nonzero number d such ∞ ∞ that ∑ n = 0 c n d n diverges. Let S be the set of values of x for which ∑ n = 0 c n x n converges. The set S does not include any x with x > d , since Theorem 18 implies the series diverges at all such values. So the set S is bounded. By the Completeness Property of the Real Numbers (Appendix A.9) S has a least upper bound R. (This is the smallest number with the property that all elements of S are less than or equal to R.) Since we are not in Case 3, the series converges at some number b ≠ 0 and, by Theorem 18, also on the open interval (− b , b ). Therefore, R > 0. If x < R then there is a number c in S with x < c < R, since otherwise R would not be the least upper bound for S. The series converges at c since c ∈ S, so by Theorem 18 the series converges absolutely at x. Proof ∞

xn

9.7  Power Series

591

Now suppose x > R. If the series converges at x, then Theorem 18 implies it converges absolutely on the open interval (− x , x ) , so that S contains this interval. Since R is an upper bound for S, it follows that x ≤ R, which is a contradiction. So if x > R, then the series diverges. This proves the theorem for power series centered at a = 0. For a power series centered at an arbitrary point x = a, set t = x − a and repeat the argument above, replacing x with t. Since t = 0 when x = a, convergence of the series ∞ ∑ n = 0 c n t n ∞on a radius R open interval centered at t = 0 corresponds to convergence of the series ∑ n = 0 c n ( x − a ) n on a radius R open interval centered at x = a. R is called the radius of convergence of the power series, and the interval of radius R centered at x = a is called the interval of convergence. The interval of convergence may be open, closed, or half-open, depending on the particular series. At points x with x − a < R, the series converges absolutely. If the series converges for all values of x, we say its radius of convergence is infinite. If it converges only at x = a, we say its radius of convergence is zero.

How to Test a Power Series for Convergence

1. Use the Ratio Test or the Root Test to find the largest open interval where the series converges absolutely, x−a < R

a − R < x < a + R.

or

2. If R is finite, test for convergence or divergence at each endpoint, as in Examples 3a and b. 3. If R is finite, the series diverges for x − a > R.

Operations on Power Series On the intersection of their intervals of convergence, two power series can be added and subtracted term by term just like series of constants (Theorem 8). They can be multiplied just as we multiply polynomials, but we often limit the computation of the product to the first few terms, which are the most important. The following result gives a formula for the coefficients in the product, but we omit the proof. (Power series can also be divided in a way similar to division of polynomials, but we do not give a formula for the general coefficient here.)

THEOREM 19—Series Multiplication for Power Series

If A( x ) = and

∞

∞

∑ n = 0 a n x n and B( x ) = ∑ n = 0 bn x n converge absolutely for

c n = a 0 b n + a1b n −1 + a 2 b n − 2 +  + a n −1b1 + a n b 0 =

x < R,

n

∑ a k bn−k , k=0

∞

then ∑ n = 0 c n x n converges absolutely to A( x ) B( x ) for x < R: ⎛ ∞ ⎞⎛ ∞ ⎞ ⎜⎜ ∑ a n x n ⎟⎟⎜⎜ ∑ b n x n ⎟⎟ = ⎟ ⎟⎠ ⎜⎝ ⎠⎜⎝ n = 0 n=0

∞

∑ c n x n.

n=0

Finding the general coefficient c n in the product of two power series can be tedious, and the term may be unwieldy. The following computation provides an illustration of a

592

Chapter 9 Infinite Sequences and Series

product where we find the first few terms by multiplying the terms of the second series by each term of the first series: n +1 ⎞ ⎛ ∞ n ⎟⎞ ⎛ ∞ ⎟⎟ ⎜⎜ ∑ x ⎟ ⋅ ⎜⎜ ∑ ( −1) n x ⎟⎠ ⎜⎝ ⎟⎠ ⎜⎝ n + 1 n=0 n=0

(

)

x2 x3 = (1 + x + x 2 + ) x − + − 2 3

(

Multiply second series …

) (

) (

)

x2 x3 x3 x4 x4 x5 = x− + −  + x2 − + −  + x3 − + − + 2  3 2  3 2  3       by x

by 1

5x 3 x2 = x+ + + . 2 6

by x 2 and gather the first three powers.

We can also substitute a function f ( x ) for x in a convergent power series. ∞

∑ n = 0 a n x n converges absolutely for x < R, and f is a ∞ continuous function, then ∑ n = 0 a n ( f ( x )) n converges absolutely on the set of If

THEOREM 20

points x that satisfy f ( x ) < R.

∞

∑ n = 0 x n converges absolutely for x < 1, it fol∞ lows from Theorem 20 that 1 (1 − 4 x 2 ) = ∑ n = 0 (4 x 2 ) n converges absolutely when x For example, since 1 (1 − x ) =

satisfies 4 x 2 < 1, or, equivalently, when x < 1 2. Theorem 21 says that a power series can be differentiated term by term at each interior point of its interval of convergence. A proof of a restricted case of the theorem is outlined in Exercise 66.

THEOREM 21—Term-by-Term Differentiation

If ∑ c n ( x − a ) n has radius of convergence R > 0, it defines a function f (x) =

∞

∑ c n ( x − a )n

on the interval

a − R < x < a + R.

n=0

This function f has derivatives of all orders inside the interval, and we obtain the derivatives by differentiating the original series term by term: f ′( x ) =

∞

∑ nc n ( x − a )n−1, n =1

f ′′( x ) =

∞

∑ n ( n − 1) c n ( x − a ) n − 2 ,

n=2

and so on. Each of these derived series converges at every point of the interval a − R < x < a + R.

EXAMPLE 4

Find series for f ′( x ) and f ′′( x ) if f (x) = =

1 = 1 + x + x2 + x3 + x4 +  + xn +  1− x ∞

∑ x n,

n=0

−1 < x < 1.

9.7  Power Series

593

Solution We differentiate the power series on the right term by term:

f ′( x ) = =

(1

1 = 1 + 2 x + 3 x 2 + 4 x 3 +  + nx n −1 +  − x )2

∞

∑ nx n−1,

−1 < x < 1;

n =1

f ′′( x ) = =

(1

2 = 2 + 6 x + 12 x 2 +  + n ( n − 1) x n − 2 +  − x )3

∞

∑ n ( n − 1) x n−2,

−1 < x < 1.

n=2

Caution Term-by-term differentiation might not work for series that are not power series. For example, the trigonometric series ∞

(

∑ sin nn2 ! x

)

n =1

converges for all x. But if we differentiate term by term, we get the series ∞

n! cos ( n! x ) , n2 n =1

∑

which diverges for all x. These are not power series since they are not sums of positive integer powers of x. It is also true that a power series can be integrated term by term throughout its interval of convergence. The proof is outlined in Exercise 67.

THEOREM 22—Term-by-Term Integration

Suppose that f (x) =

∞

∑ c n ( x − a )n

n=0

converges for a − R < x < a + R (where R > 0). Then ∞

∑ cn

(x

n=0

− a ) n +1 n+1

converges for a − R < x < a + R and

∫

f ( x ) dx =

∞

∑ cn

n=0

(x

− a ) n +1 +C n+1

for a − R < x < a + R.

EXAMPLE 5 f (x) =

Identify the function ∞ ( − 1) n x 2 n + 1

∑

n=0

2n + 1

= x−

x3 x5 + − , 3 5

−1 ≤ x ≤ 1.

Solution We differentiate the original series term by term and get

f ′( x ) = 1 − x 2 + x 4 − x 6 + ,

−1 < x < 1.

   Theorem 21

594

Chapter 9 Infinite Sequences and Series

This is a geometric series with first term 1 and ratio −x 2, so 1 1 . = 1 − ( −x 2 ) 1 + x2

f ′( x ) =

We can now integrate f ′( x ) = 1 (1 + x 2 ) to get

∫ ∞

= arctan x + C.

The series for f ( x ) is zero when x = 0, so C = 0. Hence

The Number π as a Series π = arctan 1 = 4

dx

∫ 1 + x2

f ′( x ) dx =

( − 1) n

f (x) = x −

∑ 2n + 1 n=0

x3 x5 x7 + − +  = arctan x , 3 5 7

−1 < x < 1.

(6)

It can be shown that the series also converges to arctan x at the endpoints x = ±1, but we omit the proof. Notice that the original series in Example 5 converges at both endpoints of the original interval of convergence, but Theorem 22 can guarantee only the convergence of the integrated series inside the interval. EXAMPLE 6

The series 1 = 1 − t + t2 − t3 +  1+ t

converges on the open interval −1 < t < 1. Therefore, x 1 t2 t3 t4 ⎤ dt = t − + − +  ⎥ ∫0 1 + t ⎦0 2 3 4 2 3 4 x x x = x− + − + , 2 3 4

ln (1 + x ) =

x

   Theorem 22

or ln (1 + x ) =

Alternating Harmonic Series Sum ln 2 =

∞

∑

EXERCISES

∞

∞

∞

∑ xn

2.

n=0

∑ (−1)n ( 4 x + 1)n

∞

4.

n=0

5.

∑

n=0

(x

− 2) 10 n

∑

n =1 ∞

n

6.

7.

−1 < x < 1.

( 3x

− 2 )n n

∞

9.

∑n n =1

11.

8.

∞

13.

∑4 n =1 ∞

15.

∑

n=0

n

∑

( − 1) n ( x

xn

n! n x 2n

n xn n2 + 3

∞

10.

∑

+ 2 )n

n

n =1

xn n 3n

∞ ( −1)

∑

n=0

∑ ( 2 x )n n=0

∞

n

∑ nnx+ 2 n=0

∑ ( x + 5 )n n=0

∞

∞

,

9.7

Intervals of Convergence

3.

n

xn

It can also be shown that the series converges at x = 1 to the number ln 2, but that was not guaranteed by the theorem. A proof of this is outlined in Exercise 63.

n

In Exercises 1–36, (a) find the series’ radius and interval of convergence. For what values of x does the series converge (b) absolutely, (c) conditionally? 1.

∑

n =1

( − 1) n −1

n =1

∞ ( −1) n −1

(x

n =1

− 1) n n

∞

3n x n n = 0 n!

12.

∑

14.

∑

∞

n =1

16.

− 1) n n 33n

(x

∞ ( −1) n

∑

n=0

x n +1 n +3

595

9.7  Power Series

∞

17.

∑n

(x

n=0 ∞

19.

∞

+ 3)n 5n

18.

∞

20.

3n

n=0

47.

n=0

nx n

∑

∞

n

∑ 4 n ( nnx2 + 1)

⎛ x ⎞ − 1⎟⎟ ⎠ 2 n=0

∑ ⎜⎜⎝ ∞

∑ n n ( 2 x + 5 )n

49.

n =1

∑(x n=0

2

+1 3

)

∞

n

48.

∑ ( ln x )n n=0 ∞

n

50.

∑(x n=0

2

−1 2

)

n

∞

21.

∑ ( 2 + (−1)n ) ⋅ ( x + 1)n−1

Using the Geometric Series

n =1

22.

∞ ( −1) n 3 2 n ( x

∑

3n

n =1 ∞

23.

∑ (1 + 1n )

xn

24.

∞

∑ nnx n

26.

n =1

∞ ( −1)

∑

∞

xn 30. ∑ n ln n n=2 (4x

n =1

28.

∑ (−2 )n ( n + 1)( x − 1)n

54. a. Find the interval of convergence of the power series

n=0

∞

∑ 4 n8+ 2 x n.

Get the information you need about    ∑ 1 ( n( ln n )2 ) from Section 9.3, Exercise 61.

n=0

b. Represent the power series in part (a) as a power series about x = 3 and identify the interval of convergence of the new series. (Later in the chapter you will understand why the new interval of convergence does not necessarily include all of the numbers in the original interval of convergence.)

Get the information you need about          ∑ 1 ( n ln n ) from Section 9.3, Exercise 60. ∞

− 5 ) 2 n +1 n3 2

32.

∑

+ 1) n +1 2n + 2

( 3x

n =1

∞

33.

53. Represent the function g( x ) in Exercise 52 as a power series about x = 5, and find the interval of convergence.

∞

n

n2 n

xn 29. ∑ 2 n = 2 n ( ln n )

∞

+ 2)

(x

∞

∑

∑ n! ( x − 4 ) n n=0

n +1

n =1

31.

∑ ( ln n ) x n n =1

∞

27.

52. Use a geometric series to represent each of the given functions as a power series about x = 0, and find their intervals of convergence. 5 3 a. f ( x ) = b. g( x ) = 3−x x−2

∞

n

n =1

25.

51. In Example 2 we represented the function f ( x ) = 2 x as a power series about x = 2. Use a geometric series to represent f ( x ) as a power series about x = 1, and find its interval of convergence.

− 2 )n

∞

∑ 2 ⋅ 4 ⋅ 61 ( 2n ) x n

34.

n =1

3 ⋅ 5 ⋅ 7  ( 2n + 1) n +1 x n2 ⋅ 2n n =1

∑

∞

36.

∑(

n+1−

n =1

n )( x

−

∞

∑ 3 ⋅ 6 ⋅n9!  3n x n

∞

38.

n =1 ∞

39.

(

)2

∑ 2 nn( !2n )! x n

∞

n =1 ∞

41.

⎛ 2 ⋅ 4 ⋅ 6  ( 2n ) ⎞ 2

∑ ⎜⎜⎜⎝ 2 ⋅ 5 ⋅ 8  ( 3n − 1) ⎟⎟⎟⎠

xn

n =1

40.

∑ nn! xn

n

)

∞

∑ ( n +n 1 )

n2

(Hint: Apply the Root Test.) In Exercises 43–50, use Theorem 20 to find the series’ interval of convergence and, within this interval, the sum of the series as a function of x. ∞

43.

∑

∞

xn

44.

n=0 ∞

45.

∑

n=0

∑

(e x

− 4)

(x

+ 1) 2 n 9n

n

n=0

(x

− 1) 2 n 4n

+

x3 x5 x7 x9 x 11 + − + − + 3! 5! 7! 9! 11!

converges to sin x for all x. a. Find the first six terms of a series for cos x. For what values of x should the series converge? b. By replacing x by 2x in the series for sin x, find a series that converges to sin 2x for all x.

xn

n =1

3n

n

57. The series

n =1

n =1

42.

n

56. If you integrate the series in Exercise 55 term by term, what new series do you get? For what values of x does the new series converge, and what is another name for its sum?

sin x = x −

∑ n!(nn++ab!)! x n, where a, b are positive integers (

( ) ( x − 3)

1 1 1( x − 3) + ( x − 3)2 +  + − 4 2 2

converge? What is its sum? What series do you get if you differentiate the given series term by term? For what values of x does the new series converge? What is its sum?

3)n

In Exercises 37–42, find the series’ radius of convergence. 37.

55. For what values of x does the series 1−

1+ 2 + 3++ n xn 35. ∑ 2 2 2 2 n =1 1 + 2 + 3 +  + n ∞

Theory and Examples

∞

46.

∑

n=0

c. Using the result in part (a) and series multiplication, calculate the first six terms of a series for 2 sin x cos x. Compare your answer with the answer in part (b). 58. The series ex = 1 + x +

x2 x3 x4 x5 + + + + 2! 3! 4! 5!

converges to e x for all x. a. Find a series for ( d dx ) e x. Do you get the series for e x ? Explain your answer.

596

Chapter 9 Infinite Sequences and Series

b. Find a series for ∫ e x dx. Do you get the series for e x ? Explain your answer. c. Replace x by −x in the series for e x to find a series that converges to e −x for all x. Then multiply the series for e x and e −x to find the first six terms of a series for e −x ⋅ e x.

and lim ( h 2 n − ln 2n ) = γ ,

n →∞

where γ is Euler’s constant. c. Use these facts to show that ∞

59. The series tan x = x +

x3 2 5 17 7 62 9 + x + x + x + 3 15 315 2835

converges to tan x for −π 2 < x < π 2. a. Find the first five terms of the series for ln sec x . For what values of x should the series converge? b. Find the first five terms of the series for values of x should this series converge?

sec 2

x. For what

( − 1) n +1

n

n =1

= lim s 2 n = ln 2. n →∞

64. Assume that the series ∑ a n x n converges for x = 4 and diverges for x = 7. Answer true (T), false (F), or not enough information given (N) for the following statements about the series. a. Converges absolutely for x = −4 b. Diverges for x = 5 c. Converges absolutely for x = −8.5

c. Check your result in part (b) by squaring the series given for sec x in Exercise 60. 60. The series sec x = 1 +

∑

d. Converges for x = −2 e. Diverges for x = 8 f. Diverges for x = −6

x2 2

+

5 4 61 6 277 8 x + x + x + 24 720 8064

converges to sec x for −π 2 < x < π 2. a. Find the first five terms of a power series for the function ln sec x + tan x . For what values of x should the series converge?

g. Converges absolutely for x = 0 h. Converges absolutely for x = −7.1 65. Assume that the series ∑ a n ( x − 2 ) n converges for x = −1 and diverges for x = 6. Answer true (T), false (F), or not enough information given (N) for the following statements about the series. a. Converges absolutely for x = 1

b. Find the first four terms of a series for sec x tan x. For what values of x should the series converge?

b. Diverges for x = −6

c. Check your result in part (b) by multiplying the series for sec x by the series given for tan x in Exercise 59.

d. Converges for x = 0

61. Uniqueness of convergent power series ∞

∞

a. Show that if two power series ∑ n = 0 a n x n and ∑ n = 0 b n x n are convergent and equal for all values of x in an open interval (−c, c ), then a n = b n for every n. (Hint: Let ∞ ∞ f ( x ) = ∑ n = 0 a n x n = ∑ n = 0 b n x n. Differentiate term by term to show that a n and b n both equal f ( n ) (0) ( n!).) ∞

b. Show that if ∑ n = 0 a n x n = 0 for all x in an open interval (−c, c ), then a n = 0 for every n. ∞

62. The sum of the series ∑ n= 0 ( n 2 2 n ) To find the sum of this series, express 1 (1 − x ) as a geometric series, differentiate both sides of the resulting equation with respect to x, multiply both sides of the result by x, differentiate again, multiply by x again, and set x equal to 1 2. What do you get? 63. The sum of the alternating harmonic series This exercise will show that ∞

∑

n =1

( − 1) n +1

n

= ln 2.

Let h n be the nth partial sum of the harmonic series, and let s n be the nth partial sum of the alternating harmonic series.

c. Diverges for x = 2 e. Converges absolutely for x = 5 f. Diverges for x = 4.9 g. Diverges for x = 5.1 h. Converges absolutely for x = 4 Assume that a = 0 c in Theorem 21, and assume further that L = lim n +1 exists. n →∞ c n If L ≠ 0, then set R = 1 L , while if L = 0, then set R = ∞.

66. Proof of a special case of Theorem 21

The Ratio Test implies that f ( x ) = − R < x < R. Let g( x ) =

lim ( h n − ln n ) = γ

∑ nc n x n−1.

This exercise will

prove that f is differentiable and that f ′( x ) = g( x ), that is f ( x + h ) − f (x) = g( x ) for any x with −R < x < R. lim h→ 0 h Throughout parts (a)–(h), assume that −R < x < R and that h is small enough that −R < x + h < R. a. Use the Ratio Test to prove that the series defining  g( x ) converges. b. Apply the Mean Value Theorem to the function p n ( x ) = x n to show that (x

+ h )n − x n = n (t n ) n−1 h

for some t n between x and x + h for n = 1, 2, 3, … . c. Show

b. Use the results in Exercise 63 in Section 9.3 to conclude that n →∞

converges for

n=0

∞

n =1

a. Use mathematical induction or algebra to show that s 2n = h2n − hn .

∞

∑ cn x n

g( x ) −

f ( x + h ) − f (x) = h

∞

∑ nc n ( x n−1 − ( t n )n−1 ). n=2

597

9.8  Taylor and Maclaurin Series

∞

g. Show that ∑ n = 2 n ( n − 1) c n α n−2 converges.

d. Apply the Mean Value Theorem to the function p n−1 ( x ) = x n−1 to show that

h. Let h → 0 in part (f) to conclude that

x n−1 − (t n ) n−1 = ( n − 1)( d n−1 ) n−2 x − tn

lim

h→ 0

for some d n−1 between x and t n for n = 2, 3, 4, … .

67. Proof of Theorem 22

e. Explain why x − t n < h and why

and assume that f ( x ) =

d n−1 ≤ α = max { x , x + h }.

Let g( x ) =

f. Show that g( x ) −

f ( x + h ) − f (x) ≤ h h

f ( x + h ) − f (x) = g( x ). h

∞

Assume that a = 0 in Theorem 22 ∞

∑ c n x n converges for −R
0 , then f ′( x ) = 2 x −3e −x −2 = p1 (1 x ) f ( x ), while for x = 0 ,

The approximation error is zero at x = a.

The error is negligible when compared to E (x) = 0,    ( x − a ) n. − a )n

f ′(0) = lim+ h→ 0

then

f (1 t ) f (h) − 0 = lim = 0. t →∞ 1 t h−0

b. Show that if x > 0 , then

f ″( a ) ( x − a )2 +  2! f ( n ) (a) ( x − a ) n. + n!

g ( x ) = f ( a ) + f ′( a ) ( x − a ) +

f ′′( x ) = ( 4 x −6 − 6 x −3 ) e −x −2 = p 2 (1 x ) f ( x ), and f ′′(0) = 0 . c. For n ≥ 2 , recursively define polynomials p n by p n +1 (t ) = 2t 3 p n (t ) − t 2 p′n (t ). Use mathematical induction to prove that p n has degree 3n , f ( n ) ( x ) = p n (1 x ) f ( x ) for x > 0 , and f ( n ) (0) = 0.

Thus, the Taylor polynomial Pn ( x ) is the only polynomial of degree less than or equal to n whose error is both zero at x = a and negligible when compared with ( x − a ) n.

9.9 Convergence of Taylor Series In the last section we asked when a Taylor series for a function can be expected to converge to the function that generates it. The finite-order Taylor polynomials that approximate the Taylor series provide estimates for the generating function. In order for these estimates to be useful, we need a way to control the possible errors we may encounter when approximating a function with its finite-order Taylor polynomials. How do we bound such possible errors? We answer the question in this section with the following theorem. THEOREM 23—Taylor’s Theorem

If f and its first n derivatives f ′, f ′′, … , f ( n ) are continuous on the closed interval between a and b, and f ( n ) is differentiable on the open interval between a and b, then there exists a number c between a and b such that f ′′(a) ( b − a )2 +  2! f ( n ) (a) f ( n +1) (c) ( b − a )n + ( b − a ) n +1. + ( n + 1)! n!

f (b) = f (a) + f ′(a)( b − a ) +

Taylor’s Theorem is a generalization of the Mean Value Theorem (Exercise 49), and we omit its proof here. When we apply Taylor’s Theorem, we usually want to hold a fixed and treat b as an independent variable. Taylor’s formula is easier to use in circumstances like these if we change b to x. Here is a version of the theorem with this change. Taylor’s Formula

If f has derivatives of all orders in an open interval I containing a, then for each positive integer n and for each x in I, f ( x ) = f (a) + f ′(a)( x − a ) + +

f ′′(a) ( x − a )2 +  2!

f ( n ) (a) ( x − a ) n + R n ( x ), n!

(1)

where Rn (x) =

f ( n +1) (c) ( x − a ) n +1 ( n + 1)!

for some  c  between  a  and  x.

(2)

9.9  Convergence of Taylor Series

603

When we state Taylor’s theorem this way, it says that for each x ∈ I, f ( x ) = Pn ( x ) + R n ( x ). The function R n ( x ) is determined by the value of the ( n + 1) st derivative f ( n +1) at a point c that depends on both a and x, and that lies somewhere between them. For any value of n we want, the equation gives both a polynomial approximation of f of that order and a formula for the error involved in using that approximation over the interval I. Equation (1) is called Taylor’s formula. The function R n ( x ) is called the remainder of order n or the error term for the approximation of f by Pn ( x ) over I. If R n ( x ) → 0 as n → ∞ for all x ∈ I, we say that the Taylor series generated by f at x = a converges to f on I, and we write f (x) =

∞

∑

k=0

f ( k ) (a) ( x − a ) k. k!

Often we can estimate R n without knowing the value of c, as the following example illustrates. EXAMPLE 1 Show that the Taylor series generated by f ( x ) = e x at x = 0 converges to f ( x ) for every real value of x. Solution The function has derivatives of all orders throughout the interval I = ( −∞, ∞ ). Equations (1) and (2) with f ( x ) = e x and a = 0 give

ex = 1 + x +

x2 xn ++ + Rn (x) n! 2!

  

Polynomial from Section 9.8, Example 2

and Rn (x) =

ec x n +1 ( n + 1)!

for some  c  between 0 and x.

Since e x is an increasing function of x , e c lies between e 0 = 1 and e x. When x is negative, so is c, and e c < 1. When x is zero, e x = 1 so that R n ( x ) = 0. When x is positive, so is c, and e c < e x. Thus, for R n ( x ) given as above, Rn (x) ≤

x n +1 ( n + 1)!

when  x ≤ 0,

x n +1 ( n + 1)!

when  x > 0.

e c < 1 since c < 0

and Rn (x) < e x

   e c < e x  since c < x

Finally, because x n +1 = 0 n →∞ ( n + 1)! lim

for every x ,

Section 9.1, Theorem 5

we see that lim R n ( x ) = 0, and therefore the Taylor series converges to e x for every x. n →∞ Thus, ex =

∞

∑

k=0

The Number e as a Series e =

xk x2 xk = 1+ x + ++ + . k! 2! k!

We can use the result of Example 1 with x = 1 to write

∞

∑ n1! n=0

e = 1+1+

1 1 ++ + R n (1), 2! n!

(3)

604

Chapter 9 Infinite Sequences and Series

where, for some c between 0 and 1, R n (1) = e c

(n

1 3 . < ( n + 1)! + 1)!

   e c < e 1 < 3

Estimating the Remainder It is often possible to estimate R n ( x ) as we did in Example 1. This method of estimation is so convenient that we state it as a theorem for future reference.

THEOREM 24—The Remainder Estimation Theorem

If there is a positive constant M such that f ( n +1) (t ) ≤ M for all t between x and a, inclusive, then the remainder term R n ( x ) in Taylor’s Theorem satisfies the inequality Rn (x) ≤ M

x − a n +1 . ( n + 1)!

If this inequality holds for every n, and the other conditions of Taylor’s Theorem are satisfied by f , then the Taylor series converges to f ( x ).

The next two examples use Theorem 24 to show that the Taylor series generated by the sine and cosine functions do in fact converge to the functions themselves. Show that the Taylor series for sin x at x = 0 converges for all x.

EXAMPLE 2

Solution The function and its derivatives are

f (x) =

sin x ,

f ″( x ) =

− sin x ,

 f

(2 k ) ( x )

  

= ( −1) sin x , k

f ′( x ) =

cos x ,

f ′′′( x ) =

− cos x ,

 f

( 2 k + 1)

( x ) = (−1) k cos x ,

so f (2 k ) (0) = 0

and

f ( 2 k +1) (0) = (−1) k.

The series has only odd-powered terms, and for n = 2 k + 1, Taylor’s Theorem gives sin x = x −

( − 1) k x 2 k + 1 x3 x5 + −+ + R 2 k +1 ( x ). ( 2 k + 1)! 3! 5!

All the derivatives of sin x have absolute values less than or equal to 1, so we can apply the Remainder Estimation Theorem with M = 1 to obtain R 2 k +1 ( x ) ≤ 1 ⋅

x 2k +2 . ( 2 k + 2 )!

From Theorem 5, Rule 6, we have ( x 2 k + 2 ( 2 k + 2 )!) → 0 as k → ∞, whatever the value of x, so R 2 k +1 ( x ) → 0 and the Maclaurin series for sin x converges to sin x for every x. Thus, sin x = x −

x3 3!

+

x5 5!

−

x7 7!

+

sin x =

∞ ( − 1) k

∑

k=0

x 2 k +1 x3 x5 x7 = x− + − + . ( 2 k + 1)! 3! 5! 7!

(4)

605

9.9  Convergence of Taylor Series

EXAMPLE 3 Show that the Taylor series for cos x at x = 0 converges to cos x for every value of x. Solution We add the remainder term to the Taylor polynomial for cos x (Section 9.8, Example 3) to obtain Taylor’s formula for cos x with n = 2 k :

cos x = 1 −

x2 x4 x 2k + −  + ( − 1) k + R 2 k ( x ). 2! 4! (2 k )!

Because the derivatives of the cosine have absolute value less than or equal to 1, the Remainder Estimation Theorem with M = 1 gives R2k (x) ≤ 1 ⋅

x 2 k +1 . ( 2 k + 1)!

For every value of x, R 2 k ( x ) → 0 as k → ∞. Therefore, the series converges to cos x for every value of x. Thus,

cos x = 1 −

x2 2!

+

x4 4!

−

x6 6!

cos x =

+

∞ ( − 1) k

∑

k=0

x 2k x2 x4 x6 = 1− + − + . (2 k )! 2! 4! 6!

(5)

Using Taylor Series Since every Taylor series is a power series, the operations of adding, subtracting, and multiplying Taylor series are all valid on the intersection of their intervals of convergence.

EXAMPLE 4 Using known series, find the first few terms of the Taylor series for the given function by using power series operations. (a)

1 ( 2 x + x cos x ) 3

(b) e x cos x

Solution

(a)

(

)

x2 x4 x 2k 1 2 1 + −  + (−1) k + ( 2 x + x cos x ) = x + x 1 − 3 3 3 2! 4! (2 k )! =

Taylor series for cos x

2 1 x3 x5 x3 x5 + − = x − + − x+ x− 3 3 3! 3 ⋅ 4! 6 72

x x x x x + + + )(1 − + − ) series by each term ( 2! 3! 4! 2! 4! of the second series. x x x x x x x = (1 + x + + + + ) − ( + + + + ) 2! 3! 4! 2! 2! 2!2! 2!3! x x x +( + + + ) +  4! 4! 2!4!

(b) e x cos x = 1 + x +

2

3

4

2

3

4

2

2

4

=1+ x −

Multiply the first

4

3

4

5

5

6

x3 x4 − + 3 6

By Theorem 20, if the Taylor series generated by f at x = 0, ∞

∑

k=0

f ( k ) (0) k f ′′(0) 2 f ( n ) (0) n x = f (0) + f ′(0) x + x ++ x + , k! 2! n!

(6)

606

Chapter 9 Infinite Sequences and Series

converges absolutely for x < R, and if u is a continuous function, then the series ∞

f ( k ) (0) (u( x )) k k!

∑

k=0

(7)

obtained by substituting u( x ) for x in the Taylor series of Formula (6) converges absolutely on the set of points x where u( x ) < R. If u( x ) = cx m , where c ≠ 0 and m is a positive integer, then the series of Formula (7) is a power series and can be shown to be the Taylor series generated by f (u( x )) at x = 0. For instance, we can find the Taylor series for cos 2x by substituting 2x for x in the Taylor series for cos x: cos 2 x =

∞ ( −1) k (2 x ) 2 k

∑

(2 k )!

k=0

=1− =

= 1−

(2 x ) 2 (2 x ) 4 (2 x ) 6 + − + 2! 4! 6!

22 x 2 24 x 4 26 x 6 + − + 2! 4! 6!

∞

2k

2k

∑ (−1) k 2(2kx)!

Eq. (5) with 2 x  for  x

   

.

k=0

By Theorem 20, this new Taylor series converges for all x. EXAMPLE 5 For what values of x can we replace sin x by x − ( x 3 3!) and obtain an error whose magnitude is no greater than 3 × 10 −4 ? Solution Here we can take advantage of the fact that the Taylor series for sin x is an alternating series for every nonzero value of x. According to the Alternating Series Estimation Theorem (Section 9.6), the error in truncating

sin x = x −

x3 x5 x7 + − + 3! 5! 7!

after ( x 3 3!) is no greater than x5 x5 . = 5! 120 Therefore, the error will be less than or equal to 3 × 10 −4 if x5 < 3 × 10 −4 120

x
b is nearly the same. The Taylor polynomial Pn ( x ) = f (a) + f ′(a)( x − a ) +

f ′′(a) ( f ( n ) (a) ( x − a )2 +  + x − a )n 2! n!

and its first n derivatives match the function f and its first n derivatives at x = a. We do not disturb that matching if we add another term of the form K ( x − a ) n +1, where K is any constant, because such a term and its first n derivatives are all equal to zero at x = a. The new function φ n ( x ) = Pn ( x ) + K ( x − a ) n +1 and its first n derivatives still agree with f and its first n derivatives at x = a. We now choose the particular value of K that makes the curve y = φ n ( x ) agree with the original curve y = f ( x ) at x = b. In symbols, f (b) = Pn (b) + K ( b − a ) n +1,

or

K =

f (b) − Pn (b) . ( b − a ) n +1

(8)

With K defined by Equation (8), the function F ( x) = f ( x) − φn ( x) measures the difference between the original function f and the approximating function φ n for each x in [ a, b ]. We now use Rolle’s Theorem (Section 4.2). First, because F (a) = F (b) = 0 and both F and F ′ are continuous on [ a, b ], we know that F ′(c1 ) = 0

for some c1  in  ( a, b ).

Next, because F ′(a) = F ′(c1 ) = 0 and both F ′ and F ′′ are continuous on [ a, c1 ], we know that F ″(c 2 ) = 0

for some c 2  in  ( a, c1 ).

Rolle’s Theorem, applied successively to F ′′, F ′′′, … , F ( n −1), implies the existence of c3

in  ( a, c 2 )

such that F ′′′(c 3 ) = 0,

c4

in  ( a, c 3 )

such that F (4) (c 4 ) = 0,

cn

 in  ( a, c n −1 )

such that F ( n ) (c n ) = 0.

Finally, because F ( n ) is continuous on [ a, c n ] and differentiable on ( a, c n ) , and F ( n ) (a) = F ( n ) (c n ) = 0, Rolle’s Theorem implies that there is a number c n +1 in ( a, c n ) such that F ( n +1) ( c n +1 ) = 0. (9) If we differentiate F ( x ) = f ( x ) − Pn ( x ) − K ( x − a ) n +1 a total of n + 1 times, we get F ( n +1) ( x ) = f ( n +1) ( x ) − 0 − ( n + 1)! K .

(10)

Equations (9) and (10) together give K =

f ( n +1) (c) ( n + 1)!

for some number  c = c n +1  in  ( a, b ).

Equations (8) and (11) give f (b) = Pn (b) + This concludes the proof.

f ( n +1) (c) ( b − a ) n +1. ( n + 1)!

(11)

608

Chapter 9 Infinite Sequences and Series

EXERCISES

9.9

Finding Taylor Series

Use substitution (as in Formula (7)) to find the Taylor series at x = 0 of the functions in Exercises 1–12. 1. e −5 x 4. sin

( π2x )

7. ln (1 + x 2 ) 10.

2. e −x 2

3. 5 sin (−x )

5. cos 5 x 2

6. cos ( x

8. arctan (3 x 4 )

1 2−x

11. ln ( 3 + 6 x )

9.

2)

1 1+

12. e −x

3 4

x3

2 + ln 5

Use power series operations to find the Taylor series at x = 0 for the functions in Exercises 13–30. 13. xe x 16. sin x − x +

14. x 2 sin x x3 17. x cos π x 3!

15.

x2 − 1 + cos x 2

18. x 2 cos ( x 2 )

19. cos 2 x (Hint: cos 2 x = (1 + cos 2 x ) 2.) 20. sin 2 x 23.

(1

1 − x )2

26. sin x ⋅ cos x 29.

21. 24.

x2 1 − 2x (1

22. x ln (1 + 2 x )

2 − x )3

27. e x +

x (  ln 1 + x 2 ) 3

1 1+ x

25. x arctan x 2 28. cos x − sin x

30. ln (1 + x ) − ln (1 − x )

Find the first four nonzero terms in the Maclaurin series for the functions in Exercises 31–38. ln (1 + x ) 31. e x sin x 33. ( arctan x ) 2 32. 1− x 34. cos 2 x ⋅ sin x 37. cos ( e x − 1)

35. e sin x

36. sin ( tan −1 x ) 38. cos x + ln ( cos x )

Error Estimates

39. Estimate the error if P3 ( x ) = x − ( x 3 6 ) is used to estimate the value of sin x at x = 0.1. 40. Estimate the error if P4 ( x ) = 1 + x + ( x 2 2 ) + ( x 3 6 ) + ( x 4 24 ) is used to estimate the value of e x at x = 1 2. 41. For approximately what values of x can you replace sin x by x − ( x 3 6 ) with an error of magnitude no greater than 5 × 10 −4 ? Give reasons for your answer. 42. If cos x is replaced by 1 − ( x 2 2 ) and x < 0.5, what estimate can be made of the error? Does 1 − ( x 2 2 ) tend to be too large, or too small? Give reasons for your answer. 43. How close is the approximation sin x = x when x < 10 −3 ? For which of these values of x is x < sin x ? 44. The estimate 1 + x = 1 + ( x 2 ) is used when x is small. Estimate the error when x < 0.01. 45. The approximation e x = 1 + x + ( x 2 2 ) is used when x is small. Use the Remainder Estimation Theorem to estimate the error when x < 0.1.

46. (Continuation of Exercise 45.) When x < 0, the series for e x is an alternating series. Use the Alternating Series Estimation Theorem to estimate the error that results from replacing e x by 1 + x + ( x 2 2 ) when −0.1 < x < 0. Compare your estimate with the one you obtained in Exercise 45. Theory and Examples

47. Use the identity sin 2 x = (1 − cos 2 x ) 2 to obtain the Maclaurin series for sin 2 x. Then differentiate this series to obtain the Maclaurin series for 2 sin x cos x. Check that this is the series for sin 2x. 48. (Continuation of Exercise 47.) Use the identity cos 2 x = cos 2 x + sin 2 x to obtain a power series for cos 2 x. 49. Taylor’s Theorem and the Mean Value Theorem Explain how the Mean Value Theorem (Section 4.2, Theorem 4) is a special case of Taylor’s Theorem. 50. Linearizations at inflection points Show that if the graph of a twice-differentiable function f ( x ) has an inflection point at x = a, then the linearization of f at x = a is also the quadratic approximation of f at x = a. This explains why tangent lines fit so well at inflection points. 51. The (second) second derivative test Use the equation f ′′(c 2 ) ( x − a )2 f ( x ) = f (a) + f ′(a)( x − a ) + 2 to establish the following test. Let f have continuous first and second derivatives and suppose that f ′(a) = 0. Then a. f has a local maximum at a if f ′′ ≤ 0 throughout an interval whose interior contains a; b. f has a local minimum at a if f ′′ ≥ 0 throughout an interval whose interior contains a. 52. A cubic approximation Use Taylor’s formula with a = 0 and n = 3 to find the standard cubic approximation of f ( x ) = 1 (1 − x ) at x = 0. Give an upper bound for the magnitude of the error in the approximation when x ≤ 0.1. 53. a. Use Taylor’s formula with n = 2 to find the quadratic approximation of f ( x ) = (1 + x ) k at x = 0 (k a constant). b. If k = 3, for approximately what values of x in the interval [ 0, 1 ] will the error in the quadratic approximation be less than 1 100? 54. Improving approximations of π a. Let P be an approximation of π accurate to n decimals. Show that P + sin P gives an approximation correct to 3n decimals. (Hint: Let P = π + x.) T b. Try it with a calculator. 55. The Taylor series generated by f ( x ) = ∞

∑ n= 0 a n x n

∞

∑ n= 0 a n x n is ∞ A function defined by a power series ∑ n = 0 a n x n

with a radius of convergence R > 0 has a Taylor series that converges to the function at every point of (−R , R ). Show this by ∞ showing that the Taylor series generated by f ( x ) = ∑ n = 0 a n x n ∞ is the series ∑ n = 0 a n x n itself. An immediate consequence of this is that series like x4 x6 x8 x sin x = x 2 − + − + 3! 5! 7!

9.10  Applications of Taylor Series

609

Step 1: Plot the function over the specified interval.

and x 2e x

=

x2

+

x3

x4 x5 + + + , 2! 3!

Step 2: Find the Taylor polynomials P1 ( x ), P2 ( x ), and P3 ( x ) at x = 0.

obtained by multiplying Taylor series by powers of x, as well as series obtained by integration and differentiation of convergent power series, are themselves the Taylor series generated by the functions they represent.

Step 3: Calculate the ( n + 1) st derivative f ( n +1) (c) associated with the remainder term for each Taylor polynomial. Plot the derivative as a function of c over the specified interval and estimate its maximum absolute value, M.

56. Taylor series for even functions and odd functions (Continu∞ ation of Section 9.7, Exercise 61.) Suppose that f ( x ) = ∑ n = 0 a n x n converges for all x in an open interval (−R , R ). Show that

Step 4: Calculate the remainder R n ( x ) for each polynomial. Using the estimate M from Step 3 in place of f ( n +1) (c), plot R n ( x ) over the specified interval. Then estimate the values of x that answer question (a).

a. If f is even, then a1 = a 3 = a 5 =  = 0, i.e., the Taylor series for f at x = 0 contains only even powers of x.

Step 5: Compare your estimated error with the actual error E n ( x ) = f ( x ) − Pn ( x ) by plotting E n ( x ) over the specified interval. This will help you answer question (b).

b. If f is odd, then a 0 = a 2 = a 4 =  = 0, i.e., the Taylor series for f at x = 0 contains only odd powers of x. COMPUTER EXPLORATIONS

Taylor’s formula with n = 1 and a = 0 gives the linearization of a function at x = 0. With n = 2 and n = 3, we obtain the standard quadratic and cubic approximations. In these exercises we explore the errors associated with these approximations. We seek answers to two questions:

Step 6: Graph the function and its three Taylor approximations together. Discuss the graphs in relation to the information discovered in Steps 4 and 5. 1 , 1+ x

57. f ( x ) =

x ≤

58. f ( x ) = (1 + x )3 2, −

1 ≤ x ≤ 2 2

a. For what values of x can the function be replaced by each approximation with an error less than 10 −2 ?

59. f ( x ) =

b. What is the maximum error we can expect if we replace the function by each approximation over the specified interval?

60. f ( x ) = ( cos x )( sin 2 x ) ,

Using a CAS, perform the following steps to aid in answering questions (a) and (b) for the functions and intervals in Exercises 57–62.

61. f ( x ) =

x , x2 + 1 e −x

62. f ( x ) = e

x 3

3 4

x ≤ 2 x ≤ 2

cos 2 x ,

x ≤1

sin 2 x ,

x ≤ 2

9.10 Applications of Taylor Series We can use Taylor series to solve problems that would otherwise be intractable. For example, many functions have antiderivatives that cannot be expressed using familiar functions. In this section we show how to evaluate integrals of such functions by giving them as Taylor series. We also show how to use Taylor series to evaluate limits that lead to indeterminate forms and how Taylor series can be used to extend the exponential function from real to complex numbers. We begin with a discussion of the binomial series, which comes from the Taylor series of the function f ( x ) = (1 + x ) m , and we conclude the section with Table 9.1, which lists some commonly used Taylor series.

The Binomial Series for Powers and Roots The Taylor series generated by f ( x ) = (1 + x ) m , when m is constant, is 1 + mx +

m ( m − 1) 2 m ( m − 1)( m − 2 ) 3 x + x + 2! 3! m ( m − 1)( m − 2 )  ( m − k + 1) k + x + . k!

(1)

610

Chapter 9 Infinite Sequences and Series

This series, called the binomial series, converges absolutely for x < 1. To derive the series, we first list the function and its derivatives: f ( x ) = (1 + x ) m f ′( x ) = m (1 + x ) m −1 f ′′( x ) = m ( m − 1)(1 + x ) m − 2 f ′′′( x ) = m ( m − 1)( m − 2 )(1 + x ) m −3 f

(k) (x)

 = m ( m − 1)( m − 2 )  ( m − k + 1)(1 + x ) m − k.

We then evaluate these at x = 0 and substitute into the Taylor series formula to obtain Series (1). If m is an integer greater than or equal to zero, the series stops after ( m + 1) terms because the coefficients from k = m + 1 on are zero. If m is not a positive integer or zero, the series is infinite and converges for x < 1. To see why, let u k be the term involving x k. Then apply the Ratio Test for absolute convergence to see that u k +1 m−k x → x = uk k +1

as  k → ∞.

Our derivation of the binomial series shows only that it is generated by (1 + x ) m and converges for x < 1. The derivation does not show that the series converges to (1 + x ) m. It does, but we leave the proof to Exercise 58. The following formulation gives a succinct way to express the series.

The Binomial Series

For −1 < x < 1, (1

+ x )m = 1 +

∞

⎛m⎞

∑ ⎜⎜⎜⎝ k ⎟⎟⎟⎠ x k , k =1

where we define ⎛ m ⎞⎟ ⎜⎜ ⎟ = m, ⎜⎝ 1 ⎠⎟

⎛ m ⎞⎟ ⎜⎜ ⎟ = m ( m − 1) , ⎜⎝ 2 ⎟⎠ 2!

and ( )( ) ( ) ⎛ m ⎞⎟ ⎜⎜ ⎟ = m m − 1 m − 2  m − k + 1 ⎜⎝ k ⎟⎠ k!

EXAMPLE 1

for k ≥ 3.

If m = −1, then ⎛ −1⎞⎟ ⎜⎜ ⎟ = −1, ⎜⎝ 1⎟⎠

⎛ −1 ⎞⎟ ⎜⎜ ⎟ = −1( −2 ) = 1, ⎜⎝ 2 ⎟⎠ 2!

and ( )( ) ( ) ⎛ −1⎞⎟ ⎜⎜ ⎟ = −1 −2 −3  −1 − k + 1 = ( −1) k k ! = (−1) k. ⎜⎝ k ⎟⎠ k! k!

( )

With these coefficient values and with x replaced by −x, the binomial series formula gives the familiar geometric series (1

+ x )−1 = 1 +

∞

∑ (−1) k x k k =1

= 1 − x + x 2 − x 3 +  + (−1) k x k +  .

9.10  Applications of Taylor Series

611

EXAMPLE 2 We know from Section 3.11, Example 1, that 1 + x ≈ 1 + ( x 2 ) for x small. With m = 1 2, the binomial series gives quadratic and higher-order approximations as well, along with error estimates that come from the Alternating Series Estimation Theorem:

(1

+

x )1 2

( )( )

( )( )( )

1 1 1 1 3 − − − x 2 2 2 2 2 x3 2 =1+ + x + 2 2! 3! 1 1 3 5 − − − 2 2 2 2 x4 +  + 4! x x2 x3 5x 4 =1+ − + − + . 2 8 16 128

( )( )( )( )

Substitution for x gives still other approximations. For example,

1−

1 − x2 ≈ 1 −

x2 x4 − 2 8

1 1 1 ≈ 1− − 2 2x 8x x

for 

for  x 2  small 1  small,  that is,   x  large. x

Evaluating Nonelementary Integrals Sometimes we can use a familiar Taylor series to find the sum of a given power series in terms of a known function. For example, x2 −

(x 2 ) 3 (x 2 ) 5 (x 2 ) 7 x6 x 10 x 14 + − +  = (x 2 ) − + − +  = sin x 2. 3! 5! 7! 3! 5! 7!

Additional examples are provided in Exercises 59–62. Taylor series can be used to express nonelementary integrals in terms of series. Integrals like ∫ sin x 2 dx arise in the study of the diffraction of light. Express ∫ sin x 2 dx as a power series.

EXAMPLE 3

Solution From the series for sin x, we substitute x 2 for x to obtain

sin x 2 = x 2 −

x6 x 10 x 14 x 18 + − + − . 3! 5! 7! 9!

Therefore,

∫ sin x 2 dx

= C+

x3 x7 x 11 x 15 x 19 − + − + − . 3 7 ⋅ 3! 11 ⋅ 5! 15 ⋅ 7! 19 ⋅ 9!

1

Estimate ∫ 0 sin x 2 dx with an error of less than 0.001.

EXAMPLE 4

Solution From the indefinite integral in Example 3, we easily find that 1

∫0 sin x 2 dx

=

1 1 1 1 1 − + − + − . 3 7 ⋅ 3! 11 ⋅ 5! 15 ⋅ 7! 19 ⋅ 9!

The series on the right-hand side alternates, and we find by numerical evaluations that 1 ≈ 0.00076 11 ⋅ 5!

612

Chapter 9 Infinite Sequences and Series

is the first term to be numerically less than 0.001. The sum of the preceding two terms gives 1

∫0 sin x 2 dx

≈

1 1 − ≈ 0.310. 3 42

With two more terms, we could estimate 1

∫0 sin x 2 dx

≈ 0.310268

with an error of less than 10 −6. With only one term beyond that, we have 1

∫0 sin x 2 dx

≈

1 1 1 1 1 − + − + ≈ 0.310268303, 3 42 1320 75600 6894720

with an error of about 1.08 × 10 −9. Guaranteeing this accuracy with the error formula for the Trapezoidal Rule would require using about 8000 subintervals.

Arctangents In Section 9.7, Example 5, we found a series for arctan x by differentiating to get d 1 arctan x = = 1 − x2 + x4 − x6 +  dx 1 + x2 and then integrating to get x3 x5 x7 + − + . 3 5 7

arctan x = x −

However, we did not prove the term-by-term integration theorem on which this conclusion depended. We now derive the series again by integrating both sides of the finite formula ( − 1) n + 1 t 2 n + 2 1 , = 1 − t 2 + t 4 − t 6 +  + ( − 1) n t 2 n + 2 1+ t 1 + t2

(2)

in which the last term comes from adding the remaining terms as a geometric series with first term a = ( −1) n +1 t 2 n + 2 and ratio r = −t 2. Integrating both sides of Equation (2) from t = 0 to t = x gives arctan x = x −

x3 x5 x7 x 2 n +1 + − +  + ( − 1) n + R n ( x ), 3 5 7 2n + 1

where Rn (x) =

x ( − 1) n + 1 t 2 n + 2

∫0

1 + t2

dt.

The denominator of the integrand is greater than or equal to 1; hence Rn (x) ≤

∫0

x

t 2 n + 2 dt =

x 2n + 3 . 2n + 3

If x ≤ 1, the right side of this inequality approaches zero as n → ∞. Therefore, lim R n ( x ) = 0 if x ≤ 1 and

n →∞

arctan x =

∞ ( −1) n

∑

n=0

x 2 n +1 , 2n + 1

x ≤ 1.

x3 x5 x7 arctan x = x − + − + , 3 5 7

(3) x ≤ 1.

9.10  Applications of Taylor Series

613

We take this route instead of finding the Taylor series directly because the formulas for the higher-order derivatives of arctan x are unmanageable. When we put x = 1 in Equation (3), we get Leibniz’s formula: ( − 1) n π 1 1 1 1 = 1− + − + −+ + . 3 5 7 9 2n + 1 4

Because this series converges very slowly, it is not used in approximating π to many decimal places. The series for arctan x converges most rapidly when x is near zero. For that reason, people who use the series for arctan x to compute π use various trigonometric identities. For example, if α = arctan

1 2

and

1 β = arctan , 3

then tan ( α + β ) =

1 tan α + tan β + = 2 1 − tan α tan β 1−

1 3 1 6

π = 1 = tan , 4

and therefore, π 1 1 = α + β = arctan + arctan . 2 3 4 Now Equation (3) may be used with x = 1 2 to evaluate arctan (1 2 ) and with x = 1 3 to give arctan (1 3 ). The sum of these results, multiplied by 4, gives π.

Evaluating Indeterminate Forms We can sometimes evaluate indeterminate forms by expressing the functions involved as Taylor series.

EXAMPLE 5

Evaluate lim

x →1

ln x . x −1

Solution We represent ln x as a Taylor series in powers of x − 1. This can be accomplished by calculating the Taylor series generated by ln x at x = 1 directly or by replacing x by x − 1 in the series for ln (1 + x ) in Section 9.7, Example 6. Either way, we obtain

ln x = ( x − 1) −

1( x − 1) 2 +  , 2

from which we find that lim

x →1

(

)

ln x 1 = lim 1 − ( x − 1) +  = 1. x → 1 x −1 2

Of course, this particular limit can be evaluated just as well using l’Hôpital’s Rule.

EXAMPLE 6

Evaluate sin x − tan x . x→0 x3

lim

614

Chapter 9 Infinite Sequences and Series

Solution The Taylor series for sin x and tan x, to terms in x 5, are

sin x = x −

x3 x5 + − , 3! 5!

tan x = x +

x3 2x 5 + + . 3 15

Subtracting the series term by term, it follows that sin x − tan x = −

(

)

x3 x5 x2 1 − −  = x3 − − − . 2 8 2 8

Division of both sides by x 3 and taking limits then give sin x − tan x 1 x2 1 = lim − − − = − . 3 x→0 x→0 x 2 8 2

(

lim

)

If we apply series to calculate lim A (1 sin x ) − (1 x ) B , we not only find the limit sucx→0

cessfully but also discover an approximation formula for csc x.

⎛ 1 1⎞ Find lim ⎜⎜ − ⎟⎟⎟. x→0 ⎜ x⎠ ⎝ sin x

EXAMPLE 7

Solution Using algebra and the Taylor series for sin x, we have

( (

) )

x3 x5 x− x− + − x − sin x 1 1 3! 5! − = = 3 5 x x sin x x x sin x x⋅ x− + − 3! 5! =

( 3!1 − x5! + ) = x ⋅ 3!1 − x5! + . x x 1− x (1 − + ) + 3! 3!

x3

2

2

2

2

2

Therefore, 1 x2 ⎛ − +  ⎞⎟⎟ ⎜⎜ ⎛ 1 ⎞⎟ 1 ⎟⎟ = 0. 3! 5! ⎜ ⎜ − ⎟⎟ = lim ⎜ x ⋅ lim ⎜ ⎟⎟ x2 x→0 ⎜ x→0 ⎜ x⎠ ⎝ sin x ⎟⎟ ⎜⎜ − +  1 ⎝ ⎠ 3! From the quotient on the right, we can see that if x is small, then 1 1 1 x − ≈ x⋅ = sin x x 3! 6

or

csc x ≈

1 x + . 6 x

Euler’s Identity A complex number is a number of the form a + bi, where a and b are real numbers and i = −1 (see Chapter 18). If we substitute x = iθ (with θ real) in the Taylor series for e x and use the relations i 2 = −1,

i 3 = i 2i = −i,

i 4 = i 2i 2 = 1,

i 5 = i 4 i = i,

615

9.10  Applications of Taylor Series

and so on to simplify the result, we obtain iθ i 2θ 2 i 3θ 3 i 4θ 4 i 5θ 5 i 6θ 6 + + + + + + 1! 2! 3! 4! 5! 6! θ2 θ4 θ6 θ3 θ5 = 1− + − + +i θ− + −  = cos θ + i sin θ. 2! 4! 6! 3! 5!

e iθ = 1 +

(

) (

)

This does not prove that e iθ = cos θ + i sin θ because we have not yet defined what it means to raise e to an imaginary power. Rather, it tells us how to define e iθ so that its properties are consistent with the properties of the exponential function for real numbers.

DEFINITION

For any real number θ, e iθ = cos θ + i sin θ.

(4)

Equation (4), called Euler’s identity, enables us to define e a + bi to be e a ⋅ e bi for any complex number a + bi. So e a + ib = e a ( cos b + i sin b ). One consequence of this identity is the equation e iπ = −1. When written in the form e iπ + 1 = 0, this equation combines five of the most important constants in mathematics. TABLE 9.1 Frequently Used Taylor Series

1 = 1 + x + x2 +  + xn +  = 1− x

∞

∑ x n, ∞

1 = 1 − x + x 2 −  + ( −x ) n +  = 1+ x ex = 1 + x +

sin x = x − cos x = 1 −

x2 xn ++ += 2! n!

x 1. p n =1 n

46. Prove that ∑

Power Series In Exercises 47–56, (a) find the series’ radius and interval of convergence. Then identify the values of x for which the series converges (b) absolutely and (c) conditionally. ∞

47. 49.

n =1

+ 4 )n n3 n

∞

n −1

∑ ∑

(x

( − 1)

∞

− 1)

∞

n

50.

∞

n

∑ nx n ∑

(n

n=0

52. +

3n

∑ ( csch n ) x n n =1

(n

+ 1)( 2 x + 1) n ( 2n + 1) 2 n

n

∑ xn n =1

1) x 2 n−1

54.

− 1) 2 n +1 2n + 1

∞ ( −1) n ( x

∑

n=0

∞

55.

∑

− 1) 2 n − 2 ( 2n − 1)!

(x

n=0

n =1

53.

∑

n =1

n2

n =1

51.

( 3x

∞

48.

∞

56.

∑ ( coth n ) x n n =1

∞

1 n

n =1

1 n2 − 1

∑n

∑ e −−n 1+ 1

∞

∞

∞

40.

( ) + ( ) − ( ) + ( 13 ) 2

1 3

∞

43.

n =1

Determining Convergence of Series Which of the series in Exercises 25–44 converge absolutely, which converge conditionally, and which diverge? Give reasons for your answers.

n n

∑ 2n3n

n=0

n=3

∞

∑ e −n

42.

+ 1) 2n 2 + n − 1

n =1

1 n ( n + 1)( n + 2 )

∑

( − 1) n ( n 2

∞

38.

n!

∞

∞

n =1

23.

∑ n ( n−+2 1) n=2

∞

∑

n =1

∞

20.

n=3

21.

37.

∑

n =1

(−3 ) n

(−4 ) n

Convergent Series Find the sums of the series in Exercises 19–24.

19.

∞

n

12. a n = 1 +

18. a n =

+

36.

n =1

∞

1)

∞

∑ n n+! 1

41. 1 −

6. a n = sin nπ

n n + ln n 9. a n = n 11. a n =

2. a n =

∞

35.

4. a n = 1 + ( 0.9 ) n

ln ( n 2 )

7. a n =

619

+1

∞

34.

∑

n =1

( − 1) n 3n 2

n3 + 1

Maclaurin Series Each of the series in Exercises 57–62 is the value of the Taylor series at x = 0 of a function f ( x ) at a particular point. What function and what point? What is the sum of the series?

57. 1 − 58.

1 1 1 + −  + (−1) n n +  4 4 16

2n 2 4 8 − + −  + (−1) n−1 n +  3 18 81 n3

59. π −

π3 π5 π 2 n +1 + −  + (−1) n + ( 2n + 1)! 3! 5!

60. 1 −

π2 π4 π 2n + −  + (−1) n 2 n + 3 (2n)! 9 ⋅ 2! 81 ⋅ 4!

620

Chapter 9 Infinite Sequences and Series

61. 1 + ln 2 + 62.

( ln 2 ) 2 ( ln 2 ) n ++ + 2! n!

1 1 1 − + − 3 9 3 45 3 1   + (−1) n−1 2 n −1 +  ( 2n − 1)( 3 )

67. cos ( x

66. sin 53

)

69. e ( πx 2 )

∞

2 ⋅ 5 ⋅ 8 ⋅  ⋅ ( 3n − 1) n x . 2 ⋅ 4 ⋅ 6 ⋅  ⋅ (2n) n =1

∑

88. Find the radius of convergence of the series ∞

Find Taylor series at x = 0 for the functions in Exercises 63–70. 1 1 64. 63. 1 − 2x 1 + x3 65. sin πx

87. Find the radius of convergence of the series

2x 3

n =1

89. Find a closed-form formula for the nth partial sum of the series ∞ ∑ n= 2 ln(1 − (1 n 2 )) and use it to determine the convergence or divergence of the series. ∞

x3 68. cos 5

90. Evaluate ∑ k = 2 (1 ( k 2 − 1)) by finding the limits as n → ∞ of the series’ nth partial sum.

70. e −x 2

91. a. Find the interval of convergence of the series y = 1+

Taylor Series In Exercises 71–74, find the first four nonzero terms of the Taylor series generated by f at x = a.

71. f ( x ) =

3 + x2

at

x = 2

73. f ( x ) = 1 ( x + 1) at

x = 3

75.

∫0

77.

∫0

12

1 ⋅ 4 ⋅ 7 ⋅  ⋅ ( 3n − 2 ) 3n x + . (3n)!

d 2y = x a y + b, dx 2

x = a > 0

1

dx

76.

∫0 x sin ( x 3 ) dx

tan −1 x dx x

78.

∫0

e

−x 3

1 3 1 6 x + x + 6 180

b. Show that the function defined by the series satisfies a differential equation of the form

Nonelementary Integrals Use series to approximate the values of the integrals in Exercises 75–78 with an error of magnitude less than 10 −8. (The answer section gives the integrals’ values rounded to ten decimal places.) 12

+

x = −1

72. f ( x ) = 1 (1 − x ) at

74. f ( x ) = 1 x at

3 ⋅ 5 ⋅ 7 ⋅  ⋅ ( 2n + 1)

∑ 4 ⋅ 9 ⋅ 14 ⋅  ⋅ ( 5n − 1)( x − 1)n.

1 64

arctan x dx x

Using Series to Find Limits In Exercises 79–84:

a. Use power series to evaluate the limit. T b. Then use a grapher to support your calculation. 7 sin x e θ − e −θ − 2θ 80. lim 79. lim 2 x x→0 e θ→0 θ − sin θ −1 ⎛ 1 1⎞ − 2 ⎟⎟⎟ 81. lim ⎜⎜ t→0 ⎜ t ⎠ ⎝ 2 − 2 cos t

82. lim

1 − cos 2 z 83. lim z → 0 ln (1 − z ) + sin z

y2 84. lim y → 0 cos y − cosh y

( sin h ) h − cos h h→ 0 h2

and find the values of the constants a and b. 92. a. Find the Maclaurin series for the function x 2 (1 + x ). b. Does the series converge at x = 1? Explain. ∞

∞

(

)

T 86. Compare the accuracies of the approximations sin x ≈ x and sin x ≈ 6 x ( 6 + x 2 ) by comparing the graphs of f ( x ) = sin x − x and g( x ) = sin x − ( 6 x ( 6 + x 2 )). Describe what you find.

∞

94. If ∑ n =1 a n and ∑ n =1 b n are divergent series of nonnegative ∞ numbers, can anything be said about ∑ n =1 a n b n? Give reasons for your answer. 95. Prove that the sequence { x n } and the series both converge or both diverge.

∞

∑ k =1 ( x k +1 − x k )

∞

96. Prove that ∑ n =1 ( a n (1 + a n )) converges if a n > 0 for all n and ∞

∑ n=1 a n converges.

97. Suppose that a1 , a 2 , a 3 , … , a n are positive numbers satisfying the following conditions: i) a1 ≥ a 2 ≥ a 3 ≥  ; ii) the series a 2 + a 4 + a 8 + a16 +  diverges. Show that the series

Theory and Examples

85. Use a series representation of sin 3x to find values of r and s for which sin 3 x r lim + 2 + s = 0. x→0 x3 x

∞

93. If ∑ n =1 a n and ∑ n =1 b n are convergent series of nonnegative ∞ numbers, can anything be said about ∑ n =1 a n b n? Give reasons for your answer.

a a1 a + 2 + 3 + 1 2 3 diverges. 98. Use the result in Exercise 97 to show that 1+

∞

∑ n ln1 n n=2

diverges.

Chapter 9  Additional and Advanced Exercises

∞

∞

n =1

n =1

99. Show that if a n > 0 and ∑ a n converges, then ∑

an converges. n

∞

102. Consider the convergent series ∑

n =1

a. b1 = 1, b n +1 = (−1)

n

n+1 b 3n + 2 n

n b = ln n n

b. b1 = 3, b n +1

101. Assume that b n > 0 and

∞

∑ bn converges. What, if anything,

( − 1) n

, where c is a constant. e n + e cn What should c be so that the first 10 terms of the series estimate the sum of the entire series with an error of less than 0.00001? n =1

∞

100. Determine whether ∑ b n converges or diverges.

621

103. Assume that the following sequence has a limit L. Find the value of L. 4 1 3, (4(4 1 3 )) 1 3, (4(4(4 1 3 )) 1 3 ) 1 3, (4(4(4(4 1 3 )) 1 3 ) 1 3 ) 1 3, … 104. Consider the infinite sequence of shaded right triangles in the accompanying diagram. Compute the total area of the triangles. 1

n =1

can be said about the following series? ∞

a.

∑ tan (bn )

1

n =1

1

∞

b.

− 2

∑ ln(1 + bn )

1

− 4

1

− 2

n =1

1

− 4

∞

c.

∑ ln( 2 + bn )

1

− 8

n =1

CHAPTER 9

Additional and Advanced Exercises

Determining Convergence of Series ∞

Which of the series ∑ n =1 a n defined by the formulas in Exercises 1–4 converge, and which diverge? Give reasons for your answers. ∞

1.

∑ ( 3n − 12 )n+(1 2) n =1

3.

∑ (−1)n tanh n

∞

2.

( tan −1 n ) 2 ∑ n2 + 1 n =1

4.

∑

∞

∞

n =1

log n ( n!) n3 n=2

∞

Which of the series ∑ n =1 a n defined by the formulas in Exercises 5–8 converge, and which diverge? Give reasons for your answers. n ( n + 1) 5. a1 = 1, a n +1 = an ( n + 2 )( n + 3 ) (Hint: Write out several terms, see which factors cancel, and then generalize.) n 6. a1 = a 2 = 7, a n +1 = a n if n ≥ 2 ( n − 1)( n + 1) 7. a1 = a 2 = 1, a n +1 =

1 1 + an

if n ≥ 2

8. a n = 1 3 n if n is odd, a n = n 3 n if n is even

point where we know the values of f and its derivatives. We also need a to be close enough to the values of f in which we are interested to make ( x − a ) n +1 so small we can neglect the remainder. In Exercises 9–14, what Taylor series would you choose to represent the function near the given value of x? (There may be more than one good answer.) Write out the first four nonzero terms of the series you choose. 9. cos x 11.

ex

near

near

13. cos x

x =1

near

f ′′(a) ( x − a )2 +  2! f ( n +1) (c) ( x − a ) n +1 + ( n + 1)!

f

( n ) (a)

n!

(x

− a )n

expresses the value of f at x in terms of the values of f and its derivatives at x = a. In numerical computations, we therefore need a to be a

12. ln x

x = 69

near near

14. arctan x

x = 6.3 x = 1.3

near

x = 2

Theory and Examples

15. Let a and b be constants with 0 < a < b. Does the sequence {( a n + b n )1 n } converge? If it does converge, what is the limit? 16. Find the sum of the infinite series 2 3 7 2 3 7 2 + 2 + 3 + 4 + 5 + 6 + 7 10 10 10 10 10 10 10 3 7 + 8 + 9 + . 10 10

1+

17. Evaluate ∞

n +1

∑ ∫n

Taylor’s formula f ( x ) = f ( a ) + f ′( a ) ( x − a ) +

10. sin x

x = 0.4

Choosing Centers for Taylor Series

+

1

− 8

n=0

1 dx. 1 + x2

18. Find all values of x for which ∞

n

∑ ( n + 1)(nx2 x + 1)n n =1

converges absolutely.

622

Chapter 9 Infinite Sequences and Series

T 19. a. Does the value of cos ( a n ) ⎞⎟ ⎛ lim ⎜⎜1 − ⎟⎟⎠ , a constant, n

⎝ n →∞ ⎜

(Hint: First show that ln (1 − a n ) ≤ a n (1 − a n ).)

appear to depend on the value of a? If so, how?

29. Nicole Oresme’s Theorem Prove Nicole Oresme’s Theorem:

b. Does the value of

1+ n

cos ( a n ) ⎞⎟ ⎛ lim ⎜⎜1 − ⎟ , a and b constant, b ≠ 0, bn ⎟⎠

1+

appear to depend on the value of b? If so, how? c. Use calculus to confirm your findings in parts (a) and (b). ∞

20. Show that if  ∑ n =1 a n converges, then ∞

n =1

1 + sin a n 2

)

1 1 n ⋅ 2 + ⋅ 3 +  + n−1 +  = 4. 2 4 2

(Hint: Differentiate both sides of the equation 1 (1 − x ) =

⎝ n →∞ ⎜

∑(

n

∞

∑ n=1 x n.)

ex − 1 . x b. By differentiating the series in part (a) term by term, show ∞ n = 1. that  ∑ ( ) n =1 n + 1 !

30. a. Find a power series representation of 

31. a. Find a power series representation of e −x 2 .

converges. 21. Find a value for the constant b that will make the radius of convergence of the power series

b. By differentiating the series in part (a) twice term-by-term, ∞ 2n + 1 = 1. show that  ∑ (−1) n +1 n 2 n! n =1 32. a. Show that

∞

bn x n n = 2 ln n

∑

∞

∑n

equal to 5.

(n

∞

∑ x n+1

twice, multiplying the result by x, and then replacing x by 1 x .

is finite, and evaluate the limit.

x =

cos (ax ) − b = −1. x→0 2x 2

+ 1) . xn

(n

33. Quality control

25. Raabe’s (or Gauss’s) Test The following test, which we state without proof, is an extension of the Ratio Test. ∞ Raabe’s Test: If  ∑ n =1 u n is a series of positive constants and there exist constants C, K, and N such that un f (n) C = 1+ + 2 , n n u n +1 ∞

where f (n) < K for n ≥ N, then ∑ n =1 u n converges if C > 1 and diverges if C ≤ 1. Show that the results of Raabe’s Test agree with what you ∞ ∞ know about the series ∑ n =1 (1 n 2 ) and ∑ n =1 (1 n ). ∞

26. (Continuation of Exercise 25.) Suppose that the terms of  ∑ n =1 u n are defined recursively by the formulas − 1) 2 u . (2n)( 2n + 1) n ( 2n

Apply Raabe’s Test to determine whether the series converges. 27. Suppose that ∑ n =1 a n converges, a n ≠ 1, and a n > 0 for all n. b. Does ∑ n =1 a n (1 − a n ) converge? Explain.

∞

∑n n =1

lim

a. Show that ∑ n =1 a n2 converges.

x2 1− x

b. Use part (a) to find the real solution greater than 1 of the equation

24. Find values of a and b for which

∞

=

n =1

sin (ax ) − sin x − x lim x→0 x3

2x 2 ( x − 1) 3

=

for x > 1 by differentiating the identity

23. Find the value of a for which the limit

∞

+ 1)

xn

n =1

22. How do you know that the functions sin x, ln x, and e x are not polynomials? Give reasons for your answer.

u1 = 1, u n +1 =

converges, and if

∞

0 < a n < 1 for all n, show that ∑ n =1 ln (1 − a n ) converges.

n

∞

∞

∑ n=1 a n

28. (Continuation of Exercise 27.) If

a. Differentiate the series 1 = 1 + x + x2 +  + xn +  1− x to obtain a series for 1 (1 − x ) 2. b. In one throw of two dice, the probability of getting a roll of 7 is p = 1 6. If you throw the dice repeatedly, the probability that a 7 will appear for the first time at the nth throw is q n−1 p, where q = 1 − p = 5 6. The expected number of throws until ∞ a 7 first appears is ∑ n =1 nq n−1 p. Find the sum of this series. c. As an engineer applying statistical control to an industrial operation, you inspect items taken at random from the assembly line. You classify each sampled item as either “good” or “bad.” If the probability of an item’s being good is p and of an item’s being bad is q = 1 − p, the probability that the first bad item found is the nth one inspected is p n−1q. The average number inspected up to and including the first bad item found ∞ is ∑ n =1 np n−1q. Evaluate this sum, assuming 0 < p < 1. 34. Expected value Suppose that a random variable X may assume the values 1, 2, 3, … , with probabilities p1 , p 2 , p 3 , … , where p k is the probability that X equals k ( k = 1, 2, 3, …). Suppose also that p k ≥ 0 and that

∞

∑ k =1 p k

= 1. The expected value of X,

Chapter 9  Technology Application Projects

∞

∑ k =1 kp k , provided∞the series converges. In each of the following cases, show that ∑ k =1 p k = 1

denoted by E ( X ), is the number

and find E ( X ) if it exists. (Hint: See Exercise 33.) a. p k = 2 −k

b. p k =

5 k −1 6k

36. Time between drug doses (Continuation of Exercise 35.) If a drug is known to be ineffective below a concentration C L and harmful above some higher concentration C H , we need to find values of C 0 and t 0 that will produce a concentration that is safe (not above C H ) but effective (not below C L ). See the accompanying figure. We therefore want to find values for C 0 and t 0 for which R = C L and C 0 + R = C H .

1 1 1 = − k ( k + 1) k k +1

T 35. Safe and effective dosage The concentration in the blood resulting from a single dose of a drug normally decreases with time as the drug is eliminated from the body. Doses may therefore need to be repeated periodically to keep the concentration from dropping below some particular level. One model for the effect of repeated doses gives the residual concentration just before the ( n + 1)st dose as

C Concentration in blood

c. p k =

R n = C 0 e −kt 0 + C 0 e −2 kt 0 +  + C 0 e −nkt 0 ,

Concentration (mgmL)

where C 0 = the change in concentration achievable by a single dose ( mg mL ) , k = the elimination constant (h −1 ), and t 0 = time between doses (h). See the accompanying figure. C1 = C0 + C0 e−k t 0

R3 t

t0 Time (h)

a. Write R n in closed from as a single fraction, and find R = lim R n . n →∞

b. Calculate R1 and R10 for C 0 = 1 mg mL , k = 0.1 h −1, and t 0 = 10 h. How good an estimate of R is R10? c. If k = 0.01 h −1 and t 0 = 10 h, find the smallest n such that R n > (1 2 ) R. Use C 0 = 1 mg mL. (Source: Prescribing Safe and Effective Dosage, B. Horelick and S. Koont, COMAP, Inc., Lexington, MA.)

CHAPTER 9

t0

Lowest effective level

Time

t

1 CH ln . k CL

To reach an effective level rapidly, one might administer a “loading” dose that would produce a concentration of C H mg mL. This could be followed every t 0 hours by a dose that raises the concentration by C 0 = C H − C L mg mL.

R1 = C0 e−k t 0 0

CL

t0 =

Rn R2

C0

Thus C 0 = C H − C L . When these values are substituted in the equation for R obtained in part (a) of Exercise 35, the resulting equation simplifies to

Cn−1

C2

Highest safe level

CH

0

C

C0

623

a. Verify the preceding equation for t 0 . b. If k = 0.05 h −1 and the highest safe concentration is e times the lowest effective concentration, find the length of time between doses that will ensure safe and effective concentrations. c. Given C H = 2 mg mL , C L = 0.5 mg mL , and k = 0.02 h −1, determine a scheme for administering the drug. d. Suppose that k = 0.2 h −1 and that the smallest effective concentration is 0.03 mg mL. A single dose that produces a concentration of 0.1 mg mL is administered. About how long will the drug remain effective?

Technology Application Projects

Mathematica/Maple Projects

Projects can be found within MyLab Math. • Bouncing Ball The model predicts the height of a bouncing ball, and the time until it stops bouncing. • Taylor Polynomial Approximations of a Function A graphical animation shows the convergence of the Taylor polynomials to functions having derivatives of all orders over an interval in their domains.

10

Parametric Equations and Polar Coordinates OVERVIEW In this chapter we study new ways to describe curves in the plane. Instead of considering a curve as the graph of a function or equation, we think of it as the path of a moving particle whose position is changing over time. Then each of the x- and y-coordinates of the particle’s position becomes a function of a third variable t. We can also change the way in which points in the plane themselves are described by using polar coordinates rather than the rectangular or Cartesian system. Both of these new tools are useful for describing motion, like that of planets and satellites, or projectiles moving in the plane or in space.

10.1 Parametrizations of Plane Curves Parametric Equations

Position of particle at time t

( f (t), g(t))

Figure 10.1 shows the path of a moving particle in the xy-plane. Notice that the path fails the vertical line test, so it cannot be described as the graph of a function of the variable x. However, we can describe the path by a pair of equations, x = f (t ) and y = g(t ), where f and g are continuous functions. In the study of motion, t usually denotes time. Equations like these can describe more general curves than those described by a single function, and they provide not only the graph of the path traced out but also the location of the particle ( x ,  y ) = ( f (t ),  g(t ) ) at any time t. DEFINITION

If x and y are given as functions x = f (t ),

FIGURE 10.1

The curve or path traced by a particle moving in the xy-plane is not always the graph of a function or single equation.

y = g(t )

over an interval I of t-values, then the set of points ( x ,  y ) = ( f (t ),  g(t ) ) defined by these equations is a parametric curve. The equations are parametric equations for the curve. The variable t is a parameter for the curve, and its domain I is the parameter interval. If I is a closed interval, a ≤ t ≤ b, the point ( f (a),  g(a) ) is the initial point of the curve and ( f (b),  g(b) ) is the terminal point. When we give parametric equations and a parameter interval for a curve, we say that we have parametrized the curve. The equations and interval together constitute a parametrization of the curve. A given curve can be represented by different sets of parametric equations. (See Exercises 29 and 30.) EXAMPLE 1

Sketch the curve defined by the parametric equations x = sin

624

πt , 2

y = t,

0 ≤ t ≤ 6.

625

10.1  Parametrizations of Plane Curves

Solution We make a table of values (Table 10.1), plot the points ( x ,  y ), and draw a smooth curve through them (Figure 10.2). If we think of the curve as the path of a moving particle, the particle starts at time t = 0 at the initial point ( 0, 0 ) and then moves upward in a wavy path until at time t = 6 it reaches the terminal point ( 0, 6 ). The direction of motion is shown by the arrows in Figure 10.2. TABLE 10.1 Values of x = sin

and y = t for selected values of t.

St 2

y

(0, 6) t=6 6

t

x

y

5

0 1 2 3 4 5 6

0 1 0 −1 0 1 0

0 1 2 3 4 5 6

4

(−1, 3) t=3

(1, 5) t=5

(0, 4) t=4

3

(0, 2) t=2 2 (1, 1) t=1

1 (0, 0) t=0 −1

0

1

x

FIGURE 10.2

The curve given by the πt and 2 y = t (Example 1).

parametric equations x = sin

EXAMPLE 2

Sketch the curve defined by the parametric equations x = t 2,

y = t + 1,

−∞ < t < ∞.

Solution We make a table of values (Table 10.2), plot the points ( x ,  y ), and draw a smooth curve through them (Figure 10.3). We think of the curve as the path that a particle moves along the curve in the direction of the arrows. Although the time intervals in the table are equal, the consecutive points plotted along the curve are not at equal arc length distances. The reason for this is that the particle slows down as it gets nearer to the y-axis along the lower branch of the curve as t increases, and then speeds up after reaching the y-axis at ( 0, 1) and moving along the upper branch. Since the interval of values for t is all real numbers, there is no initial point or terminal point for the curve. TABLE 10.2 Values of x = t 2 and

y = t + 1 for selected values of t.

y

t

x

y

−3 −2 −1 0 1 2 3

9 4 1 0 1 4 9

−2 −1 0 1 2 3 4

t=3 t=2 (4, 3)

t=1 (1, 2) t = 0 (0, 1) (1, 0) (4, −1) t = −1 t = −2

(9, 4)

x (9, −2) t = −3

FIGURE 10.3 The curve given by the parametric equations x = t 2 and y = t + 1 (Example 2).

626

Chapter 10 Parametric Equations and Polar Coordinates

For this example we can use algebraic manipulation to eliminate the parameter t and obtain an algebraic equation for the curve in terms of x and y alone. We solve y = t + 1 for t and substitute the resulting equation t = y − 1 into the equation for x, which yields x = t 2 = ( y − 1) 2 = y 2 − 2 y + 1. The equation x = y 2 − 2 y + 1 represents a parabola, as displayed in Figure 10.3. It is sometimes quite difficult, or even impossible, to eliminate the parameter from a pair of parametric equations, as we did here. y

EXAMPLE 3

t=p 2

x + 2

y2

=1

(a) x = cos t , (b) x = a cos t ,

P(cos t, sin t) t

t=p

t=0 (1, 0)

0

y = sin t , y = a sin t ,

The equations x = cos t and y = sin t describe motion on the circle x 2 + y 2 = 1. The arrow shows the direction of increasing t (Example 3). FIGURE 10.4

y = x 2, x ≥ 0

(a) Since x 2 + y 2 = cos 2 t + sin 2 t = 1, the parametric curve lies along the unit circle x 2 + y 2 = 1. As t increases from 0 to 2π , the point ( x ,  y ) = ( cos t , sin t ) starts at (1, 0 ) and traces the entire circle once counterclockwise (Figure 10.4). (b) For x = a cos t , y = a sin t , we have x 2 + y 2 = a 2 cos 2 t + a 2 sin 2 t = a 2 . The parametrization describes a motion that begins at the point ( a, 0 ) and traverses the circle x 2 + y 2 = a 2 once counterclockwise, returning to ( a, 0 ) at t = 2π. The graph is a circle centered at the origin with radius r = a and coordinate points ( a cos t ,  a sin t ). EXAMPLE 4 The position P ( x ,  y ) of a particle moving in the xy-plane is given by the equations and parameter interval

t = 4 (2, 4)

x = t=1

x

2

The equations x = t and y = t and the interval t ≥ 0 describe the path of a particle that traces the right-hand half of the parabola y = x 2 (Example 4). y = x2

(−2, 4) t=2

P(t, t 2 ) t=1

EXAMPLE 5

(1, 1)

FIGURE 10.6

Thus, the particle’s position coordinates satisfy the equation y = x 2 , so the particle moves along the parabola y = x 2 . It would be a mistake, however, to conclude that the particle’s path is the entire parabola y = x 2 ; it is only half the parabola. The particle’s x-coordinate is never negative. The particle starts at ( 0, 0 ) when t = 0 and rises into the first quadrant as t increases (Figure 10.5). The parameter interval is [ 0, ∞ ), and there is no terminal point. The graph of any function y = f ( x ) can always be given a natural parametrization x = t and y = f (t ). The domain of the parameter in this case is the same as the domain of the function f .

(2, 4)

0

t ≥ 0.

y = t = ( t ) = x 2.

FIGURE 10.5

t = −2

y = t,

Solution We try to identify the path by eliminating t between the equations x = t and y = t , which might produce a recognizable algebraic relation between x and y. We find that

(1, 1)

y

t,

Identify the path traced by the particle and describe the motion.

P(" t, t)

0 Starts at t=0

0 ≤ t ≤ 2π. 0 ≤ t ≤ 2π.

Solution

x

t = 3p 2

y

Graph the parametric curves

x

The path defined by x = t ,   y = t 2 , −∞ < t < ∞ is the entire parabola y = x 2 (Example 5).

A parametrization of the graph of the function f ( x ) = x 2 is given by x = t,

y = f (t ) = t 2 ,

−∞ < t < ∞.

When t ≥ 0 , this parametrization gives the same path in the xy-plane as we had in Example 4. However, since the parameter t here can now also be negative, we obtain the left-hand part of the parabola as well; that is, we have the entire parabolic curve. For this parametrization, there is no starting point and no terminal point (Figure 10.6).

10.1  Parametrizations of Plane Curves

627

Notice that a parametrization also specifies when a particle moving along the curve is located at a specific point along the curve. In Example 4, the point ( 2, 4 ) is reached when t  4; in Example 5, it is reached “earlier” when t  2. You can see the implications of this aspect of parametrizations when considering the possibility of two objects coming into collision: they have to be at the exact same location point P ( x ,  y ) for some (possibly different) values of their respective parameters. We will say more about this aspect of parametrizations when we study motion in Chapter 12.

EXAMPLE 6 slope m.

Find a parametrization for the line through the point ( a,  b ) having

Solution A Cartesian equation of the line is y − b = m ( x − a ). If we define the parameter t by t = x − a , we find that x = a + t and y − b = mt. That is,

x = a + t,

y = b + mt ,

−∞ < t < ∞

parametrizes the line. This parametrization differs from the one we would obtain by the natural parametrization in Example 5 when t  x. However, both parametrizations describe the same line. TABLE 10.3 Values of x = t + ( 1 t )

and y = t − ( 1 t ) for selected values of t.

EXAMPLE 7

Sketch and identify the path traced by the point P ( x ,  y ) if 1 x = t+ , t

t

1t

x

y

0.1 0.2 0.4 1.0 2.0 5.0 10.0

10.0 5.0 2.5 1.0 0.5 0.2 0.1

10.1 5.2 2.9 2.0 2.5 5.2 10.1

9.9 4.8 2.1 0.0 1.5 4.8 9.9

1 y = t− , t

Solution We make a brief table of values in Table 10.3, plot the points, and draw a smooth curve through them, as we did in Example 1. Next we eliminate the parameter t from the equations. The procedure is more complicated than in Example 2. Taking the difference between x and y as given by the parametric equations, we find that

(

x−y = t+

(

y

5

−10

(10.1, 9.9) t=5 (5.2, 4.8)

FIGURE 10.7

The curve for x = t + (1 t ) ,   y = t − (1 t ) ,   t > 0 in Example 7. (The part shown is for 0.1 b t b 10.)

) (

)

1 1 + t− = 2t. t t

(x

− y )( x + y ) =

( 2t )( 2t ) = 4.

Expanding the expression on the left-hand side, we obtain a standard equation for a hyperbola (Section 10.6): x

(10.1, −9.9) t = 0.1

)

We can then eliminate the parameter t by multiplying these last equations together:

t = 10

t=2 t = 1 (2.5, 1.5) 5 10 0 (2, 0) (2.9, −2.1) t = 0.4 (5.2, −4.8) −5 t = 0.2

) (

1 1 2 − t− = . t t t

If we add the two parametric equations, we get x+ y = t+

10

t > 0.

x 2 − y 2 = 4.

(1)

Thus the coordinates of all the points P ( x ,  y ) described by the parametric equations satisfy Equation (1). However, Equation (1) does not require that the x-coordinate be positive. So there are points ( x ,  y ) on the hyperbola that do not satisfy the parametric equation x = t + (1 t ),  t > 0. In fact, the parametric equations do not yield any points on the left branch of the hyperbola given by Equation (1), points where the x-coordinate would be negative. For small positive values of t, the path lies in the fourth quadrant and rises into the first quadrant as t increases, crossing the x-axis when t  1 (see Figure 10.7). The parameter domain is ( 0, ∞ ) and there is no starting point or terminal point for the path.

628

Chapter 10 Parametric Equations and Polar Coordinates

Examples 4, 5, and 6 illustrate that a given curve, or portion of it, can be represented by different parametrizations. In the case of Example 7, we can also represent the righthand branch of the hyperbola by the parametrization x =

4 + t 2,

y = t,

−∞ < t < ∞,

which is obtained by solving Equation (1) for x p 0 and letting y be the parameter. Still another parametrization for the right-hand branch of the hyperbola given by Equation (1) is Q Q x = 2 sec t , y = 2 tan t , − < t < . 2 2 This parametrization follows from the trigonometric identity sec 2 t − tan 2 t = 1, because HISTORICAL BIOGRAPHY Christiaan Huygens (1629–1695) Huygens was born in the Hague, Netherlands. He studied mathematics at the University of Leiden. Huygens was a follower of Descartes. He published his important geometrical results in Theoremata de quadratura hyperboles, ellipses et circuli and De circuli magnitudine inventa (1654). Later, he considered the subject of probability and published Tractatus de ratiociniis in aleae ludo (1657). To know more, visit the companion Website.

Guard cycloid

Guard cycloid

As t runs between Q 2 and Q 2,  x  sec t remains positive and y  tan t runs between −∞ and d, so P traverses the hyperbola’s right-hand branch. It comes in along the branch’s lower half as t → 0 −, reaches ( 2, 0 ) at t  0, and moves out into the first quadrant as t increases steadily toward Q 2. This is the same branch of the hyperbola shown in Figure 10.7.

Cycloids The problem with a pendulum clock whose bob swings in a circular arc is that the frequency of the swing depends on the amplitude of the swing. The wider the swing, the longer it takes the bob to return to center (its lowest position). This does not happen if the bob can be made to swing in a curve called a cycloid. In 1673, Christiaan Huygens designed a pendulum clock whose bob would swing in a cycloid, a curve we define in Example 8. He hung the bob from a fine wire constrained by guards that caused it to draw up as it swung away from center (Figure 10.8). We describe the path parametrically in the next example. EXAMPLE 8 A wheel of radius a rolls along a horizontal straight line. Find parametric equations for the path traced by a point P on the wheel’s circumference. The path is called a cycloid.

Cycloid

FIGURE 10.8

In Huygens’ pendulum clock, the bob swings in a cycloid, so the frequency is independent of the amplitude. y

x 2 − y 2 = 4 sec 2 t − 4 tan 2 t = 4 ( sec 2 t − tan 2 t ) = 4.

Solution We take the line to be the x-axis, mark a point P on the wheel, start the wheel with P at the origin, and roll the wheel to the right. As parameter, we use the angle t through which the wheel turns, measured in radians. Figure 10.9 shows the wheel a short while later when its base lies at units from the origin. The wheel’s center C lies at ( at ,  a ) and the coordinates of P are

x = at + a cos R,

P(x, y) = (at + a cos u, a + a sin u)

t

0

FIGURE 10.9

at

To express R in terms of t, we observe that t + θ = 3π 2 in the figure, so that

a u C(at, a) M

y = a + a sin R.

θ = x

The position of P ( x ,  y ) on the rolling wheel at angle t (Example 8).

3π − t. 2

This makes cos θ = cos

( 32π − t ) = −sin t,

sin θ = sin

( 32π − t ) = −cos t.

The equations we seek are x = at − a sin t ,

y = a − a cos t.

These are usually written with the a factored out: x = a ( t − sin t ),

y = a (1 − cos t ).

Figure 10.10 shows the first arch of the cycloid and part of the next.

(2)

10.1  Parametrizations of Plane Curves

y

Brachistochrones and Tautochrones (x, y) t a

O

2pa

x

FIGURE 10.10

The cycloid curve x = a ( t − sin t ),   y = a (1 − cos t ), for t ≥ 0. O

a

2a

pa

2pa

x

P(at − a sin t, a − a cos t)

a

629

2a B(ap, 2a)

If we turn Figure 10.10 upside down, Equations (2) still apply and the resulting curve (Figure 10.11) has two interesting physical properties. The first relates to the origin O and the point B at the bottom of the first arch. Among all smooth curves joining these points, the cycloid is the curve along which a frictionless bead, subject only to the force of gravity, will slide from O to B the fastest. This makes the cycloid a brachistochrone (“brah-kisstoe-krone”), or shortest-time curve for these points. The second property is that even if you start the bead partway down the curve toward B, it will still take the bead the same amount of time to reach B. This makes the cycloid a tautochrone (“taw-toe-krone”), or same-time curve for O and B. Are there any other brachistochrones joining O and B, or is the cycloid the only one? We can formulate this as a mathematical question in the following way. At the start, the kinetic energy of the bead is zero since its velocity (speed) is zero. The work done by gravity in moving the bead from ( 0, 0 ) to any other point ( x ,  y ) in the plane is mgy, and this must equal the change in kinetic energy. (See Exercise 25 in Section 6.5.) That is, mgy =

y

FIGURE 10.11

When Figure 10.10 is turned upside down, the y-axis points downward, indicating the direction of the gravitational force. Equations (2) still describe the curve parametrically.

1 1 mυ 2 − m(0) 2 . 2 2

Thus, the speed of the bead when it reaches ( x ,  y ) has to be υ = ds = dT

2 gy . That is,

ds is the arc length differential along the bead’s path, and T represents time.

2 gy

or dT =

ds = 2 gy

1 + ( dy dx ) 2 dx . 2 gy

(3)

The time T f it takes the bead to slide along a particular path y = f ( x ) from O to B( aπ, 2a ) is O

Tf =

B

FIGURE 10.12

The cycloid is the unique curve that minimizes the time it takes for a frictionless bead to slide from point O to point B. x A B

C

y

FIGURE 10.13

Beads released simultaneously on the upside-down cycloid at O, A, and C will reach B at the same time.

1 + ( dy dx ) 2 dx. 2 gy

(4)

What curves y = f ( x ), if any, minimize the value of this integral? At first sight, we might guess that the straight line joining O and B would give the shortest time, but perhaps not. There might be some advantage in having the bead fall vertically at first to build up its speed faster. With a higher speed, the bead could travel a longer path and still reach B first. Indeed, this is the right idea. The solution, from a branch of mathematics known as the calculus of variations, is that the original cycloid from O to B is the one and only brachistochrone for O and B (Figure 10.12).

cycloid

O

x = aπ

∫x =0

In the next section we show how to find the arc length differential ds for a parametrized curve. Once we know how to find ds, we can calculate the time given by the righthand side of Equation (4) for the cycloid. This calculation gives the amount of time it takes a frictionless bead to slide down the cycloid to B after it is released from rest at O. The time turns out to be equal to π a g, where a is the radius of the wheel defining the particular cycloid. Moreover, if we start the bead at some lower point on the cycloid, corresponding to a parameter value t 0 > 0, we can integrate the parametric form of ds 2 gy in Equation (3) over the interval [ t 0 ,  π ] to find the time it takes the bead to reach the point B. That calculation results in the same time T = π a g. It takes the bead the same amount of time to reach B no matter where it starts, which makes the cycloid a tautochrone. Beads starting simultaneously from O, A, and C in Figure 10.13, for instance, will all reach B at exactly the same time. This is the reason why Huygens’ pendulum clock in Figure 10.8 is independent of the amplitude of the swing.

630

Chapter 10 Parametric Equations and Polar Coordinates

EXERCISES

10.1 y

C.

Finding Cartesian from Parametric Equations

Exercises 1–18 give parametric equations and parameter intervals for the motion of a particle in the xy-plane. Identify the particle’s path by finding a Cartesian equation for it. Graph the Cartesian equation. (The graphs will vary with the equation used.) Indicate the portion of the graph traced by the particle and the direction of motion.

y

D.

3 2 2 1

1. x = 3t , y = 9t 2, −∞ < t < ∞

1

2. x = − t , y = t , t ≥ 0 3. x = 2t − 5, y = 4 t − 7, −∞ < t < ∞

−2

−1

1

x

2

1

x

2

4. x = 3 − 3t , y = 2t , 0 ≤ t ≤ 1 5. x = cos 2t , y = sin 2t , 0 ≤ t ≤ π

y

E.

6. x = cos( π − t ), y = sin ( π − t ), 0 ≤ t ≤ π 7. x = 4 cos t , y = 2 sin t , 0 ≤ t ≤ 2π

3

8. x = 4 sin t , y = 5 cos t , 0 ≤ t ≤ 2π

2

−3 −2 −1 −1

11. x = t 2 , y = t 6 − 2t 4 , −∞ < t < ∞ t t−2 12. x = , y = , −1 < t < 1 t −1 t +1 14. x =

1

2

x

3

−1

x

1

−2 −1

−3

1 − t2, − 1 ≤ t ≤ 0

t + 1, y =

1

1

π π ≤ t ≤ 2 2 10. x = 1 + sin t , y = cos t − 2, 0 ≤ t ≤ π 9. x = sin t , y = cos 2t , −

13. x = t , y =

y

F.

In Exercises 25–28, use the given graphs of x = f (t ) and y = g(t ) to sketch the corresponding parametric curve in the xy-plane.

t, t ≥ 0

15. x = sec 2 t − 1, y = tan t , −π 2 < t < π 2

25.

16. x = − sec t , y = tan t , −π 2 < t < π 2 17. x = − cosh t , y = sinh t , −∞ < t < ∞

x

y

18. x = 2 sinh t , y = 2 cosh t , −∞ < t < ∞ 1

1

In Exercises 19–24, match the parametric equations with the parametric curves labeled A through F.

f

g

19. x = 1 − sin t , y = 1 + cos t 20. x = cos t , y = 2 sin t

t

1

1 1 t cos t , y = t sin t 4 4 22. x = t , y = t cos t

−1

1

t

21. x =

23. x = ln t , y = 3e

−1

−t 2

24. x = cos t , y = sin 3t y

A.

26.

y

B.

3

x

y

1

2

1 f

2

g

1 −1

1

2

3

x

−2

2

x

−1

1

t

−1

1

−2 −3

−2

−1

−1

t

10.1  Parametrizations of Plane Curves

27.

x

1

1 −2

37. Find parametric equations and a parameter interval for the motion of a particle starting at the point ( 2, 0 ) and tracing the top half of the circle x 2 + y 2 = 4 four times.

y

2

f

−1

−1

1

2

−1

t

2

1 −1

t

631

38. Find parametric equations and a parameter interval for the motion of a particle that moves along the graph of y = x 2 in the following way: Beginning at ( 0, 0 ) it moves to ( 3, 9 ), and then it travels back and forth from ( 3, 9 ) to (−3, 9 ) infinitely many times. 39. Find parametric equations for the semicircle

g

x 2 + y 2 = a 2, y > 0,

−2

using as parameter the slope t = dy dx of the tangent line to the curve at ( x ,  y ). 40. Find parametric equations for the circle

28. x

y

4 f

using as parameter the arc length s measured counterclockwise from the point ( a, 0 ) to the point ( x ,  y ).

g 2

−2

x 2 + y 2 = a 2,

2

−1

1 1

2

t

−2

−1

1

−2

−1

−4

−2

2

t

41. Find a parametrization for the line segment joining points ( 0, 2 ) and ( 4, 0 ) using the angle θ in the accompanying figure as the parameter. y

2 (x, y)

Finding Parametric Equations

29. Find parametric equations and a parameter interval for the motion of a particle that starts at ( a, 0 ) and traces the circle x2 + y2 = a2 a. once clockwise. b. once counterclockwise. c. twice clockwise.

u 4

0

x

42. Find a parametrization for the curve y = x with terminal point ( 0, 0 ) using the angle θ in the accompanying figure as the parameter.

d. twice counterclockwise. y

(There are many ways to do these, so your answers may not be the same as the ones at the back of the text.)

y = "x

30. Find parametric equations and a parameter interval for the motion of a particle that starts at ( a, 0 ) and traces the ellipse (x 2 a2 ) + ( y2 b2 ) = 1

(x, y)

a. once clockwise. b. once counterclockwise.

u

x

0

c. twice clockwise. d. twice counterclockwise. (As in Exercise 29, there are many correct answers.) In Exercises 31–36, find a parametrization for the curve.

43. Find a parametrization for the circle ( x − 2 ) 2 + y 2 = 1 starting at (1, 0 ) and moving clockwise once around the circle, using the central angle θ in the accompanying figure as the parameter.

31. the line segment with endpoints (−1, −3 ) and ( 4, 1)

y

32. the line segment with endpoints (−1, 3 ) and ( 3, −2 ) 33. the lower half of the parabola x − 1 = y 2

(x, y)

1 1

34. the left half of the parabola y = x 2 + 2 x 35. the ray (half line) with initial point ( 2, 3 ) that passes through the point (−1, −1) 36. the ray (half line) with initial point (−1, 2 ) that passes through the point ( 0, 0 )

u 0

1

2

3

x

632

Chapter 10 Parametric Equations and Polar Coordinates

44. Find a parametrization for the circle x 2 + y 2 = 1 starting at (1, 0 ) and moving counterclockwise to the terminal point ( 0, 1), using the angle θ in the accompanying figure as the parameter.

47. As the point N moves along the line y = a in the accompanying figure, P moves in such a way that OP = MN. Find parametric equations for the coordinates of P as functions of the angle t that the line ON makes with the positive y-axis.

y

y

(0, 1)

A(0, a)

(x, y)

N

1 u (1, 0)

–2

M

x

t P O

45. The witch of Maria Agnesi The bell-shaped witch of Maria Agnesi can be constructed in the following way. Start with a circle of radius 1, centered at the point ( 0, 1), as shown in the accompanying figure. Choose a point A on the line y = 2 and connect it to the origin with a line segment. Call the point where the segment crosses the circle B. Let P be the point where the vertical line through A crosses the horizontal line through B. The witch is the curve traced by P as A moves along the line y = 2. Find parametric equations and a parameter interval for the witch by expressing the coordinates of P in terms of t, the radian measure of the angle that segment OA makes with the positive x-axis. The following equalities (which you may assume) will help. b. y = 2 − AB sin t

a. x = AQ c. AB ⋅ OA = ( AQ ) 2

Q

(0, 1)

49. Find the point on the parabola x = t ,   y = t 2 , −∞ < t < ∞, closest to the point ( 2, 1 2 ). (Hint: Minimize the square of the distance as a function of t.) 50. Find the point on the ellipse x = 2 cos t ,   y = sin t ,  0 ≤ t ≤ 2π , closest to the point ( 3 4 , 0 ). (Hint: Minimize the square of the distance as a function of t.)

b. 0 ≤ t ≤ π c. −π 2 ≤ t ≤ π 2 x

46. Hypocycloid When a circle rolls on the inside of a fixed circle, any point P on the circumference of the rolling circle describes a hypocycloid. Let the fixed circle be x 2 + y 2 = a 2 , let the radius of the rolling circle be b, and let the initial position of the tracing point P be A( a, 0 ). Find parametric equations for the hypocycloid, using as the parameter the angle θ from the positive x-axis to the line joining the circles’ centers. In particular, if b = a 4 , as in the accompanying figure, show that the hypocycloid is the astroid y = a

x = 4 cos t , y = 2 sin t , over

a. 0 ≤ t ≤ 2π

P(x, y)

B

O

x = a

Distance Using Parametric Equations

51. Ellipse

A

t

cos 3 θ,

48. Trochoids A wheel of radius a rolls along a horizontal straight line without slipping. Find parametric equations for the curve traced out by a point P on a spoke of the wheel b units from its center. As parameter, use the angle θ through which the wheel turns. The curve is called a trochoid, which is a cycloid when b = a.

T GRAPHER EXPLORATIONS Using a parametric equation grapher, graph the equations over the given intervals in Exercises 51–58.

y y=2

x

sin 3 θ.

52. Hyperbola branch x = sec t (enter as 1 cos(t )), y = tan t (enter as sin (t ) cos(t )), over a. −1.5 ≤ t ≤ 1.5 b. −0.5 ≤ t ≤ 0.5 c. −0.1 ≤ t ≤ 0.1 53. Parabola 54. Cycloid

x = 2t + 3, y = t 2 − 1, −2 ≤ t ≤ 2 x = t − sin t , y = 1 − cos t , over

a. 0 ≤ t ≤ 2π b. 0 ≤ t ≤ 4 π c. π ≤ t ≤ 3π 55. Deltoid

y

x = 2 cos t + cos 2t , y = 2 sin t − sin 2t , 0 ≤ t ≤ 2π What happens if you replace 2 with −2 in the equations for x and y? Graph the new equations and find out.

u O

C b P

56. A nice curve A(a, 0) x

x = 3 cos t + cos 3t , y = 3 sin t − sin 3t , 0 ≤ t ≤ 2π What happens if you replace 3 with −3 in the equations for x and y? Graph the new equations and find out.

633

10.2  Calculus with Parametric Curves

58. a. x = 6 cos t + 5 cos 3t , y = 6 sin t − 5 sin 3t , 0 ≤ t ≤ 2π

57. a. Epicycloid x = 9 cos t − cos 9t , y = 9 sin t − sin 9t , 0 ≤ t ≤ 2π

b. x = 6 cos 2t + 5 cos 6t , y = 6 sin 2t − 5 sin 6t , 0 ≤ t ≤ π

b. Hypocycloid x = 8 cos t + 2 cos 4 t , y = 8 sin t − 2 sin 4 t , 0 ≤ t ≤ 2π c. Hypotrochoid x = cos t + 5 cos 3t , y = 6 cos t − 5 sin 3t , 0 ≤ t ≤ 2π

c. x = 6 cos t + 5 cos 3t , y = 6 sin 2t − 5 sin 3t , 0 ≤ t ≤ 2π d. x = 6 cos 2t + 5 cos 6t , y = 6 sin 4 t − 5 sin 6t , 0 ≤ t ≤ π

10.2 Calculus with Parametric Curves In this section we use calculus to study parametric curves. Specifically, we find slopes, lengths, and areas associated with parametrized curves.

Tangent Lines and Areas A parametrized curve x = f (t ) and y = g(t ) is differentiable at t if f and g are each differentiable at t. At a point on a differentiable parametrized curve where y is also a differentiable function of x, the derivatives dy dt ,  dx dt , and dy dx are related by the Chain Rule: dy dy dx = ⋅ . dt dx dt If dx dt ≠ 0, we may divide both sides of this equation by dx dt to solve for dy dx. Parametric Formula for dy dx

If all three derivatives exist and dx dt ≠ 0, then dy dt dy = . dx dx dt

(1)

If parametric equations define y as a twice-differentiable function of x, we can apply Equation (1) to the function dy dx = y′ to calculate d 2 y dx 2 as a function of t: dy′ dt d 2y d = . ( y′) = 2 dx dx dx dt

Eq. (1) with y′ in place of y

Parametric Formula for d 2 y dx 2

If the equations x = f (t ),  y = g(t ) define y as a twice-differentiable function of x, then at any point where dx dt ≠ 0 and y′ = dy dx,

y

dy′ dt d 2y = . 2 dx dx dt

2

1

0

t= p 4 (" 2, 1)

1

FIGURE 10.14

x 2 x = sec t, y = tan t, –p 0 throughout the interval. At a point where a curve stops and then doubles back on itself, either the curve fails to be differentiable or both derivatives must simultaneously equal zero. We will examine this phenomenon in Chapter 12, where we study tangent vectors to curves. If there are two different parametrizations for a curve C whose length we want to find, it does not matter which one we use. However, the parametrization we choose must meet the conditions stated in the definition of the length of C (see Exercise 41 for an example).

EXAMPLE 4 Using the definition, find the length of the circle of radius r defined parametrically by x = r cos t

y = r sin t ,

and

0 ≤ t ≤ 2Q.

Solution As t varies from 0 to 2Q, the circle is traversed exactly once, so the circumference is 2Q

∫0

L =

( ) dx dt

2

+

( ) dy dt

2

dt.

We find dx = −r sin t , dt

dy = r cos t dt

and

( dxdt )

2

+

( dydt )

2

= r 2 ( sin 2 t + cos 2 t ) = r 2 .

Therefore, the total arc length is 2Q

∫0

L = EXAMPLE 5

2Q

r 2 dt = r ⎡⎢ t ⎤⎥ = 2Qr. ⎣ ⎦0

Find the length of the astroid (Figure 10.15) x = cos 3 t ,

y = sin 3 t ,

0 ≤ t ≤ 2Q.

Solution Because of the curve’s symmetry with respect to the coordinate axes, its length is four times the length of the first-quadrant portion. We have

x = cos 3 t ,

( dxdt )

2

( dydt )

2

( dxdt ) + ( dydt ) 2

y = sin 3 t

= [ 3 cos 2 t (− sin t ) ]2 = 9 cos 4 t sin 2 t

2

= [ 3 sin 2 t ( cos t ) ]2 = 9 sin 4 t cos 2 t =

9 cos 2 t sin 2 t ( cos 2 t + sin 2 t )

=

9 cos 2 t sin 2 t

 1

= 3 cos t sin t = 3 cos t sin t.

cos t sin t ≥ 0 for 0 ≤ t ≤ Q 2

10.2  Calculus with Parametric Curves

637

Therefore, Length of first-quadrant portion = =

Q 2

∫0

3 cos t sin t dt

3 Q2 sin 2t dt 2 ∫0

cos t sin t = (1 2 )sin 2t

Q 2 3 3 = − cos 2t ⎤⎥ = . ⎦0 4 2

The length of the astroid is four times this: 4 ( 3 2 ) = 6.

EXAMPLE 6

Find the perimeter of the ellipse

y2 x2 + 2 = 1, where a  b  0. 2 a b

Solution Parametrically, we represent the ellipse by the equations x  a sin t and y = b cos t , 0 ≤ t ≤ 2Q. Then

( dxdt )

2

+

( dydt )

2

= a 2 cos 2 t + b 2 sin 2 t = a 2 − ( a 2 − b 2 )sin 2 t

b2 a2 (eccentricity, not the number 2.71828 . . . )

e =

= a 2 [ 1 − e 2 sin 2 t ].

1−

From Equation (3), the perimeter is given by P = 4a ∫

Q 2 0

1 − e 2 sin 2 t dt.

The integral for P is nonelementary and is known as the complete elliptic integral of the second kind. We can compute its value to within any degree of accuracy using infinite series in the following way. From the binomial expansion for 1  x 2 in Section 9.10, we have 1 − e 2 sin 2 t = 1 −

1 2 1 4 e sin 2 t − e sin 4 t − ". 2 2⋅4

e sin t ≤ e < 1

Then, to each term in this last expression, we apply the integral Formula 157 (at the back Q 2 of the text) for ¨ 0 sin n t dt when n is even, which yields the perimeter P = 4a ∫

Q 2 0

1 − e 2 sin 2 t dt

( )( 12 ⋅ Q2 ) − ( 2 1⋅ 4 e )( 21 ⋅⋅ 34 ⋅ Q2 ) − ( 2 1⋅ 4⋅ 3⋅ 6 e )( 21 ⋅⋅ 34 ⋅⋅ 56 ⋅ Q2 ) − "⎤⎥⎦

Q 1 = 4 a ⎡⎢ − e 2 ⎣2 2 ⎡ 1 = 2Qa ⎢ 1 − ⎢⎣ 2

( )e

4

2

2

−

( 21 ⋅⋅ 34 )

2

6

(

e4 1⋅3⋅5 − 3 2⋅4⋅6

)

2

⎤ e6 − "⎥ . ⎥⎦ 5

Since e  1, the series on the right-hand side converges by comparison with the geometric ∞ series ∑ n =1 (e 2 ) n . We do not have an explicit value for P, but we can estimate it as closely as we like by summing finitely many terms from the infinite series.

638

Chapter 10 Parametric Equations and Polar Coordinates

Length of a Curve y = f ( x) We will show that the length formula in Section 6.3 is a special case of Equation (3). Given a continuously differentiable function y = f ( x ),   a ≤ x ≤ b, we can assign x = t as a parameter. The graph of the function f is then the curve C defined parametrically by x = t

y = f (t ),

and

a ≤ t ≤ b,

which is a special case of what we have considered in this chapter. We have dx =1 dt

dy = f ′(t ). dt

and

From Equation (1), dy dt dy = = f ′(t ), dx dx dt giving

( dxdt )

2

+

( ) dy dt

2

= 1 + [ f ′(t ) ]2 = 1 + [ f ′( x ) ]2 .

t = x

Substitution into Equation (3) gives exactly the arc length formula for the graph of y = f ( x ) that we found in Section 6.3.

The Arc Length Differential As in Section 6.3, we define the arc length function for a parametrically defined curve x = f (t ) and y = g(t ),   a ≤ t ≤ b, by s (t ) =

t

∫a

[ f ′( z ) ] + [ g′( z ) ] dz. 2

2

Then, by the Fundamental Theorem of Calculus, ds = dt

[ f ′(t ) ] + [ g′(t ) ] 2

2

=

( dxdt )

2

+

( dydt ) . 2

The differential of arc length is ds =

( dxdt )

2

+

( dydt )

2

dt.

(4)

Equation (4) is often abbreviated as ds =

dx 2 + dy 2 .

Just as in Section 6.3, we can integrate the differential ds between appropriate limits to find the total length of a curve. Here’s an example where we use the arc length differential to find the centroid of an arc. EXAMPLE 7

Find the centroid of the first-quadrant arc of the astroid in Example 5.

Solution We take the curve’s density to be δ = 1 and calculate the curve’s mass and moments about the coordinate axes as we did in Section 6.6. The distribution of mass is symmetric about the line y = x, so x = y. A typical segment of the curve (Figure 10.18) has mass

dm = 1 ⋅ ds =

( dxdt )

2

+

( dydt )

2

dt = 3 cos t sin t dt.

From Example 5

10.2  Calculus with Parametric Curves

639

The curve’s mass is M =

π 2

∫0

dm =

π 2

∫0

3 cos t sin t dt =

3 . 2

Again from Example 5

y

The curve’s moment about the x-axis is B(0, 1) ~~ (x, y) = (cos 3 t, sin3 t) x~ ds ~ y

Mx =

π 2

∫ y dm = ∫0

= 3∫

C

0

A(1, 0)

x

sin 3 t ⋅ 3 cos t sin t dt

π 2

sin 4 t cos t dt = 3 ⋅

0

sin 5 t ⎤ π 2 3 = . ⎥ 5 ⎦0 5

It follows that y =

FIGURE 10.18

The centroid C of the astroid arc in Example 7.

35 Mx 2 = = . M 32 5

The centroid is the point ( 2 5, 2 5 ). EXAMPLE 8 Find the time Tc it takes for a frictionless bead to slide along the cycloid x = a ( t − sin t ),  y = a (1 − cos t ) from t = 0 to  t = π (see Figure 10.13). Solution From Equation (3) in Section 10.1, we want to find the time

Tc =

t=π

∫t = 0

ds .  2 gy

We need to express ds parametrically in terms of the parameter t. For the cycloid, dx dt = a (1 − cos t ) and dy dt = a sin t , so

( dxdt )

ds =

2

+

( dydt )

2

dt

=

a 2 (1 − 2 cos t + cos 2 t + sin 2 t ) dt

=

a 2 ( 2 − 2 cos t ) dt.

Substituting for ds and y in the integrand, it follows that π

a 2 ( 2 − 2 cos t ) dt 2 ga (1 − cos t )

π

a a dt = π . g g

Tc =

∫0

=

∫0

y = a (1 − cos t )

This is the amount of time it takes the frictionless bead to slide down the cycloid to B after it is released from rest at O (see Figure 10.13).

Areas of Surfaces of Revolution In Section 6.4 we found integral formulas for the area of a surface when a curve is revolved about a coordinate axis. Specifically, we found that the surface area is S = ∫ 2π y ds for revolution about the x-axis, and S = ∫ 2π x ds for revolution about the y-axis. If the curve is parametrized by the equations x = f (t ) and y = g(t ),   a ≤ t ≤ b, where f and g are continuously differentiable and ( f ′) 2 + ( g′) 2 > 0 on [ a,  b ], then the arc length differential ds is given by Equation (4). This observation leads to the following formulas for area of surfaces of revolution for smooth parametrized curves.

640

Chapter 10 Parametric Equations and Polar Coordinates

Area of Surface of Revolution for Parametrized Curves

If a smooth curve x = f (t ),   y = g(t ),   a ≤ t ≤ b, is traversed exactly once as t increases from a to b, then the areas of the surfaces generated by revolving the curve about the coordinate axes are as follows. 1. Revolution about the x-axis ( y ≥ 0 ): S =

b

∫a

2π y

( dxdt )

2

+

( dydt )

2

+

( dydt )

2

dt

(5)

dt

(6)

2. Revolution about the y-axis ( x ≥ 0 ): S =

b

∫a

2π x

( dxdt )

2

As with length, we can calculate surface area from any convenient parametrization that meets the stated criteria. Circle x = cos t y = 1 + sin t 0 ≤ t ≤ 2p

y

EXAMPLE 9 The standard parametrization of the circle of radius 1 centered at the point ( 0, 1) in the xy-plane is x = cos t , (0, 1)

0 ≤ t ≤ 2π.

Use this parametrization to find the area of the surface swept out by revolving the circle about the x-axis (Figure 10.19). x

FIGURE 10.19

In Example 9 we calculate the area of the surface of revolution swept out by this parametrized curve.

EXERCISES

y = 1 + sin t ,

Solution We evaluate the formula b

S =

∫a

=

∫0

2π y

2π

= 2π ∫

( dxdt ) + ( dydt ) 2

2

Eq. (5) for revolution about the x-axis; y = 1 + sin t ≥ 0

dt

2π (1 + sin t )   (− sin t ) 2 + ( cos t ) 2 dt 2π 0

 1

(1 + sin t ) dt

⎤ 2π ⎡ = 2π ⎢ t − cos t ⎥ = 4 π 2 . ⎢⎣ ⎥⎦ 0

10.2

Tangent Lines to Parametrized Curves

In Exercises 1–14, find an equation for the line tangent to the curve at the point defined by the given value of t. Also, find the value of d 2 y dx 2 at this point.

5. x = t , y = 6. x =

sec 2

t, t = 1 4

t − 1, y = tan t , t = −π 4

7. x = sec t , y = tan t , t = π 6

1. x = 2 cos t , y = 2 sin t , t = π 4

8. x = − t + 1, y =

3t , t = 3

2. x = sin 2πt , y = cos 2πt , t = −1 6

9. x = 2t 2 + 3, y = t 4 , t = −1

3. x = 4 sin t , y = 2 cos t , t = π 4

10. x = 1 t , y = −2 + ln t , t = 1

4. x = cos t , y =

11. x = t − sin t , y = 1 − cos t , t = π 3

3 cos t , t = 2π 3

10.2  Calculus with Parametric Curves

12. x = cos t , y = 1 + sin t , t = π 2 1 t 13. x = , y = , t = 2 t +1 t −1 14. x = t + e t , y = 1 − e t , t = 0

Centroids

37. Find the coordinates of the centroid of the curve x = cos t + t sin t , y = sin t − t cos t , 0 ≤ t ≤ π 2. 38. Find the coordinates of the centroid of the curve

Implicitly Defined Parametrizations

Assuming that the equations in Exercises 15–20 define x and y implicitly as differentiable functions x = f (t ),   y = g(t ), find the slope of the curve x = f (t ),   y = g(t ) at the given value of t. 15. x 3 + 2t 2 = 9, 2 y 3 − 3t 2 = 4, t = 2 16. x =

5−

t , y ( t − 1) =

t, t = 4

17. x + 2 x 3 2 = t 2 + t , y t + 1 + 2t y = 4, t = 0 18. x sin t + 2 x = t , t sin t − 2t = y, t = π

x = e t cos t , y = e t sin t , 0 ≤ t ≤ π. 39. Find the coordinates of the centroid of the curve x = cos t , y = t + sin t , 0 ≤ t ≤ π. T 40. Most centroid calculations for curves are done with a calculator or computer that has an integral evaluation program. As a case in point, find, to the nearest hundredth, the coordinates of the centroid of the curve

19. x = t 3 + t , y + 2t 3 = 2 x + t 2 , t = 1

x = t 3 , y = 3t 2 2, 0 ≤ t ≤

20. t = ln ( x − t ), y = te t , t = 0

41. Length is independent of parametrization To illustrate the fact that the numbers we get for length do not depend on the way we parametrize our curves (except for the mild restrictions preventing doubling back mentioned earlier), calculate the length of the semicircle y = 1 − x 2 with these two different parametrizations:

21. Find the area under one arch of the cycloid x = a ( t − sin t ), y = a (1 − cos t ). 22. Find the area enclosed by the y-axis and the curve x = t − t 2 , y = 1 + e −t .

a. x = cos 2t , y = sin 2t , 0 ≤ t ≤ π 2.

23. Find the area enclosed by the ellipse x = a cos t , y = b sin t , 0 ≤ t ≤ 2π. 24. Find the area under y = x 3 over [ 0, 1 ] using the following parametrizations. a. x = t 2 , y = t 6 b. x = t 3 , y = t 9

b. x = sin πt , y = cos πt , −1 2 ≤ t ≤ 1 2. 42. a. Show that the Cartesian formula

Find the lengths of the curves in Exercises 25–30. 25. x = cos t , y = t + sin t , 0 ≤ t ≤ π 26. x =

y =

2, 0 ≤ t ≤

L =

3

27. x = t 2 2, y = ( 2t + 1)3 2 3, 0 ≤ t ≤ 4

y = cos t , 0 ≤ t ≤ π 3

0 ≤ t ≤ π2

b

∫a

( dxdt )

2

+

( dydt )

2

dt.

b. x = y 3 2 , 0 ≤ y ≤ 4 3 3 c. x = y 2 3 , 0 ≤ y ≤ 1 2 43. The curve with parametric equations x = (1 + 2 sin θ ) cos θ, y = (1 + 2 sin θ ) sin θ

Surface Area

Find the areas of the surfaces generated by revolving the curves in Exercises 31–34 about the indicated axes. 31. x = cos t , y = 2 + sin t , 0 ≤ t ≤ 2π; x -axis 32. x = ( 2 3 ) t 3 2 , y = 2 t , 0 ≤ t ≤ 33. x = t +

⎛ dx ⎞ 2 1 + ⎜⎜ ⎟⎟⎟ dy ⎝ dy ⎠

d

Use this result to find the length of each curve.

28. x = ( 2t + 3 )3 2 3, y = t + t 2 2, 0 ≤ t ≤ 3 30. x = ln ( sec t + tan t ) − sin t 29. x = 8 cos t + 8t sin t y = 8 sin t − 8t cos t ,

∫c

L =

for the length of the curve x = g( y ),   c ≤ y ≤ d (Section 6.3, Equation 4), is a special case of the parametric length formula

Lengths of Curves

3t 2

3.

Theory and Examples

Area

t 3,

641

2, y = ( t 2 2 ) +

3; y-axis

2t , − 2 ≤ t ≤

is called a limaçon and is shown in the accompanying figure. Find the points ( x ,  y ) and the slopes of the tangent lines at these points for a. θ = 0.

b. θ = π 2.

c. θ = 4 π 3.

y

2; y-axis

34. x = ln ( sec t + tan t ) − sin t ,   y = cos t ,  0 ≤ t ≤ π 3;   x -axis 35. A cone frustum The line segment joining the points ( 0, 1) and ( 2, 2 ) is revolved about the x-axis to generate a frustum of a cone. Find the surface area of the frustum using the parametrization x = 2t ,   y = t + 1,  0 ≤ t ≤ 1. Check your result with the geometry formula: Area = π ( r1 + r2 )( slant height ).

3

36. A cone The line segment joining the origin to the point ( h,  r ) is revolved about the x-axis to generate a cone of height h and base radius r. Find the cone’s surface area with the parametric equations x = ht ,   y = rt ,  0 ≤ t ≤ 1. Check your result with the geometry formula: Area = πr ( slant height ).

1

2

−1

1

x

642

Chapter 10 Parametric Equations and Polar Coordinates

44. The curve with parametric equations

48. Volume Find the volume swept out by revolving the region bounded by the x-axis and one arch of the cycloid

x = t , y = 1 − cos t , 0 ≤ t ≤ 2Q is called a sinusoid and is shown in the accompanying figure. Find the point ( x ,  y ) where the slope of the tangent line is a. largest.

b. smallest.

x = t − sin t , y = 1 − cos t about the x-axis. 49. Find the volume swept out by revolving the region bounded by the x-axis and the graph of

y

x = 2t , y = t ( 2 − t )

2

about the x-axis. x

2p

0

50. Find the volume swept out by revolving the region bounded by the y-axis and the graph of x = t (1 − t ) , y = 1 + t 2

T The curves in Exercises 45 and 46 are called Bowditch curves or Lissajous figures. In each case, find the point in the interior of the first quadrant where the tangent line to the curve is horizontal, and find the equations of the two tangent lines at the origin. 45.

y

46.

y

x = sin t y = sin 2t

−1

1

1

x

1

COMPUTER EXPLORATIONS

In Exercises 51–54, use a CAS to perform the following steps for the given curve over the closed interval. a. Plot the curve together with the polygonal path approximations for n  2, 4, 8 partition points over the interval. (See Figure 10.16.)

x = sin 2t y = sin 3t

−1

about the y-axis.

x

−1

47. Cycloid a. Find the length of one arch of the cycloid x = a ( t − sin t ), y = a (1 − cos t ). b. Find the area of the surface generated by revolving one arch of the cycloid in part (a) about the x-axis for a  1.

b. Find the corresponding approximation to the length of the curve by summing the lengths of the line segments. c. Evaluate the length of the curve using an integral. Compare your approximations for n  2, 4, 8 with the actual length given by the integral. How does the actual length compare with the approximations as n increases? Explain your answer. 1 1 51. x = t 3 , y = t 2 , 0 ≤ t ≤ 1 3 2 52. x = 2t 3 − 16t 2 + 25t + 5, y = t 2 + t − 3, 0 ≤ t ≤ 6 53. x = t − cos t , y = 1 + sin t , −Q ≤ t ≤ Q 54. x = e t cos t , y = e t sin t , 0 ≤ t ≤ Q

10.3 Polar Coordinates In this section we study polar coordinates and their relation to Cartesian coordinates. You will see that polar coordinates are very useful for calculating many multiple integrals studied in Chapter 14. They are also useful in describing the paths of planets and satellites.

Definition of Polar Coordinates P(r, u) r Origin (pole)

u O

FIGURE 10.20

Initial ray

x

To define polar coordinates for the plane, we start with an origin, called the pole, and an initial ray.

To define polar coordinates, we first fix an origin O (called the pole) and an initial ray from O (Figure 10.20). Usually the positive x-axis is chosen as the initial ray. Then each point P can be located by assigning to it a polar coordinate pair ( r, R ) in which r gives the directed distance from O to P, and R gives the directed angle from the initial ray to ray OP. So we label the point P as P ( r, R ) / \

Directed distance from  O  to  P

Directed angle from initial ray to  OP

10.3  Polar Coordinates

P a2, pb = P a2, −11pb 6 6 11p − 6 u = p6

x

O

Initial ray u=0

FIGURE 10.21

Polar coordinates are not

643

As in trigonometry, θ is positive when measured counterclockwise and negative when measured clockwise. The angle associated with a given point is not unique. A point in the plane has just one pair of Cartesian coordinates, but it has infinitely many pairs of polar coordinates. For instance, the point 2 units from the origin along the ray θ = π 6 has polar coordinates r = 2,  θ = π 6. It also has coordinates r = 2,  θ = −11π 6 (Figure 10.21). In some situations we allow r to be negative. That is why we use directed distance in defining P ( r,  θ ). The point P ( 2, 7π 6 ) can be reached by turning 7π 6 radians counterclockwise from the initial ray and going forward 2 units (Figure 10.22). It can also be reached by turning π 6 radians counterclockwise from the initial ray and going backward 2 units. So the point also has polar coordinates r = −2,  θ = π 6.

unique.

EXAMPLE 1 u = p6

7p6

p6 u=0

O

x

Find all the polar coordinates of the point P ( 2,  π 6 ).

Solution We sketch the initial ray of the coordinate system, draw the ray from the origin that makes an angle of π 6 radians with the initial ray, and mark the point ( 2,  π 6 ) (Figure 10.23). We then find the angles for the other coordinate pairs of P in which r = 2 and r = −2. For r = 2, the complete list of angles is

π π π π , ± 2π , ± 4π, ± 6π , . . . . 6 6 6 6

P a2, 7pb = P a–2, pb 6 6

For r = −2, the angles are

FIGURE 10.22

Polar coordinates can have negative r-values.

−

5p p a2, b = a–2, – b 6 6 7p6

The corresponding coordinate pairs of P are

= a–2, 7pb 6 etc.

p 6

( 2,  π6 + 2nπ ), x

–5p6

The point P ( 2,  π 6 ) has infinitely many polar coordinate pairs (Example 1).

FIGURE 10.23

r=a 0a0 O

x

(−2, −56π + 2nπ ),

circle is r = a.

The polar equation for a

n = 0, ±1, ±2, . . . .

When n = 0, the formulas give ( 2,  π 6 ) and (−2, −5π 6 ). When n = 1, they give ( 2, 13π 6 ) and (−2, 7π 6 ), and so on.

Polar Equations and Graphs If we hold r fixed at a constant value r = a ≠ 0, the point P ( r, θ ) will lie a units from the origin O. As θ varies over any interval of length 2π, P then traces a circle of radius a centered at O (Figure 10.24). If we hold θ fixed at a constant value θ = θ 0 and let r vary between −∞ and ∞, the point P ( r, θ ) traces the line through O that makes an angle of measure θ 0 with the initial ray. (See Figure 10.22 for an example.) EXAMPLE 2

FIGURE 10.24

n = 0, ±1, ±2, . . . 

and Initial ray

O

5π 5π 5π 5π , − ± 2π , − ± 4π, − ± 6π , . . . . 6 6 6 6

A circle or line can have more than one polar equation.

(a) r = 1 and r = −1 are equations for the circle of radius 1 centered at O. (b) θ = π 6,  θ = 7π 6, and θ = −5π 6 are equations for the line in Figure 10.23. Equations of the form r = a and θ = θ 0 , and inequalities such as r ≤ a and 0 ≤ θ ≤ π, can be combined to define regions, segments, and rays.

644

Chapter 10 Parametric Equations and Polar Coordinates

y

(a)

p 1 ≤ r ≤ 2, 0 ≤ u ≤ 2

EXAMPLE 3 conditions.

Graph the sets of points whose polar coordinates satisfy the following

(a) 1 ≤ r ≤ 2 0

1

2

x

(b) −3 ≤ r ≤ 2

y

(b)

2 0

u =p , 4 p −3 ≤ r ≤ 2 4 x

3

(c)

and

2π 5π ≤ θ ≤ 3 6

θ =

π 2

π 4

(no restriction on  r )

Solution The graphs are shown in Figure 10.25.

Relating Polar and Cartesian Coordinates

2p 3

(c)

0 ≤ θ ≤

and

y

5p 6

2p 5p ≤u≤ 3 6

When we use both polar and Cartesian coordinates in a plane, we place the two origins together and let the initial polar ray be the positive x-axis. The ray θ = π 2,  r > 0, becomes the positive y-axis (Figure 10.26). The two coordinate systems are then related by the following equations.

x

0

Equations Relating Polar and Cartesian Coordinates

Polar to Cartesian: FIGURE 10.25

The graphs of typical inequalities in r and θ (Example 3).

x = r cos θ,

y = r sin θ

r 2 = x 2 + y2,

tan θ =

Cartesian to Polar: y x

y Ray u = p 2 P(x, y) = P(r, u) r

Common origin 0

y

u x

u = 0, r ≥ 0 x Initial ray

The first two of these equations uniquely determine the Cartesian coordinates x and y given  the polar coordinates r and θ. On the other hand, if x and y are given and ( x , y ) ≠ ( 0, 0 ), the third equation gives two possible choices for r (a positive and a negative value). For each of these r values, there is a unique θ ∈ [ 0, 2π ) satisfying the first two equations, each then giving a polar coordinate representation of the Cartesian point ( x ,  y ). The other polar coordinate representations for the point can be determined from these two, as in Example 1. EXAMPLE 4 Here are some plane curves expressed in terms of both polar coordinate and Cartesian coordinate equations.

FIGURE 10.26

The usual way to relate polar and Cartesian coordinates.

y

Polar equation

Cartesian equivalent

r cos θ = 2

x = 2

r2 x 2 + ( y − 3) 2 = 9 or r = 6 sin u

(0, 3)

0

FIGURE 10.27

r2

cos θ sin θ = 4

cos 2 θ

−

r 2 sin 2 θ

=1

xy = 4 x2

− y2 = 1

r = 1 + 2r cos θ

y 2 − 3x 2 − 4 x − 1 = 0

r = 1 − cos θ

x 4 + y 4 + 2 x 2 y 2 + 2 x 3 + 2 xy 2 − y 2 = 0

Some curves are more simply expressed with polar coordinates; others are not. x

The circle in Example 5.

EXAMPLE 5

Find a polar equation for the circle x 2 + ( y − 3 ) 2 = 9 (Figure 10.27).

Solution We apply the equations relating polar and Cartesian coordinates:

10.3  Polar Coordinates

645

x 2 + ( y − 3)2 = 9 x 2 + y 2 − 6y + 9 = 9

Expand ( y − 3 ) 2.

x 2 + y 2 − 6y = 0 r2

Cancelation

− 6r sin θ = 0

x 2 + y 2 = r 2 ,  y = r sin θ

r = 0 or r − 6 sin θ = 0 r = 6 sin θ

Includes both possibilities

EXAMPLE 6 Replace the following polar equations by equivalent Cartesian equations and identify their graphs. (a) r cos θ = −4 (b) r 2 = 4r cos θ (c) r =

4 2 cos θ − sin θ

Solution We use the substitutions r cos θ = x ,  r sin θ = y, and  r 2 = x 2 + y 2 .

(a) r cos θ = −4 The Cartesian equation:

r cos θ = −4 x = −4

The graph:

Substitute.

Vertical line through x = −4 on the x-axis

(b) r 2 = 4r cos θ The Cartesian equation:

r 2 = 4r cos θ x 2 + y 2 = 4x

Substitute.

x 2 − 4x + y 2 = 0 x 2 − 4x + 4 + y 2 = 4 (x

The graph: (c) r =

Factor.

Circle, radius 2, center ( h,  k ) = ( 2, 0 )

4 2 cos θ − sin θ

The Cartesian equation:

The graph:

EXERCISES

− 2 )2 + y 2 = 4

Complete the square.

r ( 2 cos θ − sin θ ) = 4 2r cos θ − r sin θ = 4

Multiply by r.

2x − y = 4

Substitute.

y = 2x − 4

Solve for y.

Line, slope m = 2, y-intercept b = −4

10.3 2. Which polar coordinate pairs label the same point?

Polar Coordinates

1. Which polar coordinate pairs label the same point?

a. (−2,  π 3 )

b. ( 2, −π 3 )

c.

f. ( 2, −2π 3 )

a. ( 3, 0 )

b. (−3, 0 )

c. ( 2, 2π 3 )

d. ( r,  θ + π )

e. (−r,  θ )

d. ( 2, 7π 3 )

e. (−3,  π )

f. ( 2,  π 3 )

g. (−r,  θ + π )

h. (−2, 2π 3 )

g. (−3, 2π )

h. (−2, −π 3 )

( r,  θ )

646

Chapter 10 Parametric Equations and Polar Coordinates

3. Plot the following points, given in polar coordinates. Then find all the polar coordinates of each point. a. ( 2,  π 2 )

b.

c. (−2,  π 2 )

d. (−2, 0 )

( 2, 0 )

19. θ = π 2, r ≥ 0

20. θ = π 2, r ≤ 0

21. 0 ≤ θ ≤ π , r = 1

22. 0 ≤ θ ≤ π , r = −1

23. π 4 ≤ θ ≤ 3π 4 , 0 ≤ r ≤ 1

4. Plot the following points, given in polar coordinates. Then find all the polar coordinates of each point.

24. −π 4 ≤ θ ≤ π 4 , −1 ≤ r ≤ 1 25. −π 2 ≤ θ ≤ π 2, 1 ≤ r ≤ 2

a. ( 3,  π 4 )

b. (−3,  π 4 )

26. 0 ≤ θ ≤ π 2, 1 ≤ r ≤ 2

c. ( 3, −π 4 )

d. (−3, −π 4 )

Polar to Cartesian Equations

Polar to Cartesian Coordinates

5. Find the Cartesian coordinates of the points in Exercise 1. 6. Find the Cartesian coordinates of the following points, given in polar coordinates.

Replace the polar equations in Exercises 27–52 with equivalent Cartesian equations. Then describe or identify the graph. 27. r cos θ = 2

28. r sin θ = −1

29. r sin θ = 0

30. r cos θ = 0 32. r = −3 sec θ 34. r sin θ = r cos θ

a. ( 2,  π 4 )

b. (1, 0 )

31. r = 4 csc θ

c. ( 0,  π 2 )

d. (− 2,  π 4 )

33. r cos θ + r sin θ = 1

tan −1 ( 4

e. (−3, 5π 6 )

f. ( 5,

3 ))

g. (−1, 7π )

h. ( 2 3, 2π 3 )

Cartesian to Polar Coordinates

7. Find the polar coordinates, 0 ≤ θ < 2π and r ≥ 0, of the following points given in Cartesian coordinates. a. (1, 1)

b. (−3, 0 )

c. ( 3, −1)

d. (−3, 4 )

8. Find the polar coordinates, −π ≤ θ < π and r ≥ 0, of the following points given in Cartesian coordinates. a. (−2, −2 )

b.

( 0, 3 )

c. (− 3, 1)

d.

( 5, − 12 )

9. Find the polar coordinates, 0 ≤ θ < 2π and r ≤ 0, of the following points given in Cartesian coordinates. a. ( 3, 3 )

b. (−1, 0 )

c. (−1,  3 )

d.

( 4, − 3 )

10. Find the polar coordinates, −π ≤ θ < π and r ≤ 0, of the following points given in Cartesian coordinates. a. (−2, 0 ) c.

( 0, − 3 )

b. (1, 0 ) ⎛ 3 1 ⎞⎟ ,  d. ⎜⎜ ⎝ 2 2 ⎟⎟⎠

Graphing Sets of Polar Coordinate Points

Graph the sets of points whose polar coordinates satisfy the equations and inequalities in Exercises 11–26. 11. r = 2

12. 0 ≤ r ≤ 2

13. r ≥ 1

14. 1 ≤ r ≤ 2

15. 0 ≤ θ ≤ π 6, r ≥ 0

16. θ = 2π 3, r ≤ −2

17. θ = π 3, −1 ≤ r ≤ 3

18. θ = 11π 4 , r ≥ −1

35.

r2

=1

37. r =

36. r 2 = 4r sin θ

5 sin θ − 2 cos θ

38. r 2 sin 2θ = 2

39. r = cot θ csc θ 41. r = csc θ e

40. r = 4 tan θ sec θ

r cos θ

42. r sin θ = ln r + ln cos θ

43. r 2 + 2r 2 cos θ sin θ = 1

44. cos 2 θ = sin 2 θ

45. r 2 = −4r cos θ

46. r 2 = −6r sin θ

47. r = 8 sin θ

48. r = 3 cos θ

49. r = 2 cos θ + 2 sin θ π 51. r sin θ + = 2 6

50. r = 2 cos θ − sin θ 2π 52. r sin −θ = 5 3

(

)

(

)

Cartesian to Polar Equations

Replace the Cartesian equations in Exercises 53–66 with equivalent polar equations. 53. x = 7

54. y = 1

56. x − y = 3

57.

x2

+

55. x = y

y2

= 4

58. x 2 − y 2 = 1

y2 x2 + =1 9 4 61. y 2 = 4 x

60. xy = 2

63. x 2 + ( y − 2 ) = 4

64. ( x − 5 ) 2 + y 2 = 25

65. ( x − 3 ) 2 + ( y + 1) 2 = 4

66. ( x + 2 ) 2 + ( y − 5 ) 2 = 16

59.

2

62. x 2 + xy + y 2 = 1

67. Find all polar coordinates of the origin. 68. Vertical and horizontal lines a. Show that every vertical line in the xy-plane has a polar equation of the form r = a sec θ. b. Find the analogous polar equation for horizontal lines in the xy-plane.

10.4 Graphing Polar Coordinate Equations It is often helpful to graph an equation expressed in polar coordinates in the Cartesian xy-plane. This section describes some techniques for graphing these equations using symmetries and tangent lines to the graph.

647

10.4  Graphing Polar Coordinate Equations

y

Symmetry (r, u)

The following list shows how to test for three standard types of symmetries when using polar coordinates. These symmetries are illustrated in Figure 10.28. x

0

(r, −u) or (−r, p − u) (a) About the x-axis (r, p − u) y or (−r, −u)

(r, u)

Symmetry Tests for Polar Graphs in the Cartesian xy-Plane

1. Symmetry about the x-axis: If the point ( r,  θ ) lies on the graph, then the point ( r, −θ ) or ( −r,  π − θ ) lies on the graph (Figure 10.28a). 2. Symmetry about the y-axis: If the point ( r,  θ ) lies on the graph, then the point ( r,  π − θ ) or ( −r, −θ ) lies on the graph (Figure 10.28b). 3. Symmetry about the origin: If the point ( r,  θ ) lies on the graph, then the point ( −r,  θ ) or ( r,  θ + π ) lies on the graph (Figure 10.28c).

x

0

Slope The slope of a polar curve r = f (θ) in the xy-plane is dy dx, but this is not given by the formula r ′ = df dθ. To see why, think of the graph of f as the graph of the parametric equations

(b) About the y-axis y

x = r cos θ = f (θ) cos θ,

(r, u)

0

x

(−r, u) or (r, u + p) (c) About the origin

FIGURE 10.28

Three tests for symmetry in polar coordinates.

y = r sin θ = f (θ)sin θ.

If f is a differentiable function of θ, then so are x and y, and when dx dθ ≠ 0, we can calculate dy dx from the parametric formula dy dθ dy = dx dx dθ

Section 10.2, Eq. (1) with t = θ

d ( f (θ)sin θ) d = θ d ( f (θ) cos θ) dθ

Substitute

df sin θ + f (θ) cos θ d = θ df cos θ − f (θ)sin θ dθ

Product Rule for derivatives

Therefore, we see that dy dx is not the same as df dθ. Slope of the Curve r = f (T ) in the Cartesian xy-Plane

dy dx

( r ,  θ )

=

f ′(θ)sin θ + f (θ) cos θ , f ′(θ) cos θ − f (θ)sin θ

(1)

provided dx dθ ≠ 0 at ( r,  θ ). If the curve r = f (θ) passes through the origin at θ = θ 0 , then f (θ 0 ) = 0, and the slope equation gives dy dx

( 0,  θ 0 )

=

f ′(θ 0 )sin θ 0 = tan θ 0 . f ′(θ 0 ) cos θ 0

That is, the slope at ( 0,  θ 0 ) is tan θ 0 . The reason we say “slope at ( 0,  θ 0 )” and not just “slope at the origin” is that a polar curve may pass through the origin (or any point) more than once, with different slopes at different θ-values. This is not the case in our first example, however.

648

Chapter 10 Parametric Equations and Polar Coordinates

T

r = 1 − cos T

0 π 3 π 2

0 1 2

2π 3 π

3 2 2

Graph the curve r = 1 − cos θ in the Cartesian xy-plane.

EXAMPLE 1

Solution The curve is symmetric about the x-axis because (r,

θ ) on the graph ⇒ r = 1 − cos θ ⇒ r = 1 − cos( −θ ) ⇒ ( r , −θ ) on the graph.

1

(a)

3 2p a , b y 2 3 p a1, 2b 3 2 (p, 2)

2

1

1 p a , b 2 3 x

0

r = 1 − cos u

2

Graph the curve r 2 = 4 cos θ in the Cartesian xy-plane.

EXAMPLE 2

( r ,  θ ) on the graph

1 p a , b 2 3 x

0

a1,

3p b 2

(c)

FIGURE 10.29

The steps in graphing the cardioid r = 1 − cos θ (Example 1). The arrow shows the direction of increasing θ.

cos θ = cos(−θ )

⇒ ( r , −θ ) on the graph. The curve is also symmetric about the origin because ( r ,  θ ) on the graph

⇒ r 2 = 4 cos θ ⇒ (−r ) 2 = 4 cos θ

1 5p a2 , 3 b 3 4p a , b 2 3

⇒ r 2 = 4 cos θ ⇒ r 2 = 4 cos( −θ )

p a1, 2b

1 (p, 2)

As θ increases from 0 to π , cos θ decreases from 1 to −1, and r = 1 − cos θ increases from a minimum value of 0 to a maximum value of 2. As θ continues on from π to 2π , cos θ increases from −1 back to 1, and r decreases from 2 back to 0. The curve starts to repeat when θ = 2π because the cosine has period 2π. The curve leaves the origin with slope tan(0) = 0 and returns to the origin with slope tan ( 2π ) = 0. We make a table of values from θ = 0 to θ = π , plot the points, draw a smooth curve through them with a horizontal tangent line at the origin, and reflect the curve across the x-axis to complete the graph (Figure 10.29). The curve is called a cardioid because of its heart shape.

Solution The equation r 2 = 4 cos θ requires cos θ ≥ 0, so we get the entire graph by running θ from −π 2 to π 2. The curve is symmetric about the x-axis because

(b)

3 2p y a , b 2 3

cos θ = cos(−θ )

⇒ (−r ,  θ ) on the graph. Together, these two symmetries imply symmetry about the y-axis. The curve passes through the origin when θ = −π 2 and θ = π 2. It has a vertical tangent line both times because tan θ is infinite. For each value of θ in the interval between −π 2 and π 2, the formula r 2 = 4 cos θ gives two values of r: r = ±2 cos θ . We make a short table of values, plot the corresponding points, and use information about symmetry and tangent lines to guide us in connecting the points with a smooth curve (Figure 10.30). T

cos T

r = ±2 cos T

0

1

±2

π 6 π ± 4 π ± 3 π ± 2

3 2 1 2 1 2

≈ ±1.9

±

y

2

≈ ±1.7

r 2 = 4 cos u

2 0

≈ ±1.4

0

0 (a)

Loop for r = −2" cos u,

−p ≤ u ≤ p 2 2

Loop for r = 2" cos u,

−p ≤ u ≤ p 2 2

(b)

The graph of r 2 = 4 cos θ. The arrows show the direction of increasing θ. The values of r in the table are rounded (Example 2).

FIGURE 10.30

x

10.4  Graphing Polar Coordinate Equations

Converting a Graph from the rT -Plane to the xy-Plane

(a) r 2

One way to graph a polar equation r = f (θ) in the xy-plane is to make a table of ( r,  θ ) -values, plot the corresponding points there, and connect them in order of increasing θ. This can work well if enough points have been plotted to reveal all the loops and dimples in the graph. Another method of graphing follows.

r = sin 2u 2

1

p 2

3p 2

2p

p

p 4

0

649

u

1. First graph the function r = f (θ) in the Cartesian r θ-plane. 2. Then use that Cartesian graph as a “table” and guide to sketch the polar coordinate graph in the xy-plane.

−1 No square roots of negative numbers

This method is sometimes better than simple point plotting because the first Cartesian graph shows at a glance where r is positive, where negative, and where nonexistent, as well as where r is increasing and where it is decreasing. Here is an example.

(b) r r = +" sin 2u

1

p p 2

0 −1

3p 2

u

± parts from square roots

y r 2 = sin 2u 0

x

Graphing Polar Curves Parametrically For complicated polar curves, we may need to use a graphing calculator or computer to graph the curve. If the device does not plot polar graphs directly, we can convert r = f (θ) into parametric form using the equations

USING TECHNOLOGY

To plot r = f (θ) in the Cartesian rθ-plane in (b), we first plot r 2 = sin 2θ in the r 2θ-plane in (a) and then ignore the values of θ for which sin 2θ is negative. The radii from the sketch in (b) cover the polar graph of the lemniscate in (c) twice (Example 3). FIGURE 10.31

EXERCISES

Graph the lemniscate curve r 2 = sin 2θ in the Cartesian xy-plane.

Solution For this example it will be easier to first plot r 2 , instead of r, as a function of θ in the Cartesian r 2θ-plane (see Figure 10.31a). We pass from there to the graph of r = ± sin 2θ in the r θ-plane (Figure 10.31b), and then draw the polar graph (Figure 10.31c). The graph in Figure 10.31b “covers” the final polar graph in Figure 10.31c twice. We could have managed with either loop alone, with the two upper halves, or with the two lower halves. The double covering does no harm, however, and we actually learn a little more about the behavior of the function this way.

r = −" sin 2u

(c)

EXAMPLE 3

x = r cos θ = f (θ) cos θ,

y = r sin θ = f (θ) sin θ.

Then we use the device to draw a parametrized curve in the Cartesian xy-plane.

10.4

Symmetries and Polar Graphs

Identify the symmetries of the curves in Exercises 1–12. Then sketch the curves in the xy-plane.

Graph the lemniscates in Exercises 13–16. What symmetries do these curves have? 13. r 2 = 4 cos 2θ

14. r 2 = 4 sin 2θ

= − sin 2θ

16. r 2 = − cos 2θ

1. r = 1 + cos θ

2. r = 2 − 2 cos θ

3. r = 1 − sin θ

4. r = 1 + sin θ

Slopes of Polar Curves in the xy-Plane

5. r = 2 + sin θ

6. r = 1 + 2 sin θ

Find the slopes of the curves in Exercises 17–20 at the given points. Sketch the curves along with their tangent lines at these points.

7. r = sin ( θ 2 )

8. r = cos( θ 2 )

17. Cardioid r = −1 + cos θ; θ = ±π 2

9. r 2 = cos θ 11.

r2

= − sin θ

10. r 2 = sin θ 12.

r2

= − cos θ

15.

r2

18. Cardioid r = −1 + sin θ; θ = 0,   π 19. Four-leaved rose r = sin 2θ; θ = ±π 4 ,   ±3π 4 20. Four-leaved rose r = cos 2θ; θ = 0,   ±π 2,   π

650

Chapter 10 Parametric Equations and Polar Coordinates

Concavity of Polar Curves in the xy-Plane

In Exercises 31 and 32, sketch the region defined by the inequality.

Equation (1) gives the formula for the derivative y′ of a polar curve dy′ dθ d 2y r = f (θ). The second derivative is 2 = (see Equation (2) in dx dθ dx Section 10.2). Find the slope and concavity of the curves in Exercises 21–24 at the given points.

31. 0 ≤ r ≤ 2 − 2 cos θ

21. r = sin θ, θ = π 6,  π 3

22. r = e θ , θ = 0,  π

23. r = θ, θ = 0, π 2

24. r = 1 θ, θ = −π , 1

32. 0 ≤ r 2 ≤ cos θ

T 33. Which of the following has the same graph as r = 1 − cos θ ? a. r = −1 − cos θ b. r = 1 + cos θ Confirm your answer with algebra. T 34. Which of the following has the same graph as r = cos 2θ ? b. r = − cos( θ 2 ) a. r = − sin ( 2θ + π 2 ) Confirm your answer with algebra.

Graphing Limaçons

Graph the limaçons in Exercises 25–28. Limaçon (“lee-ma-sahn”) is Old French for “snail.” You will understand the name when you graph the limaçons in Exercise 25. Equations for limaçons have the form r = a ± b cos θ or r = a ± b sin θ. There are four basic shapes. 25. Limaçons with an inner loop 1 1 b. r = + sin θ a. r = + cos θ 2 2 26. Cardioids a. r = 1 − cos θ

T 35. A rose within a rose Graph the equation r = 1 − 2 sin 3θ. T 36. The nephroid of Freeth Graph the nephroid of Freeth: θ r = 1 + 2 sin . 2 T 37. Roses Graph the roses r = cos mθ for m = 1 3, 2, 3, and 7. T 38. Spirals Polar coordinates are just the thing for defining spirals. Graph the following spirals. a. r = θ

b. r = −1 + sin θ

27. Dimpled limaçons 3 a. r = + cos θ 2

b. r =

b. r = −θ c. A logarithmic spiral: r = e θ 10

3 − sin θ 2

d. A hyperbolic spiral: r = 8 θ

28. Oval limaçons

e. An equilateral hyperbola: r = ±10

a. r = 2 + cos θ

b. r = −2 + sin θ

Graphing Polar Regions and Curves in the xy-Plane

29. Sketch the region defined by the inequalities −1 ≤ r ≤ 2 and −π 2 ≤ θ ≤ π 2.

θ

(Use different colors for the two branches.) T 39. Graph the equation r = sin ( 87 θ ) for 0 ≤ θ ≤ 14 π . T 40. Graph the equation r = sin 2 ( 2.3θ ) + cos 4 ( 2.3θ )

30. Sketch the region defined by the inequalities 0 ≤ r ≤ 2 sec θ and −π 4 ≤ θ ≤ π 4.

for 0 ≤ θ ≤ 10 π .

10.5 Areas and Lengths in Polar Coordinates This section shows how to calculate areas of plane regions and lengths of curves in polar coordinates. y

Area in the Plane

u = b Δuk S

rn

The region OTS in Figure 10.32 is bounded by the rays θ = α and θ = β and the curve r = f (θ). We approximate the region with n nonoverlapping fan-shaped circular sectors based on a partition P of angle TOS. The typical sector has radius rk = f (θ k ) and central angle of radian measure Δθ k . Its area is Δθ k 2π times the area of a circle of radius rk , or

( f (uk ), uk) rk r = f (u)

Ak =

r2 r1 uk O

FIGURE 10.32

T

1 2 1 r Δθ k = ( f (θ k ) ) 2 Δθ k . 2 k 2

The area of region OTS is approximately

u = a

n

x

To derive a formula for the area of region OTS, we approximate the region with fan-shaped circular sectors.

∑ Ak k =1

=

n

∑ 12 ( f (θ k ) )2 Δθ k . k =1

If f is continuous, we expect the approximations to improve as the norm of the partition P goes to zero, where the norm of P is the largest value of Δθ k . We are therefore led to the following formula for the region’s area.

651

10.5  Areas and Lengths in Polar Coordinates

y

n

u

x

O

A =

The area differential dA for the curve r = f (θ).

r = 2(1 + cos u) P(r, u)

4

1 2 r dθ 2

1 2 1 r dθ = ( f (θ) ) 2 dθ. 2 2

In the area formula above, we assumed that r ≥ 0 and that the region does not sweep out an angle of more than 2π. This avoids issues with negatively signed areas or with regions that overlap themselves. More general regions can usually be handled by subdividing them into regions of this type if necessary.

u = 0, 2p

0

β

∫α

This is the integral of the area differential (Figure 10.33) dA =

r

1 ( f (θ) ) 2 d θ. 2

Area of the Fan-Shaped Region Between the Origin and the Curve r = f (T ) when α ≤ θ ≤ β,   r ≥ 0, and β − α ≤ 2 π

FIGURE 10.33

2

β

∫α

P(r, u)

r

y

=

k =1

dA = 1 r 2 du 2 du

1

∑ 2 ( f (θ k ) )2 Δθ k P →0

A = lim

x

EXAMPLE 1 Find the area of the region in the xy-plane enclosed by the cardioid r = 2(1 + cos θ ).

−2

FIGURE 10.34

The cardioid in

Example 1.

Solution We graph the cardioid (Figure 10.34) and determine that the radius OP sweeps out the region exactly once as θ runs from 0 to 2π. The area is therefore θ = 2π

∫θ = 0

2π

1 2 r dθ = 2

∫0

=

∫0

=

∫0

=

∫0

2π

2π

2π

1 ⋅ 4 (1 + cos θ ) 2 d θ 2 2(1 + 2 cos θ + cos 2 θ ) d θ 2θ ( 2 + 4 cos θ + 2 ⋅ 1 + cos ) dθ 2

( 3 + 4 cos θ + cos 2θ ) d θ

sin 2θ ⎤ 2 π ⎡ = ⎢ 3θ + 4 sin θ + ⎥ = 6π − 0 = 6π. 2 ⎦0 ⎣ To find the area of a region like the one in Figure 10.35, which lies between two polar curves r1 = r1 (θ) and r2 = r2 (θ) from θ = α to θ = β , we subtract the integral of (1 2 )r 12 dθ from the integral of (1 2 )r 22 dθ. This leads to the following formula.

y u=b r2

Area of the Region 0 ≤ r1 (θ ) ≤ r ≤ r2 (θ ),   α ≤ θ ≤ β, and β − α ≤ 2 π r1 u=a

0

FIGURE 10.35

A =

β

∫α

1 2 r dθ − 2 2

β

∫α

1 2 r dθ = 2 1

β

∫α

1( 2 r − r 12 ) dθ 2 2

(1)

x

The area of the shaded region is calculated by subtracting the area of the region between r1 and the origin from the area of the region between r2 and the origin.

EXAMPLE 2 Find the area of the region that lies inside the circle r = 1 and outside the cardioid r = 1 − cos θ. Solution We sketch the region to determine its boundaries and find the limits of integration (Figure 10.36). The outer curve is r2 = 1, the inner curve is r1 = 1 − cos θ, and θ runs from −π 2 to π 2. The area, from Equation (1), is

652

Chapter 10 Parametric Equations and Polar Coordinates

y r1 = 1 − cos u

Upper limit u = p2

A=

r2 = 1

π 2

= 2∫

u x

0

Lower limit u = −p2

FIGURE 10.36

The region and limits of integration in Example 2.

1

∫−π 2 2 ( r 22 π 2 0

π 2

=

∫0

=

∫0

π 2

− r 12 ) d θ

Eq. (1)

1( 2 r − r 12 ) d θ 2 2

Symmetry

(1 − (1 − 2 cos θ + cos 2 θ )) d θ ( 2 cos θ − cos 2 θ ) d θ =

π 2

∫0

r2 = 1 and  r1 = 1 − cos θ

2θ ( 2 cos θ − 1 + cos ) dθ 2

sin 2θ ⎤ π 2 ⎡ θ π = ⎢ 2 sin θ − − = 2− . ⎥ 2 4 ⎦0 4 ⎣ The fact that we can represent a point in different ways in polar coordinates requires that we take extra care in deciding when a point lies on the graph of a polar equation and in determining the points at which polar graphs intersect. (We needed intersection points in Example 2.) In Cartesian coordinates, we can always find the points where two curves cross by solving their equations simultaneously. In polar coordinates, the story is different. Simultaneous solution may reveal some intersection points without revealing others, so it is sometimes difficult to find all points of intersection of two polar curves. One way to identify all the points of intersection is to graph the equations. EXAMPLE 3 Find all of the points where the curve r = 2 cos( θ 3 ) intersects the circle of radius 2 centered at the origin.

y 2

("2 , 3p/4)

r = 2 cos (u/3)

1 r =" 2

−1

("2 , 5p/4)

1

2

−1

FIGURE 10.37

The curves r = 2 cos( θ 3 ) and r = 2 intersect at two points (Example 3).

x

Solution Note that the function r = 2 cos( θ 3 ) takes both positive and negative values. Therefore, when we look for the points where this curve intersects the circle, it is important to take into account that the circle is described both by the equation r = 2 and by the equation r = − 2. Solving 2 cos( θ 3 ) = 2 for θ yields

2 cos( θ 3 ) =

2, cos( θ 3 ) =

2 2,

θ 3 = π 4, θ = 3π 4.

This gives us one point, ( 2, 3π 4 ), where the two curves intersect. However, as we can see by looking at the graphs in Figure 10.37, there is a second intersection point. To find the second point, we solve 2 cos( θ 3 ) = − 2 for θ: 2 cos( θ 3 ) = − 2, cos( θ 3 ) = − 2 2,

θ 3 = 3π 4, θ = 9π 4.

The second intersection point is located at (− 2, 9π 4 ). We can specify this point in polar coordinates using a positive value of r and an angle between 0 and 2π. In polar coordinates, adding multiples of 2π to θ gives a second description of the same point in the plane. Similarly, changing the sign of r, while at the same time adding or subtracting π to or from θ, also gives a description of the same point. So in polar coordinates, (− 2, 9π 4 ) describes the same point in the plane as (− 2,  π 4 ) and also as ( 2, 5π 4 ). The second intersection point is located at ( 2, 5π 4 ).

Length of a Polar Curve We can obtain a polar coordinate formula for the length of a curve r = f (θ),  α ≤ θ ≤ β , by parametrizing the curve as x = r cos θ = f (θ) cos θ,

y = r sin θ = f (θ)sin θ,

α ≤ θ ≤ β.

(2)

653

10.5  Areas and Lengths in Polar Coordinates

The parametric length formula, Equation (3) from Section 10.2, then gives the length as dx ∫ ( dθ ) β

L =

α

2

+

( ddyθ )

2

d θ.

This equation becomes L =

β

∫α

( ddrθ )

r2 +

2

dθ

when Equations (2) are substituted for x and y (Exercise 29).

Length of a Polar Curve

If r = f (θ) has a continuous first derivative for α ≤ θ ≤ β , and if the point P ( r, θ ) traces the curve r = f (θ) exactly once as θ runs from α to β , then the length of the curve is L =

y

r

1 u

2

0

x

∫α

( ddrθ )

r2 +

2

d θ.

(3)

Find the length of the cardioid r = 1 − cos θ.

EXAMPLE 4

r = 1 − cos u P(r, u)

β

Solution We sketch the cardioid to determine the limits of integration (Figure 10.38). The point P ( r, θ ) traces the curve once, counterclockwise as θ runs from 0 to 2π, so these are the values we take for α and β. With

dr = sin θ, dθ

r = 1 − cos θ, we have FIGURE 10.38

Calculating the length of a cardioid (Example 4).

r2 +

( ddrθ )

2

= (1 − cos θ ) 2 + ( sin θ ) 2 = 1 − 2 cos θ +  cos 2 θ + sin 2 θ = 2 − 2 cos θ 1

and β

L=

∫α

=

∫0

=

∫0

=

∫0

2π

2π

2π

r2 +

( ddrθ )

4 sin 2 2 sin

2

θ dθ 2

dθ =

2π

∫0

2 − 2 cos θ d θ

1 − cos θ = 2 sin 2 ( θ 2 )

θ dθ 2

θ 2 sin d θ 2

sin ( θ 2 ) ≥ 0 for 0 ≤ θ ≤ 2π

θ 2π = ⎡⎢ −4 cos ⎤⎥ = 4 + 4 = 8. ⎣ 2⎦0

654

Chapter 10 Parametric Equations and Polar Coordinates

EXERCISES

10.5 15. Inside the circle r = −2 cos θ and outside the circle r = 1

Finding Polar Areas

Find the areas of the regions in Exercises 1–8.

16. Inside the circle r = 6 and above the line r = 3 csc θ

1. Bounded by the spiral r = θ for 0 ≤ θ ≤ π

17. Inside the circle r = 4 cos θ and to the right of the vertical line r = sec θ

y

18. Inside the circle r = 4 sin θ and below the horizontal line r = 3 csc θ

p p a2 , 2b

r=u

(p, p)

19. a. Find the area of the shaded region in the accompanying figure. y

x

0

r = tan u –p < u < p 2 2 (1, p4)

2. Bounded by the circle r = 2 sin θ for π 4 ≤ θ ≤ π 2

r = (" 22) csc u

y −1

p a2 , b 2 r = 2 sin u

0

x

1

b. It looks as if the graph of r = tan θ,   −π 2 < θ < π 2, could be asymptotic to the lines x = 1 and x = −1. Is it? Give reasons for your answer.

u= p 4

20. The area of the region that lies inside the cardioid curve r = cos θ + 1 and outside the circle r = cos θ is not

x

0

1 2π ⎡ ( cos θ + 1) 2 − cos 2 θ ⎤⎦ d θ = π. 2 ∫0 ⎣

3. Inside the oval limaçon r = 4 + 2 cos θ

Why not? What is the area? Give reasons for your answers.

4. Inside the cardioid r = a (1 + cos θ ), a > 0 5. Inside one leaf of the four-leaved rose r = cos 2θ

Finding Lengths of Polar Curves

6. Inside one leaf of the three-leaved rose r = cos 3θ

Find the lengths of the curves in Exercises 21–28. 21. The spiral r = θ 2 , 0 ≤ θ ≤

y

22. The spiral r =

eθ

5

2, 0 ≤ θ ≤ π

23. The cardioid r = 1 + cos θ 24. The curve r = a sin 2 ( θ 2 ), 0 ≤ θ ≤ π , a > 0

r = cos 3u 1

25. The parabolic segment r = 6 (1 + cos θ ), 0 ≤ θ ≤ π 2

x

26. The parabolic segment r = 2 (1 − cos θ ), π 2 ≤ θ ≤ π 27. The curve r = cos 3 ( θ 3 ), 0 ≤ θ ≤ π 4 28. The curve r =

7. Inside one loop of the lemniscate r 2 = 4 sin 2θ 8. Inside the six-leaved rose r 2 = 2 sin 3θ

x = f (θ) cos θ, y = f (θ)sin θ

Find the areas of the regions in Exercises 9–18. 9. Shared by the circles r = 2 cos θ and r = 2 sin θ

(Equations 2 in the text) transform

10. Shared by the circles r = 1 and r = 2 sin θ

L =

11. Shared by the circle r = 2 and the cardioid r = 2(1 − cos θ ) 12. Shared by the cardioids r = 2(1 + cos θ ) and r = 2(1 − cos θ ) 13. Inside the lemniscate r 2 = 6 cos 2θ and outside the circle r =

1 + sin 2θ , 0 ≤ θ ≤ π 2

29. The length of the curve r = f (θ ),   α ≤ θ ≤ β Assuming that the necessary derivatives are continuous, show how the substitutions

3

14. Inside the circle r = 3a cos θ and outside the cardioid r = a (1 + cos θ ),   a > 0

( dx dθ )

β

∫α

2

+

( ddyθ )

2

dθ

into L =

β

∫α

r2 +

( ddrθ )

2

d θ.

10.6  Conic Sections

Use this formula to find the average value of r with respect to R over the following curves ( a > 0 ).

30. Circumferences of circles As usual, when faced with a new formula, it is a good idea to try it on familiar objects to be sure it gives results consistent with past experience. Use the length formula in Equation (3) to calculate the circumferences of the following circles ( a > 0 ). a. r  a

b. r  a cos R

c. r  a sin R

Theory and Examples

31. Average value If f is continuous, the average value of the polar coordinate r over the curve r = f (θ),   α ≤ θ ≤ β , with respect to R is given by the formula rav =

655

a. The cardioid r = a (1 − cos R ) b. The circle r  a c. The circle r = a cos θ, −π 2 ≤ θ ≤ π 2 32. r  f (R )  vs.  r  2 f (R ) Can anything be said about the relative lengths of the curves r = f (θ),   α ≤ θ ≤ β , and r = 2 f (θ),   α ≤ θ ≤ β ? Give reasons for your answer.

β 1 f (θ) d θ. ∫ β−α α

10.6 Conic Sections In this section we define and review parabolas, ellipses, and hyperbolas geometrically and derive their standard Cartesian equations. These curves are called conic sections or conics because they are formed by cutting a double cone with a plane (Figure 10.39). This

Circle: plane perpendicular to cone axis

Ellipse: plane oblique to cone axis

Parabola: plane parallel to side of cone

Hyperbola: plane parallel to cone axis

(a)

HISTORICAL BIOGRAPHY Gregory St. Vincent

(1584–1667) Born in Belgium, St. Vincent studied mathematics at Douai. He made important contributions to the development of calculus, and his books were read by the next generation of mathematicians as they connected ideas and refined the concepts of calculus. To know more, visit the companion Website.

Point: plane through cone vertex only

Single line: plane tangent to cone

Pair of intersecting lines

(b)

FIGURE 10.39

The standard conic sections (a) are the curves in which a plane cuts a double cone. Hyperbolas come in two parts, called branches. The point and lines obtained by passing the plane through the cone’s vertex (b) are degenerate conic sections.

656

Chapter 10 Parametric Equations and Polar Coordinates

geometric method was the only way that conic sections could be described by Greek mathematicians, since they did not have our tools of Cartesian or polar coordinates. In the next section we express the conics in polar coordinates.

Parabolas DEFINITIONS A set that consists of all the points in a plane equidistant from a given fixed point and a given fixed line in the plane is a parabola. The fixed point is the focus of the parabola. The fixed line is the directrix. y

x 2 = 4py Focus p

The vertex lies halfway between p directrix and focus. Directrix: y = −p

FIGURE 10.40

F(0, p) P(x, y)

Q(x, −p)

x

If the focus F lies on the directrix L, the parabola is the line through F perpendicular to L. We consider this to be a degenerate case and assume henceforth that F does not lie on L. A parabola has its simplest equation when its focus and directrix straddle one of the coordinate axes. For example, suppose that the focus lies at the point F ( 0,  p ) on the positive y-axis and that the directrix is the line y = − p (Figure 10.40). In the notation of the figure, a point P ( x ,  y ) lies on the parabola if and only if PF = PQ. From the distance formula,

L

The standard form of the parabola x 2 = 4 py,   p > 0.

PF =

(x

− 0 )2 + ( y − p )2 =

x 2 + ( y − p )2

PQ =

(x

− x ) 2 + ( y − ( − p )) 2 =

(y

+ p )2 .

When we equate these expressions, square, and simplify, we get y =

x2 4p

x 2 = 4 py.

or

(1)

Standard form

These equations reveal the parabola’s symmetry about the y-axis. We call the y-axis the axis of the parabola (short for “axis of symmetry”). The point where a parabola crosses its axis is the vertex. The vertex of the parabola x 2 = 4 py lies at the origin (Figure 10.40). The positive number p is the parabola’s focal length. If the parabola opens downward, with its focus at ( 0, − p ) and its directrix the line y = p, then Equations (1) become y = −

x2 4p

and

x 2 = −4 py.

By interchanging the variables x and y, we obtain similar equations for parabolas opening to the right or to the left (Figure 10.41). y

y

Directrix x = −p

y2 = −4px

y2 = 4px

Vertex

Vertex Focus 0 F( p, 0)

(a)

FIGURE 10.41

y2

Directrix x=p

= −4 px.

x

Focus F(−p, 0) 0

(b)

(a) The parabola y 2 = 4 px. (b) The parabola

x

10.6  Conic Sections

EXAMPLE 1

657

Find the focus and directrix of the parabola y 2 = 10 x.

Solution We find the value of p in the standard equation y 2 = 4 px :

4 p = 10,

p =

so

10 5 = . 4 2

Then we find the focus and directrix for this value of p: ( p, 0 )

Focus: Directrix:

=

( 52 , 0 )

x = −p

or

5 x = − . 2

Ellipses Vertex

Focus Center

Focus

Vertex

Focal axis

FIGURE 10.42

Points on the focal axis

DEFINITIONS An ellipse is the set of points in a plane whose distances from two fixed points in the plane have a constant sum. The two fixed points are the foci of the ellipse. The line through the foci of an ellipse is the ellipse’s focal axis. The point on the axis halfway between the foci is the center. The points where the focal axis and ellipse cross are the ellipse’s vertices (Figure 10.42).

of an ellipse. y

If the foci are F1 (−c, 0 ) and F2 ( c, 0 ) (Figure 10.43), and PF1 + PF2 is denoted by 2a, then the coordinates of a point P on the ellipse satisfy the equation

b P(x, y) Focus F1(−c, 0)

Focus 0 Center F2(c, 0)

a

(x

x

+ c )2 + y 2 +

(x

− c ) 2 + y 2 = 2a.

To simplify this equation, we move the second radical to the right-hand side, square, isolate the remaining radical, and square again, obtaining y2 x2 + 2 = 1. 2 a a − c2

FIGURE 10.43

The ellipse defined by the equation PF1 + PF2 = 2a is the graph of the equation ( x 2 a 2 ) + ( y 2 b 2 ) = 1, where b 2 = a 2 − c 2 .

(2)

Since PF1 + PF2 is greater than the length F1F2 (by the triangle inequality for triangle PF1F2), the number 2a is greater than 2c. Accordingly, a > c and the number a 2 − c 2 in Equation (2) is positive. The algebraic steps leading to Equation (2) can be reversed to show that every point P whose coordinates satisfy an equation of this form with 0 < c < a also satisfies the equation PF1 + PF2 = 2a. A point therefore lies on the ellipse if and only if its coordinates satisfy Equation (2). If we let b denote the positive square root of a 2 − c 2 , b =

a2 − c2 ,

(3)

then a 2 − c 2 = b 2 and Equation (2) takes the form y2 x2 + = 1. a2 b2

(4)

Equation (4) reveals that this ellipse is symmetric with respect to the origin and both coordinate axes. It lies inside the rectangle bounded by the lines x = ±a and y = ±b. It crosses the axes at the points ( ±a, 0 ) and ( 0, ±b ). The tangents at these points are perpendicular to the axes because dy b 2x = − 2 , dx a y

Obtained from Eq. (4) by implicit differentiation

which is zero if x = 0 and infinite if y = 0.

658

Chapter 10 Parametric Equations and Polar Coordinates

The major axis of the ellipse in Equation (4) is the line segment of length 2a joining the points ( ±a, 0 ). The minor axis is the line segment of length 2b joining the points ( 0, ±b ). The number a itself is the semimajor axis, the number b the semiminor axis. The number c, found from Equation (3) as c =

a2 − b2,

is the center-to-focus distance of the ellipse. If a = b then the ellipse is a circle. EXAMPLE 2

The ellipse y2 x2 + =1 16 9

shown in Figure 10.44 has

y

2 x2 + y = 1 16 9 Vertex (−4, 0)

(0, 3) Vertex (4, 0)

Focus

Focus

(−" 7, 0) 0

(5)

(" 7, 0) Center

x

Semimajor axis: a =

16 = 4,

Center-to-focus distance: c = Foci:

( ±c, 0 )

Vertices: Center:

Semiminor axis: b = 16 − 9 =

9 = 3,

7,

= ( ± 7, 0 ),

( ±a, 0 )

= ( ±4, 0 ),

( 0, 0 ).

If we interchange x and y in Equation (5), we have the equation

(0, −3)

y2 x2 + = 1. 9 16

FIGURE 10.44

An ellipse with its major axis horizontal (Example 2).

(6)

The major axis of this ellipse is now vertical instead of horizontal, with the foci and vertices on the y-axis. We can determine which way the major axis runs simply by finding the intercepts of the ellipse with the coordinate axes. The longer of the two axes of the ellipse is the major axis. Standard-Form Equations for Ellipses Centered at the Origin

Foci on the x-axis:

y2 x2 + 2 = 1 (a > b) 2 a b Center-to-focus distance: c = Foci: ( ±c, 0 ) Vertices: x2

a2 − b2

( ±a, 0 )

y2

+ 2 = 1 (a > b) b2 a Center-to-focus distance: c = a 2 − b 2 Foci: ( 0, ±c ) Vertices: ( 0, ±a ) In each case, a is the semimajor axis and b is the semiminor axis. Foci on the y-axis:

Hyperbolas Vertices Focus

Focus Center Focal axis

FIGURE 10.45

of a hyperbola.

Points on the focal axis

DEFINITIONS A hyperbola is the set of points in a plane whose distances from

two fixed points in the plane have a constant difference. The two fixed points are the foci of the hyperbola. The line through the foci of a hyperbola is the focal axis. The point on the axis halfway between the foci is the hyperbola’s center. The points where the focal axis and hyperbola cross are the vertices (Figure 10.45).

10.6  Conic Sections

y x = −a

If the foci are F1 (−c, 0 ) and F2 ( c, 0 ) (Figure 10.46) and the constant difference is 2a, then a point ( x ,  y ) lies on the hyperbola if and only if

x=a

(x

P(x, y)

F1(−c, 0)

0

659

F2(c, 0)

+ c )2 + y 2 −

(x

− c ) 2 + y 2 = ±2a.

(7)

To simplify this equation, we move the second radical to the right-hand side, square, isolate the remaining radical, and square again, obtaining

x

y2 x2 + = 1. a2 a2 − c2

FIGURE 10.46

Hyperbolas have two branches. For points on the right-hand branch of the hyperbola shown here, PF1 − PF2 = 2a. For points on the lefthand branch, PF2 − PF1 = 2a. We then let b = c 2 − a 2 .

(8)

So far, this looks just like the equation for an ellipse. But now a 2 − c 2 is negative because 2a, being the difference of two sides of triangle PF1F2 , is less than 2c, the third side. The algebraic steps leading to Equation (8) can be reversed to show that every point P whose coordinates satisfy an equation of this form with 0 < a < c also satisfies Equation (7). A point therefore lies on the hyperbola if and only if its coordinates satisfy Equation (8). If we let b denote the positive square root of c 2 − a 2 , b =

c2 − a2,

(9)

then a 2 − c 2 = −b 2 and Equation (8) takes the compact form y2 x2 − = 1. a2 b2

(10)

The differences between Equation (10) and the equation for an ellipse (Equation (4)) are the minus sign and the new relation c 2 = a 2 + b 2.

From Eq. (9)

Like the ellipse, the hyperbola is symmetric with respect to the origin and coordinate axes. It crosses the x-axis at the points ( ±a, 0 ). The tangents at these points are vertical because dy b 2x = 2 dx a y

Obtained from Eq. (10) by implicit differentiation

and this is infinite when y = 0. The hyperbola has no y-intercepts; in fact, no part of the curve lies between the lines x = −a and x = a. The lines b y = ± x a are the two asymptotes of the hyperbola defined by Equation (10). The fastest way to find the equations of the asymptotes is to replace the 1 in Equation (10) by 0 and solve the new equation for y: y2 y2 x2 x2 b − = 1 → − = 0 → y = ± x. 2 2 2 2 a b a b a   

y y = − "5 x 2

hyperbola

y = "5 x 2 2 x2 − y = 1 4 5

F(−3, 0) −2

F(3, 0) 2

x

EXAMPLE 3

The equation y2 x2 − =1 4 5

Foci:

( ±c, 0 )

Center: The hyperbola and its asymptotes in Example 3.

(11)

is Equation (10) with a 2 = 4 and b 2 = 5 (Figure 10.47). We have Center-to-focus distance: c =

FIGURE 10.47

asymptotes

0 for 1

a2 + b2 =

= ( ±3, 0 ), Vertices:

( ±a, 0 )

( 0, 0 ),

Asymptotes:

y2 x2 − = 0 or 4 5

y = ±

5 x. 2

4 + 5 = 3, = ( ±2, 0 ),

660

Chapter 10 Parametric Equations and Polar Coordinates

If we interchange x and y in Equation (11), the foci and vertices of the resulting hyperbola will lie along the y-axis. We still find the asymptotes in the same way as before, but now their equations will be y = ±2 x 5.

Standard-Form Equations for Hyperbolas Centered at the Origin

Foci   on   the   x - axis: 

y2 x2 − 2 =1 2 a b

Center-to-focus distance: c = Foci:

a2 + b2

( ±c, 0 )

Vertices:

Center-to-focus distance: c =

Vertices:

y2 x2 − 2 = 0 or 2 a b

b y = ± x a

a2 + b2

( 0, ±c )

Foci:

( ±a, 0 )

Asymptotes:

y2 x2 − 2 =1 2 a b

Foci on the y-axis: 

( 0, ±a )

y2 x2 − 2 = 0 or 2 a b

Asymptotes:

a y = ± x b

Notice the difference in the asymptote equations (b a in the first, a b in the second).

We shift conics using the principles reviewed in Section 1.2, replacing x by x + h and y by y + k. EXAMPLE 4 Show that the equation x 2 − 4 y 2 + 2 x + 8 y − 7 = 0 represents a hyperbola. Find its center, asymptotes, and foci. Solution We reduce the equation to standard form by completing the square in x and y as follows: ( x 2 + 2 x ) − 4( y 2 − 2 y ) = 7 ( x 2 + 2 x + 1) − 4 ( y 2 − 2 y + 1) = 7 + 1 − 4 (x

+ 1) 2 ( − y − 1) 2 = 1. 4

This is the standard form Equation (10) of a hyperbola with x replaced by x + 1 and y replaced by y − 1. The hyperbola is shifted one unit to the left and one unit upward, and it has center x + 1 = 0 and y − 1 = 0, or x = −1 and y = 1. Moreover, a 2 = 4,

b 2 = 1,

c 2 = a 2 + b 2 = 5,

so the asymptotes are the two lines x +1 ( − y − 1) = 0 2

and

x +1 ( + y − 1) = 0, 2

or y −1 = ± The shifted foci have coordinates (−1 ±

1( x + 1). 2

5,  1).

661

10.6  Conic Sections

10.6

EXERCISES Identifying Graphs

Match the parabolas in Exercises 1–4 with the following equations: x2

= 2 y,

x2

= − 6 y,

y2

= 8x,

y2

= − 4 x.

Then find each parabola’s focus and directrix. 1.

y

13. y = 4 x 2

15. x = −3 y 2

16. x = 2 y 2

Ellipses

17. 16 x 2 + 25 y 2 = 400 19. x

x

y

3.

+

y2

= 2

20. 2 x 2 + y 2 = 4

21. 3 x 2 + 2 y 2 = 6

22. 9 x 2 + 10 y 2 = 90

23. 6 x 2 + 9 y 2 = 54

24. 169 x 2 + 25 y 2 = 4225

26. Foci: ( 0, ±4 ) Vertices: ( 0, ±5 ) x

Match each conic section in Exercises 5–8 with one of these equations: y2 x2 + = 1, 4 9 y2 − x 2 = 1, 4

x2 + y 2 = 1,   2 y2 x2 − = 1. 4 9

y

6.

y

x

x

Hyperbolas

Exercises 27–34 give equations for hyperbolas. Put each equation in standard form and find the hyperbola’s asymptotes. Then sketch the hyperbola. Include the asymptotes and foci in your sketch. 27. x 2 − y 2 = 1

28. 9 x 2 − 16 y 2 = 144

29. y 2 − x 2 = 8

30. y 2 − x 2 = 4

−

32. y 2 − 3 x 2 = 3

31.

Then find the conic section’s foci and vertices. If the conic section is a hyperbola, find its asymptotes as well.

8x 2

2y 2

= 16

33. 8 y 2 − 2 x 2 = 16

y

8.

34. 64 x 2 − 36 y 2 = 2304

Exercises 35–38 give information about the foci, vertices, and asymptotes of hyperbolas centered at the origin of the xy-plane. In each case, find the hyperbola’s standard-form equation from the information given. 35. Foci: ( 0, ± 2 ) Asymptotes: y = ± x 37. Vertices: ( ±3,  0 )

7.

18. 7 x 2 + 16 y 2 = 112

25. Foci: ( ± 2, 0 ) Vertices: ( ±2, 0 )

x

5.

2x 2

Exercises 25 and 26 give information about the foci and vertices of ellipses centered at the origin of the xy-plane. In each case, find the ellipse’s standard-form equation from the given information.

y

4.

14. y = −8 x 2

Exercises 17–24 give equations for ellipses. Put each equation in standard form. Then sketch the ellipse. Include the foci in your sketch.

y

2.

12. y 2 = −2 x

4 Asymptotes: y = ± x 3

36. Foci: ( ±2, 0 ) Asymptotes: y = ±

1 x 3

38. Vertices: ( 0, ±2 ) 1 Asymptotes: y = ± x 2

y

Shifting Conic Sections

You may wish to review Section 1.2 before solving Exercises 39–56. x

x

39. The parabola y 2 = 8 x is shifted down 2 units and right 1 unit to generate the parabola ( y + 2 ) 2 = 8( x − 1). a. Find the new parabola’s vertex, focus, and directrix. b. Plot the new vertex, focus, and directrix, and sketch in the parabola.

Parabolas

Exercises 9–16 give equations of parabolas. Find each parabola’s focus and directrix. Then sketch the parabola. Include the focus and directrix in your sketch. 9. y 2 = 12 x

10. x 2 = 6 y

11. x 2 = −8 y

40. The parabola x 2 = −4 y is shifted left 1 unit and up 3 units to generate the parabola ( x + 1) 2 = −4 ( y − 3 ). a. Find the new parabola’s vertex, focus, and directrix. b. Plot the new vertex, focus, and directrix, and sketch in the parabola.

662

Chapter 10 Parametric Equations and Polar Coordinates

41. The ellipse ( x 2 16 ) + ( y 2 9 ) = 1 is shifted 4 units to the right and 3 units up to generate the ellipse (x

( y − 3)2 − 4 )2 + = 1. 16 9

a. Find the foci, vertices, and center of the new ellipse. b. Plot the new foci, vertices, and center, and sketch in the new ellipse. 42. The ellipse ( x 2 9 ) + ( y 2 25 ) = 1 is shifted 3 units to the left and 2 units down to generate the ellipse (x

( y + 2 )2 + 3)2 + = 1. 9 25

a. Find the foci, vertices, and center of the new ellipse. b. Plot the new foci, vertices, and center, and sketch in the new ellipse.

(x2

( y2

16 ) − 9 ) = 1 is shifted 2 units to the 43. The hyperbola right to generate the hyperbola − 2) − = 1. 16 9 2

(x

y2

a. Find the center, foci, vertices, and asymptotes of the new hyperbola. b. Plot the new center, foci, vertices, and asymptotes, and sketch in the hyperbola.

( y2

4) − 44. The hyperbola generate the hyperbola (y

(x2

5 ) = 1 is shifted 2 units down to

+ 2 )2 x2 − = 1. 4 5

a. Find the center, foci, vertices, and asymptotes of the new hyperbola. b. Plot the new center, foci, vertices, and asymptotes, and sketch in the hyperbola.

Exercises 53–56 give equations for hyperbolas and tell how many units up or down and to the right or left each hyperbola is to be shifted. Find an equation for the new hyperbola, and find the new center, foci, vertices, and asymptotes. 53.

y2 x2 − = 1,  right 2,  up 2 4 5

54.

y2 x2 − = 1,  left 2,  down 1 16 9

55. y 2 − x 2 = 1,  left 1,  down 1 56.

y2 − x 2 = 1,  right 1,  up 3 3

Find the center, foci, vertices, asymptotes, and radius, as appropriate, of the conic sections in Exercises 57–68. 57. x 2 + 4 x + y 2 = 12 58. 2 x 2 + 2 y 2 − 28 x + 12 y + 114 = 0 59. x 2 + 2 x + 4 y − 3 = 0 61.

x2

+

5y 2

+ 4x = 1

45.

= 4 x , left 2,  down 3

46. y 2 = −12 x , right 4,  up 3 47. x 2 = 8 y, right 1,  down 7 48. x 2 = 6 y, left 3,  down 2 Exercises 49–52 give equations for ellipses and tell how many units up or down and to the right or left each ellipse is to be shifted. Find an equation for the new ellipse, and find the new foci, vertices, and center.

64. 4 x 2 + y 2 + 8 x − 2 y = −1 65. x 2 − y 2 − 2 x + 4 y = 4 67.

2x 2

−

y2

66. x 2 − y 2 + 4 x − 6 y = 6

+ 6y = 3

68. y 2 − 4 x 2 + 16 x = 24

Theory and Examples

69. If lines are drawn parallel to the coordinate axes through a point P on the parabola y 2 = kx ,   k > 0, the parabola partitions the rectangular region bounded by these lines and the coordinate axes into two smaller regions, A and B. a. If the two smaller regions are revolved about the y-axis, show that they generate solids whose volumes have the ratio 4:1. b. What is the ratio of the volumes generated by revolving the regions about the x-axis? y y 2 = kx P

A B

x

0

70. Suspension bridge cables hang in parabolas The suspension bridge cable shown in the accompanying figure supports a uniform load of w newtons per horizontal meter. It can be shown that if H is the horizontal tension of the cable at the origin, then the curve of the cable satisfies the equation

x2

dy w = x. dx H

x2

Show that the cable hangs in a parabola by solving this differential equation subject to the initial condition that y = 0 when x = 0.

y2 49. + = 1, left 2,  down 1 6 9 50.

62. 9 x 2 + 6 y 2 + 36 y = 0

63. x 2 + 2 y 2 − 2 x − 4 y = −1

Exercises 45–48 give equations for parabolas and tell how many units up or down and to the right or left each parabola is to be shifted. Find an equation for the new parabola, and find the new vertex, focus, and directrix. y2

60. y 2 − 4 y − 8 x − 12 = 0

2

+ y 2 = 1, right 3,  up 4

y

y2 + = 1, right 2,  up 3 51. 3 2 x2

52.

y2 x2 + = 1, left 4,  down 5 16 25

Bridge cable 0

x

10.7  Conics in Polar Coordinates

663

71. The width of a parabola at the focus Show that the number 4p is the width of the parabola x 2 = 4 py ( p > 0 ) at the focus by showing that the line y = p cuts the parabola at points that are 4p units apart.

80. Surface area The curve y = x 2 + 1,  0 ≤ x ≤ 2, which is part of the upper branch of the hyperbola y 2 − x 2 = 1, is revolved about the x-axis to generate a surface. Find the area of the surface.

72. The asymptotes of ( x 2 a 2 ) − ( y 2 b 2 ) = 1 Show that the vertical distance between the line y = ( b a ) x and the upper half of the right-hand branch y = ( b a ) x 2 − a 2 of the hyperbola ( x 2 a 2 ) − ( y 2 b 2 ) = 1 approaches 0 by showing that

81. The reflective property of parabolas The accompanying figure shows a typical point P ( x 0 ,  y 0 ) on the parabola y 2 = 4 px. The line L is tangent to the parabola at P. The parabola’s focus lies at F ( p, 0 ). The ray L′ extending from P to the right is parallel to the x-axis. We show that light from F to P will be reflected out along L′ by showing that β equals α. Establish this equality by taking the following steps.

lim

x →∞

( ba x − ba

)

x2 − a2 =

b lim ( x − a x →∞

x 2 − a 2 ) = 0.

Similar results hold for the remaining portions of the hyperbola and the lines y = ±( b a ) x. 73. Area Find the dimensions of the rectangle of largest area that can be inscribed in the ellipse x 2 + 4 y 2 = 4 with its sides parallel to the coordinate axes. What is the area of the rectangle? 74. Volume Find the volume of the solid generated by revolving the region enclosed by the ellipse 9 x 2 + 4 y 2 = 36 about the (a) x-axis, (b) y-axis. 75. Volume The “triangular” region in the first quadrant bounded by the x-axis, the line x = 4, and the hyperbola 9 x 2 − 4 y 2 = 36 is revolved about the x-axis to generate a solid. Find the volume of the solid. 76. Tangents Show that the tangents to the curve any point on the line x = − p are perpendicular.

y2

a. Show that tan β = 2 p y 0 . b. Show that tan φ = y 0 ( x 0 − p ). c. Use the identity tan α =

tan φ − tan β 1 + tan φ tan β

to show that tan α = 2 p y 0 . Since α and β are both acute, tan β = tan α implies β = α. This reflective property of parabolas is used in applications like car headlights, radio telescopes, and satellite TV dishes. y L

= 4 px from b

P(x 0 , y0) b

77. Tangents Find equations for the tangents to the circle ( x − 2 ) 2 + ( y − 1) 2 = 5 at the points where the circle crosses the coordinate axes.

y0

a f

78. Volume The region bounded on the left by the y-axis, on the right by the hyperbola x 2 − y 2 = 1, and above and below by the lines y = ±3 is revolved about the y-axis to generate a solid. Find the volume of the solid.

0

L′

x

F( p, 0)

y 2 = 4 px

79. Centroid Find the centroid of the region that is bounded below by the x-axis and above by the ellipse ( x 2 9 ) + ( y 2 16 ) = 1.

10.7 Conics in Polar Coordinates Polar coordinates are especially important in astronomy and astronautical engineering because satellites, moons, planets, and comets all move approximately along ellipses, parabolas, and hyperbolas that can be described with a single relatively simple polar coordinate equation. We develop that equation here after first introducing the idea of a conic section’s eccentricity. The eccentricity reveals the conic section’s type (circle, ellipse, parabola, or hyperbola) and the degree to which it is “squashed” or flattened.

Eccentricity Although the center-to-focus distance c does not appear in the standard Cartesian equation y2 x2 + = 1, a2 b2

(a

> b)

for an ellipse, we can still determine c from the equation c = a 2 − b 2 . If we fix a and vary c over the interval 0 ≤ c ≤ a, the resulting ellipses will vary in shape. They are circles if c = 0 (so that a = b) and flatten, becoming more oblong, as c increases. If c = a, the foci and vertices overlap and the ellipse degenerates into a line segment. Thus we are led to consider the ratio e = c a. We use this ratio for hyperbolas as well, except in this

664

Chapter 10 Parametric Equations and Polar Coordinates

case c equals a 2 + b 2 instead of a 2 − b 2 . We refer to this ratio as the eccentricity of the ellipse or hyperbola.

DEFINITION

The eccentricity of the ellipse ( x 2 a 2 ) + ( y 2 b 2 ) = 1  ( a > b ) is

y Directrix x = −c

e =

D

P(x, y)

e =

PF = 1 ⋅ PD

Directrix 2 x = ae

b

F2(c, 0)

F1(−c, 0)

PF1 = e ⋅ PD1 ,

x

0

D2

P(x, y) −b c = ae a

a e

FIGURE 10.49

The foci and directrices of the ellipse ( x 2 a 2 ) + ( y 2 b 2 ) = 1. Directrix 1 corresponds to focus F1 and directrix 2 to focus F2 . y

D1

F1(−c, 0)

Directrix 2 x = ae D2

0 a e

x

a c = ae

PF2 = e ⋅ PD 2 .

PF1 = e ⋅ PD1

and

PF2 = e ⋅ PD 2 .

The foci and directrices of the hyperbola ( x 2 a 2 ) − ( y 2 b 2 ) = 1. No matter where P lies on the hyperbola, PF1 = e ⋅ PD1 and PF2 = e ⋅ PD 2 .

(3)

Here P is any point on the hyperbola, F1 and F2 are the foci, and D1 and D 2 are the points nearest P on the directrices (Figure 10.50). In both the ellipse and the hyperbola, the eccentricity is the ratio of the distance between the foci to the distance between the vertices (because c a = 2c 2a). distance between foci distance between vertices

In an ellipse, the foci are closer together than the vertices and the ratio is less than 1. In a hyperbola, the foci are farther apart than the vertices and the ratio is greater than 1. The “focus–directrix” equation PF = e ⋅ PD unites the parabola, ellipse, and hyperbola in the following way. Suppose that the distance PF of a point P from a fixed point F (the focus) is a constant multiple of its distance from a fixed line (the directrix). That is, suppose PF = e ⋅ PD, where e is the constant of proportionality. Then the path traced by P is

FIGURE 10.50

(2)

Here, e is the eccentricity, P is any point on the ellipse, F1 and F2 are the foci, and D1 and D 2 are the points on the directrices nearest P (Figure 10.49). In both Equations (2) the directrix and focus must correspond; that is, if we use the distance from P to F1 , we must also use the distance from P to the directrix at the same end of the ellipse. The directrix x = −a e corresponds to F1 (−c, 0 ), and the directrix x = a e corresponds to F2 ( c, 0 ). As with the ellipse, it can be shown that the lines x = ±a e act as directrices for the hyperbola and that

Eccentricity =

P(x, y)

F2(c, 0)

(1)

for any point P on it, where F is the focus and D is the point nearest P on the directrix. For an ellipse, it can be shown that the equations that replace Equation (1) are

y

Directrix 1 x = − ae

a2 + b2 . a

Whereas a parabola has one focus and one directrix, each ellipse has two foci and two directrices. These are the lines perpendicular to the major axis at distances ±a e from the center. From Figure 10.48 we see that a parabola has the property

The distance from the focus F to any point P on a parabola equals the distance from P to the nearest point D on the directrix, so PF = PD.

D1

c = a

The eccentricity of a parabola is e = 1.

FIGURE 10.48

Directrix 1 a x = −e

a2 − b2 . a

The eccentricity of the hyperbola ( x 2 a 2 ) − ( y 2 b 2 ) = 1 is x

0 F(c, 0)

c = a

(a) a parabola if e = 1, (b) an ellipse of eccentricity e if e < 1, and (c) a hyperbola of eccentricity e if e > 1.

(4)

665

10.7  Conics in Polar Coordinates

As e increases ( e → 1− ), ellipses become more oblong, and ( e → ∞ ) hyperbolas flatten toward two lines parallel to the directrix. There are no coordinates in Equation (4), and when we try to translate it into Cartesian coordinate form, it translates in different ways depending on the size of e. However, as we are about to see, in polar coordinates the equation PF = e ⋅ PD translates into a single equation regardless of the value of e. Given the focus and corresponding directrix of a hyperbola centered at the origin and with foci on the x-axis, we can use the dimensions shown in Figure 10.50 to find e. Knowing e, we can derive a Cartesian equation for the hyperbola from the equation PF = e ⋅ PD, as in the next example. We can find equations for ellipses centered at the origin and with foci on the x-axis in a similar way, using the dimensions shown in Figure 10.49. y x=1 D(1, y)

2 x2 − y = 1 3 6

EXAMPLE 1 Find a Cartesian equation for the hyperbola centered at the origin that has a focus at ( 3, 0 ) and the line x = 1 as the corresponding directrix. Solution We first use the dimensions shown in Figure 10.50 to find the hyperbola’s eccentricity. The focus is (see Figure 10.51)

P(x, y)

( c, 0 )

0 1

F(3, 0)

x

= ( 3, 0 ),

c = 3.

so

Again from Figure 10.50, the directrix is the line x =

a = 1, e

so

a = e.

When combined with the equation e = c a that defines eccentricity, these results give FIGURE 10.51

The hyperbola and directrix in Example 1.

c 3 so e 2 = 3 and e = 3. = , a e Knowing e, we can now derive the equation we want from the equation PF = e ⋅ PD. In the coordinates of Figure 10.51, we have e =

PF = e ⋅ PD (x

− 3) + ( y − x 2 − 6x + 9 + y 2 2x 2 − y 2 y2 x2 − 3 6 2

0) 2

= 3 x −1 = 3( x 2 − 2 x + 1) =6

Eq. (4) e =

3

Square both sides. Simplify.

= 1.

Polar Equations To find a polar equation for an ellipse, parabola, or hyperbola, we place one focus at the origin and the corresponding directrix to the right of the origin along the vertical line x = k (Figure 10.52). In polar coordinates, this makes

Directrix P Focus at origin

PF = r

D

r

and

F B r cos u

n

ctio

c se

ni Co

FIGURE 10.52

k

x

PD = k − FB = k − r cos θ. The conic’s focus–directrix equation PF = e ⋅ PD then becomes

x=k

If a conic section is put in the position with its focus placed at the origin and a directrix perpendicular to the initial ray and right of the origin, we can find its polar equation from the conic’s focus–directrix equation.

r = e ( k − r cos θ ), which can be solved for r to obtain the following expression. Polar Equation for a Conic with Eccentricity e

r =

ke , 1 + e cos θ

where x = k > 0 is the vertical directrix.

(5)

666

Chapter 10 Parametric Equations and Polar Coordinates

r=

ke 1 + e cos u

EXAMPLE 2 Here are polar equations for three conics. The eccentricity values identifying the conic are the same for both polar and Cartesian coordinates.

Focus at origin

e =

x

ellipse

r =

k 2 + cos θ

Directrix x = k (a)

e = 1:

parabola

r =

k 1 + cos θ

ke 1 − e cos u

e = 2:

hyperbola

r =

2k 1 + 2 cos θ

r=

You may see variations of Equation (5), depending on the location of the directrix. If the directrix is the line x = −k to the left of the origin (the origin is still a focus), we replace Equation (5) with

Focus at origin x Directrix x = −k (b) r=

1 : 2

r =

ke 1 + e sin u y

ke . 1 − e cos θ

Directrix y = k

The denominator now has a (−) instead of a (+). If the directrix is either of the lines y = k or y = −k, the equations have sines in them instead of cosines, as shown in Figure 10.53.

(c)

EXAMPLE 3 x = 2.

Focus at origin

r=

ke 1 − e sin u y Focus at origin

Find an equation for the hyperbola with eccentricity 3 2 and directrix

Solution We use Equation (5) with k = 2 and e = 3 2:

r = Directrix y = −k (d)

EXAMPLE 4

2( 3 2 ) 1 + ( 3 2 ) cos θ

or

r =

Find the directrix of the parabola r =

FIGURE 10.53

Equations for conic sections with eccentricity e > 0 but different locations of the directrix. The graphs here show a parabola, so e = 1.

6 . 2 + 3 cos θ

25 . 10 + 10 cos θ

Solution We divide the numerator and denominator by 10 to put the equation in standard polar form:

r =

52 . 1 + cos θ

r =

ke 1 + e cos θ

This is the equation Directrix x=k Focus at Center origin

x

From the ellipse diagram in Figure 10.54, we see that k is related to the eccentricity e and the semimajor axis a by the equation

ea a a e

FIGURE 10.54

with k = 5 2 and e = 1. The equation of the directrix is x = 5 2.

In an ellipse with semimajor axis a, the focus–directrix distance is k = ( a e ) − ea, so ke = a (1 − e 2 ).

k =

a − ea. e

From this, we find that ke = a (1 − e 2 ). Replacing ke in Equation (5) by a (1 − e 2 ) gives the standard polar equation for an ellipse.

667

10.7  Conics in Polar Coordinates

y

Polar Equation for the Ellipse with Eccentricity e and Semimajor Axis a P(r, u)

a (1 − e 2 ) 1 + e cos θ

r =

r

(6)

P0(r0 , u0 )

r0 u

Notice that when e = 0, Equation (6) becomes r = a, which represents a circle.

L

u0

x

O

Lines

FIGURE 10.55

We can obtain a polar equation for line L by reading the relation r0 = r cos( θ − θ 0 ) from the right triangle OP0 P.

Suppose the perpendicular from the origin to line L meets L at the point P0 ( r0 ,  θ 0 ), with r0 ≥ 0 (Figure 10.55). Then, if P ( r,  θ ) is any other point on L, the points P ,  P0 , and O are the vertices of a right triangle, from which we can read the relation r0 = r cos( θ − θ 0 ).

The Standard Polar Equation for Lines If the point P0 ( r0 ,  θ 0 ) is the foot of the perpendicular from the origin to the line L,

and r0 ≥ 0, then an equation for L is

r cos( θ − θ 0 ) = r0 .

(7)

For example, if θ 0 = π 3 and r0 = 2, we find that

(

(

y P(r, u)

1 3 r cos θ + r sin θ = 2, 2 2

a r

O

u0 u

FIGURE 10.56

r0

) )

π =2 3 π π r cos θ cos + sin θ sin =2 3 3 r cos θ −

or

x+

3 y = 4.

P0(r0 , u0 )

Circles x

We can get a polar equation for this circle by applying the Law of Cosines to triangle OP0 P.

To find a polar equation for the circle of radius a centered at P0 ( r0 ,  θ 0 ), we let P ( r,  θ ) be a point on the circle and apply the Law of Cosines to triangle OP0 P (Figure 10.56). This gives a 2 = r0 2 + r 2 − 2r0 r cos( θ − θ 0 ). If the circle passes through the origin, then r0 = a and this equation simplifies to a 2 = a 2 + r 2 − 2ar cos( θ − θ 0 ) r 2 = 2ar cos( θ − θ 0 ) r = 2a cos( θ − θ 0 ). If the circle’s center lies on the positive x-axis, θ 0 = 0, and we get the further simplification r = 2a cos θ.

(8)

If the center lies on the positive y-axis, θ = π 2, cos( θ − π 2 ) = sin θ, and the equation r = 2a cos( θ − θ 0 ) becomes r = 2a sin θ.

(9)

Equations for circles through the origin centered on the negative x- and y-axes can be obtained by replacing r with −r in the above equations.

668

Chapter 10 Parametric Equations and Polar Coordinates

EXAMPLE 5 Here are several polar equations given by Equations (8) and (9) for circles through the origin and having centers that lie on the x- or y-axis.

EXERCISES

Radius

Center (polar coordinates)

Polar equation

3

( 3, 0 )

r = 6 cos θ

2

( 2,  π 2 )

r = 4 sin θ

12

(−1 2, 0 )

r = − cos θ

1

(−1,  π 2 )

r = −2 sin θ

10.7

Ellipses and Eccentricity

21. 8 x 2 − 2 y 2 = 16

22. y 2 − 3 x 2 = 3

In Exercises 1–8, find the eccentricity of the ellipse. Then find and graph the ellipse’s foci and directrices.

23. 8 y 2 − 2 x 2 = 16

24. 64 x 2 − 36 y 2 = 2304

1. 16 x 2 + 25 y 2 = 400 3.

2x 2

+

y2

= 2

2. 7 x 2 + 16 y 2 = 112 4. 2 x 2 + y 2 = 4

5. 3 x 2 + 2 y 2 = 6

6. 9 x 2 + 10 y 2 = 90

7. 6 x 2 + 9 y 2 = 54

8. 169 x 2 + 25 y 2 = 4225

Exercises 25–28 give the eccentricities and the vertices or foci of hyperbolas centered at the origin of the xy-plane. In each case, find the hyperbola’s standard-form equation in Cartesian coordinates. 25. Eccentricity: Vertices:

Exercises 9–12 give the foci or vertices and the eccentricities of ellipses centered at the origin of the xy-plane. In each case, find the ellipse’s standard-form equation in Cartesian coordinates. 9. Foci: ( 0, ±3 ) Eccentricity: 0.5 11. Vertices: ( 0, ±70 ) Eccentricity: 0.1

10. Foci: ( ±8, 0 ) Eccentricity: 0.2 12. Vertices: ( ±10, 0 ) Eccentricity: 0.24

Exercises 13–16 give foci and corresponding directrices of ellipses centered at the origin of the xy-plane. In each case, use the dimensions in Figure 10.49 to find the eccentricity of the ellipse. Then find the ellipse’s standard-form equation in Cartesian coordinates. 13. Focus:

( 5, 0 )

Directrix: 15. Focus:

9 x = 5

( −4, 0 )

14. Focus:

( 4, 0 )

Directrix: 16. Focus:

16 x = 3

(− 2, 0 )

3

26. Eccentricity:

( 0, ± 1)

Vertices:

27. Eccentricity: Foci:

3

( ±3, 0 )

28. Eccentricity: Foci:

2

( ±2, 0 )

1.25

( 0, ± 5 )

Eccentricities and Directrices

Exercises 29–36 give the eccentricities of conic sections with one focus at the origin along with the directrix corresponding to that focus. Find a polar equation for each conic section. 29. e = 1, x = 2

30. e = 1, y = 2

31. e = 5, y = −6

32. e = 2, x = 4

33. e = 1 2, x = 1

34. e = 1 4 , x = −2

35. e = 1 5, y = −10

36. e = 1 3, y = 6

Parabolas and Ellipses

Sketch the parabolas and ellipses in Exercises 37–44. Include the directrix that corresponds to the focus at the origin. Label the vertices with appropriate polar coordinates. Label the centers of the ellipses as well. 37. r =

1 1 + cos θ

38. r =

6 2 + cos θ

Hyperbolas and Eccentricity

39. r =

25 10 − 5 cos θ

40. r =

4 2 − 2 cos θ

In Exercises 17–24, find the eccentricity of the hyperbola. Then find and graph the hyperbola’s foci and directrices.

41. r =

400 16 + 8 sin θ

42. r =

12 3 + 3 sin θ

43. r =

8 2 − 2 sin θ

44. r =

4 2 − sin θ

Directrix:

x = −16

Directrix:

x = −2 2

17. x 2 − y 2 = 1

18. 9 x 2 − 16 y 2 = 144

19. y 2 − x 2 = 8

20. y 2 − x 2 = 4

Chapter 10  Questions to Guide Your Review

Lines

71. r = 1 (1 − sin θ )

72. r = 1 (1 + cos θ )

Sketch the lines in Exercises 45–48 and find Cartesian equations for them. π 3π 45. r cos θ − = 2 46. r cos θ + =1 4 4

73. r = 1 (1 + 2 sin θ )

74. r = 1 (1 + 2 cos θ )

(

(

)

47. r cos θ −

)

2π = 3 3

(

)

(

π = 2 3

48. r cos θ +

)

75. Perihelion and aphelion A planet travels about its sun in an ellipse whose semimajor axis has length a. (See accompanying figure.) a. Show that r = a (1 − e ) when the planet is closest to the sun and that r = a (1 + e ) when the planet is farthest from the sun. b. Use the data in the table in Exercise 76 to find how close each planet in our solar system comes to the sun and how far away each planet gets from the sun.

Find a polar equation in the form r cos( θ − θ 0 ) = r0 for each of the lines in Exercises 49–52. 49.

2x +

2y = 6

50.

51. y = −5

3x − y = 1

Aphelion (farthest from sun)

52. x = −4

Planet

Circles

Sketch the circles in Exercises 53–56. Give polar coordinates for their centers and identify their radii. 53. r = 4 cos θ

54. r = 6 sin θ

55. r = −2 cos θ

56. r = −8 sin θ

Perihelion (closest to sun)

u

a

Sun

Find polar equations for the circles in Exercises 57–64. Sketch each circle in the coordinate plane and label it with both its Cartesian and polar equations. 57. ( x − 6 ) 2 + y 2 = 36

58. ( x + 2 ) 2 + y 2 = 4

59. x 2 + ( y − 5 ) 2 = 25

60. x 2 + ( y + 7 ) 2 = 49

61. x 2 + 2 x + y 2 = 0

62. x 2 − 16 x + y 2 = 0

63. x 2 + y 2 + y = 0

64. x 2 + y 2 −

4 y = 0 3

Examples of Polar Equations

T Graph the lines and conic sections in Exercises 65–74.

76. Planetary orbits Use the data in the table below and Equation (6) to find polar equations for the orbits of the planets. Planet

Semimajor axis (astronomical units)

Eccentricity

Mercury

0.3871

0.2056

Venus

0.7233

0.0068

Earth

1.000

0.0167

Mars

1.524

0.0934

Jupiter

5.203

0.0484

9.539

0.0543

66. r = 4 sec ( θ + π 6 )

Saturn

67. r = 4 sin θ

68. r = −2 cos θ

Uranus

19.18

0.0460

69. r = 8 ( 4 + cos θ )

70. r = 8 ( 4 + sin θ )

Neptune

30.06

0.0082

65. r = 3 sec ( θ − π 3 )

CHAPTER 10

669

Questions to Guide Your Review

1. What is a parametrization of a curve in the xy-plane? Does a function y = f ( x ) always have a parametrization? Are parametrizations of a curve unique? Give examples.

6. How do you find the length of a smooth parametrized curve x = f (t ),   y = g(t ),   a ≤ t ≤ b? What does smoothness have to do with length? What else do you need to know about the parametrization in order to find the curve’s length? Give examples.

2. Give some typical parametrizations for lines, circles, parabolas, ellipses, and hyperbolas. How might the parametrized curve differ from the graph of its Cartesian equation?

7. What is the arc length function for a smooth parametrized curve? What is its arc length differential?

3. What is a cycloid? What are typical parametric equations for cycloids? What physical properties account for the importance of cycloids?

8. Under what conditions can you find the area of the surface generated by revolving a curve x = f (t ),   y = g(t ),   a ≤ t ≤ b, about the x-axis? the y-axis? Give examples.

4. What is the formula for the slope dy dx of a parametrized curve x = f (t ),   y = g(t )? When does the formula apply? When can you expect to be able to find d 2 y dx 2 as well? Give examples.

9. What are polar coordinates? What equations relate polar coordinates to Cartesian coordinates? Why might you want to change from one coordinate system to the other?

5. How can you sometimes find the area bounded by a parametrized curve and one of the coordinate axes?

10. What consequence does the lack of uniqueness of polar coordinates have for graphing? Give an example.

670

Chapter 10 Parametric Equations and Polar Coordinates

11. How do you graph equations in polar coordinates? Include in your discussion symmetry, slope, behavior at the origin, and the use of Cartesian graphs. Give examples. 12. How do you find the area of a region 0 ≤ r1 (θ) ≤ r ≤ r2 (θ), α ≤ θ ≤ β , in the polar coordinate plane? Give examples. 13. Under what conditions can you find the length of a curve r = f (θ),   α ≤ θ ≤ β , in the polar coordinate plane? Give an example of a typical calculation. 14. What is a parabola? What are the Cartesian equations for parabolas whose vertices lie at the origin and whose foci lie on the coordinate axes? How can you find the focus and directrix of such a parabola from its equation?

How can you find the foci, vertices, and directrices of such an ellipse from its equation? 16. What is a hyperbola? What are the Cartesian equations for hyperbolas centered at the origin with foci on one of the coordinate axes? How can you find the foci, vertices, and directrices of such an ellipse from its equation? 17. What is the eccentricity of a conic section? How can you classify conic sections by eccentricity? How does eccentricity change the shape of ellipses and hyperbolas? 18. Explain the equation PF = e ⋅ PD. 19. What are the standard equations for lines and conic sections in polar coordinates? Give examples.

15. What is an ellipse? What are the Cartesian equations for ellipses centered at the origin with foci on one of the coordinate axes?

CHAPTER 10

Practice Exercises

Identifying Parametric Equations in the Plane Exercises 1–6 give parametric equations and parameter intervals for the motion of a particle in the xy-plane. Identify the particle’s path by finding a Cartesian equation for it. Graph the Cartesian equation and indicate the direction of motion and the portion traced by the particle.

Lengths of Curves Find the lengths of the curves in Exercises 13–19.

13. y = x 1 2 − (1 3 ) x 3 2 , 1 ≤ x ≤ 4 14. x = y 2 3 , 1 ≤ y ≤ 8

1. x = t 2, y = t + 1, −∞ < t < ∞

15. y = ( 5 12 ) x 6 5 − ( 5 8 ) x 4 5 , 1 ≤ x ≤ 32

2. x =

16. x = ( y 3 12 ) + (1 y ), 1 ≤ y ≤ 2

t, y = 1 −

t, t ≥ 0

3. x = (1 2 ) tan t , y = (1 2 )sec t , −π 2 < t < π 2

17. x = 5 cos t − cos 5t , y = 5 sin t − sin 5t , 0 ≤ t ≤ π 2

4. x = −2 cos t , y = 2 sin t , 0 ≤ t ≤ π

18. x = t 3 − 6t 2 , y = t 3 + 6t 2 , 0 ≤ t ≤ 1 3π 19. x = 3 cos θ, y = 3 sin θ, 0 ≤ θ ≤ 2 20. Find the length of the enclosed loop x = t 2 ,   y = ( t 3 3 ) − t shown here. The loop starts at t = − 3 and ends at t = 3.

5. x = − cos t , y = cos 2 t , 0 ≤ t ≤ π 6. x = 4 cos t , y = 9 sin t , 0 ≤ t ≤ 2π Finding Parametric Equations and Tangent Lines

7. Find parametric equations and a parameter interval for the motion of a particle in the xy-plane that traces the ellipse 16 x 2 + 9 y 2 = 144 once counterclockwise. (There are many ways to do this.)

y t >0

1

8. Find parametric equations and a parameter interval for the motion of a particle that starts at the point (−2, 0 ) in the xy-plane and traces the circle x 2 + y 2 = 4 three times clockwise. (There are many ways to do this.) t=0

In Exercises 9 and 10, find an equation for the line in the xy-plane that is tangent to the curve at the point corresponding to the given value of t. Also, find the value of d 2 y dx 2 at this point.

t = ±" 3 0

1

2

4

x

9. x = (1 2 ) tan t , y = (1 2 )sec t , t = π 3 10. x = 1 + 1 t 2 , y = 1 − 3 t , t = 2 11. Eliminate the parameter to express the curve in the form y = f ( x ). a. x = 4 t 2 , y = t 3 − 1

−1

t 0

Chapter 10  Additional and Advanced Exercises

Suppose the equation of the curve is given in the form r = f (θ), where f (θ) is a differentiable function of θ. Then

Polar Coordinates

17. a. Find an equation in polar coordinates for the curve x = e 2 t cos t , y = e 2 t sin t , −∞ < t < ∞.

x = r cos θ and

b. Find the length of the curve from t = 0 to t = 2π.

dx dr = −r sin θ + cos θ , dθ dθ dy dr = r cos θ + sin θ . dθ dθ

Exercises 19–22 give the eccentricities of conic sections with one focus at the origin of the polar coordinate plane, along with the directrix for that focus. Find a polar equation for each conic section. 20. e = 1, r cos θ = −4

21. e = 1 2, r sin θ = 2

22. e = 1 3, r sin θ = −6

y = r sin θ

tan φ − tan θ . 1 + tan φ tan θ

tan ψ = tan ( φ − θ ) =

23. Epicycloids When a circle rolls externally along the circumference of a second, fixed circle, any point P on the circumference of the rolling circle describes an epicycloid, as shown here. Let the fixed circle have its center at the origin O and have radius a.

Furthermore, tan φ =

y P

dy dθ dy = dx dx dθ

because tan φ is the slope of the curve at P. Also, C

tan θ = u O

A(a, 0)

x

24. Find the centroid of the region enclosed by the x-axis and the cycloid arch

dy dθ y dy dx − x −y dx dθ x θ d dθ . tan ψ = = dy dx y dy dθ x + y 1+ dθ dθ x dx dθ

The Angle Between the Radius Vector and the Tangent Line to a Polar Coordinate Curve In Cartesian coordinates, when we want to discuss the direction of a curve at a point, we use the angle φ measured counterclockwise from the positive x-axis to the tangent line. In polar coordinates, it is more convenient to calculate the angle ψ from the radius vector to the tangent line (see the accompanying figure). The angle φ can then be calculated from the relation φ = θ + ψ,

(1)

which comes from applying the Exterior Angle Theorem to the triangle in the accompanying figure. y

c

r

Similarly, the denominator is x

dy dr dx + y = r . dθ dθ dθ

When we substitute these into Equation (4), we obtain tan ψ =

r . dr dθ

(5)

25. Show, by reference to a figure, that the angle β between the tangents to two curves at a point of intersection may be found from the formula tan β =

f

dy dx −y = r 2. dθ dθ

x

This is the equation we use for finding ψ as a function of θ.

r = f (u)

u

(4)

The numerator in the last expression in Equation (4) is found from Equations (2) and (3) to be

x = a ( t − sin t ), y = a (1 − cos t ); 0 ≤ t ≤ 2π.

P(r, u)

y . x

Hence

Let the radius of the rolling circle be b and let the initial position of the tracing point P be A( a, 0 ). Find parametric equations for the epicycloid, using as the parameter the angle θ from the positive x-axis to the line through the circles’ centers.

0

(3)

Since ψ = φ − θ from (1),

Theory and Examples

b

(2)

are differentiable functions of θ with

18. Find the length of the curve r = 2 sin 3 ( θ 3 ),  0 ≤ θ ≤ 3π, in the polar coordinate plane.

19. e = 2, r cos θ = 2

673

x

tan ψ2 − tan ψ1 . 1 + tan ψ2 tan ψ1

When will the two curves intersect at right angles? 26. Find the value of tan ψ for the curve r = sin 4 ( θ 4 ).

(6)

674

Chapter 10 Parametric Equations and Polar Coordinates

27. Find the angle between the radius vector to the curve r = 2a sin 3θ and its tangent when θ = π 6. T 28. a. Graph the hyperbolic spiral rθ = 1. What appears to happen to ψ as the spiral winds in around the origin?

29. The circles r = 3 cos θ and r = sin θ intersect at the point ( 3 2,  π 3 ). Show that their tangents are perpendicular there. 30. Find the angle at which the cardioid r = a (1 − cos θ ) crosses the ray θ = π 2.

b. Confirm your finding in part (a) analytically.

CHAPTER 10

Technology Application Projects

Mathematica/Maple Projects

Projects can be found within MyLab Math. • Radar Tracking of a Moving Object Part I: Convert from polar to Cartesian coordinates. • Parametric and Polar Equations with a Figure Skater Part I: Visualize position, velocity, and acceleration to analyze motion defined by parametric equations. Part II: Find and analyze the equations of motion for a figure skater tracing a polar plot.

11

Vectors and the Geometry of Space OVERVIEW In this chapter we begin the study of multivariable calculus. To apply calculus in many real-world situations, we introduce three-dimensional coordinate systems and vectors. We establish coordinates in space by adding a third axis that measures distance above and below the xy-plane. Then we define vectors, which provide simple ways to introduce equations for lines, planes, curves, and surfaces in space.

11.1 Three-Dimensional Coordinate Systems

z

z = constant

(0, 0, z)

(0, y, z) (x, 0, z) 0

P(x, y, z) (0, y, 0) y

(x, 0, 0) x

x = constant

y = constant (x, y, 0)

FIGURE 11.1 The Cartesian coordinate system is right-handed.

To locate a point in space, we use three mutually perpendicular coordinate axes, arranged as in Figure 11.1. The axes shown there make a right-handed coordinate frame. When you hold your right hand so that the fingers curl from the positive x-axis toward the positive y-axis, your thumb points along the positive z-axis. So when you look down on the xyplane from the positive direction of the z-axis, positive angles in the plane are measured counterclockwise from the positive x-axis and around the positive z-axis. (In a left-handed coordinate frame, the z-axis would point downward in Figure 11.1, and angles in the plane would be positive when measured clockwise from the positive x-axis. Right-handed and left-handed coordinate frames are not equivalent.) The Cartesian coordinates ( x ,  y,  z ) of a point P in space are the values at which the planes through P perpendicular to the axes cut the axes. Cartesian coordinates for space are also called rectangular coordinates because the axes that define them meet at right angles. Points on the x-axis have y- and z-coordinates equal to zero. That is, they have coordinates of the form ( x, 0, 0 ). Similarly, points on the y-axis have coordinates of the form ( 0,  y, 0 ), and points on the z-axis have coordinates of the form ( 0, 0,  z ). The planes determined by the coordinate axes are the xy-plane, whose standard equation is z = 0; the yz-plane, whose standard equation is x = 0; and the xz-plane, whose standard equation is y = 0. They meet at the origin ( 0, 0, 0 ) (Figure 11.2), which is also identified by 0 or the letter O. The three coordinate planes x = 0,  y = 0, and z = 0 divide space into eight cells called octants. The octant in which the point coordinates are all positive is called the first octant; there is no convention for numbering the other seven octants. The points in a plane perpendicular to the x-axis all have the same x-coordinate, this being the number at which that plane cuts the x-axis. The y- and z-coordinates can be any numbers. Similarly, the points in a plane perpendicular to the y-axis have a common y-coordinate, and the points in a plane perpendicular to the z-axis have a common zcoordinate. To write equations for these planes, we name the common coordinate’s value. The plane x = 2 is the plane perpendicular to the x-axis at x = 2. The plane y = 3 is the plane perpendicular to the y-axis at y = 3. The plane z = 5 is the plane perpendicular to the z-axis at z = 5. Figure 11.3 shows the planes x = 2,  y = 3, and z = 5, together with their intersection point ( 2, 3, 5 ).

675

676

Chapter 11 Vectors and the Geometry of Space

z z

(0, 0, 5)

(2, 3, 5) Line y = 3, z = 5

xz-plane: y = 0

Plane z = 5 Line x = 2, z = 5

yz-plane: x = 0 Plane x = 2

xy-plane: z = 0

Plane y = 3

Origin

0

(0, 3, 0)

(2, 0, 0) (0, 0, 0)

y

y x

x

Line x = 2, y = 3

The planes x = 0,   y = 0, and z = 0 divide space into eight octants. FIGURE 11.2

The planes x = 2,   y = 3, and z = 5 determine three lines through the point ( 2, 3, 5 ). FIGURE 11.3

The planes x = 2 and y = 3 in Figure 11.3 intersect in a line parallel to the z-axis. This line is described by the pair of equations x = 2,  y = 3. A point ( x ,  y,  z ) lies on the line if and only if x = 2 and y = 3. Similarly, the line of intersection of the planes y = 3 and z = 5 is described by the equation pair y = 3,  z = 5. This line runs parallel to the x-axis. The line of intersection of the planes x = 2 and z = 5, parallel to the y-axis, is described by the equation pair x = 2,  z = 5. In the following examples, we match coordinate equations and inequalities with the sets of points they define in space.

EXAMPLE 1

We interpret these equations and inequalities geometrically.

(a) z ≥ 0 (b) x = −3 (c) z = 0,  x ≤ 0,  y ≥ 0 (d) x ≥ 0,  y ≥ 0,  z ≥ 0 (e) −1 ≤ y ≤ 1

z The circle x 2 + y 2 = 4, z = 3

(f) y = −2,  z = 2

(0, 2, 3) (2, 0, 3) The plane z=3 (0, 2, 0)

EXAMPLE 2 (2, 0, 0) x

y x 2 + y 2 = 4, z = 0

The circle x 2 + y 2 = 4 in the plane z = 3 (Example 2). FIGURE 11.4

The half-space consisting of the points on and above the xy-plane. The plane perpendicular to the x-axis at x = −3. This plane lies parallel to the yz-plane and 3 units behind it. The second quadrant of the xy-plane. The first octant. The slab between the planes y = −1 and y = 1 (planes included). The line in which the planes y = −2 and z = 2 intersect. Alternatively, the line through the point ( 0, −2, 2 ) parallel to the x-axis.

What points ( x ,  y,  z ) satisfy the equations x 2 + y2 = 4

and

z = 3?

Solution The points lie in the horizontal plane z = 3 and, in this plane, make up the circle x 2 + y 2 = 4. We call this set of points “the circle x 2 + y 2 = 4 in the plane z = 3” or, more simply, “the circle x 2 + y 2 = 4,  z = 3” (Figure 11.4).

677

11.1  Three-Dimensional Coordinate Systems

Distance and Spheres in Space The formula for the distance between two points in the xy-plane extends to points in space.

The Distance Between P1( x 1 , y1 , z 1 ) and P2 ( x 2 , y 2 , z 2 )

P1P2 = z P2(x2, y2, z2)

P1(x1, y1, z1)

( x 2 − x 1 ) 2 + ( y 2 − y1 ) 2 + ( z 2 − z 1 ) 2

Proof We construct a rectangular box with faces parallel to the coordinate planes and the points P1 and P2 at opposite corners of the box (Figure 11.5). If A( x 2 ,  y1 ,  z 1 ) and B( x 2 ,  y 2 ,  z 1 ) are the vertices of the box indicated in the figure, then the three box edges P1 A,  AB, and BP2 have lengths P1 A = x 2 − x 1 ,

0

B(x2, y2, z1) y

A(x2, y1, z1)

P1P2

We find the distance between P1 and P2 by applying the Pythagorean theorem to the right triangles P1 AB and P1BP2 .

BP2 = z 2 − z 1 .

Because triangles P1BP2 and P1 AB are both right-angled, two applications of the Pythagorean theorem give

x

FIGURE 11.5

AB = y 2 − y1 ,

2

= P1B

2

+ BP2

2

and

P1B

2

= P1 A

2

+ AB

2

(see Figure 11.5). So P1P2

2

= P1B

2

+ BP2

= P1 A

2

+ AB

= x 2 − x1

2

2 2

+ BP2

+ y 2 − y1

2 2

Substitute  P1B

+ z 2 − z1

2

= P1 A

2

+ AB 2.

2

= ( x 2 − x 1 ) 2 + ( y 2 − y1 ) 2 + ( z 2 − z 1 ) 2 . Therefore, P1P2 = EXAMPLE 3

( x 2 − x 1 ) 2 + ( y 2 − y1 ) 2 + ( z 2 − z 1 ) 2 .

The distance between P1 ( 2, 1, 5 ) and P2 (−2, 3, 0 ) is P1P2 =

(−2

− 2 ) 2 + ( 3 − 1) 2 + ( 0 − 5 ) 2

=

16 + 4 + 25

=

45 ≈ 6.708.

We can use the distance formula to write equations for spheres in space (Figure 11.6). A point P ( x ,  y,  z ) lies on the sphere of radius a centered at P0 ( x 0 ,  y 0 ,  z 0 ) precisely when P0 P = a, or

z P0(x0, y0, z0)

( x − x 0 )2 + ( y − y 0 )2 + ( z − z 0 )2 = a 2 .

P(x, y, z) a

The Standard Equation for the Sphere of Radius a and Center ( x 0 , y 0 , z 0 ) ( x − x 0 )2 + ( y − y 0 )2 + ( z − z 0 )2 = a 2

0 y x

FIGURE 11.6

The sphere of radius a centered at the point ( x 0 ,  y 0 ,  z 0 ).

EXAMPLE 4

Find the center and radius of the sphere x 2 + y 2 + z 2 + 3 x − 4 z + 1 = 0.

Solution We find the center and radius of a sphere the way we find the center and radius of a circle: Complete the squares on the x-, y-, and z-terms as necessary and write each

678

Chapter 11 Vectors and the Geometry of Space

quadratic as a squared linear expression. Then, from the equation in standard form, read off the center and radius. For this sphere, we have x 2 + y 2 + z 2 + 3x − 4 z + 1 = 0 ( x 2 + 3 x ) + y 2 + ( z 2 − 4 z ) = −1

⎛ 2 ⎜⎜ x + 3 x + 3 ⎝ 2

( ) ⎞⎟⎟⎠ + y 2

2

⎛ −4 + ⎜⎜ z 2 − 4 z + ⎝ 2

( x + 32 )

( ) ⎞⎟⎟⎠ = −1 + ( 32 )

2

2

+ y 2 + ( z − 2 )2 =

2

+

( −24 )

2

21 . 4

From this standard form, we read that x 0 = −3 2,  y 0 = 0,  z 0 = 2, and a  The center is (−3 2, 0, 2 ). The radius is 21 2.

21 2.

EXAMPLE 5 Here are some geometric interpretations of inequalities and equations involving spheres. (a) x 2 + y 2 + z 2 < 4

The interior of the sphere x 2 + y 2 + z 2 = 4.

(b) x 2 + y 2 + z 2 ≤ 4

The solid ball bounded by the sphere x 2 + y 2 + z 2 = 4. Alternatively, the sphere x 2 + y 2 + z 2 = 4 together with its interior.

(c) x 2 + y 2 + z 2 > 4

The exterior of the sphere x 2 + y 2 + z 2 = 4.

(d) x 2 + y 2 + z 2 = 4,  z ≤ 0

The lower hemisphere cut from the sphere x 2 + y 2 + z 2 = 4 by the xy-plane (the plane z  0).

Just as polar coordinates give another way to locate points in the xy-plane (Section 10.3), alternative coordinate systems, different from the Cartesian coordinate system developed here, exist for three-dimensional space. We examine two of these coordinate systems in Section 14.7.

EXERCISES

11.1

Geometric Interpretations of Equations

Geometric Interpretations of Inequalities and Equations

In Exercises 1–16, give a geometric description of the set of points in space whose coordinates satisfy the given pairs of equations. 1. x  2, y  3

2. x = −1, z = 0

In Exercises 17–24, describe the sets of points in space whose coordinates satisfy the given inequalities or combinations of equations and inequalities.

3. y  0, z  0

4. x  1, y  0

17. a. x ≥ 0, y ≥ 0, z = 0 b. x ≥ 0, y ≤ 0, z = 0

5.

x2

+

y2

= 4, z = 0

7. x 2 + z 2 = 4, y = 0

6.

x2

+

y2

= 4, z = −2

8. y 2 + z 2 = 1, x = 0

9. x 2 + y 2 + z 2 = 1, x = 0 10. x 2 + y 2 + z 2 = 25, y = −4 11. x 2 + y 2 + ( z + 3 ) 2 = 25, z = 0

18. a. 0 b x b 1

b. 0 b x b 1, 0 b y b 1

c. 0 b x b 1, 0 b y b 1, 0 b z b 1 19. a. x 2 + y 2 + z 2 ≤ 1

b. x 2 + y 2 + z 2 > 1

20. a. x 2 + y 2 ≤ 1, z = 0

b. x 2 + y 2 ≤ 1, z = 3

c. x 2 + y 2 ≤ 1, no restriction on z

12. x 2 + ( y − 1) 2 + z 2 = 4, y = 0

21. a. 1 ≤ x 2 + y 2 + z 2 ≤ 4

b. x 2 + y 2 + z 2 ≤ 1, z ≥ 0

13. x 2 + y 2 = 4, z = y

22. a. x  y, z  0

b. x  y, no restriction on z

14. x 2 + y 2 + z 2 = 4, y = x

23. a. y p

b. x b y 2 , 0 b z b 2

15. y  x 2 , z  0

24. a. z = 1 − y, no restriction on x

16. z 

y2,

x 1

x 2,

z p 0

b. z  y 3 , x  2

11.1  Three-Dimensional Coordinate Systems

679

In Exercises 25–30, find the distance between points P1 and P2 .

46. The solid cube in the first octant bounded by the coordinate planes and the planes x = 2,   y = 2, and z = 2

25. P1 (1, 1, 1),

P2 ( 3, 3, 0 )

47. The half-space consisting of the points on and below the xy-plane

26. P1 (−1, 1, 5 ),

P2 ( 2, 5, 0 )

48. The upper hemisphere of the sphere of radius 1 centered at the origin

Distance

27. P1 (1, 4, 5 ),

P2 ( 4, −2, 7 )

28. P1 ( 3, 4, 5 ),

P2 ( 2, 3, 4 )

29. P1 ( 0, 0, 0 ),

P2 ( 2, −2, −2 )

49. The (a) interior and (b) exterior of the sphere of radius 1 centered at the point (1, 1, 1)

30. P1 ( 5, 3, −2 ), P2 ( 0, 0, 0 ) 31. Find the distance from the point ( 3, −4, 2 ) to the a. xy-plane

b. yz-plane

c. xz-plane

32. Find the distance from the point (−2, 1, 4 ) to the a. plane x = 3

b. plane y = −5

c. plane z = −1

50. The closed region bounded by the spheres of radius 1 and radius 2 centered at the origin. (Closed means the spheres are to be included. Had we wanted the spheres left out, we would have asked for the open region bounded by the spheres. This is analogous to the way we use closed and open to describe intervals: closed means endpoints included, open means endpoints left out. Closed sets include boundaries; open sets leave them out.)

33. Find the distance from the point ( 4, 3, 0 ) to the a. x-axis

b. y-axis

c. z-axis

34. Find the distance from the

Spheres

Find the center C and the radius a for the spheres in Exercises 51–60. 51. ( x + 2 ) 2 + y 2 + ( z − 2 ) 2 = 8

a. x-axis to the plane z = 3.

(

b. origin to the plane 2 = z − x.

52. ( x − 1) 2 + y +

c. point ( 0, 4, 0 ) to the plane y = x.

53. ( x −

In Exercises 35–44, describe the given set with a single equation or with a pair of equations. 35. The plane perpendicular to the b. y-axis at ( 0, −1, 0 )

a. x-axis at ( 3, 0, 0 ) c. z-axis at ( 0, 0, −2 )

36. The plane through the point ( 3, −1, 2 ) perpendicular to the a. x-axis

b. y-axis

c. z-axis

37. The plane through the point ( 3, −1, 1) parallel to the a. xy-plane

b. yz-plane

c. xz-plane

1 2

)

2

2

2

2) + (y −

(

)

+ ( z + 3 ) 2 = 25 2) + (z +

(

2

2) = 2

)

1 2 1 2 16 + z− = 3 3 9 55. x 2 + y 2 + z 2 + 4 x − 4 z = 0

54. x 2 + y +

56. x 2 + y 2 + z 2 − 6 y + 8 z = 0 57. 2 x 2 + 2 y 2 + 2 z 2 + x + y + z = 9 58. 3 x 2 + 3 y 2 + 3z 2 + 2 y − 2 z = 9 59. x 2 + y 2 + z 2 − 4 x + 6 y − 10 z = 11 60. ( x − 1) 2 + ( y − 2 ) 2 + ( z + 1) 2 = 103 + 2 x + 4 y − 2 z Find equations for the spheres whose centers and radii are given in Exercises 61–64.

38. The circle of radius 2 centered at ( 0, 0, 0 ) and lying in the a. xy-plane

b. yz-plane

39. The circle of radius 2 centered at ( 0, 2, 0 ) and lying in the a. xy-plane

a. xy-plane

( − 3, 4, 1)

b. yz-plane

41. The line through the point a. x-axis

c. plane y = 2

b. yz-plane

40. The circle of radius 1 centered at parallel to the

Center

c. xz-plane

c. xz-plane

(1, 3, − 1) parallel

b. y-axis

and lying in a plane

Radius

61. (1, 2, 3 ) 62. ( 0, − 1, 5 ) 63.

(−1,  12 , − 23 )

64. ( 0, − 7, 0 )

14 2 4 9 7

to the

c. z-axis

42. The set of points in space equidistant from the origin and the point ( 0, 2, 0 ) 43. The circle in which the plane through the point (1, 1, 3 ) perpendicular to the z-axis meets the sphere of radius 5 centered at the origin 44. The set of points in space that lie 2 units from the point ( 0, 0, 1) and, at the same time, 2 units from the point ( 0, 0, −1)

Theory and Examples

65. Find a formula for the distance from the point P ( x ,  y,  z ) to the a. x-axis.

b. y-axis.

c. z-axis.

66. Find a formula for the distance from the point P ( x ,  y,  z ) to the a. xy-plane.

b. yz-plane.

c. xz-plane.

67. Find the perimeter of the triangle with vertices A(−1, 2, 1), B (1, −1, 3 ), and C ( 3, 4, 5 ). 68. Show that the point P ( 3, 1, 2 ) is equidistant from the points A( 2, −1, 3 ) and B ( 4, 3, 1).

Write inequalities to describe the sets in Exercises 45–50.

69. Find an equation for the set of all points equidistant from the planes y = 3 and y = −1.

45. The slab bounded by the planes z = 0 and z = 1 (planes included)

70. Find an equation for the set of all points equidistant from the point ( 0, 0, 2 ) and the xy-plane.

Inequalities to Describe Sets of Points

680

Chapter 11 Vectors and the Geometry of Space

71. Find the point on the sphere x 2 + ( y − 3 ) 2 + ( z + 5 ) 2 = 4 nearest b. the point ( 0, 7, −5 ).

a. the xy-plane.

74. Find an equation for the set of points equidistant from the y-axis and the plane z  6. 75. Find an equation for the set of points equidistant from the

72. Find the point equidistant from the points ( 0, 0, 0 ), ( 0, 4, 0 ), ( 3, 0, 0 ), and ( 2, 2, − 3 ). 73. Find an equation for the set of points equidistant from the point ( 0, 0, 2 ) and the x-axis.

a. xy-plane and the yz-plane. b. x-axis and the y-axis. 76. Find all points that simultaneously lie 3 units from each of the points ( 2, 0, 0 ),   ( 0, 2, 0 ), and ( 0, 0, 2 ).

11.2 Vectors Some of the things we measure are determined simply by their magnitudes. To record mass, length, or time, for example, we need only write down a number and name an appropriate unit of measure. We need more information to describe a force, displacement, or velocity. To describe a force, we need to record the direction in which it acts as well as how large it is. To describe a body’s displacement, we have to say in what direction it moved as well as how far. To describe a body’s velocity, we have to know its direction of motion, as well as how fast it is going. In this section we show how to represent things that have both magnitude and direction in the plane or in space.

Component Form Terminal point B

AB

Initial point A

FIGURE 11.7

The directed line segment

AB is called a vector.

A quantity such as force, displacement, or velocity is called a vector and is represented by a directed line segment (Figure 11.7). The arrow points in the direction of the action and its length gives the magnitude of the action in terms of a suitably chosen unit. For example, a force vector points in the direction in which the force acts and its length is a measure of the force’s strength; a velocity vector points in the direction of motion and its length is the speed of the moving object. Figure 11.8 displays the velocity vector v at a specific location for a particle moving along a path in the plane or in space. (This application of vectors is studied in Chapter 12.) y

z v

v

y

(a) two dimensions

B A

0

x

0

x

y

(b) three dimensions

FIGURE 11.8

The velocity vector of a particle moving along a path (a) in the plane (b) in space. The arrowhead on the path indicates the direction of motion of the particle.

D C P O

x F

E

FIGURE 11.9 The four arrows in the plane (directed line segments) shown here have the same length and direction. They therefore represent the same vector, and







we write AB  CD  OP  EF.



The vector represented by the directed line segment AB has initial point A and terminal point B, and its length is denoted by AB . Two vectors are equal if they have the same length and direction. DEFINITIONS

The arrows we use when we draw vectors are understood to represent the same vector if they have the same length, are parallel, and point in the same direction (Figure 11.9) regardless of the initial point.

11.2  Vectors

z P(x 1, y1, z 1)

0

Q(x 2 , y 2 , z 2 )

Position vector of PQ v = ⟨v1, v 2, v3⟩ v2

x

(v1, v 2 , v3)

v3 v1

y



A vector PQ in standard position has its initial point at the origin.

The directed line segments PQ and v are parallel and have the same length.

681

In texts, vectors are usually written in lowercase boldface letters—for example u, v, and w. Sometimes we use uppercase boldface letters, such as F, to denote a force vector. In handwritten form, it is customary to draw small arrows above the letters—for example K K K K u ,  V ,  w, and F. We need a way to represent vectors algebraically so that we can be more precise about

the direction of a vector. Let v  PQ. There is one directed line segment equal to PQ whose initial point is the origin (Figure 11.10). It is the representative of v in standard position and is the vector we normally use to represent v. We can specify v by writing the coordinates of its terminal point ( V1 ,  V 2 ,  V 3 ) when v is in standard position. If v is a vector in the plane, its terminal point ( V1 ,  V 2 ) has two coordinates.

FIGURE 11.10

If v is a two-dimensional vector in the plane equal to the vector with initial point at the origin and terminal point ( V1 ,  V 2 ), then the component form of v is

DEFINITION

v = 〈V1 ,  V 2 〉. If v is a three-dimensional vector equal to the vector with initial point at the origin and terminal point ( V1 ,  V 2 ,  V 3 ), then the component form of v is v = 〈V1 ,  V 2 ,  V 3 〉.

HISTORICAL BIOGRAPHY Carl Friedrich Gauss (1777–1855) Gauss was born in Brunswick, Germany. The list of Gauss’s accomplishments in science and mathematics is astonishing, ranging from the invention of the electric telegraph (with Wilhelm Weber in 1833) to the development of a theory of planetary orbits and the development of an accurate theory of non-Euclidean geometry. To know more, visit the companion Website.

Thus a two-dimensional vector is an ordered pair v = 〈V1 ,  V 2 〉 of real numbers, and a three-dimensional vector is an ordered triple v = 〈V1 ,  V 2 ,  V 3 〉 of real numbers. The numbers V1 ,  V 2 , and V 3 are the components of v.

If v = 〈V1 ,  V 2 ,  V 3 〉 is represented by the directed line segment PQ, where the initial point is P ( x 1 ,  y1 ,  z 1 ) and the terminal point is Q ( x 2 ,  y 2 ,  z 2 ), then x 1 + V1 = x 2 , y1 + V 2 = y 2 , and z 1 + V 3 = z 2 (see Figure 11.10). Thus V1 = x 2 − x 1 , V 2 = y 2 − y1 , and V 3 = z 2 − z 1 are the components of PQ. In summary, given the points P ( x 1 ,  y1 ,  z 1 ) and Q ( x 2 ,  y 2 ,  z 2 ), the standard position vector v = 〈V1 ,  V 2 ,  V 3 〉 equal to PQ is v = 〈 x 2 − x 1 ,  y 2 − y1 ,  z 2   −   z 1 〉. If v is two-dimensional with P ( x 1 ,  y1 ) and Q ( x 2 ,  y 2 ) as points in the plane, then v = 〈 x 2 − x 1 ,  y 2 − y1 〉. There is no third component for planar vectors. With this understanding, we will develop the algebra of three-dimensional vectors and simply drop the third component when the vector is two-dimensional (a planar vector). Two vectors are equal if and only if their standard position vectors are identical. Thus 〈u1 ,  u 2 ,  u 3 〉 and 〈V1 ,  V 2 ,  V 3 〉 are equal if and only if u1  V1 ,  u 2  V 2 , and u 3  V 3 . The magnitude or length of the vector PQ is the length of any of its equivalent directed line segment representations. In particular, if v = 〈 x 2 − x 1 ,  y 2 − y1 ,  z 2 − z 1 〉 is

the component form of PQ, then the distance formula gives the magnitude or length of v, denoted by the symbol v or (in some texts) &v&.

The magnitude or length of the vector v = 〈v 1 , v 2 , v 3 〉 = 〈 x 2 −x 1 , y 2 − y1 , z 2 −z 1 〉 is the nonnegative number v =

V1 2 + V 2 2 + V 3 2 =

( x 2 − x 1 ) 2 + ( y 2 − y1 ) 2 + ( z 2 − z 1 ) 2

(see Figure 11.10).

The only vector with length 0 is the zero vector 0 = 〈0, 0〉 or 0 = 〈0, 0, 0〉. This vector is also the only vector with no specific direction.

682

Chapter 11 Vectors and the Geometry of Space

EXAMPLE 1 Find the (a) component form and (b) length of the vector with initial point P (−3, 4, 1) and terminal point Q (−5, 2, 2 ). Solution

 (a) The vector v = PQ has components υ1 = x 2 − x 1 = −5 − (−3 ) = −2,

υ 2 = y 2 − y1 = 2 − 4 = −2,

and υ 3 = z 2 − z 1 = 2 − 1 = 1.  The component form of PQ is y

v = 〈−2, −2, 1〉.  (b) The length, or magnitude, of v = PQ is

F = ⟨a, b⟩ 45˚

v =

(−2 ) 2

+ (−2 ) 2 + (1) 2 =

9 = 3.

x

FIGURE 11.11

The force pulling the cart forward is represented by the vector F whose horizontal component is the effective force (Example 2).

EXAMPLE 2 A small cart is being pulled along a smooth horizontal floor with a 20-N force F making a 45° angle to the floor (Figure 11.11). What is the effective force moving the cart forward? Solution The effective force is the horizontal component of F = 〈a,  b〉, given by

a = F cos 45 ° = (20)

( 22 ) ≈ 14.14 N.

Notice that F is a two-dimensional vector.

y ⟨u1 + y1, u 2 + y2⟩

u+v

Vector Algebra Operations Two principal operations involving vectors are vector addition and scalar multiplication. A scalar is simply a real number; we call it a scalar when we want to draw attention to the differences between numbers and vectors. Scalars can be positive, negative, or zero and are used to “scale” a vector by multiplication.

y2

v y1

u u2

x

u1

0

DEFINITIONS

(a)

Let u = 〈u1 ,  u 2 ,  u 3 〉 and v = 〈υ1 ,  υ 2 ,  υ 3 〉 be vectors with k a

scalar. y

Addition:

u + v = 〈u1 + υ1 ,  u 2 + υ 2 ,  u 3 + υ 3 〉

Scalar   multiplication: ku = 〈ku1 ,  ku 2 ,  ku 3 〉

u+v v u x

0 (b)

FIGURE 11.12

(a) Geometric interpretation of the vector sum. (b) The parallelogram law of vector addition in which both vectors are in standard position.

We add vectors by adding the corresponding components of the vectors. We multiply a vector by a scalar by multiplying each component by the scalar. The definitions also apply to planar vectors, except in that case there are only two components, 〈u1 ,  u 2 〉 and 〈υ1 ,  υ 2 〉. The definition of vector addition is illustrated geometrically for planar vectors in Figure 11.12a, where the initial point of one vector is placed at the terminal point of the other. Another interpretation is shown in Figure 11.12b. In this parallelogram law of addition, the sum, called the resultant vector, is the diagonal of the parallelogram. In physics, forces add vectorially, as do velocities, accelerations, and so on. So the force acting on a particle subject to two gravitational forces, for example, is obtained by adding the two force vectors.

11.2  Vectors

683

Figure 11.13 displays a geometric interpretation of the product ku of the scalar k and vector u. If k > 0, then ku has the same direction as u; if k < 0, then the direction of ku is opposite to that of u. Comparing the lengths of u and ku, we see that

1.5u

ku = u

−2u

2u (a)

=

( ku1 ) 2 + ( ku 2 ) 2 + ( ku 3 ) 2 =

k 2 u1 2 + u 2 2 + u 3 2 = k u .

The length of ku is the absolute value of the scalar k times the length of u. The vector = −u has the same length as u but points in the opposite direction. For k ≠ 0, we often express the scalar multiple (1 k ) u as u k . The difference u − v of two vectors is defined by

( − 1) u

y 2u

u

0

k 2 ( u1 2 + u 2 2 + u 3 2 )

u − v = u + (−v ). x

If u = 〈u1 ,  u 2 ,  u 3 〉 and v = 〈υ1 ,  υ 2 ,  υ 3 〉, then u − v = 〈u1 − υ1 ,  u 2 − υ 2 ,  u 3 − υ 3 〉. Note that ( u − v ) + v = u, so adding the vector ( u − v ) to v gives u (Figure 11.14a). Figure 11.14b shows the difference u − v as the sum u + (−v ).

−u (b)

FIGURE 11.13

(a) Scalar multiples of u. (b) Scalar multiples of a vector u in standard position.

u−v

EXAMPLE 3 (a) 2u + 3v

Let u = 〈−1, 3, 1〉 and v = 〈4, 7, 0〉. Find the components of (b) u − v

(c)

1 u. 2

Solution

(a) 2u + 3v = 2〈−1, 3, 1〉 + 3〈4, 7, 0〉 = 〈−2, 6, 2〉 + 〈12, 21, 0〉 = 〈10, 27, 2〉

v

(b) u − v = 〈−1, 3, 1〉 − 〈4, 7, 0〉 = 〈−1 − 4, 3 − 7, 1 − 0〉 = 〈−5, −4, 1〉 u

(c)

(a)

( )

( )

( )

1 1 3 1 1 2 3 2 1 2 1 u = − ,  ,  = − + + = 11. 2 2 2 2 2 2 2 2 Vector operations have many of the properties of ordinary arithmetic.

Properties of Vector Operations

v

Let u, v, w be vectors and a, b be scalars. u −v u + (−v) (b)

FIGURE 11.14 (a) The vector u − v, when added to v, gives u. (b) u − v = u + (−v ).

1. 3. 5. 7.

u+v = v+u u+0 = u 0u = 0

a(b u) = (ab)u 9. + b ) u = au + bu

2. ( u + v ) + w = u + ( v + w ) 4. u + (−u) = 0 6. 1u = u 8. a ( u + v ) = a u + a v

(a

These properties are readily verified using the definitions of vector addition and multiplication by a scalar. For instance, to establish Property 1, we have u + v = 〈u1 ,  u 2 ,  u 3 〉 + 〈υ1 ,  υ 2 ,  υ 3 〉 = 〈u1 + υ1 ,  u 2 + υ 2 ,  u 3 + υ 3 〉

Definition of vector addition

= 〈υ1 + u1 ,  υ 2 + u 2 ,  υ 3 + u 3 〉

Commutativity of real numbers (in each component)

= 〈υ1 ,  υ 2 ,  υ 3 〉 + 〈u1 ,  u 2 ,  u 3 〉 = v + u.

Definition of vector addition

When three or more space vectors lie in the same plane, we say they are coplanar vectors. For example, the vectors u, v, and u + v are always coplanar.

684

Chapter 11 Vectors and the Geometry of Space

z

Unit Vectors

OP2 = x 2 i + y2 j + z 2 k P2(x2, y 2 , z 2 )

A vector v of length 1 is called a unit vector. The standard unit vectors are i = 〈1, 0, 0〉,

k O i

Any vector v = 〈V1 ,  V 2 ,  V 3 〉 can be written as a linear combination of the standard unit vectors as follows:

P1P2

j

v = 〈V1 ,  V 2 ,  V 3 〉 = 〈V1 , 0, 0〉 + 〈0,  V 2 , 0〉 + 〈0, 0,  V 3 〉 y

x

j = 〈0, 1, 0〉, and k = 〈0, 0, 1〉.

= V1〈1, 0, 0〉 + V 2 〈0, 1, 0〉 + V 3〈0, 0, 1〉

P1(x 1, y 1 , z 1 )

= V1i + V 2 j + V 3k.

OP1 = x 1i + y 1 j + z 1k

FIGURE 11.15 The vector from P1 to P2

is P1 P2 = ( x 2 − x 1 ) i + ( y 2 − y1 ) j + ( z 2 − z 1 ) k.

We call the scalar (or number) V1 the i-component of the vector v, V 2 the j-component, and V 3 the k-component. As shown in Figure 11.15, the component form of the vector from P1 ( x 1 ,  y1 ,  z 1 ) to P2 ( x 2 ,  y 2 ,  z 2 ) is

P1P2 = ( x 2 − x 1 ) i + ( y 2 − y1 ) j + ( z 2 − z 1 ) k. If v v 0, then its length v is not zero and 1 1 v  v  1. v v That is, if the vector v is not the zero vector, then v v is a unit vector in the direction of v, and it is also called the direction of v. EXAMPLE 4 P2 ( 3, 2, 0 ).

Find a unit vector u in the direction of the vector from P1 (1, 0, 1) to



Solution We write P1P2 as a linear combination of the standard unit vectors and then divide it by its length:

P1P2 = ( 3 − 1) i + ( 2 − 0 ) j + ( 0 − 1) k = 2 i + 2 j − k

P1P2 = ( 2 ) 2 + ( 2 ) 2 + (−1) 2 = 4 + 4 + 1 = 9 = 3

P1P2 2i + 2 j − k 2 2 1

= u = = i + j − k. 3 3 3 3 P1P2

This unit vector u is the direction of P1P2 . EXAMPLE 5 If v = 3i − 4 j is a velocity vector, express v as a product of its magnitude (its speed) times its direction. Solution Speed is the magnitude (length) of v: HISTORICAL BIOGRAPHY Hermann Grassmann (1809–1877) Grassmann was born in Prussia (modern-day Poland) and attended the University of Berlin. However, his study of mathematics and physics was done on his own. In 1844, he published Die lineale Ausdehnungslehre, which contained new concepts in geometric calculus. He introduced the n-dimensional vector space and new concepts and structures in linear algebra. To know more, visit the companion Website.

v =

( 3)2

+ (−4 ) 2 =

9 + 16 = 5.

The unit vector v v is the direction of v: 3i − 4 j v 3 4 = = i − j. v 5 5 5 So v = 3i − 4 j = 5 Length (speed)

(35i −45 j).

Direction of motion

11.2  Vectors

685

In summary, we can express any nonzero vector v in terms of its two important features, v length and direction, by writing v = v . v

If v ≠ 0, then v is a unit vector called the direction of v; v v 2. the equation v = v expresses v as its length times its direction. v 1.

EXAMPLE 6 A force of 6 newtons is applied in the direction of the vector v = 2 i + 2 j − k. Express the force F as a product of its magnitude and direction. Solution The force vector has magnitude 6 and direction

F=6 =6

P1(x 1, y1, z 1)

v , so v

2i + 2 j − k 2i + 2 j − k v = 6 = 6 2 2 2 v 3 2 + 2 + (−1)

( 23 i + 23 j − 13 k ).

Midpoint of a Line Segment x + x 2 y1 + y 2 z 1 + z 2 b Ma1 , , 2 2 2

P2(x 2, y 2 , z 2 )

Vectors are often useful in geometry. For example, the coordinates of the midpoint of a line segment are found by averaging.

The midpoint M of the line segment joining points P1 ( x 1 ,  y1 ,  z 1 ) and P2 ( x 2 ,  y 2 ,  z 2 ) is the point

(x

O

1

)

+ x 2 y1 + y 2 z 1 + z 2 ,  ,  . 2 2 2

FIGURE 11.16

The coordinates of the midpoint are the averages of the coordinates of P1 and P2 .

To see why, observe (Figure 11.16) that     1  1  OM = OP1 + ( P1P2 ) = OP1 + ( OP2 − OP1 ) 2 2       1 = ( OP1 + OP2 ) 2 =

EXAMPLE 7

x1 + x 2 y + y2 z + z2 i+ 1 j+ 1 k. 2 2 2

The midpoint of the segment joining P1 ( 3, −2, 0 ) and P2 ( 7, 4, 4 ) is

( 3 +2 7 ,  −2 2+ 4 ,  0 +2 4 ) = ( 5, 1, 2 ).

686

Chapter 11 Vectors and the Geometry of Space

N

Applications An important application of vectors occurs in navigation.

v 30˚ 110

u+v

θ

E

u

800 NOT TO SCALE

FIGURE 11.17

Vectors representing the velocities of the airplane u and tailwind v in Example 8.

EXAMPLE 8 A jet airliner, flying due east at 800 kmh in still air, encounters a 110  kmh tailwind blowing in the direction 60n north of east. The airplane holds its compass heading due east but, because of the wind, acquires a new ground speed and direction. What are they? Solution If u is the velocity of the airplane alone and v is the velocity of the tailwind, then  u  800 and  v  110 (Figure 11.17). The velocity of the airplane with respect to the ground is given by the magnitude and direction of the resultant vector u v. If we let the positive x-axis represent east and the positive y-axis represent north, then the component forms of u and v are

u = 〈800,  0〉

v = 〈110 cos 60 °,  110 sin 60 °〉 = 〈55,  55 3〉.

and

Therefore, u + v = 〈855, 55 3〉 = 855i + 55 3 j u+v =

2

855 2 + ( 55 3 ) ≈ 860.3

and R = tan −1

55 3 ≈ 6.4 °. 855

The new ground speed of the airplane is about 860.3 kmh, and its new direction is about 6.4n north of east. Another important application occurs in physics and engineering when several forces are acting on a single object.

40°

55° F1 F2 55°

Solution The force vectors F1 and F2 have magnitudes F1 and F2 and components that are measured in newtons. The resultant force is the sum F1 F2, which must be equal in magnitude to the weight vector w but acting in the opposite (or upward) direction (see Figure 11.18b). It follows from the figure that

40° 75 (a)

F1 = 〈− F1 cos 55 °, F1 sin 55 °〉

F = F1+ F2 = ⟨0, 75⟩ F1

EXAMPLE 9 A 75-N weight is suspended by two wires, as shown in Figure 11.18a. Find the forces F1 and F2 acting in both wires.

F2 = 〈 F2 cos 40 °, F2 sin 40 °〉.

Since F1 + F2 = 〈0, 75〉, the resultant vector leads to the system of equations − F1 cos 55 ° + F2 cos 40 ° = 0

0 F1 0 0 F2 0 55°

40°

(b)

FIGURE 11.18

F1 sin 55 ° + F2 sin 40 ° = 75.

F2

w = ⟨0, −75⟩

Example 9.

and

The suspended weight in

Solving for F2 in the first equation and substituting the result into the second equation, we get F2 =

F1 cos 55 ° cos 40 °

and

F1 sin 55 ° +

F1 cos 55 ° sin 40 ° = 75. cos 40 °

It follows that F1 =

75 ≈ 57.67 N sin 55 ° + cos 55 ° tan 40 °

11.2  Vectors

687

and F2 =

F1 cos 55 ° cos 55 ° 75 = ≈ 43.18 N. cos 40 ° sin 55 ° + cos 55 ° tan 40 ° cos 40 °

The force vectors are then F1 = 〈− F1 cos 55 °,  F1 sin 55 °〉 ≈ 〈−33.08, 47.24〉 and F2 = 〈 F2 cos 40 °,  F2 sin 40 °〉 ≈ 〈33.08, 27.76〉.

Vectors in n Dimensions So far in this section, we introduced two- and three-dimensional vectors. We extend these notions by considering an n-dimensional vector v = 〈υ1 , υ 2 , … , υ n 〉 (an n-tuple of real numbers). We define 1. the magnitude or length of v:

v =

υ12 + υ 22 +  + υ n2 ;

2. addition of two vectors: 〈u1 , u 2 , … , u n 〉 + 〈υ1 , υ 2 , … , υ n 〉 = 〈u1 + υ1 , u 2 + υ 2 , … , u n + υ n 〉; 3. scalar multiplication of a vector by a real number: k 〈υ1 , υ 2 , … , υ n 〉 = 〈k υ1 , k υ 2 , … , k υ n 〉. The Properties of Vector Operations stated earlier in this section hold for n-dimensional vectors. When discussing vectors in this text, we will usually focus on the cases where n = 2 and n = 3 since these correspond to vectors in the plane and vectors in three-dimensional space, which are used in many applications (for instance, such vectors can represent velocity or force). Vectors with more than three components do arise in applications, but in those contexts they do not correspond to line segments in the plane or in threedimensional space. EXAMPLE 10 The components of the vector u = 〈70.9, 63.2, 77.2, 47.1, 55.6〉 contain the average temperatures (in degrees Fahrenheit) recorded between the years 2006 and 2010 in Houston, Los Angeles, Miami, Minneapolis, and New York. Likewise, the vectors v = 〈71.2, 64.3, 78.0, 47.1, 56.0〉 and w = 〈72.5, 64.7, 78.6, 47.6, 56.5〉 represent average temperatures in the same cities over the periods 2011 through 2015 and 2016 through 2020. Find the vector whose components are the average temperatures between the year 2006 and the year 2020 in these cities. Solution To find the vector containing average temperatures, we add scalar multiples of the three vectors (in other words, we evaluate a linear combination):

1 1 1 u+ v+ w 3 3 3 1 1 = 〈70.9, 63.2, 77.2, 47.1, 55.6〉 + 〈71.2, 64.3, 78.0, 47.1, 56.0〉 3 3 1 + 〈72.5, 64.7, 78.6, 47.6, 56.5〉 3 ≈ 〈71.5, 64.1, 77.9, 47.3, 56.0〉.

a=

688

Chapter 11 Vectors and the Geometry of Space

EXAMPLE 11 A rectangular grayscale image m pixels (dots) wide and n pixels tall can be represented on a computer as a vector with m ⋅ n components. A common format assigns an integer between 0 and 255 to each pixel, with 255 corresponding to the highest intensity (white), 0 to the lowest (black), and intermediate values to various shades of gray. Using this format, the 400 × 400 image containing all white pixels is represented by the 160,000-dimensional vector w = 〈255, … , 255〉. Figures 11.19a and b can be similarly represented by vectors u and v, respectively. In parts c, d, and e of Figure 11.19, we show different linear combinations of the vectors u and v. Notice how changing the scalar in front of each vector affects the resulting image. In Figure 11.19f, the image corresponds to the difference w − v , effectively inverting the grayscale in that image.

(a)

(b)

(a)

(d)

(c)

(b)

(e)

(d)

(c)

(f)

(e)

(f)

FIGURE 11.19 Each 400 × 400 pixel image corresponds to a 160,000-dimensional vector: (a) u; (b) v; (c) 14 u + 43 v ; (d) 12 u + 12 v; (e) 43 u + 14 v ; (f) w − v.

EXERCISES

11.2

Vectors in the Plane

In Exercises 1–8, let u = 〈3, −2〉 and v = 〈−2, 5〉. Find the (a) component form and (b) magnitude (length) of the vector. 1. 3u

2. −2v

3. u + v 5. 2u − 3v 3 4 7. u + v 5 5

4. u − v 6. −2u + 5v 5 12 8. − u + v 13 13

11.2  Vectors

In Exercises 9–16, find the component form of the vector.  9. The vector PQ, where P = (1, 3 ) and Q = ( 2, −1)  10. The vector OP, where O is the origin and P is the midpoint of segment RS, where R = ( 2, −1) and S = (−4, 3 ) 11. The vector from the point A = ( 2, 3 ) to the origin   12. The sum of AB and CD, where A = (1, −1),   B = ( 2, 0 ), C = (−1, 3 ), and D = (−2, 2 ) 13. The unit vector that makes an angle θ = 2π 3 with the positive x-axis 14. The unit vector that makes an angle θ = −3π 4 with the positive x-axis 15. The unit vector obtained by rotating the vector 〈0, 1〉 by 120 ° counterclockwise about the origin 16. The unit vector obtained by rotating the vector 〈1, 0〉 by 135 ° counterclockwise about the origin

31. Find the vectors whose lengths and directions are given. Try to do the calculations without writing. Length

Direction

a. 2

i

b.

In Exercises 17–22, express each vector in the form w = w1i + w 2 j + w 3k.  17. P1P2 if P1 is the point ( 5, 7, −1) and P2 is the point ( 2, 9, −2 )  18. P1P2 if P1 is the point (1, 2, 0 ) and P2 is the point (−3, 0, 5 )  19. AB if A is the point (−7, −8, 1) and B is the point (−10, 8, 1)  20. AB if A is the point (1, 0, 3 ) and B is the point (−1, 4, 5 )

3

1 c. 2

−k 3 j+ 5 6 i− 7

d. 7

4 k 5 2 3 j+ k 7 7

32. Find the vectors whose lengths and directions are given. Try to do the calculations without writing. Length

Direction

a. 7

−j

3 4 − i− k 5 5 13 3 4 12 i− j− k c. 12 13 13 13 1 1 1 i+ j− k d. a > 0 2 3 6 33. Find a vector of magnitude 7 in the direction of v = 12 i − 5k. b.

Vectors in Space

689

2

21. 5u − v if u = 〈1, 1, −1〉 and v = 〈2, 0, 3〉

34. Find a vector of magnitude 3 in the direction opposite to the direction of v = (1 2 ) i − (1 2 ) j − (1 2 ) k.

22. −2u + 3v if u = 〈−1, 0, 2〉 and v = 〈1, 1, 1〉

Direction and Midpoints

Geometric Representations

In Exercises 23 and 24, copy vectors u, v, and w head to tail as needed to sketch the indicated vector. a. u + v

23.

b. u + v + w

w

c. u − v

u

d. u − w v

24.

35. P1 (−1, 1, 5 ) P2 ( 2, 5, 0 ) 36. P1 (1, 4, 5 ) P2 ( 4, −2, 7 ) 37. P1 ( 3, 4, 5 ) P2 ( 2, 3, 4 ) 38. P1 ( 0, 0, 0 ) P2 ( 2, −2, −2 )  39. If AB = i + 4 j − 2k and B is the point ( 5, 1, 3 ), find A.  40. If AB = −7 i + 3 j + 8k and A is the point (−2, −3, 6 ), find B. Theory and Applications

v

a. u − v u

b. u − v + w c. 2u − v

w

d. u + v + w

Length and Direction

In Exercises 25–30, express each vector as a product of its length and direction. 25. 2 i + j − 2k 27. 5k 29.

 In Exercises 35–38, find a. the direction of P1P2 and b. the midpoint of line segment P1P2 .

1 1 1 i− j− k 6 6 6

26. 9 i − 2 j + 6k 3 4 28. i + k 5 5 j i k 30. + + 3 3 3

41. Linear combination Let u = 2 i + j, v = i + j, w = i − j. Find scalars a and b such that u = a v + b w.

and

42. Linear combination Let u = i − 2 j, v = 2 i + 3 j, and w = i + j. Write u = u 1 + u 2 , where u 1 is parallel to v and u 2 is parallel to w. (See Exercise 41.) 43. Linear combination Let u = 〈1, 2, 1〉, v = 〈1, −1, −1〉, w = 〈1, 1, −1〉, and z = 〈2, −3, −4〉. Find scalars a, b, and c such that z = a u + b v + cw. 44. Linear combination Let u = 〈1, 2, 2〉, v = 〈1, −1, −1〉, w = 〈1, 3, −1〉, and z = 〈2, 11, 8〉. Write z = u 1 + u 2 + u 3 , where u 1 is parallel to u,   u 2 is parallel to v, and u 3 is parallel to w. What are u 1 ,   u 2 ,   u 3? When solving Exercises 45–50, you may need to use a calculator or a computer. 45. Velocity An airplane is flying in the direction 25° west of north at 800 km h. Find the component form of the velocity of the airplane, assuming that the positive x-axis represents due east and the positive y-axis represents due north.

690

Chapter 11 Vectors and the Geometry of Space

46. (Continuation of Example 8.) What speed and direction should the jetliner in Example 8 have in order for the resultant vector to be 800 km/h due east?

52. Use similar triangles to find the coordinates of the point Q that divides the segment from P1 ( x 1 ,  y1 ,  z 1 ) to P2 ( x 2 ,  y 2 ,  z 2 ) into two lengths whose ratio is p q = r.

47. Consider a 100-N weight suspended by two wires as shown in the accompanying figure. Find the magnitudes and components of the force vectors F1 and F2 .

53. Medians of a triangle Suppose that A, B, and C are the corner points of the thin triangular plate of constant density shown here.

30°

45°

F1

F2

a. Find the vector from C to the midpoint M of side AB. b. Find the vector from C to the point that lies two-thirds of the way from C to M on the median CM. c. Find the coordinates of the point in which the medians of ΔABC intersect. According to Exercise 27, Section 6.6, this point is the plate’s center of mass. (See the figure.)

100

z C(1, 1, 3)

48. Consider a 50-N weight suspended by two wires as shown in the accompanying figure. If the magnitude of vector F1 is 35 N, find angle α and the magnitude of vector F2 . a

60°

c.m.

F1

y B(1, 3, 0)

F2 x

M A(4, 2, 0)

50

49. Consider a w-N weight suspended by two wires as shown in the accompanying figure. If the magnitude of vector F2 is 100 N, find w and the magnitude of vector F1. 40°

35°

F1

F2

w

50. Consider a 25-N weight suspended by two wires as shown in the accompanying figure. If the magnitudes of vectors F1 and F2 are both 75 N, then angles α and β are equal. Find α. a

b

F1

F2 25

51. Location A bird flies from its nest 5 km in the direction 60° north of east, where it stops to rest on a tree. It then flies 10 km in the direction due southeast and lands atop a telephone pole. Place an xy-coordinate system so that the origin is the bird’s nest, the x-axis points east, and the y-axis points north.

54. Find the vector from the origin to the point of intersection of the medians of the triangle whose vertices are A(1, −1, 2 ), B ( 2, 1, 3 ), and C (−1, 2, −1). 55. Let ABCD be a general, not necessarily planar, quadrilateral in space. Show that the two segments joining the midpoints of opposite sides of ABCD bisect each other. (Hint: Show that the segments have the same midpoint.) 56. Vectors are drawn from the center of a regular n-sided polygon in the plane to the vertices of the polygon. Show that the sum of the vectors is zero. (Hint: What happens to the sum if you rotate the polygon about its center?) 57. Suppose that A, B, and C are vertices of a triangle and that a, b, and care,   respectively,  the midpoints of the opposite sides. Show that Aa + Bb + Cc = 0. 58. Unit vectors in the plane Show that a unit vector in the plane can be expressed as u = ( cos θ ) i + ( sin θ ) j, obtained by rotating i through an angle θ in the counterclockwise direction. Explain why this form gives every unit vector in the plane. 59. Consider a triangle whose vertices are A( 2, −3, 4 ),   B (1, 0, −1), and C ( 3, 1, 2 ).       b. Find BA + AC + CB. a. Find AB + BC + CA. n-Dimensional Vectors

In Exercises 60–65, let u = 〈2, −3, 0, 1〉 and v = 〈0, −4, −1, 3〉. Find the (a) component form and (b) magnitude (length) of the vector. 60. 2u

61. −4 v

a. At what point is the tree located?

62. u + v

63. v − u

b. At what point is the telephone pole?

64. 3u − 2v

65. 4 u + 3v

11.3  The Dot Product

691

11.3 The Dot Product F

u

v

Length = 0 F 0 cos u

FIGURE 11.20

The magnitude of the force F in the direction of vector v is the length F cos θ of the projection of F onto v.

If a force F is applied to a particle moving along a path, we often need to know the magnitude of the force in the direction of motion. If v is parallel to the tangent line to the path at the point where F is applied, then we want the magnitude of F in the direction of v. Figure 11.20 shows that the scalar quantity we seek is the length F cos θ, where θ is the angle between the two vectors F and v. In this section we show how to calculate easily the angle between two vectors directly from their components. A key part of the calculation is an expression called the dot product. Dot products are also called inner or scalar products because the product results in a scalar, not a vector. After investigating the dot product, we apply it to finding the projection of one vector onto another (as displayed in Figure 11.20) and to finding the work done by a constant force acting through a displacement.

Angle Between Vectors When two nonzero vectors u and v are placed so their initial points coincide, they form an angle θ of measure 0 ≤ θ ≤ π (Figure 11.21). If the vectors do not lie along the same line, the angle θ is measured in the plane containing both of them. If they do lie along the same line, the angle between them is 0 if they point in the same direction and π if they point in opposite directions. The angle θ is the angle between u and v. Theorem 1 gives a formula to determine this angle. u

THEOREM 1—Angle Between Two Vectors

u

The angle θ between two nonzero vectors u = 〈u1 ,  u 2 ,  u 3 〉 and v = 〈υ1 ,  υ 2 ,  υ 3 〉 is given by

v

⎛ u υ + u 2 υ 2 + u 3υ 3 ⎞⎟ θ = arccos ⎜⎜ 1 1 ⎟⎟. ⎜⎝ u v ⎠

FIGURE 11.21

The angle between u and v given by Theorem 1 lies in the interval [ 0,  π ].

We use the law of cosines to prove Theorem 1, but before doing so, we focus attention on the expression u1υ1 + u 2 υ 2 + u 3υ 3 in the calculation for θ. This expression is the sum of the products of the corresponding components of the vectors u and v.

The dot product u ⋅ v (“u dot v”) of vectors u = 〈u1 ,  u 2 ,  u 3 〉 and v = 〈υ1 ,  υ 2 ,  υ 3 〉 is the scalar

DEFINITION

u ⋅ v = u 1υ 1 + u 2 υ 2 + u 3 υ 3 .

EXAMPLE 1

We illustrate the definition.

(a) 〈1, −2, −1〉 ⋅ 〈−6, 2, −3〉 = (1)(−6 ) + (−2 )( 2 ) + (−1)(−3 ) = −6 − 4 + 3 = −7 (b)

( 12 i + 3 j + k ) ⋅ ( 4 i − j + 2k ) = ( 12 )( 4 ) + ( 3)(−1) + (1)( 2 ) = 1 The dot product of a pair of two-dimensional vectors is defined in a similar fashion: 〈u1 ,  u 2 〉 ⋅ 〈υ1 ,  υ 2 〉 = u1υ1 + u 2 υ 2 .

We will see throughout the remainder of this text that the dot product is a key tool for many important geometric and physical calculations in space (and the plane).

692

Chapter 11 Vectors and the Geometry of Space

Proof of Theorem 1 Applying the law of cosines (Equation (8), Section 1.3) to the triangle in Figure 11.22, we find that w

= u

2

+ v

2

− 2 u v cos θ

2 u v cos θ = u

2

+ v

2

− w 2.

w

u

u

2

Law of cosines

Because w = u − v, the component form of w is 〈u1 − υ1 ,  u 2 − υ 2 ,  u 3 − υ 3 〉. So

v

u

2

= ( u1 2 + u 2 2 + u 3 2 ) = u1 2 + u 2 2 + u 3 2

v

2

= ( υ1 2 + υ 2 2 + υ 3 2 ) = υ1 2 + υ 2 2 + υ 3 2

FIGURE 11.22

The parallelogram law of addition of vectors gives w = u − v.

w

2

2

2

=

(

( u 1 − υ1 ) 2 + ( u 2 − υ 2 ) 2 + ( u 3 − υ 3 ) 2

)

2

= ( u 1 − υ1 ) 2 + ( u 2 − υ 2 ) 2 + ( u 3 − υ 3 ) 2 = u1 2 − 2u1υ1 + υ1 2 + u 2 2 − 2u 2 υ 2 + υ 2 2 + u 3 2 − 2u 3υ 3 + υ 3 2 and u

2

+ v

− w

2

2

= 2( u1υ1 + u 2 υ 2 + u 3υ 3 ).

Therefore, 2 u v cos θ = u

2

+ v

2

− w

2

= 2( u1υ1 + u 2 υ 2 + u 3υ 3 )

u v cos θ = u1υ1 + u 2 υ 2 + u 3υ 3 cos θ =

u1υ1 + u 2 υ 2 + u 3υ 3 . u v

⎛ u υ + u 2 υ 2 + u 3υ 3 ⎞⎟ Thus, for 0 ≤ θ ≤ π , we have θ = arccos ⎜⎜ 1 1 ⎟⎟. ⎜⎝ u v ⎠

Dot Product and Angles

⎛ u ⋅ v ⎞⎟ The angle between two nonzero vectors u and v is θ = arccos ⎜⎜ . ⎜⎝ u v ⎟⎟⎠ The dot product of two vectors u and v is given by u ⋅ v = u v cos θ.

Find the angle between u = i − 2 j − 2k and v = 6 i + 3 j + 2k.

EXAMPLE 2

Solution We use the formula above:

u ⋅ v = (1)( 6 ) + (−2 )( 3 ) + (−2 )( 2 ) = 6 − 6 − 4 = −4

y B(3, 5)

u

C(5, 2) x

1

FIGURE 11.23

Example 3.

( 1) 2

+ (−2 ) 2 + (−2 ) 2 =

v =

( 6 )2

+ ( 3)2 + ( 2 )2 =

⎛ u ⋅ v ⎞⎟ θ = arccos ⎜⎜ = arccos ⎜⎝ u v ⎟⎟⎠

1 A

u =

The triangle in

9 = 3

49 = 7

( 3−47 ) ≈ 1.76 radians or 100.98°. ( )( )

The angle formula applies to two-dimensional vectors as well. Note that the angle θ is acute if u ⋅ v > 0 and obtuse if u ⋅ v < 0. EXAMPLE 3 Find the angle θ in the triangle ABC determined by the vertices A = ( 0, 0 ),  B = ( 3, 5 ), and C = ( 5, 2 ) (Figure 11.23).

11.3  The Dot Product

693

  Solution The angle θ is the angle between the vectors CA and CB. The component forms of these two vectors are   CA = 〈−5, −2〉 and CB = 〈−2, 3〉. First we calculate the dot product and magnitudes of these two vectors.   CA ⋅ CB = (−5 )(−2 ) + (−2 )( 3 ) = 4  CA = ( −5 ) 2 + ( −2 ) 2 = 29  CB = (−2 ) 2 + ( 3 ) 2 = 13 Then, applying the angle formula, we have   ⎞ ⎞⎟ ⎛ ⎛ CA ⋅ CB 4 ⎟ θ = arccos ⎜⎜   ⎟⎟⎟ = arccos ⎜⎜ ⎜⎝ ( 29 )( 13 ) ⎟⎟⎠ ⎝⎜ CA CB ⎠⎟ ≈ 78.1 ° or 1.36 radians.

Orthogonal Vectors Two nonzero vectors u and v are perpendicular if the angle between them is π 2. For such vectors, we have u ⋅ v = 0 because cos( π 2 ) = 0. The converse is also true. If u and v are nonzero vectors with u ⋅ v = u v cos θ = 0, then cos θ = 0 and θ = arccos 0 = π 2. The following definition also allows for one or both of the vectors to be the zero vector.

DEFINITION Vectors u and v are orthogonal if u ⋅ v = 0.

EXAMPLE 4

To determine if two vectors are orthogonal, calculate their dot product.

(a) u = 〈3, −2〉 and v = 〈4, 6〉 are orthogonal because u ⋅ v = ( 3 )( 4 ) + (−2 )( 6 ) = 0. (b) u = 3i − 2 j + k and v = 2 j + 4 k are orthogonal because u ⋅ v = ( 3 )( 0 ) + (−2 )( 2 ) + (1)( 4 ) = 0. (c) 0 is orthogonal to every vector u because 0 ⋅ u = 〈0, 0, 0〉 ⋅ 〈u1 ,  u 2 ,  u 3 〉 = ( 0 )(u1 ) + ( 0 )(u 2 ) + ( 0 )(u 3 ) = 0.

Dot Product Properties and Vector Projections The dot product obeys many of the laws that hold for ordinary products of real numbers (scalars).

Properties of the Dot Product

If u, v, and w are any vectors and c is a scalar, then 1. u ⋅ v = v ⋅ u 3. u ⋅ ( v + w ) = u ⋅ v + u ⋅ w 5. 0 ⋅ u = 0.

2. ( cu ) ⋅ v = u ⋅ ( cv ) = c ( u ⋅ v ) 4. u ⋅ u = u 2

694

Chapter 11 Vectors and the Geometry of Space

Q

Proofs of Properties 1 and 3 The properties are easy to prove using the definition. For instance, here are the proofs of Properties 1 and 3. 1. u ⋅ v = u1υ1 + u 2 υ 2 + u 3υ 3 = υ1u1 + υ 2u 2 + υ 3u 3 = v ⋅ u

u

3. u ⋅ ( v + w ) = 〈u1 ,  u 2 ,  u 3 〉 ⋅ 〈υ1 + w1 ,  υ 2 + w 2 ,  υ 3 + w 3 〉

v R

P

S

Q

= u 1 ( υ1 + w 1 ) + u 2 ( υ 2 + w 2 ) + u 3 ( υ 3 + w 3 ) = u1υ1 + u1w1 + u 2 υ 2 + u 2 w 2 + u 3υ 3 + u 3 w 3 = ( u 1υ 1 + u 2 υ 2 + u 3 υ 3 ) + ( u 1 w 1 + u 2 w 2 + u 3 w 3 ) =u⋅v+u⋅w

u v P

R

FIGURE 11.24

S

The vector projection

We now return to the problem of projecting one vector onto another, posed in the   = onto a nonzero vector v = PS opening to this section. The vector projection of u PQ  (Figure 11.24) is the vector PR determined by dropping a perpendicular from Q to the line PS. The notation for this vector is

of u onto v.

(“ the vector projection of  u  onto  v” ).

proj v u

If u represents a force, then proj v u represents the effective force in the direction of v (Figure 11.25). If the angle θ between u and v is acute, proj v u has length u cos θ and direction v v (Figure 11.26). If θ is obtuse, cos θ < 0 and proj v u has length − u cos θ and direction −v v . In both cases,

Force = u v

projv u

proj v u = ( u cos θ )

v v

( u v⋅ v ) vv u⋅v v. =( v )

FIGURE 11.25

=

If we pull on the box with force u, the effective force moving the box forward in the direction v is the projection of u onto v.

u cos θ =

u v cos θ u⋅v = v v

2

u

u

u u

proj v u

proj v u

v

Length = 0 u 0 cos u (a)

v

Length = −0 u 0 cos u (b)

FIGURE 11.26 The length of proj v u is (a) u cos θ if cos θ ≥ 0 and (b) − u cos θ if cos θ < 0.

The number u cos θ is called the scalar component of u in the direction of v. To summarize,

The vector projection of u onto v is the vector proj v u =

( uv⋅ v ) v = ( u v⋅ v ) vv . 2

(1)

The scalar component of u in the direction of v is the scalar u cos θ =

u⋅v v = u⋅ . v v

(2)

11.3  The Dot Product

695

Note that both the vector projection of u onto v and the scalar component of u in the direction of v depend only on the direction of the vector v, not on its length. This is because in both cases we take the dot product of u with the direction vector v v , which is the direction of v, and for the projection we go on to multiply the result by the direction vector. EXAMPLE 5 Find the vector projection of u = 6 i + 3 j + 2k onto v = i − 2 j − 2k and the scalar component of u in the direction of v. Solution We find proj v u from Equation (1):

6 − 6 − 4( u⋅v u⋅v v = v = i − 2 j − 2k ) v2 v⋅v 1+ 4 + 4 4 4 8 8 = − ( i − 2 j − 2k ) = − i + j + k. 9 9 9 9

proj v u =

We find the scalar component of u in the direction of v from Equation (2):

(

v 1 2 2 = ( 6 i + 3 j + 2k ) ⋅ i − j − k v 3 3 3 4 4 =2−2− = − . 3 3

u cos θ = u ⋅

)

Equations (1) and (2) also apply to two-dimensional vectors. We demonstrate this in the next example. EXAMPLE 6 Find the vector projection of a force F = 5 i + 2 j onto v = i − 3 j and the scalar component of F in the direction of v. Solution The vector projection is

proj v F = =

( Fv⋅ v ) v = ( Fv ⋅⋅ vv ) v 2

5 − 6( 1 i − 3 j) = − ( i − 3 j) 1+ 9 10

=−

1 3 i+ j. 10 10

The scalar component of F in the direction of v is F cos θ = EXAMPLE 7 proj v u.

F⋅v 5−6 1 . = = − v 1+ 9 10

Verify that the vector u − proj v u is orthogonal to the projection vector

Solution The vector proj v u =

( uv⋅ v ) v is parallel to v. So it suffices to show that the 2

vector u − proj v u is orthogonal to v. We verify orthogonality by showing that the dot product of u − proj v u with v is zero:

( u − proj v u ) ⋅ v = u ⋅ v −

( uv⋅ v v ) ⋅ v 2

u⋅v( v ⋅ v) v2 u⋅v 2 =u⋅v− v v2 = u ⋅ v − u ⋅ v = 0. =u⋅v−

Definition of proj v u Dot product property (2) v⋅v = v

2

696

Chapter 11 Vectors and the Geometry of Space

u − projv u

Example 7 verifies that the vector u  proj v u is orthogonal to the projection vector proj v u (which has the same direction as v). So the equation

u

u = proj v u + ( u − proj v u ) =

( uv⋅ v ) v + ( u − ( uv⋅ v ) v ) 2

2

 Orthogonal to v

 Parallel to v

expresses u as a sum of orthogonal vectors (see Figure 11.27).

projv u

Work

v

FIGURE 11.27

The vector u is the sum of two perpendicular vectors: a vector proj v u, parallel to v, and a vector u  proj v u, perpendicular to v.

F P

D

Q

u 0 F 0 cos u

FIGURE 11.28

The work done by a constant force F during a displacement D is ( F cos R ) D , which is the dot product F ¸ D.

In Chapter 6, we calculated the work done by a constant force of magnitude F in moving an object through a distance d as W  Fd. That formula holds only if the force is directed

along the line of motion. If a force F moving an object through a displacement D  PQ has some other direction, the work is performed by the component of F in the direction of D. If R is the angle between F and D (Figure 11.28), then ⎛ scalar component of  F ⎞⎟ ⎟⎟( length of  D ) Work = ⎜⎜⎜ ⎝ in the direction of  D ⎟⎠ = ( F cos R ) D = F ⋅ D.

DEFINITION The work done by a constant force F acting through a displacement

D  PQ is

W = F ⋅ D.

EXAMPLE 8 If F  40 N (newtons), D  3 m, and R = 60 °, the work done by F in acting from P to Q is Work = F ⋅ D = F D cos R

Definition

= ( 40 )( 3 ) cos 60 °

Given values

= (120 )(1 2 ) = 60 J ( joules ). We encounter more challenging work problems in Chapter 15 when we learn to find the work done by a variable force along a more general path in space.

The Dot Product of Two n-Dimensional Vectors If u = 〈u1 , u 2 , ! , u n 〉 and v = 〈V1 , V 2 , ! , V n 〉 are n-dimensional vectors, then we define the dot product to be u ⋅ v = u 1V 1 + u 2 V 2 + " + u n V n . As for two- and three-dimensional vectors, the dot product is calculated by adding the products of the corresponding components of the two vectors. This generalized dot product can be shown to satisfy the Properties of the Dot Product that were introduced earlier in this section, and similar terminology is used. If u and v are n-dimensional vectors, then 1. u and v are said to be orthogonal if u ⋅ v = 0 , 2. the vector projection of u onto v is proj v u =

u⋅v v, and v2

697

11.3  The Dot Product

⎛ u ⋅ v ⎞⎟ 3. the angle between the vectors u and v is defined as θ = arccos⎜⎜ . (The Cauchy⎜⎝ u v ⎟⎟⎠ Schwarz inequality, u ⋅ v ≤ u v , stated in Exercise 27 can be extended to u⋅v n-dimensional vectors. This guarantees that is within the interval [ −1, 1 ].) u v EXAMPLE 9 An automobile assembly plant makes four different car models. The components of the vector u = 〈36, 50, 24, 10〉 indicate the plant’s output of each model per hour, whereas the revenue per vehicle (in US dollars) of each model is represented by the vector v = 〈24,000, 31,000, 39,000, 52,000〉. Calculate the dot product u ⋅ v and explain the significance of the value that was obtained. Solution

u ⋅ v = ( 36 )( 24,000 ) + ( 50 )( 31,000 ) + ( 24 )( 39,000 ) + (10 )( 52,000 ) = 3,870,000. The value $3,870,000 represents the total hourly revenue.

EXERCISES

11.3

For some exercises, a calculator may be helpful when expressing answers in decimal form.

15. Direction angles and direction cosines The direction angles α,  β , and γ of a vector v = a i + b j + ck are defined as follows:

Dot Product and Projections

α is the angle between v and the positive x-axis ( 0 ≤ α ≤ π ).

In Exercises 1–8, find

β is the angle between v and the positive y-axis ( 0 ≤ β ≤ π ).

a. v ⋅ u,   v ,   u

γ is the angle between v and the positive z-axis ( 0 ≤ γ ≤ π ).

b. the cosine of the angle between v and u c. the scalar component of u in the direction of v

z

d. the vector proj v u. 1. v = 2 i − 4 j +

5k, u = −2 i + 4 j −

5k

2. v = ( 3 5 ) i + ( 4 5 ) k, u = 5 i + 12 j 3. v = 10 i + 11 j − 2k, u = 3 j + 4 k g

4. v = 2 i + 10 j − 11k, u = 2 i + 2 j + k 0

5. v = 5 j − 3k, u = i + j + k 6. v = −i + j, u =

2i +

3 j + 2k

7. v = 5 i + j, u = 2 i + 17 j 1 1 1 1 ,  , u = , − 8. v = 2 3 2 3

a

v b y

x

Angle Between Vectors

Find the angles between the vectors in Exercises 9–12 to the nearest hundredth of a radian. 9. u = 2 i + j, v = i + 2 j − k 10. u = 2 i − 2 j + k, v = 3i + 4 k 11. u =

3i − 7 j, v =

12. u = i +

2j −

3i + j − 2k

2k, v = −i + j + k

13. Triangle Find the measures of the angles of the triangle whose vertices are A = (−1, 0 ),   B = ( 2, 1), and C = (1, −2 ). 14. Rectangle Find the measures of the angles between the diagonals of the rectangle whose vertices are A = (1, 0 ),   B = ( 0, 3 ), C = ( 3, 4 ), and D = ( 4, 1).

a. Show that cos α =

a , v

cos β =

b , v

cos γ =

c , v

and cos 2 α + cos 2 β + cos 2 γ = 1. These cosines are called the direction cosines of v. b. Unit vectors are built from direction cosines Show that if v = a i + b j + ck is a unit vector, then a, b, and c are the direction cosines of v. 16. Water main construction A water main is to be constructed with a 20% grade in the north direction and a 10% grade in the

698

Chapter 11 Vectors and the Geometry of Space

east direction. Determine the angle θ required in the water main for the turn from north to east.

25. Projectile motion A gun with muzzle velocity of 400 m s is fired at an angle of 8° above the horizontal. Find the horizontal and vertical components of the velocity. 26. Inclined plane Suppose that a box is being towed up an inclined plane as shown in the figure. Find the force w needed to make the component of the force parallel to the inclined plane equal to 2.5 N.

u East

rth

No

w

For Exercises 17 and 18, find the acute angle between the given lines by using vectors parallel to the lines.

33°

17. y = x , y = 2 x + 3

15°

18. 2 − x + 2 y = 0, 3 x − 4 y = −12 Theory and Examples

19. Sums and differences In the accompanying figure, it looks as if v 1 + v 2 and v 1 − v 2 are orthogonal. Is this mere coincidence, or are there circumstances under which we may expect the sum of two vectors to be orthogonal to their difference? Give reasons for your answer.

v1 + v 2

v1

−v 2

30. Cancelation in dot products In real-number multiplication, if uυ1 = uυ 2 and u ≠ 0, we can cancel the u and conclude that υ1 = υ 2 . Does the same rule hold for the dot product? That is, if u ⋅ v 1 = u ⋅ v 2 and u ≠ 0, can you conclude that v 1 = v 2? Give reasons for your answer. 31. If u and v are orthogonal, show that proj v u = 0.

C v

O

u

B

21. Diagonals of a rhombus Show that the diagonals of a rhombus (parallelogram with sides of equal length) are perpendicular. 22. Perpendicular diagonals Show that squares are the only rectangles with perpendicular diagonals. 23. When parallelograms are rectangles Prove that a parallelogram is a rectangle if and only if its diagonals are equal in length. (This fact is often exploited by carpenters.) 24. Diagonal of parallelogram Show that the indicated diagonal of the parallelogram determined by vectors u and v bisects the angle between u and v if u = v . u v

28. Dot multiplication is positive definite Show that dot multiplication of vectors is positive definite; that is, show that u ⋅ u ≥ 0 for every vector u and that u ⋅ u = 0 if and only if u = 0. 29. Orthogonal unit vectors If u 1 and u 2 are orthogonal unit vectors and v = a u 1 + b u 2 , find v ⋅ u 1.

20. Orthogonality on a circle Suppose that AB is the diameter of a circle with center O and that  C is apoint on one of the two arcs joining A and B. Show that CA and CB are orthogonal.

−u

b. Under what circumstances, if any, does u ⋅ v equal u v ? Give reasons for your answer.

v2

v1 − v 2

A

27. a. Cauchy-Schwarz inequality Since u ⋅ v = u v cos θ, show that the inequality u ⋅ v ≤ u v holds for any vectors u and v.

32. A force F = 2 i + j − 3k is applied to a spacecraft with velocity vector v = 3i − j. Express F as a sum of a vector parallel to v and a vector orthogonal to v. Equations for Lines in the Plane

33. Line perpendicular to a vector Show that v = a i + b j is perpendicular to the line ax + by = c . (Hint: For a and b nonzero, establish that the slope of the vector v is the negative reciprocal of the slope of the given line. Also verify the statement when a = 0 or b = 0.) 34. Line parallel to a vector Show that the vector v = a i + b j is parallel to the line bx − ay = c. (Hint: For a and b nonzero, establish that the slope of the line segment representing v is the same as the slope of the given line. Also verify the statement when a = 0 or b = 0.) In Exercises 35–38, use the result of Exercise 33 to find an equation for the line through P perpendicular to v. Then sketch the line. Include v in your sketch as a vector starting at the origin. 35. P ( 2, 1), v = i + 2 j

36. P (−1, 2 ), v = −2 i − j

37. P (−2, −7 ), v = −2 i + j

38. P (11, 10 ), v = 2 i − 3 j

In Exercises 39–42, use the result of Exercise 34 to find an equation for the line through P parallel to v. Then sketch the line. Include v in your sketch as a vector starting at the origin. 39. P (−2, 1), v = i − j

40. P ( 0, −2 ), v = 2 i + 3 j

41. P (1, 2 ), v = −i − 2 j

42. P (1, 3 ), v = 3i − 2 j

11.4  The Cross Product

Work

44. Locomotive The Union Pacific’s Big Boy locomotive could pull 6000-tonne trains with a tractive effort (pull) of 602,148 N. At this level of effort, about how much work did Big Boy do on the (approximately straight) 605-km journey from San Francisco to Los Angeles? 45. Inclined plane How much work does it take to slide a crate 20 m along a loading dock by pulling on it with a 200-N force at an angle of 30° from the horizontal? 46. Sailboat The wind passing over a boat’s sail exerted a 1000 N magnitude force F as shown here. How much work did the wind perform in moving the boat forward 1 km? Answer in joules.

n1

n2

43. Work along a line Find the work done by a force F = 5 i (magnitude 5 N) in moving an object along the line from the origin to the point (1, 1) (distance in meters).

699

L2 L2

u

v2 u

u

v1 L1

L1

Use this fact and the results of Exercise 33 or 34 to find the acute angles between the lines in Exercises 47–52. 47. 3 x + y = 5, 2 x − y = 4 48. y = 49.

3x − 1, y = − 3x + 2

3x − y = −2, x −

50. x +

3 y = 1, (1 −

3y = 1 3 ) x + (1 +

3)y = 8

51. 3 x − 4 y = 3, x − y = 7 52. 12 x + 5 y = 1, 2 x − 2 y = 3 Dot Products of n-Dimensional Vectors

In Exercises 53–56, (a) find u ⋅ v and (b) determine whether the vectors u and v are orthogonal.

60° 1000-N magnitude force

53. u = 〈3, 2, −4, 0〉,   v = 〈1, 0, 0, 2〉 F

54. u = 〈−2, 1, 1, 2〉,   v = 〈−1, 2, −2, −1〉

Angles Between Lines in the Plane

The acute angle between intersecting lines that do not cross at right angles is the same as the angle determined by vectors normal to the lines or by vectors parallel to the lines.

55. u = 〈6, 3, 0, 1, −2〉,   v = 〈0, 2, −7, 0, 3〉 56. u = 〈4, 2, −3, −2, 1, 5〉,   v = 〈3, −3, 2, −2, 1, −1〉

11.4 The Cross Product u×v

v

n

u

In studying lines in the plane, when we needed to describe how a line was tilting, we used the notions of slope and angle of inclination. In space, we want a way to describe how a plane is tilting. We accomplish this by multiplying two vectors in the plane together to get a third vector perpendicular to the plane. The direction of this third vector tells us the “inclination” of the plane. The product we use to multiply the vectors together is the vector or cross product, the second of the two vector multiplication methods. The cross product gives us a simple way to find a variety of geometric quantities, including volumes, areas, and perpendicular vectors. We study the cross product in this section.

u

The Cross Product of Two Vectors in Space FIGURE 11.29

u × v.

The construction of

We start with two nonzero vectors u and v in space. Two vectors are parallel if one is a nonzero multiple of the other. If u and v are not parallel, they determine a plane. The vectors in this plane are linear combinations of u and v, so they can be written as a sum a u + b v. We select the unit vector n perpendicular to the plane by the right-hand rule. This means that we choose n to be the unit normal vector that points the way your right thumb points when your fingers curl through the angle θ from u to v (Figure 11.29). Then we define a new vector as follows.

DEFINITION

The cross product u × v (“u cross v”) is the vector u × v = ( u v sin θ )   n.

700

Chapter 11 Vectors and the Geometry of Space

Unlike the dot product, the cross product is a vector. For this reason it is also called the vector product of u and v, and can be applied only to vectors in space. The vector u q v is orthogonal to both u and v because it is a scalar multiple of n. There is a straightforward way to calculate the cross product of two vectors from their components. The method does not require that we know the angle between them (as suggested by the definition), but we postpone that calculation momentarily so we can focus first on the properties of the cross product. Because the sines of 0 and Q are both zero, it makes sense to define the cross product of two parallel nonzero vectors to be 0. If one or both of u and v are zero, we also define u q v to be zero. This way, the cross product of two vectors u and v is zero if and only if u and v are parallel or one or both of them are zero.

Parallel Vectors

Nonzero vectors u  and  v  are parallel if and only if  u × v = 0 . The cross product obeys the following laws.

Properties of the Cross Product

If u, v, and w are any vectors and r, s are scalars, then 1. ( ru ) × ( sv ) = ( rs )( u × v ) 3. v × u = −( u × v ) 5. 0 × u = 0

v u

−n u

v×u

FIGURE 11.30

The construction of

v q u.

2. u × ( v + w ) = u × v + u × w 4. ( v + w ) × u = v × u + w × u 6. u × ( v × w ) = ( u ⋅ w ) v − ( u ⋅ v ) w

To visualize Property 3, for example, notice that when the fingers of your right hand curl through the angle R from v to u, your thumb points the opposite way; the unit vector we choose in forming v q u is the negative of the one we choose in forming u q v (Figure 11.30). Property 1 can be verified by applying the definition of cross product to both sides of the equation and comparing the results. Property 2 is proved in Appendix A.9. Property 4 follows by multiplying both sides of the equation in Property 2 by 1 and reversing the order of the products using Property 3. Property 5 is a definition. As a rule, cross product multiplication is not associative so ( u × v ) × w does not generally equal u × ( v × w ). (See Additional Exercise 17.) When we apply the definition and Property 3 to calculate the pairwise cross products of i, j, and k, we find (Figure 11.31) i × j = −( j × i ) = k j × k = −( k × j ) = i k × i = −( i × k ) = j

z k=i×j

and i × i = j × j = k × k = 0.

j=k×i y

u q v Is the Area of a Parallelogram Because n is a unit vector, the magnitude of u q v is

x i=j×k

FIGURE 11.31

The pairwise cross products of i, j, and k.

u × v = u v sin R n = u v sin R.

11.4  The Cross Product

Area = base · height = 0 u0 · 0 v 0 0 sin u 0 = 0u × v0

701

This is the area of the parallelogram determined by u and v (Figure 11.32), u being the base of the parallelogram and v sin θ being the height.

v

Determinant Formula for u × v

h = 0 v 0 0 sin u 0

Our next objective is to calculate u × v from the components of u and v relative to a Cartesian coordinate system. Suppose that

u u

FIGURE 11.32

The parallelogram determined by u and v.

u = u1i + u 2 j + u 3 k

and

v = υ1i + υ 2 j + υ 3k.

Then the distributive laws and the rules for multiplying i, j, and k tell us that u × v = ( u 1 i + u 2 j + u 3 k ) × ( υ1 i + υ 2 j + υ 3 k ) = u1υ1i × i + u1υ 2 i × j + u1υ 3 i × k + u 2 υ1 j × i + u 2 υ 2 j × j + u 2 υ 3 j × k + u 3υ1k × i + u 3υ 2 k × j + u 3υ 3k × k = ( u 2 υ 3 − u 3υ 2 ) i − ( u1υ 3 − u 3υ1 ) j + ( u1υ 2 − u 2 υ1 ) k. The component terms in the last line are hard to remember, but they are the same as the terms in the expansion of the symbolic determinant

Determinants 2 × 2 and 3 × 3 determinants are evaluated as follows: a b c d

= ad − bc

a1 a 2 a 3 b1 b 2 b3 c1 c 2 c 3 −a 2

i j k u1 u 2 u 3 . υ1 υ 2 υ 3

b1 b3 c1 c 3

= a1

b 2 b3 c2 c3

+ a3

b1 b 2 c1 c 2

So we restate the calculation in the following easy-to-remember form.

Calculating the Cross Product as a Determinant

If u = u1i + u 2 j + u 3k and v = υ1i + υ 2 j + υ 3k, then u×v =

i j k u1 u 2 u 3 . υ1 υ 2 υ 3

z R(−1, 1, 2)

Find u × v and v × u if u = 2 i + j + k and v = −4 i + 3 j + k.

EXAMPLE 1

Solution We expand the symbolic determinant. 0 P(1, −1, 0)

x

i j k y

Q(2, 1, –1)

  The vector PQ × PR is perpendicular to the plane of triangle PQR (Example 2). The area of triangle PQR is   half of PQ × PR (Example 3).

u×v =

2 1 1 −4 3 1

=

2 1 1 1 2 1 i− j+ k 3 1 −4 3 −4 1

= −2 i − 6 j + 10 k v × u = −( u × v ) = 2 i + 6 j − 10 k

Property 3

FIGURE 11.33

EXAMPLE 2 Find a vector perpendicular to the plane of P (1, −1, 0 ),  Q ( 2, 1, −1), and R (−1, 1, 2 ) (Figure 11.33).

702

Chapter 11 Vectors and the Geometry of Space

  Solution The vector PQ × PR is perpendicular to the plane because it is perpendicular to both vectors. In terms of components,  PQ = ( 2 − 1) i + (1 + 1) j + ( −1 − 0 ) k = i + 2 j − k  PR = ( −1 − 1) i + (1 + 1) j + ( 2 − 0 ) k = −2 i + 2 j + 2k   PQ × PR =

i j k 1 2 −1 = −2 2 2

2 −1 1 −1 1 2 i− j+ k 2 2 −2 2 −2 2

= 6 i + 6k. EXAMPLE 3 Find the area of the triangle with vertices P (1, −1, 0 ),  Q ( 2, 1, −1), and R (−1, 1, 2 ) (Figure 11.33). Solution The area of the parallelogram determined by P, Q, and R is   PQ × PR = 6 i + 6k Values from Example 2

=

( 6 )2

+ ( 6 )2 =

2 ⋅ 36 = 6 2.

The triangle’s area is half of this, or 3 2. EXAMPLE 4 Find a unit vector perpendicular to the plane of P (1, −1, 0 ), Q ( 2, 1, −1), and R (−1, 1, 2 ).   Solution Since PQ × PR is perpendicular to the plane, its direction n is a unit vector perpendicular to the plane. Taking values from Examples 2 and 3, we have   PQ × PR 6 i + 6k 1 1 n =   = = i+ k. PQ × PR 6 2 2 2 For ease in calculating the cross product using determinants, we usually write vectors in the form v = υ1i + υ 2 j + υ 3k rather than as ordered triples v = 〈υ1 ,  υ 2 ,  υ 3 〉.

Torque When we turn a bolt by applying a force F to a wrench (Figure 11.34), we produce a torque that causes the bolt to rotate. The torque vector points in the direction of the axis of the bolt according to the right-hand rule (so the rotation is counterclockwise when viewed from the tip of the vector). The magnitude of the torque depends on how far out on the wrench the force is applied and on how much of the force is perpendicular to the wrench at the point of application. The number we use to measure the torque’s magnitude is the product of the length of the lever arm r and the scalar component of F perpendicular to r. In the notation of Figure 11.34,

n

Torque

Magnitude of torque vector = r F sin θ,

r Component of F perpendicular to r. Its length is 0 F 0 sin u.

FIGURE 11.34

or r × F . If we let n be a unit vector along the axis of the bolt in the direction of the torque, then a complete description of the torque vector is r × F, or u F

The torque vector describes the tendency of the force F to drive the bolt forward.

Torque vector = r × F = ( r F sin θ ) n. Recall that we defined u × v to be 0 when u and v are parallel. This is consistent with the torque interpretation as well. If the force F in Figure 11.34 is parallel to the wrench, meaning that we are trying to turn the bolt by pushing or pulling along the line of the wrench’s handle, the torque produced is zero.

11.4  The Cross Product

3 m bar P

Q 70° 20 N magnitude force

F

703

EXAMPLE 5 The magnitude of the torque generated by force F at the pivot point P in Figure 11.35 is



PQ × F = PQ F sin 70 ° ≈ ( 3 )( 20 )( 0.94 ) ≈ 56.4 N ⋅ m. In this example, the torque vector is pointing out of the page toward you.

FIGURE 11.35

The magnitude of the torque exerted by F at P is about 56.4 N · m (Example 5). The bar rotates counterclockwise around P.

Triple Scalar or Box Product The product ( u × v ) ⋅ w is called the triple scalar product of u, v, and w (in that order). As you can see from the formula (u

× v ) ⋅ w = u × v w cos R ,

the absolute value of this product is the volume of the parallelepiped (parallelogram-sided box) determined by u, v, and w (Figure 11.36). The number u q v is the area of the base parallelogram. The number w cos R is the parallelepiped’s height. Because of this geometry, ( u × v ) ⋅ w is also called the box product of u, v, and w.

u×v w Height = 0 w0 0 cos u 0

u v

Area of base = 0 u × v0

u Volume = area of base · height = 0 u × v 0 0 w 0 0 cos u 0 = 0 (u × v) · w 0

FIGURE 11.36

The number ( u × v ) ⋅ w is the volume of a parallelepiped.

By treating the planes of v and w and of w and u as the base planes of the parallelepiped determined by u, v, and w, we see that (u

The dot and cross may be interchanged in a triple scalar product without altering its value.

× v ) ⋅ w = ( v × w ) ⋅ u = ( w × u ) ⋅ v.

Since the dot product is commutative, we also have (u

× v ) ⋅ w = u ⋅ ( v × w ).

The triple scalar product can be evaluated as a determinant: (u

⎛ u2 u3 u1 u 3 u1 u 2 ⎞⎟ × v ) ⋅ w = ⎜⎜⎜ j+ k ⎟⎟ ⋅ w i− V1 V 3 V1 V 2 ⎟⎟⎠ ⎜⎝ V 2 V 3 = w1

u2 u3 u1 u 3 u1 u 2 − w2 + w3 V2 V3 V1 V 3 V1 V 2

u1 u 2 u 3 = V1 V 2 V 3 . w1 w 2 w 3

704

Chapter 11 Vectors and the Geometry of Space

Calculating the Triple Scalar Product as a Determinant

u1 u 2 u 3 V1 V 2 V 3 w1 w 2 w 3

(u

w1 w 2 w 3 = − V1 V 2 V 3 u1 u 2 u 3

Any two rows of a matrix can be interchanged without changing the absolute value of the determinant. So we can take the vectors u, v, w in any order when calculating the absolute value of the triple product. (u

u1 u 2 u 3 V1 V 2 V 3 w1 w 2 w 3

× v) ⋅ w =

EXAMPLE 6 Find the volume of the box (parallelepiped) that is determined by u = i + 2 j − k,  v = −2 i + 3k, and w = 7 j − 4 k. Solution Using the rule for calculating a 3 q 3 determinant, we find

× v) ⋅ w =

1 2 −1 −2 0 3 0 7 −4

= ( 1)

0 3 −2 3 − ( 2) 7 −4 0 −4

+ ( − 1)

−2 0 0 7

= −23.

The volume is ( u × v ) ⋅ w = 23 units cubed.

EXERCISES

11.4

Cross Product Calculations

Triple Scalar Products

In Exercises 1–8, find the length and direction (when defined) of u q v and v q u.

In Exercises 19–22, verify that

1. u = 2 i − 2 j − k, v = i − k 2. u = 2 i + 3 j, v = −i + j

(u

and find the volume of the parallelepiped (box) determined by u, v, and w.

3. u = 2 i − 2 j + 4 k, v = −i + j − 2k 4. u = i + j − k, v = 0 5. u = 2 i, v = −3 j 6. u = i × j, v = j × k 7. u = −8 i − 2 j − 4 k, v = 2 i + 2 j + k 8. u =

3 1 i − j + k, v = i + j + 2k 2 2

In Exercises 9–14, sketch the coordinate axes and then include the vectors u, v, and u q v as vectors starting at the origin. 9. u  i, v  j

10. u = i − k, v = j

11. u = i − k, v = j + k

12. u = 2 i − j, v = i + 2 j

13. u = i + j, v = i − j

14. u = j + 2k, v = i

Triangles in Space

× v) ⋅ w = (v × w) ⋅ u = (w × u) ⋅ v

u

v

w

19. 2i

2j

2k

20. i − j + k

2 i + j − 2k

−i + 2 j − k

21. 2 i j

2i − j + k

i 2k

22. i + j − 2k

i  k

2 i + 4 j − 2k

Theory and Examples

23. Parallel and perpendicular vectors Let u = 5 i − j + k, v = j − 5k,   w = −15 i + 3 j − 3k. Which vectors, if any, are (a) perpendicular? (b) Parallel? Give reasons for your answers. 24. Parallel and perpendicular vectors Let u = i + 2 j − k , v = −i + j + k , w = i + k, r = −( Q 2 ) i − Q j + ( Q 2 ) k . Which vectors, if any, are (a) perpendicular? (b) Parallel? Give reasons for your answers.

In Exercises 15–18, a. Find the area of the triangle determined by the points P, Q, and R. b. Find a unit vector perpendicular to plane PQR. 15. P (1, −1, 2 ), Q ( 2, 0, −1), R ( 0, 2, 1)

In Exercises 25 and 26,

find the magnitude of the torque exerted by F on the bolt at P if PQ  20 cm and F  15 N. Answer in newtonmeters. 26.

25. F

16. P (1, 1, 1), Q ( 2, 1, 3 ), R ( 3, −1, 1) 60°

17. P ( 2, −2, 1), Q ( 3, −1, 2 ), R ( 3, −1, 1) 18. P (−2, 2, 0 ), Q ( 0, 1, −1), R (−1, 2, −2 )

P

F

135° Q

Q P

11.5  Lines and Planes in Space

27. Which of the following are always true, and which are not always true? Give reasons for your answers. a. u =

u⋅u

Area of a Parallelogram

Find the areas of the parallelograms whose vertices are given in Exercises 35–40.

b. u ⋅ u = u

35. A(1, 0 ), B ( 0, 1), C (−1, 0 ), D ( 0, −1)

c. u × 0 = 0 × u = 0

36. A( 0, 0 ), B ( 7, 3 ), C ( 9, 8 ), D ( 2, 5 ) 37. A(−1, 2 ), B ( 2, 0 ), C ( 7, 1), D ( 4, 3 )

d. u × (−u ) = 0

38. A(−6, 0 ), B (1, −4 ), C ( 3, 1), D (−4, 5 )

e. u × v = v × u

39. A( 0, 0, 0 ), B ( 3, 2, 4 ), C ( 5, 1, 4 ), D ( 2, −1, 0 )

f. u × ( v + w ) = u × v + u × w g.

(u

40. A(1, 0, −1), B (1, 7, 2 ) , C ( 2, 4, −1) , D ( 0, 3, 2 )

× v) ⋅ v = 0

h. ( u × v ) ⋅ w = u ⋅ ( v × w )

Area of a Triangle

28. Which of the following are always true, and which are not always true? Give reasons for your answers.

Find the areas of the triangles whose vertices are given in Exercises 41–47. 41. A( 0, 0 ), B (−2, 3 ), C ( 3, 1)

a. u ⋅ v = v ⋅ u

42. A(−1, −1), B ( 3, 3 ), C ( 2, 1)

b. u × v = −( v × u )

43. A(−5, 3 ), B (1, −2 ), C ( 6, −2 )

c. (−u ) × v = −( u × v ) d. ( cu ) ⋅ v = u ⋅ ( cv ) = c ( u ⋅ v ) ( any number  c ) e. c ( u × v ) = ( cu ) × v = u × ( cv ) ( any number  c ) f. u ⋅ u = u

705

2

44. A(−6, 0 ), B (10, −5 ), C (−2, 4 ) 45. A(1, 0, 0 ), B ( 0, 2, 0 ), C ( 0, 0, −1) 46. A( 0, 0, 0 ), B (−1, 1, − 1), C ( 3, 0, 3 )

g.

(u

× u) ⋅ u = 0

47. A(1, −1, 1), B ( 0, 1, 1), C (1, 0, −1)

h.

(u

× v) ⋅ u = v ⋅ (u × v)

48. Find the volume of a parallelepiped with one of its eight vertices at A( 0, 0, 0 ) and three adjacent vertices at B (1, 2, 0 ), C ( 0, −3, 2 ), and D ( 3, −4, 5 ).

29. Given nonzero vectors u, v, and w, use dot product and cross product notation, as appropriate, to describe the following. a. The vector projection of u onto v b. A vector orthogonal to u and v c. A vector orthogonal to u × v and w

49. Triangle area Find a 2 × 2 determinant formula for the area of the triangle in the xy-plane with vertices at ( 0, 0 ), ( a1 ,  a 2 ), and ( b1 ,  b 2 ). Explain your work.

e. A vector orthogonal to u × v and u × w

50. Triangle area Find a concise 3 × 3 determinant formula that gives the area of a triangle in the xy-plane having vertices ( a1 ,  a 2 ) , ( b1 ,  b 2 ) , and ( c1 ,  c 2 ).

f. A vector of length u in the direction of v

Volume of a Tetrahedron

d. The volume of the parallelepiped determined by u, v, and w

30. Compute ( i × j ) × j and i × ( j × j ). What can you conclude about the associativity of the cross product? 31. Let u, v, and w be vectors. Which of the following make sense, and which do not? Give reasons for your answers. a.

(u

× v) ⋅ w

c. u × ( v × w )

b. u × ( v ⋅ w ) d. u ⋅ ( v ⋅ w )

Using the methods of Section 6.1, where volume is computed by integrating cross-sectional area, it can be shown that the volume of 1 a tetrahedron formed by three vectors is equal to the volume of the 6 parallelepiped formed by the three vectors. Find the volumes of the tetrahedra whose vertices are given in Exercises 51–54. 51. A( 0, 0, 0 ), B ( 2, 0, 0 ), C ( 0, 3, 0 ), D ( 0, 0, 4 )

32. Cross products of three vectors Show that except in degenerate cases, ( u × v ) × w lies in the plane of u and v, whereas u × ( v × w ) lies in the plane of v and w. What are the degenerate cases?

52. A( 0, 0, 0 ), B (1, 0, 2 ), C ( 0, 2, 1), D ( 3, 4, 0 )

33. Cancelation in cross products If u × v = u × w and u ≠ 0, then does v = w? Give reasons for your answer.

In Exercises 55–57, determine whether the given points are coplanar.

34. Double cancelation If u ≠ 0 and if u × v = u × w and u ⋅ v = u ⋅ w, then does v = w? Give reasons for your answer.

56. A( 0, 0, 4 ), B ( 6, 2, 0 ), C ( 2, −1, 1), D (−3, −4, 3 )

53. A(1, −1, 0 ), B ( 0, 2, −2 ), C (−3, 0, 3 ), D ( 0, 4, 4 ) 54. A(−1, 2, 3 ), B ( 2, 0, 1), C (1, −3, 2 ), D (−2, 1, −1) 55. A(1, 1, 1), B (−1, 0, 4 ), C ( 0, 2, 1), D ( 2, −2, 3 ) 57. A( 0, 1, 2 ), B (−1, 1, 0 ), C ( 2, 0, −1), D (1, −1, 1)

11.5 Lines and Planes in Space This section shows how to use scalar and vector products to write equations for lines, line segments, and planes in space. We will use these representations throughout the rest of the text in studying the calculus of curves and surfaces in space.

706

Chapter 11 Vectors and the Geometry of Space

z

Lines and Line Segments in Space

P0(x 0 , y0 , z 0 ) P(x, y, z)

L v 0

y

In the plane, a line is determined by a point and a number giving the slope of the line. In space a line is determined by a point and a vector giving the direction of the line. Suppose that L is a line in space passing through a point P0 ( x 0 ,  y 0 ,  z 0 ) parallel to  a vector v = υ1i + υ 2 j + υ 3k. Then Lis the set of all points P ( x ,  y,  z ) for which P0 P is parallel to v (Figure 11.37). Thus, P0 P = tv for some scalar parameter t. The value of t ( ) depends on the location of the point  P along the line, and the domain of t is −∞, ∞ . The expanded form of the equation P0 P = tv is ( x − x 0 ) i + ( y − y 0 ) j + ( z − z 0 ) k = t ( υ1 i + υ 2 j + υ 3 k ) ,

x

FIGURE 11.37

A point P lies on L  through P0 parallel to v if and only if P0 P is a scalar multiple of v.

which can be rewritten as xi + yj + zk = x 0 i + y 0 j + z 0 k + t ( υ1i + υ 2 j + υ 3k ).

(1)

If r(t ) is the position vector of a point P ( x ,  y,  z ) on the line and r0 is the position vector of the point P0 ( x 0 ,  y 0 ,  z 0 ), then Equation (1) gives the following vector form for the equation of a line in space.

Vector Equation for a Line

A vector equation for the line L through P0( x 0 , y0 , z0 ) parallel to a nonzero vector v is r ( t ) = r0 + t v ,

−∞ < t < ∞,

(2)

where r is the position vector of a point P ( x ,  y,  z ) on L and r0 is the position vector of P0 ( x 0 ,  y 0 ,  z 0 ). Equating the corresponding components of the two sides of Equation (1) gives three scalar equations involving the parameter t: x = x 0 + t υ1 ,

y = y 0 + tυ 2 ,

z = z 0 + tυ 3.

These equations give us the standard parametrization of the line for the parameter interval −∞ < t < ∞.

Parametric Equations for a Line

The standard parametrization of the line through P0( x 0 , y0 , z0 ) parallel to a nonzero vector v = υ1i + υ 2 j + υ 3k is

z

x = x 0 + t υ1 , y = y 0 + t υ 2 , z = z 0 + t υ 3 , −∞ < t < ∞. 4

P0(–2, 0, 4) t=0

2

t=1 4 8

4 x

EXAMPLE 1 Find parametric equations for the line through (−2, 0, 4 ) parallel to v = 2 i + 4 j − 2k (Figure 11.38).

P1(0, 4, 2)

0 2

t=2 v = 2i + 4j − 2k

FIGURE 11.38

(3)

y P2(2, 8, 0)

Selected points and parameter values on the line in Example 1. The arrows show the direction of increasing t.

Solution With P0 ( x 0 ,  y 0 ,  z 0 ) equal to ( −2, 0, 4 ) and v 1i + v 2 j + v 3k equal to 2 i + 4 j − 2k, Equations (3) become

x = − 2 + 2t , EXAMPLE 2 Q (1, −1, 4 ).

y = 4t ,

z = 4 − 2 t.

Find parametric equations for the line through P (−3, 2, −3 ) and

Solution The vector  PQ = (1 − (−3 )) i + (−1 − 2 ) j + ( 4 − (−3 )) k = 4 i − 3 j + 7k

11.5  Lines and Planes in Space

707

is parallel to the line, and Equations (3) with ( x 0 ,  y 0 ,  z 0 ) = (−3, 2, −3 ) give x = −3 + 4 t ,

y = 2 − 3t ,

z = − 3 + 7 t.

We could have chosen Q (1, −1, 4 ) as the “base point” and written x = 1 + 4t ,

y = −1 − 3t ,

z = 4 + 7 t.

These equations serve as well as the first; they simply place you at a different point on the line for a given value of t. Notice that parametrizations are not unique. Not only can the “base point” change, but so can the parameter. The equations x = −3 + 4 t 3 ,  y = 2 − 3t 3 , and z = −3 + 7t 3 also parametrize the line in Example 2. To parametrize a line segment joining two points, we first parametrize the line through the points. We then find the t-values for the endpoints and restrict t to lie in the closed interval bounded by these values. The line equations, together with this added restriction, parametrize the segment. EXAMPLE 3 Parametrize the line segment joining the points P (−3, 2, −3 ) and ( ) Q 1, −1, 4 (Figure 11.39).

z

t=1

Solution We begin with equations for the line through P and Q, taking them, in this case, from Example 2:

−3

x = −3 + 4 t ,

0 2

FIGURE 11.39

z = − 3 + 7 t.

We observe that the point y

x

y = 2 − 3t ,

t=0 P(−3, 2, −3)

Example 3 derives a parametrization of line segment PQ. The arrow shows the direction of increasing t.

( x ,  y,  z )

= (−3 + 4 t , 2 − 3t , −3 + 7t )

on the line passes through P (−3, 2, −3 ) at t = 0 and Q (1, −1, 4 ) at t = 1. We add the restriction 0 ≤ t ≤ 1 to parametrize the segment: x = −3 + 4 t ,

y = 2 − 3t ,

z = − 3 + 7t ,

0 ≤ t ≤ 1.

The vector form (Equation (2)) for a line in space is more revealing if we think of a line as the path of a particle starting at position P0 ( x 0 ,  y 0 ,  z 0 ) and moving in the direction of vector v. Rewriting Equation (2), we have r ( t ) = r0 + t v v . v = r0 + t  › v 

(4)

›

1

›

−1

›

Q(1, −1, 4)

Initial position

Time

Speed Direction

In other words, the position of the particle at time t is its initial position plus its distance moved ( speed × time ) in the direction v v of its straight-line motion. EXAMPLE 4 A helicopter is to fly directly from a helipad at the origin in the direction of the point (1, 1, 1) at a speed of 60 m s. What is the position of the helicopter after 10 s? Solution We place the origin at the starting position (helipad) of the helicopter. Then the unit vector

u =

1 1 1 i+ j+ k 3 3 3

708

Chapter 11 Vectors and the Geometry of Space

gives the flight direction of the helicopter. From Equation (4), the position of the helicopter at any time t is r (t ) = r0 + t ( speed ) u = 0 + t (60)

( 13 i +

1 1 j+ k 3 3

)

= 20 3 t ( i + j + k ). When t = 10 s, r (10) = 200 3 ( i + j + k ) = 〈200 3, 200 3, 200 3〉. After 10 s of flight from the origin toward (1, 1, 1), the helicopter is located at the point ( 200 3, 200 3, 200 3 ) in space. It has traveled a distance of ( 60 m s )(10 s ) = 600 m, which is the length of the vector r(10). S

The Distance from a Point to a Line in Space 0 PS 0 sin u

u P

v

To find the distance from a point S to a line that passes through  a point P parallel to a vector v, we find the absolute value of the scalar component of PS in the direction of a vector normal to the line (Figure 11.40). In the notation of the figure, the absolute value of the    PS v sin θ PS × v scalar component is PS sin θ, which is = . v v

FIGURE 11.40

The distance from S to the line through P parallel to v is  PS sin θ, where θ is the angle between  PS and v.

Distance from a Point S to a Line Through P Parallel to v

 PS × v v

d =

EXAMPLE 5

(5)

Find the distance from the point S (1, 1, 5 ) to the line L:

x = 1 + t,

y = 3 − t,

z = 2 t.

Solution We see from the equations for L that L passes through P (1, 3, 0 ) parallel to v = i − j + 2k. With  PS = (1 − 1) i + (1 − 3 ) j + ( 5 − 0 ) k = −2 j + 5k

and

Equation (5) gives

i  PS × v = 0

−2

5 = i + 5 j + 2k ,

1

−1

2

 PS × v d = = v

j

k

1 + 25 + 4 = 1+1+ 4

30 = 6

5.

An Equation for a Plane in Space A plane in space is determined by knowing a point on the plane and its “tilt” or orientation. This “tilt” is defined by specifying a vector that is perpendicular, or normal, to the plane.

11.5  Lines and Planes in Space

n

Plane M P(x, y, z)

P0(x 0 , y0 , z 0 )

709

Suppose that plane M passes through a point P0 ( x 0 ,  y 0 ,  z 0 ) and is normal to the nonzero vector n = Ai + Bj + C k. A vector from P0 to any point  P on the plane is orthogonal to n. ( x ,  y,  z ) for which P0 P is orthogonal to n (Figure 11.41). Then M is the set of all points P  Thus, the dot product n ⋅ P0 P = 0. This equation is equivalent to ( Ai

+ Bj + C k ) ⋅ [ ( x − x 0 ) i + ( y − y 0 ) j + ( z − z 0 ) k ] = 0,

so the plane M consists of the points ( x ,  y,  z ) satisfying A( x − x 0 ) + B( y − y 0 ) + C ( z − z 0 ) = 0.

FIGURE 11.41

The standard equation for a plane in space is defined in terms of a vector normal to the plane: A point P lies in the plane through P0 normal to n if and  only if n ⋅ P0 P = 0.

Equation for a Plane

The plane through P0 ( x 0 , y 0 , z 0 ) normal to a nonzero vector n = Ai + Bj + C k has  Vector   equation: n ⋅ P0 P = 0 Component   equation:

A( x − x 0 ) + B ( y − y 0 ) + C ( z − z 0 ) = 0

Component   equation   simplified:

Ax + By + Cz = D, where D = Ax 0 + By 0 + Cz 0

EXAMPLE 6 Find an equation for the plane through P0 (−3, 0, 7 ) perpendicular to n = 5 i + 2 j − k. Solution The component equation is

5( x − ( −3 )) + 2( y − 0 ) + (−1)( z − 7 ) = 0. Simplifying, we obtain 5 x + 15 + 2 y − z + 7 = 0 5 x + 2 y − z = −22. Notice in Example 6 how the components of n = 5 i + 2 j − k became the coefficients of x, y, and z in the equation 5 x + 2 y − z = −22. The vector n = Ai + Bj + C k is normal to the plane Ax + By + Cz = D. EXAMPLE 7 C ( 0, 3, 0 ).

Find an equation for the plane through A( 0, 0, 1),  B( 2, 0, 0 ), and

Solution We find a vector normal to the plane and use it with one of the points (it does not matter which) to write an equation for the plane. The cross product

i   AB × AC = 2

j

k

0

−1 = 3i + 2 j + 6k

0

3

−1

is normal to the plane. We substitute the components of this vector and the coordinates of A( 0, 0, 1) into the component form of the equation to obtain 3( x − 0 ) + 2( y − 0 ) + 6( z − 1) = 0 3 x + 2 y + 6 z = 6.

Lines of Intersection Just as lines are parallel if and only if they have the same direction, two planes are parallel if and only if their normals are parallel, or n 1 = kn 2 for some scalar k. Two planes that are not parallel intersect in a line.

710

Chapter 11 Vectors and the Geometry of Space

EXAMPLE 8 Find a vector parallel to the line of intersection of the planes 3 x − 6 y − 2 z = 15 and 2 x + y − 2 z = 5. Solution The line of intersection of two planes is perpendicular to both planes’ normal vectors n 1 and n 2 (Figure 11.42) and therefore parallel to n 1 × n 2 . Turning this around, n 1 × n 2 is a vector parallel to the planes’ line of intersection. In our case, N

AN PL

j

n1 × n 2 = 3

−6

2

1

E1

n1 × n2

i

PL A

n1

E

2

n2

FIGURE 11.42

How the line of intersection of two planes is related to the planes’ normal vectors (Example 8).

k −2 = 14 i + 2 j + 15k. −2

Any nonzero scalar multiple of n 1 × n 2 will do as well. EXAMPLE 9 Find parametric equations for the line in which the planes 3 x − 6 y − 2 z = 15 and 2 x + y − 2 z = 5 intersect. Solution We find a vector parallel to the line and a point on the line and use Equations (3). Example 8 identifies v = 14 i + 2 j + 15k as a vector parallel to the line. To find a point on the line, we can take any point common to the two planes. Substituting z = 0 in the plane equations and solving for x and y simultaneously identifies one of these points as ( 3, − 1, 0 ). The line is

x = 3 + 14 t ,

y = − 1 + 2t ,

z = 15t.

The choice z = 0 is arbitrary, and we could have chosen z = 1 or z = −1 just as well. Or we could have let x = 0 and solved for y and z. The different choices would simply give different parametrizations of the same line. Sometimes we want to know where a line and a plane intersect. For example, if we are looking at a flat plate and a line segment passes through it, we may be interested in knowing what portion of the line segment is hidden from our view by the plate. This application is used in computer graphics (Exercise 78). EXAMPLE 10

Find the point where the line x =

8 + 2t , 3

y = − 2t ,

z = 1+ t

intersects the plane 3 x + 2 y + 6 z = 6. Solution The point

( 83 + 2t, −2t, 1 + t ) lies in the plane if its coordinates satisfy the equation of the plane—that is, if

( 83 + 2t ) + 2(−2t ) + 6(1 + t ) = 6

3

8 + 6t − 4 t + 6 + 6t = 6 8t = − 8 t = −1. The point of intersection is ( x ,  y,  z ) t =−1

=

( 83 − 2, 2, 1 − 1) = ( 23 , 2, 0 ).

711

11.5  Lines and Planes in Space

The Distance from a Point to a Plane If P is a point on a plane with a normal n, then the distance from any point S to the plane is the length of the vector projection of PS onto n, as given in the following formula.

Distance from a Point S to a Plane Through a Point P with a Normal n

 n d = PS ⋅ n

EXAMPLE 11

(6)

Find the distance from S (1, 1, 3 ) to the plane 3 x + 2 y + 6 z = 6.

Solution We find a point P in the plane and calculate the length of the vector projection  of PS onto a vector n normal to the plane (Figure 11.43). The coefficients in the equation 3 x + 2 y + 6 z = 6 give

n = 3i + 2 j + 6k. z n = 3i + 2j + 6k S(1, 1, 3)

3x + 2y + 6z = 6

(0, 0, 1) Distance from S to the plane

0

P(0, 3, 0)

(2, 0, 0)

y

x

FIGURE 11.43

The distance from S to the plane is the  length of the vector projection of PS onto n (Example 11).

The points on the plane easiest to find from the plane’s equation are the intercepts. If we take P to be the y-intercept ( 0, 3, 0 ), then  PS = (1 − 0 ) i + (1 − 3 ) j + ( 3 − 0 ) k = i − 2 j + 3k, n =

( 3)2

+ ( 2 )2 + ( 6 )2 =

49 = 7.

Therefore, the distance from S to the plane is  n d = PS ⋅ n

 Length of proj n PS

(

3 2 6 = ( i − 2 j + 3k ) ⋅ i + j + k 7 7 7 3 4 18 17 . = − + = 7 7 7 7

)

Angles Between Planes The angle between two intersecting planes is defined to be the acute angle between their normal vectors (Figure 11.44).

712

Chapter 11 Vectors and the Geometry of Space

EXAMPLE 11 Find the angle between the planes 3 x − 6 y − 2 z = 15 and 2 x + y − 2 z = 5.

u n2

n1

Solution The vectors

n 1 = 3i − 6 j − 2k ,

n 2 = 2 i + j − 2k

are normals to the planes. The angle between them is ⎛ n ⋅ n2 θ = arccos ⎜⎜ 1 ⎜⎝ n 1 n 2

u

= arccos FIGURE 11.44

The angle between two planes is obtained from the angle between their normals.

EXERCISES

⎞⎟ ⎟⎟ ⎠

( 214 ) ≈ 1.38 radians.

About 79 degrees

11.5

Lines and Line Segments

Find parametric equations for the lines in Exercises 1–12. 1. The line through the point P ( 3, −4, −1) parallel to the vector i+ j+k 2. The line through P (1, 2, −1) and Q (−1, 0, 1) 3. The line through P (−2, 0, 3 ) and Q ( 3, 5, −2 ) 4. The line through P (1, 2, 0 ) and Q (1, 1, −1) 5. The line through the origin parallel to the vector 2 j + k 6. The line through the point ( 3, −2, 1) parallel to the line x = 1 + 2t ,   y = 2 − t ,   z = 3t 7. The line through (1, 1, 1) parallel to the z-axis 8. The line through ( 2, 4, 5 ) perpendicular to the plane 3 x + 7 y − 5 z = 21 9. The line through x + 2 y + 2 z = 13

( 0, − 7, 0 )

perpendicular to the plane

10. The line through ( 2, 3, 0 ) perpendicular to the vectors u = i + 2 j + 3k and v = 3i + 4 j + 5k 11. The x-axis

12. The z-axis

25. The plane through P0 ( 2, 4, 5 ) perpendicular to the line x = 5 + t , y = 1 + 3t , z = 4 t 26. The plane through A(1, −2, 1) perpendicular to the vector from the origin to A. 27. Find the point of intersection of the lines x = 2t + 1, y = 3t + 2, z = 4 t + 3, and x = s + 2, y = 2s + 4, z = −4 s − 1, and then find the plane determined by these lines. 28. Find the point of intersection of the lines x = t ,   y = −t + 2, z = t + 1, and x = 2s + 2, y = s + 3, z = 5s + 6, and then find the plane determined by these lines. In Exercises 29 and 30, find the plane containing the intersecting lines. 29. L1: x = −1 + t , y = 2 + t , z = 1 − t; −∞ < t < ∞ L 2: x = 1 − 4 s, y = 1 + 2s, z = 2 − 2s; −∞ < s < ∞ 30. L1: x = t , y = 3 − 3t , z = −2 − t; −∞ < t < ∞ L 2: x = 1 + s, y = 4 + s, z = −1 + s; −∞ < s < ∞ 31. Find a plane through P0 ( 2, 1, −1) and perpendicular to the line of intersection of the planes 2 x + y − z = 3,   x + 2 y + z = 2. 32. Find a plane through the points P1 (1, 2, 3 ), and P2 ( 3, 2, 1) and perpendicular to the plane 4 x − y + 2 z = 7.

Find parametrizations for the line segments joining the points in Exercises 13–20. Draw coordinate axes and sketch each segment, indicating the direction of increasing t for your parametrization.

Distances

13. ( 0, 0, 0 ), (1, 1, 3 2 )

14. ( 0, 0, 0 ),

In Exercises 33–38, find the distance from the point to the line.

15. (1, 0, 0 ),

16. (1, 1, 0 ),

(1, 1, 0 )

17. ( 0, 1, 1),

( 0, − 1, 1)

19. ( 2, 0, 2 ),

( 0, 2, 0 )

18. ( 0, 2, 0 ), 20. (1, 0, −1),

(1, 0, 0 ) (1, 1, 1) ( 3, 0, 0 ) ( 0, 3, 0 )

33. ( 0, 0, 12 ); x = 4 t , y = −2t , z = 2t 34. ( 0, 0, 0 ); x = 5 + 3t , y = 5 + 4 t , z = −3 − 5t 35. ( 2, 1, 3 ); x = 2 + 2t , y = 1 + 6t , z = 3 36. ( 2, 1, −1); x = 2t , y = 1 + 2t , z = 2t

Planes

Find equations for the planes in Exercises 21–26.

37. ( 3, −1, 4 ); x = 4 − t , y = 3 + 2t , z = −5 + 3t

21. The plane through P0 ( 0, 2, −1) normal to n = 3i − 2 j − k

38. (−1, 4, 3 ); x = 10 + 4 t , y = −3, z = 4 t

22. The plane through (1, −1, 3 ) parallel to the plane

In Exercises 39–44, find the distance from the point to the plane.

3x + y + z = 7

39. ( 2, −3, 4 ), x + 2 y + 2 z = 13

23. The plane through (1, 1, −1),   ( 2, 0, 2 ), and ( 0, −2, 1)

40. ( 0, 0, 0 ), 3 x + 2 y + 6 z = 6

24. The plane through ( 2, 4, 5 ),   (1, 5, 7 ), and (−1, 6, 8 )

41. ( 0, 1, 1), 4 y + 3z = −12

713

11.5  Lines and Planes in Space

42. ( 2, 2, 3 ), 2 x + y + 2 z = 4

Theory and Examples

43. ( 0, −1, 0 ), 2 x + y + 2 z = 4

67. Use Equations (3) to generate a parametrization of the line through P ( 2, −4, 7 ) parallel to v 1 = 2 i − j + 3k. Then generate another parametrization of the line using the point P2 (−2, −2, 1) and the vector v 2 = −i + (1 2 ) j − ( 3 2 ) k.

44. (1, 0, −1), −4 x + y + z = 4 45. Find the distance from the plane x + 2 y + 6 z = 1 to the plane x + 2 y + 6 z = 10. 46. Find the distance from the line x = 2 + t ,   y = 1 + t , z = −(1 2 ) − (1 2 ) t to the plane x + 2 y + 6 z = 10. Angles

In Exercises 47 and 48, find the angles between the planes. 47. x + y = 1, 2 x + y − 2 z = 2 48. 5 x + y − z = 10, x − 2 y + 3z = −1 In Exercises 49 and 50, find the acute angles between the intersecting lines. 49. x = t ,   y = 2t ,   z = −t and

x = 1 − t ,   y = 5 + t ,   z = 2t

50. x = 2 + t ,   y = 4 t + 2,   z = 1 + t and x = 3t − 2,   y = −2,   z = 2 − 2t In Exercises 51 and 52, find the acute angles between the lines and planes. 51. x = 1 − t ,   y = 3t ,   z = 1 + t; 2 x − y + 3z = 6 52. x = 2,   y = 3 + 2t ,   z = 1 − 2t; x − y + z = 0 T Use a calculator to find the acute angles between the planes in Exercises 53–56 to the nearest hundredth of a radian. 53. 2 x + 2 y + 2 z = 3, 2 x − 2 y − z = 5 54. x + y + z = 1, z = 0 ( the  xy-plane ) 55. 2 x + 2 y − z = 3, x + 2 y + z = 2 56. 4 y + 3z = −12, 3 x + 2 y + 6 z = 6 Intersecting Lines and Planes

In Exercises 57–60, find the point in which the line meets the plane. 57. x = 1 − t , y = 3t , z = 1 + t; 2 x − y + 3z = 6 58. x = 2, y = 3 + 2t , z = −2 − 2t; 6 x + 3 y − 4 z = −12 59. x = 1 + 2t , y = 1 + 5t , z = 3t; x + y + z = 2 60. x = −1 + 3t , y = −2, z = 5t; 2 x − 3z = 7 Find parametrizations for the lines in which the planes in Exercises 61–64 intersect. 61. x + y + z = 1, x + y = 2 62. 3 x − 6 y − 2 z = 3, 2 x + y − 2 z = 2 63. x − 2 y + 4 z = 2, x + y − 2 z = 5 64. 5 x − 2 y = 11, 4 y − 5 z = −17

68. Use the component form to generate an equation for the plane through P1 ( 4, 1, 5 ) normal to n 1 = i − 2 j + k. Then generate another equation for the same plane using the point P2 ( 3, −2, 0 ) and the normal vector n 2 = − 2i + 2 2 j − 2k. 69. Find the points in which the line x = 1 + 2t , y = −1 − t , z = 3t meets the coordinate planes. Describe the reasoning behind your answer. 70. Find equations for the line in the plane z = 3 that makes an angle of π 6 rad with i and an angle of π 3 rad with j. Describe the reasoning behind your answer. 71. Is the line x = 1 − 2t ,   y = 2 + 5t ,   z = −3t parallel to the plane 2 x + y − z = 8? Give reasons for your answer. 72. How can you tell when two planes A1 x + B1 y + C 1z = D1 and A2 x + B 2 y + C 2 z = D 2 are parallel? Perpendicular? Give reasons for your answer. 73. Find two different planes whose intersection is the line x = 1 + t ,   y = 2 − t ,   z = 3 + 2t. Write equations for each plane in the form Ax + By + Cz = D. 74. Find a plane through the origin that is perpendicular to the plane M : 2 x + 3 y + z = 12 in a right angle. How do you know that your plane is perpendicular to M? 75. The graph of ( x a ) + ( y b ) + ( z c ) = 1 is a plane for any nonzero numbers a, b, and c. Which planes have an equation of this form? 76. Suppose L1 and L 2 are disjoint (nonintersecting) nonparallel lines. Is it possible for a nonzero vector to be perpendicular to both L1 and L 2 ? Give reasons for your answer. 77. Perspective in computer graphics In computer graphics and perspective drawing, we need to represent objects seen by the eye in space as images on a two-dimensional plane. Suppose that the eye is at E ( x 0 , 0, 0 ) as shown here and that we want to represent a point P1 ( x 1 ,  y1 ,  z 1 ) as a point on the yz-plane. We do this by projecting P1 onto the plane with a ray from E. The point P1 will be portrayed as the point P ( 0,  y,  z ). The problem for us as graphics designers is to find y and z given E and P1.   a. Write a vector equation that holds between EP and EP1. Use the equation to express y and z in terms of x 0 ,   x 1 ,   y1 , and z 1. b. Test the formulas obtained for y and z in part (a) by investigating their behavior at x 1 = 0 and x 1 = x 0 and by seeing what happens as x 0 → ∞. What do you find? z

Given two lines in space, either they are parallel, they intersect, or they are skew (lie in parallel planes). In Exercises 65 and 66, determine whether the lines, taken two at a time, are parallel, intersect, or are skew. If they intersect, find the point of intersection. Otherwise, find the distance between the two lines.

P(0, y, z) P1(x1, y1, z1) 0

65. L1: x = 3 + 2t ,   y = −1 + 4 t ,   z = 2 − t; −∞ < t < ∞ L 2: x = 1 + 4 s,   y = 1 + 2s,   z = −3 + 4 s; −∞ < s < ∞

(x1, y1, 0)

L 3: x = 3 + 2r ,   y = 2 + r ,   z = −2 + 2r; −∞ < r < ∞ 66. L1: x = 1 + 2t , y = −1 − t , z = 3t; −∞ < t < ∞ L 2: x = 2 − s, y = 3s, z = 1 + s; −∞ < s < ∞ L 3: x = 5 + 2r ,

y = 1 − r , z = 8 + 3r; −∞ < r < ∞

y

E(x 0, 0, 0) x

714

Chapter 11 Vectors and the Geometry of Space

78. Hidden lines in computer graphics Here is another typical problem in computer graphics. Your eye is at ( 4, 0, 0 ). You are looking at a triangular plate whose vertices are at (1, 0, 1),   (1, 1, 0 ), and (−2, 2, 2 ). The line segment from (1, 0, 0 ) to ( 0, 2, 2 )

passes through the plate. What portion of the line segment is hidden from your view by the plate? (This is an exercise in finding intersections of lines and planes.)

11.6 Cylinders and Quadric Surfaces z

Generating curve (in the yz-plane)

Up to now, we have studied two special types of surfaces: spheres and planes. In this section, we extend our inventory to include a variety of cylinders and quadric surfaces. Quadric surfaces are surfaces defined by second-degree equations in x, y, and z. Spheres are quadric surfaces, but there are others of equal interest that will be needed in Chapters 13–15.

Cylinders

y Lines through generating curve parallel to x-axis

x

FIGURE 11.45

A cylinder and

Suppose we are given a plane in space that contains a curve, and in addition we are given a line that is not parallel to this plane. A cylinder is a surface that is generated by moving a line that is parallel to the given line along the curve, while keeping it parallel to the given line. The curve is called a generating curve for the cylinder (Figure 11.45 illustrates this when the given plane is the yz-plane and the given line is the x-axis). In solid geometry, where cylinder means circular cylinder, the generating curves are circles, but now we allow generating curves of any kind. The cylinder in our first example is generated by a parabola.

generating curve. z Q 0 (x 0, x 02, z)

P0 (x 0, x 02, 0) 0 x y BO RA PA

L

A

FIGURE 11.46

y = x2

Every point of the cylinder in Example 1 has coordinates of the form ( x 0 ,  x 02 ,  z ).

EXAMPLE 1 Find an equation for the cylinder made by the lines parallel to the z-axis that pass through the parabola y  x 2 ,  z  0 (Figure 11.46). Solution The point P0 ( x 0 ,  x 02 , 0 ) lies on the parabola y  x 2 in the xy-plane. Then, for any value of z, the point Q ( x 0 ,  x 02 ,  z ) lies on the cylinder because it lies on the line x  x 0 ,  y  x 02 through P0 parallel to the z-axis. Conversely, any point Q ( x 0 ,  x 02 ,  z ) whose y-coordinate is the square of its x-coordinate lies on the cylinder because it lies on the line x  x 0 ,  y  x 02 through P0 parallel to the z-axis (Figure 11.46). Regardless of the value of z, therefore, the points on the surface are the points whose coordinates satisfy the equation y  x 2 . This makes y  x 2 an equation for the cylinder.

As Example 1 suggests, any curve f ( x ,  y ) = c in the xy-plane generates a cylinder parallel to the z-axis whose equation is also f ( x ,  y ) = c . For instance, the equation x 2 + y 2 = 1 corresponds to the circular cylinder made by the lines parallel to the z-axis that pass through the circle x 2 + y 2 = 1 in the xy-plane. In a similar way, any curve g ( x ,  z ) = c in the xz-plane generates a cylinder parallel to the y-axis whose space equation is also g ( x ,  z ) = c. Any curve h ( y,  z ) = c generates a cylinder parallel to the x-axis whose space equation is also h ( y,  z ) = c. The axis of a cylinder need not be parallel to a coordinate axis, however.

Quadric Surfaces A quadric surface is the graph in space of a second-degree equation in x, y, and z. We first focus on quadric surfaces given by the equation Ax 2 + By 2 + Cz 2 + Dz = E, where A, B, C, D, and E are constants. The basic quadric surfaces are ellipsoids, paraboloids, elliptical cones, and hyperboloids. Spheres are special cases of ellipsoids. We present a few examples illustrating how to sketch a quadric surface, and then we give a summary table of graphs of the basic types.

11.6  Cylinders and Quadric Surfaces

715

The ellipsoid

EXAMPLE 2

y2 x2 z2 + 2 + 2 =1 2 a b c (Figure 11.47) cuts the coordinate axes at ( ±a, 0, 0 ),  ( 0, ±b, 0 ), and ( 0, 0, ±c ). It lies within the rectangular box defined by the inequalities x ≤ a,  y ≤ b, and z ≤ c. The surface is symmetric with respect to each of the coordinate planes because each variable in the defining equation is squared.

Elliptical cross-section in the plane z = z0

z

z c y2 x2 + 2= 1 a2 b in the xy-plane The ellipse

a

ELLIPS

E

z0

y

b

y

SE

ELLIPSE

x

EL

LIP

x The ellipse x2 z2 + 2 =1 a2 c in the xz-plane

FIGURE 11.47

The ellipse

y2

b2 in the yz-plane

The ellipsoid

+

z2 =1 c2

y2 x2 z2 + 2 + 2 = 1 in Example 2 has elliptical cross-sections in each of the 2 a b c

three coordinate planes.

The curves in which the three coordinate planes cut the surface are ellipses. For example, y2 x2 + =1 a2 b2

z = 0.

when

The curve cut from the surface by the plane z = z 0 ,  z 0 < c, is the ellipse y2 x2 + = 1. 2 a 2 (1 − ( z 0 c ) ) b 2 (1 − ( z 0 c ) 2 ) If any two of the semiaxes a, b, and c are equal, the surface is an ellipsoid of revolution. If all three are equal, the surface is a sphere. EXAMPLE 3

The hyperbolic paraboloid y2 x2 z − 2 = , 2 b a c

c > 0

has symmetry with respect to the planes x = 0 and y = 0 (Figure 11.48). The crosssections in these planes are c 2 y . b2 c y = 0: the parabola z = − 2 x 2 . a

x = 0: the parabola z =

(1) (2)

Chapter 11 Vectors and the Geometry of Space

In the plane x  0, the parabola opens upward from the origin. The parabola in the plane y  0 opens downward. The parabola z = c2 y 2 in the yz-plane b z HY

Part of the hyperbola in the plane z = c

PER

y2 b2

z

2 − x2 = 1 a

BOLA

PA

The parabola z = − in the xz-plane

BOLA

y

BO

RA

x

RA

LA

Saddle point PA

716

c 2 x a2

2 y2 Part of the hyperbola x 2 − 2 = 1 a b in the plane z = −c

y x

FIGURE 11.48 The hyperbolic paraboloid ( y 2 b 2 ) − ( x 2 a 2 ) = z c ,   c > 0. The cross-sections in planes perpendicular to the z-axis above and below the xy-plane are hyperbolas. The cross-sections in planes perpendicular to the other axes are parabolas.

If we cut the surface by a plane z = z 0 > 0, the cross-section is a hyperbola, z y2 x2 − 2 = 0, 2 b a c with its focal axis parallel to the y-axis and its vertices on the parabola in Equation (1). If z 0 is negative, the focal axis is parallel to the x-axis and the vertices lie on the parabola in Equation (2). Near the origin, the surface is shaped like a saddle or mountain pass. To a person traveling along the surface in the yz-plane the origin looks like a minimum. To a person traveling the xz-plane the origin looks like a maximum. Such a point is called a saddle point of a surface. We will say more about saddle points in Section 13.7. Table 11.1 shows graphs of the six basic types of quadric surfaces. Each surface shown is symmetric with respect to the z-axis, but other coordinate axes can serve as well (with appropriate changes to the equation).

General Quadric Surfaces The quadric surfaces we have considered have symmetries relative to the x-, y-, or z-axes. The general equation of second degree in three variables x, y, z is Ax 2 + By 2 + Cz 2 + Dxy + Exz + Fyz + Gz + Hy + Iz + J = 0, where A, B, C, D, E, F, G, H, I, and J are constants. This equation leads to surfaces similar to those in Table 11.1, but in general these surfaces might be translated and rotated relative to the x-, y-, and z-axes. Terms of the type Gx, Hy, or Iz in the above formula lead to translations, which can be seen by a process of completing the squares. EXAMPLE 4

Identify the surface given by the equation x 2 + y 2 + 4 z 2 − 2 x + 4 y + 1 = 0.

Solution We complete the squares to simplify the expression:

x 2 + y 2 + 4 z 2 − 2 x + 4 y + 1 = ( x − 1) 2 − 1 + ( y + 2 ) 2 − 4 + 4 z 2 + 1 = ( x − 1) 2 + ( y + 2 ) 2 + 4 z 2 − 4.

717

11.6  Cylinders and Quadric Surfaces

TABLE 11.1 Graphs of Quadric Surfaces z

Elliptical cross-section in the plane z = z0

z c z

y2 x2 + 2 =1 a2 b in the xy-plane The ellipse

z0

The parabola z = in the xz-plane

E

ELLIPS

b

a y

SE

y

OL

x

A

LIP

ELLIPSE

AB

b

ELLIPSE

x

x2 z2 + 2 =1 a2 c in the xz-plane

The ellipse

y2

b2 in the yz-plane

+

z2 c2

y2 x2 z2 + + = 1 a2 b2 c2

y2

2 The ellipse x2 + 2 = 1 a b in the plane z = c

x

ELLIPTICAL PARABOLOID

z

HYPERBOLA

y

y x

Part of the hyperbola PSE

in the yz-plane

ELLIPTICAL CONE

y2

x2 z2 + 2 = 2 a2 b c

x y2 z2 − 2 =1 2 b c

x ELLIPSE

in the plane z = c

A

RB

HY

0

R

BOLA

PE

HY

BO

LA

BO

BOLA

y

(0, 0, c) Vertex

The hyperbola y2 z2 − 2 =1 c2 b in the yz-plane y

(0, 0, −c) Vertex

y x

y

HYPER

x

x2 =1 a2

z

PE

The hyperbola z2 x2 − =1 c2 a2 in the xz-plane

−

OL

ELLIPSE

RA

The parabola z = – c2 x2 a in the xz-plane

b

a

y2 b2

2 y2 Part of the hyperbola x2 − 2 = 1 a b in the plane z = –c

PA

x

RA

LA

Saddle point z

Part of the hyperbola

ERBOLA PA

y2 x2 The ellipse 2 + 2 = 1 a b in the plane z = c" 2

y2 x2 z2 + − 2 = 1 a2 b2 c

HYPERBOLOID OF ONE SHEET

HYP

y

LLIPS

The parabola z = c2 y2 in the yz-plane b z

z

z

HYPERBO

SE

x

ELLI

y2 x2 z + = a2 b2 c

2 2 Part of the hyperbola x2 − z2 = 1 in the xz-plane a c z y2 x2 The ellipse 2 + 2 = 2 z=c a b in the plane z = c b"2 a" 2 y2 x2 ELLIPSE The ellipse 2 + 2 = 1 a b in the xy-plane b y E a E

b

ELLIP

y

y x

LA

a

in the yz-plane

=1

ELLIPSOID

c z The line z = – y b in the yz-plane z = c

c 2 y b2

The parabola z =

EL

The ellipse

The line z = ac x in the xz-plane

z

y2 x2 + 2 =1 a2 b in the plane z = c The ellipse

z=c PA R

a

c 2 x a2

x

ELLIPSE

HYPERBOLOID OF TWO SHEETS

z2 c2

−

x2 a2

−

y2 b2

=1

HYPERBOLIC PARABOLOID

y2 x2 z − 2 = ,  c > 0 2 b a c

718

Chapter 11 Vectors and the Geometry of Space

We can rewrite the original equation as (x

( y + 2 )2 − 1) 2 z2 + + = 1. 4 4 1

This is the equation of an ellipsoid whose three semiaxes have lengths 2, 2, and 1 and which is centered at the point (1, −2, 0 ), as shown in Figure 11.49.

(y + 2)2 z2 + =1 4 1 in the plane x = 1

z

The ellipse

z

1

1 −1 0

1 z2 (x - 1)2 + =1 4 1 in the plane y = -2 The ellipse

FIGURE 11.49

EXERCISES

x

−1 0

y

−1

3 x

(x (y + + =1 4 4 in the plane z = 0 (This ellipse is a circle.) The ellipse

1)2

2)2

An ellipsoid centered at the point (1, −2, 0 ).

11.6 e.

Matching Equations with Surfaces

f.

z

In Exercises 1–12, match the equation with the surface it defines. Also, identify each surface by type (paraboloid, ellipsoid, etc.). The surfaces are labeled (a)–(l). 1. x 2 + y 2 + 4 z 2 = 10

2. z 2 + 4 y 2 − 4 x 2 = 4

3. 9 y 2 + z 2 = 16

4. y 2 + z 2 = x 2

5. x =

y2

−

6. x =

z2

7. x 2 + 2 z 2 = 8 9. x =

z2

−

−y 2

−

x x

y z

g.

z

h.

12. 9 x 2 + 4 y 2 + 2 z 2 = 36 b.

z

z y

x z

i.

x

y z

j.

y

x

x

d.

z

y

z

y

x

y

x z

k. x

y

10. z = −4 x 2 − y 2

11. x 2 + 4 z 2 = y 2

c.

z

z2

8. z 2 + x 2 − y 2 = 1

y2

a.

y

−1

x

y z

l.

y x

y

x y

11.6  Cylinders and Quadric Surfaces

719

z

Drawing

Sketch the surfaces in Exercises 13–44. CYLINDERS

r

h

13.

x2

+

y2

15.

x2

+

4z 2

= 4

14. z =

= 16

16.

4x 2

−1

y2

+ y 2 = 36

R y

ELLIPSOIDS

17. 9 x 2 + y 2 + z 2 = 9 19.

4x 2

+

+

9y 2

4z 2

18. 4 x 2 + 4 y 2 + z 2 = 16

= 36

−h

x

20. 9 x 2 + 4 y 2 + 36 z 2 = 36

r

PARABOLOIDS AND CONES

21. z = x 2 + 4 y 2 23. x = 4 − 25.

x2

+

4y2 =

y2

22. z = 8 − x 2 − y 2

−

47. Show that the volume of the segment cut from the paraboloid

24. y = 1 − x 2 − z 2

z2

z2

26.

4x 2

+

9z 2

=

y2 x2 z + 2 = 2 a b c

9y 2

by the plane z = h equals half the segment’s base times its altitude.

HYPERBOLOIDS

27. x 2 + y 2 − z 2 = 1 29.

z2

−

x2

−

y2

48. a. Find the volume of the solid bounded by the hyperboloid

28. y 2 + z 2 − x 2 = 1

=1

30.

( y2

4) −

(x2

4) −

z2

=1

y2 x2 z2 + 2 − 2 =1 2 a b c

HYPERBOLIC PARABOLOIDS

31. y 2 − x 2 = z

and the planes z = 0 and z = h,   h > 0.

32. x 2 − y 2 = z 34. 4 x 2 + 4 y 2 = z 2

b. Express your answer in part (a) in terms of h and the areas A0 and Ah of the regions cut by the hyperboloid from the planes z = 0 and z = h.

36. 16 x 2 + 4 y 2 = 1

c. Show that the volume in part (a) is also given by the formula

ASSORTED

33. z = 1 + y 2 − x 2 35. y = 37.

x2

−( x 2

+

y2

+

−

z2

z2) = 4

38.

x2

+

z2

= y

39. x 2 + z 2 = 1

40. 16 y 2 + 9 z 2 = 4 x 2

41. z = −( x 2 + y 2 )

42. y 2 − x 2 − z 2 = 1

43.

4y2

+

z2

−

4x 2

= 4

V =

h( A + 4 A m + Ah ) , 6 0

where Am is the area of the region cut by the hyperboloid from the plane z = h 2.

44. x 2 + y 2 = z

Viewing Surfaces

Theory and Examples

45. a. Express the area A of the cross-section cut from the ellipsoid x2 +

y2 z2 + =1 4 9

by the plane z = c as a function of c. (The area of an ellipse with semiaxes a and b is πab.) b. Use slices perpendicular to the z-axis to find the volume of the ellipsoid in part (a).

T Plot the surfaces in Exercises 49–52 over the indicated domains. If you can, rotate the surface into different viewing positions. 49. z = y 2 , −2 ≤ x ≤ 2, −0.5 ≤ y ≤ 2 50. z = 1 − y 2 , −2 ≤ x ≤ 2, −2 ≤ y ≤ 2 51. z = x 2 + y 2 , −3 ≤ x ≤ 3, −3 ≤ y ≤ 3 52. z = x 2 + 2 y 2

b. −1 ≤ x ≤ 1, −2 ≤ y ≤ 3

c. Now find the volume of the ellipsoid

c. −2 ≤ x ≤ 2, −2 ≤ y ≤ 2

y2 x2 z2 + 2 + 2 = 1. a2 b c Does your formula give the volume of a sphere of radius a if a = b = c? 46. The barrel shown here is shaped like an ellipsoid with equal pieces cut from the ends by planes perpendicular to the z-axis. The crosssections perpendicular to the z-axis are circular. The barrel is 2h units high, its midsection radius is R, and its end radii are both r. Find a formula for the barrel’s volume. Then check two things. First, suppose the sides of the barrel are straightened to turn the barrel into a cylinder of radius R and height 2h. Does your formula give the cylinder’s volume? Second, suppose r = 0 and h = R so the barrel is a sphere. Does your formula give the sphere’s volume?

over

a. −3 ≤ x ≤ 3, −3 ≤ y ≤ 3

d. −2 ≤ x ≤ 2, −1 ≤ y ≤ 1 COMPUTER EXPLORATIONS

Use a CAS to plot the surfaces in Exercises 53–58. Identify the type of quadric surface from your graph. 53.

y2 x2 z2 + = 1− 9 36 25

55. 5 x 2 = z 2 − 3 y 2 57.

y2 x2 z2 −1 = + 9 16 2

y2 x2 z2 − = 1− 9 9 16 y2 x2 56. = 1− +z 16 9

54.

58. y −

4 − z2 = 0

720

Chapter 11 Vectors and the Geometry of Space

CHAPTER 11

Questions to Guide Your Review

1. When do directed line segments in the plane represent the same vector? 2. How are vectors added and subtracted geometrically? How are they added and subtracted algebraically? 3. How do you find a vector’s magnitude and direction? 4. If a vector is multiplied by a positive scalar, how is the result related to the original vector? What if the scalar is zero? Negative? 5. Define the dot product (scalar product) of two vectors. Which algebraic laws are satisfied by dot products? Give examples. When is the dot product of two vectors equal to zero? 6. What geometric interpretation does the dot product have? Give examples. 7. What is the vector projection of a vector u onto a vector v? Give an example of a useful application of a vector projection. 8. Define the cross product (vector product) of two vectors. Which algebraic laws are satisfied by cross products, and which are not? Give examples. When is the cross product of two vectors equal to zero?

10. What is the determinant formula for calculating the cross product of two vectors relative to the Cartesian i, j, k-coordinate system? Use it in an example. 11. How do you find equations for lines, line segments, and planes in space? Give examples. Can you express a line in space by a single equation? A plane? 12. How do you find the distance from a point to a line in space? From a point to a plane? Give examples. 13. What are box products? What significance do they have? How are they evaluated? Give an example. 14. How do you find equations for spheres in space? Give examples. 15. How do you find the intersection of two lines in space? A line and a plane? Two planes? Give examples. 16. What is a cylinder? Give examples of equations that define cylinders in Cartesian coordinates. 17. What are quadric surfaces? Give examples of different kinds of ellipsoids, paraboloids, cones, and hyperboloids (equations and sketches).

9. What geometric or physical interpretations do cross products have? Give examples.

CHAPTER 11

Practice Exercises

Vector Calculations in Two Dimensions In Exercises 1–4, let u = 〈−3, 4〉 and v = 〈2, −5〉. Find (a) the component form of the vector and (b) its magnitude.

1. 3u − 4 v

2. u + v

3. −2u

4. 5v

In Exercises 5–8, find the component form of the vector. 5. The vector obtained by rotating 〈0, 1〉 through an angle of 2π 3 radians 6. The unit vector that makes an angle of π 6 radian with the positive x-axis 7. The vector 2 units long in the direction 4 i − j 8. The vector 5 units long in the direction opposite to the direction of (3 5)i + ( 4 5) j Express the vectors in Exercises 9–12 in terms of their lengths and directions. 9.

2i +

2j

10. −i − j

11. Velocity vector v = (−2 sin t ) i + ( 2 cos t ) j when t = π 2. 12. Velocity vector v = ( e t cos t − e t sin t ) i + ( e t sin t + e t cos t ) j when t = ln 2. Vector Calculations in Three Dimensions Express the vectors in Exercises 13 and 14 in terms of their lengths and directions.

13. 2 i − 3 j + 6k

14. i + 2 j − k

15. Find a vector 2 units long in the direction of v = 4 i − j + 4 k. 16. Find a vector 5 units long in the direction opposite to the direction of v = ( 3 5 ) i + ( 4 5 ) k. In Exercises 17 and 18, find v ,   u ,   v ⋅ u,   u ⋅ v,   v × u,   u × v, v × u , the angle between v and u, the scalar component of u in the direction of v, and the vector projection of u onto v. 17. v = i + j u = 2 i + j − 2k 18. v = i + j + 2k u = −i − k In Exercises 19 and 20, find proj v u. 19. v = 2 i + j − k u = i + j − 5k 20. u = i − 2 j v =i+ j+k In Exercises 21 and 22, draw coordinate axes and then sketch u, v, and u × v as vectors at the origin. 21. u = i, v = i + j

22. u = i − j, v = i + j

23. If v = 2,   w = 3, and the angle between v and w is π 3, find v − 2w . 24. For what value or values of a will the vectors u = 2 i + 4 j − 5k and v = −4 i − 8 j + a k be parallel?

Chapter 11  Practice Exercises

In Exercises 25 and 26, find (a) the area of the parallelogram determined by vectors u and v and (b) the volume of the parallelepiped determined by the vectors u, v, and w. 25. u = i + j − k, v = 2 i + j + k, w = −i − 2 j + 3k 26. u = i + j, v = j, w = i + j + k Lines, Planes, and Distances

27. Suppose that n is normal to a plane and that v is parallel to the plane. Describe how you would find a vector n that is both perpendicular to v and parallel to the plane.

721

46. Find an equation for the plane that passes through the point (1, 2, 3 ) parallel to u = 2 i + 3 j + k and v = i − j + 2k. 47. Is v = 2 i − 4 j + k related in any special way to the plane 2 x + y = 5? Give reasons for your answer.  plane 48. The equation n ⋅ P0 P = 0 represents the  through P0 normal to n. What set does the inequality n ⋅ P0 P > 0 represent? 49. Find the distance from the point P (1, 4, 0 ) to the plane through A( 0, 0, 0 ),   B ( 2, 0, −1), and C ( 2, −1, 0 ). 50. Find the distance from the point 2 x + 3 y + 5 z = 0.

( 2, 2, 3 )

to the plane

28. Find a vector in the plane parallel to the line ax + by = c.

51. Find a vector parallel to the plane 2 x − y − z = 4 and orthogonal to i + j + k.

In Exercises 29 and 30, find the distance from the point to the line.

52. Find a unit vector orthogonal to A in the plane of B and C if A = 2 i − j + k,   B = i + 2 j + k, and C = i + j − 2k.

29. ( 2,  2,  0 );   x = −t , y = t , z = −1 + t 30. ( 0, 4, 1);   x = 2 + t , y = 2 + t , z = t 31. Parametrize the line that passes through the point to the vector v = −3i + 7k.

(1, 2, 3 ) parallel

32. Parametrize the line segment joining the points P (1, 2, 0 ) and Q (1, 3, −1). In Exercises 33 and 34, find the distance from the point to the plane. 33. ( 6, 0, −6 ), x − y = 4 34. ( 3, 0, 10 ), 2 x + 3 y + z = 2 35. Find an equation for the plane that passes through the point ( 3, − 2, 1) normal to the vector n = 2 i + j + k. 36. Find an equation for the plane that passes through the point ( − 1, 6, 0 ) perpendicular to the line x = − 1 + t ,   y = 6 − 2t , z = 3t. In Exercises 37 and 38, find an equation for the plane through points P, Q, and R. 37. P (1, −1, 2 ), Q ( 2, 1, 3 ), R (−1, 2, −1) 38. P (1, 0, 0 ), Q ( 0, 1, 0 ), R ( 0, 0, 1) 39. Find the points in which the line x = 1 + 2t ,   y = −1 − t , z = 3t meets the three coordinate planes. 40. Find the point in which the line through the origin perpendicular to the plane 2 x − y − z = 4 meets the plane 3 x − 5 y + 2 z = 6. 41. Find the acute angle between the planes x = 7 and x + y + 2z = −3. 42. Find the acute angle between the planes x + y = 1 and y + z = 1. 43. Find parametric equations for the line in which the planes x + 2 y + z = 1 and x − y + 2 z = −8 intersect. 44. Show that the line in which the planes x + 2 y − 2 z = 5 and 5 x − 2 y − z = 0 intersect is parallel to the line x = −3 + 2t , y = 3t , z = 1 + 4 t. 45. The planes 3 x + 6 z = 1 and 2 x + 2 y − z = 3 intersect in a line. a. Show that the planes are orthogonal. b. Find equations for the line of intersection.

53. Find a vector of magnitude 2 parallel to the line of intersection of the planes x + 2 y + z − 1 = 0 and x − y + 2 z + 7 = 0. 54. Find the point in which the line through the origin perpendicular to the plane 2 x − y − z = 4 meets the plane 3 x − 5 y + 2 z = 6. 55. Find the point in which the line through P ( 3, 2, 1) normal to the plane 2 x − y + 2 z = −2 meets the plane. 56. What angle does the line of intersection of the planes 2 x + y − z = 0 and x + y + 2 z = 0 make with the positive x-axis? 57. The line L : x = 3 + 2t , y = 2t , z = t intersects the plane x + 3 y − z = −4 in a point P. Find the coordinates of P and find equations for the line in the plane through P perpendicular to L. 58. Show that for every real number k, the plane x − 2 y + z + 3 + k ( 2 x − y − z + 1) = 0 contains the line of intersection of the planes x − 2 y + z + 3 = 0 and 2 x − y − z + 1 = 0. 59. Find an equation for the plane through A(−2, 0, −3 ) and B (1, −2, 1) that lies parallel to the line through C (−2, −13 5, 26 5 ) and D (16 5, −13 5, 0 ). 60. Is the line x = 1 + 2t ,   y = −2 + 3t ,   z = −5t related in any way to the plane −4 x − 6 y + 10 z = 9? Give reasons for your answer. 61. Which of the following are equations for the plane through the points P (1, 1, −1),   Q ( 3, 0, 2 ), and R (−2, 1, 0 )? a. ( 2 i − 3 j + 3k ) ⋅ (( x + 2 ) i + ( y − 1) j + zk ) = 0 b. x = 3 − t , y = −11t , z = 2 − 3t c.

(x

+ 2 ) + 11( y − 1) = 3z

d. ( 2 i − 3 j + 3k ) × (( x + 2 ) i + ( y − 1) j + zk ) = 0 e. ( 2 i − j + 3k ) × (−3i + k ) ⋅ (( x + 2 ) i + ( y − 1) j + z k ) = 0

722

Chapter 11 Vectors and the Geometry of Space

62. The parallelogram shown here has vertices at A( 2, −1, 4 ), B (1, 0, −1),   C (1, 2, 3 ), and D. Find z D

63. Distance between skew lines Find the distance between the line L1 through the points A(1, 0, −1) and B (−1, 1, 0 ) and the line L 2 through the points C ( 3, 1, −1) and D ( 4, 5, −2 ). The distance is to be measured along the line perpendicular to the two lines. First

find a vector n perpendicular to both lines. Then project AC onto n. 64. (Continuation of Exercise 63.) Find the distance between the line through A( 4, 0, 2 ) and B ( 2, 4, 1) and the line through C (1, 3, 2 ) and D ( 2, 2, 4 ).

A(2, −1, 4)

Quadric Surfaces Identify and sketch the surfaces in Exercises 65–76.

C(1, 2, 3)

65. x 2 + y 2 + z 2 = 4 66. x 2 + ( y − 1) 2 + z 2 = 1 67. 4 x 2 + 4 y 2 + z 2 = 4

y x

68. 36 x 2 + 9 y 2 + 4 z 2 = 36

B(1, 0, −1)

69. z = −( x 2 + y 2 ) 70. y = −( x 2 + z 2 )

a. the coordinates of D.

71. x 2 + y 2 = z 2

b. the cosine of the interior angle at B.



c. the vector projection of BA onto BC.

72. x 2 + z 2 = y 2

d. the area of the parallelogram.

73. x 2 + y 2 − z 2 = 4

e. an equation for the plane of the parallelogram.

74. 4 y 2 + z 2 − 4 x 2 = 4

f. the areas of the orthogonal projections of the parallelogram on the three coordinate planes.

CHAPTER 11

76. z 2 − x 2 − y 2 = 1

Additional and Advanced Exercises

1. Submarine hunting Two surface ships on maneuvers are trying to determine a submarine’s course and speed to prepare for an aircraft intercept. As shown here, ship A is located at ( 4, 0, 0 ), whereas ship B is located at ( 0, 5, 0 ). All coordinates are given in thousands of meters. Ship A locates the submarine in the direction of the vector 2 i + 3 j − (1 3 ) k, and ship B locates it in the direction of the vector 18 i  6 j  k. Four minutes ago, the submarine was located at ( 2, −1, −1 3 ). The aircraft is due in 20 min. Assuming that the submarine moves in a straight line at a constant speed, to what position should the surface ships direct the aircraft?

2. A helicopter rescue Two helicopters, H 1 and H 2 , are traveling together. At time t  0, they separate and follow different straight-line paths given by H 1 : x = 6 + 40 t , y = −3 + 10 t , z = −3 + 2t H 2 : x = 6 + 110 t , y = −3 + 4 t , z = −3 + t. Time t is measured in hours, and all coordinates are measured in kilometers. Due to system malfunctions, H 2 stops its flight at ( 446, 13, 1) and, in a negligible amount of time, lands at ( 446, 13, 0 ). Two hours later, H 1 is advised of this fact and heads toward H 2 at 150 kmh. How long will it take H 1 to reach H 2? 3. Torque The operator’s manual for the Toro® 53-cm lawnmower says, “tighten the spark plug to 20.4 N ¸ m.” If you are installing the plug with a 26.5-cm socket wrench that places the center of your hand 23 cm from the axis of the spark plug, about how hard should you pull? Answer in newtons.

z

Ship A

75. y 2 − x 2 − z 2 = 1

Ship B (4, 0, 0)

x

NOT TO SCALE

(0, 5, 0)

y

Submarine

23 cm

Chapter 11  Additional and Advanced Exercises

4. Rotating body The line through the origin and the point A(1, 1, 1) is the axis of rotation of a rigid body rotating with a constant angular speed of 3 2 rad s. The rotation appears to be clockwise when we look toward the origin from A. Find the velocity v of the point of the body that is at the position B (1, 3, 2 ).

723

7. Determinants and planes a. Show that

z

x1 − x

y1 − y

z1 − z

x2 − x

y2 − y

z2 − z

x3 − x

y3 − y

z3 − z

= 0

is an equation for the plane through the three noncollinear points P1 ( x 1 ,  y1 ,  z 1 ),   P2 ( x 2 ,  y 2 ,  z 2 ), and P3 ( x 3 ,  y 3 ,  z 3 ). B(1, 3, 2)

A(1, 1, 1) O

b. What set of points in space is described by the equation

1

1 v

x

3 y

x

y

z

1

x1

y1

z1

1

x2

y2

z2

1

x3

y3

z3

1

 0?

8. Determinants and lines Show that the lines 5. Consider the weight suspended by two wires in each diagram. Find the magnitudes and components of vectors F1 and F2 , and angles B and C. a. 5m a

F2

F1

3m

b

x = a1s + b1 , y = a 2 s + b 2 , z = a 3s + b3 , −∞ < s < ∞ and x = c1t + d1 , y = c 2 t + d 2 , z = c 3t + d 3 , −∞ < t < ∞ intersect or are parallel if and only if

4m 100 N

b.

13 m

5m

F1

F2

c1

b1 − d1

a2

c2

b 2 − d 2 = 0.

a3

c3

b3 − d 3

9. Consider a regular tetrahedron of side length 2. a. Use vectors to find the angle R formed by the base of the tetrahedron and any one of its other edges.

b

a

a1

12 m

D 200 N

2

2

(Hint: This triangle is a right triangle.)

u

6. Consider a weight of w N suspended by two wires in the diagram, where T1 and T2 are force vectors directed along the wires. a

C

A

1 P

2

1 B

b T2

T1 b

a w

a. Find the vectors T1 and T2 and show that their magnitudes are T1 =

b. Use vectors to find the angle R formed by any two adjacent faces of the tetrahedron. This angle is commonly referred to as a dihedral angle. 10. In the figure here, D is the midpoint of side AB of triangle ABC, and E is one-third of the way between C and B. Use vectors to prove that F is the midpoint of line segment CD. C

w cos β sin ( α + β )

E

and

F

T2 =

w cos α . sin ( α + β )

b. For a fixed C , determine the value of B that minimizes the magnitude T1 . c. For a fixed B, determine the value of C that minimizes the magnitude T2 .

A

D

B

11. Use vectors to show that the distance from P1 ( x 1 ,   y1 ) to the line ax + by = c is d =

ax 1 + by1 − c . a2 + b2

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Chapter 11 Vectors and the Geometry of Space

17. Triple vector products The triple vector products ( u × v ) × w and u × ( v × w ) are usually not equal, although the formulas for evaluating them from components are similar:

12. a. Use vectors to show that the distance from P1 ( x 1 ,  y1 ,  z 1 ) to the plane Ax + By + Cz = D is d =

Ax 1 + By1 + Cz 1 − D . A2 + B2 + C 2

(u

u × ( v × w ) = ( u ⋅ w ) v − ( u ⋅ v ) w.

b. Find an equation for the sphere that is tangent to the planes x + y + z = 3 and x + y + z = 9 if the planes 2 x − y = 0 and 3 x − z = 0 pass through the center of the sphere.

Verify each formula for the following vectors by evaluating its two sides and comparing the results. u

13. a. Distance between parallel planes Show that the distance between the parallel planes Ax + By + Cz = D1 and Ax + By + Cz = D 2 is d =

D1 − D 2 . Ai + Bj + Ck

b. Find the distance between the planes 2 x + 3 y − z = 6 and 2 x + 3 y − z = 12.

× v ) × w = ( u ⋅ w ) v − ( v ⋅ w ) u.

v

w

a. 2i

2j

2k

b. i − j + k

2 i + j − 2k

−i + 2 j − k

c. 2 i + j

2i − j + k

i + 2k

d. i + j − 2k

−i − k

2 i + 4 j − 2k

18. Cross and dot products Show that if u, v, w, and r are any vectors, then a. u × ( v × w ) + v × ( w × u ) + w × ( u × v ) = 0

c. Find an equation for the plane parallel to the plane 2 x − y + 2 z = −4 if the point ( 3, 2, −1) is equidistant from the two planes.

b. u × v = ( u ⋅ v × i ) i + ( u ⋅ v × j ) j + ( u ⋅ v × k ) k c. ( u × v ) ⋅ ( w × r ) =

d. Write equations for the planes that lie parallel to, and 5 units away from, the plane x − 2 y + z = 3. 14. Prove that four points A, B, C, and  D are  coplanar (lie in a common plane) if and only if AD ⋅ ( AB × BC ) = 0. 15. The projection of a vector on a plane Let P be a plane in space and let v be a vector. The vector projection of v onto the plane P, proj P v, can be defined informally as follows. Suppose the sun is shining so that its rays are normal to the plane P. Then proj P v is the “shadow” of v onto P. If P is the plane x + 2 y + 6 z = 6 and v = i + j + k, find proj P v.

u⋅w u⋅r

v⋅w . v⋅r

19. Cross and dot products Prove or disprove the formula u × ( u × ( u × v )) ⋅ w = − u 2 u ⋅ v × w. 20. By forming the cross product of two appropriate vectors, derive the trigonometric identity sin ( A − B ) = sin A cos B − cos A sin B. 21. Use vectors to prove that ( a 2 + b 2 )( c 2 + d 2 ) ≥ ( ac + bd ) 2

16. The accompanying figure shows nonzero vectors v, w, and z, with z orthogonal to the line L, and v and w making equal angles β with L. Assuming v = w , find w in terms of v and z.

for any four numbers a, b, c, and d. (Hint: Let u = a i + b j and v = ci + d j.) 22. Dot multiplication is positive definite Show that dot multiplication of vectors is positive definite; that is, show that u ⋅ u ≥ 0 for every vector u and that u ⋅ u = 0 if and only if u = 0.

z

23. Show that u + v ≤ u + v for any vectors u and v. v b

CHAPTER 11

24. Show that w = v u + u v bisects the angle between u and v.

w b

25. Show that v u + u v and v u − u v are orthogonal. L

Technology Application Projects

Mathematica/Maple Projects

Projects can be found within MyLab Math. • Using Vectors to Represent Lines and Find Distances Parts I and II: Learn the advantages of interpreting lines as vectors. Part III: Use vectors to find the distance from a point to a line. • Putting a Scene in Three Dimensions onto a Two-Dimensional Canvas Use the concept of planes in space to obtain a two-dimensional image. • Getting Started in Plotting in 3D Part I: Use the vector definition of lines and planes to generate graphs and equations, and to compare different forms for the equations of a single line. Part II: Plot functions that are defined implicitly.

12

Vector-Valued Functions and Motion in Space OVERVIEW In this chapter we introduce the calculus of vector-valued functions. The domains of these functions are sets of real numbers, as before, but their ranges consist of vectors instead of scalars. When a vector-valued function changes, the change can occur in both magnitude and direction, so the derivative is itself a vector. The integral of a vectorvalued function is also a vector. We use the calculus of these functions to describe the paths and motions of objects moving in a plane or in space, so their velocities and accelerations are given by vectors.

12.1 Curves in Space and Their Tangents z

When a particle moves through space during a time interval I, we think of the particle’s coordinates as functions defined on I: x = f (t ), r

P( f (t), g(t), h(t))

O y x

FIGURE 12.1

The position vector

r  OP of a particle moving through space is a function of time.

y = g(t ),

z = h(t ),

t ∈ I.

(1)

The points ( x , y, z ) = ( f (t ), g(t ), h(t )), t ∈ I , make up the curve in space that we call the particle’s path. The equations and interval in Equation (1) parametrize the curve. A curve in space can also be represented in vector form. The vector

r (t ) = OP = f (t ) i + g(t ) j + h(t )k (2) from the origin to the particle’s position P ( f (t ), g(t ), h(t )) at time t is the particle’s position vector (Figure 12.1). The functions f, g, and h are the component functions (or components) of the position vector. We think of the particle’s path as the curve traced by r during the time interval I. Figure 12.2 displays several space curves generated by a computer graphing program. Equation (2) defines r as a vector function of the real variable t on the interval I. More generally, a vector-valued function or vector function on a domain set D is a rule that assigns a vector in space to each element in D. For now, the domains will be intervals of real numbers, and the graph of the function represents a curve in space. Vector functions on a domain in the plane or in space give rise to “vector fields,” which are important to the study of fluid flows, gravitational fields, and electromagnetic phenomena. We investigate vector fields and their applications in Chapter 15. Real-valued functions are often called scalar functions to distinguish them from vector functions. The components of r in Equation (2) are scalar functions of t. The domain of a vector-valued function is the common domain of its components.

725

726

Chapter 12 Vector-Valued Functions and Motion in Space

z

z

z

y

x

x

y

x

y r(t) = (sin 3t)(cos t)i + (sin3t)(sin t)j + tk z

(a)

FIGURE 12.2

EXAMPLE 1

2p

Space curves are defined by the position vectors r(t ).

Graph the vector function r (t ) = ( cos t ) i + ( sin t ) j + t k.

r

0 t t=0

r(t) = (4 + sin 20t)(cos t)i + (4 + sin 20t)(sint)j + (cos20t)k (c)

(b)

t=p

t = 2p

(1, 0, 0)

r(t) = (cos t)i + (sin t)j + (sin 2t)k

P

Solution This vector function r(t ) is defined for all real values of t. The curve traced by r winds around the circular cylinder x 2 + y 2 = 1 (Figure 12.3). The curve lies on the cylinder because the i- and j-components of r, being the x- and y-coordinates of the tip of r, satisfy the cylinder’s equation:

t=p 2

x2 + y2 = 1

x 2 + y 2 = ( cos t ) 2 + ( sin t ) 2 = 1. y

x

The curve rises as the k-component z = t increases. Each time t increases by 2π, the curve completes one turn around the cylinder. The curve is called a helix (from an ancient Greek word for “spiral”). The equations x = cos t ,

FIGURE 12.3

The helix r (t ) = ( cos t ) i + ( sin t ) j + t k (Example 1).

y = sin t ,

parametrize the helix. The domain is the largest set of points t for which all three equations are defined, or −∞ < t < ∞ for this example. Figure 12.4 shows more helices. z

z

x

z

x

x y r(t) = (cos t)i + (sin t)j + tk

FIGURE 12.4

z = t

y r(t) = (cos t)i + (sin t)j + 0.3tk

y r(t) = (cos 5t)i + (sin 5t)j + tk

Helices spiral upward around a cylinder, like coiled springs.

Limits and Continuity The way we define limits of vector-valued functions is similar to the way we define limits of real-valued functions.

12.1  Curves in Space and Their Tangents

727

Let r (t ) = f (t ) i + g(t ) j + h(t )k be a vector function with domain D, and let L be a vector. We say that r has limit L as t approaches t 0 and write lim r (t ) = L

DEFINITION

t →t0

if, for every number ε > 0, there exists a corresponding number δ > 0 such that, for all t ∈ D, r (t ) − L < ε whenever 0 < t − t 0 < δ. If L = L1i + L 2 j + L 3k, then it can be shown that lim r (t ) = L precisely when t →t0

lim f (t ) = L1 ,

lim g(t ) = L 2 ,

t →t0

t →t0

and

lim h(t ) = L 3 .

t →t0

We omit the proof. The equation To calculate the limit of a vector function, we find the limit of each component scalar function.

(

) (

) (

)

lim r (t ) = lim f (t ) i + lim g(t ) j + lim h(t ) k

t →t0

t →t0

t →t0

t →t0

(3)

provides a practical way to calculate limits of vector functions. If r (t ) = ( cos t ) i + ( sin t ) j + t k, then

EXAMPLE 2

lim r (t ) =

t→π 4

( lim cos t ) i + ( lim sin t ) j + ( lim t )k

=

t→π 4

t→π 4

t→π 4

2 π 2 i+ j + k. 2 2 4

We define continuity for vector functions the same way we define continuity for scalar functions defined over an interval. DEFINITION A vector function r(t ) is continuous at a point t = t 0 in its

domain if lim r (t ) = r (t 0 ). The function is continuous if it is continuous at t →t0

every point in its domain. From Equation (3), we see that r(t ) is continuous at t = t 0 if and only if each component function is continuous there (Exercise 45). EXAMPLE 3 (a) All the space curves shown in Figures 12.2 and 12.4 are continuous because their component functions are continuous at every value of t in (−∞, ∞ ). (b) The function g(t ) = ( cos t ) i + ( sin t ) j + ⎣ t ⎦ k is discontinuous at every integer, because the greatest integer function ⎣ t ⎦ is discontinuous at every integer.

Derivatives and Motion Suppose that r (t ) = f (t ) i + g(t ) j + h(t )k is the position vector of a particle moving along a curve in space and that f, g, and h are differentiable functions of t. Then the difference between the particle’s positions at time t and time t + Δt is the vector Δr = r ( t + Δt ) − r(t )

728

Chapter 12 Vector-Valued Functions and Motion in Space

z

(Figure 12.5a). In terms of components,

P

Δr = r ( t + Δt ) − r (t )

r(t + Δt) − r(t) Q

= [ f ( t + Δt ) i + g ( t + Δt ) j + h ( t + Δt ) k ] − [ f ( t ) i + g( t ) j + h( t ) k ] = [ f ( t + Δt ) − f (t ) ] i + [ g ( t + Δt ) − g(t ) ] j + [ h ( t + Δt ) − h(t ) ] k.

r(t) r(t + Δt) C O

y

As Δt approaches zero, three things seem to happen simultaneously. First, Q approaches P along the curve. Second, the secant line PQ seems to approach a limiting position tangent to the curve at P. Third, the quotient Δr Δt (Figure 12.5b) approaches the limit

(a) x

lim

Δt → 0

z

+ ⎡⎢ lim ⎣ Δt → 0

r′(t)

Q

r(t) r(t + Δt) − r(t) Δr = Δt Δt r(t + Δt) y

O

h ( t + Δt ) − h(t ) ⎤ ⎥⎦ k Δt

⎡df ⎤ dg dh = ⎢ ⎥ i + ⎡⎢ ⎤⎥ j + ⎡⎢ ⎤⎥ k. ⎢⎣ dt ⎥⎦ ⎣ dt ⎦ ⎣ dt ⎦

P

C

f ( t + Δt ) − f ( t ) ⎤ g ( t + Δt ) − g(t ) ⎤ Δr ⎡ = ⎡⎢ lim ⎥j ⎥ i + ⎢ Δlim Δt Δt Δt ⎣ t→0 ⎦ ⎣ Δt → 0 ⎦

(b) x

FIGURE 12.5 As Δt → 0, the point Q approaches the point P along the curve C.  In the limit, the vector PQ Δt becomes the tangent vector r ′(t ).

These observations lead us to the following definition.

DEFINITION The vector function r (t ) = f (t ) i + g(t ) j + h(t )k has a derivative (is differentiable) at t if f, g, and h have derivatives at t. The derivative is the vector function

r ′(t ) =

df dg r ( t + Δt ) − r (t ) dr dh = lim = i+ j+ k. Δt → 0 Δt dt dt dt dt

A vector function r is differentiable if it is differentiable at every point of its domain.

C1 C2

C3

C4 C5

FIGURE 12.6 A piecewise smooth curve made up of five smooth curves connected end to end in a continuous fashion. The curve here is not smooth at the points joining the five smooth curves.

The geometric significance of the definition of derivative is shown in Figure 12.5. The  points P and Q have position vectors r(t ) and r ( t + Δt ) , and the vector PQ is represented by r ( t + Δt ) − r(t ). For Δt > 0, the scalar multiple (1 Δt )( r ( t + Δt ) − r (t ) ) points in the same direction as the vector PQ. As Δt → 0, this vector approaches the vector r ′(t ), which is a vector tangent to the curve at P, as long as it is different from the zero vector 0 (Figure 12.5b). The curve traced by r is smooth if d r dt is continuous and never 0, that is, if f, g, and h have continuous first derivatives that are not simultaneously 0. We require d r dt ≠ 0 for a smooth curve to make sure the curve has a continuously turning tangent at each point. On a smooth curve, there are no sharp corners or cusps. A curve that is made up of a finite number of smooth curves pieced together in a continuous fashion is called piecewise smooth (Figure 12.6). Look once again at Figure 12.5. We drew the figure for Δt positive, so Δr points forward, in the direction of the motion. The vector Δr Δt , having the same direction as Δr, points forward too. Had Δt been negative, Δr would have pointed backward, against the direction of motion. The quotient Δr Δt , however, being a negative scalar multiple of Δr, would once again have pointed forward. No matter how Δr points, Δr Δt points forward, and we expect the vector d r dt = lim Δr Δt , when different from 0, to do the same. Δt → 0

This means that the derivative d r dt, which is the rate of change of position with respect to time, always points in the direction of motion. For a smooth curve, d r dt is never zero; the particle does not stop or reverse direction.

729

12.1  Curves in Space and Their Tangents

DEFINITIONS If r is the position vector of a particle moving along a smooth curve in space, then

v(t ) =

dr dt

is the particle’s velocity vector. If v is a nonzero vector, then it is tangent to the curve, and its direction is the direction of motion. The magnitude of v is the particle’s speed, and the derivative a = d v dt , when it exists, is the particle’s acceleration vector. In summary, v =

1. Velocity is the derivative of position: 2. Speed is the magnitude of velocity:

dr . dt

Speed = v .

3. Acceleration is the derivative of velocity:

a =

dv d 2r . = dt dt 2

4. The unit vector v v is the direction of motion at time t.

EXAMPLE 4 Find the velocity, speed, and acceleration of a particle whose motion in space is given by the position vector r (t ) = 2 cos t i + 2 sin t j + 5 cos 2 t k. Sketch the velocity vector v ( 7π 4 ). Solution The velocity and acceleration vectors at time t are

v(t ) = r ′(t ) = −2 sin t i + 2 cos t j − 10 cos t sin t k z

= −2 sin t i + 2 cos t j − 5 sin 2t k,

7p r′ a b 4

a (t ) = r ′′(t ) = −2 cos t i − 2 sin t j − 10 cos 2t k, y

and the speed is v(t ) =

t = 7p 4

(−2 sin t ) 2 + ( 2 cos t ) 2 + (−5 sin 2t ) 2 =

4 + 25 sin 2 2t .

When t = 7π 4 , we have x

v FIGURE 12.7 The curve and the velocity vector when t = 7π 4 for the motion given in Example 4.

( 74π ) =

2i +

2 j + 5k ,

a

( 74π ) = −

2i +

2 j,

v

( 74π )

=

29.

A sketch of the curve of motion, and the velocity vector when t = 7π 4 , can be seen in Figure 12.7.

We can express the velocity of a moving particle as the product of its speed and direction: Velocity = v

( vv ) =

( speed )( direction ).

Differentiation Rules Because the derivatives of vector functions may be computed component by component, the rules for differentiating vector functions have the same form as the rules for differentiating scalar functions.

730

Chapter 12 Vector-Valued Functions and Motion in Space

Differentiation Rules for Vector Functions

Let u and v be differentiable vector functions of t, C a constant vector, c any scalar, and f any differentiable scalar function. 1. Constant Function Rule: 2. Scalar Multiple Rules:

3. Sum Rule: 4. Difference Rule: 5. Dot Product Rule:

When you use the Cross Product Rule, remember to preserve the order of the factors. If u comes first on the left side of the equation, it must also come first on the right, or the signs will be wrong.

6. Cross Product Rule: 7. Chain Rule:

d C = 0 dt d[ cu(t ) ] = cu′(t ) dt d[ f (t )  u(t ) ] = f ′(t )  u(t ) + f (t )  u′(t ) dt d[ u(t ) + v(t ) ] = u′(t ) + v′(t ) dt d[ u(t ) − v(t ) ] = u′(t ) − v′(t ) dt d[ u(t ) ⋅ v(t ) ] = u′(t ) ⋅ v(t ) + u(t ) ⋅ v′(t ) dt d[ u(t ) × v(t ) ] = u′(t ) × v(t ) + u(t ) × v′(t ) dt d [ u( f (t )) ] = f ′(t ) u′( f (t )) dt

We will prove the product rules and the Chain Rule but will leave the rules for constants, scalar multiples, sums, and differences as exercises. Proof of the Dot Product Rule Suppose that u = u1 (t ) i + u 2 (t ) j + u 3 (t ) k and v = υ1 (t ) i + υ 2 (t ) j + υ 3 (t ) k. Then d( d u ⋅ v ) = ( u 1υ 1 + u 2 υ 2 + u 3 υ 3 ) dt dt = u1′ υ1 + u′2 υ 2 + u′3υ 3 + u1υ1′ + u 2 υ′2 + u 3υ′3 .   u′ ⋅ v u ⋅ v′ Proof of the Cross Product Rule We model the proof after the proof of the Product Rule for scalar functions. According to the definition of derivative, u ( t + h ) × v ( t + h ) − u (t ) × v(t ) d( . u × v ) = lim h → 0 dt h To change this fraction into an equivalent one that contains the difference quotients for the derivatives of u and v, we subtract and add u(t ) × v ( t + h ) in the numerator. Then d( u × v) dt = lim

h→0

u ( t + h ) × v ( t + h ) − u (t ) × v ( t + h ) + u (t ) × v ( t + h ) − u (t ) × v(t ) h

u ( t + h ) − u (t ) v ( t + h ) − v(t ) ⎤ = lim ⎡⎢ × v ( t + h ) + u (t ) × ⎥⎦ h→0 ⎣ h h = lim

h→0

u ( t + h ) − u (t ) v ( t + h ) − v(t ) . × lim v ( t + h ) + lim u(t ) × lim h→0 h→0 h→0 h h

731

12.1  Curves in Space and Their Tangents

The last of these equalities holds because the limit of the cross product of two vector functions is the cross product of their limits if the latter exist (Exercise 46). As h approaches zero, v ( t + h ) approaches v(t ) because v, being differentiable at t, is continuous at t (Exercise 47). The two fractions approach the values of d u dt and d v dt at t. In short,

(

) (

)

d( du dv . u × v) = ×v + u× dt dt dt Proof of the Chain Rule Suppose that u(s) = a(s) i + b(s) j + c(s) k is a differentiable vector function of s and that s = f (t ) is a differentiable scalar function of t. Then a, b, and c are differentiable functions of t, and the Chain Rule for differentiable real-valued functions gives d[ da db dc u (s ) ] = i+ j+ k dt dt dt dt As an algebraic convenience, we sometimes write the product of a scalar c and a vector v as vc instead of cv. This permits us, for instance, to write the Chain Rule in a familiar form:

=

da ds db ds dc ds i+ j+ k ds dt ds dt ds dt

=

ds da db dc i+ j+ k dt ds ds ds

du d u ds , = dt ds dt

=

ds d u dt ds

(

= f ′(t )  u′( f (t )).

where s = f (t ). z

) s = f (t )

Vector Functions of Constant Length When we track a particle moving on a sphere centered at the origin (Figure 12.8), the position vector has a constant length equal to the radius of the sphere. The velocity vector d r dt, tangent to the path of motion, is tangent to the sphere and hence perpendicular to r. This is always the case for a differentiable vector function of constant length: The vector and its first derivative are orthogonal. By direct calculation,

dr dt P r(t)

r (t ) ⋅ r (t ) = r (t ) y

x

FIGURE 12.8 If a particle moves on a sphere in such a way that its position r is a differentiable function of time, then r ⋅ ( d r dt ) = 0.

2

= c2

d[ r (t ) ⋅ r (t ) ] = 0 dt r ′(t ) ⋅ r(t ) + r (t ) ⋅ r ′(t ) = 0

r(t ) = c is constant. Differentiate both sides. Rule 5 with r (t ) = u(t ) = v(t )

2r ′(t ) ⋅ r (t ) = 0. Thus the vectors r ′(t ) and r(t ) are orthogonal because their dot product is 0. In summary, the following holds.

If r is a differentiable vector function of t and the length of r(t ) is constant, then r⋅

dr = 0. dt

(4)

We will use this observation repeatedly in Section 12.4. The converse is also true (see Exercise 41).

732

Chapter 12 Vector-Valued Functions and Motion in Space

EXERCISES

12.1

In Exercises 1–4, find the given limits. t 2 5 1. lim ⎡⎢ sin i + cos t j + tan t k ⎤⎥ t→π ⎣ ⎦ 3 4 2

(

) ( (

) (

16. r (t ) = ( sec t ) i + ( tan t ) j +

)

17. r (t ) = ( 2 ln ( t + 1)) i + t 2 j +

)

π 2. lim ⎡⎢ t 3 i + sin t j + ( ln ( t + 2 )) k ⎤⎥ t →−1 ⎣ ⎦ 2 ⎡ ⎛ t 2 − 1 ⎞⎟ ⎤ ⎛ t − 1 ⎞⎟ j + ( arctan t ) k ⎥ 3. lim ⎢ ⎜⎜ ⎟ i − ⎜⎜ ⎝ 1 − t ⎟⎠ t →1 ⎢⎣ ⎜ ⎥⎦ ⎝ ln t ⎟⎠ 2 3 ⎡ sin t ⎛ tan t ⎞⎟ t − 8 ⎤⎥ i + ⎜⎜ j− k 4. lim ⎢ ⎜⎝ sin 2t ⎟⎟⎠ t → 0 ⎣⎢ t t + 2 ⎥⎦

( )

(

)

In Exercises 5–8, r(t ) is the position of a particle in the xy-plane at time t. Find an equation in x and y whose graph is the path of the particle. Then find the particle’s velocity and acceleration vectors at the given value of t. 5. r (t ) = ( t + 1) i + ( t 2 − 1) j, t = 1

7. r (t ) = e t i +

8. r (t ) = ( cos 2t ) i + ( 3 sin 2t ) j, t = 0 Exercises 9–12 give the position vectors of particles moving along various curves in the xy-plane. In each case, find the particle’s velocity and acceleration vectors at the stated times and sketch them as vectors on the curve. 9. Motion on the circle x 2 + y 2 = 1 r (t ) = ( sin t ) i + ( cos t ) j; t = π 4 and π 2

) (

)

t t i + 4 sin j; t = π and 3π 2 2 2

11. Motion on the cycloid x = t − sin t,

y = 1 − cos t

r (t ) = ( t − sin t ) i + (1 − cos t ) j; t = π and 3π 2 12. Motion on the parabola y = x 2 + 1 r (t ) = ti + ( t 2 + 1) j; t = −1, 0, and 1

21. r (t ) = ( ln ( t 2 + 1)) i + ( arctan t ) j +

t 2 + 1k

4( 4 1 1 + t ) 3 2 i + (1 − t ) 3 2 j + t k 9 9 3

As mentioned in the text, the tangent line to a smooth curve r (t ) = f (t ) i + g(t ) j + h(t ) k at t = t 0 is the line that passes through the point ( f (t 0 ), g(t 0 ), h(t 0 )) parallel to v(t 0 ), the curve’s velocity vector at t 0 . In Exercises 23–26, find parametric equations for the line that is tangent to the given curve at the given parameter value t = t 0 . 23. r (t ) = ( sin t ) i + ( t 2 − cos t ) j + e t k, t 0 = 0 24. r (t ) = t 2 i + ( 2t − 1) j + t 3 k, t 0 = 2 25. r (t ) = ln t i +

t −1 j + t ln t k, t 0 = 1 t+2 π 2

In Exercises 27–30, find the value(s) of t so that the tangent line to the given curve contains the given point. 27. r (t ) = t 2 i + (1 + t ) j + ( 2t − 3 ) k; (−8, 2, −1) 28. r (t ) = ti + 3 j +

( 23 t )k; 32

( 0, 3, −8 3 )

29. r (t ) = 2ti + t 2 j − t 2 k; ( 0, − 4, 4 ) 30. r (t ) = −ti + t 2 j + ( ln t ) k; ( 2, − 5, − 3 )

Motion in Space

In Exercises 13–18, r(t ) is the position of a particle in space at time t. Find the particle’s velocity and acceleration vectors. Then find the particle’s speed and direction of motion at the given value of t. Write the particle’s velocity at that time as the product of its speed and direction. 13. r (t ) = ( t + 1) i + ( t 2 − 1) j + 2tk, t = 1 14. r (t ) = (1 + t ) i +

⎛ 2 ⎞⎟ ⎛ 2 ⎞ t − 16t 2 ⎟⎟⎟ j 20. r (t ) = ⎜⎜ t i + ⎜⎜ ⎝ 2 ⎟⎟⎠ ⎝ 2 ⎠

26. r (t ) = ( cos t ) i + ( sin t ) j + ( sin 2t ) k, t 0 =

10. Motion on the circle x 2 + y 2 = 16 r (t ) = 4 cos

3tj + t 2 k

Tangents to Curves

2 2t e j, t = ln 3 9

(

In Exercises 19–22, r(t ) is the position of a particle in space at time t. Find the angle between the velocity and acceleration vectors at time t = 0.

22. r (t ) =

t 1 1 i + j, t = − t +1 t 2

t2 k, t = 1 2

18. r (t ) = e −t i + ( 2 cos 3t ) j + ( 2 sin 3t ) k, t = 0

19. r (t ) = ( 3t + 1) i +

Motion in the Plane

6. r (t ) =

4 tk, t = π 6 3

t2 t3 j + k, t = 1 3 2

15. r (t ) = ( 2 cos t ) i + ( 3 sin t ) j + 4 tk, t = π 2

In Exercises 31–36, r(t ) is the position of a particle in space at time t. Match each position function with one of the graphs A–F. 31. r (t ) = ( t cos t ) i + ( t sin t ) j + tk 32. r (t ) = ( cos t ) i + ( sin t ) j + ( sin 2t ) k 33. r (t ) = t 2 i + ( t 2 + 1) j + t 4 k 34. r (t ) = ti + ( ln t ) j + ( sin t ) k

12.1  Curves in Space and Their Tangents

a. r (t ) = ( cos t ) i + ( sin t ) j, t ≥ 0

35. r (t ) = ti + ( cos t ) j + ( sin t ) k 36. r (t ) = ( t sin t ) i + ( t cos t ) j +

733

(t

b. r (t ) = cos (2t ) i + sin (2t ) j, t ≥ 0

)

t k +1

2

z

c. r (t ) = cos ( t − π 2 ) i + sin ( t − π 2 ) j, t ≥ 0 z

d. r (t ) = ( cos t ) i − ( sin t ) j, t ≥ 0 e. r (t ) = cos (t 2 ) i + sin (t 2 ) j, t ≥ 0 38. Motion along a circle Show that the vector-valued function r (t ) = ( 2 i + 2 j + k ) 1 1 1 1 1 + cos t i− j + sin t i+ j+ k 2 2 3 3 3

(

y x A.

B.

z

(

)

describes the motion of a particle moving in the circle of radius 1 centered at the point ( 2, 2, 1) and lying in the plane x + y − 2 z = 2.

y x

)

39. Motion along a parabola A particle moves along the top of the parabola y 2 = 2 x from left to right at a constant speed of 5 units per second. Find the velocity of the particle as it moves through the point ( 2, 2 ).

z

40. Motion along a cycloid A particle moves in the xy-plane in such a way that its position at time t is r (t ) = ( t − sin t ) i + (1 − cos t ) j.

y y

x x C.

D.

z

T a. Graph r(t ). The resulting curve is a cycloid. b. Find the maximum and minimum values of v and a . (Hint: Find the extreme values of v 2 and a 2 first, and take square roots later.) 41. Let r be a differentiable vector function of t. Show that if r ⋅ ( d r dt ) = 0 for all t, then r is constant.

z

42. Derivatives of triple scalar products a. Show that if u, v, and w are differentiable vector functions of t, then y

b. Show that

y x E.

d( du dv dw . u ⋅ v × w) = ⋅v×w + u⋅ ×w + u⋅v× dt dt dt dt

(

Theory and Examples

37. Motion along a circle Each of the following equations in parts (a)–(e) describes the motion of a particle having the same path, namely the unit circle x 2 + y 2 = 1. Although the path of each particle in parts (a)–(e) is the same, the behavior, or “dynamics,” of each particle is different. For each particle, answer the following questions. i) Does the particle have constant speed? If so, what is its constant speed? ii) Is the particle’s acceleration vector always orthogonal to its velocity vector? iii) Does the particle move clockwise or counterclockwise around the circle? iv) Is the particle initially located at the point (1, 0 )?

)

(

)

d dr d 2 r dr d 3r r⋅ × 2 = r⋅ × 3 . dt dt dt dt dt

x F.

(Hint: Differentiate on the left and look for vectors whose products are zero.) 43. Prove the two Scalar Multiple Rules for vector functions. 44. Prove the Sum and Difference Rules for vector functions. 45. Component test for continuity at a point Show that the vector function r defined by r (t ) = f (t ) i + g(t ) j + h(t ) k is continuous at t = t 0 if and only if f, g, and h are continuous at t 0 . 46. Limits of cross products of vector functions Suppose that r1 (t ) = f1 (t ) i + f 2 (t ) j + f 3 (t ) k, r2 (t ) = g1 (t ) i + g 2 (t ) j + g 3 (t )k, lim r1 (t ) = A, and lim r2 (t ) = B. Use the determinant formula t→t0

t→t0

for cross products and the Limit Product Rule for scalar functions to show that lim ( r1 (t ) × r2 (t ) ) = A × B.

t→t0

734

Chapter 12 Vector-Valued Functions and Motion in Space

47. Differentiable vector functions are continuous Show that if r (t ) = f (t ) i + g(t ) j + h(t ) k is differentiable at t = t 0 , then it is continuous at t 0 as well. 48. Constant Function Rule Prove that if u is the vector function with the constant value C, then d u dt = 0.

52. r (t ) = ( ln ( t 2 + 2 )) i + ( arctan 3t ) j + −3 ≤ t ≤ 5, t 0 = 3

t 2 + 1  k,

In Exercises 53 and 54, you will explore graphically the behavior of the helix r (t ) = ( cos at ) i + ( sin at ) j + btk

COMPUTER EXPLORATIONS

as you change the values of the constants a and b. Use a CAS to perform the steps in each exercise.

Use a CAS to perform the following steps in Exercises 49–52.

53. Set b = 1. Plot the helix r(t ) together with the tangent line to the curve at t = 3π 2 for a = 1, 2, 4, and 6 over the interval 0 ≤ t ≤ 4 π. Describe in your own words what happens to the graph of the helix and the position of the tangent line as a increases through these positive values.

a. Plot the space curve traced out by the position vector r. b. Find the components of the velocity vector d r dt . c. Evaluate d r dt at the given point t 0 and determine the equation of the tangent line to the curve at r(t 0 ). d. Plot the tangent line together with the curve over the given interval.

54. Set a = 1. Plot the helix r(t ) together with the tangent line to the curve at t = 3π 2 for b = 1 4 , 1 2, 2, and 4 over the interval 0 ≤ t ≤ 4 π. Describe in your own words what happens to the graph of the helix and the position of the tangent line as b increases through these positive values.

49. r (t ) = ( sin t − t cos t ) i + ( cos t + t sin t ) j + t 2 k, 0 ≤ t ≤ 6π , t 0 = 3π 2 50. r (t ) =

2t i + e t j + e −t k, − 2 ≤ t ≤ 3, t 0 = 1

51. r (t ) = ( sin 2t ) i + ( ln (1 + t )) j + tk, 0 ≤ t ≤ 4 π, t0 = π 4

12.2 Integrals of Vector Functions; Projectile Motion In this section we investigate integrals of vector functions and their application to motion along a path in space or in the plane.

Integrals of Vector Functions A differentiable vector function R(t ) is an antiderivative of a vector function r(t ) on an interval I if d R dt = r at each point of I. If R is an antiderivative of r on I, it can be shown, working one component at a time, that every antiderivative of r on I has the form R + C for some constant vector C (Exercise 45). The set of all antiderivatives of r on I is the indefinite integral of r on I.

The indefinite integral of r with respect to t is the set of all antiderivatives of r, denoted by

DEFINITION

∫ r(t ) dt. The usual arithmetic rules for indefinite integrals apply. EXAMPLE 1

To integrate a vector function, we integrate each of its components.

∫ (( cos t ) i +

( ∫ cos t dt ) i + ( ∫ dt ) j − ( ∫ 2t dt ) k

(1)

= ( sin t + C 1 ) i + ( t + C 2 ) j − ( t 2 + C 3 ) k

(2)

j − 2tk ) dt =

= ( sin t ) i + tj − t 2 k + C

   C = C1i + C 2 j − C 3k

12.2  Integrals of Vector Functions; Projectile Motion

735

As in the integration of scalar functions, we recommend that you skip the steps in Equations (1) and (2) and go directly to the final form. Find an antiderivative for each component and add a constant vector at the end. Definite integrals of vector functions are best defined in terms of components. The definition is consistent with how we compute limits and derivatives of vector functions. DEFINITION If the components of r (t ) = f (t ) i + g(t ) j + h(t )k are integrable over [ a, b ], then so is r, and the definite integral of r from a to b is b

∫a r(t ) dt EXAMPLE 2

=

(∫

b a

) (∫

f (t ) dt i +

b a

) (∫

g(t ) dt j +

b a

)

h(t ) dt k.

As in Example 1, we integrate each component.

∫0 (( cos t ) i + j − 2tk ) dt = ( ∫0 π

π

) ( ∫ dt ) j − ( ∫

cos t dt i +

π

0

π 0

)

2t dt k

⎡ ⎤π ⎡ ⎤π ⎡ ⎤π = ⎢ sin t ⎥ i + ⎢ t ⎥ j − ⎢ t 2 ⎥   k ⎢⎣ ⎥⎦ 0 ⎢⎣ ⎥⎦ 0 ⎢⎣ ⎥⎦ 0 ] [ ] = [ 0 − 0 i + π − 0 j −[ π 2 − 0 2 ] k = π j − π 2k The Fundamental Theorem of Calculus for continuous vector functions says that b

∫a r(t ) dt

b

⎤ = R(t ) ⎥ = R(b) − R(a), ⎦a

where R is any antiderivative of r, so that R′(t ) = r(t ) (Exercise 46). Notice that an antiderivative of a vector function is also a vector function, whereas a definite integral of a vector function is a single constant vector. EXAMPLE 3 Suppose we do not know the path of a hang glider, but only its acceleration vector a (t ) = −( 3 cos t ) i − ( 3 sin t ) j + 2k. We also know that initially (at time t = 0) the glider departed from the point ( 4, 0, 0 ) with velocity v(0) = 3 j. Find the glider’s position as a function of t. Solution Our goal is to find r(t ) knowing

The differential equation: The initial conditions:

d 2r = −( 3 cos t ) i − ( 3 sin t ) j + 2k dt 2 v(0) = 3 j and r (0) = 4 i + 0 j + 0 k. a =

Integrating both sides of the differential equation with respect to t gives v(t ) = −( 3 sin t ) i + ( 3 cos t ) j + 2t k + C 1. We use v(0) = 3 j to find C 1: 3 j = −( 3 sin 0 ) i + ( 3 cos 0 ) j + (0) k + C 1 3j = 3j + C1 C 1 = 0. The glider’s velocity as a function of time is dr = v(t ) = −( 3 sin t ) i + ( 3 cos t ) j + 2t k. dt

736

Chapter 12 Vector-Valued Functions and Motion in Space

z

Integrating both sides of this last differential equation gives r (t ) = ( 3 cos t ) i + ( 3 sin t ) j + t 2 k + C 2 . We then use the initial condition r (0)  4 i to find C 2 : 4 i = ( 3 cos 0 ) i + ( 3 sin 0 ) j + ( 0 2 )k + C 2 4 i = 3i + ( 0 ) j + ( 0 ) k + C 2 C 2 = i. The glider’s position as a function of t is

(4, 0, 0) x

r (t ) = (1 + 3 cos t ) i + ( 3 sin t ) j + t 2 k.

y

This is the path of the glider shown in Figure 12.9. Although the path resembles that of a helix due to its spiraling nature around the z-axis, it is not a helix because of the way it is rising. (We say more about this in Section 12.5.)

FIGURE 12.9

The path of the hang glider in Example 3. Although the path spirals around the z-axis, it is not a helix.

The Vector and Parametric Equations for Ideal Projectile Motion A classic example of integrating vector functions is the derivation of the equations for the motion of a projectile. In physics, projectile motion describes how an object fired at some angle from an initial position, and acted upon by only the force of gravity, moves in a vertical coordinate plane. In the classic example, we ignore the effects of any frictional drag on the object, which may vary with its speed and altitude, and also the fact that the force of gravity changes slightly with the projectile’s changing height. In addition, we ignore the long-distance effects of Earth turning beneath the projectile, such as in a rocket launch or the firing of a projectile from a cannon. Ignoring these effects gives us a reasonable approximation of the motion in most cases. To derive equations for projectile motion, we assume that the projectile behaves like a particle moving in a vertical coordinate plane and that the only force acting on the projectile during its flight is the constant force of gravity, which always points straight down. The magnitude of the gravitational acceleration g is approximately 9.8 m s 2 at sea level. We assume that the projectile is launched from the origin at time t  0 into the first quadrant with an initial velocity v 0 (Figure 12.10). If v 0 makes an angle B with the horizontal, then

y

v0

@ v0 @ sin a j

a x

@ v0 @ cos a i r = 0 at time t = 0 a = −gj

v 0 = ( v 0 cos B ) i + ( v 0 sin B ) j. If we use the simpler notation V 0 for the initial speed v 0 , then

(a)

v 0 = ( υ 0 cos α ) i + ( υ 0 sin α ) j.

(3)

The projectile’s initial position is

y (x, y)

r0 = 0 i + 0 j = 0.

v

Newton’s second law of motion says that the force acting on the projectile is equal to the projectile’s mass m times its acceleration, or m ( d 2 r dt 2 ), if r is the projectile’s position vector and t is time. If the force is solely the gravitational force mgj, then

a = −gj r = x i + yj x

0

R Horizontal range (b)

FIGURE 12.10

(a) Position, velocity, acceleration, and launch angle at t  0. (b) Position, velocity, and acceleration at a later time t.

(4)

m

d 2r = −mg j dt 2

and

d 2r = −g j, dt 2

where g is the acceleration due to gravity. We find r as a function of t by solving the following initial value problem. Differential equation: Initial conditions:

d 2r = −g j dt 2 r = r0

and

dr = v0 dt

when  t = 0

737

12.2  Integrals of Vector Functions; Projectile Motion

The first integration gives dr = −( gt ) j + v 0 . dt A second integration gives 1 r = − gt 2 j + v 0 t + r0 . 2 Substituting the values of v 0 and r0 from Equations (3) and (4) gives 1 r = − gt 2 j + ( υ 0 cos α ) ti + ( υ 0 sin α ) tj + 0 .  2 v0 t

Collecting terms, we obtain the following.

Ideal Projectile Motion Equation

(

r = ( υ 0 cos α ) ti + ( υ 0 sin α ) t −

)

1 2 gt j. 2

(5)

Equation (5) is the vector equation of the path for ideal projectile motion. The angle α is the projectile’s launch angle (firing angle, angle of elevation), and υ 0 , as we said before, is the projectile’s initial speed. The components of r give the parametric equations x = ( υ 0 cos α ) t

and

y = ( υ 0 sin α ) t −

1 2 gt , 2

(6)

where x is the distance downrange and y is the height of the projectile at time t ≥ 0.

EXAMPLE 4 A projectile is fired from the origin over horizontal ground at an initial speed of 500 m s and a launch angle of 60 °. Where will the projectile be 10 s later? Solution We use Equation (5) with υ 0 = 500, α = 60 °, g = 9.8, and t = 10 to find the projectile’s components 10 s after firing.

(

( 12 )(10 )i + ⎛⎜⎜⎝( 500 )⎛⎜⎜⎝

= ( 500 )

) ( )

1 2 gt j 2 ⎞ 3 ⎞⎟ 1 ( )( 9.8 100 )⎟⎟⎟ j ⎟⎟⎠10 − ⎠ 2 2

r = ( υ 0 cos α ) ti + ( υ 0 sin α ) t −

≈ 2500 i + 3840 j Ten seconds after firing, the projectile is about 3840 m above ground and 2500 m downrange from the origin.

Ideal projectiles move along parabolas, as we now deduce from Equations (6). If we substitute t = x ( υ 0 cos α ) from the first equation into the second, we obtain the Cartesian coordinate equation ⎛ ⎞⎟ 2 g y = −⎜⎜⎜ ⎟ x + ( tan α ) x. 2 2 ⎝ 2υ 0 cos α ⎟⎠ This equation has the form y = ax 2 + bx, so its graph is a parabola.

738

Chapter 12 Vector-Valued Functions and Motion in Space

A projectile reaches its highest point when its vertical velocity component is zero. When fired over horizontal ground, the projectile lands when its vertical component equals zero in Equation (5), and the range R is the distance from the origin to the point of impact. We summarize the results here, which you are asked to verify in Exercise 31.

Height, Flight Time, and Range for Ideal Projectile Motion

For ideal projectile motion when an object is launched from the origin over a horizontal surface with initial speed υ 0 and launch angle α: Maximum height:

y

y max =

Flight time:

t =

2υ 0 sin α g

R =

υ0 2 sin 2α g

v0

Range:

a

( υ 0 sin α ) 2 2g

(x 0 , y0 )

0

x

If we fire our ideal projectile from the point ( x 0 , y 0 ) instead of the origin (Figure 12.11), the position vector for the path of motion is

FIGURE 12.11

The path of a projectile fired from ( x 0 , y 0 ) with an initial velocity v 0 at an angle of α degrees with the horizontal.

(

r = ( x 0 + ( υ 0 cos α ) t ) i + y 0 + ( υ 0 sin α ) t −

)

1 2 gt j, 2

(7)

as you are asked to show in Exercise 33.

Projectile Motion with Wind Gusts The next example shows how to account for another force acting on a projectile due to a gust of wind. We assume that the path of the baseball in Example 5 lies in a vertical plane. EXAMPLE 5 A baseball is hit when it is 1 m above the ground. It leaves the bat with initial speed of 50 m s, making an angle of 20° with the horizontal. At the instant the ball is hit, an instantaneous gust of wind blows in the horizontal direction directly opposite the direction the ball is taking toward the outfield, adding a component of −2.5i ( m s ) to the ball’s initial velocity ( 2.5 m s = 9 km h ). (a) Find a vector equation (position vector) for the path of the baseball. (b) How high does the baseball go, and when does it reach maximum height? (c) Assuming that the ball is not caught, find its range and flight time. Solution

(a) Using Equation (3) and accounting for the gust of wind, the initial velocity of the baseball is v 0 = ( υ 0 cos α ) i + ( υ 0 sin α ) j − 2.5i = ( 50 cos 20 ° ) i + ( 50 sin 20 ° ) j − ( 2.5 ) i = ( 50 cos 20 ° − 2.5 ) i + ( 50 sin 20 ° ) j. The initial position is r0 = 0 i + 1 j. Integration of d 2 r dt 2 = −g j gives dr = −( gt ) j + v 0 . dt

12.2  Integrals of Vector Functions; Projectile Motion

739

A second integration gives 1 r = − gt 2 j + v 0 t + r0 . 2 Substituting the values of v 0 and r0 into the last equation gives the position vector of the baseball. 1 r = − gt 2 j + v 0 t + r0 2 = −4.9t 2 j + ( 50 cos 20 ° − 2.5 ) ti + ( 50 sin 20 ° ) tj + 1 j = ( 50 cos 20 ° − 2.5 ) ti + (1 + ( 50 sin 20 ° ) t − 4.9t 2 ) j. (b) The baseball reaches its highest point when the vertical component of velocity is zero, or dy = 50 sin 20 ° − 9.8t = 0. dt Solving for t we find t =

50 sin 20 ° ≈ 1.75 s. 9.8

Substituting this time into the vertical component for r gives the maximum height y max = 1 + ( 50 sin 20 ° )(1.75 ) − 4.9(1.75 ) 2 ≈ 15.9 m. That is, the maximum height of the baseball is about 15.9 m, reached about 11.75 s after leaving the bat. (c) To find when the baseball lands, we set the vertical component for r equal to 0 and solve for t: 1 + ( 50 sin 20 ° ) t − 4.9t 2 = 0 1 + (17.1) t − 4.9t 2 = 0. The solution values are about t = 3.55 s and t = −0.06 s. Substituting the positive time into the horizontal component for r, we find the range R = ( 50 cos 20 ° − 2.5 )( 3.55 ) ≈ 157.8 m. Thus, the horizontal range is about 157.8 m, and the flight time is about 3.55 s. In Exercises 41 and 42, we consider projectile motion when there is air resistance slowing down the flight.

EXERCISES

12.2

Integrating Vector-Valued Functions

Evaluate the integrals in Exercises 1–10.

∫0 [ t 3 i + 7 j + ( t + 1) k ] dt

2.

∫1

3.

∫−π 4 [ ( sin t ) i + (1 + cos t ) j + ( sec 2 t ) k ] dt

4.

∫0

5.

∫1

2

( )

⎡ ( 6 − 6t ) i + 3 t j + 4 k ⎤ dt ⎢⎣ t 2 ⎥⎦

7.

∫0 [ te t

8.

∫1

9.

∫0

10.

∫0

π4

π3

4

[ ( sec t tan t ) i + ( tan t ) j + ( 2 sin t cos t ) k ] dt

⎡ 1 i + 1 j + 1 k ⎤ dt ⎢⎣ t 5−t 2t ⎥⎦

⎡ ⎤ 2 3 i+ k ⎥ dt ⎢ ⎢⎣ 1 − t 2 1 + t 2 ⎥⎦

∫0

1

1.

1

6.

1

ln 3

π2

π4

2

i + e −t j + k ] dt

[ te t i + e t j + ln t k ] dt [ cos t i − sin 2t j + sin 2 t k ] dt [ sec t i + tan 2 t j − t sin t k ] dt

740

Chapter 12 Vector-Valued Functions and Motion in Space

Initial Value Problems

Projectile Motion

Solve the initial value problems in Exercises 11–20 for r as a vector function of t. dr 11. Differential equation: = −ti − tj − tk dt Initial condition: r (0) = i + 2 j + 3k

Projectile flights in Exercises 23–40 are to be treated as ideal unless stated otherwise. All launch angles are assumed to be measured from the horizontal. All projectiles are assumed to be launched from the origin over a horizontal surface unless stated otherwise. For some exercises, a calculator may be helpful.

12. Differential equation: Initial condition: 13. Differential equation: Initial condition: 14. Differential equation: Initial condition:

dr = (180t ) i + (180t − 16t 2 ) j dt r (0) = 100 j dr 3 1 k = ( t + 1)1 2 i + e −t j + dt t +1 2 r (0) = k dr = ( t 3 + 4 t ) i + t j + 2t 2 k dt r (0) = i + j

15. Differential equation:

( ( ))

π π dr 1 = ( tan t ) i + cos t j − ( sec 2t ) k, − < t < dt 4 4 2 r (0) = 3i − 2 j + k Initial condition: 16. Differential equation:

(

) (

) (

)

dr t t2 + 1 t2 + 4 i− j+ 2 k, t < 2 = 2 dt t +2 t−2 t +3 r (0) = i − j + k Initial condition: 17. Differential equation: Initial conditions:

d 2r

= − 32k dt 2 r (0) = 100k  and dr dt

18. Differential equation: Initial conditions:

Initial conditions:

= 8i + 8 j

d 2r

= −( i + j + k ) dt 2 r (0) = 10 i + 10 j + 10 k  and dr dt

19. Differential equation:

t=0

t=0

= 0

d 2r = e t i − e −t j + 4 e 2 t k dt 2 r (0) = 3i + j + 2k  and dr dt

t=0

= −i + 4 j

20. Differential equation: d 2r = ( sin t ) i − ( cos t ) j + ( 4 sin t cos t ) k dt 2 Initial conditions: r (0) = i − k  and dr dt

t=0

= i

23. Travel time A projectile is fired at a speed of 840 m s at an angle of 60 °. How long will it take to get 21 km downrange? 24. Range and height versus speed a. Show that doubling a projectile’s initial speed at a given launch angle multiplies its range by 4. b. By about what percentage should you increase the initial speed to double the height and range? 25. Flight time and height A projectile is fired with an initial speed of 500 m s at an angle of elevation of 45 °. a. When and how far away will the projectile strike? b. How high overhead will the projectile be when it is 5 km downrange? c. What is the greatest height reached by the projectile? 26. Throwing a baseball A baseball is thrown from the stands 9.8 m above the field at an angle of 30 ° up from the horizontal. When and how far away will the ball strike the ground if its initial speed is 9.8 m s? 27. Firing golf balls A spring gun at ground level fires a golf ball at an angle of 45 °. The ball lands 10 m away. a. What was the ball’s initial speed? b. For the same initial speed, find the two firing angles that make the range 6 m. 28. Beaming electrons An electron in a cathode-ray tube (CRT) is beamed horizontally at a speed of 5 × 10 6 m s toward the face of the tube 40 cm away. About how far will the electron drop before it hits? 29. Equal-range firing angles What two angles of elevation will enable a projectile to reach a target 16 km downrange on the same level as the gun if the projectile’s initial speed is 400 m s? 30. Finding muzzle speed Find the muzzle speed of a gun whose maximum range is 24.5 km. 31. Verify the results given in the text (following Example 4) for the maximum height, flight time, and range for ideal projectile motion. 32. Colliding marbles The accompanying figure shows an experiment with two marbles. Marble A was launched toward marble B with launch angle α and initial speed υ 0 . At the same instant, marble B was released to fall from rest at R tan α units directly above a spot R units downrange from A. The marbles were found to collide regardless of the value of υ 0 . Was this mere coincidence, or must this happen? Give reasons for your answer.

Motion Along a Straight Line B

21. At time t = 0, a particle is located at the point (1, 2, 3 ). It travels in a straight line to the point ( 4, 1, 4 ), has speed 2 at (1, 2, 3 ), and has constant acceleration 3i − j + k. Find an equation for the position vector r(t ) of the particle at time t. 22. A particle traveling in a straight line is located at the point (1, −1, 2 ) and has speed 2 at time t = 0. The particle moves toward the point ( 3, 0, 3 ) with constant acceleration 2 i + j + k. Find its position vector r(t ) at time t.

1 2 gt 2

R tan a

v0 A

a R

12.2  Integrals of Vector Functions; Projectile Motion

33. Firing from ( x 0 , y 0 )

that is elevated 14 m above the tee as shown in the diagram. Assuming that the pin, 112 m downrange, does not get in the way, where will the ball land in relation to the pin?

Derive the equations

x = x 0 + ( υ 0 cos α ) t , y = y 0 + ( υ 0 sin α ) t −

1 2 gt 2

Pin

(see Equation (7) in the text) by solving the following initial value problem for a vector r in the plane. d 2r = −g j dt 2 r (0) = x 0 i + y 0 j

Differential equation: Initial conditions:

υ ⎛ x 2 + 4 ⎜⎜ y − 0 ⎝ 4g

14 m 45°

Tee 112 m NOT TO SCALE

34. Where trajectories crest For a projectile fired from the ground at launch angle α with initial speed υ 0 , consider α as a variable and υ 0 as a fixed constant. For each α, 0 < α < π 2, we obtain a parabolic trajectory as shown in the accompanying figure. Show that the points in the plane that give the maximum heights of these parabolic trajectories all lie on the ellipse 2

Green 35.5 ms

dr (0) = ( υ 0 cos α ) i + ( υ 0 sin α ) j dt

2

741

37. Volleyball A volleyball is hit when it is 1.3 m above the ground and 4 m from a 2-m-high net. It leaves the point of impact with an initial velocity of 12 m s at an angle of 27 ° and slips by the opposing team untouched. a. Find a vector equation for the path of the volleyball. b. How high does the volleyball go, and when does it reach maximum height?

υ0 ⎞⎟ , ⎟⎟⎠ = 4g 2 4

c. Find its range and flight time.

where x ≥ 0.

d. When is the volleyball 2.3 m above the ground? How far (ground distance) is the volleyball from where it will land?

y

e. Suppose that the net is raised to 2.5 m. Does this change things? Explain.

Ellipse 1 a R, ymaxb 2

38. Shot put In Moscow in 1987, Natalya Lisouskaya set a women’s world record by putting a 4kg shot 22.63 m. Assuming that she launched the shot at a 40 ° angle to the horizontal from 2 m above the ground, what was the shot’s initial speed?

Parabolic trajectory

x

0

35. Launching downhill An ideal projectile is launched straight down an inclined plane as shown in the accompanying figure. a. Show that the greatest downhill range is achieved when the initial velocity vector bisects angle AOR. b. If the projectile were fired uphill instead of down, what launch angle would maximize its range? Give reasons for your answer.

Vertical

A

O

39. A child throws a ball with an initial speed of 18 m s at an angle of elevation of 60 ° toward a tall building that is 7 m from the child. If the child’s hand is 1.6 m from the ground, show that the ball hits the building, and find the height above the ground of the point where the ball hits the building. 40. Hitting a baseball under a wind gust A baseball is hit when it is 0.8 m above the ground. It leaves the bat with an initial velocity of 40 m s at a launch angle of 23 °. At the instant the ball is hit, an instantaneous gust of wind blows against the ball, adding a component of −4 i ( m s ) to the ball’s initial velocity. A 5-m-high fence lies 90 m from home plate in the direction of the flight. a. Find a vector equation for the path of the baseball. b. How high does the baseball go, and when does it reach maximum height?

v0

c. Find the range and flight time of the baseball, assuming that the ball is not caught.

a

d. When is the baseball 6 m high? How far (ground distance) is the baseball from home plate at that height? e. Has the batter hit a home run? Explain.

Hi

ll

Projectile Motion with Linear Drag R

36. Elevated green A golf ball is hit with an initial speed of 35.5 m s at an angle of elevation of 45 ° from the tee to a green

The main force affecting the motion of a projectile, other than gravity, is air resistance. This slowing down force is drag force, and it acts in a direction opposite to the velocity of the projectile (see accompanying figure). For projectiles moving through the air at relatively low speeds,

742

Chapter 12 Vector-Valued Functions and Motion in Space

however, the drag force is (very nearly) proportional to the speed (to the first power) and so is called linear. y

b

∫a C × r(t ) dt

Drag force Gravity

υ x = 0 (1 − e −kt ) cos α k υ0 g (1 − e −kt )( sin α ) + 2 (1 − kt − e −kt ) y= k k by solving the following initial value problem for a vector r in the plane. d 2r dr = −g j − kv = −g j − k dt 2 dt r (0) = 0 dr dt

t=0

b

∫a r(t ) dt

( any constant vector  C )

b

= C × ∫ r (t ) dt ( any constant vector  C ) a

a. Show that ur is continuous on [ a, b ] if u and r are continuous on [ a, b ].

Derive the equations

Initial conditions:

= C⋅

44. Products of scalar and vector functions Suppose that the scalar function u(t ) and the vector function r(t ) are both defined for a b t b b.

x

Differential equation:

b

∫a C ⋅ r(t ) dt and

Velocity

41. Linear drag

c. The Constant Vector Multiple Rules:

= v 0 = ( υ 0 cos α ) i + ( υ 0 sin α ) j

The drag coefficient k is a positive constant representing resistance due to air density, V 0 and B are the projectile’s initial speed and launch angle, and g is the acceleration of gravity. 42. Hitting a baseball with linear drag Consider the baseball problem in Example 5 when there is linear drag (see Exercise 41). Assume a drag coefficient k  0.12, but no gust of wind. a. From Exercise 41, find a vector form for the path of the baseball. b. How high does the baseball go, and when does it reach maximum height? c. Find the range and flight time of the baseball.

b. If u and r are both differentiable on [ a, b ], show that ur is differentiable on [ a, b ] and that dr du d (u r ) = u + r. dt dt dt 45. Antiderivatives of vector functions a. Use Corollary 2 of the Mean Value Theorem for scalar functions to show that if two vector functions R 1 (t ) and R 2 (t ) have identical derivatives on an interval I, then the functions differ by a constant vector value throughout I. b. Use the result in part (a) to show that if R(t ) is any antiderivative of r(t ) on I, then any other antiderivative of r on I equals R(t ) C for some constant vector C. 46. The Fundamental Theorem of Calculus The Fundamental Theorem of Calculus for scalar functions of a real variable holds for vector functions of a real variable as well. Prove this by using the theorem for scalar functions to show first that if a vector function r(t ) is continuous for a b t b b, then d dt

e. A 3-m-high outfield fence is 115 m from home plate in the direction of the flight of the baseball. The outfielder can jump and catch any ball up to 3.3 m off the ground to stop it from going over the fence. Has the batter hit a home run?

= r (t )

at every point t of ( a, b ). Then use the conclusion in part (b) of Exercise 45 to show that if R is any antiderivative of r on [ a, b ], then b

d. When is the baseball 9 m high? How far (ground distance) is the baseball from home plate at that height?

t

∫ a r (U ) d U

∫a r(t ) dt

= R(b) − R(a).

47. Hitting a baseball with linear drag under a wind gust Consider again the baseball problem in Example 5. This time, assume a drag coefficient of 0.08 and an instantaneous gust of wind that adds a component of −5i ( m s ) to the initial velocity at the instant the baseball is hit. a. Find a vector equation for the path of the baseball.

Theory and Examples

43. Establish the following properties of integrable vector functions. a. The Constant Scalar Multiple Rule: b

∫a

c. Find the range and flight time of the baseball.

b

kr (t ) dt = k ∫ r (t ) dt ( any scalar  k ) a

The Rule for Negatives, b

∫a (−r(t ) ) dt

= −∫ r (t ) dt , a

is obtained by taking k = −1.

∫a ( r1 (t ) ± r2 (t ) ) dt

=

b

d. When is the baseball 10 m high? How far (ground distance) is the baseball from home plate at that height? e. A 6-m-high outfield fence is 120 m from home plate in the direction of the flight of the baseball. Has the batter hit a home run? If “yes,” what change in the horizontal component of the ball’s initial velocity would have kept the ball in the park? If “no,” what change would have allowed it to be a home run?

b

b. The Sum and Difference Rules: b

b. How high does the baseball go, and when does it reach maximum height?

b

∫a r1 (t ) dt ± ∫a r2 (t ) dt

48. Height versus time Show that a projectile attains three-quarters of its maximum height in half the time it takes to reach the maximum height.

743

12.3  Arc Length in Space

12.3 Arc Length in Space In this and the next two sections, we study the mathematical features of a curve’s shape that describe the sharpness of its turning and its twisting.

Arc Length Along a Space Curve

Base point 3

0

2

1

s

–2

–1

4

FIGURE 12.12

Smooth curves can be scaled like number lines, the coordinate of each point being its directed distance along the curve from a preselected base point.

One of the features of smooth space and plane curves is that they have a measurable length. This enables us to locate points along these curves by giving their directed distance s along the curve from some base point, the way we locate points on coordinate axes by giving their directed distance from the origin (Figure 12.12). This is what we did for plane curves in Section 10.2. To measure distance along a smooth curve in space, we add a z-term to the formula we use for curves in the plane.

DEFINITION The length of a smooth curve r (t ) = x (t ) i + y(t ) j + z (t ) k, a ≤ t ≤ b, that is traced exactly once as t increases from t = a to t = b is b

L =

∫ ( dxdt ) + ( dydt ) + ( dzdt ) 2

2

2

dt.

(1)

a

Just as for plane curves, we can calculate the length of a curve in space from any convenient parametrization that meets the stated conditions. We omit the proof. The square root in Equation (1) is v , the length of a velocity vector d r dt. This enables us to write the formula for length a shorter way.

Arc Length Formula

L =

EXAMPLE 1

b

∫a

v dt

(2)

A glider is soaring upward along the helix r (t ) = ( cos t ) i + ( sin t ) j + tk.

z

How long is the glider’s path from t = 0 to t = 2π ? Solution The path segment during this time corresponds to one full turn of the helix (Figure 12.13). The length of this portion of the curve is

2p t=p

t = 2p

0 (1, 0, 0)

t=0

r

P

t=p 2

This is stands.

b

L=

∫a

=

∫0

2π

v dt =

2π

∫0

(− sin t ) 2 + ( cos t ) 2 + (1) 2 dt

2 dt = 2π 2 units of length.

2 times the circumference of the circle in the xy-plane over which the helix

y

x

FIGURE 12.13 The helix in Example 1, r (t ) = ( cos t ) i + ( sin t ) j + tk.

If we choose a base point P (t 0 ) on a smooth curve C parametrized by t, each value of t determines a point P (t ) = ( x (t ), y(t ), z (t )) on C and a “directed distance” s (t ) =

∫t

t 0

v(τ ) d τ ,

744

Chapter 12 Vector-Valued Functions and Motion in Space

z

r

0

P(t)

Base point s(t) P(t0)

x

measured along C from the base point (Figure 12.14). This is the arc length function we defined in Section 10.2 for plane curves that have no z-component. If t  t 0 , s(t ) is the distance along the curve from P (t 0 ) to P (t ). If t  t 0 , s(t ) is the negative of the distance. Each value of s determines a point on C, and this parametrizes C with respect to s. We call s an arc length parameter for the curve. The parameter’s value increases in the direction of increasing t. We will see that the arc length parameter is particularly effective for investigating the turning and twisting nature of a space curve.

y

Arc Length Parameter with Base Point P( t 0 )

s (t ) =

FIGURE 12.14

The directed distance along the curve from P (t 0 ) to any point

P (t ) is s(t ) =

∫t

t

∫t

t

[ x ′(U ) ] + [ y′(U ) ] + [ z ′(U ) ] d U = 2

2

2

0

∫t

t

v( U ) d U

(3)

0

v(U ) d U .

0

U is the Greek letter tau (rhymes with “now”)

We use the Greek letter U (“tau”) as the variable of integration in Equation (3) because the letter t is already in use as the upper limit. If a curve r(t ) is already given in terms of some parameter t, and s(t ) is the arc length function given by Equation (3), then we may be able to solve for t as a function of s: t  t (s). Then the curve can be reparametrized in terms of s by substituting for t: r  r(t (s)). The new parametrization identifies a point on the curve with its directed distance along the curve from the base point.

EXAMPLE 2 This is an example for which we can actually find the arc length parametrization of a curve. If t 0  0, then the arc length parameter along the helix r (t ) = ( cos t ) i + ( sin t ) j + tk from t 0 to t is s(t ) = = =

∫t

t

v(U ) d U

Eq. (3)

0

t

∫0

2 dU

Value from Example 1

2 t.

Solving this equation for t gives t  s 2. Substituting into the position vector r gives the following arc length parametrization for the helix:

(

r (t (s)) = cos

HISTORICAL BIOGRAPHY Josiah Willard Gibbs (1839–1903) Gibbs, born in Connecticut, USA, taught at Yale as a professor of mathematics. He made contributions to thermodynamics, electromagnetics, and statistical mechanics. For his foundational work, Gibbs is known as the father of vector analysis. To know more, visit the companion Website.

) (

)

s s s i + sin j+ k. 2 2 2

Unlike the case that appears in Example 2, the arc length parametrization is generally difficult to find analytically for a curve already given in terms of some other parameter t. Fortunately, however, we rarely need an exact formula for s(t ) or its inverse t (s).

Speed on a Smooth Curve Since the derivatives beneath the radical in Equation (3) are continuous (the curve is smooth), the Fundamental Theorem of Calculus tells us that s is a differentiable function of t with derivative ds  v(t ) . dt

(4)

12.3  Arc Length in Space

745

Although the base point P (t 0 ) plays a role in defining s in Equation (3), it plays no role in Equation (4). The rate at which a moving particle covers distance along its path is independent of how far away it is from the base point. Equation (4) says that this rate is the magnitude of v. Notice that ds dt > 0 since, by definition, v is never zero for a smooth curve. We see once again that s is an increasing function of t.

z

Unit Tangent Vector

v

T= v @v@

r

0

y

s P(t 0 )

On a smooth curve, we already know that the velocity vector v = d r dt is tangent to the curve r(t ) and that the vector v T = v is therefore a unit vector tangent to the curve, called the unit tangent vector (Figure 12.15). The unit tangent vector T for a smooth curve is a differentiable function of t whenever v is a differentiable function of t. As we will see in Section 12.5, T is one of three unit vectors in a traveling reference frame that is used to describe the motion of objects traveling in three dimensions.

x

FIGURE 12.15

We find the unit tangent vector T by dividing v by its length v .

EXAMPLE 3

Find the unit tangent vector of the curve r (t ) = (1 + 3 cos t ) i + ( 3 sin t ) j + t 2 k

representing the path of the glider in Example 3, Section 12.2. Solution In that example, we found

v =

dr = −( 3 sin t ) i + ( 3 cos t ) j + 2tk dt

and v =

9 + 4t 2 .

Thus, T =

3 sin t v = − i+ v 9 + 4t 2

3 cos t 9+

4t 2

j+

2t 9 + 4t 2

k.

For the counterclockwise motion

y

r (t ) = ( cos t ) i + ( sin t ) j

T=v

x2 + y2 = 1

around the unit circle, we see that P(x, y)

v = ( − sin t ) i + ( cos t ) j

r t 0

FIGURE 12.16

(1, 0)

x

Counterclockwise motion around the unit circle.

is already a unit vector, so T = v and T is orthogonal to r (Figure 12.16). The velocity vector is the change in the position vector r with respect to time t, but how does the position vector change with respect to arc length? More precisely, what is the derivative d r ds ? Since ds dt > 0 for the curves we are considering, s is one-to-one and has an inverse that gives t as a differentiable function of s (Section 3.8). The derivative of the inverse is 1 1 dt . = = ds dt v ds This makes r a differentiable function of s whose derivative can be calculated with the Chain Rule to be 1 v dr d r dt = = v = = T. (5) v v ds dt ds This equation says that d r ds is the unit tangent vector in the direction of the velocity vector v (Figure 12.15).

746

Chapter 12 Vector-Valued Functions and Motion in Space

EXERCISES

12.3

Finding Tangent Vectors and Lengths

In Exercises 1–8, find the curve’s unit tangent vector. Also, find the length of the indicated portion of the curve. 1. r (t ) = ( 2 cos t ) i + ( 2 sin t ) j +

5t k, 0 ≤ t ≤ π

2. r (t ) = ( 6 sin 2t ) i + ( 6 cos 2t ) j + 5tk, 0 ≤ t ≤ π 3. r (t ) = ti + ( 2 3 ) t 3 2 k, 0 ≤ t ≤ 8 4. r (t ) = ( 2 + t ) i − ( t + 1) j + tk, 0 ≤ t ≤ 3 6. r (t ) = 6t 3 i − 2t 3 j − 3t 3 k, 1 ≤ t ≤ 2 7. r (t ) = ( t cos t ) i + ( t sin t ) j + ( 2 2 3 ) t 3 2 k, 0 ≤ t ≤ π 2 ≤ t ≤ 2

9. Find the point on the curve r (t ) = ( 5 sin t ) i + ( 5 cos t ) j + 12tk at a distance 26π units along the curve from the point ( 0, 5, 0 ) in the direction corresponding to increasing t values. 10. Find the point on the curve r (t ) = (12 sin t ) i − (12 cos t ) j + 5t k at a distance 13π units along the curve from the point ( 0, −12, 0 ) in the direction corresponding to decreasing t values. Arc Length Parameter

In Exercises 11–14, find the arc length parameter along the curve from the point where t = 0 by evaluating the integral s (t ) =

t

∫0

d. Write an integral for the length of the ellipse. Do not try to evaluate the integral; it is nonelementary. T e. Numerical integrator two decimal places.

5. r (t ) = ( cos 3 t ) j + ( sin 3 t ) k, 0 ≤ t ≤ π 2

8. r (t ) = ( t sin t + cos t ) i + ( t cos t − sin t ) j,

c. Show that the acceleration vector always lies parallel to the plane (orthogonal to a vector normal to the plane). Thus, if you draw the acceleration as a vector attached to the ellipse, it will lie in the plane of the ellipse. Add the acceleration vectors for t = 0, π 2, π , and 3π 2 to your sketch.

Estimate the length of the ellipse to

18. Length is independent of parametrization To illustrate that the length of a smooth space curve does not depend on the parametrization you use to compute it, calculate the length of one turn of the helix in Example 1 with the following parametrizations. a. r (t ) = ( cos 4 t ) i + ( sin 4 t ) j + 4 t k, 0 ≤ t ≤ π 2 b. r (t ) = [ cos ( t 2 ) ] i + [ sin ( t 2 ) ] j + ( t 2 ) k, 0 ≤ t ≤ 4π c. r (t ) = ( cos t ) i − ( sin t ) j − tk, − 2π ≤ t ≤ 0 19. The involute of a circle If a string wound around a fixed circle is unwound while held taut in the plane of the circle, its end P traces an involute of the circle. In the accompanying figure, the circle in question is the circle x 2 + y 2 = 1 and the tracing point starts at (1, 0 ). The unwound portion of the string is tangent to the circle at Q, and t is the radian measure of the angle from the positive x-axis to segment OQ. Derive the parametric equations x = cos t + t sin t ,

y = sin t − t cos t , t > 0

of the point P ( x , y ) for the involute.

v(τ ) d τ

from Equation (3). Then use the formula for s(t ) to find the length of the indicated portion of the curve. 11. r (t ) = ( 4 cos t ) i + ( 4 sin t ) j + 3t k, 0 ≤ t ≤ π 2

y

12. r (t ) = ( cos t + t sin t ) i + ( sin t − t cos t ) j, π 2 ≤ t ≤ π

String

Q

13. r (t ) = ( e t cos t ) i + ( e t sin t ) j + e t   k, − ln 4 ≤ t ≤ 0 14. r (t ) = (1 + 2t ) i + (1 + 3t ) j + ( 6 − 6t ) k, −1 ≤ t ≤ 0

P(x, y) t

Theory and Examples

15. Arc length

Find the length of the curve

O

1

(1, 0)

x

r (t ) = ( 2t ) i + ( 2t ) j + (1 − t 2 ) k from ( 0, 0, 1) to ( 2, 2, 0 ). 16. Length of helix The length 2π 2 of the turn of the helix in Example 1 is also the length of the diagonal of a square 2π units on a side. Show how to obtain this square by cutting away and flattening a portion of the cylinder around which the helix winds. 17. Ellipse a. Show that the curve r (t ) = ( cos t ) i + ( sin t ) j + (1 − cos t ) k, 0 ≤ t ≤ 2π , is an ellipse by showing that it is the intersection of a right circular cylinder and a plane. Find equations for the cylinder and plane. b. Sketch the ellipse on the cylinder. Add to your sketch the unit tangent vectors at t = 0, π 2, π , and 3π 2.

20. (Continuation of Exercise 19.) Find the unit tangent vector to the involute of the circle at the point P ( x , y ). 21. Distance along a line Show that if u is a unit vector, then the arc length parameter along the line r (t ) = P0 + t u from the point P0 ( x 0 , y 0 , z 0 ) where t = 0 , is t itself. 22. Use Simpson’s Rule with n = 10 to approximate the length of arc of r (t ) = t i + t 2 j + t 3 k from the origin to the point ( 2, 4, 8 ).

747

12.4  Curvature and Normal Vectors of a Curve

12.4 Curvature and Normal Vectors of a Curve y

In this section we study how a curve turns or bends. To gain perspective, we look first at curves in the coordinate plane. Then we consider curves in space.

Curvature of a Plane Curve

T

P

T

s T

P0

x

0

FIGURE 12.17

As P moves along the curve in the direction of increasing arc length, the unit tangent vector turns. The value of d T ds at P is called the curvature of the curve at P.

As a particle moves along a smooth curve in the plane, T = d r ds turns as the curve bends. Since T is a unit vector, its length remains constant and only its direction changes as the particle moves along the curve. The rate at which T turns per unit of length along the curve is called the curvature (Figure 12.17). The traditional symbol for the curvature function is the Greek letter κ (“kappa”).

DEFINITION If T is the unit tangent vector of a smooth curve in the plane, then the curvature function of the curve is

κ =

dT . ds

If d T ds is large, T turns sharply as the particle passes through P, and the curvature at P is large. If d T ds is close to zero, T turns more slowly, and the curvature at P is smaller. If a smooth curve r(t ) is already given in terms of some parameter t other than the arc length parameter s, we can calculate the curvature as κ is the Greek letter kappa.

κ=

dT d T dt = ds dt ds

Chain Rule

dT dt

=

1 ds dt

=

1 dT . v dt

ds = v dt

Formula for Calculating Curvature

If r(t ) is a smooth curve in the plane, then the curvature is the scalar function κ =

1 dT , v dt

(1)

where T = v v is the unit tangent vector.

Testing the definition, we see in Examples 1 and 2 below that the curvature is constant for straight lines and circles.

T

FIGURE 12.18

Along a straight line, T always points in the same direction. The curvature, d T ds , is zero (Example 1).

EXAMPLE 1 A straight line is parametrized by r (t ) = C + tv for constant vectors C and v. Thus, r ′(t ) = v, and the unit tangent vector T = v v is a constant vector that always points in the same direction and has derivative 0 (Figure 12.18). It follows that, for any value of the parameter t, the curvature of the straight line is zero: κ =

1 dT 1 = 0 = 0. v dt v

748

Chapter 12 Vector-Valued Functions and Motion in Space

EXAMPLE 2

Here we find the curvature of a circle. We begin with the parametrization r (t ) = ( a cos t ) i + ( a sin t ) j

of a circle of radius a. Then v =

dr = −( a sin t ) i + ( a cos t ) j dt

v =

(−a sin t ) 2 + ( a cos t ) 2 =

a 2 = a = a.

Since a > 0, a = a.

From this we find T=

v = −( sin t ) i + ( cos t ) j v

dT = −( cos t ) i − ( sin t ) j dt dT = dt

cos 2 t + sin 2 t = 1.

Hence, for any value of the parameter t, the curvature of the circle is κ =

1 dT 1 1 1 = (1) = = . v dt a a radius

Among the vectors orthogonal to the unit tangent vector T, there is one of particular significance because it points in the direction in which the curve is turning. Since T has constant length (because its length is always 1), the derivative d T ds is orthogonal to T (Equation 4, Section 12.1). Therefore, if we divide d T ds by its length κ, we obtain a unit vector N orthogonal to T (Figure 12.19).

DEFINITION At a point where κ ≠ 0, the principal unit normal vector for a

smooth curve in the plane is N =

N = 1 dT k ds

T

T P2

P1

P0

The vector d T ds points in the direction in which T turns as the curve bends. Therefore, if we face in the direction of increasing arc length, the vector d T ds points toward the right if T turns clockwise and toward the left if T turns counterclockwise. In other words, the principal normal vector N will point toward the concave side of the curve (Figure 12.19). If a smooth curve r(t ) is already given in terms of some parameter t other than the arc length parameter s, we can use the Chain Rule to calculate N directly: N=

s N = 1 dT k ds

FIGURE 12.19 The vector dT ds, normal to the curve, always points in the direction in which T is turning. The unit normal vector N is the direction of dT ds.

1 dT . κ ds

= =

d T ds d T ds

( d T dt )( dt ds ) d T dt dt ds d T dt . d T dt

  

dt 1 = > 0 cancels. ds ds dt

This formula enables us to find N without having to find κ and s first.

749

12.4  Curvature and Normal Vectors of a Curve

Formula for Calculating N

If r(t ) is a smooth curve in the plane, then the principal unit normal is N =

d T dt , d T dt

(2)

where T = v v is the unit tangent vector.

EXAMPLE 3

Find T and N for the circular motion r (t ) = ( cos 2t ) i + ( sin 2t ) j.

Solution We first find T:

v = −( 2 sin 2t ) i + ( 2 cos 2t ) j v =

4 sin 2 2t + 4 cos 2 2t = 2 v T = = −( sin 2t ) i + ( cos 2t ) j. v

From this we find dT = −( 2 cos 2t ) i − ( 2 sin 2t ) j dt dT = dt

4 cos 2 2t + 4 sin 2 2t = 2

and N =

d T dt = −( cos 2t ) i − ( sin 2t ) j. d T dt

  

Eq. (2)

Notice that T ⋅ N = 0, verifying that N is orthogonal to T. Notice too, that for the circular motion here, N points from r(t ) toward the circle’s center at the origin.

Circle of Curvature for Plane Curves The circle of curvature or osculating circle at a point P on a plane curve where κ ≠ 0 is the circle in the plane of the curve that 1. is tangent to the curve at P (has the same tangent line the curve has) 2. has the same curvature the curve has at P 3. has center that lies toward the concave or inner side of the curve (as in Figure 12.20). Circle of curvature Center of curvature

Radius of curvature = ρ = Radius of curvature

N

FIGURE 12.20

The radius of curvature of the curve at P is the radius of the circle of curvature, which, according to Example 2, is

Curve

T P(x, y)

The center of the osculating circle at P( x , y ) lies toward the inner side of the curve.

1 . κ

To find ρ , we find κ and take the reciprocal. The center of curvature of the curve at P is the center of the circle of curvature. EXAMPLE 4 origin.

Find and graph the osculating circle of the parabola y = x 2 at the

Solution We parametrize the parabola using the parameter t = x (Section 10.1, Example 5):

r (t ) = ti + t 2 j.

750

Chapter 12 Vector-Valued Functions and Motion in Space

First we find the curvature of the parabola at the origin, using Equation (1): v = v =

dr = i + 2t j dt 1 + 4t 2

so that T =

v = (1 + 4 t 2 )−1 2 i + 2t (1 + 4 t 2 )−1 2 j. v

From this we find dT = −4 t (1 + 4 t 2 )−3 2 i + ⎡⎣ 2(1 + 4 t 2 )−1 2 − 8t 2 (1 + 4 t 2 )−3 2 ⎤⎦ j. dt At the origin, t = 0, so the curvature is κ( 0 ) =

y y=

x2

Osculating circle

=

1 dT ( ) 0 v ( 0 ) dt

Eq. (1)

1 0i + 2 j 1

= (1) 0 2 + 2 2 = 2. 1 2

0

x

1

Therefore, the radius of curvature is 1 κ = 1 2 . At the origin we have t = 0 and T = i, so N = j. Thus the center of the circle is ( 0, 1 2 ). The equation of the osculating circle is (x

(

− 0 )2 + y −

FIGURE 12.21

The osculating circle for the parabola y = x 2 at the origin (Example 4).

1 2

)

2

=

( 12 ) . 2

You can see from Figure 12.21 that the osculating circle is a better approximation to the parabola at the origin than is the tangent line approximation y = 0.

Curvature and Normal Vectors for Space Curves If a smooth curve in space is specified by the position vector r(t ) as a function of some parameter t, and if s is the arc length parameter of the curve, then the unit tangent vector T is d r ds = v v . The curvature in space is then defined to be z

κ =

t=p

0 (a, 0, 0)

t=0

r x2

(3)

just as for plane curves. The vector d T ds is orthogonal to T, and we define the principal unit normal to be

2pb t = 2p

dT 1 dT = ds v dt

+

N =

P

t=p 2

y2

=

a2

EXAMPLE 5 y

d T dt 1 dT . = d T dt κ ds

(4)

Find the curvature for the helix (Figure 12.22)

r (t ) = ( a cos t ) i + ( a sin t ) j + btk,

a, b ≥ 0,

a 2 + b 2 ≠ 0.

Solution We calculate T from the velocity vector v:

x

FIGURE 12.22

The helix

r (t ) = ( a cos t ) i + ( a sin t ) j + btk, drawn with a and b positive and  t ≥ 0 (Example 5).

v = −( a sin t ) i + ( a cos t ) j + b k v = T =

a 2 sin 2 t + a 2 cos 2 t + b 2 = v = v

a2 + b2

1 [ −( a sin t ) i + ( a cos t ) j + b k ]. a2 + b2

12.4  Curvature and Normal Vectors of a Curve

751

Then we use Equation (3): 1 dT v dt 1 1 = [ −( a cos t ) i − ( a sin t ) j ] a2 + b2 a2 + b2 a = 2 −( cos t ) i − ( sin t ) j a + b2 a a = 2 . ( cos t ) 2 + ( sin t ) 2 = 2 a + b2 a + b2 From this equation, we see that increasing b for a fixed a decreases the curvature. Decreasing a for a fixed b eventually decreases the curvature as well. If b = 0, the helix reduces to a circle of radius a, and its curvature reduces to 1 a, as it should. If a = 0, the helix becomes the z-axis, and its curvature reduces to 0, again as it should. κ=

Find N for the helix in Example 5 and describe how the vector is pointing.

EXAMPLE 6

Solution We have

dT 1 = − [ ( a cos t ) i + ( a sin t ) j ] dt a2 + b2 dT = dt N =

1 a2 + b2

a 2 cos 2 t + a 2 sin 2 t =

Example 5

a a2 + b2

d T dt d T dt

Eq. (4)

a2 + b2 1 ⋅ [ ( a cos t ) i + ( a sin t ) j ] 2 a a + b2 = −( cos t ) i − ( sin t ) j. = −

Thus, N is parallel to the xy-plane and always points toward the z-axis.

EXERCISES

12.4 6. A formula for the curvature of a parametrized plane curve

Plane Curves

Find T, N, and κ for the plane curves in Exercises 1–4. 1. r (t ) = ti + ( ln cos t ) j, −π 2 < t < π 2 2. r (t ) = ( ln sec t ) i + tj, −π 2 < t < π 2 3. r (t ) = ( 2t + 3 ) i + ( 5 − t 2 ) j 4. r (t ) = ( cos t + t sin t ) i + ( sin t − t cos t ) j, t > 0 5. A formula for the curvature of the graph of a function in the xy-plane a. The graph y = f ( x ) in the xy-plane automatically has the parametrization x = x , y = f ( x ), and the vector formula r ( x ) = xi + f ( x ) j. Use this formula to show that if f is a twice-differentiable function of x, then κ( x ) =

f ′′( x )

32.

2 ⎡ ⎤ ⎣ 1 + ( f ′( x ) ) ⎦ b. Use the formula for κ in part (a) to find the curvature of y = ln ( cos x ) , −π 2 < x < π 2. Compare your answer with the answer in Exercise 1.

c. Show that the curvature is zero at a point of inflection.

a. Show that the curvature of a smooth curve r (t ) = f (t ) i + g(t ) j defined by twice-differentiable functions x = f (t ) and y = g(t ) is given by the formula κ =

x ′y′′ − y′x ′′ [ ( x ′) 2 + ( y′) 2 ]

32

.

Apply this formula to find the curvatures of the following curves. b. r (t ) = ti + ( ln sin t ) j, 0 < t < π c. r (t ) = [ arctan ( sinh t ) ] i + ( ln cosh t ) j 7. Normals to plane curves a. Show that n(t ) = −g′(t ) i + f ′(t ) j and −n(t ) = g′(t ) i − f ′(t ) j are both normal to the curve r (t ) = f (t ) i + g(t ) j at the point ( f (t ),  g(t )). To obtain N for a particular plane curve, we can choose the one of n or −n from part (a) that points toward the concave side of the curve, and make it into a unit vector. (See Figure 12.19.) Apply this method to find N for the following curves.

752

Chapter 12 Vector-Valued Functions and Motion in Space

b. r (t ) = ti + e 2 t j c. r (t ) =

4 − t 2 i + tj, − 2 ≤ t ≤ 2

8. (Continuation of Exercise 7) a. Use the method of Exercise 7 to find N for the curve r (t ) = ti + (1 3 ) t 3 j when t < 0; when t > 0. b. Calculate N for t ≠ 0 directly from T using Equation (4) for the curve in part (a). Does N exist at t = 0? Graph the curve and explain what is happening to N as t passes from negative to positive values. Space Curves

Find T, N, and κ for the space curves in Exercises 9–16. 9. r (t ) = ( 3 sin t ) i + ( 3 cos t ) j + 4 tk 10. r (t ) = ( cos t + t sin t ) i + ( sin t − t cos t ) j + 3k 11. r (t ) =

(e t

cos t ) i +

(e t

sin t ) j + 2k

12. r (t ) = ( 6 sin 2t ) i + ( 6 cos 2t ) j + 5tk

derived in Exercise 5, expresses the curvature κ( x ) of a twicedifferentiable plane curve y = f ( x ) as a function of x. Find the curvature function of each of the curves in Exercises 23–26. Then graph f ( x ) together with κ( x ) over the given interval. You will find some surprises. 23. y = x 2, − 2 ≤ x ≤ 2

24. y = x 4 4 , − 2 ≤ x ≤ 2

25. y = sin x , 0 ≤ x ≤ 2π

26. y = e x , −1 ≤ x ≤ 2

In Exercises 27 and 28, determine the maximum curvature for the graph of each function. x 28. f ( x ) = for x > −1 27. f ( x ) = ln x x +1 29. Osculating circle Show that the center of the osculating circle for the parabola y = x 2 at the point ( a, a 2 ) is located at 1 −4 a 3, 3a 2 + . 2 30. Osculating circle Find a parametrization of the osculating circle for the parabola y = x 2 when x = 1.

(

)

13. r (t ) = ( t 3 3 ) i + ( t 2 2 ) j + k, t > 0 14. r (t ) = ( cos 3 t ) j + ( sin 3 t ) k, 0 < t < π 2

COMPUTER EXPLORATIONS

15. r (t ) = ti + ( a cosh ( t a )) k, a > 0

In Exercises 31–38 you will use a CAS to explore the osculating circle at a point P on a plane curve where κ ≠ 0. Use a CAS to perform the following steps:

16. r (t ) = ( cosh t ) i − ( sinh t ) j + tk More on Curvature

17. Show that the parabola y = ax 2,  a ≠ 0, has its largest curvature at its vertex and has no minimum curvature. (Note: Since the curvature of a curve remains the same if the curve is translated or rotated, this result is true for any parabola.) 18. Show that the ellipse x = a cos t , y = b sin t , a > b > 0, has its largest curvature on its major axis and its smallest curvature on its minor axis. (The same is true for any ellipse.) 19. Maximizing the curvature of a helix In Example 5, we found the curvature of the helix r (t ) = ( a cos t ) i + ( a sin t ) j + btk ( a, b ≥ 0 ) to be κ = a ( a 2 + b 2 ). What is the largest value κ can have for a given value of b? Give reasons for your answer. 20. Total curvature We find the total curvature of the portion of a smooth curve that runs from s = s 0 to s = s1 > s 0 by integrating κ from s 0 to s1. If the curve has some other parameter, say t, then the total curvature is K =

∫s

s1

κ ds =

0

∫t

t1 0

κ

ds dt = dt

∫t

t1

κ v dt ,

0

where t 0 and t 1 correspond to s 0 and s1. Find the total curvatures of a. The portion of the helix r (t ) = ( 3 cos t ) i + ( 3 sin t ) j + tk, 0 ≤ t ≤ 4 π. b. The parabola y = x 2, −∞ < x < ∞. 21. Find an equation for the circle of curvature of the curve r (t ) = ti + ( sin t ) j at the point ( π 2,1). (The curve parametrizes the graph of y = sin x in the xy-plane.) 22. Find an equation for the circle of curvature of the curve r (t ) = ( 2 ln t ) i − [ t + (1 t ) ] j, e −2 ≤ t ≤ e 2, at the point ( 0, − 2 ) , where t = 1.

a. Plot the plane curve given in parametric or function form over the specified interval to see what it looks like. b. Calculate the curvature κ of the curve at the given value t 0 using the appropriate formula from Exercise 5 or 6. Use the parametrization x = t and y = f (t ) if the curve is given as a function y = f ( x ). c. Find the unit normal vector N at t 0 . Notice that the signs of the components of N depend on whether the unit tangent vector T is turning clockwise or counterclockwise at t = t 0 . (See Exercise 7.) d. If C = a i + b j is the vector from the origin to the center ( a, b ) of the osculating circle, find the center C from the vector equation C = r (t 0 ) +

1 N(t 0 ). κ (t 0 )

The point P ( x 0 , y 0 ) on the curve is given by the position vector r(t 0 ). e. Plot implicitly the equation ( x − a ) 2 + ( y − b ) 2 = 1 κ 2 of the osculating circle. Then plot the curve and osculating circle together. You may need to experiment with the size of the viewing window, but be sure the axes are equally scaled. 31. r (t ) = ( 3 cos t ) i + ( 5 sin t ) j, 0 ≤ t ≤ 2π , t 0 = π 4 32. r (t ) = ( cos 3 t ) i + ( sin 3 t ) j, 0 ≤ t ≤ 2π , t 0 = π 4 33. r (t ) = t 2 i + ( t 3 − 3t ) j, − 4 ≤ t ≤ 4, t 0 = 3 5 3t 34. r (t ) = ( t 3 − 2t 2 − t ) i + j, − 2 ≤ t ≤ 5, t 0 = 1 1 + t2 35. r (t ) = ( 2t − sin t ) i + ( 2 − 2 cos t ) j, 0 ≤ t ≤ 3π , t 0 = 3π 2 36. r (t ) = ( e −t cos t ) i + ( e −t sin t ) j, 0 ≤ t ≤ 6π, t 0 = π 4

T The formula κ( x ) =

37. y = x 2 − x , − 2 ≤ x ≤ 5, x 0 = 1

f ′′( x ) 2 ⎡ ⎤ ⎣ 1 + ( f ′( x ) ) ⎦

32

,

38. y = x (1 − x ) 2 5, −1 ≤ x ≤ 2, x 0 = 1 2

753

12.5  Tangential and Normal Components of Acceleration

12.5 Tangential and Normal Components of Acceleration z

B=T×N N = 1 dT k ds r

P

T = dr ds y

s

If you are flying in an airplane that is traveling along a curve in space, the Cartesian i, j, and k coordinate system for representing the vectors describing your motion may not be very relevant to you. Vectors that are likely to be more important are those representing your forward direction (the unit tangent vector T) and the direction in which your path is turning (the unit normal vector N), along with a third unit vector perpendicular to the other two. Expressing the acceleration vector along the curve as a linear combination of these three mutually orthogonal unit vectors traveling with the motion (Figure 12.23) can reveal much about the nature of your path and your motion along it.

P0 x

The TNB Frame

FIGURE 12.23

The binormal vector of a curve in space is B = T × N, which is a unit vector that is orthogonal to both T and N (Figure 12.24). Together T, N, and B define a moving righthanded vector frame that plays a significant role in analyzing the paths of particles moving through space. It is called the Frenet (“fre-nay”) frame (after Jean-Frédéric Frenet, 1816–1900), or the TNB frame.

The TNB frame of mutually orthogonal unit vectors traveling along a curve in space.

B

Tangential and Normal Components of Acceleration P

When an object is accelerated by gravity, brakes, or rocket motors, we often need to know how much of the acceleration acts in the direction of motion, which is the direction of the tangent vector T. We can calculate this using the Chain Rule to rewrite v as

N

T

The vectors T, N, and B (in that order) make a right-handed frame of mutually orthogonal unit vectors in space.

dr d r ds ds = = T . dt ds dt dt

v =

FIGURE 12.24

Then we differentiate both ends of this string of equalities to get a=

(

)

dv d ds d 2s ds d T T T+ = = 2 dt dt dt dt dt dt

(

(

)

=

d 2s ds d T ds d 2s ds ds T+ T+ κN = 2 dt dt ds dt dt 2 dt dt

=

d 2s ds T+κ dt 2 dt

DEFINITION

( )

)

  

dT = κN ds

2

N.

If the acceleration vector is written as a = a T T + a N N,

then aT =

a N aN = k Q

ds R dt

2

T 2 a T = d 2s dt

s P0

FIGURE 12.25

The tangential and normal components of acceleration. The acceleration a always lies in the plane of T and N and is orthogonal to B.

d 2s d v = dt 2 dt

and

aN = κ

(1)

( dsdt )

2

= κ v

2

(2)

are the tangential and the normal scalar components of acceleration. Notice that the binormal vector B does not appear in Equation (1). No matter how the path of the moving object we are watching may appear to twist and turn in space, the acceleration a always lies in the plane of T and N and therefore is orthogonal to B. The equation also tells us exactly how much of the acceleration takes place tangent to the motion ( d 2 s dt 2 ) and how much takes place normal to the motion ⎡⎣ κ ( ds dt ) 2 ⎤⎦ (Figure 12.25). What information can we discover from Equations (2)? By definition, acceleration a is the rate of change of velocity v, and in general, both the length and direction of v change as an object moves along its path. The tangential component of acceleration a T measures

754

Chapter 12 Vector-Valued Functions and Motion in Space

the rate of change of the length of v (that is, the change in the speed). The normal component of acceleration a N is proportional to the rate of change of the direction of v. Notice that the normal scalar component of the acceleration is the curvature times the square of the speed. This explains why you have to hold on when your car makes a sharp (large κ), high-speed (large v ) turn. If you double the speed of your car, you will experience four times the normal component of acceleration for the same curvature. If an object moves in a circle at a constant speed, d 2 s dt 2 is zero and all the acceleration points along N toward the circle’s center. If the object is speeding up or slowing down, a has a positive or negative tangential component (Figure 12.26). To calculate a N , we usually use the formula a N = a 2 − a T 2 , which comes from 2 solving the equation a = a ⋅ a = a T 2 + a N 2 for a N (which, unlike the tangential component, cannot be negative). With this formula, we can find a N without having to calculate κ first.

d 2s T dt 2 P

a

0 v0 2 k0 v0 2 N = r N

C

FIGURE 12.26

The tangential and normal components of the acceleration of an object that is speeding up as it moves counterclockwise around a circle of radius ρ.

Formula for Calculating the Normal Component of Acceleration

aN =

a

2

− aT 2

(3)

Without finding T and N, write the acceleration of the motion

EXAMPLE 1

r (t ) = ( cos t + t sin t ) i + ( sin t − t cos t ) j, T

in the form a = a T T + a N N. (The path of the motion is the involute of the circle in Figure 12.27. See also Section 12.3, Exercise 19.)

a P(x, y)

tN

y

Solution We use the first of Equations (2) to find a T :

ing

Str

v =

r

Q

(1, 0)

x

x2 + y2 = 1

dr = (− sin t + sin t + t cos t ) i + ( cos t − cos t + t sin t ) j dt = ( t cos t ) i + ( t sin t ) j v =

t O

t > 0

aT =

t 2 cos 2 t + t 2 sin 2 t =

t2 = t = t

t > 0

d d v = (t ) = 1. dt dt

Eq. (2)

Knowing a T , we use Equation (3) to find a N : a = ( cos t − t sin t ) i + ( sin t + t cos t ) j a

FIGURE 12.27

The tangential and normal components of the acceleration of the motion r (t ) = ( cos t + t sin t ) i + ( sin t − t cos t ) j, for t > 0. If a string wound around a fixed circle is unwound while held taut in the plane of the circle, its end P traces an involute of the circle (Example 1).

2

= t2 + 1

aN = =

a

2

  After some algebra

− aT 2

( t 2 + 1) − ( 1) =

t 2 = t.

We then use Equation (1) to write a = a T T + a N N = (1) T + (t )N = T + tN.

Torsion How does d B ds behave in relation to T, N, and B? From the rule for differentiating a cross product in Section 12.1, we have dB d(T × N) dT dN . = = ×N + T× ds ds ds ds

755

12.5  Tangential and Normal Components of Acceleration

Since N is the direction of d T ds , ( d T ds ) × N = 0 and dB dN dN . = 0 + T× = T× ds ds ds From this we see that d B ds is orthogonal to T, since a cross product is orthogonal to its factors. Since d B ds is also orthogonal to B (the latter has constant length), it follows that d B ds is orthogonal to the plane of B and T. In other words, d B ds is parallel to N, so d B ds is a scalar multiple of N. In symbols, dB = −τ N. ds The negative sign in this equation is traditional. The scalar τ is called the torsion along the curve. Notice that dB ⋅ N = −τ N ⋅ N = −τ (1) = −τ . ds We use this equation for our next definition.

DEFINITION

Let B = T × N. The torsion function of a smooth curve is τ = −

Binormal Rectifying plane

Normal plane

B P T

Unit tangent

N

Principal normal

Osculating plane

dB ⋅ N. ds

(4)

Unlike the curvature κ, which is never negative, the torsion τ may be positive, negative, or zero. The three planes determined by T, N, and B are named and shown in Figure 12.28. The curvature κ = d T ds can be thought of as the rate at which the normal plane turns as the point P moves along its path. Similarly, the torsion τ = −( d B ds ) ⋅ N is the rate at which the osculating plane turns about T as P moves along the curve. Torsion measures how the curve twists. Look at Figure 12.29. If P is a train climbing up a curved track, the rate at which the headlight turns from side to side per unit distance is the curvature of the track. The rate at which the engine tends to twist out of the plane formed by T and N is the torsion. It can be shown that a space curve is a helix if and only if it has constant nonzero curvature and constant nonzero torsion.

FIGURE 12.28

The names of the three planes determined by T, N, and B.

dB ds

The torsion at P is −(dBds) ∙ N.

B s increases s=0

FIGURE 12.29

N

The curvature at P is @ (dTds)@ . T

P

Every moving body travels with a TNB frame that characterizes the geometry of its path of motion.

756

Chapter 12 Vector-Valued Functions and Motion in Space

Formulas for Computing Curvature and Torsion We now give easy-to-use formulas for computing the curvature and torsion of a smooth curve. From Equations (1) and (2), we have v×a =

( dsdt T ) × ⎡⎢⎢⎣ ddt s T + κ ( dsdt ) 2

2

(

)

⎤ N⎥ ⎥⎦

( ) (T × N)

ds d 2 s ( ds T × T) + κ = dt dt 2 dt =κ

2

3

( dsdt ) B. 3

v = dr dt = (ds dt )T

T × T = 0 and T×N = B

It follows that ds 3 B = κ v 3. dt Solving for κ gives the following formula. v×a = κ

  

ds = v dt

and

B =1

Vector Formula for Curvature

κ =

v×a v3

(5)

Equation (5) calculates the curvature, a geometric property of the curve, from the velocity and acceleration of any vector representation of the curve in which v is different from zero. From any formula for motion along a curve, no matter how variable the motion may be (as long as v is never zero), we can calculate a geometric property of the curve that seems to have nothing to do with the way the curve is parametrically defined. The most widely used formula for torsion, derived in more advanced texts, is given in a determinant form.

Formula for Torsion

τ =

Newton’s Dot Notation for Derivatives The dots in Equation (6) denote differentiation with respect to t, one derivative for each dot. Thus, x (“x dot”) means dx dt , x (“x double dot”) means d 2 x dt 2, x (“x triple dot”) means d 3 x dt 3. and  Similarly, y = dy dt, and so on.

x y z x y z  x  y  z v×a

2

( if  v

× a ≠ 0)

(6)

This formula calculates the torsion directly from the derivatives of the component functions x = f (t ), y = g(t ), z = h(t ) that make up r. The determinant’s first row comes from v, the second row comes from a, and the third row comes from a = d a dt . This formula for torsion is traditionally written using Newton’s dot notation for derivatives.

EXAMPLE 2 helix

Use Equations (5) and (6) to find the curvature κ and torsion τ for the

r (t ) = ( a cos t ) i + ( a sin t ) j + bt k,

a, b ≥ 0,

a 2 + b 2 ≠ 0.

12.5  Tangential and Normal Components of Acceleration

757

Solution We calculate the curvature with Equation (5):

v = −( a sin t ) i + ( a cos t ) j + b k a = −( a cos t ) i − ( a sin t ) j i v × a = −a sin t

j

k

a cos t

b

−a cos t −a sin t 0 = ( ab sin t ) i − ( ab cos t ) j + a 2 k v×a a 2b 2 + a 4 a a2 + b2 a = = . 3 3 2 32 = v a2 + b2 (a2 + b2 ) (a2 + b2 )

κ=

(7)

Notice that Equation (7) agrees with the result in Example 5 in Section 12.4, where we calculated the curvature directly from its definition. To evaluate Equation (6) for the torsion, we find the entries in the determinant by differentiating r with respect to t. We already have v and a, and da a = = ( a sin t ) i − ( a cos t ) j. dt Hence,

τ =

x

y

z

−a sin t

a cos t b

x

y

z

−a cos t

−a sin t 0

 x  y  z v×a

2

=

a sin t −a cos t 0

(a

a2 + b2 )

2

  Value of  v × a  from Eq. (7)

b ( a 2 cos 2 t + a 2 sin 2 t ) a2(a2 + b2 ) b = 2 . a + b2

=

From this last equation we see that the torsion of a helix about a circular cylinder is constant. In fact, constant curvature and constant torsion characterize the helix among all curves in space.

Computation Formulas for Curves in Space

Unit tangent vector: Principal unit normal vector: Binormal vector: Curvature:

v v d T dt N = d T dt B = T×N dT v×a κ = = ds v3 T =

Torsion:

τ = −

Tangential and normal scalar components of acceleration:

a

dB ⋅N = ds

x

y

z

x

y

z

 x  y  z v×a

2

= a TT + a NN

aT =

d v dt

aN = κ v

2

=

a

2

− aT 2

758

Chapter 12 Vector-Valued Functions and Motion in Space

EXERCISES

12.5

Finding Tangential and Normal Components

Theory and Examples

In Exercises 1–4, write a in the form a = a T T + a N N without finding T and N.

25. Show that κ and τ are both zero for the line

1. r (t ) = ( 2t + 3 ) i + ( t 2 − 1) j 2. r (t ) = 3t 2 i + 2t 3 j

26. Show that a moving particle will move in a straight line if the normal component of its acceleration is zero.

3. r (t ) = ( a cos t ) i + ( a sin t ) j + bt k 4. r (t ) = (1 + 3t ) i + ( t − 2 ) j − 3t k In Exercises 5–10, write a in the form a = a T T + a N N at the given value of t without finding T and N. 5. r (t ) = e t i + e −t j, t = 0 6. r (t ) = cos(t 2 ) i + sin (t 2 ) j, t =

r (t ) = ( x 0 + At ) i + ( y 0 + Bt ) j + ( z 0 + Ct ) k.

27. A sometime shortcut to curvature If you already know a N and v , then the formula a N = κ v 2 gives a convenient way to find the curvature. Use it to find the curvature and radius of curvature of the curve r (t ) = ( cos t + t sin t ) i + ( sin t − t cos t ) j, t > 0.

π

1 2

7. r (t ) = ( t + 1) i + 2tj + t 2 k, t = 1 8. r (t ) = ( t cos t ) i + ( t sin t ) j + t 2 k, t = 0

(Take a N and v from Example 1.) 28. What can be said about the torsion of a smooth plane curve r (t ) = f (t ) i + g(t ) j ? Give reasons for your answer.

12. r (t ) = ( cos t ) i + ( sin t ) j + t k, t = 0

29. Differentiable curves with zero torsion lie in planes That a sufficiently differentiable curve with zero torsion lies in a plane is a special case of the fact that a particle whose velocity remains perpendicular to a fixed vector C moves in a plane perpendicular to C. This, in turn, can be viewed as the following result. Suppose r (t ) = f (t ) i + g(t ) j + h(t ) k is twice differentiable for all t in an interval [ a, b ], that r = 0 when t = a, and that v ⋅ k = 0 for all t in [ a, b ]. Show that h(t ) = 0 for all t in [ a, b ]. (Hint: Start with a = d 2 r dt 2 and apply the initial conditions in reverse order.)

In Exercises 9–16 of Section 12.4, you found T, N, and κ. Now, in the following Exercises 13–20, find B and τ for these space curves.

30. A formula that calculates τ from B and v If we start with the definition τ = −( d B ds ) ⋅ N and apply the Chain Rule to rewrite d B ds as

9. r (t ) = t 2 i + ( t + (1 3 ) t 3 ) j + ( t − (1 3 ) t 3 ) k, t = 0 10. r (t ) = ( e t cos t ) i + ( e t sin t ) j +

2 e t k, t = 0

Finding the TNB Frame

In Exercises 11 and 12, find r, T, N, and B at the given value of t. Then find equations for the osculating, normal, and rectifying planes at that value of t. 11. r (t ) = ( cos t ) i + ( sin t ) j − k, t = π 4

13. r (t ) = ( 3 sin t ) i + ( 3 cos t ) j + 4 tk 14. r (t ) = ( cos t + t sin t ) i + ( sin t − t cos t ) j + 3k 15. r (t ) = ( e t cos t ) i + ( e t sin t ) j + 2k 16. r (t ) = ( 6 sin 2t ) i + ( 6 cos 2t ) j + 5tk 17. r (t ) = ( t 3 3 ) i + ( t 2 2 ) j + k, t > 0

dB d B dt dB 1 = = , ds dt ds dt v we arrive at the formula τ = −

18. r (t ) = ( cos 3 t ) j + ( sin 3 t ) k, 0 < t < π 2 19. r (t ) = ti + ( a cosh ( t a )) k, a > 0

(

)

1 dB ⋅N . v dt

Use the formula to find the torsion of the helix in Example 2.

20. r (t ) = ( cosh t ) i − ( sinh t ) j + tk COMPUTER EXPLORATIONS Physical Applications

21. The speedometer on your car reads a steady 35 km/h. Could you be accelerating? Explain.

Rounding the answers to four decimal places, use a CAS to find v, a, speed, T, N, B, κ,  τ , and the tangential and normal components of acceleration for the curves in Exercises 31–34 at the given values of t.

22. Can anything be said about the acceleration of a particle that is moving at a constant speed? Give reasons for your answer.

32. r (t ) = ( e t cos t ) i + ( e t sin t ) j + e t k, t = ln 2

23. Can anything be said about the speed of a particle whose acceleration is always orthogonal to its velocity? Give reasons for your answer. 24. An object of mass m travels along the parabola y = x 2 with a constant speed of 10 units s. What is the force on the object due to its acceleration at ( 0, 0 )? at ( 2 1 2, 2 )? Write your answers in terms of i and j. (Remember Newton’s law, F = ma.)

31. r (t ) = ( t cos t ) i + ( t sin t ) j + tk, t = 33. r (t ) = ( t − sin t ) i + (1 − cos t ) j +

3

−t k, t = −3π

34. r (t ) = ( 3t − t 2 ) i + ( 3t 2 ) j + ( 3t + t 3 ) k, t = 1

759

12.6  Velocity and Acceleration in Polar Coordinates

12.6 Velocity and Acceleration in Polar Coordinates In this section we derive equations for velocity and acceleration in polar coordinates. These equations are useful for calculating the paths of planets and satellites in space, and we use them to examine Kepler’s three laws of planetary motion.

Motion in Polar and Cylindrical Coordinates

y

When a particle at P ( r, θ ) moves along a curve in the polar coordinate plane, we express its position, velocity, and acceleration in terms of the moving unit vectors

uu P(r, u)

r

u r = ( cos θ ) i + ( sin θ ) j,

u θ = −( sin θ ) i + ( cos θ ) j, (1)  shown in Figure 12.30. The vector u r points along the position vector OP, so r = ru r . The vector u θ , orthogonal to u r , points in the direction of increasing θ. We find from Equations (1) that

ur

u

x

O

FIGURE 12.30

The length of r is the positive polar coordinate r of the point P. Thus u r , which is r r , is also r r . Equations (1) express u r and u θ in terms of i and j.

du r = −( sin θ ) i + ( cos θ ) j = u θ dθ du θ = −( cos θ ) i − ( sin θ ) j = −u r . dθ We next differentiate u r and u θ with respect to t to find how they change with time. The Chain Rule gives u r =

y v

x

O

du θ  θ = −θu r . dθ

(2)

d v = r = ( ru r ) = r u r + r u r = r u r + rθu θ . dt

P(r, u)

u

u θ =

Hence, we can express the velocity vector in terms of u r and u θ as

. ruuu . r ur

r

du r  θ = θu θ , dθ

See Figure 12.31. As in the previous section, we use Newton’s dot notation for time derivatives to keep the formulas as simple as we can: u r means d u r dt, θ means dθ dt , and so on. The acceleration is a = v = ( ru r + ru r ) + ( rθu θ + r θu θ + r θu θ ).

FIGURE 12.31

In polar coordinates, the velocity vector is v = r u r + r θ u θ .

When Equations (2) are used to evaluate u r and u θ and the components are separated, the equation for acceleration in terms of u r and u θ becomes a = ( r − rθ 2 ) u r + ( r θ + 2rθ ) u θ .

z

To extend these equations of motion to space, we add zk to the right-hand side of the equation r = ru r . Then, in these cylindrical coordinates, we have k

uu

r = r ur + z k

zk u r ur

Position:

ur

Velocity: y

x

Acceleration:

r = ru r + zk v = ru r + rθu θ + zk a = ( r − rθ 2 ) u r + ( rθ + 2rθ ) u θ + zk

FIGURE 12.32

Position vector and basic unit vectors in cylindrical coordinates. Notice that r ≠ r if z ≠ 0 since r = r 2 + z2.

The vectors u r , u θ , and k make a right-handed frame (Figure 12.32) in which u r × u θ = k,

uθ × k = ur,

k × u r = u θ.

(3)

760

Chapter 12 Vector-Valued Functions and Motion in Space

Planets Move in Planes Newton’s law of gravitation says that if r is the radius vector from the center of a sun of mass M to the center of a planet of mass m, then the force F of the gravitational attraction between the planet and sun is

M r 0 r0

r

m r F = − GmM 0 r0 2 0 r0

F = −

FIGURE 12.33

The force of gravity is directed along the line joining the centers of mass.

GmM r r2 r

(Figure 12.33). The number G is the universal gravitational constant. If we measure mass in kilograms, force in newtons, and distance in meters, G is about 6.6738 × 10 −11 Nm 2 kg −2 . Combining the gravitation law with Newton’s second law, F  mr, for the force acting on the planet gives mr = −

GmM r , r2 r

r = − GM2 r . r r The planet is therefore accelerated toward the sun’s center of mass at all times. Since r is a scalar multiple of r, we have r × r = 0. From this last equation, . C=r×r r

d( r × r ) =  r × r + r × r = r × r = 0.  dt

Sun

0

It follows that

Planet . r

r × r = C FIGURE 12.34

A planet that obeys Newton’s laws of gravitation and motion travels in the plane through its sun’s center  of mass perpendicular to C = r × r.

(4)

for some constant vector C. Equation (4) tells us that r and r always lie in a plane perpendicular to C. Hence, the planet moves in a fixed plane through the center of mass of its sun (Figure 12.34). We next see how Kepler’s laws describe the motion in a precise way.

Kepler’s First Law (Ellipse Law) Kepler’s first law says that a planet’s path is an ellipse with its sun at one focus. The eccentricity of the ellipse is e =

r0 V 0 2 − 1, GM

(5)

and the polar equation (see Section 10.7 Equation (5)) is r = Planet r

(1 + e ) r0 . 1 + e cos R

(6)

Here V 0 is the speed when the planet is positioned at its minimum distance r0 from the sun. We omit the lengthy proof. The sun’s mass M is 1.99 q 10 30 kg.

Sun

Kepler’s Second Law (Equal Area Law) FIGURE 12.35

The line joining a planet to its sun sweeps over equal areas in equal times.

Kepler’s second law says that the radius vector from the sun to a planet (the vector r in our model) sweeps out equal areas in equal times, as displayed in Figure 12.35. In that figure, we assume the plane of the planet is the xy-plane, so the unit vector in the direction of C is k.

12.6  Velocity and Acceleration in Polar Coordinates

761

We introduce polar coordinates in the plane, choosing as initial line R  0, the direction r when r  r is a minimum value. Then at t  0, we have r (0)  r0 being a minimum so, r

t=0

=

dr dt

t=0

= 0 and υ 0 = v

t=0

= ⎡⎣ rθ ⎤⎦ t = 0 .

  

Eq. (3), z  0

To derive Kepler’s second law, we use Equation (3) to evaluate the cross product C = r × r from Equation (4): C = r × r = r × v = ru r × (ru r + rRu R ) = rr ( u r × u r ) + r (rR)( u r × u R )   = r (rR) k.

  

Eq. (3), z  0

k

0

(7)

Setting t equal to zero shows that C = ⎡⎣ r (rθ) ⎤⎦ t = 0 k = r0 υ 0 k. Substituting this value for C in Equation (7) gives r0 υ 0 k  r 2θk, or r 2θ  r0 υ 0 . This is where the area comes in. The area differential in polar coordinates is dA 

1 2 r dR 2

(Section 10.5). Accordingly, dA dt has the constant value dA 1 1  r 2θ  r0 υ 0 . dt 2 2

(8)

So dA dt is constant, giving Kepler’s second law. HISTORICAL BIOGRAPHY Johannes Kepler (1571–1630) The German astronomer, mathematician, and physicist Johannes Kepler was the first scientist to demand physical explanations of celestial phenomena. His three laws of planetary motion, the results of a lifetime of work, changed astronomy and played a crucial role in the development of Newtonian physics and calculus.

Kepler’s Third Law (Time–Distance Law) The time T it takes a planet to go around its sun once is the planet’s orbital period. Kepler’s third law says that T and the orbit’s semimajor axis a are related by the equation T2 4Q 2 .  a3 GM Since the right-hand side of this equation is constant within a given solar system, the ratio of T 2 to a 3 is the same for every planet in the system. Here is a partial derivation of Kepler’s third law. The area enclosed by the planet’s elliptical orbit is calculated as follows:

To know more, visit the companion Website.

T

Area =

∫0

=

∫0

=

1 Tr V . 2 0 0

T

dA 1 r V dt 2 0 0

  

Eq. (8)

If b is the semiminor axis, the area of the ellipse is Qab, so T =

2πab 2πa 2 1 − e2. = r0 υ 0 r0 υ 0

  

For any ellipse, b = a 1 − e 2 .

(9)

It remains only to express a and e in terms of r0 , V 0 , G , and M. Equation (5) does this for e. For a, we observe that setting R equal to Q in Equation (6) gives rmax = r0

1+e . 1−e

762

Chapter 12 Vector-Valued Functions and Motion in Space

Planet

Hence, from Figure 12.36,

r rmax

Sun

2a = r0 + rmax = r0

2r0 GM 2r0 = . 1−e 2GM − r0 υ 0 2

(10)

Squaring both sides of Equation (9) and substituting the results of Equations (5) and (10) produce Kepler’s third law (Exercise 11).

FIGURE 12.36

The length of the major axis of the ellipse is 2a = r0 + rmax .

12.6

EXERCISES

In Exercises 1–7, find the velocity and acceleration vectors in terms of u r and u θ . 1. r = θ

and

dθ = 2 dt

1 θ

and

dθ = t2 dt

2. r =

3. r = a (1 − cos θ ) and 4. r = a sin 2θ 5. r = e aθ

and

and

12. Do the data in the accompanying table support Kepler’s third law? Give reasons for your answer.

dθ = 2t dt

dθ = 2 dt and θ = 2t

8. Type of orbit For what values of υ 0 in Equation (5) is the orbit in Equation (6) a circle? An ellipse? A parabola? A hyperbola? 9. Circular orbits Show that a planet in a circular orbit moves with a constant speed. (Hint: This is a consequence of one of Kepler’s laws.) 10. Suppose that r is the position vector of a particle moving along a plane curve and dA dt is the rate at which the vector sweeps out area. Without introducing coordinates, and assuming the necessary derivatives exist, give a geometric argument based on increments and limits for the validity of the equation dA 1 = r × r . dt 2

CHAPTER 12

Semimajor axis a ( 10 10 m )

Period T (years)

Mercury

    5.79

  0.241

Venus

  10.81

  0.615

Mars

  22.78

  1.881

Saturn

142.70

29.457

Planet

dθ = 3 dt

6. r = a (1 + sin t ) and θ = 1 − e −t 7. r = 2 cos 4 t

11. Kepler’s third law Complete the derivation of Kepler’s third law (the part following Equation (10)).

13. Earth’s major axis Estimate the length of the major axis of Earth’s orbit if its orbital period is 365.256 days. 14. Estimate the length of the major axis of the orbit of Uranus if its orbital period is 84 years. 15. The eccentricity of Earth’s orbit is e = 0.0167, so the orbit is nearly circular, with radius approximately 150 × 10 6 km . Find, in units of km 2 s, the rate dA dt satisfying Kepler’s second law. 16. Jupiter’s orbital period Estimate the orbital period of Jupiter, assuming that a = 77.8 × 10 10 m. 17. Mass of Jupiter Io is one of the moons of Jupiter. It has a semimajor axis of 0.042 × 10 10 m and an orbital period of 1.769 days. Use these data to estimate the mass of Jupiter. 18. Distance from Earth to the moon The period of the moon’s rotation around Earth is 2.36055 × 10 6 s. Estimate the distance to the moon.

Questions to Guide Your Review

1. State the rules for differentiating and integrating vector functions. Give examples. 2. How do you define and calculate the velocity, speed, direction of motion, and acceleration of a body moving along a sufficiently differentiable space curve? Give an example. 3. What is special about the derivatives of vector functions of constant length? Give an example. 4. What are the vector and parametric equations for ideal projectile motion? How do you find a projectile’s maximum height, flight time, and range? Give examples.

5. How do you define and calculate the length of a segment of a smooth space curve? Give an example. What mathematical assumptions are involved in the definition? 6. How do you measure distance along a smooth curve in space from a preselected base point? Give an example. 7. What is a differentiable curve’s unit tangent vector? Give an example. 8. Define curvature, circle of curvature (osculating circle), center of curvature, and radius of curvature for twice-differentiable curves in the plane. Give examples. What curves have zero curvature? Constant curvature?

Chapter 12  Practice Exercises

9. What is a plane curve’s principal normal vector? When is it defined? Which way does it point? Give an example. 10. How do you define N and κ for curves in space? How are these quantities related? Give examples. 11. What is a curve’s binormal vector? Give an example. How is this vector related to the curve’s torsion? Give an example.

CHAPTER 12

763

12. What formulas are available for writing a moving object’s acceleration as a sum of its tangential and normal components? Give an example. Why might one want to write the acceleration this way? What if the object moves at a constant speed? At a constant speed around a circle? 13. State Kepler’s laws.

Practice Exercises

Motion in the Plane In Exercises 1 and 2, graph the curves and sketch their velocity and acceleration vectors at the given values of t. Then write a in the form a = a T T + a N N without finding T and N, and find the value of κ at the given values of t.

y

1. r (t ) = ( 4 cos t ) i + ( 2 sin t ) j, t = 0 and  π 4

C

2. r (t ) = ( 3 sec t ) i + ( 3 tan t ) j, t = 0 3. The position of a particle in the plane at time t is

r

t j. 1 + t2

0

r =

1 i+ 1 + t2

Find the particle’s highest speed. 4. Suppose r (t ) = ( e t cos t ) i + ( e t sin t ) j. Show that the angle between r and a never changes. What is the angle? 5. Finding curvature At point P, the velocity and acceleration of a particle moving in the plane are v = 3i + 4 j and a = 5 i + 15 j. Find the curvature of the particle’s path at P. 6. Find the point on the curve y = e x where the curvature is greatest. 7. A particle moves around the unit circle in the xy-plane. Its position at time t is r = xi + yj, where x and y are differentiable functions of t. Find dy dt if v ⋅ i = y. Is the motion clockwise or counterclockwise? 8. You send a message through a pneumatic tube that follows the curve 9 y = x 3 (distance in meters). At the point ( 3, 3 ), v ⋅ i = 4 and a ⋅ i = −2. Find the values of v ⋅ j and a ⋅ j at ( 3, 3 ). 9. Characterizing circular motion A particle moves in the plane so that its velocity and position vectors are always orthogonal. Show that the particle moves in a circle centered at the origin. 10. Speed along a cycloid A circular wheel with radius 1 m and center C rolls to the right along the x-axis at a half-turn per second. (See the accompanying figure.) At time t seconds, the position vector of the point P on the wheel’s circumference is r = ( πt − sin πt ) i + (1 − cos πt ) j. a. Sketch the curve traced by P during the interval 0 ≤ t ≤ 3. b. Find v and a at t = 0, 1, 2, and 3 and add these vectors to your sketch. c. At any given time, what is the forward speed of the topmost point of the wheel? Of C?

pt

1

P

x

Projectile Motion

11. Shot put A shot leaves the thrower’s hand 2 m above the ground at a 45 ° angle at 14 m s. Where is it 3 s later? 12. Javelin A javelin leaves the thrower’s hand 2.5 m above the ground at a 45 ° angle at 24 m s. How high does it go? 13. A golf ball is hit with an initial speed υ 0 at an angle α to the horizontal from a point that lies at the foot of a straight-sided hill that is inclined at an angle φ to the horizontal, where 0 0, is a sphere of radius c centered at the origin. Figure 13.8 shows a cutaway view of three of these spheres. The level surface x 2 + y 2 + z 2 = 0 consists of the origin alone. We are not graphing the function here; we are looking at level surfaces in the function’s domain. The level surfaces show how the function’s values change as we move through its domain. If we remain on a sphere of radius c centered at the origin, the function maintains a constant value, namely c. If we move from a point on one sphere to a point on another, the function’s value changes. It increases if we move away from the origin and decreases if we move toward the origin. The way the values change depends on the direction we take. The dependence of change on direction is important. We return to it in Section 13.5.

x

The definitions of interior, boundary, open, closed, bounded, and unbounded for regions in space are similar to those for regions in the plane. To accommodate the extra dimension, we use solid balls of positive radius instead of disks. (a) Interior point (x0, y0, z0)

z

y x

(b) Boundary point

FIGURE 13.9 Interior points and boundary points of a region in space. As with regions in the plane, a boundary point need not belong to the space region R.

DEFINITIONS A point ( x 0 , y 0 , z 0 ) in a region R in space is an interior point of R if it is the center of a solid ball that lies entirely in R (Figure 13.9a). A point ( x 0 , y 0 , z 0 ) is a boundary point of R if every solid ball centered at ( x 0 , y 0 , z 0 ) contains points that lie outside of R as well as points that lie inside R (Figure 13.9b). The interior of R is the set of interior points of R. The boundary of R is the set of boundary points of R. A region is open if it consists entirely of interior points. A region is closed if it contains its entire boundary. A region is bounded if it lies inside a solid ball of finite radius; otherwise, the region is unbounded.

Examples of open sets in space include the interior of a sphere, the open half-space z > 0, the first octant (where x, y, and z are all positive), and space itself. Examples of closed sets in space include lines, planes, and the closed half-space z ≥ 0. A solid sphere

772

Chapter 13 Partial Derivatives

with part of its boundary removed or a solid cube with a missing face, edge, or corner point is neither open nor closed. Functions of more than three independent variables are also important. For example, a model that measures temperature in the atmosphere may depend not only on the location of the point P ( x , y, z ) in space, but also on the time t when it is measured, so we would write T = f ( x , y, z , t ).

Computer Graphing Three-dimensional graphing software makes it possible to graph functions of two variables. We can often get information more quickly from a graph than from a formula, since the surfaces reveal increasing and decreasing behavior, and high points or low points. EXAMPLE 5 The temperature w beneath the Earth’s surface is a function of the depth x beneath the surface and the time t of the year. If we measure x in meters and t as the number of days elapsed from the expected date of the yearly highest surface temperature, we can model the variation in temperature with the function

w

w = cos (1.7 × 10 −2 t − 0.6 x ) e −0.6 x . x

8

(The temperature at 0 m is scaled to vary from +1 to −1, so that the variation at x meters can be interpreted as a fraction of the variation at the surface.) Figure 13.10 shows a graph of the function. At a depth of 5 m, the variation (change in vertical amplitude in the figure) is about 5% of the surface variation. At 8 m, there is almost no variation during the year. The graph also shows that the temperature 5 m below the surface is about half a year out of phase with the surface temperature. When the temperature is lowest on the surface (late January, say), it is at its highest 5 m below. Five meters below the ground, the seasons are reversed.

5 t

FIGURE 13.10

This graph shows the seasonal variation of the temperature below ground as a fraction of surface temperature (Example 5).

Figure 13.11 shows computer-generated graphs of a number of functions of two variables together with their level curves. z z z

y x y

x

y

x

y

y

y x

x

x

(a) z = sin x + 2 sin y FIGURE 13.11

(b) z = ( 4 x 2 + y 2 ) e −x 2 − y 2

(c) z = xye − y 2

Computer-generated graphs and level curves of typical functions of two variables.

13.1  Functions of Several Variables

13.1

EXERCISES

Domain, Range, and Level Curves

In Exercises 1–4, find the specific function values. 1. f ( x , y ) = x 2 + xy 3 a. f ( 0, 0 )

b. f (−1, 1)

c. f ( 2, 3 )

d. f (−3, − 2 )

( π6 ) 1 f ( π, ) 4

( 12π ) π f (− , −7 ) 2

a. f 2,

b. f −3,

c.

d. x−y y2 + z 2

(

1 c. f 0, − , 0 3

(

)

4. f ( x , y, z ) =

18. f ( x , y ) =

19. f ( x , y ) = 4 x 2 + 9 y 2

20. f ( x , y ) = x 2 − y 2

21. f ( x , y ) = xy

22. f ( x , y ) = y x 2

)

d. f ( 2, 2, 100 )

24. f ( x , y ) =

9 − x 2 − y2

25. f ( x , y ) = ln ( x 2 + y 2 )

26. f ( x , y ) = e −( x

27. f ( x , y ) = sin −1 ( y − x )

28. f ( x , y ) = tan −1

2 + y2 )

( xy )

29. f ( x , y ) = ln ( x 2 + y 2 − 1) 30. f ( x , y ) = ln ( 9 − x 2 − y 2 )

Exercises 31–36 show level curves for six functions. The graphs of these functions are given on the next page (items a–f ), as are their equations (items g–l). Match each set of level curves with the appropriate graph and the appropriate equation. 31.

49 − x 2 − y 2 − z 2

a. f ( 0, 0, 0 )

b. f ( 2, − 3, 6 )

c. f (−1, 2, 3 )

d. f

32. y

y

( 42 , 52 , 62 ) x

x

In Exercises 5–12, find and sketch the domain for each function. 5. f ( x , y ) =

1 16 − x 2 − y 2

y−x

Matching Surfaces with Level Curves

1 1 b. f 1, , − 2 4

a. f ( 3, −1, 2 )

17. f ( x , y ) = y − x

23. f ( x , y ) =

2. f ( x , y ) = sin ( xy)

3. f ( x , y, z ) =

773

y−x−2

6. f ( x , y ) = ln ( x 2 + y 2 − 4 ) 7. f ( x , y ) = 8. f ( x , y ) =

(y

− 1)( y + 2 ) − x )( y − x 3 )

x2

sin ( xy) + y 2 − 25

(x

33.

34. y

y

9. f ( x , y ) = cos −1 ( y − x 2 ) 10. f ( x , y ) = ln ( xy + x − y − 1) 11. f ( x , y ) =

( x 2 − 4 )( y 2 − 9 )

x

x

1 12. f ( x , y ) = ln ( 4 − x 2 − y 2 ) In Exercises 13–16, find and sketch the level curves f ( x , y ) = c on the same set of coordinate axes for the given values of c. We refer to these level curves as a contour map. 13. f ( x , y ) = x + y − 1, c = −3, − 2, −1, 0, 1, 2, 3

35.

36. y

y

14. f ( x , y ) = x 2 + y 2, c = 0, 1, 4, 9, 16, 25 15. f ( x , y ) = xy, c = −9, − 4, −1, 0, 1, 4, 9 16. f ( x , y ) =

25 − x 2 − y 2 , c = 0, 1, 2, 3, 4

In Exercises 17–30, (a) find the function’s domain, (b) find the function’s range, (c) describe the function’s level curves, (d) find the boundary of the function’s domain, (e) determine whether the domain is an open region, a closed region, or neither, and (f) decide whether the domain is bounded or unbounded.

x

x

774

Chapter 13 Partial Derivatives

a.

z

z

f.

v x y

x

b.

z

x

y

g. z = −

xy 2 x 2 + y2

i. z = ( cos x )( cos y ) e − c.

h. z = y 2 − y 4 − x 2 x 2 + y2 4

j. z = e − y cos x

z

l. z =

k. z =

xy ( x 2 − y 2 ) x 2 + y2

1 4x 2 + y 2

Functions of Two Variables

Display the values of the functions in Exercises 37–48 in two ways: (a) by sketching the surface z = f ( x , y ) and (b) by drawing an assortment of level curves in the function’s domain. Label each level curve with its function value. 37. f ( x , y ) = y 2 x

d.

y

39. f ( x , y ) =

x2

+

41. f ( x , y ) =

x2

−y

38. f ( x , y ) =

x

40. f ( x , y ) =

x 2 + y2

42. f ( x , y ) = 4 − x 2 − y 2

43. f ( x , y ) = 4 x 2 + y 2

44. f ( x , y ) = 6 − 2 x − 3 y

45. f ( x , y ) = 1 − y

46. f ( x , y ) = 1 − x − y

47. f ( x , y ) =

z

y2

x 2 + y 2 + 4 48. f ( x , y ) =

x 2 + y2 − 4

Finding Level Curves

In Exercises 49–52, find an equation for, and sketch the graph of, the level curve of the function f ( x , y ) that passes through the given point. 49. f ( x , y ) = 16 − x 2 − y 2 , ( 2 2, 2 ) 50. f ( x , y ) =

x 2 − 1, (1, 0 )

51. f ( x , y ) =

x + y 2 − 3, ( 3, −1) 2y − x 52. f ( x , y ) = , (−1, 1) x + y+1

y x

Sketching Level Surfaces

In Exercises 53–60, sketch a typical level surface for the function. 53. f ( x , y, z ) = x 2 + y 2 + z 2 54. f ( x , y, z ) = ln ( x 2 + y 2 + z 2 ) e.

z

55. f ( x , y, z ) = x + z

56. f ( x , y, z ) = z

57. f ( x , y, z ) = x 2 + y 2

58. f ( x , y, z ) = y 2 + z 2

59. f ( x , y, z ) = z − x 2 − y 2 60. f ( x , y, z ) = ( x 2 25 ) + ( y 2 16 ) + ( z 2 9 ) Finding Level Surfaces y x

In Exercises 61–64, find an equation for the level surface of the function through the given point. 61. f ( x , y, z ) =

x − y − ln z , ( 3, −1, 1)

13.2  Limits and Continuity in Higher Dimensions

62. f ( x , y, z ) = ln ( x 2 + y + z 2 ) , (−1, 2, 1)

71. f ( x , y ) = sin ( x + 2 cos y ), − 2π ≤ x ≤ 2π , − 2π ≤ y ≤ 2π , P ( π , π )

+ + (1, −1, 2 ) x−y+z , (1, 0, − 2 ) 64. g ( x , y, z ) = 2x + y − z 63. g ( x , y, z ) =

x2

y2

z2,

72. f ( x , y ) = e ( x − y )  sin ( x 2 + y 2 ) , 0 ≤ x ≤ 2π , − 2π ≤ y ≤ π , P ( π , −π ) 0.1

In Exercises 65–68, find and sketch the domain of f . Then find an equation for the level curve or surface of the function passing through the given point. 65. f ( x , y ) =

∞

⎛ ⎞

n

∑ ⎜⎜⎝ xy ⎟⎟⎟⎠ , ∞

∑

n=0

(x

∫x

y

COMPUTER EXPLORATIONS

Use a CAS to perform the following steps for each of the functions in Exercises 69–72.

74. x 2 + z 2 = 1

3z 2

Parametrized Surfaces Just as you describe curves in the plane parametrically with a pair of equations x = f (t ), y = g(t ) defined on some parameter interval I, you can sometimes describe surfaces in space with a triple of equations x = f ( u, υ ) , y = g ( u, υ ) , z = h ( u, υ ) defined on some parameter rectangle a ≤ u ≤ b, c ≤ υ ≤ d. Many computer algebra systems permit you to plot such surfaces in parametric mode. (Parametrized surfaces are discussed in detail in Section 15.5.) Use a CAS to plot the surfaces in Exercises 77–80. Also plot several level curves in the xy-plane. 77. x = u cos υ, y = u sin υ, z = u, 0 ≤ u ≤ 2, 0 ≤ υ ≤ 2π

a. Plot the surface over the given rectangle. b. Plot several level curves in the rectangle. c. Plot the level curve of f through the given point. y 69. f ( x , y ) = x sin + y sin 2 x , 0 ≤ x ≤ 5π, 0 ≤ y ≤ 5π , 2 P ( 3π , 3π ) x 2 + y2 8

y2

( )

+ y )n , ( ln 4, ln 9, 2 ) n! z n

70. f ( x , y ) = ( sin x )( cos y ) e 0 ≤ y ≤ 5π , P ( 4 π , 4 π )

73. 4 ln ( x 2 + y 2 + z 2 ) = 1

− =1 x 76. sin − ( cos y ) x 2 + z 2 = 2 2

dθ , ( 0, 1) 1 − θ2 y z dt dθ 68. g ( x , y, z ) = ∫ +∫ , ( 0, 1, 3 ) x 1 + t2 0 4 − θ2 67. f ( x , y ) =

Use a CAS to plot the implicitly defined level surfaces in Exercises 73–76. 75. x +

(1, 2 )

n=0

66. g ( x , y, z ) =

775

, 0 ≤ x ≤ 5π ,

78. x = u cos υ, y = u sin υ, z = υ, 0 ≤ u ≤ 2, 0 ≤ υ ≤ 2π 79. x = ( 2 + cos u ) cos υ, y = ( 2 + cos u ) sin υ, z = sin u, 0 ≤ u ≤ 2π , 0 ≤ υ ≤ 2π 80. x = 2 cos u cos υ, y = 2 cos u sin υ, z = 2 sin u, 0 ≤ u ≤ 2π , 0 ≤ υ ≤ π

13.2 Limits and Continuity in Higher Dimensions In this section we develop limits and continuity for multivariable functions. The theory is similar to that developed for single-variable functions, but since we now have more than one independent variable, there is additional complexity that requires some new ideas.

Limits for Functions of Two Variables If the values of f ( x , y ) lie arbitrarily close to a fixed real number L for all points ( x , y ) sufficiently close to a point ( x 0 , y 0 ), we say that f approaches the limit L as ( x , y ) approaches ( x 0 , y 0 ). This is similar to the informal definition for the limit of a function of a single variable. Notice, however, that when ( x 0 , y 0 ) lies in the interior of f ’s domain, ( x , y ) can approach ( x 0 , y 0 ) from any direction, not just from the left or the right. For the limit to exist, the same limiting value must be obtained whatever direction of approach is taken. We illustrate this issue in several examples following the definition. DEFINITION Suppose that every open circular disk centered at ( x 0 , y 0 ) contains a point in the domain of f other than ( x 0 , y 0 ) itself. We say that a function f ( x , y ) approaches the limit L as ( x , y ) approaches ( x 0 , y 0 ), and write

lim

( x , y )→ ( x 0 , y 0 )

f ( x, y ) = L,

if, for every number ε > 0, there exists a corresponding number δ > 0 such that for all ( x , y ) in the domain of f , f ( x, y ) − L < ε

whenever

0
0 is chosen. If we let δ equal this ε, we see that if 0
0 be given, but arbitrary. We want to find a δ > 0 such that

z

1

Find

1

4 xy 2 −0 < ε + y2

whenever

x2

y

0
0 such that for all r and θ,

Give reasons for your answer.

2 xy −

r→0

For instance,

tell you anything about tan −1 xy lim ? ( x , y )→( 0, 0 ) xy

f ( x , y ) = lim f ( r cos θ, r sin θ ) = L.

xy

r cos 3 θ = r cos 3 θ ≤ r ⋅ 1 = r , ?

the implication holds for all r and θ if we take δ = ε. In contrast,

Give reasons for your answer. x2

x2 r 2 cos 2 θ = = cos 2 θ 2 + y r2

784

Chapter 13 Partial Derivatives

takes on all values from 0 to 1 regardless of how small r is, so that lim x 2 ( x 2 + y 2 ) does not exist.

( x , y )→( 0, 0 )

In each of these instances, the existence or nonexistence of the limit as r → 0 is fairly clear. Shifting to polar coordinates does not always help, however, and may even tempt us to false conclusions. For example, the limit may exist along every straight line (or ray) θ = constant and yet fail to exist in the broader sense. Example 5 illustrates this point. In polar coordinates, f ( x , y ) = ( 2 x 2 y ) ( x 4 + y 2 ) becomes f ( r cos θ, r sin θ ) =

r cos θ sin 2θ r 2 cos 4 θ + sin 2 θ

for r ≠ 0. If we hold θ constant and let r → 0, the limit is 0. On the path y = x 2, however, we have r sin θ = r 2 cos 2 θ and r cos θ sin 2θ f ( r cos θ, r sin θ ) = 2 r cos 4 θ + ( r cos 2 θ ) 2 =

2r cos 2 θ sin θ r sin θ = 2 = 1. 2r 2 cos 4 θ r cos 2 θ

In Exercises 65–70, find the limit of f as ( x , y ) → ( 0, 0 ) or show that the limit does not exist. 65. f ( x , y ) =

x 3 − xy 2 x 2 + y2

y2 67. f ( x , y ) = 2 x + y2

⎛ x 3 − y 3 ⎞⎟ 66. f ( x , y ) = cos ⎜⎜ 2 ⎝ x + y 2 ⎟⎟⎠ 2x 68. f ( x , y ) = 2 x + x + y2

⎛ x + y ⎞⎟ 69. f ( x , y ) = tan −1 ⎜⎜ 2 ⎝ x + y 2 ⎟⎟⎠ 70. f ( x , y ) =

Using the Limit Definition

Each of Exercises 73–78 gives a function f ( x , y ) and a positive number ε. In each exercise, show that there exists a δ > 0 such that for all ( x , y ), f ( x , y ) − f ( 0, 0 ) < ε.

x 2 + y2 < δ ⇒ 73. f ( x , y ) =

+

x2

y2,

ε = 0.01

74. f ( x , y ) = y ( x 2 + 1),

ε = 0.05

75. f ( x , y ) = ( x + y ) ( x 2 + 1) , ε = 0.01 76. f ( x , y ) = ( x + y ) ( 2 + cos x ), ε = 0.02 77. f ( x , y ) = 78. f ( x , y ) =

xy 2 + y2

and

f ( 0, 0 ) = 0, ε = 0.04

x3 + y4 x 2 + y2

and

f ( 0, 0 ) = 0, ε = 0.02

x2

Each of Exercises 79–82 gives a function f ( x , y, z ) and a positive number ε. In each exercise, show that there exists a δ > 0 such that for all ( x , y, z ), x 2 + y2 + z 2 < δ ⇒

f ( x , y, z ) − f ( 0, 0, 0 ) < ε.

79. f ( x , y, z ) = x 2 + y 2 + z 2 , ε = 0.015 80. f ( x , y, z ) = xyz , ε = 0.008 81. f ( x , y, z ) =

x+ y+z , ε = 0.015 x 2 + y2 + z 2 + 1

82. f ( x , y, z ) = tan 2 x + tan 2 y + tan 2 z , ε = 0.03 83. Show that f ( x , y, z ) = x + y − z is continuous at every point ( x 0 , y 0 , z 0 ).

x 2 − y2 x 2 + y2

84. Show that f ( x , y, z ) = x 2 + y 2 + z 2 is continuous at the origin.

In Exercises 71 and 72, define f ( 0, 0 ) in a way that extends f to be continuous at the origin. ⎛ 3 x 2 − x 2 y 2 + 3 y 2 ⎞⎟ 71. f ( x , y ) = ln ⎜⎜ ⎟⎟⎠ ⎝ x 2 + y2 72. f ( x , y ) =

3x 2 y + y2

x2

13.3 Partial Derivatives The calculus of several variables is similar to single-variable calculus applied to several variables, one at a time. When we hold all but one of the independent variables of a function constant and differentiate with respect to that one variable, we get a “partial” derivative. This section shows how partial derivatives are defined and interpreted geometrically, and how to calculate them by applying the familiar rules for differentiating functions of a single variable. The idea of differentiability for functions of several variables requires more than the existence of the partial derivatives, because a point can be approached from many different directions. However, we will see that differentiable functions of several variables behave similarly to differentiable single-variable functions. In particular, they are continuous and can be well approximated by linear functions.

Partial Derivatives of a Function of Two Variables If ( x 0 , y 0 ) is a point in the domain of a function f ( x , y ), the vertical plane y = y 0 will cut the surface z = f ( x , y ) in the curve z = f ( x , y 0 ) (Figure 13.16). This curve is the graph

13.3  Partial Derivatives

785

z Vertical axis in the plane y = y0 P(x0, y0, f (x0, y0)) z = f (x, y)

The curve z = f (x, y0) in the plane y = y0 Tangent line 0

x0

y0 x (x0 + h, y0)

(x0, y0)

y

Horizontal axis in the plane y = y0

The intersection of the plane y = y 0 with the surface z = f ( x , y ) , viewed from above the first quadrant of the xy-plane.

FIGURE 13.16

of the function z = f ( x , y 0 ) in the plane y = y 0 . The horizontal coordinate in this plane is x; the vertical coordinate is z. The y-value is held constant at y 0, so y is not a variable. We define the partial derivative of f with respect to x at the point ( x 0 , y 0 ) as the ordinary derivative of f ( x , y 0 ) with respect to x at the point x = x 0 . To distinguish partial derivatives from ordinary derivatives, we use the symbol ∂ rather than the d previously used. In the definition, h represents a real number, positive or negative.

DEFINITION

( x 0 , y 0 ) is

The partial derivative of f ( x, y ) with respect to x at the point ∂f ∂x

( x 0 , y0 )

= lim

h→0

f ( x 0 + h, y 0 ) − f ( x 0 , y 0 ) , h

provided the limit exists.

The partial derivative of f ( x , y ) with respect to x at the point ( x 0 , y 0 ) is the same as the ordinary derivative of f ( x , y 0 ) at the point x 0: ∂f ∂x

( x 0 , y0 )

=

d f ( x, y0 ) dx

x= x0

.

A variety of notations are used to denote the partial derivative at a point ( x 0 , y 0 ) , including ∂f ( x , y ), ∂x 0 0

f x ( x 0 , y 0 ),

and

∂z ∂x

.

( x 0 , y0 )

When we do not specify a specific point ( x 0 , y 0 ) at which the partial derivative is being evaluated, then the partial derivative becomes a function whose domain is the points where the partial derivative exists. Notations for this function include ∂f , ∂x

fx,

and

∂z . ∂x

786

Chapter 13 Partial Derivatives

The slope of the curve z = f ( x , y 0 ) at the point P ( x 0 , y 0 , f ( x 0 , y 0 )) in the plane y = y 0 is the value of the partial derivative of f with respect to x at ( x 0 , y 0 ). (In Figure 13.16 this slope is negative.) The tangent line to the curve at P is the line in the plane y = y 0 that passes through P with this slope. The partial derivative ∂ f ∂ x at ( x 0 , y 0 ) gives the rate of change of f with respect to x when y is held fixed at the value y 0 . The definition of the partial derivative of f ( x , y ) with respect to y at a point ( x 0 , y 0 ) is similar to the definition of the partial derivative of f with respect to x. We hold x fixed at the value x 0 and take the ordinary derivative of f ( x 0 , y ) with respect to y at y 0 . DEFINITION

( x 0 , y 0 ) is ∂f ∂y

The partial derivative of f ( x, y ) with respect to y at the point =

( x 0 , y0 )

d f ( x 0 , y) dy

= lim y = y0

h→0

f ( x 0 , y0 + h ) − f ( x 0 , y0 ) , h

provided the limit exists.

Vertical axis in the plane x = x0

The slope of the curve z = f ( x 0 , y ) at the point P ( x 0 , y 0 , f ( x 0 , y 0 )) in the vertical plane x = x 0 (Figure 13.17) is the partial derivative of f with respect to y at ( x 0 , y 0 ). The tangent line to the curve at P is the line in the plane x = x 0 that passes through P with this slope. The partial derivative gives the rate of change of f with respect to y at ( x 0 , y 0 ) when x is held fixed at the value x 0 . The partial derivative with respect to y is denoted the same way as the partial derivative with respect to x:

z

Tangent line P(x0, y0, f (x0, y0))

∂f ( x , y ), ∂y 0 0

z = f(x, y) 0 x0

y0

x (x0, y0)

y

f y ( x 0 , y 0 ),

FIGURE 13.17

f y.

Notice that we now have two tangent lines associated with the surface z = f ( x , y ) at the point P ( x 0 , y 0 , f ( x 0 , y 0 )) (Figure 13.18). Is the plane they determine tangent to the surface at P? We will see that it is for the differentiable functions defined at the end of this section, and we will learn how to find the tangent plane in Section 13.6. First we have to better understand partial derivatives.

(x0, y0 + k) The curve z = f(x0, y) in the plane x = x0

∂f , ∂y

z Horizontal axis in the plane x = x0

The intersection of the plane x = x 0 with the surface z = f ( x , y ) , viewed from above the first quadrant of the xy-plane.

This tangent line P(x0, y0, f (x0, y0)) has slope fy (x0, y0). The curve z = f (x0, y) in the vertical plane x = x0

This tangent line has slope fx(x0, y0). The curve z = f (x, y0) in the vertical plane y = y0 z = f (x, y)

x y = y0

FIGURE 13.18

(x0, y0)

x = x0

y

Figures 13.16 and 13.17 combined. The tangent lines at the point ( x 0 , y 0 , f ( x 0 , y 0 )) determine a plane that, in this picture at least, appears to be tangent to the surface.

13.3  Partial Derivatives

787

Calculations The definitions of ∂ f ∂ x and ∂ f ∂ y give us two different ways of differentiating f at a point: with respect to x in the usual way while treating y as a constant, and with respect to y in the usual way while treating x as a constant. As the following examples show, the values of these partial derivatives are usually different at a given point ( x 0 , y 0 ). EXAMPLE 1

Find the values of ∂ f ∂ x and ∂ f ∂ y at the point ( 4, −5 ) if f ( x , y ) = x 2 + 3 xy + y − 1.

Solution To find ∂ f ∂ x , we treat y as a constant and differentiate with respect to x:

∂f ∂( 2 = x + 3 xy + y − 1) = 2 x + 3 ⋅ 1 ⋅ y + 0 − 0 = 2 x + 3 y. ∂x ∂x The value of ∂ f ∂ x at ( 4, −5 ) is 2( 4 ) + 3(−5 ) = −7. To find ∂ f ∂ y , we treat x as a constant and differentiate with respect to y: ∂f ∂ 2 ( x + 3 xy + y − 1) = 0 + 3 ⋅ x ⋅ 1 + 1 − 0 = 3 x + 1. = ∂y ∂y The value of ∂ f ∂ y at ( 4, −5 ) is 3( 4 ) + 1 = 13. EXAMPLE 2

Find ∂ f ∂ y as a function if f ( x , y ) = y sin xy.

Solution We treat x as a constant and f as a product of y and sin xy:

∂f ∂ ∂ ∂ = ( y sin xy ) = y sin xy + ( sin xy ) ( y) ∂y ∂y ∂y ∂y ∂ = ( y cos xy ) ( xy) + sin xy = xy cos xy + sin xy. ∂y EXAMPLE 3

Find f x and f y as functions if f ( x, y ) =

2y . y + cos x

Solution We treat f as a quotient. With y held constant, we use the quotient rule to get

∂ ∂ ( y + cos x ) ( 2 y ) − 2 y ( y + cos x ) ⎞⎟ 2y ∂ ⎛⎜ ∂ x ∂ x fx = ⎟ = ⎜ ∂ x ⎜⎝ y + cos x ⎟⎠ ( y + cos x ) 2 =

2 y sin x ( y + cos x )( 0 ) − 2 y ( − sin x ) . = ( y + cos x ) 2 ( y + cos x ) 2

With x held constant and again applying the quotient rule, we get ⎞⎟ 2y ∂ ⎛⎜ fy = ⎟ = ⎜⎜ ∂ y ⎝ y + cos x ⎟⎠ =

( y + cos x )

∂( ) ∂ 2 y − 2 y ( y + cos x ) ∂y ∂y ( y + cos x ) 2

2 cos x ( y + cos x )( 2 ) − 2 y (1) = . 2 ( y + cos x ) ( y + cos x ) 2

Implicit differentiation works for partial derivatives the way it works for ordinary derivatives, as the next example illustrates.

788

Chapter 13 Partial Derivatives

Find ∂ z ∂ x assuming that the equation

EXAMPLE 4

yz − ln z = x + y defines z as a function of the two independent variables x and y and the partial derivative exists. Solution We differentiate both sides of the equation with respect to x, holding y constant and treating z as a differentiable function of x:

∂y ∂ ∂ ∂x + ( yz ) − ln z = ∂x ∂x ∂x ∂x y

∂z 1 ∂z − =1+ 0 z ∂x ∂x

  

With y  constant, 

∂ ∂z ( yz ) = y . ∂x ∂x

( y − 1z ) ∂∂xz = 1 ∂z z = . ∂x yz − 1

EXAMPLE 5 The plane x = 1 intersects the paraboloid z = x 2 + y 2 in a parabola. Find the slope of the tangent line to the parabola at (1, 2, 5 ) (Figure 13.19).

z Plane x=1 Surface z = x2 + y 2

Tangent line

(1, 2, 5)

Solution The parabola lies in a plane parallel to the yz-plane, and the slope is the value of the partial derivative ∂ z ∂ y at (1, 2 ):

∂z ∂y

= (1, 2 )

∂( 2 x + y2 ) = 2y = 2( 2 ) = 4. ∂y (1, 2 ) (1, 2 )

As a check, we can treat the parabola as the graph of the single-variable function z = (1) 2 + y 2 = 1 + y 2 in the plane x = 1 and ask for the slope at y = 2. The slope, calculated now as an ordinary derivative, is dz dy

= y=2

d( 1 + y2 ) dy

= 2y y=2

y=2

= 4.

2

1

y x

FIGURE 13.19

x=1

The tangent line to the curve of intersection of the plane x = 1 and the surface z = x 2 + y 2 at the point (1, 2, 5 ) (Example 5).

Functions of More Than Two Variables The definitions of the partial derivatives of functions of more than two independent variables are similar to the definitions for functions of two variables. They are ordinary derivatives with respect to one variable, taken while the other independent variables are held constant.

EXAMPLE 6

If x, y, and z are independent variables and f ( x , y, z ) = x sin ( y + 3z ) ,

then ∂f ∂ ∂ = [ x sin ( y + 3z ) ] = x sin ( y + 3z ) ∂z ∂z ∂z ∂ = x cos ( y + 3z ) ( y + 3z ) ∂z = 3 x cos ( y + 3z ).

x held constant Chain rule y held constant

13.3  Partial Derivatives

789

EXAMPLE 7 If resistors of R1 , R 2 , and R 3 ohms are connected in parallel to make an R-ohm resistor, the value of R can be found from the equation

R1 R2

1 1 1 1 = + + R R1 R2 R3

R3

(Figure 13.20). Find the value of ∂ R ∂ R 2 when R1 = 30, R 2 = 45, and R 3 = 90 ohms. Solution To find ∂ R ∂ R 2 , we treat R1 and R 3 as constants and, using implicit differentiation, differentiate both sides of the equation with respect to R 2 :

+ −

∂ 1 ∂ ⎛⎜ 1 1 1 ⎞⎟ = + + ⎟ ⎜ ⎜ ∂ R2 R ∂ R 2 ⎝ R1 R2 R 3 ⎟⎠

( )

FIGURE 13.20

Resistors arranged this way are said to be connected in parallel (Example 7). Each resistor lets a portion of the current through. Their equivalent resistance R is calculated with the formula 1 1 1 1 . = + + R R1 R2 R3

−

1 ∂R 1 =0− +0 R 2 ∂ R2 R2 2 ⎛ R ⎞⎟ 2 R2 ∂R ⎜ ⎟ . = = ⎜⎝ R ⎟⎠ R2 2 ∂ R2 2

When R1 = 30, R 2 = 45, and R 3 = 90, 1 1 1 1 3+2+1 6 1 = = , = + + = R 30 45 90 90 90 15 so R = 15 and

( )

()

∂R 15 2 1 2 1 = = = . ∂ R2 45 3 9 Thus at the given values, a small change in the resistance R 2 leads to a change in R about one-ninth as large.

Partial Derivatives and Continuity A function f ( x , y ) can have partial derivatives with respect to both x and y at a point without the function being continuous there. This is different from functions of a single variable, where the existence of a derivative implies continuity. If the partial derivatives of f ( x , y ) exist and are continuous throughout a disk centered at ( x 0 , y 0 ) , however, then f is continuous at ( x 0 , y 0 ) , as we see at the end of this section. z=

0, xy ≠ 0 1, xy = 0

z

EXAMPLE 8

Let

L1

⎧ 0, xy ≠ 0 f ( x , y ) = ⎪⎨ ⎪⎪⎩ 1, xy = 0

1

(Figure 13.21). 0 L2 x

y

FIGURE 13.21

The graph of

⎧⎪ 0, xy ≠ 0 f ( x, y ) = ⎨ ⎪⎪⎩ 1, xy = 0 consists of the lines L1 and L 2 (lying 1 unit above the xy-plane) and the four open quadrants of the xy-plane. The function has partial derivatives at the origin but is not continuous there (Example 8).

(a) Find the limit of f as ( x , y ) approaches ( 0, 0 ) along the line y = x. (b) Find the limit of f as ( x , y ) approaches ( 0, 0 ) along the line y = 0. (c) Prove that f is not continuous at the origin. (d) Show that both partial derivatives ∂ f ∂ x and ∂ f ∂ y exist at the origin. Solution

(a) Since f ( x , y ) is zero at every point on the line y = x (except at the origin), we have lim

( x , y )→( 0, 0 )

f ( x, y )

y= x

=

lim

( x , y )→( 0, 0 )

0 = 0.

(b) Since f ( x , y ) takes the constant value 1 at every point on the line y = 0, we have lim

( x , y )→( 0, 0 )

f ( x, y )

y= 0

=

lim

( x , y )→( 0, 0 )

1 = 1.

790

Chapter 13 Partial Derivatives

(c) By the two-path test, f has no limit as ( x , y ) approaches ( 0, 0 ). Consequently, f is not continuous at ( 0, 0 ). (d) To find s f s x at ( 0, 0 ), we hold y fixed at y  0. Then f ( x , y ) = 1 for all x, and the graph of f is the line L1 in Figure 13.21. The slope of this line at any x is ∂ f ∂ x = 0. In particular, ∂ f ∂ x = 0 at ( 0, 0 ). Similarly, s f s y is the slope of line L 2 at any y, so ∂ f ∂ y = 0 at ( 0, 0 ). What Example 8 suggests is that we need a stronger requirement for differentiability in higher dimensions than the mere existence of the partial derivatives. We define differentiability for functions of two variables (which is somewhat more complicated than for single-variable functions) at the end of this section and then revisit the connection to continuity.

Second-Order Partial Derivatives When we differentiate a function f ( x , y ) twice, we produce its second-order derivatives. These derivatives are usually denoted by s2 f  or   f xx , sx 2 s2 f  or   f yx , sx sy

s2 f  or   f yy , sy2 s2 f  or   f xy . sy sx

and

The defining equations are

( )

∂2 f ∂2 f ∂ ∂f ∂ ⎛⎜ ∂ f ⎞⎟ = = , ⎟, ∂x 2 ∂x ∂x ∂x ∂y ∂ x ⎜⎝ ∂ y ⎟⎠ and so on. Notice the order in which the mixed partial derivatives are taken:

HISTORICAL BIOGRAPHY Pierre-Simon Laplace (1749–1827) Mathematician and astronomer, Laplace was born in Normandy, France. He was among the most influential scientists of his time and was called the Newton of France for contributions to the understanding of the solar system’s stability. Laplace also generalized the laws of mechanics for their application to the motion and properties of the heavenly bodies. To know more, visit the companion Website.

∂2 f ∂x ∂y

Differentiate first with respect to y, then with respect to x.

f yx = ( f y ) x

Means the same thing

EXAMPLE 9

If f ( x , y ) = x cos y + ye x , find the second-order derivatives s2 f , sx 2

s2 f , sy sx

s2 f , sy2

and

s2 f . sx sy

Solution The first step is to calculate both first partial derivatives.

∂f ∂ = ( x cos y + ye x ) ∂y ∂y = −x sin y + e x

∂f ∂ = ( x cos y + ye x ) ∂x ∂x = cos y + ye x

Now we find both partial derivatives of each first partial:

( )

∂2 f ∂ ⎛⎜ ∂ f ⎞⎟ = ⎟ = − sin y + e x ∂ x ⎜⎝ ∂ y ⎟⎠ ∂x ∂y

( )

∂2 f ∂ ⎛⎜ ∂ f ⎞⎟ = ⎟ = −x cos y. 2 ∂y ∂ y ⎜⎝ ∂ y ⎟⎠

∂2 f ∂ ∂f = − sin y + e x = ∂y ∂x ∂y ∂x ∂2 f ∂ ∂f = = ye x . 2 ∂x ∂x ∂x

The Mixed Derivative Theorem You may have noticed that the “mixed” second-order partial derivatives s2 f sy sx

and

s2 f sx sy

13.3  Partial Derivatives

791

in Example 9 are equal. This is not a coincidence. They must be equal whenever f , f x , f y , f xy , and f yx are continuous, as stated in the following theorem. However, the mixed derivatives can be different when the continuity conditions are not satisfied (see Exercise 82).

THEOREM 2—The Mixed Derivative Theorem

If f ( x , y ) and its partial derivatives f x , f y , f xy , and f yx are defined throughout an open region containing a point ( a, b ) and are all continuous at ( a, b ), then f xy ( a, b ) = f yx ( a, b ).

HISTORICAL BIOGRAPHY Alexis Clairaut (1713–1765) Alexis Clairaut was a mathematical genius, who was called to visit the Academy of Sciences in Paris when he was only 12 years old. n a study published in 1743, the Clairaut proposition postulates in a simple way the dependency of the geometrical flattening ratio on the relationship between the gravity and the centrifugal force. To know more, visit the companion Website.

Theorem 2 is also known as Clairaut’s Theorem, after the French mathematician Alexis Clairaut, who discovered it. A proof is given in Appendix A.10. Theorem 2 says that to calculate a mixed second-order derivative, we may differentiate in either order, provided the continuity conditions are satisfied. This ability to proceed in different order sometimes simplifies our calculations. EXAMPLE 10

Find 

s2w   if sx sy w = xy +

ey . +1

y2

Solution The symbol s 2 w s x s y tells us to differentiate first with respect to y and then with respect to x. However, if we interchange the order of differentiation and differentiate first with respect to x, we get the answer more quickly. In two steps,

∂w = y ∂x

and

∂2w = 1. ∂y ∂x

If we differentiate first with respect to y, we obtain ∂ 2 w ∂ x ∂ y = 1 as well, but with more work. We can differentiate in either order because the conditions of Theorem 2 hold for w at all points (x 0 , y 0 ).

Partial Derivatives of Still Higher Order Although we will deal mostly with first- and second-order partial derivatives, because these appear the most frequently in applications, there is no theoretical limit to how many times we can differentiate a function as long as the derivatives involved exist. Thus, we get third- and fourth-order derivatives denoted by symbols like ∂3 f = f yyx , ∂x ∂y2 ∂4 f = f yyxx , ∂x 2 ∂y 2 and so on. As with second-order derivatives, the order of differentiation is immaterial as long as all the derivatives through the order in question are continuous. EXAMPLE 11

Find f yxyz if f ( x , y, z ) = 1 − 2 xy 2 z + x 2 y.

Solution We first differentiate with respect to the variable y, then x, then y again, and finally with respect to z: f y = −4 xyz + x 2

f yx = −4 yz + 2 x f yxy = −4 z f yxyz = −4.

792

Chapter 13 Partial Derivatives

Differentiability The concept of differentiability for functions of several variables is more complicated than for single-variable functions, because a point in the domain can be approached from many directions and along any path, not just from the left or from the right. The existence of both partial derivatives at a point ( x 0 , y 0 ) is not by itself even enough to show continuity at ( x 0 , y 0 ), as we saw in Example 8. The differentiability of f is instead based on the idea that a linear function gives a good model of a differentiable function near a point. In Section 3.11, we saw that a differentiable function f can be approximated near a point x 0 by its linearization, L ( x ) = f ( x 0 ) + f ′( x 0 ) ( x − x 0 ). This formula allows us to find a linear function L, a function whose graph is a straight line, such that L closely approximates f near x 0. This can be done whenever f is differentiable, even when f itself is described by a very complicated formula. Approximations are much more useful and meaningful when they are accompanied by information on their accuracy. In Section 3.11, Equation (1), we saw that a differentiable function f satisfies f ( x ) − f ( x 0 ) = f ′( x 0 )( x − x 0 ) + ε ( x − x 0 ) , where ε → 0 as x → x 0 . Framed in terms of approximating f by L, this becomes f ( x ) − L( x ) = ε( x − x 0 ),

(1)

where again ε → 0 as x → x 0 . Rather than being a consequence of the definition, the differentiability for a function of two variables f ( x , y ) is defined to mean that f can be approximated by a linear function. The approximating linear function L ( x , y ) for f ( x , y ) near the point ( x 0 , y 0 ) takes the form L ( x , y ) = f ( x 0 , y 0 ) + f x ( x 0 , y 0 )( x − x 0 ) + f y ( x 0 , y 0 )( y − y 0 ) , and the graph of L is a plane, called the tangent plane, that approximates the graph of f near ( x 0 , y 0 ). Notice that L ( x 0 , y 0 ) = f ( x 0 , y 0 ), so the functions L and f coincide at ( x 0 , y 0 ). Moreover the partial derivatives of L and f are also equal at ( x 0 , y 0 ). We will study tangent planes in detail in Section 13.6. We now specify how closely f is approximated by L at ( x 0 , y 0 ). Extending the formula for single variable functions in Equation (1), we require that the difference between f and L satisfies f ( x , y ) − L ( x , y ) = ε1 ( x − x 0 ) + ε 2 ( y − y 0 ) , (2) where both ε1 → 0 and ε 2 → 0 as ( x , y ) → ( x 0 , y 0 ). If we insert the formula for L ( x , y ) into Equation (2) we see that f ( x , y ) − f ( x 0 , y 0 ) = f x ( x 0 , y 0 )( x − x 0 ) + f y ( x 0 , y 0 )( y − y 0 ) + ε1 ( x − x 0 ) + ε 2 ( y − y 0 ). Setting Δx = x − x 0 , Δy = y − y 0 , and Δz = f ( x , y ) − f ( x 0 , y 0 ), we get Δz = f x ( x 0 , y 0 ) Δx + f y ( x 0 , y 0 ) Δy + ε1 Δx + ε 2 Δy. Based on these ideas, we now state the formal definition of differentiability, which captures the idea that  f  is well approximated by L. DEFINITION A function z = f ( x , y ) is differentiable at ( x 0 , y 0 ) if both

f x ( x 0 , y 0 ) and f y ( x 0 , y 0 ) exist and if Δz = f ( x , y ) − f ( x 0 , y 0 ) satisfies Δ z = f x ( x 0 , y 0 ) Δ x + f y ( x 0 , y 0 ) Δ y + ε1 Δ x + ε 2 Δ y ,

where Δx = x − x 0 , Δy = y − y 0 , and both ε1 → 0 and ε 2 → 0 as ( x , y ) → ( x 0 , y 0 ). We call the function f differentiable if it is differentiable at every point in its domain, and we then say that its graph is a smooth surface. Differentiability is defined in a similar way for functions of more than two variables.

13.3  Partial Derivatives

793

The following theorem (proved in Appendix A.10) and its accompanying corollary tell us that functions with continuous first partial derivatives at ( x 0 , y 0 ) are differentiable there, and they are closely approximated locally by a linear function. We study this approximation in Section 13.6. THEOREM 3—The Increment Theorem for Functions of Two Variables

Suppose that the first partial derivatives of f ( x , y ) are defined throughout an open region R containing the point ( x 0 , y 0 ) and that f x and f y are continuous at ( x 0 , y 0 ). Then the change Δz = f ( x 0 + Δx , y 0 + Δy ) − f ( x 0 , y 0 ) in the value of f that results from moving from ( x 0 , y 0 ) to another point ( x 0 + Δx , y 0 + Δy ) in R satisfies an equation of the form Δ z = f x ( x 0 , y 0 ) Δ x + f y ( x 0 , y 0 ) Δ y + F1 Δ x + F 2 Δ y , in which each of F1 , F 2 l 0 as both Δx , Δy → 0. In many cases the partial derivatives are defined and continuous at every point in the domain of f . We then have the following Corollary. Corollary of Theorem 3

If the partial derivatives f x and f y of a function f ( x , y ) are continuous throughout an open region R, then f is differentiable at every point of R. If z = f ( x , y ) is differentiable, then the definition of differentiability ensures that Δz = f ( x 0 + Δx , y 0 + Δy ) − f ( x 0 , y 0 ) approaches 0 as %x and %y approach 0. This tells us that a function of two variables is continuous at every point where it is differentiable. THEOREM 4—Differentiable Implies Continuous

If a function f ( x , y ) is differentiable at ( x 0 , y 0 ) , then f is continuous at ( x 0 , y 0 ). As we can see from Corollary 3 and Theorem 4, a function f ( x , y ) must be continuous at a point ( x 0 , y 0 ) if f x and f y are continuous throughout an open region containing ( x 0 , y 0 ). Remember, however, that it is still possible for a function of two variables to be discontinuous at a point where its first partial derivatives exist, as we saw in Example 8. Existence alone of the partial derivatives at that point is not enough, but continuity of the partial derivatives guarantees differentiability.

13.3

EXERCISES

Calculating First-Order Partial Derivatives

In Exercises 1–22, find s f s x and s f s y. 1. f ( x , y ) = 2 x 2 − 3 y − 4

2. f ( x , y ) = x 2 − xy + y 2

3. f ( x , y ) = ( x 2 − 1)( y + 2 )

7. f ( x , y ) =

x2

+

y2

9. f ( x , y ) = 1 ( x + y )

6. f ( x , y ) = ( 2 x − 3 y )3 8. f ( x , y ) =

16. f ( x , y ) = e xy ln y

17. f ( x , y ) =

18. f ( x , y ) = cos 2 ( 3 x − y 2 )

(x3

+ ( y 2 ))

22. f ( x , y ) = 23

10. f ( x , y ) = x ( x 2 + y 2 ) tan −1 ( y

11. f ( x , y ) = ( x + y ) ( xy − 1)

12. f ( x , y ) =

x)

13. f ( x , y ) = e ( x + y +1)

14. f ( x , y ) = e −x sin ( x + y )

sin 2 ( x

− 3y)

19. f ( x , y ) = x y 21. f ( x , y ) =

4. f ( x , y ) = 5 xy − 7 x 2 − y 2 + 3 x − 6 y + 2 5. f ( x , y ) = ( xy − 1) 2

15. f ( x , y ) = ln ( x + y )

∫x

20. f ( x , y ) = log y x y

g(t ) dt ( g  continuous for all  t )

∞

∑ ( xy) n

(

xy < 1)

n=0

In Exercises 23–34, find f x , f y , and f z . 23. f ( x , y, z ) = 1 + xy 2 − 2 z 2 25. f ( x , y, z ) = x −

y2 + z 2

24. f ( x , y, z ) = xy + yz + xz

794

Chapter 13 Partial Derivatives

62. The fifth-order partial derivative ∂ 5 f ∂ x 2 ∂ y 3 is zero for each of the following functions. To show this as quickly as possible, which variable would you differentiate with respect to first: x or y? Try to answer without writing anything down.

26. f ( x , y, z ) = ( x 2 + y 2 + z 2 )−1 2 27. f ( x , y, z ) = arcsin ( xyz ) 28. f ( x , y, z ) = arcsec ( x + yz ) 29. f ( x , y, z ) = ln ( x + 2 y + 3z )

a. f ( x , y ) = y 2 x 4e x + 2

30. f ( x , y, z ) = yz ln ( xy) 31. f ( x , y, z ) = e −( x

b. f ( x , y ) = y 2 + y ( sin x − x 4 )

2 + y2 +z 2 )

c. f ( x , y ) = x 2 + 5 xy + sin x + 7e x

32. f ( x , y, z ) = e −xyz

d. f ( x , y ) = x e y

33. f ( x , y, z ) = tanh ( x + 2 y + 3z ) 34. f ( x , y, z ) = sinh ( xy − z 2 )

Using the Partial Derivative Definition

In Exercises 35–40, find the partial derivative of the function with respect to each variable. 35. f ( t , α ) = cos ( 2πt − α ) υ 2e ( 2 u υ )

36. g ( u, υ ) =

38. g ( r , θ, z ) = r (1 − cos θ ) − z 39. Work done by the heart (Section 3.11, Exercise 59) W ( P , V , δ , υ, g ) = PV +

V δυ 2 2g

(Section 4.6, Exercise 61)

A ( c, h , k , m , q ) =

hq km + cm + 2 q

Calculating Second-Order Partial Derivatives

Find all the second-order partial derivatives of the functions in Exercises 41–54. 41. f ( x , y ) = x + y + xy

42. f ( x , y ) = sin xy

43. g ( x , y ) = x 2 y + cos y + y sin x 44. h ( x , y ) = xe y + y + 1

45. r ( x , y ) = ln ( x + y )

46. s ( x , y ) = arctan ( y x )

47. w = x 2 tan ( xy)

48. w = ye x− y 50. w = 2 x + y

49. w = x sin ( x 2 y)

x 2 −  y

52. g ( x , y ) = cos 54. z = xe

− sin 3 y 53. z = x sin ( 2 x −

y2 )

x y2

In Exercises 55–60, verify that w xy = w yx . 55. w = ln ( 2 x + 3 y ) 57. w =

+

x 2y3

56. w = +

2 x + 3 y − 1,

∂f ∂x

∂f ∂x

and

and

∂f ∂y

∂f ∂y ∂f ∂y

at (1, 2 ) at (−2, 1)

at (−2, 3 )

3 4 ⎪⎧⎪ sin ( x + y ) , ( x , y ) ≠ ( 0, 0 ) 2 2 ⎪ 66. f ( x , y ) = ⎨ x + y ⎪⎪ ( x , y ) = ( 0, 0 ) , ⎪⎩ 0, ∂f ∂f and at ( 0, 0 ) ∂x ∂y

67. Three variables Let w = f ( x , y, z ) be a function of three independent variables and write the formal definition of the partial derivative ∂ f ∂ z at ( x 0 , y 0 , z 0 ). Use this definition to find ∂ f ∂ z at (1, 2, 3 ) for f ( x , y, z ) = x 2 yz 2 . 68. Three variables Let w = f ( x , y, z ) be a function of three independent variables and write the formal definition of the partial derivative ∂ f ∂ y at ( x 0 , y 0 , z 0 ). Use this definition to find ∂ f ∂ y at (−1, 0, 3 ) for f ( x , y, z ) = −2 xy 2 + yz 2 .

69. Find the value of ∂ z ∂ x at the point (1, 1, 1) if the equation xy + z 3 x − 2 yz = 0 defines z as a function of the two independent variables x and y and the partial derivative exists. xz + y ln x − x 2 + 4 = 0

ex

+ x ln y + y ln x

x 3y4

58. w = x sin y + y sin x + xy 3x − y x2 59. w = 3 60. w = y x + y 61. Which order of differentiation enables one to calculate f xy faster: x first or y first? Try to answer without writing anything down. a. f ( x , y ) = x sin y + e y b. f ( x , y ) = 1 x c. f ( x , y ) = y + ( x y ) d. f ( x , y ) = y + x 2 y + 4 y 3 − ln ( y 2 + 1) e. f ( x , y ) = x 2 + 5 xy + sin x + 7e x f. f ( x , y ) = x ln xy

65. f ( x , y ) =

and

70. Find the value of ∂ x ∂ z at the point (1, −1, −3 ) if the equation

Mixed Partial Derivatives

xy 2

∂f ∂x

63. f ( x , y ) = 1 − x + y − 3 x 2 y,

Differentiating Implicitly

51. f ( x , y ) = x 2 y 3 − x 4 + y 5 x2

In Exercises 63–66, use the limit definition of partial derivative to compute the partial derivatives of the functions at the specified points.

64. f ( x , y ) = 4 + 2 x − 3 y − xy 2,

37. h ( ρ , φ, θ ) = ρ sin φ cos θ

40. Wilson lot size formula

2 2

defines x as a function of the two independent variables y and z and the partial derivative exists. Exercises 71 and 72 are about the triangle shown here. B c a C

b

A

71. Express A implicitly as a function of a, b, and c and calculate ∂ A ∂ a and ∂ A ∂ b. 72. Express a implicitly as a function of A, b, and B and calculate ∂ a ∂ A and ∂ a ∂ B.

13.3  Partial Derivatives

73. Two dependent variables Express υ x in terms of u and y if the equations x = υ ln u and y = u ln υ define u and υ as functions of the independent variables x and y, and if υ x exists. (Hint: Differentiate both equations with respect to x and solve for υ x by eliminating u x . ) 74. Two dependent variables Find ∂ x ∂u and ∂ y ∂u if the equations u = x 2 − y 2 and υ = x 2 − y define x and y as functions of the independent variables u and υ, and the partial derivatives exist. (See the hint in Exercise 73.) Then let s = x 2 + y 2 and find ∂ s ∂u. Theory and Examples

75. Let f ( x , y ) = 2 x + 3 y − 4. Find the slope of the line tangent to this surface at the point ( 2, −1) and lying in a. the plane x = 2 b. the plane y = −1. 76. Let f ( x , y ) = x 2 + y 3. Find the slope of the line tangent to this surface at the point (−1, 1) and lying in a. the plane x = −1 b. the plane y = 1. In Exercises 77–80, find a function z = f ( x , y ) whose partial derivatives are as given, or explain why this is impossible. 77.

∂f ∂f = 3x 2 y 2 − 2 x , = 2x 3 y + 6 y ∂x ∂y

78.

∂f ∂f = 2 xe xy 2 + x 2 y 2e xy 2 + 3, = 2 x 3 ye xy 2 − e y ∂x ∂y

79.

2y ∂f ∂f 2x = , = ( x + y )2 ( x + y )2 ∂y ∂x

795

Show that each function in Exercises 83–90 satisfies a Laplace equation. 83. f ( x , y, z ) = x 2 + y 2 − 2 z 2 84. f ( x , y, z ) = 2 z 3 − 3( x 2 + y 2 ) z 85. f ( x , y ) = e −2 y cos 2 x 86. f ( x , y ) = ln x 2 + y 2 87. f ( x , y ) = 3 x + 2 y − 4 x 88. f ( x , y ) = arctan y 89. f ( x , y, z ) = ( x 2 + y 2 + z 2 )−1 2 90. f ( x , y, z ) = e 3 x + 4 y cos 5 z The wave equation If we stand on an ocean shore and take a snapshot of the waves, the picture shows a regular pattern of peaks and valleys in an instant of time. We see periodic vertical motion in space, with respect to distance. If we stand in the water, we can feel the rise and fall of the water as the waves go by. We see periodic vertical motion in time. In physics, this beautiful symmetry is expressed by the one-dimensional wave equation ∂2w ∂2w , = c2 2 ∂t ∂x 2 where w is the wave height, x is the distance variable, t is the time variable, and c is the velocity with which the waves are propagated. w

∂f ∂f 80. = xy cos ( xy ) + sin ( xy ) , = x cos ( xy ) ∂x ∂y y ≥ 0 ⎧⎪ y 3 , 81. Let f ( x , y ) = ⎪⎨ 2 ⎩⎪⎪− y , y < 0. Find f x , f y , f xy ,  and  f yx , and state the domain for each partial derivative. ⎧ x 2 − y2 ⎪ , if  ( x ,   y ) ≠ 0, ⎪ xy 2 2 82. Let f ( x , y ) = ⎪ ⎨ x + y ⎪ ⎪ if  ( x ,   y ) = 0. ⎪ ⎩ 0, ∂f ∂f a. Show that ( x , 0 ) = x for all x, and ( 0, y ) = − y for ∂y ∂x all y. ∂2 f ∂2 f ( 0, 0 ) ≠ ( 0, 0 ). b. Show that ∂x ∂y ∂y ∂x The three-dimensional Laplace equation ∂2 f ∂2 f ∂2 f + + = 0 2 2 ∂x ∂y ∂z 2 is satisfied by steady-state temperature distributions T = f ( x , y, z ) in space, by gravitational potentials, and by electrostatic potentials. The two-dimensional Laplace equation ∂2 f ∂2 f + = 0, ∂x 2 ∂y2 obtained by dropping the ∂ 2 f ∂ z 2 term from the previous equation, describes potentials and steady-state temperature distributions in a plane. The plane may be treated as a thin slice of the solid perpendicular to the z-axis.

0 x

In our example, x is the distance across the ocean’s surface, but in other applications, x might be the distance along a vibrating string, distance through air (sound waves), or distance through space (light waves). The number c varies with the medium and type of wave. Show that the functions in Exercises 91–97 are all solutions of the wave equation. 91. w = sin ( x + ct )

92. w = cos ( 2 x + 2ct )

93. w = sin ( x + ct ) + cos ( 2 x + 2ct ) 94. w = ln ( 2 x + 2ct )

95. w = tan ( 2 x − 2ct )

96. w = 5 cos( 3 x + 3ct ) + e x + ct 97. w = f (u), where f is a differentiable function of u, and u = a ( x + ct ) , where a is a constant 98. Does a function f ( x , y ) with continuous first partial derivatives throughout an open region R have to be continuous on R? Give reasons for your answer. 99. If a function f ( x , y ) has continuous second partial derivatives throughout an open region R, must the first-order partial derivatives of f be continuous on R? Give reasons for your answer.

796

Chapter 13 Partial Derivatives

100. The heat equation An important partial differential equation that describes the distribution of heat in a region at time t can be represented by the one-dimensional heat equation ∂f ∂2 f = . ∂t ∂x 2

⎧⎪ 0, x 2 < y < 2 x 2 102. Let f ( x , y ) = ⎨ ⎪⎪⎩1, otherwise. Show that f x ( 0, 0 ) and f y ( 0, 0 ) exist, but f is not differentiable at ( 0, 0 ). 103. The Korteweg–de Vries equation

Show that u ( x , t ) = sin (α x ) ⋅ e −βt satisfies the heat equation for constants α and β . What is the relationship between α and β for this function to be a solution? ⎧⎪ xy 2 , ( x , y ) ≠ ( 0, 0 ) ⎪ 101. Let f ( x , y ) = ⎪⎨ x 2 + y 4 ⎪⎪ ( x , y ) = ( 0, 0 ). ⎪⎩ 0, Show that f x ( 0, 0 ) and f y ( 0, 0 ) exist, but f is not differentiable at ( 0, 0 ). (Hint: Use Theorem 4 and show that f is not continuous at ( 0, 0 ).)

This nonlinear differential equation, which describes wave motion on shallow water surfaces, is given by u t + u xxx + 12u u x = 0. Show that u ( x , t ) = sech 2 ( x − t ) satisfies the Korteweg–de Vries equation. 1 satisfies the equation 104. Show that T = x 2 + y2 T xx + T yy = T 3 .

13.4 The Chain Rule The Chain Rule for functions of a single variable studied in Section 3.6 says that if w = f ( x ) is a differentiable function of x, and x = g(t ) is a differentiable function of t, then w is a differentiable function of t, and dw dt can be calculated by the formula dw dw dx . = dt dx dt To find dw dt, we read down the route from w to t, multiplying derivatives along the way. Chain Rule Dependent variable

w = f (x) dw dx

Intermediate variable

x

Functions of Two Variables The Chain Rule formula for a differentiable function w = f ( x , y ) when x = x (t ) and y = y(t ) are both differentiable functions of t is given in the following theorem.

dx dt t dw dw dx = dt dx dt

For this composite function w(t ) = f ( g(t )), we can think of t as the independent variable and x = g(t ) as the “intermediate variable” because t determines the value of x that in turn gives the value of w from the function f . We display the Chain Rule in a “dependency diagram” in the margin. Such diagrams capture which variables depend on which. For functions of several variables the Chain Rule has more than one form, which depends on how many independent and intermediate variables are involved. However, once the variables are taken into account, the Chain Rule works in the same way we just discussed.

Independent variable

THEOREM 5—Chain Rule for Functions of One Independent Variable and Two Intermediate Variables

If w = f ( x , y ) is differentiable and if x = x (t ), y = y(t ) are differentiable functions of t, then the composition w = f ( x (t ), y(t )) is a differentiable function of t and dw = f x ( x (t ),  y(t ) ) x ′(t ) + f y ( x (t ),  y(t ) ) y′(t ), dt or ∂ f dx ∂ f dy dw + = . dt ∂ x dt ∂ y dt

13.4  The Chain Rule

∂ f ∂w , , f indicates the partial ∂x ∂x x derivative of f with respect to x. Each of

797

Proof The proof consists of showing that if x and y are differentiable at t = t 0 , then w is differentiable at t 0 and dy dw ∂w dx ∂w (t ) = ( P ) (t ) + ( P ) (t ), dt 0 ∂ x 0 dt 0 ∂ y 0 dt 0 where P0 = ( x (t 0 ), y(t 0 ) ). Let Δx , Δy, and Δw be the increments that result from changing t from t 0 to t 0 + Δt. Since f is differentiable (see the definition in Section 13.3), Δw =

∂w ∂w ( P ) Δx + ( P ) Δ y + ε1 Δ x + ε 2 Δ y , ∂y 0 ∂x 0

where ε1 , ε 2 → 0 as Δx , Δy → 0. To find dw dt , we divide this equation through by Δt and let Δt approach zero (therefore, Δx and Δy approach zero as well since the fact that x (t ) and y(t ) are differentiable implies that they are continuous). The division gives Δy Δy ∂w Δx ∂w Δx Δw = (P ) + (P ) + ε1 + ε2 . ∂ x 0 Δt ∂ y 0 Δt Δt Δt Δt Letting Δt approach zero gives

To remember the Chain Rule, picture the diagram below. To find dw dt, start at w and read down each route to t, multiplying derivatives along the way. Then add the products.

dw Δw (t ) = lim Δt → 0 Δ t dt 0 dy dy dx dx ∂w ∂w ( P ) (t ) + ( P ) (t ) + 0 ⋅ (t ) + 0 ⋅ (t 0 ). = dt 0 dt ∂ x 0 dt 0 ∂ y 0 dt 0 Often we write ∂ w ∂ x for the partial derivative ∂ f ∂ x , so we can rewrite the Chain Rule in Theorem 5 in the form

Chain Rule w = f (x, y)

'w 'x

'w 'y y Intermediate variables

x dx dt

dw ∂ w dx ∂ w dy + = . dt ∂ x dt ∂ y dt

Dependent variable

dy dt

t dw 'w dx 'w dy = + dt 'x dt 'y dt

Independent variable

However, the meaning of the dependent variable w is different on each side of the preceding equation. On the left-hand side, it refers to the composite function w = f ( x (t ), y(t ) ) as a function of the single variable t. On the right-hand side, it refers to the function w = f ( x , y ) as a function of the two variables x and y. Moreover, the single derivatives dw dt , dx dt, and dy dt are being evaluated at a point t 0 , whereas the partial derivatives ∂ w ∂ x and ∂ w ∂ y are being evaluated at the point ( x 0 , y 0 ) , with x 0 = x (t 0 ) and y 0 = y(t 0 ). With that understanding, we will use both of these forms interchangeably throughout the text whenever no confusion will arise. The dependency diagram on the preceding page provides a convenient way to remember the Chain Rule. The “true” independent variable in the composite function is t, whereas x and y are intermediate variables (controlled by t) and w is the dependent variable. A more precise notation for the Chain Rule shows where the various derivatives in Theorem 5 are evaluated: ∂f ∂f dy dw dx (t 0 ) = ( x 0 , y 0 ) (t 0 ) + ( x 0 , y 0 ) (t 0 ), dt ∂x dt ∂y dt or, using another notation, dw dt

t0

=

∂f ∂x

dx ( x 0 , y 0 ) dt

t0

+

∂f ∂y

dy . ( x 0 , y 0 ) dt t 0

798

Chapter 13 Partial Derivatives

EXAMPLE 1

Use the Chain Rule to find the derivative of w = xy

with respect to t along the path x = cos t , y = sin t. What is the derivative’s value at t = π 2? Solution We apply the Chain Rule to find dw dt as follows:

∂ w dy dw ∂ w dx = + dt ∂ x dt ∂ y dt ∂ ( xy) d ∂ ( xy) d = ( cos t ) + ( sin t ) ∂ x dt ∂ y dt = ( y)(− sin t ) + ( x )( cos t ) = ( sin t )( − sin t ) + ( cos t )( cos t ) = − sin 2 t + cos 2 t = cos 2t. In this example, we can check the result with a more direct calculation. As a function of t, w = xy = cos t sin t =

1 sin 2t , 2

so

(

)

dw d 1 1 = sin 2t = ( 2 cos 2t ) = cos 2t. dt dt 2 2 In either case, at the given value of t, dw dt

t=π 2

( π2 ) = cos π = −1.

= cos 2

Functions of Three Variables You can probably predict the Chain Rule for functions of three intermediate variables, as it involves adding the expected third term to the two-variable formula. Here we have three routes from w to t instead of two, but finding dw dt is still the same. Read down each route, multiplying derivatives along the way; then add.

THEOREM 6—Chain Rule for Functions of One Independent Variable and Three Intermediate Variables

If w = f ( x , y, z ) is differentiable and x, y, and z are differentiable functions of t, then w is a differentiable function of t, and

Chain Rule w = f (x, y, z)

'w 'x

'w 'z

'w 'y

x

y dx dt

dy dt

dw ∂ w dx ∂ w dy ∂ w dz = . + + ∂ x dt ∂ y dt ∂ z dt dt

Dependent variable

z

Intermediate variables

dz dt

Independent t variable dw 'w dx 'w dy 'w dz = + + 'x dt 'y dt 'z dt dt

The proof is identical to the proof of Theorem 5, except that there are now three intermediate variables instead of two. The dependency diagram we use for remembering the new equation is similar as well, with three routes from w to t. EXAMPLE 2

Find dw dt if w = xy + z ,

x = cos t ,

y = sin t ,

z = t.

13.4  The Chain Rule

799

In this example the values of w(t) are changing along the path of a helix (Section 12.1) as t changes. What is the derivative’s value at t = 0? Solution Using the Chain Rule for three intermediate variables, we have

dw ∂ w dz ∂ w dx ∂ w dy = + + dt ∂ z dt ∂ x dt ∂ y dt = ( y) (− sin t ) + ( x ) ( cos t ) + (1)(1) = ( sin t )(− sin t ) + ( cos t )( cos t ) + 1

Substitute for intermediate variables.

= − sin 2 t + cos 2 t + 1 = 1 + cos 2t , so dw dt

t=0

= 1 + cos ( 0 ) = 2.

For a physical interpretation of change along a curve, think of an object whose position is changing with time t. If w = T ( x , y, z ) is the temperature at each point ( x , y, z ) along a curve C with parametric equations x = x (t ), y = y(t ), and z = z (t ), then the composite function w = T ( x (t ), y(t ), z (t ) ) represents the temperature relative to t along the curve. The derivative dw dt is then the instantaneous rate of change of temperature due to the motion along the curve, as calculated in Theorem 6.

Functions Defined on Surfaces If we are interested in the temperature w = f ( x , y, z ) at points ( x , y, z ) on Earth’s surface, we might prefer to think of x, y, and z as functions of the variables r and s that give the points’ longitudes and latitudes. If x = g ( r , s ) , y = h ( r , s ) , and z = k ( r , s ) , we could then express the temperature as a function of r and s with the composite function w = f ( g ( r , s ) , h ( r , s ) , k ( r , s )). Under the conditions stated below, w has partial derivatives with respect to both r and s that can be calculated in the following way.

THEOREM 7—Chain Rule for Two Independent Variables and Three Intermediate Variables

Suppose that w = f ( x , y, z ) , x = g ( r , s ) , y = h ( r , s ) , and z = k ( r , s ). If all four functions are differentiable, then w has partial derivatives with respect to r and s, given by the formulas ∂w ∂w ∂x ∂w ∂ y ∂w ∂z = + + ∂r ∂ x ∂r ∂ y ∂r ∂ z ∂r ∂w ∂w ∂x ∂w ∂ y ∂w ∂z = + + . ∂s ∂ x ∂s ∂ y ∂s ∂ z ∂s

The first of these equations can be derived from the Chain Rule in Theorem 6 by holding s fixed and treating r as t. The second can be derived in the same way, holding r fixed and treating s as t. The dependency diagrams for both equations are shown in Figure 13.22.

800

Chapter 13 Partial Derivatives

w = f (x, y, z)

w = f (x, y, z)

Dependent variable

w

'w 'x

f Intermediate variables

x

y g

Independent variables

h

x

z k

y

'x 'r

r, s

'y 'r

x

z

'x 's

'z 'r

'w 'z y

'y 's

z

'z 's

(c)

(b)

(a)

'w 'y

s 'w 'w 'x 'w 'y 'w 'z = + + 's 'x 's 'y 's 'z 's

r 'w 'w 'x 'w 'y 'w 'z = + + 'r 'x 'r dy 'r 'z 'r

w = f( g(r, s), h(r, s), k(r, s))

FIGURE 13.22

'w 'x

'w 'z

'w 'y

Composite function and dependency diagrams for Theorem 7.

EXAMPLE 3

Express ∂ w ∂r and ∂ w ∂ s in terms of r and s if

w = x + 2 y + z 2,

x =

r , s

y = r 2 + ln s,

z = 2r.

Solution Using the formulas in Theorem 7, we find

∂w ∂w ∂x ∂w ∂ y ∂w ∂z = + + ∂r ∂ x ∂r ∂ y ∂r ∂ z ∂r 1 = ( 1) + ( 2 )( 2r ) + ( 2 z )( 2 ) s 1 1 = + 4r + ( 4r )( 2 ) = + 12r s s

()

Substitute for intermediate variable z.

∂w ∂w ∂x ∂w ∂ y ∂w ∂z = + + ∂s ∂ x ∂s ∂ y ∂s ∂ z ∂s r 1 2 r = ( 1) − 2 + ( 2 ) + ( 2 z )( 0 ) = − 2 . s s s s

(

)

()

Chain Rule

If f is a function of two intermediate variables instead of three, each equation in Theorem 7 becomes correspondingly one term shorter.

w = f (x, y)

'w 'x

'w 'y

x

y

'x 'r

If w = f ( x , y ) , x = g ( r , s ) , and y = h ( r , s ) , then ∂w ∂x ∂w ∂ y ∂w = + ∂r ∂ x ∂r ∂ y ∂r

'y 'r

and

∂w ∂w ∂x ∂w ∂ y = + . ∂s ∂ x ∂s ∂ y ∂s

r

'w 'w 'x 'w 'y = + 'r 'x 'r 'y 'r

FIGURE 13.23

Figure 13.23 shows the dependency diagram for the first of these equations. The diagram for the second equation is similar; just replace r with s.

Dependency diagram for

the equation ∂w ∂ x ∂w ∂ y ∂w = + . ∂ x ∂r ∂ y ∂r ∂r

EXAMPLE 4

Express ∂ w ∂r and ∂ w ∂ s in terms of r and s if w = x 2 + y 2,

x = r − s,

y = r + s.

801

13.4  The Chain Rule

Solution The preceding discussion gives the following.

∂w ∂w ∂x ∂w ∂ y = + ∂r ∂ x ∂r ∂ y ∂r

∂w ∂w ∂x ∂w ∂ y = + ∂s ∂ x ∂s ∂ y ∂s

= (2 x ) (1) + (2 y) (1)

= (2 x ) (−1) + (2 y) (1)

= 2( r − s ) + 2( r + s )

= −2 ( r − s ) + 2 ( r + s )

= 4r

= 4s

Substitute for the intermediate variables.

If f is a function of a single intermediate variable x, our equations are even simpler.

If w = f ( x ) and x = g ( r , s ) , then dw ∂ x ∂w = dx ∂r ∂r

Chain Rule

and

∂w dw ∂ x = . ∂s dx ∂ s

w = f (x) dw dx

In this case, we use the ordinary (single-variable) derivative, dw dx. The dependency diagram is shown in Figure 13.24. x

'x 'r

'x 's

r

Implicit Differentiation Revisited The two-variable Chain Rule in Theorem 5 leads to a formula that takes some of the algebra out of implicit differentiation. Suppose that

s

'w dw 'x = 'r dx 'r

1. The function F ( x , y ) is differentiable and 2. The equation F ( x , h( x ) ) = 0 defines y implicitly as a differentiable function of x, say y = h( x ).

'w dw 'x = 's dx 's

FIGURE 13.24

Dependency diagram for differentiating f as a composite function of r and s with one intermediate variable. w = F(x, y)

'w = F x 'x

Fy = 'w 'y

Since w = F ( x , h( x ) ) = 0, the derivative dw dx must be zero. Computing the derivative from the Chain Rule (dependency diagram in Figure 13.25), we find dy dw dx = Fx + Fy dx dx dx dy = Fx ⋅ 1 + Fy ⋅ . dx

0=

If Fy = ∂ w ∂ y ≠ 0, we can solve this equation for dy dx to get

y = h(x)

x dx =1 dx

dy = h′(x) dx

x dy dw = Fx • 1 + Fy • dx dx

FIGURE 13.25

Dependency diagram for differentiating w = F ( x , y ) with respect to x. Setting dw dx = 0 leads to a simple computational formula for implicit differentiation (Theorem 8).

Theorem 5 with t = x and f = F

F dy = − x. Fy dx We state this result formally.

THEOREM 8—A Formula for Implicit Differentiation

Suppose that F ( x , y ) is differentiable and that the equation F ( x , y ) = 0 defines y as a differentiable function of x. Then, at any point where Fy ≠ 0, F dy = − x. Fy dx

(1)

802

Chapter 13 Partial Derivatives

EXAMPLE 5

Use Theorem 8 to find dy dx if y 2 − x 2 − sin xy = 0.

Solution Take F ( x , y ) = y 2 − x 2 − sin xy. Then

−2 x − y cos xy 2 x + y cos xy F dy = = − x = − . 2 y − x cos xy Fy 2 y − x cos xy dx This calculation is significantly shorter than a single-variable calculation using implicit differentiation.

The result in Theorem 8 is easily extended to three variables. Suppose that the equation F ( x , y, z ) = 0 defines the variable z implicitly as a function z = f ( x , y ). Then, for all ( x , y ) in the domain of f , we have F ( x , y, f ( x , y )) = 0. Assuming that F and f are differentiable functions, we can use the Chain Rule to differentiate the equation F ( x , y, z ) = 0 with respect to the independent variable x: ∂F ∂ x ∂F ∂ y ∂F ∂z + + ∂x ∂x ∂y ∂x ∂z ∂x ∂z = Fx ⋅ 1 + Fy ⋅ 0 + Fz ⋅ , ∂x

0=

y is constant when we differentiate with respect to x.

so Fx + Fz

∂z = 0. ∂x

A similar calculation for differentiating with respect to the independent variable y gives Fy + Fz

∂z = 0. ∂y

Whenever Fz ≠ 0, we can solve these last two equations for the partial derivatives of z = f ( x , y ) to obtain

F ∂z = − x ∂x Fz

and

Fy ∂z = − . ∂y Fz

(2)

An important result from advanced calculus, called the Implicit Function Theorem, states the conditions for which our results in Equations (2) are valid. If the partial derivatives Fx , Fy , and Fz are continuous throughout an open region R in space containing the point ( x 0 , y 0 , z 0 ) , and if for some constant c, F ( x 0 , y 0 , z 0 ) = c and Fz ( x 0 , y 0 , z 0 ) ≠ 0, then the equation F ( x , y, z ) = c defines z implicitly as a differentiable function of x and y near ( x 0 , y 0 , z 0 ) , and the partial derivatives of z are given by Equations (2).

∂z ∂z and at ( 0, 0, 0 ) if x 3 + z 2 + ye xz + z cos y = 0. ∂x ∂y Solution Let F ( x , y, z ) = x 3 + z 2 + ye xz + z cos y. Then EXAMPLE 6

Find

Fx = 3 x 2 + zye xz ,

Fy = e xz − z sin y,

and

Fz = 2 z + xye xz + cos y.

13.4  The Chain Rule

803

Since F ( 0, 0, 0 ) = 0, Fz ( 0, 0, 0 ) = 1 ≠ 0, and all first partial derivatives are continuous, the Implicit Function Theorem says that F ( x , y, z ) = 0 defines z as a differentiable function of x and y near the point ( 0, 0, 0 ). From Equations (2), F 3 x 2 + zye xz ∂z = − x = − ∂x Fz 2 z + xye xz + cos y

and

Fy e xz − z sin y ∂z = − = − . 2 z + xye xz + cos y ∂y Fz

and

∂z 1 = − = −1. ∂y 1

At ( 0, 0, 0 ) we find ∂z 0 = − = 0 ∂x 1

Functions of Many Variables We have seen several different forms of the Chain Rule in this section, but each one is just a special case of one general formula. When solving particular problems, it may help to draw the appropriate dependency diagram by placing the dependent variable on top, the intermediate variables in the middle, and the selected independent variable at the bottom. To find the derivative of the dependent variable with respect to the selected independent variable, start at the dependent variable and read down each route of the dependency diagram to the independent variable, calculating and multiplying the derivatives along each route. Then add the products found for the different routes. In general, suppose that w = f ( x 1 , x 2 , … , x n ) is a differentiable function of the intermediate variables x 1 , x 2 , … , x n (a finite set) and that x 1 , x 2 , … , x n are differentiable functions of the independent variables t 1 , t 2 , …, t m (another finite set). Then w is a differentiable function of the variables t 1 , t 2 , … , t m , and the partial derivatives of w with respect to these variables are given by equations of the form ∂ w ∂ x1 ∂w ∂x 2 ∂w ∂x n ∂w = + ++ ∂ x1 ∂t i ∂ x 2 ∂t i ∂ x n ∂t i ∂t i

for i = 1, 2, … , m.

One way to remember this equation is to think of the right-hand side as the dot product of two n-dimensional vectors: ∂w = ∂t i

∂x ∂ x1 ∂ x 2 ∂w ∂w ∂w , ,…, , ,…, n . ⋅ ∂t i ∂t i ∂t i ∂ x1 ∂ x 2 ∂xn       Derivatives of w  with respect to the intermediate variables

Derivatives of the intermediate variables with respect to the selected independent variable

The first vector describes how w changes in various directions, while the second vector indicates the velocity vector of x( t i ) = x 1 ( t i ), x 2 ( t i ), …, x n ( t i ) . These concepts will be studied further in the next section.

EXERCISES

13.4

Chain Rule: One Independent Variable

In Exercises 1–6, (a) express dw dt as a function of t, both by using the Chain Rule and by expressing w in terms of t and differentiating directly with respect to t. Then (b) evaluate dw dt at the given value of t. 1. w =

x2

+

y2,

x = cos t ,

y = sin t; t = π

2. w = x 2 + y 2, x = cos t + sin t , y = cos t − sin t; t = 0 y x 3. w = + , x = cos 2 t , y = sin 2 t , z = 1 t ; t = 3 z z 4. w = ln ( x 2 + y 2 + z 2 ) , x = cos t , t = 3

y = sin t , z = 4 t ;

5. w = 2 ye x − ln z , x = ln ( t 2 + 1), t =1 6. w = z − sin xy, x = t ,

y = tan −1 t , z = e t ;

y = ln t , z = e t −1; t = 1

Chain Rule: Two and Three Independent Variables

In Exercises 7 and 8, (a) express ∂ z ∂u and ∂ z ∂ υ as functions of u and υ both by using the Chain Rule and by expressing z directly in terms of u and υ before differentiating. Then (b) evaluate ∂ z ∂u and ∂ z ∂ υ at the given point ( u, υ ). 7. z = 4e x ln y, x = ln ( u cos υ ) , y = u sin υ; ( u, υ ) = ( 2, π 4 )

804

Chapter 13 Partial Derivatives

8. z = tan −1 ( x y ) , x = u cos υ, y = u sin υ; ( u, υ ) = (1.3, π 6 ) In Exercises 9 and 10, (a) express ∂ w ∂u and ∂ w ∂ υ as functions of u and υ both by using the Chain Rule and by expressing w directly in terms of u and υ before differentiating. Then (b) evaluate ∂ w ∂u and ∂ w ∂ υ at the given point (u, υ). 9. w = xy + yz + xz , x = u + υ, y = u − υ, z = uυ; ( u, υ ) = (1 2, 1) 10. w = ln ( x 2 + y 2 + z 2 ) , x = ue υ sin u, z = ue υ ; ( u, υ ) = (−2, 0 )

y = ue υ cos u,

In Exercises 11 and 12, (a) express ∂u ∂ x , ∂u ∂ y , and ∂u ∂ z as functions of x, y, and z both by using the Chain Rule and by expressing u directly in terms of x, y, and z before differentiating. Then (b) evaluate ∂u ∂ x , ∂u ∂ y , and ∂u ∂ z at the given point ( x , y, z ). p−q , p = x + y + z, q = x − y + z, 11. u = q−r r = x + y − z; ( x , y, z ) = ( 3, 2, 1)

27. x 2 + xy + y 2 − 7 = 0, (1, 2 ) 28. xe y + sin xy + y − ln 2 = 0, ( 0, ln 2 ) 29. ( x 3 − y 4 ) 6 + ln ( x 2 + y ) = 1, (−1, 0 ) 30. xe x 2 y − ye x = x + y − 2, (1, 1) Find the values of ∂ z ∂ x and ∂ z ∂ y at the points in Exercises 31–34. 31. z 3 − xy + yz + y 3 − 2 = 0, (1, 1, 1) 1 1 1 32. + + − 1 = 0, ( 2, 3, 6 ) x y z 33. sin ( x + y ) + sin ( y + z ) + sin ( x + z ) = 0, ( π, π, π ) 34. xe y + ye z + 2 ln x − 2 − 3 ln 2 = 0, (1, ln 2, ln 3 ) Finding Partial Derivatives at Specified Points

35. Find ∂ w ∂r when r = 1, s = −1 if w = ( x + y + z ) 2, x = r − s, y = cos ( r + s ) , z = sin ( r + s ). 36. Find ∂ w ∂ υ when u = −1, υ = 2 x = υ 2 u , y = u + υ, z = cos u.

if

w = xy + ln z , w = x 2 + ( y x ),

12. u = e qr sin −1 p, p = sin x , q = z 2 ln y, r = 1 z ; ( x , y, z ) = ( π 4 ,1 2, −1 2 )

37. Find ∂ w ∂ υ when u = 0, υ = 0 x = u − 2υ + 1, y = 2u + υ − 2.

Using a Dependency Diagram

38. Find ∂ z ∂u when u = 0, υ = 1 if z = sin xy + x sin y, x = u 2 + υ 2, y = uυ.

In Exercises 13–24, draw a dependency diagram and write a Chain Rule formula for each derivative. 13. 14. 15.

16.

17. 18. 19. 20. 21. 22.

23. 24.

dz for z = f ( x , y ) , x = g(t ), y = h(t ) dt dz for z = f ( u, υ, w ) , u = g(t ), υ = h(t ), w = k (t ) dt ∂w ∂w and for w = h ( x , y, z ) , x = f ( u, υ ) , y = g ( u, υ ) , ∂u ∂υ z = k ( u, υ ) ∂w ∂w and for w = f ( r , s, t ) , r = g ( x , y ) , s = h ( x , y ) , ∂x ∂y t = k ( x, y ) ∂w ∂w and for w = g ( x , y ) , x = h ( u, υ ), y = k ( u, υ ) ∂u ∂υ ∂w ∂w and for w = g ( u, υ ) , u = h ( x , y ) , υ = k ( x , y ) ∂x ∂y ∂z ∂z and for z = f ( x , y ) , x = g ( t , s ) , y = h ( t , s ) ∂t ∂s ∂y for y = f (u), u = g ( r , s ) ∂r ∂w ∂w and for w = g(u), u = h ( s, t ) ∂s ∂t ∂w for w = f ( x , y, z , υ ) , x = g ( p, q ) , y = h ( p, q ) , ∂p z = j ( p, q ) , υ = k ( p, q ) ∂w ∂w and for w = f ( x , y ) , x = g(r ), y = h(s) ∂r ∂s ∂w for w = g ( x , y ) , x = h ( r , s, t ) , y = k ( r , s, t ) ∂s

Implicit Differentiation

Assuming that the equations in Exercises 25–30 define y as a differentiable function of x, use Theorem 8 to find the value of dy dx at the given point. 25. x 3 − 2 y 2 + xy = 0, (1, 1) 26. xy + y 2 − 3 x − 3 = 0, (−1, 1)

if

39. Find ∂ z ∂u and ∂ z ∂ υ when u = ln 2, υ = 1 if z = 5 tan −1 x and x = e u + ln υ. 40. Find ∂ z ∂u and ∂ z ∂ υ when u = 1, υ = −2 if z = ln q and q = υ + 3 tan −1 u. Theory and Examples

41. Assume that w = f ( s 3 + t 2 ) and f ′( x ) = e x. Find

(

∂w ∂w . and ∂t ∂s

)

∂f s and ( x , y ) = xy, , t ∂x 2 ∂f ∂w ∂w x . and ( x, y ) = . Find ∂t ∂s 2 ∂y 43. Assume that z = f ( x , y ) , x = g(t ), y = h(t ), f x ( 2, −1) = 3, and f y ( 2, −1) = −2. If g(0) = 2, h(0) = −1, g′(0) = 5, and dz . h′(0) = −4, find dt t = 0

42. Assume

that

w = f ts 2,

44. Assume that z = f ( x , y ) 2, x = g(t ), y = h(t ), f x (1, 0 ) = −1, f y (1, 0 ) = 1, and f (1, 0 ) = 2. If g(3) = 1, h(3) = 0, dz g′(3) = −3, and h′(3) = 4, find . dt t = 3 45. Assume that z = f ( w), w = g ( x , y ) , x = 2r 3 − s 2, and y = re s . If g x ( 2, 1) = −3, g y ( 2, 1) = 2, f ′(7) = −1, and ∂z ∂z . and g ( 2, 1) = 7, find ∂r r =1, s = 0 ∂ s r =1, s = 0 46. Assume that z = ln ( f ( w) ) ,  w = g ( x , y ) , x = r − s , and y = r 2 s . If g x ( 2, −9 ) = −1, g y ( 2, −9 ) = 3, f ′(−2) = 2, ∂z f (−2) = 5, and g ( 2, −9 ) = −2, find and ∂r r = 3, s =−1 ∂z . ∂ s r = 3, s =−1 47. Changing voltage in a circuit The voltage V in a circuit that satisfies the law V = IR is slowly dropping as the battery wears out. At the same time, the resistance R is increasing as the resistor heats up. Use the equation dV ∂V dI ∂V dR = + dt ∂ I dt ∂ R dt

13.4  The Chain Rule

to find how the current is changing at the instant when R  600 ohms, I  0.04 amp, dR dt  0.5 ohm s, and dV dt = −0.01 volt s. V + −

805

a. Find where the maximum and minimum temperatures on the circle occur by examining the derivatives dT dt and d 2T dt 2. b. Suppose that T = 4 x 2 − 4 xy + 4 y 2 . Find the maximum and minimum values of T on the circle. 56. Temperature on an ellipse Let T = g ( x , y ) be the temperature at the point ( x , y ) on the ellipse

Battery

x = 2 2 cos t ,

I

y =

2 sin t ,

0 ≤ t ≤ 2Q ,

and suppose that

48. Changing dimensions in a box The lengths a, b, and c of the edges of a rectangular box are changing with time. At the instant in question, a  1 m, b  2 m, c  3 m, da dt  db dt  1 m s, and dc dt = −3 m s. At what rates are the box’s volume V and surface area S changing at that instant? Are the box’s interior diagonals increasing in length or decreasing? 49. If f ( u, V, w ) is differentiable and u = x − y, V = y − z , and w = z − x, show that ∂f ∂f ∂f + + = 0. ∂x ∂y ∂z 50. Polar coordinates Suppose that we substitute polar coordinates x  r cos R and y  r sin R in a differentiable function w = f ( x , y ). a. Show that ∂w = f x cos R + f y sin R ∂r and 1 ∂w = − f x sin R + f y cos R. r ∂R

∂T = x. ∂y

∂T = y, ∂x

R

a. Locate the maximum and minimum temperatures on the ellipse by examining dT dt and d 2T dt 2 . b. Suppose that T = xy − 2. Find the maximum and minimum values of T on the ellipse. 57. The temperature T = T ( x , y ) in nC at point ( x , y ) satisfies T x (1, 2 ) = 3 and T y (1, 2 ) = −1. If x = e 2 t −2 cm and y = 2 + ln t cm, find the rate at which the temperature T changes when t  1 s. 58. A bug crawls on the surface z = x 2 − y 2 directly above a path in the xy-plane given by x  f (t ) and y  g(t ) . If f (2) = 4, f ′(2) = −1, g(2) = −2, and g′(2) = −3, then at what rate is the bug’s elevation z changing when t  2? Differentiating Integrals Under mild continuity restrictions, it is true that if F (x) = then F ′( x ) =

b

∫a

b

∫a

g ( t , x ) dt ,

g x ( t , x ) dt. Using this fact and the Chain Rule, we

can find the derivative of

b. Solve the equations in part (a) to express f x and f y in terms of s w sr and s w sR .

F (x) =

∫a

f (x)

g ( t , x ) dt

by letting

c. Show that ( fx )2 + ( fy)2 =

( ∂∂wr )

2

+

( )

1 ∂w 2 . r 2 ∂R

51. Laplace equations Show that if w = f ( u, V ) satisfies the Laplace equation f uu + f VV = 0 and if u = ( x 2 − y 2 ) 2 and V  xy, then w satisfies the Laplace equation w xx + w yy = 0. 52. Laplace equations Let w = f (u) + g(V), where u = x + iy, V = x − iy, and i = −1. Show that w satisfies the Laplace equation w xx + w yy = 0 if all the necessary functions are differentiable. 53. Extreme values on a helix Suppose that the partial derivatives of a function f ( x , y, z ) at points on the helix x  cos t , y  sin t , z  t are f x = cos t ,

f y = sin t ,

f z = t 2 + t − 2.

At what points on the curve, if any, can f take on extreme values? 54. A space curve Let w  x 2e 2 y cos 3z. Find the value of dw dt at the point (1, ln 2, 0 ) on the curve x = cos t , y = ln ( t + 2 ) , z  t. 55. Temperature on a circle Let T = f ( x , y ) be the temperature at the point ( x , y ) on the circle x = cos t , y = sin t , 0 ≤ t ≤ 2Q, and suppose that ∂T = 8 x − 4 y, ∂x

∂T = 8 y − 4 x. ∂y

G ( u, x ) =

u

∫a

g ( t , x ) dt ,

where u  f ( x ). Find the derivatives of the functions in Exercises 59 and 60. x2

59. F ( x ) =

∫0

60. F ( x ) =

∫x

1 2

t 4 + x 3 dt t 3 + x 2 dt

61. Water is flowing into a tank in the form of a right-circular cylinder at the rate of ( 4 5 ) Q m 3 min . The tank is stretching in such a way that even though it remains cylindrical, its radius is increasing at the rate of 0.002 m min. How fast is the surface of the water rising when the radius is 2 m and the volume of water in the tank is 20 Q m 3? 62. Suppose f is a differentiable function of x, y, and z and u = f ( x , y, z ). Then if x  r sin φ cos θ, y  r sin φ sin θ , and z  r cos G , express su sr, su sG, and su sR in terms of su s x , su s y, and su s z . 63. At a given instant, the length of one leg of a right triangle is 10 m, and it is increasing at the rate of 1 m min, and the length of the other leg of the right triangle is 12 m, and it is decreasing at the rate of 2 m min. Find the rate of change of the measure of the acute angle opposite the leg of length 12 m at the given instant.

806

Chapter 13 Partial Derivatives

13.5 Directional Derivatives and Gradient Vectors If you look at the map (Figure 13.26) showing contours within Yosemite National Park in California, you will notice that the streams flow perpendicular to the contours. The streams are following paths of steepest descent so the waters reach lower elevations as quickly as possible. Therefore, the fastest instantaneous rate of change in a stream’s elevation above sea level has a particular direction. In this section, you will see why this direction, called the “downhill” direction, is perpendicular to the contours.

Directional Derivatives in the Plane FIGURE 13.26

Contours within Yosemite National Park in California show streams, which follow paths of steepest descent, running perpendicular to the contours. (Source: Yosemite National Park Map from U.S. Geological Survey, http://www.usgs.gov)

y Line x = x 0 + su 1, y = y 0 + su 2

We know from Section 13.4 that if f ( x , y ) is differentiable, then the rate at which f changes with respect to t along a differentiable curve x = g(t ), y = h(t ) is df ∂ f dx ∂ f dy = + . dt ∂ x dt ∂ y dt At any point P0 ( x 0 , y 0 ) = P0 ( g(t 0 ),   h(t 0 ) ) , this equation gives the rate of change of f with respect to increasing t and therefore depends, among other things, on the direction of motion along the curve. If the curve is a straight line and t is the arc length parameter along the line measured from P0 in the direction of a given unit vector u, then df dt is the rate of change of f with respect to distance in its domain in the direction of u. By varying u, we find the rates at which f changes with respect to distance as we move through P0 in different directions. We now define this idea more precisely. Suppose that the function f ( x , y ) is defined throughout a region R in the xy-plane, that P0 ( x 0 , y 0 ) is a point in R, and that u = u1i + u 2 j is a unit vector. Then the equations

u = u1i + u 2 j

x = x 0 + su1 , y = y 0 + su 2

Direction of increasing s

parametrize the line through P0 parallel to u. If the parameter s measures arc length from P0 in the direction of u, we find the rate of change of f at P0 in the direction of u by calculating d f ds at P0 (Figure 13.27).

R P0(x 0, y0 ) 0

x

FIGURE 13.27 The rate of change of f in the direction of u at a point P0 is the rate at which f changes along this line at P0 .

DEFINITION The derivative of f at P0 ( x 0 , y 0 ) in the direction of the unit vector u = u1 i + u 2 j is the number

( dfds )

u, P0

= lim s→0

f ( x 0 + su1 , y 0 + su 2 ) − f ( x 0 , y 0 ) , s

(1)

provided the limit exists.

The directional derivative defined by Equation (1) is also denoted by D u f ( P0 )

or

Du f

P0 .

“The derivative of f    in the direction of u, evaluated at P0”

The partial derivatives f x ( x 0 , y 0 ) and f y ( x 0 , y 0 ) are the directional derivatives of f at P0 in the i and j directions. This observation can be seen by comparing Equation (1) to the definitions of the two partial derivatives given in Section 13.3. EXAMPLE 1

Using the definition, find the derivative of f ( x , y ) = x 2 + xy

at P0 (1, 2 ) in the direction of the unit vector u = (1

2 ) i + (1

2 ) j.

13.5  Directional Derivatives and Gradient Vectors

807

Solution Applying the definition in Equation (1), we obtain

( dfds )

u, P0

= lim s→0

=

= = =

f ( x 0 + su1 , y 0 + su 2 ) − f ( x 0 , y 0 ) s

(

Eq. (1)

)

1 1 ,2 + s ⋅ − f (1, 2 ) 2 2 Substitute. lim s→0 s s 2 s s 1+ 2+ + 1+ − (1 2 + 1 ⋅ 2 ) 2 2 2 lim s→0 s 2 s s2 2s 3s 1+ + + 2+ + −3 2 2 2 2 lim s→0 s 5s 2 +s 5 5 = lim +s = lim 2 . s→0 s→0 s 2 2 f 1+ s⋅

(

) (

(

)(

)

) (

)

(

)

The rate of change of f ( x , y ) = x 2 + xy at P0 (1, 2 ) in the direction u is 5

2.

Interpretation of the Directional Derivative The equation z = f ( x , y ) represents a surface S in space. If z 0 = f ( x 0 , y 0 ) , then the point P ( x 0 , y 0 , z 0 ) lies on S. The vertical plane that passes through P and P0 ( x 0 , y 0 ) parallel to u intersects S in a curve C (Figure 13.28). The rate of change of f in the direction of u is the slope of the tangent to C at P in the right-handed system formed by the vectors u and k. z Surface S: z = f (x, y)

f (x0 + su1, y0 + su2) − f (x0, y0)

Tangent line

Q

P(x0, y0, z0) s

y

C (x0 + su1, y0 + su2)

x

u = u1i + u2 j

P0(x0, y0)

FIGURE 13.28

The slope of the trace curve C at P0 is lim slope ( PQ); this is the directional derivative

Q→ P

( dfds )

u , P0

= Du f

. P0

When u = i, the directional derivative at P0 is ∂ f ∂ x evaluated at ( x 0 , y 0 ). When u = j, the directional derivative at P0 is ∂ f ∂ y evaluated at ( x 0 , y 0 ). The directional derivative generalizes the two partial derivatives. We can now ask for the rate of change of f in any direction u, not just in the directions i and j. For a physical interpretation of the directional derivative, suppose that T = f ( x , y ) is the temperature at each point ( x , y ) over a region in the plane. Then f ( x 0 , y 0 ) is the

808

Chapter 13 Partial Derivatives

temperature at the point P0 ( x 0 , y 0 ), and D u f P is the instantaneous rate of change of the 0 temperature at P0 stepping off in the direction u.

Calculation and Gradients We now develop an efficient formula to calculate the directional derivative for a differentiable function f . We begin with the line x = x 0 + su1 , y = y 0 + su 2 ,

(2)

through P0 ( x 0 , y 0 ) , parametrized with the arc length parameter s increasing in the direction of the unit vector u = u1i + u 2 j. Then, by the Chain Rule we find

( dfds )

u, P0

= =

∂f ∂x ∂f ∂x

P0

P0

∂f dx + ∂y ds ∂f ∂y

  u1 +

P0

dy ds

Chain Rule for differentiable f

u2

From Eqs. (2), dx ds = u1 and dy ds = u 2

P0

⎡∂ f ⎤ ∂f i+ j⎥ = ⎢ ⎢ ∂ x P0 ∂ y P0 ⎥⎦ ⎣   Gradient of  f  at P0

⎡ ⎤ ⋅ ⎢⎢ u1i + u 2 j ⎥⎥ . ⎣⎦

(3)

Direction u

Equation (3) says that the derivative of a differentiable function f in the direction of u at P0 is the dot product of u with a special vector, which we now define.

DEFINITION

The gradient vector (or gradient) of f ( x , y ) is the vector ∇f =

∂f ∂f i+ j. ∂x ∂y

The value of the gradient vector obtained by evaluating the partial derivatives at a point P0 ( x 0 , y 0 ) is written ∇f

or

P0

∇f ( x 0 , y 0 ).

The notation ∇f is read “grad f ” as well as “gradient of f ” and “del f .” The symbol ∇ by itself is read “del.” Another notation for the gradient is grad f . Using the gradient notation, we restate Equation (3) as a theorem.

THEOREM 9—The Directional Derivative Is a Dot Product

If f ( x , y ) is differentiable in an open region containing P0 ( x 0 , y 0 ) , then

( dfds )

u, P0

= ∇f

P0

⋅ u,

(4)

the dot product of the gradient ∇f at P0 with the vector u. In brief, D u f = ∇f ⋅ u.

EXAMPLE 2 Find the derivative of f ( x , y ) = xe y + cos ( xy) at the point ( 2, 0 ) in the direction of v = 3i − 4 j. Solution Recall that the direction of a vector v is the unit vector obtained by dividing v by its length:

u =

v v 3 4 = = i − j. v 5 5 5

809

13.5  Directional Derivatives and Gradient Vectors

The partial derivatives of f are everywhere continuous and at ( 2, 0 ) are given by

y ∇f = i + 2j

2

f x ( 2, 0 ) = ( e y − y sin ( xy) )

1

0 −1

= e0 − 0 = 1 ( 2, 0 )

f y ( 2, 0 ) = ( xe y − x sin ( xy) ) 1 P0 (2, 0)

3

4

u = 3i − 4j 5 5

Picture ∇f as a vector in the domain of f . The figure shows a number of level curves of f . The rate at which f changes at ( 2, 0 ) in the direction u is ∇f ⋅ u = −1, which is the component of ∇f in the direction of unit vector u (Example 2). FIGURE 13.29

x

= 2e 0 − 2 ⋅ 0 = 2. ( 2, 0 )

The gradient of f at ( 2, 0 ) is ∇f

( 2, 0 )

= f x ( 2, 0 ) i + f y ( 2, 0 ) j = i + 2 j

(Figure 13.29). The derivative of f at ( 2, 0 ) in the direction of v is therefore Du f

( 2, 0 )

= ∇f

( 2, 0 )

⋅u

= ( i + 2 j) ⋅

Eq. (4) with the D u f

P0

notation

( 35 i − 45 j) = 35 − 85 = −1.

Evaluating the dot product in the brief version of Equation (4) gives D u f = ∇f ⋅ u = ∇f u cos θ = ∇f cos θ, where θ is the angle between the vectors u and ∇f , and reveals the following properties.

Properties of the Directional Derivative Du f = ∇f ⋅ u = ∇f cos θ

1. The function f increases most rapidly when cos θ = 1, which means that θ = 0 and u is the direction of ∇f . That is, at each point P in its domain, f increases most rapidly in the direction of the gradient vector ∇f at P. The derivative in this direction is D u f = ∇f cos (0) = ∇f . 2. Similarly, f decreases most rapidly in the direction of −∇f . The derivative in this direction is D u f = ∇f cos (π ) = − ∇f . 3. Any direction u orthogonal to a gradient ∇f ≠ 0 is a direction of zero change in f because θ then equals π 2 and D u f = ∇f cos ( π 2 ) = ∇f ⋅ 0 = 0.

As we discuss later, these properties hold in three dimensions as well as two. EXAMPLE 3

Find the directions in which f ( x , y ) = ( x 2 2 ) + ( y 2 2 )

(a) increases most rapidly at the point (1, 1), and (b) decreases most rapidly at (1, 1). (c) What are the directions of zero change in f at (1, 1)? Solution

(a) The function increases most rapidly in the direction of ∇f at (1, 1). The gradient there is ∇f

(1, 1)

= ( xi + yj )

(1, 1)

= i + j.

Its direction is u =

i+ j = i+ j

i+ j ( 1) 2

+

( 1) 2

=

1 1 i+ j. 2 2

810

Chapter 13 Partial Derivatives

(b) The function decreases most rapidly in the direction of −∇f at (1, 1), which is

z = f (x, y) y2 x2 = + 2 2

z

−u = −

1 1 i− j. 2 2

(c) The directions of zero change at (1, 1) are the directions orthogonal to ∇f : (1, 1, 1)

n = − 1

−∇f

1

x

Most rapid decrease in f Most rapid increase in f

(1, 1)

y Zero change in f

∇f = i + j

1 1 i+ j 2 2

Gradients and Tangents to Level Curves If a differentiable function f ( x , y ) has a constant value c along a smooth curve r = g(t ) i + h(t ) j (making the curve part of a level curve of f ), then f ( g(t ), h(t ) ) = c. Differentiating both sides of this equation with respect to t leads to the equations

The direction in which f ( x , y ) increases most rapidly at (1, 1) is the direction of ∇f (1,1) = i + j. It corresponds to the direction of steepest ascent on the surface at (1, 1, 1) (Example 3).

d d f ( g(t ),   h(t ) ) = (c) dt dt ∂ f dg ∂ f dh + = 0 ∂ x dt ∂ y dt

(

When it is nonzero, the gradient of a differentiable function of two variables at a point is always normal to the function’s level curve through that point.

)

dr dt

∇f

FIGURE 13.31

(5)

Chain Rule

⎛∂ f ∂ f ⎞⎟ dg dh ⎜⎜ i+ j⎟ ⋅ i+ j = 0. ⎝ ∂x dt ∂ y ⎟⎠ dt      

The level curve f (x, y) = f (x 0 , y 0 )

∇f (x 0 , y 0 )

1 1 i− j. 2 2

See Figure 13.30.

FIGURE 13.30

(x 0 , y 0 )

−n =

and

Assuming the gradient of f is a nonzero vector, Equation (5) says that ∇f is normal to the tangent vector d r dt, so it is normal to the curve. This is seen in Figure 13.31. At every point ( x 0 , y 0 ) in the domain of a differentiable function f ( x , y ) where the gradient of f is a nonzero vector, this vector is normal to the level curve through ( x 0 , y 0 ) (Figure 13.31). Equation (5) validates our observation that streams flow perpendicular to the contours in topographical maps (see Figure 13.26). Since the downflowing stream will reach its destination in the fastest way, it must flow in the direction of the negative gradient vectors from Property 2 for the directional derivative. Equation (5) tells us these directions are perpendicular to the level curves. This observation also enables us to find equations for tangent lines to level curves. They are the lines normal to the gradients. The line through a point P0 ( x 0 , y 0 ) normal to a nonzero vector N = Ai + Bj has the equation A( x − x 0 ) + B ( y − y 0 ) = 0 (Exercise 39). If N is the gradient ∇f ( x , y ) = f x ( x 0 , y 0 ) i + f y ( x 0 , y 0 ) j,   and this gra0 0 dient is not the zero vector, then this equation gives the following formula. Equation for the Tangent Line to a Level Curve f x ( x 0 , y 0 )( x − x 0 ) + f y ( x 0 , y 0 )( y − y 0 ) = 0

EXAMPLE 4

Find an equation for the tangent to the ellipse x2 + y2 = 2 4

(Figure 13.32) at the point (−2, 1).

(6)

13.5  Directional Derivatives and Gradient Vectors

y

Solution The ellipse is a level curve of the function x − 2y = −4

∇f (−2, 1) = −i + 2j

−2 −1

1 0

f ( x, y ) =

x2 + y2 = 2 4

"2 (−2, 1)

811

1

2

2" 2

x2 + y 2. 4

The gradient of f at (−2, 1) is ∇f

x

(−2, 1)

=

( 2x i + 2 yj)

(−2, 1)

= −i + 2 j.

Because this gradient vector is nonzero, the tangent to the ellipse at (−2, 1) is the line ( − 1)( x

FIGURE 13.32

We can find the tangent to the ellipse ( x 2 4 ) + y 2 = 2 by treating the ellipse as a level curve of the function f ( x , y ) = ( x 2 4 ) + y 2 (Example 4).

+ 2 ) + ( 2 )( y − 1) = 0

Eq. (6)

x − 2 y = −4.

Simplify.

If we know the gradients of two functions f and g, we automatically know the gradients of their sum, difference, constant multiples, product, and quotient. You are asked to establish the following rules in Exercise 40. Notice that these rules have the same form as the corresponding rules for derivatives of single-variable functions.

Algebra Rules for Gradients

1. Sum Rule:

∇ ( f + g ) = ∇f + ∇g

2. Difference Rule:

∇ ( f − g ) = ∇f − ∇g

3. Constant Multiple Rule:

∇( kf ) = k∇f

4. Product Rule:

∇( fg) = f ∇g + g∇f ⎪⎫ ⎪⎪ ⎛ f ⎟⎞ ⎬ ∇ − ∇ g f f g ⎪⎪ ∇⎜⎜ ⎟⎟ = ⎝g⎠ g2 ⎪⎭

5. Quotient Rule:

EXAMPLE 5

( any number  k )

Scalar multipliers on left of gradients

We illustrate two of the rules with f ( x, y ) = x − y

g( x, y ) = 3y

∇f = i − j

∇g = 3 j.

We have 1. ∇( f − g ) = ∇ ( x − 4 y ) = i − 4 j = ∇f − ∇g 2. ∇( fg) = ∇( 3 xy − 3 y 2 ) = 3 yi + ( 3 x − 6 y ) j and f ∇g + g∇f = ( x − y ) 3 j + 3 y ( i − j ) = 3 yi + ( 3 x − 6 y ) j.

Rule 2

Substitute. Simplify.

We have therefore verified that for this example, ∇( fg) = f ∇g + g∇f .

Functions of Three Variables For a differentiable function f ( x , y, z ) and a unit vector u = u1i + u 2 j + u 3k in space, we have ∇f =

∂f ∂f ∂f i+ j+ k ∂x ∂y ∂z

and D u f = ∇f ⋅ u =

∂f ∂f ∂f u + u + u . ∂y 2 ∂z 3 ∂x 1

812

Chapter 13 Partial Derivatives

The directional derivative can once again be written in the form D u f = ∇f ⋅ u = ∇f u cos θ = ∇f cos θ, so the properties listed earlier for functions of two variables extend to three variables. At any given point, f increases most rapidly in the direction of ∇f and decreases most rapidly in the direction of −∇f . In any direction orthogonal to ∇f , the derivative is zero. EXAMPLE 6 (a) Find the derivative of f ( x , y, z ) = x 3 − xy 2 − z at P0 (1, 1, 0 ) in the direction of v = 2 i − 3 j + 6k. (b) In what directions does f change most rapidly at P0 , and what are the rates of change in these directions? Solution

(a) The direction of v is obtained by dividing v by its length: v =

+ (−3 ) 2 + ( 6 ) 2 = 2 3 6 v u= = i − j + k. 7 7 7 v ( 2 )2

49 = 7

The partial derivatives of f at P0 are f x = ( 3x 2 − y 2 )

= 2,

f y = −2 xy

(1, 1, 0 )

(1, 1, 0 )

= −2,

f z = −1

= −1. (1, 1, 0 )

The gradient of f at P0 is ∇f

(1, 1, 0 )

= 2 i − 2 j − k.

The derivative of f at P0 in the direction of v is therefore Du f

(1, 1, 0 )

(

2 3 6 ⋅ u = ( 2i − 2 j − k ) ⋅ i − j + k 7 7 7 4 6 6 4 = + − = . 7 7 7 7 = ∇f

(1, 1, 0 )

)

(b) The function increases most rapidly in the direction of ∇f = 2 i − 2 j − k and decreases most rapidly in the direction of −∇f . The rates of change in the directions are, respectively, ∇f =

( 2 )2

+ ( − 2 ) 2 + ( − 1) 2 =

9 = 3

and

− ∇f = −3.

Functions of More Than Three Variables The gradient of a differentiable function of n variables f ( x 1 , x 2 , … , x n ) is ∇f =

∂f ∂f ∂f , ,, . ∂ x1 ∂ x 2 ∂xn

If u = 〈u1 , u 2 , … , u n 〉 is an n-dimensional vector such that u12 + u 22 +  + u n2 = 1 (so u is a unit vector since u = u12 + u 22 +  + u n2 = 1), then the directional derivative of f in the direction of u is D u f = ∇f ⋅ u =

∂f ∂f ∂f u1 + u2 +  + u . ∂x2 ∂xn n ∂ x1

13.5  Directional Derivatives and Gradient Vectors

EXAMPLE 7 w

z

813

The volume of the solid shown in Figure 13.33 consisting of a tetrahexy w dron on top of a triangular prism is given by f ( x , y, z , w ) = z+ . 2 3

(

)

(a) Calculate the derivative of f ( x , y, z , w ) at the point P0 ( 6, 5, 8, 4 ) in the direction of v = 〈1, −1, −1, 1〉. (b) What is the geometric significance of the value obtained in part (a)? Solution

y x

(a) The direction of v is the unit vector u=

FIGURE 13.33

A tetrahedron on top of a triangular prism (Example 7).

=

1 v v ( − 1) 2

1 〈1, −1, −1, 1〉 + ( − 1) 2 + 1 2

+ 1 = 〈1, −1, −1, 1〉 2 1 1 1 1 = ,− ,− , . 2 2 2 2 The four partial derivatives of f at the point P0 are 12

( ) x w = (z + ) 2 3

fx = fy

y w z+ 2 3

xy 2 xy fw = 6 fz =

( 6, 5, 8, 4 )

( 6, 5, 8, 4 )

( 6, 5, 8, 4 )

( 6, 5, 8, 4 )

=

70 , 3

= 28,

= 15, = 5.

The gradient of f at P0 is ∇f

( 6, 5, 8, 4 )

=

70 , 28, 15, 5 . 3

The derivative of f at ( 6, 5, 8, 4 ) in the direction of v is Du f

P0

= ∇f

P0

⋅u

70 1 1 1 1 , 28, 15, 5 ⋅ , − , − , 3 2 2 2 2 70 1 1 1 1 = + ( 28 ) − + (15 ) − + ( 5 ) 3 2 2 2 2 22 =− . 3 =

( )( )

( )

( )

( )

(b) Geometrically, this means that if the dimensions are x = 6, y = 5, z = 8, and w = 4, and the dimensions are changed by moving at unit speed so that x and w increase at the same rate while both y and z decrease at that rate, then the volume of the solid decreases at the rate of 7 1 3.

The Chain Rule for Paths If r (t ) = x (t ) i + y(t ) j + z (t )k is a smooth path C, and w = f ( r(t ) ) is a scalar function evaluated along C, then according to the Chain Rule, Theorem 6 in Section 13.4, dw ∂ w dx ∂ w dy ∂ w dz . = + + ∂ x dt ∂ y dt ∂ z dt dt

814

Chapter 13 Partial Derivatives

The partial derivatives on the right-hand side of the above equation are evaluated along the curve r(t ), and the derivatives of the intermediate variables are evaluated at t. If we express this equation using vector notation, we have The Derivative Along a Path

d f ( r (t ) ) = ∇f ( r (t ) ) ⋅ r ′(t ). dt

(7)

What Equation (7) says is that the derivative of the composite function f ( r(t ) ) is the “derivative” (gradient) of the outside function f , evaluated at r(t ), “times” (dot product) the derivative of the inside function r. This is analogous to the “Outside-Inside” Rule for derivatives of composite functions studied in Section 3.6. That is, the multivariable Chain Rule for paths has exactly the same form as the rule for single-variable differential calculus when appropriate interpretations are given to the meanings of the terms and operations involved.

13.5

EXERCISES Calculating Gradients

In Exercises 1–6, find the gradient of the function at the given point. Then sketch the gradient, together with the level curve that passes through the point. 1. f ( x , y ) = y − x , ( 2, 1)

2. f ( x , y ) = ln ( x 2 + y 2 ) , (1, 1)

3. g ( x , y ) = xy 2, ( 2, −1)

4. g ( x , y ) =

5. f ( x , y ) =

y2 x2 , ( 2, 1) − 2 2

x , ( 4, − 2 ) y

In Exercises 7–10, find ∇f at the given point. 7. f ( x , y, z ) = x 2 + y 2 − 2 z 2 + z ln x , (1, 1, 1) 8. f ( x , y, z ) = 2 z 3 − 3( x 2 + y 2 ) z + arctan xz , (1, 1, 1) 9. f ( x , y, z ) =

(x2

+

y2

+

z 2 )−1 2

21. f ( x , y, z ) = ( x y ) − yz , P0 ( 4, 1, 1) 22. g ( x , y, z ) = xe y + z 2 , P0 (1, ln 2, 1 2 ) 23. f ( x , y, z ) = ln xy + ln yz + ln xz , P0 (1, 1, 1) 24. h ( x , y, z ) = ln ( x 2 + y 2 − 1) + y + 6 z , P0 (1, 1, 0 ) Tangent Lines to Level Curves

In Exercises 25–28, sketch the curve f ( x , y ) = c, together with ∇f and the tangent line at the given point. Then write an equation for the tangent line.

2 x + 3 y , (−1, 2 )

6. f ( x , y ) = tan −1

20. f ( x , y ) = x 2 y + e xy sin y, P0 (1, 0 )

+ ln ( xyz ), (−1, 2, − 2 )

10. f ( x , y, z ) = e x + y cos z + ( y + 1) arcsin x , ( 0, 0, π 6 ) Finding Directional Derivatives

In Exercises 11–18, find the derivative of the function at P0 in the direction of v. 11. f ( x , y ) = 2 xy − 3 y 2, P0 ( 5, 5 ) , v = 4 i + 3 j 12. f ( x , y ) = 2 x 2 + y 2, P0 (−1, 1) , v = 3i − 4 j x−y 13. g ( x , y ) = , P0 (1, −1) , v = 12 i + 5 j xy + 2 14. h ( x , y ) = arctan ( y x ) + 3 arcsin ( xy 2 ) , P0 (1, 1) , v = 3i − 2 j 15. f ( x , y, z ) = xy + yz + zx , P0 (1, −1, 2 ) , v = 3i + 6 j − 2k 16. f ( x , y, z ) = x 2 + 2 y 2 − 3z 2, P0 (1, 1, 1) , v = i + j + k 17. g ( x , y, z ) = 3e x cos yz , P0 ( 0, 0, 0 ), v = 2 i + j − 2k 18. h ( x , y, z ) = cos xy + e yz + ln zx , P0 (1, 0, 1 2 ) , v = i + 2 j + 2k In Exercises 19–24, find the directions in which the functions increase most rapidly, and the directions in which they decrease most rapidly, at P0 . Then find the derivatives of the functions in these directions. 19. f ( x , y ) = x 2 + xy + y 2, P0 (−1, 1)

25. x 2 + y 2 = 4, ( 2, 2 )

26. x 2 − y = 1, ( 2, 1)

27. xy = −4, ( 2, −2 )

28. x 2 − xy + y 2 = 7, (−1, 2 )

Theory and Examples

29. Let f ( x , y ) = x 2 − xy + y 2 − y. Find the directions u and the values of D u f (1, −1) for which a. D u f (1, −1) is largest

b. D u f (1, −1) is smallest

c. D u f (1, −1) = 0

d. D u f (1, −1) = 4

e. D u f (1, −1) = −3 30. Let f ( x , y ) =

(

)

(x (x

− y) . Find the directions u and the values of + y)

1 3 D u f − , for which 2 2 1 3 a. D u f − , is largest 2 2 1 3 c. D u f − , = 0 2 2 1 3 e. D u f − , =1 2 2

( ( (

) ) )

( (

) )

1 3 b. D u f − , is smallest 2 2 1 3 d. D u f − , = −2 2 2

31. Zero directional derivative In what direction is the derivative of f ( x , y ) = xy + y 2 at P ( 3, 2 ) equal to zero? 32. Zero directional derivative In what directions is the derivative of f ( x , y ) = ( x 2 − y 2 ) ( x 2 + y 2 ) at P (1, 1) equal to zero? 33. Is there a direction u in which the rate of change of f ( x , y ) = x 2 − 3 xy + 4 y 2 at P (1, 2 ) equals 14? Give reasons for your answer.

815

13.6  Tangent Planes and Differentials

34. Changing temperature along a circle Is there a direction u in which the rate of change of the temperature function T ( x , y, z ) = 2 xy − yz (temperature in degrees Celsius, distance in meters) at P (1, −1, 1) is −3 °C m ? Give reasons for your answer. 35. The derivative of f ( x , y ) at P0 (1, 2 ) in the direction of i + j is 2 2 and in the direction of −2 j is −3. What is the derivative of f in the direction of −i − 2 j ? Give reasons for your answer. 36. The derivative of f ( x , y, z ) at a point P is greatest in the direction of v = i + j − k. In this direction, the value of the derivative is 2 3. a. What is ∇f at P? Give reasons for your answer. b. What is the derivative of f at P in the direction of i + j? 37. Directional derivatives and scalar components How is the derivative of a differentiable function f ( x , y, z ) at a point P0 in the direction of a unit vector u related to the scalar component of ∇f P in the direction of u? Give reasons for your answer. 0

38. Directional derivatives and partial derivatives Assuming that the necessary derivatives of f ( x , y, z ) are defined, how are D i f , D j f , and D k f related to f x , f y , and f z ? Give reasons for your answer. 39. Lines in the xy-plane Show that A( x − x 0 ) + B ( y − y 0 ) = 0 is an equation for the line in the xy-plane through the point ( x 0 , y 0 ) normal to the vector N = Ai + Bj. 40. The algebra rules for gradients Given a constant k and the gradients ∂f ∇f = i+ ∂x ∂g ∇g = i+ ∂x

∂f ∂f j+ k, ∂y ∂z ∂g ∂g j+ k, ∂y ∂z

establish the algebra rules for gradients.

In Exercises 41–44, find a parametric equation for the line that is perpendicular to the graph of the given equation at the given point. 41. x 2 + y 2 = 25, (−3, 4 ) 42. x 2 + xy + y 2 = 3, ( 2, −1) 43. x 2 + y 2 + z 2 = 14, ( 3, −2, 1) 44. z = x 3 − xy 2, (−1, 1, 0 ) Gradients and Directional Derivatives for Functions of More Than Three Variables

In Exercises 45–48, find ∇f at the given point. x y − x 2 z 3, ( 2, 4, −1, 3 ) w 46. f ( x , y, z , w ) = x 3 sin y + w 2 cos z , (−2, π, 0, 3 ) π 47. f ( x , y, z , s, t ) = e y ln s + x 2 t tan z , 3, 0, , e, 5 4 ( x 2 + y 2 ) arctan t 48. f ( x , y, z , s, t ) = , (−2, 1, −1, 2, 1) z 2s 45. f ( x , y, z , w ) =

(

)

In Exercises 49–52, find the derivative of the function at P0 in the direction of v. w ln x 49. f ( x , y, z , w ) = 2 3 , P0 ( e 2, −2, 1, −3 ) , v = 〈−1, 2, −2, 4〉 y z 50. f ( x , y, z , w ) = ( x − y ) 2 + e z − w , P0 ( 4, 2, 3,1), v = 〈1, 0, −2, 2〉 51. f ( x , y, z , s, t ) = s arcsin ( x + y ) − t 2 arctan ( x − z ), 1 P0 0, , −1, 1, −1 , v = 〈−1, 1, 0, 3, 5〉 2 π π st 52. f ( x , y, z , s, t ) = sin tx + cos sy − , P0 , , 2, 5, 1 , z 4 6 v = 〈−3, 2, −2, 2, 2〉

(

)

(

)

13.6 Tangent Planes and Differentials In single-variable differential calculus, we saw how the derivative defined the tangent line to the graph of a differentiable function at a point on the graph. The tangent line then provided for a linearization of the function at the point. In this section, we will see analogously how the gradient defines the tangent plane to the level surface of a function w = f ( x , y, z ) at a point on the surface. The tangent plane then provides for a linearization of f at the point and defines the total differential of the function. ∇f

Tangent Planes and Normal Lines

v2 P0

If r (t ) = x (t ) i + y(t ) j + z (t )k is a smooth curve on the level surface f ( x , y, z ) = c of a differentiable function f , we found in Equation (7) of the last section that

v1 f (x, y, z) = c

The gradient ∇f is orthogonal to the velocity vector of every smooth curve in the surface through P0 . The velocity vectors at P0 therefore lie in a common plane, which we call the tangent plane at P0 . FIGURE 13.34

d f ( r (t ) ) = ∇f ( r (t ) ) ⋅ r ′(t ). dt Since f is constant along the curve r, the derivative on the left-hand side of the equation is 0, so the gradient ∇f is orthogonal to the curve’s velocity vector r ′. Now let us restrict our attention to the curves that pass through a point P0 (Figure 13.34). All the velocity vectors at P0 are orthogonal to ∇f at P0 , so the curves’ tangent lines all lie in the plane through P0 normal to ∇f . (assuming it is a nonzero vector). We now define this plane.

816

Chapter 13 Partial Derivatives

DEFINITIONS The tangent plane to the level surface f ( x , y, z ) = c of a differentiable function f at a point P0 where the gradient is not zero is the plane through P0 normal to ∇f P0 .

The normal line of the surface at P0 is the line through P0 parallel to ∇f

P0

.

The results of Section 11.5 imply that the tangent plane and normal line satisfy the following equations, as long as the gradient at the point P0 is not the zero vector.

Tangent Plane to f ( x, y, z ) = c at P0 ( x 0 , y 0 , z 0 )

f x ( P0 ) ( x − x 0 ) + f y ( P0 ) ( y − y 0 ) + f z ( P0 ) ( z − z 0 ) = 0

(1)

Normal Line to f ( x, y, z ) = c at P0 ( x 0 , y 0 , z 0 )

x = x 0 + f x ( P0 ) t ,

z P0(1, 2, 4)

EXAMPLE 1 The surface x2 + y2 + z − 9 = 0

z = z 0 + f z ( P0 ) t

(2)

Find the tangent plane and normal line of the level surface f ( x , y, z ) = x 2 + y 2 + z − 9 = 0

   A circular paraboloid

at the point P0 (1, 2, 4 ). Normal line

Solution The surface is shown in Figure 13.35. The tangent plane is the plane through P0 perpendicular to the gradient of f at P0 . The gradient is

∇f

Tangent plane 1

y = y 0 + f y ( P0 ) t ,

P0

= ( 2 xi + 2 yj + k )

(1, 2, 4 )

= 2 i + 4 j + k.

The tangent plane is therefore the plane

2 y

x

2( x − 1) + 4 ( y − 2 ) + ( z − 4 ) = 0,

2 x + 4 y + z = 14.

The line normal to the surface at P0 is x = 1 + 2t ,

FIGURE 13.35

The tangent plane and normal line to this level surface at P0 (Example 1).

or

y = 2 + 4t ,

z = 4 + t.

To find an equation for the plane tangent to a smooth surface z = f ( x , y ) at a point P0 ( x 0 , y 0 , z 0 ) where z 0 = f ( x 0 , y 0 ) , we first observe that the equation z = f ( x , y ) is equivalent to f ( x , y ) − z = 0. The surface z = f ( x , y ) is therefore the zero level surface of the function F ( x , y, z ) = f ( x , y ) − z. The partial derivatives of F are ∂ ( f ( x, y ) − z ) = f x − 0 = f x ∂x ∂ Fy = ( f ( x, y ) − z ) = f y − 0 = f y ∂y ∂ Fz = ( f ( x , y ) − z ) = 0 − 1 = −1. ∂z Fx =

The formula Fx ( P0 ) ( x − x 0 ) + Fy ( P0 ) ( y − y 0 ) + Fz ( P0 ) ( z − z 0 ) = 0 for the plane tangent to the level surface at P0 therefore reduces to f x ( x 0 , y 0 ) ( x − x 0 ) + f y ( x 0 , y 0 ) ( y − y 0 ) − ( z − z 0 ) = 0.

817

13.6  Tangent Planes and Differentials

Plane Tangent to a Surface z = f ( x, y ) at ( x 0 , y 0 , f ( x 0 , y 0 ))

The plane tangent to the surface z = f ( x , y ) of a differentiable function f at the point P0 ( x 0 , y 0 , z 0 ) = ( x 0 , y 0 , f ( x 0 , y 0 )) is f x ( x 0 , y 0 ) ( x − x 0 ) + f y ( x 0 , y 0 ) ( y − y 0 ) − ( z − z 0 ) = 0.

EXAMPLE 2

(3)

Find the plane tangent to the surface z = x cos y − ye x at ( 0, 0, 0 ).

Solution We calculate the partial derivatives of f ( x , y ) = x cos y − ye x and use Equation (3):

f x ( 0, 0 ) = ( cos y − ye x )

= 1− 0 ⋅1 = 1 ( 0, 0 )

f y ( 0, 0 ) = (−x sin y − e x )

= 0 − 1 = −1. ( 0, 0 )

The tangent plane is therefore 1 ⋅ ( x − 0 ) − 1 ⋅ ( y − 0 ) − ( z − 0 ) = 0,

  

Eq. (3)

or x − y − z = 0.

EXAMPLE 3

z

The plane x + z− 4 = 0 g(x, y, z)

The surfaces f ( x , y, z ) = x 2 + y 2 − 2 = 0

   A cylinder

and g ( x , y, z ) = x + z − 4 = 0

   A plane

meet in an ellipse E (Figure 13.36). Find parametric equations for the line tangent to E at the point P0 (1, 1, 3 ). ∇g The ellipse E

Solution The tangent line is orthogonal to both ∇f and ∇g at P0 , and therefore parallel to v = ∇f × ∇g. The components of v and the coordinates of P0 give us equations for the line. We have

(1, 1, 3) ∇ f × ∇g

∇f

∇f

(1, 1, 3 )

= ( 2 xi + 2 yj )

∇g (1, 1, 3) = ( i + k ) y x

The cylinder x2 + y2 − 2 = 0

(1, 1, 3 )

(1, 1, 3 )

= 2i + 2 j

= i+k i

j

k

v = ( 2 i + 2 j ) × ( i + k ) = 2 2 0 = 2 i − 2 j − 2k. 1 0

1

f(x, y, z)

FIGURE 13.36

This cylinder and plane intersect in an ellipse E (Example 3).

The tangent line to the ellipse of intersection is x = 1 + 2t ,

y = 1 − 2t ,

z = 3 − 2 t.

818

Chapter 13 Partial Derivatives

Estimating Change in a Specific Direction The directional derivative plays a role similar to that of an ordinary derivative when we want to estimate how much the value of a function f changes if we move a small distance ds from a point P0 to another point nearby. If f were a function of a single variable, we would have df = f ′( P0 ) ds.

  

Ordinary derivative × increment

For a function of two or more variables, we use the formula df = ( ∇f

P0

⋅ u ) ds,

  

Directional derivative × increment

where u is the direction of the motion away from P0 .

Estimating the Change in f in a Direction u

To estimate the change in the value of a differentiable function f when we move a small distance ds from a point P0 in a particular direction u, use this formula: df = ( ∇f P ⋅ u ) ds 0 

Directional Distance derivative increment

EXAMPLE 4

Estimate how much the value of f ( x , y, z ) = y sin x + 2 yz

will change if the point P ( x , y, z ) moves 0.1 unit from P0 ( 0, 1, 0 ) straight toward P1 ( 2, 2, −2 ). Solution We first find the derivative of f at P0 in the direction of the vector  P0 P1 = 2 i + j − 2k. The direction of this vector is   P0 P1 PP 2 1 2  = 0 1 = i + j − k. u =  3 3 3 3 P0 P1

z 2

∇f

The gradient of f at P0 is 1

x

2

1

P1(2, 2, −2)

∇f P0

0

1

2

= A ( y cos x ) i + ( sin x + 2 z ) j + 2 ykB

( 0, 1, 0 )

= i + 2k .

Therefore, y

–2

As P ( x , y, z ) moves off the level surface at P0 by 0.1 unit directly toward P1, the function f changes value by approximately −0.067 unit (Example 4). FIGURE 13.37

( 0, 1, 0 )

∇f

P0

(

)

2 1 2 2 4 2 ⋅ u = ( i + 2k ) ⋅ i + j − k = − = − . 3 3 3 3 3 3

The change df in f that results from moving ds = 0.1 unit away from P0 in the direction of u is approximately df = ( ∇f

P0

( 23 )( 0.1) ≈ −0.067 unit.

⋅ u ) (ds) = −

See Figure 13.37.

How to Linearize a Function of Two Variables Functions of two variables can be quite complicated, and we sometimes need to approximate them with simpler ones that give the accuracy required for specific applications without being so difficult to work with. We do this in a way that is similar to the way we find linear replacements for functions of a single variable (Section 3.11).

13.6  Tangent Planes and Differentials

A point near (x 0 , y 0 )

819

Suppose the function we wish to approximate is z = f ( x , y ) near a point ( x 0 , y 0 ) at which we know the values of f , f x , and f y and at which f is differentiable. If we move from ( x 0 , y 0 ) to a nearby point ( x , y ) by increments Δx = x − x 0 and Δy = y − y 0 (see Figure 13.38), then the definition of differentiability from Section 13.3 shows that the change

(x, y)

Δy = y − y0

f ( x , y ) − f ( x 0 , y 0 ) = f x ( x 0 , y 0 ) Δ x + f y ( x 0 , y 0 ) Δ y + ε1 Δ x + ε 2 Δ y ,

A point where f is differentiable

where ε1 , ε 2 → 0 as Δx , Δy → 0. If the increments Δx and Δy are small, the products ε1 Δx and ε 2 Δy will eventually be smaller still, and we have the approximation

(x 0 , y0 ) Δx = x − x 0

f ( x , y ) ≈ f ( x 0 , y 0 ) + f x ( x 0 , y 0 ) ( x − x 0 ) + f y ( x 0 , y 0 ) ( y − y 0 ).  L( x , y )

FIGURE 13.38

If f is differentiable In other words, as long as Δx and Δy are small, f will have approximately the same value at ( x 0 , y 0 ) , then the value of f at as the linear function L. point ( x , y ) nearby is approximately f ( x 0 , y 0 ) + f x ( x 0 , y 0 )Δx + f y ( x 0 , y 0 )Δy.

The linearization of a function f ( x , y ) at a point ( x 0 , y 0 ) where f is differentiable is the function

DEFINITIONS

L ( x , y ) = f ( x 0 , y 0 ) + f x ( x 0 , y 0 ) ( x − x 0 ) + f y ( x 0 , y 0 ) ( y − y 0 ). The approximation f ( x, y ) ≈ L ( x, y ) is the standard linear approximation of f at ( x 0 , y 0 ).

From Equation (3), we find that the plane z = L ( x , y ) is tangent to the surface z = f ( x , y ) at the point ( x 0 , y 0 ). Thus, the linearization of a function of two variables is a tangent-plane approximation in the same way that the linearization of a function of a single variable is a tangent-line approximation. (See Exercise 57.)

z

EXAMPLE 5

z = f(x, y)

Find the linearization of f ( x , y ) = x 2 − xy +

1 2 y +3 2

at the point ( 3, 2 ). (3, 2, 8)

Solution We first evaluate f , f x , and f y at the point ( x 0 , y 0 ) = ( 3, 2 ):

(

f ( 3, 2 ) = x 2 − xy + x

4

3

2

1 3 L(x, y)

4

y

1 2 y +3 2

)

( 3, 2 )

= 8

(

)

( 3, 2 )

(

)

( 3, 2 )

f x ( 3, 2 ) =

∂ 2 1 x − xy + y 2 + 3 ∂x 2

f y ( 3, 2 ) =

1 ∂ 2 x − xy + y 2 + 3 2 ∂y

= ( 2x − y ) = ( −x + y )

( 3, 2 )

( 3, 2 )

= 4 = −1,

which yields FIGURE 13.39

The tangent plane L ( x , y ) represents the linearization of f ( x , y ) in Example 5.

L ( x , y ) = f ( x 0 , y 0 ) + f x ( x 0 , y 0 )( x − x 0 ) + f y ( x 0 , y 0 )( y − y 0 ) = 8 + ( 4 )( x − 3 ) + ( −1)( y − 2 ) = 4 x − y − 2. The linearization of f at ( 3, 2 ) is L ( x , y ) = 4 x − y − 2 (see Figure 13.39).

820

Chapter 13 Partial Derivatives

y

When we approximate a differentiable function f ( x , y ) by its linearization L ( x , y ) at ( x 0 , y 0 ) , an important question is how accurate the approximation might be. If we can find a common upper bound M for f xx , f yy , and f xy on a rectangle R centered at ( x 0 , y 0 ) (Figure 13.40), then we can bound the error E throughout R by using a simple formula. The error is defined by E ( x , y ) = f ( x , y ) − L ( x , y ).

k h (x 0 , y0 ) R

0

FIGURE 13.40

x

The rectangular region R: x − x 0 ≤ h, y − y 0 ≤ k in the xy-plane.

The Error in the Standard Linear Approximation

If f has continuous first and second partial derivatives throughout an open set containing a rectangle R centered at ( x 0 , y 0 ), and if M is any upper bound for the values of f xx , f yy , and f xy on R, then the error E ( x , y ) incurred in replacing f ( x , y ) on R by its linearization L ( x, y ) = f ( x 0 , y0 ) + f x ( x 0 , y0 ) ( x − x 0 ) + f y ( x 0 , y0 ) ( y − y0 ) satisfies the inequality E ( x, y ) ≤

1 M ( x − x 0 + y − y 0 )2 . 2

To make E ( x , y ) small for a given M, we just make x − x 0 and y − y 0 small.

Differentials Recall from Section 3.11 that for a function of a single variable, y = f ( x ), we defined the change in f as x changes from a to a + Δx by Δf = f ( a + Δx ) − f ( a) and the differential of f as d f = f ′(a)Δx. We now consider the differential of a function of two variables. Suppose a differentiable function f ( x , y ) and its partial derivatives exist at a point ( x 0 , y 0 ). If we move to a nearby point ( x 0 + Δx , y 0 + Δy ) , the change in f is Δf = f ( x 0 + Δx , y 0 + Δy ) − f ( x 0 , y 0 ). A straightforward calculation based on the definition of L ( x , y ), using the notation x − x 0 = Δx and y − y 0 = Δy, shows that the corresponding change in L is ΔL = L ( x 0 + Δx , y 0 + Δy ) − L ( x 0 , y 0 ) = f x ( x 0 , y 0 ) Δx + f y ( x 0 , y 0 ) Δy. The differentials dx and dy are independent variables, so they can be assigned any values. Often we take dx = Δx = x − x 0 , and dy = Δy = y − y 0 . We then have the following definition of the differential or total differential of f .

If we move from ( x 0 , y 0 ) to a point ( x 0 + dx , y 0 + dy ) nearby, the resulting change

DEFINITION

df = f x ( x 0 , y 0 ) dx + f y ( x 0 , y 0 ) dy in the linearization of f is called the total differential of f .

13.6  Tangent Planes and Differentials

821

EXAMPLE 6 Suppose that a cylindrical can is designed to have a radius of 1 cm and a height of 5 cm, but that the radius and height are off by the amounts dr = +0.03 and dh = −0.1. Estimate the resulting absolute change in the volume of the can. Solution To estimate the absolute change in V = πr 2 h, we use

ΔV ≈ dV = Vr ( r0 , h 0 ) dr + Vh ( r0 , h 0 ) dh. With Vr = 2πrh and Vh = πr 2, we get dV = 2πr0 h 0 dr + πr0 2 dh = 2π (1)( 5 )( 0.03 ) + π (1) 2 (−0.1) = 0.3π − 0.1π = 0.2π ≈ 0.63 cm 3 .

EXAMPLE 7 Your company manufactures stainless steel right circular cylindrical molasses storage tanks that are 2.5 m high with a radius of 0.5 m. How sensitive are the tanks’ volumes to small variations in height and radius? Solution With V = πr 2 h, the total differential gives the approximation for the change in volume as

dV = Vr ( 0.5, 2.5 ) dr + Vh ( 0.5, 2.5 ) dh = ( 2πrh ) r=5 r = 25 h=5

FIGURE 13.41

dr + ( πr 2 )

dh ( 0.5, 2.5 )

= 2.5π dr + 0.25π dh.

h = 25

(a)

( 0.5, 2.5 )

(b)

The volume of cylinder (a) is more sensitive to a small change in r than it is to an equally small change in h. The volume of cylinder (b) is more sensitive to small changes in h than it is to small changes in r (Example 7).

Thus, a 1-unit change in r will change V by about 2.5π units. A 1-unit change in h will change V by about 0.25π units. The tank’s volume is 10 times more sensitive to a small change in r than it is to a small change of equal size in h. As a quality control engineer concerned with being sure the tanks have the correct volume, you would want to pay special attention to their radii. In contrast, if the values of r and h are reversed to make r = 2.5 and h = 0.5, then the total differential in V becomes dV = ( 2πrh )

( 2.5, 0.5 )

dr + ( πr 2 )

( 2.5, 0.5 )

dh = 2.5π dr + 6.25π dh.

Now the volume is more sensitive to changes in h than to changes in r (Figure 13.41). The general rule is that functions are most sensitive to small changes in the variables that generate the largest partial derivatives.

Functions of More Than Two Variables Analogous results hold for differentiable functions of more than two variables. 1. The linearization of f ( x , y, z ) at a point P0 ( x 0 , y 0 , z 0 ) is L ( x , y, z ) = f ( P0 ) + f x ( P0 ) ( x − x 0 ) + f y ( P0 ) ( y − y 0 ) + f z ( P0 ) ( z − z 0 ). 2. Suppose that R is a closed rectangular solid centered at P0 and lying in an open region on which the second partial derivatives of f are continuous. Suppose also that f xx , f yy , f zz , f xy , f xz , and f yz are all less than or equal to M throughout R. Then the error E ( x , y, z ) = f ( x , y, z ) − L ( x , y, z ) in the approximation of f by L is bounded throughout R by the inequality E ≤

1 M ( x − x 0 + y − y 0 + z − z 0 ) 2. 2

822

Chapter 13 Partial Derivatives

3. If the second partial derivatives of f are continuous and if x, y, and z change from x 0 , y 0 , and z 0 by small amounts dx, dy, and dz, the total differential d f = f x ( P0 ) dx + f y ( P0 ) dy + f z ( P0 ) dz gives a good approximation of the resulting change in f . Find the linearization L ( x , y, z ) of

EXAMPLE 8

f ( x , y, z ) = x 2 − xy + 3 sin z at the point ( x 0 , y 0 , z 0 ) = ( 2, 1, 0 ). Find an upper bound for the error incurred in replacing f by L on the rectangular region R:

x − 2 ≤ 0.01,

y − 1 ≤ 0.02,

z ≤ 0.01.

Solution Routine calculations give

f ( 2, 1, 0 ) = 2,

f x ( 2, 1, 0 ) = 3,

f y ( 2, 1, 0 ) = −2,

f z ( 2, 1, 0 ) = 3.

Thus, L ( x , y, z ) = 2 + 3( x − 2 ) + ( −2 ) ( y − 1) + 3( z − 0 ) = 3 x − 2 y + 3z − 2. Since f xx = 2,

f yy = 0,

f zz = −3 sin z ,

f xy = −1,

f xz = 0,

f yz = 0,

and −3 sin z ≤ 3 sin 0.01 ≈ 0.03, we may take M = 2 as a bound on the second partials. Hence, the error incurred by replacing f by L on R satisfies E ≤

EXERCISES

13.6

Tangent Planes and Normal Lines to Surfaces

In Exercises 1–10, find equations for the (a) tangent plane and (b) normal line at the point P0 on the given surface. 1. x 2 + y 2 + z 2 = 3, P0 (1, 1, 1) 2. x 2 + y 2 − z 2 = 18, P0 ( 3, 5, −4 ) 3. 2 z − x 2 = 0, P0 ( 2, 0, 2 ) 4. x 2 + 2 xy − y 2 + z 2 = 7, P0 (1, −1, 3 ) 5. cos π x − x 2 y + e xz + yz = 4, P0 ( 0, 1, 2 ) 6. x 2 − xy − y 2 − z = 0, P0 (1, 1, −1) 7. x + y + z = 1, P0 ( 0, 1, 0 ) 8. x 2 + y 2 − 2 xy − x + 3 y − z = −4, P0 ( 2, −3, 18 ) 9. x ln y + y ln z = x , P0 (1, 1, e ) 10. ye x + ze y 2 = z , 

P0 ( 0, 0, 1)

In Exercises 11–14, find an equation for the plane that is tangent to the given surface at the given point. 11. z = ln ( x 2 + y 2 ) , (1, 0, 0 ) 12. z = e −( x

2 + y2 )

1 ( )( 2 0.01 + 0.02 + 0.01) 2 = 0.0016. 2

, ( 0, 0, 1)

13. z =

y − x , (1, 2, 1)

14. z = 4 x 2 + y 2 , (1, 1, 5 ) Tangent Lines to Intersecting Surfaces

In Exercises 15–20, find parametric equations for the line tangent to the curve of intersection of the surfaces at the given point. 15. Surfaces: x + y 2 + 2 z = 4, x = 1

(1, 1, 1) Point: 16. Surfaces: xyz = 1, x 2 + 2 y 2 + 3z 2 = 6 (1, 1, 1) Point: 17. Surfaces: x 2 + 2 y + 2 z = 4, y = 1 (1, 1, 1 2 ) Point: 18. Surfaces: x + y 2 + z = 2, y = 1 (1 2, 1, 1 2 ) Point: 19. Surfaces: x 3 + 3 x 2 y 2 + y 3 + 4 xy − z 2 = 0, x 2 + y 2 + z 2 = 11

(1, 1, 3 ) Point: 20. Surfaces: x 2 + y 2 = 4, x 2 + y 2 − z = 0 Point:

( 2, 2, 4 )

823

13.6  Tangent Planes and Differentials

wind speed. The precise formula, updated by the National Weather Service in 2001 and based on modern heat transfer theory, a human face model, and skin tissue resistance, is (after unit conversion)

Estimating Change

21. By about how much will f ( x , y, z ) = ln x 2 + y 2 + z 2 change if the point P ( x , y, z ) moves from P0 ( 3, 4, 12 ) a distance of ds = 0.1 unit in the direction of 3i + 6 j − 2k ? 22. By about how much will

W = W ( υ, T ) = 13.13 + 0.6215 T − 11.36 υ 0.16 + 0.396 T ⋅ υ 0.16, where T is air temperature in °C and υ is wind speed in km/h. A partial wind chill chart is given.

f ( x , y, z ) = e x cos yz

T(°C)

change as the point P ( x , y, z ) moves from the origin a distance of ds = 0.1 unit in the direction of 2 i + 2 j − 2k ?

5

23. By about how much will g ( x , y, z ) = x + x cos z − y sin z + y change if the point P ( x , y, z ) moves from P0 ( 2, −1, 0 ) a distance of ds = 0.2 unit toward the point P1 ( 0, 1, 2 )?

υ (km/ h)

24. By about how much will h ( x , y, z ) = cos (π xy) + xz 2 change if the point P ( x , y, z ) moves from P0 (−1, −1, −1) a distance of ds = 0.1 unit toward the origin? 25. Temperature change along a circle Suppose that the Celsius temperature at the point( x , y ) in the xy-plane is T ( x , y ) = x sin 2 y and that distance in the xy-plane is measured in meters. A particle is moving clockwise around the circle of radius 1 m centered at the origin at the constant rate of 2 m s. a. How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point P (1 2, 3 2 )?

a. How fast is the temperature experienced by the particle changing in degrees Celsius per meter when the particle is at the point P ( 8, 6, −4 )? b. How fast is the temperature experienced by the particle changing in degrees Celsius per second at P? Finding Linearizations

In Exercises 27–32, find the linearization L ( x , y ) of the function at each point. 27. f ( x , y ) = x 2 + y 2 + 1 at

a. ( 0, 0 ),

b. (1, 1)

28. f ( x , y ) =

at

a. ( 0, 0 ),

b. (1, 2 )

29. f ( x , y ) = 3 x − 4 y + 5 at

a. ( 0, 0 ),

b. (1, 1)

30. f ( x , y ) = x 3 y 4

at

a. (1, 1),

b. ( 0, 0 )

31. f ( x , y ) = e x cos y

at

a. ( 0, 0 ),

b. ( 0, π 2 )

32. f ( x , y ) =

at

a. ( 0, 0 ),

b. (1, 2 )

(x

+ y+

e 2 y− x

2 )2

33. Wind chill factor Wind chill, a measure of the apparent temperature felt on exposed skin, is a function of air temperature and

−10

−15

−20

−25

10

2.7 −3.3

−9.3 −15.2 −21.2 −27.2 −33.1

20

1.1 −5.2 −11.5 −17.8 −24.1 −30.4 −36.7

30

0.1 −6.4 −13.0 −19.5 −26.0 −32.5 −39.0

40

−0.7 −7.4 −14.0 −20.7 −27.4 −34.1 −40.8

50

−1.3 −8.1 −14.9 −21.7 −28.5 −35.4 −42.2

60

−1.8 −8.7 −15.7 −22.6 −29.5 −36.4 −43.3

a. Use the table to find W ( 30, −5 ) , W ( 50, −25 ) , and W ( 30, −10 ) . b. Use the formula to find W (15, −40 ) , W ( 80, −40 ) , and W ( 90, 0 ) . c. Find the linearization L ( υ, T ) of the function W ( υ, T ) at the point ( 40, −10 ). d. Use L ( υ, T ) in part (c) to estimate the following wind chill values.   i) W ( 39, −9 )

b. How fast is the temperature experienced by the particle changing in degrees Celsius per second at P? 26. Changing temperature along a space curve The Celsius temperature in a region in space is given by T ( x , y, z ) = 2 x 2 − xyz. A particle is moving in this region and its position at time t is given by x = 2t 2, y = 3t , z = −t 2, where time is measured in seconds and distance in meters.

−5

0

ii) W ( 42, −12 )

iii) W (10, −25 ) (Explain why this value is much different from the value found in the table.) 34. Find the linearization L ( υ, T ) of the function W ( υ, T ) in Exercise 31 at the point ( 50, −20 ). Use it to estimate the following wind chill values. a. W ( 49, −22 ) b. W ( 53, −19 ) c. W ( 60, −30 ) Bounding the Error in Linear Approximations

In Exercises 35–40, find the linearization L ( x , y ) of the function f ( x , y ) at P0 . Then find an upper bound for the magnitude E of the error in the approximation f ( x , y ) ≈ L ( x , y ) over the rectangle R. 35. f ( x , y ) = x 2 − 3 xy + 5 at R:

x − 2 ≤ 0.1,

P0 ( 2, 1) ,

y − 1 ≤ 0.1

36. f ( x , y ) = (1 2 ) x 2 + xy + (1 4 ) y 2 + 3 x − 3 y + 4 at P0 ( 2, 2 ), R:

x − 2 ≤ 0.1,

y − 2 ≤ 0.1

37. f ( x , y ) = 1 + y + x cos y at R:

x ≤ 0.2,

P0 ( 0, 0 ),

y ≤ 0.2

(Use cos y ≤ 1 and sin y ≤ 1 in estimating E.) 38. f ( x , y ) = xy 2 + y cos ( x − 1) at R:

x − 1 ≤ 0.1,

y − 2 ≤ 0.1

P0 (1, 2 ) ,

824

Chapter 13 Partial Derivatives

39. f ( x , y ) = e x cos y at R:

x ≤ 0.1,

P0 ( 0, 0 ) ,

a. Show that

y ≤ 0.1

⎛ R ⎞2 ⎛ R ⎞2 dR = ⎜⎜ ⎟⎟⎟ dR1 + ⎜⎜ ⎟⎟⎟ dR 2 . ⎝ R1 ⎠ ⎝ R2 ⎠

(Use e x ≤ 1.11 and cos y ≤ 1 in estimating E.) 40. f ( x , y ) = ln x + ln y at R:

x − 1 ≤ 0.2,

P0 (1, 1) ,

y − 1 ≤ 0.2

Linearizations for Three Variables

Find the linearizations L(x, y, z) of the functions in Exercises 41–46 at the given points.

b. You have designed a two-resistor circuit, like the one shown, to have resistances of R1 = 100 ohms and R 2 = 400 ohms, but there is always some variation in manufacturing, and the resistors received by your firm will probably not have these exact values. Will the value of R be more sensitive to variation in R1 or to variation in R 2? Give reasons for your answer.

41. f ( x , y, z ) = xy + yz + xz at a. (1, 1, 1) 42. f ( x , y, z ) =

b. (1, 0, 0 ) x2

+

y2

a. (1, 1, 1) 43. f ( x , y, z ) =

+

z2

c. ( 0, 0, 0 )

+ V −

at

b. ( 0, 1, 0 )

c. (1, 0, 0 )

b. (1, 1, 0 )

c. (1, 2, 2 )

c. In another circuit like the one shown, you plan to change R1 from 20 to 20.1 ohms and R 2 from 25 to 24.9 ohms. By about what percentage will this change R?

44. f ( x , y, z ) = ( sin xy ) z at a. ( π 2, 1, 1)

b. ( 2, 0, 1)

45. f ( x , y, z ) = e x + cos ( y + z ) at π a. ( 0, 0, 0 ) b. 0, , 0 2 46. f ( x , y, z ) = tan −1 ( xyz ) at

(

a. (1, 0, 0 )

)

(

π π c. 0, , 4 4

b. (1, 1, 0 )

)

c. (1, 1, 1)

In Exercises 47–50, find the linearization L(x, y, z) of the function f (x, y, z) at P0 . Then find an upper bound for the magnitude of the error E in the approximation f ( x , y, z ) ≈ L ( x , y, z ) over the region R. 47. f ( x , y, z ) = xz − 3 yz + 2 at x − 1 ≤ 0.01,

P0 (1, 1, 2 ) ,

y − 1 ≤ 0.01,

48. f ( x , y, z ) = x 2 + xy + yz + (1 4 ) z 2 R:

x − 1 ≤ 0.01,

y − 1 ≤ 0.01,

49. f ( x , y, z ) = xy + 2 yz − 3 xz at R:

x − 1 ≤ 0.01,

50. f ( x , y, z ) = R:

R2

x 2 + y 2 + z 2 at

a. (1, 0, 0 )

R:

R1

2 cos x

x ≤ 0.01,

sin ( y

y ≤ 0.01,

+

z)

at

at

b. What ratio of dx to dy will make df equal zero at (1, 0 )? 55. Value of a 2 × 2 determinant If a is much greater than b , c , and d , to which of a, b, c, and d is the value of the determinant f ( a , b , c, d ) =

P0 (1, 1, 2 ) ,

z − 2 ≤ 0.08 z ≤ 0.01 P0 ( 0, 0, π 4 ) ,

z − π 4 ≤ 0.01

Estimating Error; Sensitivity to Change

51. Estimating maximum error Suppose that T is to be found from the formula T = x ( e y + e − y ) , where x and y are found to be 2 and ln 2 with maximum possible errors of dx = 0.1 and dy = 0.02. Estimate the maximum possible error in the computed value of T. 52. Variation in electrical resistance The resistance R produced by wiring resistors of R1 and R 2 ohms in parallel (see accompanying figure) can be calculated from the formula 1 1 1 . = + R R1 R2

54. a. Around the point (1, 0 ), is f ( x , y ) = x 2 ( y + 1) more sensitive to changes in x or to changes in y? Give reasons for your answer.

z − 2 ≤ 0.02

P0 (1, 1, 0 ) ,

y − 1 ≤ 0.01,

53. You plan to calculate the area of a long, thin rectangle from measurements of its length and width. Which dimension should you measure more carefully? Give reasons for your answer.

a b c d

most sensitive? Give reasons for your answer. 56. The Wilson lot size formula The Wilson lot size formula in economics says that the most economical quantity Q of goods (radios, shoes, brooms, whatever) for a store to order is given by the formula Q = 2 KM h , where K is the cost of placing the order, M is the number of items sold per week, and h is the weekly holding cost for each item (cost of space, utilities, security, and so on). To which of the variables K, M, and h is Q most sensitive near the point ( K 0 , M 0 , h 0 ) = ( 2, 20, 0.05 )? Give reasons for your answer. Theory and Examples

57. The linearization of f ( x, y ) is a tangent-plane approximation Show that the tangent plane at the point P0 ( x 0 , y 0 , f ( x 0 , y 0 )) on the surface z = f ( x , y ) defined by a differentiable function f is the plane f x ( x 0 , y 0 )( x − x 0 ) + f y ( x 0 , y 0 )( y − y 0 ) − ( z − f ( x 0 , y 0 )) = 0, or z = f ( x 0 , y 0 ) + f x ( x 0 , y 0 )( x − x 0 ) + f y ( x 0 , y 0 )( y − y 0 ).

13.7  Extreme Values and Saddle Points

Thus, the tangent plane at P0 is the graph of the linearization of f at P0 (see accompanying figure).

z = f (x, y) z

(x0,

825

60. Normal curves A smooth curve is normal to a surface f ( x , y, z ) = c at a point of intersection if the curve’s velocity vector is a nonzero scalar multiple of ‹f at the point. Show that the curve 1 r (t ) = t i + t   j − ( t + 3 ) k 4 is normal to the surface x 2 + y 2 − z = 3 when t  1. 61. Consider a closed rectangular box with a square base, as shown in the figure. Assume x is measured with an error of at most 0.5% and y is measured with an error of at most 0.75%, so we have dx x  0.005 and dy y  0.0075.

y0, f (x0, y0)) z = L(x, y) y

(x0, y0)

y

x

58. Change along the involute of a circle Find the derivative of f ( x , y ) = x 2 + y 2 in the direction of the unit tangent vector of the curve r (t ) = ( cos t + t sin t ) i + ( sin t − t cos t ) j,

t > 0.

59. Tangent curves A smooth curve is tangent to the surface at a point of intersection if its velocity vector is orthogonal to ‹f there. Show that the curve r (t ) = is tangent to the surface

ti + +

x2

y2

t j + ( 2t − 1) k − z = 1 when t  1.

x x

a. Use a differential to estimate the relative error dV V in computing the box’s volume V. b. Use a differential to estimate the relative error dS S in computing the box’s surface area S. Hint for b:

4 x 2 + 4 xy 4 x 2 + 8 xy = 2 and ≤ 2 2 x + 4 xy 2 x 2 + 4 xy 2x 2

4 xy 2 x 2 + 4 xy ≤ = 1. + 4 xy 2 x 2 + 4 xy

13.7 Extreme Values and Saddle Points HISTORICAL BIOGRAPHY Siméon-Denis Poisson (1781–1840) French mathematician Poisson studied with Lagrange and Laplace at the École polytechnique.and did so well that he was made an assistant professor upon his graduation. In 1806, he replaced Fourier as the professor of mathematics. Poisson’s early work in mechanics appeared in his first volume of Traité de mécanique (1811), where he applied mathematics to applications in physics and mechanics, including elasticity and vibrations. To know more, visit the companion Website.

Continuous functions of two variables assume extreme values on closed, bounded domains (see Figures 13.42 and 13.43). We see in this section that we can narrow the search for these extreme values by examining the functions’ first partial derivatives. A function of two variables can assume extreme values only at boundary points of the domain or at interior domain points where both first partial derivatives are zero or where one or both of the first partial derivatives fail to exist. However, the vanishing of derivatives at an interior point ( a, b ) does not always signal the presence of an extreme value. The surface that is the graph of the function might be shaped like a saddle right above ( a, b ) and cross its tangent plane there.

Local Extreme Values for Functions of Two Variables To find the local extreme values of a function of a single variable, we look for points where the graph has a horizontal tangent line. At such points, we then look for local maxima, local minima, and points of inflection. For a function f ( x , y ) of two variables, we look for points where the surface z = f ( x , y ) has a horizontal tangent plane. At such points, we then look for local maxima, local minima, and saddle points. We begin by defining maxima and minima. DEFINITIONS

Let f ( x , y ) be defined on a region R containing the point ( a, b ).

Then

x y

FIGURE 13.42

The function

z = ( cos x )( cos y ) e −

x 2 + y2

has a maximum value of 1 and a minimum value of about 0.067 on the square region x b 3Q 2, y b 3Q 2.

1. f ( a, b ) is a local maximum value of f if f ( a, b ) ≥ f ( x , y ) for all domain points ( x , y ) in an open disk centered at ( a, b ). f ( a, b ) is an absolute maximum value of f on R if f ( a, b ) ≥ f ( x , y ) for all domain points ( x , y ) in R. 2. f ( a, b ) is a local minimum value of f if f ( a, b ) ≤ f ( x , y ) for all domain points ( x , y ) in an open disk centered at ( a, b ). f ( a, b ) is an absolute minimum value of f on R if f ( a, b ) ≤ f ( x , y ) for all domain points ( x , y ) in R.

826

Chapter 13 Partial Derivatives

Local maxima correspond to mountain peaks on the surface z = f ( x , y ), and local minima correspond to valley bottoms (Figure 13.44). At such points the tangent planes, when they exist, are horizontal. Local extrema are also called relative extrema. As with functions of a single variable, the key to identifying the local extrema is the First Derivative Theorem, which we next state and prove.

z

Absolute maximum No greater value of f anywhere. Also a local maximum.

x y

Local maximum No greater value of f nearby.

FIGURE 13.43

z =

The “roof surface”

1 ( x − y − x − y) 2

has a maximum value of 0 and a minimum value of −a on the square region x ≤ a, y ≤ a.

Local minimum No smaller value of f nearby.

Absolute minimum No smaller value of f anywhere. Also a local minimum.

FIGURE 13.44

A local maximum occurs at a mountain peak, and a local minimum occurs at a valley low point.

THEOREM 10—First Derivative Theorem for Local Extreme Values

If f ( x , y ) has a local maximum or minimum value at an interior point ( a, b ) of its domain and if the first partial derivatives exist there, then f x ( a, b ) = 0 and f y ( a, b ) = 0.

z

0f =0 0x z = f(x, y) 0f =0 0y

0 a

g(x) = f(x, b) b

FIGURE 13.45

If we substitute the values f x ( a, b ) = 0 and f y ( a, b ) = 0 into the equation

h(y) = f(a, y) y

x

Proof If f has a local extremum at ( a, b ), then the function g( x ) = f ( x , b ) has a local extremum at x = a (Figure 13.45). Therefore, g′(a) = 0 (Chapter 4, Theorem 2). Now g′(a) = f x ( a, b ) , so f x ( a, b ) = 0. A similar argument with the function h( y) = f ( a, y ) shows that f y ( a, b ) = 0.

(a, b, 0)

If a local maximum of f occurs at x = a,  y = b, then the first partial derivatives f x ( a, b ) and f y ( a, b ) are both zero.

f x ( a , b ) ( x − a ) + f y ( a ,   b ) ( y − b ) − ( z − f ( a , b )) = 0 for the tangent plane to the surface z = f ( x , y ) at ( a, b ), the equation reduces to 0 ⋅ ( x − a ) + 0 ⋅ ( y − b ) − z + f ( a, b ) = 0, or z = f ( a, b ). Thus, Theorem 10 says that the surface does indeed have a horizontal tangent plane at a local extremum, provided there is a tangent plane there. DEFINITION An interior point of the domain of a function f ( x , y ) where both

f x and f y are zero or where one or both of f x and f y do not exist is a critical point of f .

13.7  Extreme Values and Saddle Points

z

827

Theorem 10 says that the only points where a function f ( x , y ) can assume extreme values are critical points and boundary points. As with differentiable functions of a single variable, not every critical point gives rise to a local extremum. A differentiable function of a single variable might have a point of inflection. A differentiable function of two variables might have a saddle point, with the graph of f crossing the tangent plane defined there. y

DEFINITION A differentiable function f ( x , y ) has a saddle point at a critical

x z=

point ( a, b ) if in every open disk centered at ( a, b ) there are domain points ( x , y ) where f ( x , y ) > f ( a, b ) and domain points ( x , y ) where f ( x , y ) < f ( a, b ). The corresponding point ( a, b, f ( a, b )) on the surface z = f ( x , y ) is called a saddle point of the surface (Figure 13.46).

xy (x 2 − y 2 ) x2 + y2 z

EXAMPLE 1

Solution The domain of f is the entire plane (so there are no boundary points) and the partial derivatives f x = 2 x and f y = 2 y − 4 exist everywhere. Therefore, local extreme values can occur only where

y

x

Find the local extreme values of f ( x , y ) = x 2 + y 2 − 4 y + 9.

f x = 2x = 0

EXAMPLE 2 Saddle points at the

origin. z

15 10 5 x

2

1

f y = 2 y − 4 = 0.

The only possibility is the point ( 0, 2 ), where the value of f is 5. Since f ( x , y ) = x 2 + ( y − 2 ) 2 + 5 is never less than 5, we see that the critical point ( 0, 2 ) gives a local minimum (Figure 13.47).

z = y2 − y4 − x2

FIGURE 13.46

and

1

FIGURE 13.47

2

3

4

y

The graph of the function f ( x , y ) = + y 2 − 4 y + 9 is a paraboloid which has a local minimum value of 5 at the point ( 0, 2 ) (Example 1). x2

Find the local extreme values (if any) of f ( x , y ) = y 2 − x 2 .

Solution The domain of f is the entire plane (so there are no boundary points) and the partial derivatives f x = −2 x and f y = 2 y exist everywhere. Therefore, local extrema can occur only at the origin ( 0, 0 ), where f x = 0 and f y = 0. The value of f at the origin is 0. However, away from the origin along the positive x-axis, f has the value f ( x , 0 ) = −x 2 < 0; along the positive y-axis, f has the value f ( 0, y ) = y 2 > 0. Therefore, every open disk in the xy-plane centered at ( 0, 0 ) contains points where the function is positive and points where it is negative. The function has a saddle point at the origin and no local extreme values (Figure 13.48a). Figure 13.48b displays the level curves (they are hyperbolas) of f and shows the function decreasing and increasing in an alternating fashion among the groupings of hyperbolas.

That f x = f y = 0 at an interior point ( a, b ) of R does not guarantee that f has a local extreme value there. If f and its first and second partial derivatives are continuous on R, however, we may be able to learn more from the following theorem. THEOREM 11—Second Derivative Test for Local Extreme Values

Suppose that f ( x , y ) and its first and second partial derivatives are continuous throughout a disk centered at ( a, b ) and that f x ( a, b ) = f y ( a, b ) = 0. Then i) f has a local maximum at ( a, b ) if f xx < 0 and f xx f yy − f xy 2 > 0 at ( a, b ). ii) f has a local minimum at ( a, b ) if f xx > 0 and f xx f yy − f xy 2 > 0 at ( a, b ). iii) f has a saddle point at ( a, b ) if f xx f yy − f xy 2 < 0 at ( a, b ). iv) the test is inconclusive at ( a, b ) if f xx f yy − f xy 2 = 0 at ( a, b ). In this case, we must find some other way to determine the behavior of f at ( a, b ). The expression f xx f yy − f xy 2 is called the discriminant or Hessian of f . It is sometimes easier to remember it in determinant form, f xx f yy − f xy 2 =

f xx

f xy

f xy

f yy

.

828

Chapter 13 Partial Derivatives

The discriminant is the determinant of the Hessian matrix of f ,

z

⎡ f xx H f ( x, y ) = ⎢ ⎢⎣ f yx

z = y2 − x 2

y x

(a) y

f inc

Note that by Theorem 2, we have f xy ( a, b ) = f yx ( a, b ) at any point ( a, b ) satisfying the assumptions of Theorem 11. Theorem 11 says that if the discriminant is positive at the point ( a, b ), then the surface curves the same way in all directions: downward if f xx < 0, giving rise to a local maximum, and upward if f xx > 0, giving a local minimum. On the other hand, if the discriminant is negative at ( a, b ), then the surface curves up in some directions and down in others, so we have a saddle point. EXAMPLE 3

3

f xy ⎤ ⎥. f yy ⎥⎦

Find the local extreme values of the function f ( x , y ) = xy − x 2 − y 2 − 2 x − 2 y + 4.

1 −1

Saddle point f dec −3 −1

−3

f dec

Solution The function is defined and differentiable for all x and y, and its domain has no boundary points. The function therefore has extreme values only at the points where f x and f y are simultaneously zero. This leads to

x 1

f x = y − 2 x − 2 = 0,

3

f y = x − 2 y − 2 = 0,

or x = y = −2.

f inc

(b) FIGURE 13.48

(a) The origin is a saddle point of the function f ( x , y ) = y 2 − x 2 . There are no local extreme values (Example 2). (b) Level curves for the function f in Example 2.

Therefore, the point (−2, − 2 ) is the only point where f may take on an extreme value. To see whether it does so, we calculate f xx = −2,

f yy = −2,

f xy = 1.

The discriminant of f at ( a, b ) = ( −2, − 2 ) is f xx f yy − f xy 2 = (−2 )(−2 ) − (1) 2 = 4 − 1 = 3. The combination f xx < 0

and

f xx f yy − f xy 2 > 0

tells us that f has a local maximum at (−2, − 2 ). The value of f at this point is f (−2, −2 ) = 8. EXAMPLE 4

Find the local extreme values of f ( x , y ) = 3 y 2 − 2 y 3 − 3 x 2 + 6 xy.

Solution Since f is differentiable everywhere, it can assume extreme values only where

f x = 6 y − 6x = 0

6x − 6x 2 + 6x = 0

10

f xx = −6, 2

or

6 x ( 2 − x ) = 0.

The two critical points are therefore ( 0, 0 ) and ( 2, 2 ). To classify the critical points, we calculate the second derivatives:

5

3

f y = 6 y − 6 y 2 + 6 x = 0.

From the first of these equations we find x = y, and substitution for y into the second equation then gives

z

1

and

3

y

f yy = 6 − 12 y,

f xy = 6.

The discriminant is given by f xx f yy − f xy 2 = (−36 + 72 y ) − 36 = 72( y − 1).

At the critical point ( 0, 0 ) we see that the value of the discriminant is the negative number −72, so the function has a saddle point at the origin. At the critical point ( 2, 2 ) we see that the discriminant has the positive value 72. Combining this result with the negative value of FIGURE 13.49 The surface z = 3 y 2 − 2 y 3 − 3 x 2 + 6 xy has a saddle the second partial f xx = −6, Theorem 11 says that the critical point ( 2, 2 ) gives a local maximum value of f ( 2, 2 ) = 12 − 16 − 12 + 24 = 8. A graph of the surface is shown point at the origin and a local maximum at in Figure 13.49. the point ( 2, 2 ) (Example 4). x

829

13.7  Extreme Values and Saddle Points

EXAMPLE 5 Find the critical points of the function f ( x , y ) = 10 xye −( x + y ) and use the Second Derivative Test to classify each point as one where a saddle, local minimum, or local maximum occurs. 2

2

Solution First we find the partial derivatives f x and f y and set them simultaneously to zero in seeking the critical points:

f x = 10 ye −( x

2 + y2 )

− 20 x 2 ye −( x

2 + y2 )

= 10 y (1 − 2 x 2 ) e −( x

f y = 10 xe −( x

2 + y2 )

− 20 xy 2e −( x

2 + y2 )

= 0 ⇒ y = 0 or 1 − 2 x 2 = 0,  

2 + y2 )

= 10 x (1 − 2 y 2 ) e −( x

2 + y2 )

= 0 ⇒ x = 0 or 1 − 2 y 2 = 0.

Since both partial derivatives are continuous everywhere, the only critical points are

( 0, 0 ) ,

( 12 , 12 ), (− 12 , 12 ), ( 12 , − 12 ), and (− 12 , − 12 ).

Next we calculate the second partial derivatives in order to evaluate the discriminant at each critical point: f xx = −20 xy (1 − 2 x 2 ) e −( x

2 + y2 )

f xy = f yx = 10 (1 − 2 x 2 ) e

−( x 2 + y 2 )

f yy = −20 xy (1 − 2 y 2 ) e

−( x 2 + y 2 )

− 40 xye −( x

2 + y2 )

= −20 xy ( 3 − 2 x 2 ) e −( x

− 20 y 2 (1 − 2 x 2 ) e

− 40 xye

−( x 2 + y 2 )

2 + y2 )

, 

= 10 (1 − 2 x 2 )(1 − 2 y 2 ) e −( x

−( x 2 + y 2 )

= −20 xy ( 3 − 2 y 2 ) e

−( x 2 + y 2 )

2 + y2 )

, 

.

The following table summarizes the values needed by the Second Derivative Test. Critical Point

f xx

( 0, 0 ) z z = 10xye−(x

2

+ y 2)

y

x

FIGURE 13.50

in Example 5.

A graph of the function

( 12 , 12 ) (− 12 , 12 ) ( 12 , − 12 ) (− 12 , − 12 )

−

−

f xy

f yy

Discriminant D

0

−100

20 e

400 e2

  0

20 e

400 e2

20 e

  0

20 e

400 e2

20 e

  0

20 e

400 e2

0

10

20 e

  0

20 e

−

−

From the table we find that D < 0 at the critical point ( 0, 0 ), giving a saddle; D > 0 and f xx < 0 at the critical points (1 2 , 1 2 ) and (−1 2 , −1 2 ), giving local maximum values there; and D > 0 and f xx > 0 at the critical points (−1 2 , 1 2 ) and (1 2 , −1 2 ), each giving local minimum values. A graph of the surface is shown in Figure 13.50.

Absolute Maxima and Minima on Closed Bounded Regions We organize the search for the absolute extrema of a continuous function f ( x , y ) on a closed and bounded region R into three steps. 1. List the interior points of R where f may have local maxima and minima and evaluate f at these points. These are the critical points of f . 2. List the boundary points of R where f has local maxima and minima and evaluate f at these points. We show how to do this in the next example. 3. Look through the lists for the maximum and minimum values of f . These will be the absolute maximum and minimum values of f on R.

830

Chapter 13 Partial Derivatives

EXAMPLE 6

Find the absolute maximum and minimum values of f ( x, y ) = 2 + 2x + 4 y − x 2 − y 2

on the triangular region in the first quadrant bounded by the lines x = 0, y = 0, and y = 9 − x. Solution Since f is differentiable, the only places where f can assume these values are points inside the triangle where f x = f y = 0 and points on the boundary (Figure 13.51a).

(a) Interior points. For these we have f x = 2 − 2 x = 0,

f y = 4 − 2 y = 0,

yielding the single point ( x , y ) = (1, 2 ). The value of f there is f (1, 2 ) = 7. (b) Boundary points. We take the triangle one side at a time: i) On the segment OA we always have y = 0. Therefore, we can regard f ( x , y ) as being solely a function of x on this segment. That is, on this segment we want to consider the function g( x ) = f ( x , 0 ) = 2 + 2 x − x 2

y

for 0 ≤ x ≤ 9. Its extreme values (as we know from Chapter 4) may occur at the endpoints

B(0, 9) y=9−x (4, 5)

x=0

y=0

g(0) = f ( 0, 0 ) = 2

x = 9

where

g(9) = f ( 9, 0 ) = 2 + 18 − 81 = −61

x

A(9, 0)

g(1) = f (1, 0 ) = 3. ii) On the segment OB we always have x = 0. Therefore, on this segment we can regard f ( x , y ) as being solely a function of y, and so we consider the function

(a) z y 9

where

or at the interior points where g′( x ) = 2 − 2 x = 0. The only interior point where g′( x ) = 0 is x = 1, where

(1, 2) O

x = 0

h( y) = f ( 0, y ) = 2 + 4 y − y 2

(1, 2, 7) 6

3

6

3

9

x

on the closed interval [ 0, 9 ]. Its extreme values can occur at the endpoints or at interior points where h′( y) = 0. Since h′( y) = 4 − 2 y, the only interior point where h′( y) = 0 occurs at ( 0, 2 ), with h(2) = 6. So the candidates for this segment are h(0) = f ( 0, 0 ) = 2, h(9) = f ( 0, 9 ) = −43, and h(2) = f ( 0, 2 ) = 6. iii) We have already accounted for the values of f at the endpoints of AB, so we need only look at the interior points of the line segment AB. On this segment we have y = 9 − x , so we consider the function

−20

k ( x ) = f ( x , 9 − x ) = 2 + 2 x + 4 ( 9 − x ) − x 2 − ( 9 − x ) 2 = −43 + 16 x − 2 x 2 . Setting k ′( x ) = 16 − 4 x = 0 gives

−40

x = 4. At this value of x, (9, 0, −61)

−60 (b)

FIGURE 13.51

(a) This triangular region is the domain of the function in Example 6. (b) The graph of the function in Example 6. The blue points are the candidates for maxima or minima.

y = 9−4 = 5

and

k (4) = f ( 4, 5 ) = −11.

Summary We list all the function value candidates: 7, 2, − 61, 3, − 43, 6, −11. The maximum is 7, which f assumes at (1, 2 ). The minimum is −61, which f assumes at ( 9, 0 ). See Figure 13.51b. Solving extreme value problems with algebraic constraints on the variables usually requires the method of Lagrange multipliers, which is introduced in the next section. But sometimes we can solve such problems directly, as in the next example.

13.7  Extreme Values and Saddle Points

Girth = perimeter around here

z x

FIGURE 13.52

y

831

EXAMPLE 7 A delivery company accepts only rectangular boxes the sum of whose length and girth (perimeter of a cross-section) does not exceed 270 cm. Find the dimensions of an acceptable box of largest volume. Solution Let x, y, and z represent the length, width, and height of the rectangular box, respectively. Then the girth is 2 y 2 z. We want to maximize the volume V  xyz of the box (Figure 13.52) satisfying x + 2 y + 2 z = 270 (the largest box accepted by the delivery company). Thus, we can write the volume of the box as a function of two variables:

V ( y, z ) = ( 270 − 2 y − 2 z ) yz

The box in Example 7.

= 270 yz − 2 y 2 z − 2 yz 2 .

V = xyz and x = 270 − 2 y − 2 z

Setting the first partial derivatives equal to zero, V y ( y, z ) = 270 z − 4 yz − 2 z 2 = ( 270 − 4 y − 2 z ) z = 0 Vz ( y, z ) = 270 y − 2 y 2 − 4 yz = ( 270 − 2 y − 4 z ) y = 0, gives the critical points ( 0, 0 ), ( 0, 135 ), (135, 0 ), and ( 45, 45 ). The volume is zero at ( 0, 0 ), ( 0, 135 ), and (135, 0 ), which are not maximum values. At the point ( 45, 45 ), we apply the Second Derivative Test (Theorem 11): V yy = −4 z,

V zz = −4 y,

V yz = 270 − 4 y − 4 z.

Then V yyVzz − V yz 2 = 16 yz − 4 (135 − 2 y − 2 z ) 2 . Thus, V yy ( 45, 45 ) = −4 ( 45 ) < 0 and

(V yyVzz − V yz 2 )

= 16( 45 )( 45 ) − 4 (−45 ) 2 > 0, ( 45, 45 )

so ( 45, 45 ) gives a maximum volume. The dimensions of the package are x = 270 − 2( 45 ) − 2( 45 ) = 90 cm, y = 45 cm, and z  45 cm. The maximum volume is V = ( 90 )( 45 )( 45 ) = 182, 250 cm 3, or 182.25 liters. Despite the power of Theorem 11, we urge you to remember its limitations. It does not apply to boundary points of a function’s domain, where it is possible for a function to have extreme values along with nonzero derivatives. Also, it does not apply to points where either f x or f y fails to exist.

Summary of Max-Min Tests

The extreme values of f ( x , y ) can occur only at i) boundary points of the domain of f ii) critical points (interior points where f x  f y  0 or points where f x or f y fails to exist) If the first- and second-order partial derivatives of f are continuous throughout a disk centered at a point ( a, b ) and if f x ( a, b ) = f y ( a, b ) = 0, then the nature of f ( a, b ) can be tested with the Second Derivative Test: i) ii) iii) iv)

f xx  0 and f xx f yy − f xy 2 > 0 at ( a, b )   ⇒   local   maximum f xx  0 and f xx f yy − f xy 2 > 0 at ( a, b )   ⇒   local   minimum f xx f yy − f xy 2 < 0 at ( a, b )   ⇒   saddle   point f xx f yy − f xy 2 = 0 at ( a, b )   ⇒   test   is   inconclusive

832

Chapter 13 Partial Derivatives

Finding maximum and minimum values for functions of more than two variables is an important problem with many important applications, from machine learning to making economic predictions. The problem becomes much harder as the number of variables increases. The process of finding extrema for functions with high-dimensional domains is discussed in Appendices B.2 and B.3.

13.7

EXERCISES

Finding Local Extrema

Find all the local maxima, local minima, and saddle points of the functions in Exercises 1–30. 1. f ( x , y ) = x 2 + xy + y 2 + 3 x − 3 y + 4 2. f ( x , y ) = 2 xy − 5 x 2 − 2 y 2 + 4 x + 4 y − 4 3. f ( x , y ) =

x2

5. f ( x , y ) = 2 xy −

−

+ 3x + 4

b

∫a

7. f ( x , y ) = 2 x 2 + 3 xy + 4 y 2 − 5 x + 2 y 8. f ( x , y ) = x 2 − 2 xy + 2 y 2 − 2 x + 2 y + 1

12. f ( x , y ) = 1 −

3

−

b

∫a

− 16 x − 31 + 1 − 8 x

14. f ( x , y ) = x 3 + 3 xy + y 3 15. f ( x , y ) = 6 x 2 − 2 x 3 + 3 y 2 + 6 xy

Find the temperatures at the hottest and coldest points on the plate.

17. f ( x , y ) = x 3 + 3 xy 2 − 15 x + y 3 − 15 y

42. Find the critical point of

18. f ( x , y ) = 2 x 3 + 2 y 3 − 9 x 2 + 3 y 2 − 12 y

f ( x , y ) = xy + 2 x − ln x 2 y

19. f ( x , y ) = 4 xy − x 4 − y 4

in the open first quadrant ( x > 0, y > 0 ) and show that f takes on a minimum there.

20. f ( x , y ) = x 4 + y 4 + 4 xy 1 1 1 22. f ( x , y ) = + xy + 21. f ( x , y ) = 2 x + y2 − 1 x y 24. f ( x , y ) = e 2 x cos y 23. f ( x , y ) = y sin x 27. f ( x , y ) =

+

dx

T ( x , y ) = x 2 + 2 y 2 − x.

16. f ( x , y ) = x 3 + y 3 + 3 x 2 − 3 y 2 − 8

Theory and Examples

43. Find the maxima, minima, and saddle points of f ( x , y ), if any, given that

26. f ( x , y ) = e y − ye x y2 )

13

( 24 − 2 x − x 2 )

41. Temperatures A flat circular plate has the shape of the region x 2 + y 2 ≤ 1. The plate, including the boundary where x 2 + y 2 = 1, is heated so that the temperature at the point ( x , y ) is

13. f ( x , y ) = x 3 − y 3 − 2 xy + 6

e −y ( x 2

( 6 − x − x 2 ) dx

has its largest value.

x 2 + y2

25. f ( x , y ) = e x 2 + y 2 −4 x

a. f x = 2 x − 4 y

28. f ( x , y ) = e x ( x 2 − y 2 )

b. f x = 2 x − 2

29. f ( x , y ) = 2 ln x + ln y − 4 x − y

c. f x = 9 x 2 − 9

30. f ( x , y ) = ln ( x + y ) + x 2 − y

and and and

f y = 2y − 4x f y = 2y − 4 f y = 2y + 4

Describe your reasoning in each case.

Finding Absolute Extrema

In Exercises 31–38, find the absolute maxima and minima of the functions on the given domains. 31. f ( x , y ) = 2 x 2 − 4 x + y 2 − 4 y + 1 on the closed triangular plate bounded by the lines x = 0, y = 2, y = 2 x in the first quadrant

44. The discriminant f xx f yy − f xy 2 is zero at the origin for each of the following functions, so the Second Derivative Test fails there. Determine whether the function has a maximum, a minimum, or neither at the origin by imagining what the surface z = f ( x , y ) looks like. Describe your reasoning in each case.

32. D ( x , y ) = x 2 − xy + y 2 + 1 on the closed triangular plate in the first quadrant bounded by the lines x = 0, y = 4, y = x

a. f ( x , y ) = x 2 y 2

b. f ( x , y ) = 1 − x 2 y 2

c. f ( x , y ) = xy 2

d. f ( x , y ) = x 3 y 2

+ on the closed triangular plate bounded by the 33. f ( x , y ) = lines x = 0, y = 0, y + 2 x = 2 in the first quadrant

e. f ( x , y ) =

f. f ( x , y ) = x 4 y 4

x2

y2

34. T ( x , y ) = x 2 + xy + y 2 − 6 x 0 ≤ x ≤ 5, − 3 ≤ y ≤ 3

plate

40. Find two numbers a and b with a ≤ b such that

10. f ( x , y ) = x 2 + 2 xy 11. f ( x , y ) =

rectangular

has its largest value.

9. f ( x , y ) = x 2 − y 2 − 2 x + 4 y + 6 8y 2

the

39. Find two numbers a and b with a ≤ b such that

6. f ( x , y ) = x 2 − 4 xy + y 2 + 6 y + 2

56 x 2

on the rectangular plate

38. f ( x , y ) = 4 x − 8 xy + 2 y + 1 on the triangular plate bounded by the lines x = 0, y = 0, x + y = 1 in the first quadrant

4. f ( x , y ) = 5 xy − 7 x 2 + 3 x − 6 y + 2 2y 2

36. f ( x , y ) = 48 xy − 32 x 3 − 24 y 2 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 on 37. f ( x , y ) = ( 4 x − x 2 ) cos y 1 ≤ x ≤ 3, −π 4 ≤ y ≤ π 4

+ xy + 3 x + 2 y + 5 x2

35. T ( x , y ) = x 2 + xy + y 2 − 6 x + 2 on the rectangular plate 0 ≤ x ≤ 5, − 3 ≤ y ≤ 0

on

the

rectangular

plate

x 3y3

45. Show that ( 0, 0 ) is a critical point of f ( x , y ) = x 2 + kxy + y 2 no matter what value the constant k has. (Hint: Consider two cases: k = 0 and k ≠ 0. )

833

13.7  Extreme Values and Saddle Points

46. For what values of the constant k does the Second Derivative Test guarantee that f ( x , y ) = x 2 + kxy + y 2 will have a saddle point at ( 0, 0 )? A local minimum at ( 0, 0 )? For what values of k is the Second Derivative Test inconclusive? Give reasons for your answers. 47. If f x ( a, b ) = f y ( a, b ) = 0, must f have a local maximum or minimum value at ( a, b )? Give reasons for your answer. 48. Can you conclude anything about f ( a, b ) if f and its first and second partial derivatives are continuous throughout a disk centered at the critical point ( a, b ) and f xx ( a, b ) and f yy ( a, b ) differ in sign? Give reasons for your answer.

Extreme Values on Parametrized Curves To find the extreme values of a function f ( x , y ) on a curve x = x (t ), y = y(t ), we treat f as a function of the single variable t and use the Chain Rule to find where df dt is zero. As in any other single-variable case, the extreme values of f are then found among the values at a. The critical points (points where df dt is zero or fails to exist), and b. The endpoints of the parameter domain. In Exercises 63–66, find the absolute maximum and minimum values of the following functions on the given curves.

49. Among all the points on the graph of z = 10 − x 2 − y 2 that lie above the plane x + 2 y + 3z = 0, find the point farthest from the plane.

63. Functions:

50. Find the point on the graph of z = x 2 + y 2 + 10 nearest the plane x + 2 y − z = 0.

Curves:

51. Find the point on the plane 3 x + 2 y + z = 6 that is nearest the origin. 52. Find the minimum distance from the point ( 2, −1, 1) to the plane x + y − z = 2. 53. Find three numbers whose sum is 9 and whose sum of squares is a minimum.

a. f ( x , y ) = x + y

b. g ( x , y ) = xy

c. h ( x , y ) = 2 x 2 + y 2 i) The semicircle x 2 + y 2 = 4, ii) The quarter circle

x2

+

y2

= 4,

y ≥ 0 x ≥ 0,

y ≥ 0

Use the parametric equations x = 2 cos t , y = 2 sin t. 64. Functions: a. f ( x , y ) = 2 x + 3 y c. h ( x , y ) =

x2

+

b. g ( x , y ) = xy

3y 2

54. Find three positive numbers whose sum is 3 and whose product is a maximum.

Curves:

55. Find the maximum value of s = xy + yz + xz where x + y + z = 6.

ii) The quarter ellipse ( x 2 9 ) + ( y 2 4 ) = 1, x ≥ 0,

56. Find the minimum distance from the cone z = point (−6, 4, 0 ).

x 2 + y 2 to the

57. Find the dimensions of the rectangular box of maximum volume that can be inscribed inside the sphere x 2 + y 2 + z 2 = 4. 58. Among all closed rectangular boxes of volume 27 cm 3, what is the smallest surface area? m2

of mate59. You are to construct an open rectangular box from 12 rial. What dimensions will result in a box of maximum volume? + + 2 xy − x − y + 1 60. Consider the function f ( x , y ) = over the square 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. x2

y2

a. Show that f has an absolute minimum along the line segment 2 x + 2 y = 1 in this square. What is the absolute minimum value? b. Find the absolute maximum value of f over the square. 61. Find the point on the graph of y 2 − xz 2 = 4 nearest the origin. 62. A rectangular box is inscribed in the region in the first octant bounded above by the plane with x-intercept 6, y-intercept 6, and z-intercept 6. z

i) The semiellipse ( x 2 9 ) + ( y 2 4 ) = 1,

y ≥ 0 y ≥ 0

Use the parametric equations x = 3 cos t , y = 2 sin t. 65. Function: f ( x , y ) = xy Curves: i) The line x = 2t ,

y = t +1

ii) The line segment x = 2t ,

y = t + 1, −1 ≤ t ≤ 0

iii) The line segment x = 2t ,

y = t + 1, 0 ≤ t ≤ 1

66. Functions: a. f ( x , y ) = x 2 + y 2

b. g ( x , y ) = 1 ( x 2 + y 2 )

Curves: i) The line x = t ,

y = 2 − 2t

ii) The line segment x = t ,

y = 2 − 2t , 0 ≤ t ≤ 1

67. Least squares and regression lines When we try to fit a line y = mx + b to a set of numerical data points ( x 1 , y1 ), ( x 2 , y 2 ), … , ( x n , y n ), we usually choose the line that minimizes the sum of the squares of the vertical distances from the points to the line. In theory, this means finding the values of m and b that minimize the value of the function w = ( mx 1 + b − y1 ) 2 +  + ( mx n + b − y n ) 2.

(1)

(See the accompanying figure.) Show that the values of m and b that do this are

6 (x, y, z)

6

m =

6

y

x

a. Find an equation for the plane. b. Find the dimensions of the box of maximum volume.

b =

Q∑ x k R Q∑ yk R − n ∑ x k yk Q∑ x k R − n ∑ x k 2 2

1 Q ∑ y k − m ∑ x k R, n

,

(2) (3)

with all sums running from k = 1 to k = n. Many scientific calculators have these formulas built in, enabling you to find m and b with only a few keystrokes after you have entered the data.

834

Chapter 13 Partial Derivatives

The line y = mx + b determined by these values of m and b is called the least squares line, regression line, or trend line for the data under study. Finding a least squares line lets you

a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function’s first partial derivatives and use the CAS equation solver to find the critical points. How are the critical points related to the level curves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer.

1. summarize data with a simple expression, 2. predict values of y for other, experimentally untried values of x, 3. handle data analytically. y

d. Calculate the function’s second partial derivatives and find the discriminant f xx f yy  f xy 2 .

Pn(xn , yn )

e. Using the max-min tests, classify the critical points found in part (c). Are your findings consistent with your discussion in part (c)?

y = mx + b

P1(x1, y1 ) P2(x 2 , y2 )

0

71. f ( x , y ) = x 2 + y 3 − 3 xy, − 5 ≤ x ≤ 5, − 5 ≤ y ≤ 5 x

In Exercises 68–70, use Equations (2) and (3) to find the least squares line for each set of data points. Then use the linear equation you obtain to predict the value of y that would correspond to x  4. 68. (−2, 0 ), ( 0, 2 ), ( 2, 3 ) 69. (−1, 2 ), ( 0, 1), ( 3, − 4 )

72. f ( x , y ) = x 3 − 3 xy 2 + y 2 , − 2 ≤ x ≤ 2, − 2 ≤ y ≤ 2 73. f ( x , y ) = x 4 + y 2 − 8 x 2 − 6 y + 16, − 3 ≤ x ≤ 3, −6 ≤ y ≤ 6 74. f ( x , y ) = 2 x 4 + y 4 − 2 x 2 − 2 y 2 + 3, −3 2 ≤ x ≤ 3 2, −3 2 ≤ y ≤ 3 2 75. f ( x , y ) = 5 x 6 + 18 x 5 − 30 x 4 + 30 xy 2 − 120 x 3, − 4 ≤ x ≤ 3, − 2 ≤ y ≤ 2 ⎧⎪ x 5 ln ( x 2 + y 2 ) , ( x , y ) ≠ ( 0, 0 ) 76. f ( x , y ) = ⎪⎨ ( x , y ) = ( 0, 0 ) , ⎩⎪⎪ 0, − 2 ≤ x ≤ 2, − 2 ≤ y ≤ 2

70. ( 0, 0 ) , (1, 2 ) , ( 2, 3 ) COMPUTER EXPLORATIONS

In Exercises 71–76, you will explore functions to identify their local extrema. Use a CAS to perform the following steps:

13.8 Lagrange Multipliers HISTORICAL BIOGRAPHY Joseph Louis Lagrange (1736–1813) Lagrange was born in Turin, Italy. He enjoyed studying mathematics, despite his father’s wish that he study law. Lagrange’s mathematical contributions began as early as 1754 with the discovery of the calculus of variations and continued with applications to mechanics in 1756. To know more, visit the companion Website.

Sometimes we need to find the extreme values of a function whose domain is constrained to lie within some particular subset of the plane—for example, a disk, a closed triangular region, or along a curve. We saw an instance of this situation in Example 6 of the previous section. Here we explore a powerful method for finding extreme values of constrained functions: the method of Lagrange multipliers.

Constrained Maxima and Minima To gain some insight, we first consider a problem where a constrained minimum can be found by eliminating a variable. EXAMPLE 1 Find the point p( x , y, z ) on the plane 2 x + y − z − 5 = 0 that is closest to the origin. Solution The problem asks us to find the minimum value of the function

OP = ( x − 0 ) 2 + ( y − 0 ) 2 + ( z − 0 ) 2 = x 2 + y 2 + z 2

subject to the constraint that 2 x + y − z − 5 = 0.

Since OP has a minimum value wherever the function f ( x , y, z ) = x 2 + y 2 + z 2

13.8  Lagrange Multipliers

835

has a minimum value, we may solve the problem by finding the minimum value of f ( x , y, z ) subject to the constraint 2 x + y − z − 5 = 0 (thus avoiding square roots). If we regard x and y as the independent variables in this equation and write z as z = 2 x + y − 5, our problem reduces to finding the points ( x , y ) at which the function h ( x , y ) = f ( x , y, 2 x + y − 5 ) = x 2 + y 2 + ( 2 x + y − 5 ) 2 has its minimum value or values. Since the domain of h is the entire xy-plane, the First Derivative Theorem of Section 13.7 tells us that any minima that h might have must occur at points where h x = 2 x + 2( 2 x + y − 5 )( 2 ) = 0,

h y = 2 y + 2( 2 x + y − 5 ) = 0.

This leads to 10 x + 4 y = 20,

4 x + 4 y = 10,

which has the solution x =

5 , 3

y =

5 . 6

We may apply a geometric argument together with the Second Derivative Test to show that these values minimize h. The z-coordinate of the corresponding point on the plane z = 2 x + y − 5 is z = 2

( 53 ) + 56 − 5 = − 56 .

Therefore, the point we seek is Closest point:

P

( 53 , 56 , −56 ).

6 ≈ 2.04.

The distance from P to the origin is 5

Attempts to solve a constrained maximum or minimum problem by substitution, as we might call the method of Example 1, do not always go smoothly. z

EXAMPLE 2 Find the points on the hyperbolic cylinder x 2 − z 2 − 1 = 0 that are closest to the origin. Solution 1 The cylinder is shown in Figure 13.53. We seek the points on the cylinder closest to the origin. These are the points whose coordinates minimize the value of the function

f ( x , y, z ) = x 2 + y 2 + z 2

((−1, − 0, 0) (1, 0, 0) y

  

Square of the distance

subject to the constraint that x 2 − z 2 − 1 = 0. If we regard x and y as independent variables in the constraint equation, then z 2 = x 2 − 1,

x x 2 − z2 = 1

and the values of f ( x , y, z ) = x 2 + y 2 + z 2 on the cylinder are given by the function h ( x , y ) = x 2 + y 2 + ( x 2 − 1) = 2 x 2 + y 2 − 1.

FIGURE 13.53

The hyperbolic cylinder x 2 − z 2 − 1 = 0 in Example 2.

To find the points on the cylinder whose coordinates minimize f , we look for the points in the xy-plane whose coordinates minimize h. The only extreme value of h occurs where hx = 4x = 0

and

h y = 2 y = 0,

836

Chapter 13 Partial Derivatives

The hyperbolic cylinder x2 − z2 = 1 On this part,

z

x = " z2 + 1.

x

On this part, x = −" z2 + 1.

−1

1

x=1

x = −1

y

that is, at the point ( 0, 0 ). But there are no points on the cylinder where both x and y are zero. What went wrong? What happened is that the First Derivative Theorem found (as it should have) the point in the domain of h where h has a minimum value. We, on the other hand, want the points on the cylinder where h has a minimum value. Although the domain of h is the entire xyplane, the domain from which we can select the first two coordinates of the points ( x , y, z ) on the cylinder is restricted to the projection, or “shadow” of the cylinder on the xy-plane; it does not include the band between the lines x = −1 and x = 1 (Figure 13.54). We can avoid this problem if we treat y and z as independent variables (instead of x and y) and express x in terms of y and z as x 2 = z 2 + 1. With this substitution, f ( x , y, z ) = x 2 + y 2 + z 2 becomes

FIGURE 13.54

The region in the xyk ( y, z ) = ( z 2 + 1) + y 2 + z 2 = 1 + y 2 + 2 z 2 plane from which the first two coordinates and we look for the points where k takes on its smallest value. The domain of k in the yzof the points ( x , y, z ) on the hyperbolic 2 2 cylinder x − z = 1 are selected excludes plane now matches the domain from which we select the y- and z-coordinates of the points ( x , y, z ) on the cylinder. Hence, the points that minimize k in the plane will have correthe band −1 < x < 1 in the xy-plane sponding points on the cylinder. The smallest values of k occur where (Example 2).

k y = 2y = 0

k z = 4 z = 0,

and

or where y = z = 0. This leads to x 2 = z 2 + 1 = 1, x2

−

z2

−1=0

x = ±1.

The corresponding points on the cylinder are ( ±1, 0, 0 ). We can see from the inequality

z x +y +z −a =0 2

2

2

2

k ( y, z ) = 1 + y 2 + 2 z 2 ≥ 1 that the points ( ±1, 0, 0 ) give a minimum value for k. We can also see that the minimum distance from the origin to a point on the cylinder is 1 unit.

y

Solution 2 Another way to find the points on the cylinder closest to the origin is to imagine a small sphere centered at the origin expanding like a soap bubble until it just touches the cylinder (Figure 13.55). At each point of contact, the cylinder and sphere have the same tangent plane and normal line. Therefore, if the sphere and cylinder are represented as the level surfaces obtained by setting

f ( x , y, z ) = x 2 + y 2 + z 2 − a 2

x

FIGURE 13.55

A sphere expanding like a soap bubble centered at the origin until it just touches the hyperbolic cylinder x 2 − z 2 − 1 = 0 (Example 2).

and

g ( x , y, z ) = x 2 − z 2 − 1

equal to 0, then the gradients ∇f and ∇g will be parallel where the surfaces touch. At any point of contact, we should therefore be able to find a scalar λ (“lambda”) such that ∇f = λ∇g, or 2 xi + 2 yj + 2 zk = λ ( 2 xi − 2 zk ).

λ is the Greek letter lambda.

Thus, the coordinates x, y, and z of any point of tangency will have to satisfy the three scalar equations 2 x = 2λ x ,

2 y = 0,

2 z = −2λz.

For what values of λ will a point ( x , y, z ) whose coordinates satisfy these scalar equations also lie on the surface x 2 − z 2 − 1 = 0? To answer this question, we use our knowledge that no point on the surface has a zero x-coordinate to conclude that x ≠ 0. Hence, 2 x = 2λ x only if 2 = 2λ,

or

λ = 1.

13.8  Lagrange Multipliers

837

For λ = 1, the equation 2 z = −2λz becomes 2 z = −2 z. If this equation is to be satisfied as well, z must be zero. Since y = 0 also (from the equation 2 y = 0), we conclude that the points we seek all have coordinates of the form

( x, 0, 0 ). What points on the surface x 2 − z 2 = 1 have coordinates of this form? The answer is the points ( x, 0, 0 ) for which x 2 − ( 0 ) 2 = 1,

x 2 = 1,

or

x = ±1.

The points on the cylinder closest to the origin are the points ( ±1, 0, 0 ).

The Method of Lagrange Multipliers In Solution 2 of Example 2, we used the method of Lagrange multipliers. The method says that the local extreme values of a function f ( x , y, z ) whose variables are subject to a constraint g ( x , y, z ) = 0 are to be found on the surface g = 0 among the points where ∇f = λ∇g for some scalar λ (called a Lagrange multiplier). To explore the method further and see why it works, we first make the following observation, which we state as a theorem.

THEOREM 12—The Orthogonal Gradient Theorem

Suppose that f ( x , y, z ) is differentiable in a region whose interior contains a smooth curve C: r (t ) = x (t ) i + y(t ) j + z (t )k. If P0 is a point on C where f has a local maximum or minimum relative to its values on C, then ∇f is orthogonal to the curve’s tangent vector r ′ at P0 . Proof The values of f on C are given by the composition f ( x (t ), y(t ), z (t ) ), whose derivative with respect to t is df ∂ f dx ∂ f dy ∂ f dz = + + = ∇f ⋅ r ′. dt ∂ x dt ∂ y dt ∂ z dt At any point P0 where f has a local maximum or minimum relative to its values on the curve, df dt = 0, so ∇f ⋅ r ′ = 0. By dropping the z-terms in Theorem 12, we obtain a similar result for functions of two variables. COROLLARY At the points on a smooth curve r (t ) = x (t ) i + y(t ) j where a

differentiable function f ( x , y ) takes on its local maxima or minima relative to its values on the curve, we have ∇f ⋅ r ′ = 0.

Theorem 12 is the key to the method of Lagrange multipliers. Suppose that f ( x , y, z ) and g ( x , y, z ) are differentiable and that P0 is a point on the surface g ( x , y, z ) = 0 where f has a local maximum or minimum value relative to its other values on the surface. We assume also that ∇g ≠ 0 at points on the surface g ( x , y, z ) = 0. Then f takes on a local maximum or minimum at P0 relative to its values on every differentiable curve through P0 on the surface g ( x , y, z ) = 0. Therefore, ∇f is orthogonal to the tangent vector of every

838

Chapter 13 Partial Derivatives

such differentiable curve through P0 . Moreover, so is ∇g (because ∇g is perpendicular to the level surface g = 0, as we saw in Section 13.5). Therefore, at P0 , ∇f is some scalar multiple λ of ∇g.

The Method of Lagrange Multipliers

Suppose that f ( x , y, z ) and g ( x , y, z ) are differentiable and ∇g ≠ 0 when g ( x , y, z ) = 0. To find the local maximum and minimum values of f subject to the constraint g ( x , y, z ) = 0 (if these exist), find the values of x, y, z, and λ that simultaneously satisfy the equations ∇f = λ∇g

g ( x , y, z ) = 0.

and

(1)

If they exist, absolute extrema can be found by comparing these values of f at each critical point satisfying Equation (1). For functions of two independent variables, the condition is similar, but without the variable z.

y 2

"2

0

Some care must be used in applying this method. An extreme value may not actually exist (Exercise 45).

y2

x + =1 8 2 x

EXAMPLE 3

Find the largest and smallest values that the function f ( x , y ) = xy

2" 2

takes on the ellipse (Figure 13.56) y2 x2 + = 1. 8 2

FIGURE 13.56

Example 3 shows how to find the largest and smallest values of the product xy on this ellipse.

Solution We want to find the extreme values of f ( x , y ) = xy subject to the constraint

g( x, y ) =

y2 x2 + − 1 = 0. 8 2

To do so, we first find the values of x, y, and λ for which ∇f = λ∇g

and

g ( x , y ) = 0.

The gradient equation in Equations (1) gives yi + xj =

λ xi + λ yj, 4

from which we find y =

λ x, 4

x = λ y,

and y =

λ λ2 (λ y) = y, 4 4

  

Caution: Don’t cancel y without considering the case where y = 0.

so that y = 0 or λ = ±2. We now consider these two cases. Case 1: If y = 0, then x = y = 0. But ( 0, 0 ) is not on the ellipse. Hence, y ≠ 0. Case 2: If y ≠ 0, then λ = ±2 and x = ±2 y. Substituting this in the equation g ( x , y ) = 0 gives ( ±2 y ) 2

8

+

y2 = 1, 2

4y2 + 4y2 = 8

and

y = ±1.

839

13.8  Lagrange Multipliers

xy = −2

y

xy = 2

∇f = i + 2j 1 ∇g = i + j 2

1 0

xy = 2

x

1

x y = −2

FIGURE 13.57

When subjected to the constraint g ( x , y ) = x 2 8 + y 2 2 − 1 = 0, the function f ( x , y ) = xy takes on extreme values at the four points ( ±2, ±1). These are the points on the ellipse where ∇f (red) is a scalar multiple of ∇g (blue) (Example 3).

The function f ( x , y ) = xy therefore has critical points on the ellipse at the four points ( ±2, 1) , ( ±2, −1). The extreme values are found by examining the values of f at these four points. The absolute maximum is f (2, 1) = f (−2, −1) = 2, and the absolute minimum is f (−2, 1) = f (2, −1) = −2.

The Geometry of the Solution The level curves of the function f ( x , y ) = xy are the hyperbolas xy = c (Figure 13.57). The farther the hyperbolas lie from the origin, the larger the absolute value of f . We want to find the extreme values of f ( x , y ), given that the point ( x , y ) also lies on the ellipse x 2 + 4 y 2 = 8. Which hyperbolas intersecting the ellipse lie farthest from the origin? The hyperbolas that just graze the ellipse, the ones that are tangent to it, are farthest. At these points, any vector normal to the hyperbola is normal to the ellipse, so ∇f = yi + xj is a multiple (λ = ±2 ) of ∇g = ( x 4 ) i + yj. At the point ( 2, 1), for example, 1 i + j, 2

and

∇f = 2∇g.

1 ∇g = − i + j, 2

and

∇f = −2∇g.

∇f = i + 2 j,

∇g =

At the point (−2, 1) , ∇f = i − 2 j,

EXAMPLE 4 Find the maximum and minimum f ( x , y ) = 3 x + 4 y on the circle x 2 + y 2 = 1.

values

of

the

function

Solution We model this as a Lagrange multiplier problem with

f ( x , y ) = 3 x + 4 y,

g( x, y ) = x 2 + y 2 − 1

and look for the values of x, y, and λ that satisfy the equations ∇f = λ∇g:

3i + 4 j = 2 xλi + 2 yλ j

g ( x , y ) = 0:

x 2 + y 2 − 1 = 0.

The gradient equation implies that λ ≠ 0 and gives x =

( 23λ ) + ( λ2 ) 2

∇g = 6 i + 8 j 5 5

2 . λ

3 4 a , b 5 5 3x + 4y = 5

9 4 + 2 = 1, 4λ 2 λ

− 1 = 0,

9 + 16 = 4λ 2,

4λ 2 = 25,

and

5 λ = ± . 2

Thus, x =

3x + 4y = −5

The function f ( x , y ) = 3 x + 4 y takes on its largest value on the unit circle g ( x , y ) = x 2 + y 2 − 1 = 0 at the point ( 3 5, 4 5 ) and its smallest value at the point (−3 5, −4 5 ) (Example 4). At each of these points, ∇f is a scalar multiple of ∇g. The figure shows the gradients at the first point but not at the second.

2

so

x

FIGURE 13.58

y =

These equations tell us, among other things, that x and y have the same sign. With these values for x and y, the equation g ( x , y ) = 0 gives

5 ∇f = 3i + 4j = ∇g 2 y

x2 + y2 = 1

3 , 2λ

3 3 = ± , 2λ 5

y =

2 4 = ± , 5 λ

and f ( x , y ) = 3 x + 4 y has critical points at ( x , y ) = ±( 3 5, 4 5 ). By calculating the value of 3 x + 4 y at the points ±( 3 5, 4 5 ) , we see that its maximum and minimum values on the circle x 2 + y 2 = 1 are 3

( 35 ) + 4( 45 ) = 255 = 5

and

( 35 ) + 4(− 45 ) = − 255 = −5.

3 −

The Geometry of the Solution The level curves of f ( x , y ) = 3 x + 4 y are the lines 3 x + 4 y = c (Figure 13.58). The farther the lines lie from the origin, the larger the absolute value of f . We want to find the extreme values of f ( x , y ) given that the point ( x , y )

840

Chapter 13 Partial Derivatives

also lies on the circle x 2 + y 2 = 1. Which lines intersecting the circle lie farthest from the origin? The lines tangent to the circle are farthest. At the points of tangency, any vector normal to the line is normal to the circle, so the gradient ∇f = 3i + 4 j is a multiple (λ = ±5 2 ) of the gradient ∇g = 2 xi + 2 yj. At the point ( 3 5, 4 5 ), for example, ∇f = 3i + 4 j,

∇g =

6 8 i + j, 5 5

Many problems require us to find the extreme values of a differentiable function f ( x , y, z ) whose variables are subject to two constraints. If the constraints are g 1 ( x , y, z ) = 0

∇g1

∇g2

g1 = 0

FIGURE 13.59 The vectors ∇g1 and ∇g 2 lie in a plane perpendicular to the curve C, because ∇g1 is normal to the surface g1 = 0 and ∇g 2 is normal to the surface g 2 = 0.

pronounced “mew”.

g 2 ( x , y, z ) = 0

and

and g1 and g 2 are differentiable, with ∇g1 not parallel to ∇g 2 , we find the constrained local maxima and minima of f by introducing two Lagrange multipliers λ and μ (mu, pronounced “mew”). That is, we locate the points P ( x , y, z ) where f takes on its constrained extreme values by finding the values of x , y, z , λ, and μ that simultaneously satisfy the three equations

∇f

μ is the Greek letter mu,

5 ∇g. 2

∇f =

Lagrange Multipliers with Two Constraints

g2 = 0

C

and

∇f = λ∇g1 + μ∇g 2 ,

g1 ( x , y, z ) = 0,

g 2 ( x , y, z ) = 0

(2)

Equations (2) have a nice geometric interpretation. The surfaces g1 = 0 and g 2 = 0 (usually) intersect in a smooth curve, say C (Figure 13.59). Along this curve we seek the points where f has local maximum and minimum values relative to its other values on the curve. These are the points where ∇f is normal to C, as we saw in Theorem 12. But ∇g1 and ∇g 2 are also normal to C at these points because C lies in the surfaces g1 = 0 and g 2 = 0. Therefore, ∇f lies in the plane determined by ∇g1 and ∇g 2 , which means that ∇f = λ∇g1 + μ∇g 2 for some λ and μ. Since the points we seek also lie in both surfaces, their coordinates must satisfy the equations g1 ( x , y, z ) = 0 and g 2 ( x , y, z ) = 0, which are the remaining requirements in Equations (2). EXAMPLE 5 The plane x + y + z = 1 cuts the cylinder x 2 + y 2 = 1 in an ellipse (Figure 13.60). Find the points on the ellipse that lie closest to and farthest from the origin. z

Cylinder x2 + y2 = 1 P2

(0, 1, 0) (1, 0, 0)

y P1

x

Plane x+y+ z =1

FIGURE 13.60

On the ellipse where the plane and cylinder meet, we find the points closest to and farthest from the origin (Example 5).

Solution We find the extreme values of

f ( x , y, z ) = x 2 + y 2 + z 2

13.8  Lagrange Multipliers

841

(the square of the distance from ( x , y, z ) to the origin) subject to the constraints g 1 ( x , y, z ) = x 2 + y 2 − 1 = 0

(3)

g 2 ( x , y, z ) = x + y + z − 1 = 0.

(4)

The gradient equation in Equations (2) then gives ∇f = λ∇g1 + μ∇g 2 2 x i + 2 y j + 2 zk = λ ( 2 x i + 2 y j ) + μ ( i + j + k ) 2 xi + 2 yj + 2 zk = ( 2λ x + μ ) i + ( 2λ y + μ ) j + μk, or 2 x = 2λ x + μ,

2 y = 2λ y + μ,

2 z = μ.

(5)

The scalar equations in Equations (5) yield 2 x = 2λ x + 2 z ⇒ (1 − λ ) x = z ,   2 y = 2λ y + 2 z ⇒ (1 − λ ) y = z.

(6)

Equations (6) are satisfied simultaneously if either λ = 1 and z = 0 or λ ≠ 1 and x = y = z (1 − λ ). In the first case, where z = 0, solving Equations (3) and (4) simultaneously to find the corresponding points on the ellipse gives the two points (1, 0, 0 ) and ( 0, 1, 0 ). This makes sense when you look at Figure 13.60. In the second case, where x = y, Equations (3) and (4) give x2 + x2 − 1 = 0 2x 2

x + x + z −1= 0

=1

z = 1 − 2x

2 2 The corresponding points on the ellipse are x = ±

⎛ 2 2 P1 = ⎜⎜ , ,1 − ⎝ 2 2

⎞ 2 ⎟⎟⎟ ⎠

and

z =1∓

2.

⎛ 2 2 P2 = ⎜⎜ − ,− ,1 + ⎝ 2 2

⎞ 2 ⎟⎟⎟. ⎠

To find the points at maximum and minimum distance from the origin, we evaluate f at the four critical points (1, 0, 0 ) , ( 0, 1, 0 ) , P1, and P2. We see that f (1, 0, 0 ) = f ( 0, 1, 0 ) = 1,

f ( P1 ) = 4 − 2 2, and   f ( P2 ) = 4 + 2 2.

The largest and smallest of these give the absolute extrema. Since 1 < 4 − 2 2 < 4 + 2 2, we see that the absolute minimum value of f is 1 and is attained when f is evaluated at either (1, 0, 0 ) or ( 0, 1, 0 ). The absolute maximum value of f is 4 + 2 2 and occurs when f is evaluated at P2. The value f ( P1 ) = 4 − 2 2 is neither the largest nor the smallest among the values of f at the critical points, so f does not have an absolute extremum at P1. The points on the ellipse closest to the origin are (1, 0, 0 ) and ( 0, 1, 0 ). The point on the ellipse farthest from the origin is P2 . (See Figure 13.60.)

EXERCISES

13.8

Two Independent Variables with One Constraint

1. Extrema on an ellipse Find the points on the ellipse x 2 + 2 y 2 = 1 where f ( x , y ) = xy has its extreme values. 2. Extrema on a circle Find the extreme values of f ( x , y ) = xy subject to the constraint g ( x , y ) = x 2 + y 2 − 10 = 0.

3. Maximum on a line Find the maximum value of f ( x , y ) = 49 − x 2 − y 2 on the line x + 3 y = 10. 4. Extrema on a line Find the local extreme values of f ( x , y ) = x 2 y on the line x + y = 3.

842

Chapter 13 Partial Derivatives

5. Constrained minimum Find the points on the curve xy 2 = 54 nearest the origin. 6. Constrained minimum Find the points on the curve x 2 y = 2 nearest the origin. 7. Use the method of Lagrange multipliers to find

23. Extrema on a sphere values of

Find the maximum and minimum

f ( x , y, z ) = x − 2 y + 5 z on the sphere

x2

+ y 2 + z 2 = 30.

a. Minimum on a hyperbola The minimum value of x + y, subject to the constraints xy = 16, x > 0, y > 0.

24. Extrema on a sphere Find the points on the sphere x 2 + y 2 + z 2 = 25 where f ( x , y, z ) = x + 2 y + 3z has its maximum and minimum values.

b. Maximum on a line The maximum value of xy, subject to the constraint x + y = 16.

25. Minimizing a sum of squares Find three real numbers whose sum is 9 and the sum of whose squares is as small as possible.

Comment on the geometry of each solution.

26. Maximizing a product Find the largest product the positive numbers x, y, and z can have if x + y + z 2 = 16.

8. Extrema on a curve Find the points on the curve x 2 + xy + y 2 = 1 in the xy-plane that are nearest to and farthest from the origin. 9. Minimum surface area with fixed volume Find the dimensions of the closed right circular cylindrical can of smallest surface area whose volume is 16π cm 3 . 10. Cylinder in a sphere Find the radius and height of the open right circular cylinder of largest surface area that can be inscribed in a sphere of radius a. What is the largest surface area? 11. Rectangle of greatest area in an ellipse Use the method of Lagrange multipliers to find the dimensions of the rectangle of greatest area that can be inscribed in the ellipse x 2 16 + y 2 9 = 1 with sides parallel to the coordinate axes. 12. Rectangle of longest perimeter in an ellipse Find the dimensions of the rectangle of largest perimeter that can be inscribed in the ellipse x 2 a 2 + y 2 b 2 = 1 with sides parallel to the coordinate axes. What is the largest perimeter? 13. Extrema on a circle Find the maximum and minimum values of x 2 + y 2 subject to the constraint x 2 − 2 x + y 2 − 4 y = 0. 14. Extrema on a circle Find the maximum and minimum values of 3 x − y + 6 subject to the constraint x 2 + y 2 = 4. 15. Ant on a metal plate The temperature at a point ( x , y ) on a metal plate is T ( x , y ) = 4 x 2 − 4 xy + y 2 . An ant on the plate walks around the circle of radius 5 centered at the origin. What are the highest and lowest temperatures encountered by the ant? 16. Cheapest storage tank Your firm has been asked to design a storage tank for liquid petroleum gas. The customer’s specifications call for a cylindrical tank with hemispherical ends, and the tank is to hold 8000 m 3 of gas. The customer also wants to use the smallest amount of material possible in building the tank. What radius and height do you recommend for the cylindrical portion of the tank? Three Independent Variables with One Constraint

17. Minimum distance to a point Find the point on the plane x + 2 y + 3z = 13 closest to the point (1, 1, 1). 18. Maximum distance to a point Find the point on the sphere x 2 + y 2 + z 2 = 4 farthest from the point (1, −1, 1). 19. Minimum distance to the origin Find the minimum distance from the surface x 2 − y 2 − z 2 = 1 to the origin. 20. Minimum distance to the origin Find the point on the surface z = xy + 1 nearest the origin. 21. Minimum distance to the origin Find the points on the surface z 2 = xy + 4 closest to the origin. 22. Minimum distance to the origin face xyz = 1 closest to the origin.

Find the point(s) on the sur-

27. Rectangular box of largest volume in a sphere Find the dimensions of the closed rectangular box with maximum volume that can be inscribed in the unit sphere. 28. Box with vertex on a plane Find the volume of the largest closed rectangular box in the first octant having three faces in the coordinate planes and a vertex on the plane x a + y b + z c = 1, where a > 0, b > 0, and c > 0. 29. Hottest point on a space probe A space probe in the shape of the ellipsoid 4 x 2 + y 2 + 4 z 2 = 16 enters Earth’s atmosphere and its surface begins to heat. After 1 hour, the temperature at the point ( x , y, z ) on the probe’s surface is T ( x , y, z ) = 8 x 2 + 4 yz − 16 z + 600. Find the hottest point on the probe’s surface. 30. Extreme temperatures on a sphere Suppose that the Celsius temperature at the point ( x , y, z ) on the sphere x 2 + y 2 + z 2 = 1 is T = 400 xyz 2. Locate the highest and lowest temperatures on the sphere. 31. Cobb–Douglas production function During the 1920s, Charles Cobb and Paul Douglas modeled total production output P (of a firm, industry, or entire economy) as a function of labor hours involved x and capital invested y (which includes the monetary worth of all buildings and equipment). The Cobb–Douglas production function is given by P ( x , y ) = kx α y 1−α,   where k and α are constants representative of a particular firm or economy. a. Show that a doubling of both labor and capital results in a doubling of production P. b. Suppose a particular firm has the production function for k = 120 and α = 3 4. Assume that each unit of labor costs $250 and each unit of capital costs $400, and that the total expenses for all costs cannot exceed $100,000. Find the maximum production level for the firm. 32. (Continuation of Exercise 31.) If the cost of a unit of labor is c1 and the cost of a unit of capital is c 2, and if the firm can spend only B dollars as its total budget, then production P is constrained by c1 x + c 2 y = B. Show that the maximum production level subject to the constraint occurs at the point x =

αB c1

and

y =

(1

− α)B . c2

13.8  Lagrange Multipliers

33. Maximizing a utility function: an example from economics In economics, the usefulness or utility of amounts x and y of two capital goods G1 and G 2 is sometimes measured by a function U ( x , y ). For example, G1 and G 2 might be two chemicals a pharmaceutical company needs to have on hand, and U ( x , y ) might be the gain from manufacturing a product whose synthesis requires different amounts of the chemicals depending on the process used. If G1 costs a dollars per kilogram, G 2 costs b dollars per kilogram, and the total amount allocated for the purchase of G1 and G 2 together is c dollars, then the company’s managers want to maximize U ( x , y ) given that ax + by = c. Thus, they need to solve a typical Lagrange multiplier problem. Suppose that U ( x , y ) = xy + 2 x and that the equation ax + by = c simplifies to 2 x + y = 30. Find the maximum value of U and the corresponding values of x and y subject to this latter constraint. 34. Blood types Human blood types are classified by three gene forms A, B, and O. Blood types AA, BB, and OO are homozygous, and blood types AB, AO, and BO are heterozygous. If p, q, and r represent the proportions of the three gene forms to the population, respectively, then the Hardy–Weinberg Law asserts that the proportion Q of heterozygous persons in any specific population is modeled by Q ( p, q, r ) = 2( pq + pr + qr ) ,   subject to p + q + r = 1. Find the maximum value of Q. 35. Length of a beam In Section 4.6, Exercise 47, we posed a problem of finding the length L of the shortest beam that can reach over a wall of height h to a tall building located k units from the wall. Use Lagrange multipliers to show that 32

L = (h2 3 + k 2 3 ) . 36. Locating a radio telescope You are in charge of erecting a radio telescope on a newly discovered planet. To minimize interference, you want to place it where the magnetic field of the planet is weakest. The planet is spherical, with a radius of 6 units. Based on a coordinate system whose origin is at the center of the planet, the strength of the magnetic field is given by M ( x , y, z ) = 6 x − y 2 + xz + 60. Where should you locate the radio telescope? Extreme Values Subject to Two Constraints

37. Maximize the function f ( x , y, z ) = x 2 + 2 y − z 2 subject to the constraints 2 x − y = 0 and y + z = 0. + + subject to the 38. Minimize the function f ( x , y, z ) = constraints x + 2 y + 3z = 6 and x + 3 y + 9 z = 9. x2

y2

z2

843

42. a. Maximum on line of intersection Find the maximum value of w = xyz on the line of intersection of the two planes x + y + z = 40 and x + y − z = 0. b. Give a geometric argument to support your claim that you have found a maximum, and not a minimum, value of w. 43. Extrema on a circle of intersection Find the extreme values of the function f ( x , y, z ) = xy + z 2 on the circle in which the plane y − x = 0 intersects the sphere x 2 + y 2 + z 2 = 4. 44. Minimum distance to the origin Find the point closest to the origin on the curve of intersection of the plane 2 y + 4 z = 5 and the cone z 2 = 4 x 2 + 4 y 2 . Theory and Examples

45. The condition ∇f = λ ∇g is not sufficient Even though ∇f = λ∇g is a necessary condition for the occurrence of an extreme value of f ( x , y ) subject to the conditions g ( x , y ) = 0 and ∇g ≠ 0 , it does not in itself guarantee that one exists. As a case in point, try using the method of Lagrange multipliers to find a maximum value of f ( x , y ) = x + y subject to the constraint that xy = 16. The method will identify the two points ( 4, 4 ) and (−4, −4 ) as candidates for the location of extreme values. Yet the sum x + y has no maximum value on the hyperbola xy = 16. The farther you go from the origin on this hyperbola in the first quadrant, the larger the sum f ( x , y ) = x + y becomes. 46. A least squares plane The plane z = Ax + By + C is to be “fitted” to the following points ( x k , y k , z k ):

( 0, 0, 0 ) ,

( 0, 1, 1) ,

(1, 1, 1) ,

(1, 0, −1).

Find the values of A, B, and C that minimize 4

∑ ( Ax k k =1

+ By k + C − z k ) 2,

the sum of the squares of the deviations. 47. a. Maximum on a sphere Show that the maximum value of a 2 b 2c 2 on a sphere of radius r centered at the origin of a Cartesian abc-coordinate system is ( r 2 3 )3. b. Geometric and arithmetic means Using part (a), show that for nonnegative numbers a, b, and c, a+b+c ; 3 that is, the geometric mean of three nonnegative numbers is less than or equal to their arithmetic mean. (abc) 1 3 ≤

48. Sum of products Let a1 , a 2 , … , a n be n positive numbers. Find n n the maximum of  ∑ i =1 a i x i   subject to the constraint  ∑ i =1 x i2 = 1. COMPUTER EXPLORATIONS

In Exercises 49–54, use a CAS to perform the following steps implementing the method of Lagrange multipliers for finding constrained extrema: a. Form the function h = f − λ1g1 − λ 2 g 2 , where f is the function to optimize subject to the constraints g1 = 0 and g 2 = 0.

39. Minimum distance to the origin Find the point closest to the origin on the line of intersection of the planes y + 2 z = 12 and x + y = 6.

b. Determine all the first partial derivatives of h, including the partials with respect to λ1 and λ 2 , and set them equal to 0.

40. Find the extreme values of f ( x , y, z ) = 2 x 2 + yz on the intersection of the cylinder x 2 + z 2 = 9 and the plane y − z = 4.

c. Solve the system of equations found in part (b) for all the unknowns, including λ1 and λ 2 .

41. Extrema on a curve of intersection Find the extreme values of f ( x , y, z ) = x 2 yz + 1 on the intersection of the plane z = 1 with the sphere x 2 + y 2 + z 2 = 10.

d. Evaluate f at each of the solution points found in part (c), and select the extreme value subject to the constraints asked for in the exercise.

844

Chapter 13 Partial Derivatives

49. Minimize f ( x , y, z ) = xy + yz subject to the constraints x 2 + y 2 − 2 = 0 and x 2 + z 2 − 2 = 0. constraints

53. Minimize f ( x , y, z , w ) = x 2 + y 2 + z 2 + w 2 subject to the constraints 2 x − y + z − w − 1 = 0 and x + y − z + w − 1 = 0.

51. Maximize f ( x , y, z ) = x 2 + y 2 + z 2 subject to the constraints 2 y + 4 z − 5 = 0 and 4 x 2 + 4 y 2 − z 2 = 0.

54. Determine the distance from the line y = x + 1 to the parabola y 2 = x. (Hint: Let ( x , y ) be a point on the line and (w, z) a point on the parabola. You want to minimize ( x − w ) 2 + ( y − z ) 2. )

50. Minimize f ( x , y, z ) = xyz subject x 2 + y 2 − 1 = 0 and x − z = 0.

to

the

52. Minimize f ( x , y, z ) = x 2 + y 2 + z 2 subject to the constraints x 2 − xy + y 2 − z 2 − 1 = 0 and x 2 + y 2 − 1 = 0.

13.9 Taylor’s Formula for Two Variables In this section we use Taylor’s formula to derive the Second Derivative Test for local extreme values (Section 13.7) and the error formula for linearizations of functions of two independent variables (Section 13.6). The use of Taylor’s formula in these derivations leads to an extension of the formula that provides polynomial approximations of all orders for functions of two independent variables.

Derivation of the Second Derivative Test t=1

S(a + h, b + k)

Parametrized segment in R (a + th, b + tk), a typical point on the segment

Let f ( x , y ) have continuous first and second partial derivatives in an open region R containing a point P ( a, b ) where f x = f y = 0 (Figure 13.61). Let h and k be increments small enough to put the point S ( a + h, b + k ) and the line segment joining it to P inside R. We parametrize the segment PS as x = a + th,

y = b + tk ,

0 ≤ t ≤ 1.

If F (t ) = f ( a + th, b + tk ) , the Chain Rule gives t=0

F ′(t ) = f x P(a, b) Part of open region R

FIGURE 13.61

We begin the derivation of the Second Derivative Test at P ( a, b ) by parametrizing a typical line segment from P to a point S nearby.

dy dx + fy = hf x + kf y . dt dt

Since f x and f y are differentiable (because they have continuous partial derivatives), F ′ is a differentiable function of t and F ′′ =

∂ ∂ F ′ dx ∂ F ′ dy + = ( hf x + kf y ) ⋅ h + ∂ ( hf x + kf y ) ⋅ k ∂ x dt ∂ y dt ∂x ∂y

= h 2 f xx + 2hkf xy + k 2 f yy .

f xy = f yx

Since F and F ′ are continuous on [ 0, 1 ] and F ′ is differentiable on ( 0, 1), we can apply Taylor’s formula with n = 2 and a = 0 to obtain F (1) = F (0) + F ′(0) (1 − 0 ) + F ′′(c) = F (0) + F ′(0) +

(1

− 0 )2 2

1 F ′′(c) 2

(1)

for some c between 0 and 1. Writing Equation (1) in terms of f gives f ( a + h, b + k ) = f ( a, b ) + hf x ( a, b ) + kf y ( a, b ) +

1 2 . ( h f xx + 2hkf xy + k 2 f yy ) 2 ( a + ch , b + ck )

(2)

Since f x ( a, b ) = f y ( a, b ) = 0, this reduces to f ( a + h, b + k ) − f ( a, b ) =

1 2 . ( h f xx + 2hkf xy + k 2 f yy ) 2 ( a + ch , b + ck )

(3)

13.9  Taylor’s Formula for Two Variables

845

To determine whether f has an extremum at ( a, b ), we examine the sign of the difference f ( a + h, b + k ) − f ( a, b ). By Equation (3), this is the same as the sign of Q(c) = ( h 2 f xx + 2hkf xy + k 2 f yy )

. ( a + ch ,  b + ck )

Now, if Q(0) ≠ 0, the sign of Q(c) will be the same as the sign of Q(0) for sufficiently small values of h and k. We can predict the sign of Q(0) = h 2 f xx ( a, b ) + 2hkf xy ( a, b ) + k 2 f yy ( a, b )

(4)

from the signs of f xx and f xx f yy − f xy 2 at ( a, b ). Multiply both sides of Equation (4) by f xx and rearrange the right-hand side to get f xx Q(0) = ( hf xx + kf xy ) + ( f xx f yy − f xy 2 ) k 2. 2

(5)

From Equation (5) we see that 1. If f xx < 0 and f xx f yy − f xy 2 > 0 at ( a, b ), then Q(0) < 0 for all sufficiently small nonzero values of h and k, and f has a local maximum value at ( a, b ). 2. If f xx > 0 and f xx f yy − f xy 2 > 0 at ( a, b ), then Q(0) > 0 for all sufficiently small nonzero values of h and k, and f has a local minimum value at ( a, b ). 3. If f xx f yy − f xy 2 < 0 at ( a, b ), there are combinations of arbitrarily small nonzero values of h and k for which Q(0) > 0, and other values for which Q(0) < 0. Arbitrarily close to the point P0 ( a, b, f ( a, b )) on the surface z = f ( x , y ) there are points above P0 and points below P0 , so f has a saddle point at ( a, b ). 4. If f xx f yy − f xy 2 = 0, another test is needed. The possibility that Q(0) equals zero prevents us from drawing conclusions about the sign of Q(c).

The Error Formula for Linear Approximations We want to show that the difference E ( x , y ) between the values of a function f ( x , y ) and its linearization L ( x , y ) at ( x 0 , y 0 ) satisfies the inequality 1 M ( x − x 0 + y − y 0 ) 2. 2 The function f is assumed to have continuous second partial derivatives throughout an open set containing a closed rectangular region R centered at ( x 0 , y 0 ). The number M is an upper bound for f xx , f yy , and f xy on R. The inequality we want comes from Equation (2). We substitute x 0 and y 0 for a and b, and x − x 0 and y − y 0 for h and k, respectively, and rearrange the result as E ( x, y ) ≤

f ( x , y ) = f ( x 0 , y 0 ) + f x ( x 0 , y 0 ) ( x − x 0 ) + f y ( x 0 , y 0 )( y − y 0 )  linearization L( x , y )

+

1 . (( x − x 0 ) 2 f xx + 2( x − x 0 )( y − y 0 ) f xy + ( y − y 0 ) 2 f yy ) 2 ( x 0 + c( x − x 0 ), y 0 + c( y − y 0 ))   error E ( x , y )

This equation reveals that E ≤

1 ( x − x0 2

2

f xx + 2 x − x 0 y − y 0 f xy + y − y 0

2

f yy ).

Hence, if M is an upper bound for the values of f xx , f xy , and f yy on R, then 1 ( x − x 0 2 M + 2 x − x 0 y − y0 M + y − y0 2 1 = M ( x − x 0 + y − y 0 ) 2. 2

E ≤

2

M)

846

Chapter 13 Partial Derivatives

Taylor’s Formula for Functions of Two Variables The formulas derived earlier for F ′ and F ′′ can be obtained by applying to f ( x , y ) the differentiation operators ⎛ ∂ ∂ ⎞ ⎜⎜ h + k ⎟⎟⎟ ⎝ ∂x ∂y⎠

and

2 2 ⎛ ⎞2 ∂2 ⎜⎜ h ∂ + k ∂ ⎟⎟⎟ = h 2 ∂ + 2hk ∂ + k2 2 . 2 ⎝ ∂x ∂y⎠ ∂x ∂x ∂y ∂y

These are the first two instances of a more general formula, F ( n ) (t ) =

⎛ ⎞n dn ⎜ h ∂ + k ∂ ⎟⎟⎟ f ( x , y ) , F ( t ) = ⎜⎝ ∂ x dt n ∂y⎠

(6)

which says that applying d n dt n to F (t ) gives the same result as applying the operator n ⎛ ∂ ∂ ⎞ ⎜⎜ h + k ⎟⎟⎟ ⎝ ∂x ∂y⎠

to f ( x , y ) after expanding it by the Binomial Theorem. If the partial derivatives of f through order n + 1 are continuous throughout a rectangular region centered at ( a, b ), we may extend the Taylor formula for F (t ) to F (t ) = F (0) + F ′(0)t +

F ′′(0) 2 F ( n ) (0) ( n ) t ++ t + remainder,   2! n!

and take t = 1 to obtain F (1) = F (0) + F ′(0) +

F ′′(0) F ( n ) (0) ++ + remainder. 2! n!

When we replace the first n derivatives on the right of this last series by their equivalent expressions from Equation (6) evaluated at t = 0 and add the appropriate remainder term, we arrive at the following formula.

Taylor’s Formula for f ( x, y ) at the Point ( a, b)

Suppose f ( x , y ) and its partial derivatives through order n + 1 are continuous throughout an open rectangular region R centered at a point ( a, b ). Then, throughout R, ⎛ ⎞ f ( a + h, b + k ) = f ( a, b ) + ⎜⎜ hf x + kf y ⎟⎟ ⎟⎠ ⎜⎝ +

+

+ ( a, b )

⎞ 1 ⎛⎜ 2 2 ⎜ h f xx + 2hkf xy + k f yy ⎟⎟⎟ 2! ⎜⎝ ⎠

⎞ 1 ⎛⎜ 3 2 2 3 ⎜ h f xxx + 3h kf xxy + 3hk f xyy + k f yyy ⎟⎟⎟ 3! ⎜⎝ ⎠ n +1 ⎛ ∂ 1 ∂ ⎞⎟ ⎜⎜ h f +k ⎟ ( n + 1)! ⎜ ∂ y ⎟⎠ ⎝ ∂x

.

++ ( a, b )

( a, b ) n 1 ⎛⎜ ∂ ∂ ⎞ h + k ⎟⎟⎟ f ⎜ n! ⎝ ∂ x ∂y⎠

( a, b )

(7)

( a + ch , b + ck )

The first n derivative terms are evaluated at ( a, b ). The last term is evaluated at some point ( a + ch, b + ck ) on the line segment joining ( a, b ) and ( a + h, b + k ). If ( a, b ) = ( 0, 0 ) and we treat h and k as independent variables (denoting them now by x and y), then Equation (7) assumes the following form.

13.9  Taylor’s Formula for Two Variables

847

Taylor’s Formula for f ( x, y ) at the Origin

f ( x , y ) = f ( 0, 0 ) + xf x + yf y +

⎞ 1 ⎛⎜ 2 2 ⎜ x f xx + 2 xyf xy + y f yy ⎟⎟⎟ 2! ⎜⎝ ⎠

+

⎞ ∂n f ∂ n f ⎞⎟ 1 ⎛⎜ 3 1 ⎛ n ∂n f 2 2 3   + nx n −1 y n −1 +  + yn ⎟ ⎜⎜ x f xxx + 3 x yf xxy + 3 xy f xyy + y f yyy ⎟⎟⎟ +  + ⎜⎜⎜ x n 3! ⎝ n! ⎝ ∂ x ∂x ∂y ∂ y n ⎟⎠ ⎠

+

⎛ n +1 ∂ n +1 f ∂ n +1 f ∂ n +1 f ⎞⎟ 1 ⎜⎜ x   + ( n + 1) x n y n +  + y n +1 ⎟ 1 n + ⎜ ( n + 1)! ⎝ ∂x ∂x ∂y ∂ y n +1 ⎟⎠

(8) ( cx , cy )

The first n derivative terms are evaluated at ( 0, 0 ). The last term is evaluated at a point on the line segment joining the origin and ( x , y ). Taylor’s formula provides polynomial approximations of two-variable functions. The first n derivative terms give the polynomial; the last term gives the approximation error. The first three terms of Taylor’s formula give the function’s linearization. To improve on the linearization, we add higher-power terms. EXAMPLE 1 Find a quadratic approximation to f ( x , y ) = sin x sin y near the origin. How accurate is the approximation if x ≤ 0.1 and y ≤ 0.1? Solution We take n = 2 in Equation (8):

1 f ( x , y ) = f ( 0, 0 ) + ( xf x + yf y ) + ( x 2 f xx + 2 xyf xy + y 2 f yy ) 2 1 3 2 . + ( x f xxx + 3 x yf xxy + 3 xy 2 f xyy + y 3 f yyy ) 6 ( cx , cy ) Calculating the values of the partial derivatives, f ( 0, 0 ) = sin x sin y

= 0, ( 0, 0 )

f x ( 0, 0 ) = cos x sin y f y ( 0, 0 ) = sin x cos y

= 0, ( 0, 0 )

= 0, ( 0, 0 )

f xx ( 0, 0 ) = − sin x sin y f xy ( 0, 0 ) = cos x cos y f yy ( 0, 0 ) = − sin x sin y

= 0, ( 0, 0 )

= 1, ( 0, 0 )

= 0, ( 0, 0 )

we have the result 1 sin x sin y ≈ 0 + 0 + 0 + ( x 2 (0) + 2 xy(1) + y 2 (0) ) , or sin x sin y ≈ xy. 2 The error in the approximation is E ( x, y ) =

1 3 . ( x f xxx + 3x 2 yf xxy + 3xy 2 f xyy + y 3 f yyy ) 6 ( cx , cy )

The third derivatives never exceed 1 in absolute value because they are products of sines and cosines. Also, x ≤ 0.1 and y ≤ 0.1. Hence E ( x, y ) ≤

1 ( )3 ( 0.1 + 3( 0.1)3 + 3( 0.1)3 + ( 0.1)3 ) = 8 ( 0.1)3 ≤ 0.00134 6 6

(rounded up). The error will not exceed 0.00134 if x ≤ 0.1 and y ≤ 0.1.

848

Chapter 13 Partial Derivatives

EXERCISES

13.9

Finding Quadratic and Cubic Approximations

In Exercises 1–10, use Taylor’s formula for f ( x , y ) at the origin to find quadratic and cubic approximations of f near the origin. 1. f ( x , y ) = xe y

2. f ( x , y ) = e x cos y

3. f ( x , y ) = y sin x

4. f ( x , y ) = sin x cos y

5. f ( x , y ) =

6. f ( x , y ) = ln ( 2 x + y + 1)

ex

ln (1 + y )

7. f ( x , y ) = sin ( x 2 + y 2 )

8. f ( x , y ) = cos ( x 2 + y 2 )

9. f ( x , y ) =

1 1− x − y

10. f ( x , y ) =

1 1 − x − y + xy

11. Use Taylor’s formula to find a quadratic approximation of f ( x , y ) = cos x cos y at the origin. Estimate the error in the approximation if x ≤ 0.1 and y ≤ 0.1. 12. Use Taylor’s formula to find a quadratic approximation of e x sin y at the origin. Estimate the error in the approximation if x ≤ 0.1 and y ≤ 0.1.

13.10 Partial Derivatives with Constrained Variables In finding partial derivatives of functions like w = f ( x , y ) , we have assumed x and y to be independent. In many applications, however, this is not the case. For example, the internal energy U of a gas may be expressed as a function U = f ( P , V , T ) of pressure P, volume V, and temperature T. If the individual molecules of the gas do not interact, however, P, V, and T obey (and are constrained by) the ideal gas law PV = nRT

( n  and  R  constant ) ,

and fail to be independent. In this section we learn how to find partial derivatives in situations like this, which occur in economics, engineering, and physics.

Decide Which Variables Are Dependent and Which Are Independent If the variables in a function w = f ( x , y, z ) are constrained by a relation like the one imposed on x, y, and z by the equation z = x 2 + y 2, the geometric meanings and the numerical values of the partial derivatives of f will depend on which variables are chosen to be dependent and which are chosen to be independent. To see how this choice can affect the outcome, we consider the calculation of ∂ w ∂ x when w = x 2 + y 2 + z 2 and z = x 2 + y 2. EXAMPLE 1

Find ∂ w ∂ x if w = x 2 + y 2 + z 2 and z = x 2 + y 2 .

Solution We are given two equations in the four unknowns x, y, z, and w. Like many such systems, this one can be solved for two of the unknowns (the dependent variables) in terms of the others (the independent variables). In being asked for ∂ w ∂ x, we are told that w is to be a dependent variable and x an independent variable. The possible choices for the other variables come down to

Dependent

Independent

Choice 1:

w, z

x, y

Choice 2:

w, y

x, z

In either case, we can express w explicitly in terms of the selected independent variables. We do this by using the second equation z = x 2 + y 2 to eliminate the remaining dependent variable in the first equation. In the first case, the remaining dependent variable is z. We eliminate it from the first equation by replacing it by x 2 + y 2. The resulting expression for w is w = x 2 + y 2 + z 2 = x 2 + y 2 + ( x 2 + y 2 )2 = x 2 + y 2 + x 4 + 2x 2 y 2 + y 4

13.10  Partial Derivatives with Constrained Variables

z

849

and therefore z=

x2

,y=0

∂w = 2 x + 4 x 3 + 4 xy 2 . ∂x

z=x + 2

1

P

y2

Circle x2 + y2 = 1 in the plane z = 1

(1)

This is the formula for ∂ w ∂ x when x and y are the independent variables. In the second case, where the independent variables are x and z and the remaining dependent variable is y, we eliminate the dependent variable y in the expression for w by replacing y 2 in the second equation by z − x 2. This gives w = x 2 + y2 + z 2 = x 2 + (z − x 2 ) + z2 = z + z 2

(1, 0, 1)

and therefore 0 1

y

x

FIGURE 13.62

If P is constrained to lie on the paraboloid z = x 2 + y 2 , the value of the partial derivative of w = x 2 + y 2 + z 2 with respect to x at P depends on the direction of motion (Example 1). (1) As x changes, with y = 0, P moves up or down the surface on the parabola z = x 2 in the xz-plane with ∂ w ∂ x = 2 x + 4 x 3 . (2) As x changes, with z = 1, P moves on the circle x 2 + y 2 = 1, z = 1, and ∂ w ∂ x = 0.

∂w = 0. ∂x

(2)

This is the formula for ∂ w ∂ x when x and z are the independent variables. The formulas for ∂ w ∂ x in Equations (1) and (2) are genuinely different. We cannot change either formula into the other by using the relation z = x 2 + y 2 . There is not just one ∂ w ∂ x , there are two, and we see that the original instruction to find ∂ w ∂ x was incomplete. Which ∂ w ∂ x ? we ask. The geometric interpretations of Equations (1) and (2) help to explain why the equations differ. The function w = x 2 + y 2 + z 2 measures the square of the distance from the point (x, y, z) to the origin. The condition z = x 2 + y 2 says that the point (x, y, z) lies on the paraboloid of revolution shown in Figure 13.62. What does it mean to calculate ∂ w ∂ x at a point P ( x , y, z ) that can move only on this surface? What is the value of ∂ w ∂ x when the coordinates of P are, say, (1, 0, 1)? If we take x and y to be independent, then we find ∂ w ∂ x by holding y fixed (at y = 0 in this case) and letting x vary. Hence, P moves along the parabola z = x 2 in the xz-plane. As P moves on this parabola, w, which is the square of the distance from P to the origin, changes. We calculate ∂ w ∂ x in this case (our first solution above) to be ∂w = 2 x + 4 x 3 + 4 xy 2 . ∂x At the point P (1, 0, 1), the value of this derivative is ∂w = 2 + 4 + 0 = 6. ∂x If we take x and z to be independent, then we find ∂ w ∂ x by holding z fixed while x varies. Since the z-coordinate of P is 1, varying x moves P along a circle in the plane z = 1. As P moves along this circle, its distance from the origin remains constant, and w, being the square of this distance, does not change. That is, ∂w = 0, ∂x as we found in our second solution.

How to Find ∂w ∂x When the Variables in w = f ( x, y, z ) Are Constrained by Another Equation As we saw in Example 1, a typical routine for finding ∂ w ∂ x when the variables in the function w = f ( x , y, z ) are related by another equation has three steps. These steps apply to finding ∂ w ∂ y and ∂ w ∂ z as well.

850

Chapter 13 Partial Derivatives

1. Decide which variables are to be dependent and which are to be independent. (In practice, the decision is based on the physical or theoretical context of our work. In the exercises at the end of this section, we say which variables are which.) 2. Eliminate the other dependent variable(s) in the expression for w. 3. Differentiate as usual.

If we cannot carry out Step 2 after deciding which variables are dependent, we differentiate the equations as they are and try to solve for s w s x afterward. The next example shows how this is done.

EXAMPLE 2

Find s w s x at the point ( x , y, z ) = ( 2, −1, 1) if w = x 2 + y2 + z 2,

z 3 − xy + yz + y 3 = 1,

and x and y are the independent variables. Solution It is not convenient to eliminate z in the expression for w. We therefore differentiate both equations implicitly with respect to x, treating x and y as independent variables and w and z as dependent variables. This gives

∂w ∂z = 2 x + 2z ∂x ∂x

(3)

and 3z 2

∂z ∂z + 0 = 0. −y+ y ∂x ∂x

(4)

These equations may now be combined to express s w s x in terms of x, y, and z. We solve Equation (4) for s z s x to get y ∂z = ∂x y + 3z 2 and substitute into Equation (3) to get 2 yz ∂w = 2x + . ∂x y + 3z 2 The value of this derivative at ( x , y, z ) = ( 2, −1, 1) is HISTORICAL BIOGRAPHY Sonya Kovalevsky (1850–1891) Kovalevsky, a Russian mathematician, primarily worked on the theory of partial differential equations, and a central result on the existence of solutions still bears her name. She published numerous papers on partial differential equations, eventually gaining recognition as the first woman to be elected a member of the Russian Imperial Academy of Sciences in 1889. To know more, visit the companion Website.

∂w ∂x

( 2, −1, 1)

= 2( 2 ) +

2(−1)(1) −2 = 4+ = 3. 2 − 1 + 3 ( 1) 2

Notation To show what variables are assumed to be independent in calculating a derivative, we can use the following notation:

( ∂∂wx )

y

∂ w ∂ x with  x  and  y  independent

⎛ ∂ f ⎞⎟ ⎜⎜ ⎟⎟ ∂ f ∂ y with  y,   x ,  and t  independent. ⎝ ∂ y ⎠ x , t

13.10  Partial Derivatives with Constrained Variables

EXAMPLE 3

851

Find ( ∂ w ∂ x ) y , z if w = x 2 + y − z + sin t and x + y = t.

Solution With x, y, z independent, we have

t = x + y,

( ∂∂wx )

y, z

w = x 2 + y − z + sin ( x + y )

= 2 x + 0 − 0 + cos ( x + y )

∂( x + y) ∂x

= 2 x + cos ( x + y ).

Arrow Diagrams In solving problems like the one in Example 3, it often helps to start with an arrow diagram that shows how the variables and functions are related. If w = x 2 + y − z + sin t

x+ y = t

and

and we are asked to find ∂ w ∂ x when x, y, and z are independent, the appropriate diagram is one like this: ⎛ x ⎞⎟ ⎜⎜ ⎟ ⎜⎜ y ⎟⎟⎟ ⎜⎜ ⎟ ⎜⎝ z ⎟⎟⎠

→

Independent variables

⎛ ⎜⎜ ⎜⎜ ⎜⎜ ⎜⎜ ⎜⎜⎝

x y z t

⎞⎟ ⎟⎟ ⎟⎟ ⎟⎟ ⎟⎟ ⎟⎟ ⎠

→

Intermediate variables

w

(5)

Dependent variable

To avoid confusion between the independent and intermediate variables with the same symbolic names in the diagram, it is helpful to rename the intermediate variables (so they are seen as functions of the independent variables). Thus, let u = x , υ = y, and s = z denote the renamed intermediate variables. With this notation, the arrow diagram becomes ⎛ x ⎞⎟ ⎜⎜ ⎟ ⎜⎜ y ⎟⎟⎟ ⎜⎜ ⎟⎟ ⎜⎝ z ⎟⎠ Independent variables

→

⎛ ⎜⎜ ⎜⎜ ⎜⎜ ⎜⎜ ⎜⎝

u υ s t

⎞⎟ ⎟⎟ ⎟⎟ ⎟⎟ ⎟⎟ ⎟⎠

→

Intermediate variables and relations u=x υ =y s =z t =x +y

w

(6)

Dependent variable

The diagram shows the independent variables on the left, the intermediate variables and their relation to the independent variables in the middle, and the dependent variable on the right. The function w now becomes w = u 2 + υ − s + sin t , where u = x,

υ = y,

s = z,

and

t = x + y.

To find ∂ w ∂ x, we apply the four-variable form of the Chain Rule to w, guided by the arrow diagram in Equation (6): ∂ w ∂t ∂w ∂ w ∂u ∂ w ∂υ ∂ w ∂s = + + + ∂x ∂u ∂ x ∂υ ∂ x ∂s ∂ x ∂t ∂ x = (2u) (1) + (1)( 0 ) + (−1)( 0 ) + ( cos t )(1)

 

= 2u + cos t = 2 x + cos ( x + y ).

Substitute the original independent variables u = x and t = x + y

852

Chapter 13 Partial Derivatives

13.10

EXERCISES

Show that the equations

Finding Partial Derivatives with Constrained Variables

In Exercises 1–3, begin by drawing a diagram that shows the relations among the variables.

∂w = 2x − 1 ∂x

1. If w = x 2 + y 2 + z 2 and z = x 2 + y 2, find ⎛ ∂ w ⎟⎞ a. ⎜⎜ ⎝ ∂ y ⎟⎟⎠ z

b.

( ∂∂wz )

c. x

each give ∂ w ∂ x , depending on which variables are chosen to be dependent and which variables are chosen to be independent. Identify the independent variables in each case.

( ∂∂wz ) . y

2. If w = x 2 + y − z + sin t and x + y = t, find ⎛ ∂ w ⎟⎞ a. ⎜⎜ ⎝ ∂ y ⎟⎟⎠ x , z d.

( ∂∂wz )

⎛ ∂ w ⎟⎞ b. ⎜⎜ ⎝ ∂ y ⎟⎟⎠ z , t e.

y, t

( ∂∂wt )

f. x, z

Theory and Examples

( )

x, y

( ∂∂wt )

y, z

∂w c. ∂z

9. Establish the fact, widely used in hydrodynamics, that if f ( x , y, z ) = 0, then

( ) ( ∂∂xz )

⎛ ∂ x ⎞⎟ ∂ y ⎜⎜ ⎟⎟ ⎝ ∂ y ⎠z ∂ z

.

3. Let U = f ( P , V , T ) be the internal energy of a gas that obeys the ideal gas law PV = nRT (n and R constant). Find ∂U ∂U . a. b. ∂P V ∂T V

( )

( )

( ∂∂wx )

b.

( ∂∂wz )

5. Find

12. Suppose that f ( x , y, z , w ) = 0 and g ( x , y, z , w ) = 0 determine z and w as differentiable functions of the independent variables x and y, and suppose that

⎛ ∂ w ⎟⎞ b. ⎜⎜ ⎝ ∂ y ⎟⎟⎠ z

⎛ ∂ w ⎟⎞ a. ⎜⎜ ⎝ ∂ y ⎟⎟⎠ x

at the point ( w, x , y, z ) = ( 4, 2, 1, −1) if w = x 2 y 2 + yz − z 3

and

∂ f ∂g ∂ f ∂g − ≠ 0. ∂z ∂w ∂w ∂z

x 2 + y 2 + z 2 = 6.

6. Find ( ∂u ∂ y ) x at the point ( u, υ ) = ( 2, 1) if x = u 2 + υ 2 and y = uυ. 7. Suppose that nates. Find

x2

+

y2

=

( ∂∂rx )

r2

θ

Show that

( ) ∂z ∂x

and x = r cos θ, as in polar coordi-

and

( ∂∂rx ) .

CHAPTER 13

and

y

∂ f ∂g ∂ f ∂g − ∂ x ∂ w ∂ w ∂x = − ∂ f ∂g ∂ f ∂g − ∂z ∂w ∂w ∂z

and ∂ f ∂g ∂ f ∂g − ⎛ ∂ w ⎞⎟ ∂ z ∂ y ∂ y ∂z ⎜⎜ = − . ∂ f ∂g ∂ f ∂g ⎝ ∂ y ⎟⎟⎠ x − ∂z ∂w ∂w ∂z

y

8. Suppose that w = x 2 − y 2 + 4z + t

∂z ∂z −y = x. ∂x ∂y

∂g ∂ y ⎛ ∂ z ⎞⎟ ⎜⎜ ⎟⎟ = − . ⎝ ∂ y ⎠x ∂g ∂z

y sin z + z sin x = 0.

and

= −1.

11. Suppose that the equation g ( x , y, z ) = 0 determines z as a differentiable function of the independent variables x and y and that g z ≠ 0. Show that

y

at the point ( x , y, z ) = ( 0, 1, π ) if w = x 2 + y2 + z 2

y

10. If z = x + f (u), where u = xy, show that x

y

x

(Hint: Express all the derivatives in terms of the formal partial derivatives ∂ f ∂ x , ∂ f ∂ y , and ∂ f ∂ z.)

4. Find a.

∂w = 2x − 2 ∂x

and

x + 2 z + t = 25.

Questions to Guide Your Review

1. What is a real-valued function of two independent variables? Three independent variables? Give examples. 2. What does it mean for sets in the plane or in space to be open? Closed? Give examples. Give examples of sets that are neither open nor closed. 3. How can you display the values of a function f ( x , y ) of two independent variables graphically? How do you do the same for a function f ( x , y, z ) of three independent variables?

4. What does it mean for a function f ( x , y ) to have limit L as ( x , y ) → ( x 0 , y 0 )? What are the basic properties of limits of functions of two independent variables? 5. When is a function of two (three) independent variables continuous at a point in its domain? Give examples of functions that are continuous at some points but not others. 6. What can be said about algebraic combinations and compositions of continuous functions?

Chapter 13  Practice Exercises

7. Explain the two-path test for nonexistence of limits. 8. How are the partial derivatives ∂ f ∂ x and ∂ f ∂ y of a function f ( x , y ) defined? How are they interpreted and calculated? 9. How does the relation between first partial derivatives and continuity of functions of two independent variables differ from the relation between first derivatives and continuity for real-valued functions of a single independent variable? Give an example. 10. What is the Mixed Derivative Theorem for mixed second-order partial derivatives? How can it help in calculating partial derivatives of second and higher orders? Give examples. 11. What does it mean for a function f ( x , y ) to be differentiable? What does the Increment Theorem say about differentiability? 12. How can you sometimes decide from examining f x and f y that a function f ( x , y ) is differentiable? What is the relation between the differentiability of f and the continuity of f at a point? 13. What is the general Chain Rule? What form does it take for functions of two independent variables? Three independent variables? Functions defined on surfaces? How do you diagram these different forms? Give examples. What pattern enables one to remember all the different forms? 14. What is the derivative of a function f ( x , y ) at a point P0 in the direction of a unit vector u? What rate does it describe? What geometric interpretation does it have? Give examples. 15. What is the gradient vector of a differentiable function f ( x , y )? How is it related to the function’s directional derivatives? State the analogous results for functions of three independent variables.

853

plane and normal line at a point on a level surface of a differentiable function f ( x , y, z )? Give examples. 17. How can you use directional derivatives to estimate change? 18. How do you linearize a function f ( x , y ) of two independent variables at a point ( x 0 , y 0 )? Why might you want to do this? How do you linearize a function of three independent variables? 19. What can you say about the accuracy of linear approximations of functions of two (three) independent variables? 20. If ( x , y ) moves from ( x 0 , y 0 ) to a point ( x 0 + dx , y 0 + dy ) nearby, how can you estimate the resulting change in the value of a differentiable function f ( x , y )? Give an example. 21. How do you define local maxima, local minima, and saddle points for a differentiable function f ( x , y )? Give examples. 22. What derivative tests are available for determining the local extreme values of a function f ( x , y )? How do they enable you to narrow your search for these values? Give examples. 23. How do you find the extrema of a continuous function f ( x , y ) on a closed bounded region of the xy-plane? Give an example. 24. Describe the method of Lagrange multipliers and give examples. 25. How does Taylor’s formula for a function f ( x , y ) generate polynomial approximations and error estimates? 26. If w = f ( x , y, z ) , where the variables x, y, and z are constrained by an equation g ( x , y, z ) = 0, what is the meaning of the notation ( ∂ w ∂ x ) y? How can an arrow diagram help you calculate this partial derivative with constrained variables? Give examples.

16. How do you find the tangent line at a point on a level curve of a differentiable function f ( x , y )? How do you find the tangent

CHAPTER 13

Practice Exercises

Domain, Range, and Level Curves In Exercises 1–4, find the domain and range of the given function and identify its level curves. Sketch a typical level curve.

1. f ( x , y ) =

9x 2

+

y2

3. g ( x , y ) = 1 xy

2. f ( x , y ) =

e x+ y

4. g ( x , y ) =

x2

7. h ( x , y, z ) =

1 x 2 + y2 + z 2

6. g ( x , y, z ) = x 2 + 4 y 2 + 9 z 2 8. k ( x , y, z ) =

x2

+

y2

11. 13.

lim

( x , y )→( π , ln 2 )

e y cos x

10.

x−y x 2 − y2

12.

lim

ln x + y + z

14.

P →(1,  −1,  e )

( x , y )→( 0,0 )

x2

y −y

16.

x 2 + y2 ( x , y )→( 0,0 ) xy lim

xy ≠ 0

17. Continuous extension Let f ( x , y ) = ( x 2 − y 2 ) ( x 2 + y 2 ) for ( x , y ) ≠ ( 0, 0 ). Is it possible to define f ( 0, 0 ) in a way that makes f continuous at the origin? Why? 18. Continuous extension Let ⎪⎧⎪ sin ( x − y ) , f ( x , y ) = ⎪⎨ x + y ⎪⎪ ⎪⎩ 0,

x + y ≠ 0

( x , y ) = ( 0, 0 ).

Is f continuous at the origin? Why?

lim

( x , y )→(1,1)

lim

1 + z2 + 1

Evaluating Limits Find the limits in Exercises 9–14.

9.

15.

y≠ x 2

−y

In Exercises 5–8, find the domain and range of the given function and identify its level surfaces. Sketch a typical level surface. 5. f ( x , y, z ) = x 2 + y 2 − z

By considering different paths of approach, show that the limits in Exercises 15 and 16 do not exist.

lim

2+ y x + cos y

lim

x 3y3 − 1 xy − 1

lim

arctan ( x + y + z )

( x , y )→( 0, 0 )

( x , y )→(1,1)

P →(1,−1,−1)

Partial Derivatives In Exercises 19–24, find the partial derivative of the function with respect to each variable.

19. g ( r , θ ) = r cos θ + r sin θ 20. f ( x , y ) =

y 1 ( 2 ln x + y 2 ) + arctan 2 x

854

Chapter 13 Partial Derivatives

21. f ( R1 , R 2 , R 3 ) =

43. Directional derivatives with given values At the point (1, 2 ), the function f ( x , y ) has a derivative of 2 in the direction toward ( 2, 2 ) and a derivative of −2 in the direction toward (1, 1).

1 1 1 + + R1 R2 R3

22. h ( x , y, z ) = sin ( 2π x + y − 3z ) 23. P ( n, R , T , V ) = 24. f ( r , l , T , w ) =

a. Find f x (1, 2 ) and f y (1, 2 ).

nRT (the ideal gas law) V 1 2rl

b. Find the derivative of f at (1, 2 ) in the direction toward the point ( 4, 6 ).

T πw

Second-Order Partials Find the second-order partial derivatives of the functions in Exercises 25–28. x 25. g ( x , y ) = y + 26. g ( x , y ) = e x + y sin x y

44. Which of the following statements are true if f ( x , y ) is differentiable at ( x 0 , y 0 )? Give reasons for your answers. a. If u is a unit vector, the derivative of f at ( x 0 , y 0 ) in the direction of u is ( f x ( x 0 , y 0 ) i + f y ( x 0 , y 0 ) j ) ⋅ u. b. The derivative of f at ( x 0 , y 0 ) in the direction of u is a vector. c. The directional derivative of f at ( x 0 , y 0 ) has its greatest value in the direction of ∇f .

27. f ( x , y ) = x + xy − 5 x 3 + ln ( x 2 + 1)

d. At ( x 0 , y 0 ) , vector ∇f is normal to the curve f ( x , y ) = f ( x 0 , y 0 ).

28. f ( x , y ) = y 2 − 3 xy + cos y + 7e y Chain Rule Calculations

29. Find dw dt at t = 0 if w = sin ( xy + π ) , x = e t , and y = ln ( t + 1). 30. Find dw dt at t = 1 if w = xe y + y sin z − cos z , x = 2 t , y = t − 1 + ln t , and z = πt. 31. Find ∂ w ∂r and ∂ w ∂ s when r = π w = sin ( 2 x − y ) , x = r + sin s, y = rs.

and s = 0 if

∂ w ∂u and ∂ w ∂υ when u = υ = 0 32. Find w = ln 1 + x 2 − tan −1 x and x = 2e u cos υ.

if

33. Find the value of the derivative of f ( x , y, z ) = xy + yz + xz with respect to t on the curve x = cos t , y = sin t , z = cos 2t at t = 1. 34. Show that if w = f (s) is any differentiable function of s and if s = y + 5 x , then ∂w ∂w −5 = 0. ∂x ∂y Implicit Differentiation Assuming that the equations in Exercises 35 and 36 define y as a differentiable function of x, find the value of dy dx at point P.

35. 1 − x − y 2 − sin xy = 0, P ( 0, 1)

Gradients, Tangent Planes, and Normal Lines In Exercises 45 and 46, sketch the surface f ( x , y, z ) = c together with ∇f at the given points.

45. x 2 + y + z 2 = 0; ( 0, −1, ±1) , ( 0, 0, 0 ) 46. y 2 + z 2 = 4; ( 2, ± 2, 0 ) , ( 2, 0, ± 2 ) In Exercises 47 and 48, find an equation for the plane tangent to the level surface f ( x , y, z ) = c at the point P0 . Also, find parametric equations for the line that is normal to the surface at P0 . 47. x 2 − y − 5 z = 0, P0 ( 2, −1, 1) 48. x 2 + y 2 + z = 4, P0 (1, 1, 2 ) In Exercises 49 and 50, find an equation for the plane tangent to the surface z = f ( x , y ) at the given point. 49. z = ln ( x 2 + y 2 ) , ( 0, 1, 0 ) 50. z = 1 ( x 2 + y 2 ) , (1, 1, 1 2 ) In Exercises 51 and 52, find equations for the lines that are tangent and normal to the level curve f ( x , y ) = c at the point P0 . Then sketch the lines and level curve together with ∇f at P0 . 51. y − sin x = 1, P0 ( π , 1) y2 3 x2 − = , P0 (1, 2 ) 2 2 2

36. 2 xy + e x + y − 2 = 0, P ( 0, ln 2 )

52.

Directional Derivatives In Exercises 37–40, find the directions in which f increases and decreases most rapidly at P0 and find the derivative of f in each direction. Also, find the derivative of f at P0 in the direction of the vector v.

Tangent Lines to Curves In Exercises 53 and 54, find parametric equations for the line that is tangent to the curve of intersection of the surfaces at the given point.

37. f ( x , y ) = cos x cos y, P0 ( π 4 , π 4 ) , v = 3i + 4 j 38. f ( x , y ) = x 2e −2 y , P0 (1, 0 ) , v = i + j 39. f ( x , y, z ) = ln ( 2 x + 3 y + 6 z ) , P0 (−1, −1, 1) , v = 2 i + 3 j + 6k 40. f ( x , y, z ) = x 2 + 3 xy − z 2 + 2 y + z + 4, P0 ( 0, 0, 0 ), v = i+ j+k 41. Derivative in velocity direction Find the derivative of f ( x , y, z ) = xyz in the direction of the velocity vector of the helix r (t ) = ( cos 3t ) i + ( sin 3t ) j + 3tk at t = π 3. 42. Maximum directional derivative What is the largest value that the directional derivative of f ( x , y, z ) = xyz can have at the point (1, 1, 1)?

53. Surfaces: x 2 + 2 y + 2 z = 4, Point:

y =1

(1, 1, 1 2 )

54. Surfaces: x + y 2 + z = 2, y = 1 Point:

(1 2,1, 1 2 )

Linearizations In Exercises 55 and 56, find the linearization L ( x , y ) of the function f ( x , y ) at the point P0 . Then find an upper bound for the magnitude of the error E in the approximation f ( x , y ) ≈ L ( x , y ) over the rectangle R.

55. f ( x , y ) = sin x cos y, P0 ( π 4 , π 4 ) π π ≤ 0.1, y − ≤ 0.1 4 4 56. f ( x , y ) = xy − 3 y 2 + 2, P0 (1, 1) R:

x−

R:

x − 1 ≤ 0.1,

y − 1 ≤ 0.2

855

Chapter 13  Practice Exercises

Find the linearizations of the functions in Exercises 57 and 58 at the given points. 57. f ( x , y, z ) = xy + 2 yz − 3 xz   at  (1, 0, 0 )  and  (1, 1, 0 ) 58. f ( x , y, z ) = 2 cos x sin ( y + z ) (π 4, π 4, 0)

at

( 0, 0, π 4 )

and

Estimates and Sensitivity to Change

59. Measuring the volume of a pipeline You plan to calculate the volume inside a stretch of pipeline that is about 36 cm in diameter and 1 km long. With which measurement should you be more careful, the length or the diameter? Why? 60. Sensitivity to change Is f ( x , y ) = x 2 − xy + y 2 − 3 more sensitive to changes in x or to changes in y when it is near the point (1, 2 )? How do you know? 61. Change in an electrical circuit Suppose that the current I (amperes) in an electrical circuit is related to the voltage V (volts) and the resistance R (ohms) by the equation I = V R. If the voltage drops from 24 to 23 volts and the resistance drops from 100 to 80 ohms, will I increase or decrease? By about how much? Is the change in I more sensitive to change in the voltage or to change in the resistance? How do you know? 62. Maximum error in estimating the area of an ellipse If a = 10 cm and b = 16 cm to the nearest millimeter, what should you expect the maximum percentage error to be in the calculated area A = πab of the ellipse x 2 a 2 + y 2 b 2 = 1? 63. Error in estimating a product Let y = uυ and z = u + υ, where u and υ are positive independent variables. a. If u is measured with an error of 2% and υ with an error of 3%, about what is the percentage error in the calculated value of y? b. Show that the percentage error in the calculated value of z is less than the percentage error in the value of y. 64. Cardiac index To make different people comparable in studies of cardiac output, researchers divide the measured cardiac output by the body surface area to find the cardiac index C: C =

cardiac output . body surface area

Absolute Extrema In Exercises 71–78, find the absolute maximum and minimum values of f on the region R.

71. f ( x , y ) = x 2 + xy + y 2 − 3 x + 3 y R: The triangular region cut from the first quadrant by the line x+ y = 4 72. f ( x , y ) = x 2 − y 2 − 2 x + 4 y + 1 R: The rectangular region in the first quadrant bounded by the coordinate axes and the lines x = 4 and y = 2 73. f ( x , y ) = y 2 − xy − 3 y + 2 x R: The square region enclosed by the lines x = ±2 and y = ±2 74. f ( x , y ) = 2 x + 2 y − x 2 − y 2 R: The square region bounded by the coordinate axes and the lines x = 2, y = 2 in the first quadrant 75. f ( x , y ) = x 2 − y 2 − 2 x + 4 y R: The triangular region bounded below by the x-axis, above by the line y = x + 2, and on the right by the line x = 2 76. f ( x , y ) = 4 xy − x 4 − y 4 + 16 R: The triangular region bounded below by the line y = −2, above by the line y = x, and on the right by the line x = 2 77. f ( x , y ) = x 3 + y 3 + 3 x 2 − 3 y 2 R: The square region enclosed by the lines x = ±1 and y = ±1 78. f ( x , y ) = x 3 + 3 xy + y 3 + 1 R: The square region enclosed by the lines x = ±1 and y = ±1 Lagrange Multipliers

79. Extrema on a circle Find the extreme f ( x , y ) = x 3 + y 2 on the circle x 2 + y 2 = 1.

values

of

80. Extrema on a circle Find the extreme values of f ( x , y ) = xy on the circle x 2 + y 2 = 1. 81. Extrema in a disk Find the extreme values f ( x , y ) = x 2 + 3 y 2 + 2 y on the unit disk x 2 + y 2 ≤ 1.

of

82. Extrema in a disk Find the extreme values f ( x , y ) = x 2 + y 2 − 3 x − xy on the disk x 2 + y 2 ≤ 9.

of

The body surface area B of a person with weight w and height h is approximated by the formula

83. Extrema on a sphere Find the extreme values of f ( x , y, z ) = x − y + z on the unit sphere x 2 + y 2 + z 2 = 1.

B = 71.84 w 0.425 h 0.725,

84. Minimum distance to origin Find the points on the surface x 2 − zy = 4 closest to the origin.

which gives B in square centimeters when w is measured in kilograms and h in centimeters. You are about to calculate the cardiac index of a person 180 cm tall, weighing 70 kg, with cardiac output of 7 L min. Which will have a greater effect on the calculation, a 1-kg error in measuring the weight or a 1-cm error in measuring the height? Local Extrema Test the functions in Exercises 65–70 for local maxima and minima and saddle points. Find each function’s value at these points.

65. f ( x , y ) = x 2 − xy + y 2 + 2 x + 2 y − 4 66. f ( x , y ) =

5x 2

+ 4 xy −

2y 2

+ 4x − 4y

67. f ( x , y ) = 2 x 3 + 3 xy + 2 y 3 68. f ( x , y ) = x 3 + y 3 − 3 xy + 15 69. f ( x , y ) = x 3 + y 3 + 3 x 2 − 3 y 2 70. f ( x , y ) = x 4 − 8 x 2 + 3 y 2 − 6 y

85. Minimizing cost of a box A closed rectangular box is to have volume V cm 3 . The cost of the material used in the box is a cents cm 2 for top and bottom, b cents cm 2 for front and back, and c cents cm 2 for the remaining sides. What dimensions minimize the total cost of materials? 86. Least volume Find the plane x a + y b + z c = 1 that passes through the point ( 2, 1, 2 ) and cuts off the least volume from the first octant. 87. Extrema on curve of intersecting surfaces Find the extreme values of f ( x , y, z ) = x ( y + z ) on the curve of intersection of the right circular cylinder x 2 + y 2 = 1 and the hyperbolic cylinder xz = 1. 88. Minimum distance to origin on curve of intersecting plane and cone Find the point closest to the origin on the curve of intersection of the plane x + y + z = 1 and the cone z 2 = 2 x 2 + 2 y 2 .

856

Chapter 13 Partial Derivatives

Theory and Examples

89. Let w = f ( r , θ ) , r = x 2 + y 2 , and θ = tan −1 ( y x ). Find ∂ w ∂ x and ∂ w ∂ y, and express your answers in terms of r and θ. 90. Let z = f ( u, υ ) , u = ax + by, and υ = ax − by. Express z x and z y in terms of f u , f υ , and the constants a and b. 91. If a and b are constants, w = u 3 + tanh u + cos u, and u = ax + by, show that a

93. Angle between vectors The equations e u cos υ − x = 0 and e u sin υ − y = 0 define u and υ as differentiable functions of x and y. Show that the angle between the vectors ∂υ ∂υ i+ j ∂x ∂y

is constant. 94. Polar coordinates and second derivatives Introducing polar coordinates x = r cos θ and y = r sin θ changes f ( x , y ) to g ( r, θ ). Find the value of ∂ 2 g ∂θ 2 at the point ( r, θ ) = ( 2, π 2 ) , given that ∂f ∂f ∂2 f ∂2 f = = = =1 ∂x ∂y ∂x 2 ∂y2

(y

Find the points on the surface

x 2 + y2 + z 2

at the origin equals 1 in any direction but that f has no gradient vector at the origin. 99. Normal line through origin Show that the line normal to the surface xy + z = 2 at the point (1, 1, 1) passes through the origin. 100. Tangent plane and normal line a. Sketch the surface x 2 − y 2 + z 2 = 4. b. Find a vector normal to the surface at ( 2, −3, 3 ). Add the vector to your sketch. c. Find equations for the tangent plane and the normal line at ( 2, −3, 3 ).

101. If w = x 2e yz and z = x 2 − y 2 find ⎛ ∂ w ⎟⎞ ∂w ∂w a. ⎜⎜ b. c. ⎝ ∂ y ⎟⎟⎠ z ∂z x ∂z

( )

+ z ) 2 + ( z − x ) 2 = 16

where the normal line is parallel to the yz-plane. 96. Tangent plane parallel to xy-plane Find the points on the surface xy + yz + zx − x − z 2 = 0 where the tangent plane is parallel to the xy-plane.

CHAPTER 13

f ( x , y, z ) =

Partial Derivatives with Constrained Variables In Exercises 101 and 102, begin by drawing a diagram that shows the relations among the variables.

at that point. 95. Normal line parallel to a plane

f ( x 0 + su1 , y 0 + su 2 , z 0 + su 3 ) − f ( x 0 , y 0 , z 0 ) . s

lim

Show that the one-sided directional derivative of

92. Using the Chain Rule If w = ln ( x 2 + y 2 + 2 z ), x = r + s, y = r − s, and z = 2rs, find w r and w s by the Chain Rule. Then check your answer another way.

and

98. One-sided directional derivative in all directions, but no gradient The one-sided directional derivative of f at P ( x 0 , y 0 , z 0 ) in the direction u = u1i + u 2 j + u 3 k is the number s→0+

∂w ∂w . = b ∂y ∂x

∂u ∂u i+ j ∂x ∂y

97. When gradient is parallel to position vector Suppose that ∇f ( x , y, z ) is always parallel to the position vector xi + yj + zk. Show that f ( 0, 0, a ) = f ( 0, 0, −a ) for any a.

( ). y

102. Let U = f ( P , V , T ) be the internal energy of a gas that obeys the ideal gas law PV = nRT (n and R constant). Find a.

( ∂∂UT )

b. P

( ∂∂UV ) . T

Additional and Advanced Exercises

Partial Derivatives

1. Function with saddle at the origin If you did Exercise 64 in Section 13.2, you know that the function 2 2 ⎪⎧⎪ xy x − y , ( x , y ) ≠ ( 0, 0 ) 2 ⎪ f ( x, y ) = ⎨ x + y 2 ⎪⎪ ( x , y ) = ( 0, 0 ) ⎪⎩ 0,

(see the accompanying figure) is continuous at ( 0, 0 ). Find f xy ( 0, 0 ) and f yx ( 0, 0 ). z

2. Finding a function from second partials Find a function w = f ( x , y ) whose first partial derivatives are ∂ w ∂ x = 1 + e x cos y and ∂ w ∂ y = 2 y − e x sin y and whose value at the point ( ln 2, 0 ) is ln 2. 3. A proof of Leibniz’s Rule Leibniz’s Rule says that if f is continuous on [ a, b ] and if u( x ) and υ( x ) are differentiable functions of x whose values lie in [ a, b ], then d dx

υ( x )

∫u ( x )

f (t ) dt = f (υ( x ))

dυ du − f (u( x )) . dx dx

Prove the rule by setting g ( u, υ ) =

υ

∫u

f (t ) dt ,

u = u( x ),

υ = υ( x )

and calculating dg dx with the Chain Rule. y x

4. Finding a function with constrained second partials Suppose that f is a twice-differentiable function of r, that r = x 2 + y 2 + z 2 , and that f xx + f yy + f zz = 0.

Chapter 13  Additional and Advanced Exercises

Show that for some constants a and b,

is tangent to the surface

a f (r ) = + b. r

x 3 + y 3 + z 3 − xyz = 0

5. Homogeneous functions A function f ( x , y ) is homogeneous of degree n (n a nonnegative integer) if f ( tx , ty ) = t n f ( x , y ) for all t, x, and y. For such a function (sufficiently differentiable), prove that ∂f ∂f a. x + y = nf ( x , y ) ∂x ∂y b. x 2

( ∂∂ x f ) + 2xy⎛⎜⎜⎝ ∂∂x ∂f y ⎞⎟⎟⎟⎠ + y ⎛⎜⎜⎝ ∂∂ y f ⎞⎟⎟⎟⎠ = n n − 1 f . 2

2

2

2

2

857

(

2

)

6. Surface in polar coordinates Let ⎪⎧⎪ sin 6r , r ≠ 0 f ( r , θ ) = ⎪⎨ 6r ⎪⎪ r = 0, ⎪⎩1,

at ( 0, −1, 1). Extreme Values

11. Extrema on a surface Show that the only possible maxima and minima of z on the surface z = x 3 + y 3 − 9 xy + 27 occur at ( 0, 0 ) and ( 3, 3 ). Show that neither a maximum nor a minimum occurs at ( 0, 0 ). Determine whether z has a maximum or a minimum at ( 3, 3 ). 12. Maximum in closed first quadrant Find the maximum value of f ( x , y ) = 6 xye −( 2 x + 3 y ) in the closed first quadrant (includes the nonnegative axes). 13. Minimum volume cut from first octant Find the minimum volume for a region bounded by the planes x = 0, y = 0, z = 0 and a plane tangent to the ellipsoid

where r and θ are polar coordinates. Find a. lim f ( r , θ )

y2 x2 z2 + 2 + 2 =1 a2 b c

r→0

b. f r ( 0, 0 )

at a point in the first octant.

c. f θ ( r , θ ) , r ≠ 0.

14. Minimum distance from a line to a parabola in xy-plane By minimizing the function f ( x , y, u, υ ) = ( x − u ) 2 + ( y − υ ) 2 subject to the constraints y = x + 1 and u = υ 2, find the minimum distance in the xy-plane from the line y = x + 1 to the parabola y 2 = x.

z = f (r, u)

Theory and Examples

15. Boundedness of first partials implies continuity Prove the following theorem: If f ( x , y ) is defined in an open region R of the xy-plane and if f x and f y are bounded on R, then f ( x , y ) is continuous on R. (The assumption of boundedness is essential.)

Gradients and Tangents

7. Properties of position vectors Let r = xi + yj + zk and let r = r. a. Show that ∇r = r r . b. Show that ∇( r n ) = nr n−2 r. c. Find a function whose gradient equals r. d. Show that r ⋅ d r = r dr. e. Show that ∇( A ⋅ r ) = A for any constant vector A. 8. Gradient orthogonal to tangent Suppose that a differentiable function f ( x , y ) has the constant value c along the differentiable curve x = g(t ), y = h(t ); that is, f ( g(t ), h(t ) ) = c for all values of t. Differentiate both sides of this equation with respect to t to show that ∇f is orthogonal to the curve’s tangent vector at every point on the curve. 9. Curve tangent to a surface Show that the curve r (t ) = ( ln t ) i + ( t ln t ) j + tk is tangent to the surface xz 2 − yz + cos xy = 1 at ( 0, 0, 1). 10. Curve tangent to a surface Show that the curve r (t ) =

( t4 − 2 ) i + ( 4t − 3) j + cos t − 2 k 3

(

)

16. Suppose that r (t ) = g(t ) i + h(t ) j + k (t )k is a smooth curve in the domain of a differentiable function f ( x , y, z ). Describe the relation among df dt , ∇f , and v = d r dt . What can be said about ∇f and v at interior points of the curve where f has extreme values relative to its other values on the curve? Give reasons for your answer. 17. Finding functions from partial derivatives and g are functions of x and y such that ∂f ∂g = ∂y ∂x

and

Suppose that f

∂f ∂g , = ∂x ∂y

and suppose that ∂f = 0, ∂x

f (1, 2 ) = g (1, 2 ) = 5,

and

f ( 0, 0 ) = 4.

Find f ( x , y ) and g ( x , y ). 18. Rate of change of the rate of change We know that if f ( x , y ) is a function of two variables and if u = a i + b j is a unit vector, then D u f ( x , y ) = f x ( x , y ) a + f y ( x , y ) b is the rate of change of f ( x , y ) at ( x , y ) in the direction of u. Give a similar formula for the rate of change of the rate of change of f ( x , y ) at ( x , y ) in the direction u. 19. Path of a heat-seeking particle A heat-seeking particle has the property that at any point ( x , y ) in the plane, it moves in the direction of maximum temperature increase. If the temperature at ( x , y ) is T ( x , y ) = −e −2 y cos x , find an equation y = f ( x ) for the path of a heat-seeking particle at the point ( π 4 , 0 ).

858

Chapter 13 Partial Derivatives

20. Velocity after a ricochet A particle traveling in a straight line with constant velocity i + j − 5k passes through the point ( 0, 0, 30 ) and hits the surface z = 2 x 2 + 3 y 2 . The particle ricochets off the surface, the angle of reflection being equal to the angle of incidence. Assuming no loss of speed, what is the velocity of the particle after the ricochet? Simplify your answer. 21. Directional derivatives tangent to a surface Let S be the surface that is the graph of f ( x , y ) = 10 − x 2 − y 2 . Suppose that the temperature in space at each point (x, y, z) is T ( x , y, z ) = x 2 y + y 2 z + 4 x + 14 y + z. a. Among all the possible directions tangential to the surface S at the point ( 0, 0, 10 ), which direction will make the rate of change of temperature at ( 0, 0, 10 ) a maximum? b. Which direction tangential to S at the point (1, 1, 8 ) will make the rate of change of temperature a maximum? 22. Drilling another borehole On a flat surface of land, geologists drilled a borehole straight down and hit a mineral deposit at 300 m. They drilled a second borehole 30 m to the north of the first and hit the mineral deposit at 285 m. A third borehole 30 m east of the first borehole struck the mineral deposit at 307.5 m. The geologists have reasons to believe that the mineral deposit is in the

CHAPTER 13

shape of a dome, and for the sake of economy, they would like to find where the deposit is closest to the surface. Assuming the surface to be the xy-plane, in what direction from the first borehole would you suggest the geologists drill their fourth borehole? The one-dimensional heat equation If w ( x , t ) represents the temperature at position x at time t in a uniform wire with perfectly insulated sides, then the partial derivatives w xx and w t satisfy a differential equation of the form w xx =

1 w. c2 t

This equation is called the one-dimensional heat equation. The value of the positive constant c 2 is determined by the material from which the wire is made. 23. Find all solutions of the one-dimensional heat equation of the form w = e rt sin π x , where r is a constant. 24. Find all solutions of the one-dimensional heat equation that have the form w = e rt sin kx and satisfy the conditions that w ( 0, t ) = 0 and w ( L , t ) = 0. What happens to these solutions as t → ∞ ?

Technology Application Projects

Mathematica/Maple Projects

Projects can be found within MyLab Math. • Plotting Surfaces Efficiently generate plots of surfaces, contours, and level curves. • Exploring the Mathematics Behind Skateboarding: Analysis of the Directional Derivative The path of a skateboarder is introduced, first on a level plane, then on a ramp, and finally on a paraboloid. Compute, plot, and analyze the directional derivative in terms of the skateboarder. • Looking for Patterns and Applying the Method of Least Squares to Real Data Fit a line to a set of numerical data points by choosing the line that minimizes the sum of the squares of the vertical distances from the points to the line. • Lagrange Goes Skateboarding: How High Does He Go? Revisit and analyze the skateboarders’ adventures for maximum and minimum heights from both a graphical and analytic perspective using Lagrange multipliers.

14

Multiple Integrals

OVERVIEW In this chapter we define the double integral of a function of two variables

f ( x ,  y ) over a region in the plane as the limit of approximating Riemann sums. Just as a single integral can represent signed area, so can a double integral represent signed volume. Double integrals can be evaluated using the Fundamental Theorem of Calculus studied in Section 5.4, but now the evaluations are done twice by integrating with respect to each of the variables x and y in turn. Double integrals can be used to find areas of more general regions in the plane than those encountered in Chapter 5. Moreover, just as the Substitution Rule could simplify finding single integrals, we can sometimes use polar coordinates to simplify computing a double integral. We study more general substitutions for evaluating double integrals as well. We also define the triple integral of a function of three variables f ( x ,  y,  z ) over a region in space. Triple integrals can be used to find volumes of still more general regions in space, and their evaluation is like that of double integrals with yet a third evaluation. Cylindrical or spherical coordinates can sometimes be used to simplify the calculation of a triple integral, and we investigate those techniques. Double and triple integrals have a number of applications, such as calculating the average value of a multivariable function, and finding moments and centers of mass.

14.1 Double and Iterated Integrals over Rectangles In Chapter 5 we defined the definite integral of a function f ( x ) over an interval [ a,  b ] as a limit of Riemann sums. In this section we extend this idea to define the double integral of a function of two variables f ( x ,  y ) over a bounded rectangle R in the plane. The Riemann sums for the integral of a single-variable function f ( x ) are obtained by partitioning a finite interval into thin subintervals, multiplying the width of each subinterval by the value of f at a point c k inside that subinterval, and then adding together all the products. A similar method of partitioning, multiplying, and summing is used to construct double integrals as limits of approximating Riemann sums.

y d R

ΔAk Δyk

Double Integrals

(xk , yk ) Δxk

We begin our investigation of double integrals by considering the simplest type of planar region, a rectangle. We consider a function f ( x ,  y ) defined on a rectangular region R,

c 0

a

FIGURE 14.1

b

Rectangular grid partitioning the region R into small rectangles of area ΔAk = Δx k Δy k .

x

R: a ≤ x ≤ b , c ≤ y ≤ d . We subdivide R into small rectangles using a network of lines parallel to the x- and y-axes (Figure 14.1). The lines divide R into n rectangular pieces, where the number of such pieces n gets large as the width and height of each piece gets small. These rectangles form a partition of R. A small rectangular piece of width Δx and height Δy has area

859

860

Chapter 14 Multiple Integrals

ΔA = Δx Δy. If we number the small pieces partitioning R in some order, then their areas are given by numbers ΔA1 , ΔA2 , … , ΔAn , where ΔAk is the area of the kth small rectangle. To form a Riemann sum over R, we choose a point ( x k ,  y k ) in the kth small rectangle, multiply the value of f at that point by the area ΔAk , and add together the products: Sn =

n

∑ f ( x k ,  y k ) ΔAk . k =1

Depending on how we pick ( x k ,  y k ) in the kth small rectangle, we may get different values for S n . We are interested in what happens to these Riemann sums as the widths and heights of all the small rectangles in the partition of R approach zero. The norm of a partition P, written P , is the largest width or height of any rectangle in the partition. If P = 0.1, then all the rectangles in the partition of R have width at most 0.1 and height at most 0.1. Sometimes the Riemann sums converge as the norm of P goes to zero, which is written P → 0. The resulting limit is then written as n

∑ f ( x k ,  y k ) ΔAk . P →0 lim

k =1

As P → 0 and the rectangles get narrow and short, their number n increases, so we can also write this limit as n

∑ f ( x k ,  y k ) ΔAk ,

lim

n →∞

k =1

with the understanding that P → 0, and hence ΔAk → 0, as n → ∞. Many choices are involved in a limit of this kind. The collection of small rectangles is determined by the grid of vertical and horizontal lines that determine a rectangular partition of R. In each of the resulting small rectangles there is a choice of an arbitrary point ( x k ,  y k ) at which f is evaluated. These choices together determine a single Riemann sum. To form a limit, we repeat the whole process again and again, choosing partitions whose rectangle widths and heights both go to zero and whose number goes to infinity. When a limit of the sums S n exists, giving the same limiting value no matter what choices are made, then the function f is said to be integrable and the limit is called the double integral of f over R, which is written as

∫∫

f ( x ,  y ) dA

or

R

z = f(x, y)

f(xk , yk )

Double Integrals as Volumes d y

b R

(xk , yk )

f ( x ,  y ) dx dy.

R

It can be shown that if f ( x ,  y ) is a continuous function throughout R, then f is integrable, as in the single-variable case discussed in Chapter 5. Many discontinuous functions are also integrable, including functions that are discontinuous only on a finite number of points or smooth curves. We leave the proof of these facts to a more advanced text.

z

x

∫∫

ΔAk

FIGURE 14.2 Approximating solids with rectangular boxes leads us to define the volumes of more general solids as double integrals. The volume of the solid shown here is the double integral of f ( x ,   y ) over the base region R.

When f ( x ,  y ) is a positive function over a rectangular region R in the xy-plane, we may interpret the double integral of f over R as the volume of the three-dimensional solid region over the xy-plane bounded below by R and above by the surface z = f ( x ,  y ) (Figure 14.2). Each term f ( x k ,  y k ) ΔAk in the sum S n = ∑ f ( x k ,  y k ) ΔAk is the volume of a vertical rectangular box that approximates the volume of the portion of the solid that stands directly above the base ΔAk . The sum S n thus approximates what we want to call the total volume of the solid. We define this volume to be Volume = lim S n = n →∞

where ΔAk → 0 as n → ∞.

∫∫ R

f ( x ,   y ) dA,

14.1  Double and Iterated Integrals over Rectangles

861

As you might expect, this more general method of calculating volume agrees with the methods in Chapter 6, but we do not prove this here. Figure 14.3 shows Riemann sum approximations to the volume becoming more accurate as the number n of boxes increases.

(a) n = 16

(b) n = 64

(c) n = 256

FIGURE 14.3 As n increases, the Riemann sum approximations approach the total volume of the solid shown in Figure 14.2.

z

Fubini’s Theorem for Calculating Double Integrals Suppose that we wish to calculate the volume under the plane z = 4 − x − y over the rectangular region R: 0 ≤ x ≤ 2, 0 ≤ y ≤ 1 in the xy-plane. If we apply the method of slicing from Section 6.1, with slices perpendicular to the x-axis (Figure 14.4), then the volume is

4 z=4−x−y

x=2

∫x =0

A( x ) dx ,

(1)

where A( x ) is the cross-sectional area at x. For each value of x, we may calculate A( x ) as the integral 1 x

y

2 y=1

x

A(x) = (4 − x − y) dy Ly = 0

FIGURE 14.4

A( x ) =

To obtain the crosssectional area A( x ), we hold x fixed and integrate with respect to y.

y =1

∫ y= 0 ( 4 − x − y ) dy,

(2)

which is the area under the curve z = 4 − x − y in the plane of the cross-section at x. In calculating A( x ), x is held fixed and the integration takes place with respect to y. Combining Equations (1) and (2), we see that the volume of the entire solid is x=2

Volume =

∫x =0

=

∫x =0

x=2

A( x ) dx =

∫ x = 0 ( ∫ y= 0 ( 4 − x − y ) dy ) dx x=2

y =1

y 2 ⎤ y =1 ⎡ dx = ⎢ 4 y − xy − ⎥ 2 ⎦ y= 0 ⎣

x=2

∫x =0

( 72 − x ) dx

7 x2 ⎤2 = ⎡⎢ x − = 5. 2 ⎥⎦ 0 ⎣2 We often omit parentheses separating the two integrals in the formula above and write Volume =

2

1

∫0 ∫0 ( 4 − x − y ) dy dx.

(3)

The expression on the right, called an iterated or repeated integral, says that the volume is obtained by integrating 4 − x − y with respect to y from y = 0 to y = 1 while holding x fixed, and then integrating the resulting expression in x from x = 0 to x = 2. The limits of integration 0 and 1 are associated with y, so they are placed on the integral closest to dy. The other limits of integration, 0 and 2, are associated with the variable x, so they are placed on the outside integral symbol that is paired with dx.

862

Chapter 14 Multiple Integrals

z

What would have happened if we had calculated the volume by slicing with planes perpendicular to the y-axis (Figure 14.5)? As a function of y, the typical cross-sectional area is

4 z=4−x−y

A( y) =

x=2

∫x =0

(4

x=2 x2 − x − y ) dx = ⎢⎡ 4 x − − xy ⎥⎤ = 6 − 2 y. 2 ⎣ ⎦ x=0

(4)

The volume of the entire solid is therefore Volume =

1

⎤ ⎡ ⎢ 6 y − y 2 ⎥ = 5, ⎣ ⎦0

y

2

Volume =

x=2

x

y =1

∫ y= 0 ( 6 − 2 y ) dy =

A( y) dy =

in agreement with our earlier calculation. Again, we may give a formula for the volume as an iterated integral by writing

1 x

y =1

∫ y= 0

A(y) = (4 − x − y) dx Lx = 0

FIGURE 14.5 To obtain the crosssectional area A( y), we hold y fixed and integrate with respect to x.

1

2

∫0 ∫0 ( 4 − x − y ) dx dy.

The expression on the right says we can find the volume by integrating 4  x  y with respect to x from x  0 to x  2 as in Equation (4) and integrating the result with respect to y from y  0 to y  1. In this iterated integral, the order of integration is first x and then y, the reverse of the order in Equation (3). What do these two volume calculations with iterated integrals have to do with the double integral

∫∫

(4

− x − y ) dA

R

HISTORICAL BIOGRAPHY Guido Fubini (1879–1943) Fubini attended secondary school in Venice, Italy, where he showed that he was brilliant at mathematics. His advanced study was at the Scuola Normale Superiore di Pisa, where his doctoral thesis was in geometry. He then worked on harmonic functions in curved spaces. Fubini’s interests were wide ranging, from differential geometry to analysis and to the applications of differential equations. To know more, visit the companion Website.

over the rectangle R: 0 b x b 2, 0 b y b 1? The answer is that both iterated integrals give the value of the double integral. This is what we would reasonably expect, since the double integral measures the volume of the same region as the two iterated integrals. A theorem published in 1907 by Guido Fubini says that the double integral of any continuous function over a rectangle can be calculated as an iterated integral in either order of integration. (Fubini proved his theorem in greater generality, but this is what it says in our setting.)

THEOREM 1—Fubini’s Theorem (First Form)

If f ( x ,  y ) is continuous throughout the rectangular region R: a b x b b, c b y b d, then

∫∫ R

f ( x ,  y ) dA =

d

b

∫c ∫ a

f ( x ,   y ) dx dy =

b

∫ a ∫c

d

f ( x ,   y ) dy dx.

Fubini’s Theorem says that double integrals over rectangles can be calculated as iterated integrals. Thus, we can evaluate a double integral by integrating with respect to one variable at a time using the Fundamental Theorem of Calculus. Fubini’s Theorem also says that we may calculate the double integral by integrating in either order, a genuine convenience. When we calculate a volume by slicing, we may use either planes perpendicular to the x-axis or planes perpendicular to the y-axis. EXAMPLE 1

Calculate ∫∫R f ( x , y ) dA for

f ( x , y ) = 100 − 6 x 2 y

and

R: 0 ≤ x ≤ 2, −1 ≤ y ≤ 1.

14.1  Double and Iterated Integrals over Rectangles

z

z = 100 − 6x 2y

Solution Figure 14.6 displays the volume beneath the surface. By Fubini’s Theorem,

∫∫

100

f ( x , y ) dA =

R

=

50

2

x

2

1

2

∫−1 ∫0

(100 − 6 x 2 y ) dx dy =

1

∫−1 ( 200 − 16 y ) dy

1

R

1

2

∫0 ∫−1 (100 − 6 x 2 y ) dy dx = ∫0

y

The double integral ∫∫ R f ( x , y ) dA gives the volume under this surface over the rectangular region R (Example 1).

z = 10 + x 2 + 3y 2

∫0

=

∫0

2

y =1

⎡ ⎤ dx ⎢ 100 y − 3 x 2 y 2 ⎥ ⎣ ⎦ y =−1 [ (100 − 3 x 2 ) − ( −100 − 3 x 2 ) ] dx

200 dx = 400.

Solution The surface and volume are shown in Figure 14.7. The volume is given by the double integral

10

2

R

1

y

V = =

EXERCISES

=

1

⎡ ⎤ dx ⎢ 10 y + x 2 y + y 3 ⎥ ⎣ ⎦ y=0

1

1 2 86 ( 28 + 2 x 2 ) dx = ⎡⎢ 28 x + x 3 ⎤⎥ = .

∫0 ∫0

∫1 ∫0

3.

∫−1 ∫−1 ( x + y + 1) dx dy

2 xy dy dx

1

3

2

1

1

∫0 ∫0

( 4 − y 2 ) dy dx

y dx dy 1 + xy

7.

∫0 ∫0

9.

∫0 ∫1

ln 2

2

π2

∫−1 ∫0

⎣

13.

2

3

⎦0

3

∫0 ∫−1 ( x − y ) dy dx

4.

∫ 0 ∫ 0 (1 −

6.

1

1

3

0

4

4

1

2

)

x 2 + y2 dx dy 2

∫0 ∫−2 ( x 2 y − 2 xy ) dy dx ∫1 ∫0

e 2 x + y dy dx

10.

∫0 ∫1

y sin x dx dy

12.

2π

4

∫1 ∫1

ln x dx dy xy

e

14.

∫−1 ∫1

2

2

1

c

x ln y dy dx

1

2.

8.

ln 5

(10 + x 2 + 3 y 2 ) dy dx

14.1

1.

0

∫0 ∫0

y=2

In Exercises 1–14, evaluate the iterated integral. 4

2

(10 + x 2 + 3 y 2 ) dA =

Evaluating Iterated Integrals 2

1

∫∫ R

FIGURE 14.7 The double integral ∫∫ R f ( x ,   y ) dA gives the volume under this surface over the rectangular region R (Example 2).

11.

1

⎡ ⎤ = ⎢ 200 y − 8 y 2 ⎥ = 400. ⎣ ⎦ −1

EXAMPLE 2 Find the volume of the region bounded above by the elliptical paraboloid z = 10 + x 2 + 3 y 2 and below by the rectangle R: 0 ≤ x ≤ 1, 0 ≤ y ≤ 2.

z

5.

2

= FIGURE 14.6

x

x=2

⎡ ⎤ ∫−1 ⎢⎣100 x − 2 x 3 y ⎥⎦ x = 0 dy 1

Reversing the order of integration gives the same answer:

−1 1

863

π

∫π ∫0

( 2x + y ) dx dy xye x dy dx

( sin x + cos y ) dx dy

15. Find all values of the constant c so that ∫

0 ∫0

( 2x

+ y ) dx dy = 3.

16. Find all values of the constant c so that c

2

∫−1 ∫0 ( xy + 1) dy dx

= 4 + 4c .

Evaluating Double Integrals over Rectangles

In Exercises 17–24, evaluate the double integral over the given region R. 17.

∫∫ ( 6 y 2 − 2 x ) dA,

R: 0 ≤ x ≤ 1, 0 ≤ y ≤ 2

R

18.

⎛ x⎞

∫∫ ⎜⎜⎝ y 2 ⎟⎟⎟⎠ dA,  R

R: 0 ≤ x ≤ 4, 1 ≤ y ≤ 2

864

19.

Chapter 14 Multiple Integrals

∫∫

xy cos y dA,

∫∫

y sin ( x + y ) dA,

∫∫

e x − y dA,

∫∫

xye xy 2 dA,

R: 0 ≤ x ≤ 2, 0 ≤ y ≤ 1

∫∫

xy 3 dA, x2 + 1

R: 0 ≤ x ≤ 1, 0 ≤ y ≤ 2

31. Find the volume of the region bounded above by the plane z = 2 − x − y and below by the square R: 0 ≤ x ≤ 1, 0 ≤ y ≤ 1.

R: −1 ≤ x ≤ 1, 0 ≤ y ≤ π

R

20.

R: −π ≤ x ≤ 0, 0 ≤ y ≤ π

32. Find the volume of the region bounded above by the plane z = y 2 and below by the rectangle R: 0 ≤ x ≤ 4, 0 ≤ y ≤ 2.

R

21.

33. Find the volume of the region bounded above by the surface z = 2 sin x cos y and below by the rectangle R: 0 ≤ x ≤ π 2, 0 ≤ y ≤ π 4.

R: 0 ≤ x ≤ ln 2, 0 ≤ y ≤ ln 2

R

22.

34. Find the volume of the region bounded above by the surface z = 4 − y 2 and below by the rectangle R: 0 ≤ x ≤ 1, 0 ≤ y ≤ 2.

R

23.

R

24.

∫∫ R

y x 2y2

+1

35. Find a value of the constant k so that ∫

1

dA,

36. Evaluate∫

1

π2

−1 ∫ 0

R: 0 ≤ x ≤ 1, 0 ≤ y ≤ 1

x sin

3

∫0

kx 2 y dx dy = 1.

y dy dx.

37. Use Fubini’s Theorem to evaluate

In Exercises 25 and 26, integrate f over the given region. 25. Square f ( x , y ) = 1 ( xy ) 1≤ y ≤ 2

over

the

square

2

In Exercises 27 and 28, sketch the solid whose volume is given by the specified integral. 1

2

∫0 ∫0

( 9 − x 2 − y 2 ) dy dx

28.

3

∫0 ∫1

4

(7

1

x

∫0 ∫0 1 + xy dx dy.

1 ≤ x ≤ 2,

38. Use Fubini’s Theorem to evaluate

26. Rectangle f ( x , y ) = y cos xy over the rectangle 0 ≤ x ≤ π , 0 ≤ y ≤1

27.

2

1

3

∫0 ∫0

xe xy dx dy.

T 39. Use a software application to compute the integrals

− x − y ) dx dy

29. Find the volume of the region bounded above by the paraboloid z = x 2 + y 2 and below by the square R: −1 ≤ x ≤ 1, −1 ≤ y ≤ 1. 30. Find the volume of the region bounded above by the elliptical paraboloid z = 16 − x 2 − y 2 and below by the square R: 0 ≤ x ≤ 2, 0 ≤ y ≤ 2.

a.

1

2

∫0 ∫0

y−x dx dy ( x + y )3

b.

2

1

∫0 ∫0

y−x dy dx + y )3

(x

Explain why your results do not contradict Fubini’s Theorem. 40. If f ( x , y ) is continuous over R:  a ≤ x ≤ b,  c ≤ y ≤ d and F ( x, y ) =

x

∫ a ∫c

y

f ( u, υ ) d υ du

on the interior of R, find the second partial derivatives Fxy and Fyx .

14.2 Double Integrals over General Regions In this section we define and evaluate double integrals over bounded regions in the plane that are more general than rectangles. These double integrals are also evaluated as iterated integrals, with the main practical problem being that of determining the limits of integration. Since the region of integration may have boundaries other than line segments parallel to the coordinate axes, the limits of integration often involve variables, not just constants. ΔAk

R Δyk

Double Integrals over Bounded, Nonrectangular Regions (xk , yk )

Δxk

FIGURE 14.8 A rectangular grid partitioning a bounded, nonrectangular region into rectangular cells.

To define the double integral of a function f ( x , y ) over a bounded, nonrectangular region R, such as the one in Figure 14.8, we again begin by covering R with a grid of small rectangular cells whose union contains all points of R. This time, however, we cannot exactly fill R with a finite number of rectangles lying inside R since its boundary is curved, and some of the small rectangles in the grid lie partly outside R. A partition of R is formed by taking the rectangles that lie completely inside it, not using any that are either partly or completely outside. For commonly arising regions, more and more of R is included as the norm of a partition (the largest width or height of any rectangle used) approaches zero.

14.2  Double Integrals over General Regions

865

Once we have a partition of R, we number the rectangles in some order from 1 to n and let ΔAk be the area of the kth rectangle. We then choose a point ( x k , y k ) in the kth rectangle and form the Riemann sum Sn =

n

∑ f ( x k , y k ) ΔA k . k =1

As the norm of the partition forming S n goes to zero, P → 0, the width and height of each enclosed rectangle go to zero, their area ΔAk goes to zero, and their number goes to infinity. If f ( x , y ) is a continuous function, then these Riemann sums converge to a limiting value that is not dependent on any of the choices we made. This limit is called the double integral of f ( x , y ) over R: n

∑ f ( x k , y k ) ΔA k P →0

z = f(x, y)

z

lim

Height = f(xk, yk) 0 x y

R ΔAk

(xk, yk)

Volume = lim Σ f(xk, yk) Δ Ak =∫∫ f (x, y) dA R

z

A( x ) =

y = g2(x)

FIGURE 14.10

The area of the vertical slice shown here is A( x ). To calculate the volume of the solid, we integrate this area from x = a to x = b: A( x ) dx =

f ( x , y ) dy

1

y

y = g1(x)

R

∫a

y= g2 ( x )

∫ y= g ( x )

a

A(x)

b

R

If f ( x , y ) is positive and continuous over R, we define the volume of the solid region between R and the surface z = f ( x , y ) to be ∫∫R f ( x , y ) dA, as before (Figure 14.9). If R is a region like the one shown in the xy-plane in Figure 14.10, bounded “above” and “below” by the curves y = g 2 ( x ) and y = g1 ( x ) and on the sides by the lines x = a, x = b, we may again calculate the volume by the method of slicing. We first calculate the cross-sectional area

0

x

f ( x , y ) dA.

The nature of the boundary of R introduces issues not found in integrals over an interval. When R has a curved boundary, the n rectangles of a partition lie inside R but do not cover all of R. In order for a partition to approximate R well, the parts of R covered by small rectangles lying partly outside R must become negligible as the norm of the partition approaches zero. This property of being nearly filled in by a partition of small norm is satisfied by all the regions that we will encounter. There is no problem with boundaries made from polygons, circles, and ellipses or from continuous graphs over an interval, joined end to end. A curve with a “fractal” type of shape would be problematic, but such curves arise rarely in most applications. A careful discussion of which types of regions R can be used for computing double integrals is left to a more advanced text.

z = f (x, y) x

∫∫

Volumes

FIGURE 14.9 We define the volume of a solid with a curved base as a limit of the sums of volumes of approximating rectangular boxes.

b

=

k =1

b

g2 ( x )

∫a ∫ g ( x ) 1

f ( x , y ) dy dx.

and then integrate A( x ) from x = a to x = b to get the volume as an iterated integral: V =

b

∫a

A( x ) dx =

b

g2 ( x )

∫a ∫ g ( x )

f ( x , y ) dy dx.

(1)

1

Similarly, if R is a region like the one shown in Figure 14.11, bounded by the curves x = h 2 ( y) and x = h1 ( y) and the lines y = c and y = d, then the volume calculated by slicing is given by the iterated integral Volume =

d

h2 ( y )

∫c ∫ h ( y )

f ( x , y ) dx dy.

(2)

1

That the iterated integrals in Equations (1) and (2) both give the volume that we defined to be the double integral of f over R is a consequence of the following stronger form of Fubini’s Theorem.

866

Chapter 14 Multiple Integrals

z

THEOREM 2—Fubini’s Theorem (Stronger Form)

A( y)

z = f (x, y)

c

Let f ( x , y ) be continuous on a region R.

y d

y

1. If R is defined by a ≤ x ≤ b,  g1 ( x ) ≤ y ≤ g 2 ( x ), with g1 and g 2 continuous on [ a, b ], then

x

∫∫

x = h1(y)

f ( x , y ) dA =

FIGURE 14.11

The volume of the solid

∫c

A( y) dy =

d

h2 ( y )

∫c ∫ h ( y )

f ( x , y ) dy dx.

2. If R is defined by c ≤ y ≤ d ,  h1 ( y) ≤ x ≤ h 2 ( y), with h1 and h 2 continuous on [ c, d ], then

shown here is d

g2 ( x ) 1

R

x = h2(y)

b

∫a ∫ g ( x )

∫∫

f ( x , y ) dx dy.

f ( x , y ) dA =

d

h2 ( y )

∫c ∫ h ( y )

f ( x , y ) dx dy.

1

R

1

For a given solid, Theorem 2 says we can calculate the volume as in Figure 14.10 or in the way shown here. Both calculations have the same result.

Some iterated double integrals we will encounter later in this text will use variables of integration other than x and y. For instance, we may write u=q

υ = G 2 (u )

∫u = p ∫ υ = G ( u )

F ( u, υ ) d υ du =

1

q

G 2 (u )

∫ p ∫G ( u )

F ( u, υ ) d υ du.

1

Regardless of which specific variables of integration are used, the limits of an iterated double integral always satisfy these properties: • The limits of the outside integral are constants (they do not depend on either variable of integration), and • the limits of the inside integral are functions that may depend on the variable of the outside integral. EXAMPLE 1 Find the volume of the right prism whose base is the triangle in the xy-plane bounded by the x-axis and the lines y = x and x = 1 and whose top lies in the plane z = f ( x , y ) = 3 − x − y. Solution See Figure 14.12a. For any x between 0 and 1, y may vary from y = 0 to y = x (Figure 14.12b). Hence,

V = =

x

1

∫0 ∫0 1

∫0

(3

− x − y ) dy dx =

( 3x − 3x2 ) dx = ⎡⎢⎣ 3x2 2

2

1

∫0

−

y 2 ⎤ y= x ⎡ dx ⎢ 3 y − xy − ⎥ 2 ⎦ y= 0 ⎣

x 3 ⎤ x =1 = 1. 2 ⎥⎦ x = 0

When the order of integration is reversed (Figure 14.12c), the integral for the volume is V =

1

1

1

∫0 ∫ y ( 3 − x − y ) dx dy = ∫0

=

∫0 ( 3 − 2 − y − 3 y +

=

∫0 (

1

1

1

x =1 ⎡ 3 x − x 2 − xy ⎤ dy ⎢⎣ ⎥ 2 ⎦ x= y

)

y2 + y 2 dy 2

y 3 ⎤ y =1 5 3 ⎡5 = 1. − 4 y + y 2 dy = ⎢ y − 2 y 2 + ⎥ 2 2 2 ⎦ y= 0 ⎣2

)

The two integrals are equal, as they should be.

867

14.2  Double Integrals over General Regions

Although Fubini’s Theorem assures us that a double integral may be calculated as an iterated integral in either order of integration, the value of one integral may be easier to find than the value of the other. The next example shows how this can happen. z

y

x=1 y=x

(3, 0, 0) y=x z = f(x, y) = 3 − x − y R x

y=0 1

0 (1, 0, 2)

(b)

y

x=1 y=x

(1, 1, 1)

x=y

y

x=1

(1, 1, 0)

(1, 0, 0)

R

x=1

R x

0

y=x

1

x

(c)

(a)

FIGURE 14.12

(a) Prism with a triangular base in the xy-plane. The volume of this prism is defined as a double integral over R. To evaluate it as an iterated integral, we may integrate first with respect to y and then with respect to x, or the other way around (Example 1). (b) Integration limits of x =1

y= x

∫x = 0 ∫ y= 0

f ( x , y ) dy dx.

If we integrate first with respect to y, we integrate along a vertical line through R and then integrate from left to right to include all the vertical lines in R. (c) Integration limits of y =1

x =1

∫ y= 0 ∫x = y

f ( x , y ) dx dy.

If we integrate first with respect to x, we integrate along a horizontal line through R and then integrate from bottom to top to include all the horizontal lines in R.

EXAMPLE 2 y

Calculate

∫∫

x=1

R

y=x

1

where R is the triangle in the xy-plane bounded by the x-axis, the line y = x, and the line x = 1.

R 0

FIGURE 14.13

in Example 2.

1

sin x dA, x

x

The region of integration

Solution The region of integration is shown in Figure 14.13. If we integrate first with respect to y and next with respect to x, then because x is held fixed in the first integration, we find

∫0 ( ∫0 1

x

)

sin x dy dx = x

1

∫0

y= x

⎡ sin x ⎤ dx = ⎢y ⎥ x ⎦ y= 0 ⎣

1

∫0

sin x dx = − cos(1) + 1 ≈ 0.46.

868

Chapter 14 Multiple Integrals

y

If we reverse the order of integration and attempt to calculate x2 + y 2 = 1

1

1

1

x

(a) y Leaves at y = "1 − x 2

1 R

Finding Limits of Integration

Enters at y=1−x

We now give a procedure for finding limits of integration that applies for many regions in the plane. Regions that are more complicated, and for which this procedure fails, can often be split up into pieces on which the procedure works.

L x

0

x

1

(b)

y

Leaves at y = "1 − x 2

1 R

sin x dx dy, x

we run into a problem because ∫ (( sin x ) x ) dx cannot be expressed in terms of elementary functions (there is no simple antiderivative). There is no general rule for predicting which order of integration will be the good one in circumstances like these. If the order you first choose doesn’t work, try the other. Sometimes neither order will work, and then we may need to use numerical approximations.

x+y=1 0

1

∫0 ∫ y

R

Enters at y=1−x

Using Vertical Cross-Sections When faced with evaluating ∫∫ R f ( x , y ) dA, integrating first with respect to y and then with respect to x, do the following three steps:

1. Sketch. Sketch the region of integration and label the bounding curves (Figure 14.14a). 2. Find the y-limits of integration. Imagine a vertical line L cutting through R in the direction of increasing y. Mark the y-values where L enters and leaves. These are the y-limits of integration and are usually functions of x (instead of constants) (Figure 14.14b). 3. Find the x-limits of integration. Choose x-limits that include all the vertical lines through R. These must be constants. The integral whose region of integration is shown in Figure 14.14c is

L x

0 Smallest x is x = 0.

∫∫

x

1

x =1

R

Largest x is x = 1.

y = 1− x 2

∫ x = 0 ∫ y =1− x

f ( x , y ) dA =

f ( x , y ) dy dx.

Using Horizontal Cross-Sections To evaluate the same double integral as an iterated integral with the order of integration reversed, use horizontal lines instead of vertical lines in Steps 2 and 3 (see Figure 14.15). The integral is

(c)

FIGURE 14.14

Finding the limits of integration when integrating first with respect to y and then with respect to x.

∫∫

1

R

EXAMPLE 3

1− y 2

∫0 ∫1− y

f ( x , y ) dA =

f ( x , y ) dx dy.

Sketch the region of integration for the integral 2

2x

∫0 ∫ x

2

(4x

+ 2 ) dy dx

and write an equivalent integral with the order of integration reversed.

Largest y y is y = 1. 1

Enters at x=1−y R

y Smallest y is y = 0. 0

1

Leaves at x = "1 − y2 x

Solution The region of integration is given by the inequalities x 2 ≤ y ≤ 2 x and 0 ≤ x ≤ 2. It is therefore the region bounded by the curves y = x 2 and y = 2 x between x = 0 and x = 2 (Figure 14.16a). To find limits for integrating in the reverse order, we imagine a horizontal line passing from left to right through the region. It enters at x = y 2 and leaves at x = y. To include all such lines, we let y run from y = 0 to y = 4 (Figure 14.16b). The integral is 4

FIGURE 14.15

Finding the limits of integration when integrating first with respect to x and then with respect to y.

y

∫0 ∫ y 2 ( 4 x + 2 ) dx dy. The common value of these integrals is 8.

14.2  Double Integrals over General Regions

y

869

Properties of Double Integrals

4

Like single integrals, double integrals of continuous functions have algebraic properties that are useful in computations and applications.

(2, 4) y = 2x y = x2

If f ( x , y ) and g ( x , y ) are continuous on the bounded region R, then the following properties hold. 1. Constant Multiple: ∫∫ cf ( x , y ) dA = c ∫∫ f ( x , y ) dA ( any number  c ) R

x

2

0

R

2. Sum and Difference:

(a)

∫∫ ( f ( x ,  y ) ± g ( x ,  y )) dA = ∫∫

y

R

4

(2, 4)

f ( x , y ) dA ±

R

∫∫

g ( x , y ) dA

R

3. Domination: (a)

∫∫

f ( x , y ) dA ≥ 0

∫∫

f ( x , y ) dA ≥

f ( x , y ) ≥ 0 on R

if

R

y x= 2

x = "y

(b)

R

∫∫

g ( x , y ) dA

if

f ( x , y ) ≥ g ( x , y ) on R

R

4. Additivity: If R is the union of two nonoverlapping regions R1 and R 2, then 2 (b)

0

x

∫∫

f ( x , y ) dA =

R

FIGURE 14.16

∫∫ R1

f ( x , y ) dA +   ∫∫ f ( x , y ) dA R2

Region of integration for

Example 3.

Property 4 assumes that the region of integration R is decomposed into nonoverlapping regions R1 and R 2 with boundaries consisting of a finite number of line segments or smooth curves. Figure 14.17 illustrates an example of this property. The idea behind these properties is that integrals behave like sums. If the function f ( x , y ) is replaced by its constant multiple c f ( x , y ), then a Riemann sum for f ,

y

R1

Sn = R2 0 LL R

R = R1 ∪ R2 x

f(x, y) dA = f(x, y) dA + f (x, y) dA LL LL R2 R1

FIGURE 14.17

The Additivity Property for rectangular regions holds for regions bounded by smooth curves.

n

∑ f ( x k , y k ) ΔA k , k =1

is replaced by a Riemann sum for c f: n

∑ c f ( x k , y k ) ΔA k k =1

n

= c ∑ f ( x k , y k ) ΔAk = cS n . k =1

Taking limits as n → ∞ shows that c lim S n = c ∫∫R f dA and lim cS n = ∫∫R cf dA n →∞

n →∞

are equal. It follows that the Constant Multiple Property carries over from sums to double integrals. The other properties are also easy to verify for Riemann sums, and carry over to double integrals for the same reason. While this discussion gives the idea, an actual proof that these properties hold requires a more careful analysis of how Riemann sums converge. EXAMPLE 4 Find the volume of the wedgelike solid that lies beneath the surface z = 16 − x 2 − y 2 and above the region R bounded by the curve y = 2 x , the line y = 4 x − 2, and the x-axis. Solution Figure 14.18a shows the surface and the “wedgelike” solid whose volume we want to calculate. Figure 14.18b shows the region of integration in the xy-plane. If we integrate in the order dy dx (first with respect to y and then with respect to x), two integrations

870

Chapter 14 Multiple Integrals

z 16

will be required because y varies from y = 0 to y = 2 x for 0 ≤ x ≤ 0.5, and then varies from y = 4 x − 2 to y = 2 x for 0.5 ≤ x ≤ 1. So we choose to integrate in the order dx dy, which requires only one double integral whose limits of integration are indicated in Figure 14.18b. The volume is then calculated as the iterated integral:

z = 16 − x 2 − y 2

∫∫

(16 − x 2 − y 2 ) dA

R

2 1

y = 4x − 2

x

(a) y

y = 4x − 2

2 x=

y = 2" x

y2 4

x=

R 0

(1, 2) y+ 2 4

0.5

1

x

(b)

FIGURE 14.18

(a) The solid “wedgelike” region whose volume is found in Example 4. (b) The region of integration R showing the order dx dy.

EXERCISES

4

∫0 ∫ y

=

∫0

=

∫0 ⎢⎢⎣ 4 ( y + 2 ) −

y

y = 2" x

( y+ 2)

2

=

2

2 4

(16 − x 2 − y 2 ) dx dy

x = y+2 ⎡16 x − x 3 − xy 2 ⎤ ⎢ ⎥ 2 3 ⎣ ⎦ x=y / 4

2⎡

(

)/ 4

dx

( y + 2) y 2 + 2 )3 y6 y4 ⎤ − − 4y2 + + ⎥ dy 3 ⋅ 64 4 3 ⋅ 64 4 ⎥⎦

(y

y5 y7 ⎤ 2 63 y 2 145 y 3 49 y 4 20803 ⎡ 191y ≈ 12.4. = ⎢ + − − + + ⎥ = 32 96 768 20 1344 ⎦ 0 1680 ⎣ 24 Our development of the double integral has focused on its representation of the volume of the solid region between R and the surface z = f ( x , y ) of a positive continuous function. Just as we saw with signed area in the case of single integrals, when f ( x k , y k ) is negative, the product f ( x k , y k ) ΔAk is the negative of the volume of the rectangular box shown in Figure 14.9 that was used to form the approximating Riemann sum. So for an arbitrary continuous function f defined over R, the limit of any Riemann sum represents the signed volume (not the total volume) of the solid region between R and the surface. The double integral has other interpretations as well, and in the next section we will see how it is used to calculate the area of a general region in the plane.

14.2

Sketching Regions of Integration

Finding Limits of Integration

In Exercises 1–8, sketch the regions of integration associated with the given double integrals.

In Exercises 9–18, write an iterated integral for ∫∫R dA over the described region R using (a) vertical cross-sections, (b) horizontal cross-sections.

1. 2.

3

2x

2

x2

∫0 ∫0

9.

10. y

∫−1 ∫ x −1 f ( x , y ) dy dx 2

4

3.

∫−2 ∫ y

4.

∫0 ∫ y

5.

∫ 0 ∫e

1

2y

1

e

e2

ln x

1

arcsin y

∫1 ∫0

7.

∫0 ∫0 8

y y = 2x

y=8

x=3 x

f ( x , y ) dx dy f ( x , y ) dy dx

x

y=

x3

f ( x ,   y ) dx dy

2

6.

8.

f ( x , y ) dy dx

11.

12.

y

y y = ex

y = 3x

f ( x , y ) dy dx f ( x , y ) dx dy

x

y=1

y = x2 x

y1 3

∫0 ∫ y 4

f ( x , y ) dx dy

x x=2

14.2  Double Integrals over General Regions

13. Bounded by y =

x ,   y = 0, and x = 9

14. Bounded by y = tan x ,   x = 0, and y = 1 15. Bounded by y = e −x ,   y = 1, and x = ln 3 16. Bounded by y = 0,   x = 0,   y = 1, and y = ln x 17. Bounded by y = 3 − 2 x ,   y = x , and x = 0

47.

∫0 ∫0

21.

1

1

∫1 ∫0 ∫0 ∫ y

xy 3 dy dx

π

x2

∫0 ∫0

25.

∫1 ∫ y

4

∫3 ∫1

5

ex

1 dx dy 1 + y2

1 dy dx xy

In Exercises 27–32, sketch the region of integration and evaluate the integral. π

x

∫0 ∫0

29.

∫1 ∫1

31.

∫0 ∫0

x sin y dy dx

ln 8

1

ln y

y2

e x + y dx dy

3 y 3e xy dx dy

π

sin x

28.

∫0 ∫0

30.

∫1 ∫ y

32.

∫1 ∫0

2

y2

4

y dy dx

dx dy

x

3 y e 2

x

dy dx

In Exercises 33–36, integrate f over the given region. 33. Quadrilateral f ( x , y ) = x y over the region in the first quadrant bounded by the lines y = x ,   y = 2 x ,   x = 1, and x = 2 34. Triangle f ( x , y ) = x 2 + y 2 over the triangular region with vertices ( 0, 0 ), (1, 0 ), and ( 0, 1) 35. Triangle f ( u, υ ) = υ − u over the triangular region cut from the first quadrant of the uυ-plane by the line u + υ = 1 36. Curved region f ( s, t ) = e s ln t over the region in the first quadrant of the st-plane that lies above the curve s = ln t from t = 1 to t = 2 Each of Exercises 37–40 gives an integral over a region in a Cartesian coordinate plane. Sketch the region and evaluate the integral. 37. 38.

0

−υ

1

1− s 2

∫−2 ∫υ ∫0 ∫0

8t dt ds ( the  st -plane )

sec t

32

4−2u

∫−π 3 ∫0

40.

∫0 ∫1

ln x

3

ey

π

54.

∫0 ∫0

60.

∫0 ∫

61.

∫0 ∫ y

62.

∫0 ∫

ln 3

∫y 2

1 x 3

1 16

tan −1 y

3

x

2

2

2

4− x 2

xy dx dy

2 y 2 sin xy dy dx xe 2 y dy dx 4−y

e x 2 dx dy

e y 3 dy dx

12

2

6 x dy dx

12

∫0 ∫0

2 ln 3

4− x 2

π6

58.

∫0

3

∫0 ∫sin x xy 2 dy dx

x 2e xy dx dy

59.

8

52.

∫0 ∫ x

∫0 ∫ y

3

+ y ) dx dy

∫0 ∫−

56.

57.

1

xy dy dx

50.

sin y dy dx y

∫0 ∫ x 1

2

(x

π

3 y dx dy

14

cos(16π x 5 ) dx dy

dy dx y4 + 1

63. Square region ∫∫ R ( y − 2 x 2 ) dA where R is the region bounded by the square x + y = 1 64. Triangular region ∫∫ R xy dA where R is the region bounded by the lines y = x ,   y = 2 x , and x + y = 2 Volume Beneath a Surface z = f ( x, y )

65. Find the volume of the region bounded above by the paraboloid z = x 2 + y 2 and below by the triangle enclosed by the lines y = x ,   x = 0, and x + y = 2 in the xy-plane. 66. Find the volume of the solid that is bounded above by the cylinder z = x 2 and below by the region enclosed by the parabola y = 2 − x 2 and the line y = x in the xy-plane. 67. Find the volume of the solid whose base is the region in the xy-plane that is bounded by the parabola y = 4 − x 2 and the line y = 3 x, while the top of the solid is bounded by the plane z = x + 4.

2 dp d υ ( the  pυ -plane )

π3

39.

e

55.

Finding Regions of Integration and Double Integrals

27.

4− x 2

In Exercises 55–64, sketch the region of integration, reverse the order of integration, and evaluate the integral.

arctan y

26.

2

y dx dy

( y 2 − x ) dy dx

1

y dx dy x

y2

4− y 2

2

∫0 ∫1

x3

∫0 ∫0

dx dy

53.

2

24.

∫0 ∫0

y

y dx dy

∫1 ∫ y

x sin y dy dx

48.

2

∫1 ∫0

2y

∫0 ∫0

1− y 2

1

16 x dy dx

∫ 0 ∫e

51. 3

22.

( x + xy ) dx dy

23.

20.

9− 4 x 2

32

ln 2

46.

∫0 ∫− 1− y

In Exercises 19–26, evaluate the integral. 2x

dy dx

49.

Evaluating Iterated Integrals 2

ex

∫0 ∫1

18. Bounded by y = x 2 and y = x + 2

19.

1

45.

871

68. Find the volume of the solid in the first octant bounded by the coordinate planes, the cylinder x 2 + y 2 = 4, and the plane z + y = 3.

3 cos t du dt ( the  tu -plane ) 4 − 2u d υ du ( the  uυ -plane ) υ2

Reversing the Order of Integration

69. Find the volume of the solid in the first octant bounded by the coordinate planes, the plane x = 3, and the parabolic cylinder z = 4 − y 2.

In Exercises 41–54, sketch the region of integration, and write an equivalent double integral with the order of integration reversed.

70. Find the volume of the solid cut from the first octant by the surface z = 4 − x 2 − y.

1

4−2 x

1

y

41.

∫0 ∫2

43.

∫0 ∫ y

dy dx

dx dy

42.

∫0 ∫ y−2 dx dy

2

0

71. Find the volume of the wedge cut from the first octant by the cylinder z = 12 − 3 y 2 and the plane x + y = 2.

44.

∫0 ∫1− x

1

1− x 2

72. Find the volume of the solid cut from the square column x + y ≤ 1 by the planes z = 0 and 3 x + z = 3.

dy dx

872

Chapter 14 Multiple Integrals

73. Find the volume of the solid that is bounded on the front and back by the planes x = 2 and x = 1, on the sides by the cylinders y = ±1 x , and above and below by the planes z = x + 1 and z = 0.

85. Noncircular cylinder A solid right (noncircular) cylinder has its base R in the xy-plane and is bounded above by the paraboloid z = x 2 + y 2 . The cylinder’s volume is V =

74. Find the volume of the solid bounded on the front and back by the planes x = ±π 3, on the sides by the cylinders y = ± sec x, above by the cylinder z = 1 + y 2 , and below by the xy-plane. In Exercises 75 and 76, sketch the region of integration and the solid whose volume is given by the double integral. 75. 76.

3

2− 2 x 3

∫0 ∫0

(

16− y 2

4

∫0 ∫− 16− y

2

79. 80.

∞

1

∞

∞

∞

∞

86. Converting to a double integral Evaluate the integral 2

(Hint: Write the integrand as an integral.) 87. Maximizing a double integral What region R in the xy-plane maximizes the value of

78.

1

1

∫−1 ∫−1

1− x 2 1− x 2

( 2y

∫∫ ( 4 − x 2 − 2 y 2 ) dA ? R

Give reasons for your answer. 88. Minimizing a double integral minimizes the value of

R

Give reasons for your answer.

1

xe −( x + 2 y ) dx dy

Approximating Integrals with Finite Sums

In Exercises 81 and 82, approximate the double integral of f ( x, y ) over the region R partitioned by the given vertical lines x = a and horizontal lines y = c. In each subrectangle, use ( x k , y k ) as indicated for your approximation.

∫∫

f ( x , y ) dA ≈

R

89. Is it possible to evaluate the integral of a continuous function f ( x, y ) over a rectangular region in the xy-plane and get different answers depending on the order of integration? Give reasons for your answer. 90. How would you evaluate the double integral of a continuous function f ( x, y ) over the region R in the xy-plane enclosed by the triangle with vertices ( 0, 1), ( 2, 0 ), and (1, 2 )? Give reasons for your answer. 91. Unbounded region

n

∑ f ( x k , y k ) ΔA k

∞

82. f ( x, y ) = x + 2 y over the region R inside the circle ( x − 2 ) 2 + ( y − 3 ) 2 = 1 using the partition x = 1, 3 2, 2, 5 2, 3 and y = 2, 5 2, 3, 7 2, 4 with ( x k , y k ) the center (centroid) in the kth subrectangle (provided the subrectangle lies within R) Theory and Examples

83. Circular sector Integrate f ( x, y ) = 4 − x 2 over the smaller sector cut from the disk x 2 + y 2 ≤ 4 by the rays θ = π 6 and θ = π 2. 84. Unbounded region Integrate f ( x , y ) = 1 ⎡⎣ ( x 2 − x )( y − 1) 2 3 ⎤⎦ over the infinite rectangle 2 ≤ x < ∞,  0 ≤ y ≤ 2.

∞

Prove that

∫−∞ ∫−∞ e −x − y

k =1

81. f ( x, y ) = x + y over the region R bounded above by the semicircle y = 1 − x 2 and below by the x-axis, using the partition x = −1, −1 2, 0, 1 4, 1 2, 1 and y = 0, 1 2, 1 with ( x k , y k ) the lower left corner in the kth subrectangle (provided the subrectangle lies within R)

What region R in the xy-plane

∫∫ ( x 2 + y 2 − 9 ) dA ?

+ 1) dy dx

∫−∞ ∫−∞ ( x 2 + 1)( y 2 + 1) dx dy ∫0 ∫0

( x 2 + y 2 ) dx dy.

∫0 ( arctan π x − arctan x ) dx.

25 − x 2 − y 2 dx dy

1 dy dx x 3y

−x

2− y

2

Sketch the base region R, and express the cylinder’s volume as a single iterated integral with the order of integration reversed. Then evaluate the integral to find the volume.

)

Improper double integrals can often be computed similarly to improper integrals of one variable. The first iteration of the following improper integrals is conducted just as if they were proper integrals. One then evaluates an improper integral of a single variable by taking appropriate limits, as in Section 8.8. Evaluate the improper integrals in Exercises 77–80 as iterated integrals.

∫1 ∫e

y

1 1 1 − x − y dy dx 3 2

Integrals over Unbounded Regions

77.

1

∫0 ∫0 ( x 2 + y 2 ) dx dy + ∫1 ∫0

2

2

b

b

e −x − y b →∞ ∫−b ∫−b

dx dy = lim =4

(∫

∞ 0

2

e − x 2 dx

2

dx dy

). 2

92. Improper double integral Evaluate the improper integral 1

3

∫0 ∫0

(y

x2 dy dx. − 1) 2 3

COMPUTER EXPLORATIONS

Use a CAS double-integral evaluator to estimate the values of the integrals in Exercises 93–96. 3

x

1

1

1 dy dx xy

93.

∫1 ∫1

95.

∫0 ∫0 arctan xy dy dx

96.

∫−1 ∫0

1

1− x 2

94.

3 1 − x 2 − y 2 dy dx

1

1

∫0 ∫0 e −( x + y 2

2)

dy dx

14.3  Area by Double Integration

Use a CAS double-integral evaluator to find the integrals in Exercises 97–102. Then reverse the order of integration and evaluate, again with a CAS. 1

4

3

9

97.

∫0 ∫2 y

98.

∫0 ∫ x

99.

2

2

3

4− y 2

2

x2

2

8

∫0 ∫0

e x 2 dx dy

101.

∫1 ∫0

x cos ( y 2 ) dy dx

102.

∫1 ∫ y

4 2y

∫0 ∫ y

2

100.

873

e xy dx dy

1 dy dx x+ y 1 dx dy x2 + y2

3

( x 2 y − xy 2 ) dx dy

14.3 Area by Double Integration In this section we show how to use double integrals to calculate the areas of bounded regions in the plane, and to find the average value of a function of two variables.

Areas of Bounded Regions in the Plane If we take f ( x, y ) = 1 in the definition of the double integral over a region R in the preceding section, the Riemann sums reduce to Sn =

n

∑ f ( x k , y k ) ΔAk

=

k =1

n

∑ ΔA k .

(1)

k =1

This is simply the sum of the areas of the small rectangles in the partition of R, and it approximates what we would like to call the area of R. As the norm of a partition of R approaches zero, the height and width of all rectangles in the partition approach zero, and the coverage of R becomes increasingly complete (Figure 14.8). We define the area of R to be the limit n

∑ ΔA k P →0 lim

=

k =1

DEFINITION

∫∫

dA.

(2)

R

The area of a closed, bounded plane region R is A =

∫∫

dA.

R

As with the other definitions in this chapter, the definition here applies to a greater variety of regions than does the earlier single-variable definition of area, but it agrees with the earlier definition on regions to which they both apply. To evaluate the integral in the definition of area, we integrate the constant function f ( x, y ) = 1 over R. y 1

(1, 1)

y=x y=x

Solution We sketch the region (Figure 14.19), noting where the two curves intersect at the origin and (1, 1), and calculate the area as

y = x2

A =

y = x2 0

FIGURE 14.19

Example 1.

1

Find the area of the region R bounded by y = x and y = x 2 in the first

EXAMPLE 1 quadrant.

x

The region in

1

∫0 ∫ x

x 2

dy dx =

1

∫0

⎡ ⎤ y= x dx = ⎢ y⎥ ⎣ ⎦ y= x 2 1

1

∫0 ( x − x 2 ) dx

x2 x 3 ⎤1 1 = . = ⎡⎢ − 3 ⎥⎦ 0 6 ⎣ 2

Notice that the single-variable integral ∫0 ( x − x 2 ) dx, obtained from evaluating the inside iterated integral, is the integral for the area between these two curves using the method of Section 5.6.

874

Chapter 14 Multiple Integrals

y

EXAMPLE 2 Find the area of the region R enclosed by the parabola y = x 2 and the line y = x + 2.

y = x2 (2, 4)

y=x+2

4 "y 1 Ly – 2 L

dx dy

Solution If we divide R into the regions R1 and R 2 shown in Figure 14.20a, we may calculate the area as

(−1, 1)

1

R1

∫∫

A =

R2

∫∫

dA +

R1

"y

dx dy

L0 L–" y

x

0

dA =

R2

1

∫0 ∫−

y y

dx dy +

4

y

∫1 ∫ y−2 dx dy.

On the other hand, reversing the order of integration (Figure 14.20b) gives A =

(a)

2

∫−1 ∫ x

x+2 2

dy dx.

This second result, which requires only one integral, is simpler to evaluate, giving

y

y = x2

A = (2, 4)

y=x+2

2

x+2

2 L x −1 L

(−1, 1)

0

dy dx

⎡ ⎤ y= x+2 2

dx =

x

Calculating this area takes (a) two double integrals if the first integration is with respect to x, but (b) only one if the first integration is with respect to y (Example 2).

2

∫−1 ( x + 2 − x 2 ) dx

x2 x3 ⎤2 9 = ⎡⎢ + 2x − = . ⎣ 2 3 ⎥⎦ −1 2

EXAMPLE 3 Find the area of the playing field described by R: −2 ≤ x ≤ 2, −1 − 4 − x 2 ≤ y ≤ 1 + 4 − x 2 , using (a) Fubini’s Theorem

(b)

FIGURE 14.20

2

∫−1 ⎢⎣ y ⎥⎦ y = x

(b) simple geometry.

Solution The region R is shown in Figure 14.21a.

(a) From the symmetries observed in the figure, we see that the area of R is 4 times its area in the first quadrant. As shown in Figure 14.21b, a vertical line at x enters this part of the region at y = 0 and exits at y = 1 + 4 − x 2 . Therefore, using Fubini’s Theorem, we have A=

∫∫

dA = 4 ∫

R

= 4∫

2 0

(1 +

2

1+ 4 − x 2

0 ∫0

dy dx

4 − x 2 ) dx

x x 2 4 = 4 ⎡⎢ x + 4 − x 2 + sin −1 ⎤⎥ ⎣ 2 2 2⎦0 π = 4 2 + 0 + 2 ⋅ − 0 = 8 + 4 π. 2

(

Integral Table Formula 45

)

(b) The region R consists of a rectangle mounted on two sides by half disks of radius 2. The area can be computed by summing the area of the 4 × 2 rectangle and the area of a circle of radius 2, so A = 8 + π 2 2 = 8 + 4 π.

Average Value The average value of an integrable function of one variable on a closed interval is the integral of the function over the interval divided by the length of the interval. For an integrable function of two variables defined on a bounded region in the plane, the average value is the integral over the region divided by the area of the region. This can be visualized by thinking of the region as being the base of a tank with vertical walls around the boundary of the region, and imagining that the tank is filled with water that is sloshing around. The value

875

14.3  Area by Double Integration

f ( x, y ) is then the height of the water that is directly above the point ( x, y ). The average height of the water in the tank can be found by letting the water settle down to a constant height. This height is equal to the volume of water in the tank divided by the area of R. We therefore define the average value of an integrable function f over a region R as follows:

y 3 2

1 −2

0

2

x

Average value of  f  over  R =

1 area of  R

∫∫

f dA. 

(3)

R

−3 (a) y Leaves at y = 1 + "4 − x 2

3

If f is the temperature of a thin plate covering R, then the double integral of f over R divided by the area of R is the plate’s average temperature. If f ( x, y ) is the distance from the point ( x, y ) to a fixed point P, then the average value of f over R is the average distance of points in R from P. EXAMPLE 4 Find the average value of f ( x, y ) = x cos xy over the rectangle R: 0 ≤ x ≤ π , 0 ≤ y ≤ 1.

1

0

Solution The value of the integral of f over R is 2

Enters at y=0

π

x

1

∫0 ∫0

x cos xy dy dx =

(b)

=

FIGURE 14.21

(a) The playing field described by the region R in Example 3. (b) First quadrant of the playing field.

EXERCISES

∫0

π

π

∫0

⎡ ⎤ y =1 dx ⎢ sin xy ⎥ ⎢⎣ ⎥⎦ y = 0

∫ x cos xy dy = sin xy + C

⎤π ( sin x − 0 ) dx = − cos x ⎥ = 1 + 1 = 2. ⎥⎦ 0

The area of R is π. The average value of f over R is 2 π.

14.3

Area by Double Integrals

In Exercises 1–12, sketch the region bounded by the given lines and curves. Then express the region’s area as an iterated double integral and evaluate the integral.

curve with its equation, and give the coordinates of the points where the curves intersect. Then find the area of the region. 6

2y

13.

∫0 ∫ y

15.

∫0 ∫sin x

17.

∫−1 ∫−2 x

6. The curves y = ln x and y = 2 ln x and the line x = e, in the first quadrant

18.

∫0 ∫ x −4 dy dx + ∫0 ∫0

7. The parabolas x = y 2 and x = 2 y − y 2

Finding Average Values

1. The coordinate axes and the line x + y = 2 2. The lines x = 0,   y = 2 x, and y = 4 3. The parabola x =

−y 2

and the line y = x + 2

4. The parabola x = y − y 2 and the line y = −x 5. The curve y = e x and the lines y = 0,   x = 0, and x = ln 2

2 3

π4

dx dy

cos x

0

1− x

2

0

dy dx

dy dx +

2

1− x

4

x

3

x( 2− x )

2

y+2

14.

∫0 ∫−x

16.

∫−1 ∫ y

2

dy dx

dx dy

∫0 ∫−x 2 dy dx

2

dy dx

19. Find the average value of f ( x, y ) = sin ( x + y ) over

8. The parabolas x = y 2 − 1 and x = 2 y 2 − 2

a. the rectangle 0 ≤ x ≤ π, 0 ≤ y ≤ π .

9. The lines y = x,   y = x 3, and y = 2

b. the rectangle 0 ≤ x ≤ π, 0 ≤ y ≤ π 2.

10. The lines y = 1 − x and y = 2 and the curve y = e x

Identifying the Region of Integration

20. Which do you think will be larger, the average value of f ( x, y ) = xy over the square 0 ≤ x ≤ 1,  0 ≤ y ≤ 1, or the average value of f over the quarter circle x 2 + y 2 ≤ 1 in the first quadrant? Calculate them to find out.

The integrals and sums of integrals in Exercises 13–18 give the areas of regions in the xy-plane. Sketch each region, label each bounding

21. Find the average height of the paraboloid z = x 2 + y 2 over the square 0 ≤ x ≤ 2,  0 ≤ y ≤ 2.

11. The lines y = 2 x,   y = x 2, and y = 3 − x 12. The lines y = x − 2 and y = −x and the curve y =

x

876

Chapter 14 Multiple Integrals

22. Find the average value of f ( x, y ) = 1 ( xy ) over the square ln 2 ≤ x ≤ 2 ln 2, ln 2 ≤ y ≤ 2 ln 2.

station in each county. Assume that at time t 0 , each of the 254 weather stations recorded the local temperature. Find a formula that would give a reasonable approximation of the average temperature in Texas at time t 0 . Your answer should involve information that you would expect to be readily available in the Texas Almanac.

Theory and Examples

23. Geometric area Find the area of the region R: 0 ≤ x ≤ 2,  2 − x ≤ y ≤

4 − x2,

using (a) Fubini’s Theorem, (b) simple geometry. 24. Geometric area Find the area of the circular washer with outer radius 2 and inner radius 1, using (a) Fubini’s Theorem, (b) simple geometry. 25. Bacterium population If f ( x, y ) = (10, 000 e y ) (1 + x 2 ) represents the “population density” of a certain bacterium on the xy-plane, where x and y are measured in centimeters, find the total population of bacteria within the rectangle −5 ≤ x ≤ 5 and −2 ≤ y ≤ 0. 26. Regional population If f ( x , y ) = 100 ( y + 1) represents the population density of a planar region on Earth, where x and y are measured in kilometers, find the number of people in the region bounded by the curves x = y 2 and x = 2 y − y 2 .

28. If y = f ( x ) is a nonnegative continuous function over the closed interval a ≤ x ≤ b, show that the double integral definition of area for the closed plane region bounded by the graph of f , the vertical lines x = a and x = b, and the x-axis agrees with the definition for area beneath the curve in Section 5.3. 29. Suppose f ( x, y ) is continuous over a region R in the plane and that the area A( R) of the region is defined. If there are constants m and M such that m ≤ f ( x, y ) ≤ M for all ( x, y ) ∈ R, prove that mA( R ) ≤

∫∫

f ( x , y ) dA ≤ MA( R ).

R

30. Suppose f ( x, y ) is continuous and nonnegative over a region R in the plane with a defined area A( R). If ∫∫R f ( x, y ) dA = 0, prove that f ( x, y ) = 0 at every point ( x, y ) ∈ R.

27. Average temperature in Texas According to the Texas Almanac, Texas has 254 counties and a National Weather Service

14.4 Double Integrals in Polar Form Double integrals are sometimes easier to evaluate if we change to polar coordinates. This section shows how to accomplish the change and how to evaluate double integrals over regions whose boundaries are given by polar equations.

Integrals in Polar Coordinates When we defined the double integral of a function over a region R in the xy-plane, we began by cutting R into rectangles whose sides were parallel to the coordinate axes. These were the natural shapes to use because their sides have either constant x-values or constant y-values. In polar coordinates, the natural shape is a “polar rectangle” whose sides have constant r- and θ-values. To avoid ambiguities when describing the region of integration with polar coordinates, we use polar coordinate points ( r, θ ) where r ≥ 0. Suppose that a function f ( r, θ ) is defined over a region R that is bounded by the rays θ = α and θ = β and by the continuous curves r = g1 (θ) and r = g 2 (θ). Suppose also that 0 ≤ g1 (θ) ≤ g 2 (θ) ≤ a for every value of θ between α and β. Then R lies in a fanshaped region Q defined by the inequalities 0 ≤ r ≤ a and α ≤ θ ≤ β , where 0 ≤ β − α ≤ 2π. See Figure 14.22. We cover Q by a grid of circular arcs and rays. The arcs are cut from circles centered at the origin, with radii Δr , 2Δr , … , mΔr , where Δr = a m. The rays are given by θ = α,

θ = α + Δθ ,

θ = α + 2Δθ,

 … ,

θ = α + m′Δθ = β ,

where Δθ = ( β − α ) m′. The arcs and rays partition Q into small patches called “polar rectangles.” We number the polar rectangles that lie inside R (the order does not matter), calling their areas ΔA1 , ΔA2 , … , ΔAn . We let ( rk , θ k ) be any point in the polar rectangle whose area is ΔAk . We then form the sum Sn =

n

∑ f ( rk , θk ) ΔAk . k =1

877

14.4  Double Integrals in Polar Form

a + 2Δ u ΔAk

u =b

a + Δu

(rk , uk)

R

Δr Q

Δu

u =a

r = g2(u)

3Δr 2Δ r

r = g1(u)

Δr

u =p

r=a

u =0

0

FIGURE 14.22 The region R: g1 (θ ) ≤ r ≤ g 2 (θ ),   α ≤ θ ≤ β , is contained in the fan-shaped region Q: 0 ≤ r ≤ a,  α ≤ θ ≤ β , where 0 ≤ β − α ≤ 2π. The partition of Q by circular arcs and rays induces a partition of R.

If f is continuous throughout R, this sum will approach a limit as we refine the grid to make Δr and Δθ go to zero. The limit is the double integral of f over R. In symbols,

∫∫

lim S n =

n →∞

To evaluate this limit, we first have to write the sum S n in a way that expresses ΔAk in terms of Δr and Δθ. For convenience we choose rk to be the average of the radii of the inner and outer arcs bounding the kth polar rectangle ΔAk . The radius of the inner arc bounding ΔAk is then rk − ( Δr 2 ) (Figure 14.23). The radius of the outer arc is rk + ( Δr 2 ). The area of a wedge-shaped sector of a circle having radius r and central angle Δθ is

Δu Δr Δr ark − b 2

Δr ark + b 2 rk Δ Ak

Small sector

as can be seen by multiplying πr 2 , the area of the circle, by Δθ 2π, the fraction of the circle’s area contained in the wedge. So the areas of the circular sectors subtended by these arcs at the origin are

0

The observation that

⎛ area of ΔAk = ⎜⎜⎜ ⎝ large sector

⎞⎟ ⎛ area of ⎟⎟ − ⎜⎜ ⎠⎟ ⎜⎝ small sector

1 Δθ ⋅ r 2 , 2

A =

Large sector

FIGURE 14.23

f ( r , θ ) dA.

R

⎞⎟ ⎟⎟ ⎠

leads to the formula ΔAk = rk Δr Δθ.

(

)

2

(

)

2

Area of small sector:

1 Δr r − 2 k 2

Area of large sector:

1 Δr r + 2 k 2

Δθ Δθ.

Therefore, ΔAk = area of large sector − area of small sector =

Δθ ⎡ Δr ⎢ r + 2 ⎢⎣ k 2

(

) − (r 2

k

−

Δr 2

) ⎤⎥⎥⎦ = Δ2θ ( 2r 2

k

Δr ) = rk Δr Δθ.

Combining this result with the sum defining S n gives Sn =

n

∑ f ( rk , θk )rk Δr Δθ. k =1

As n → ∞ and the values of Δr and Δθ approach zero, these sums converge to the double integral lim S n =

n →∞

∫∫ R

f ( r , θ ) r dr d θ.

878

Chapter 14 Multiple Integrals

y

A version of Fubini’s Theorem says that the limit approached by these sums can be evaluated by repeated single integrations with respect to r and θ as

x 2 + y2 = 4

2 R

"2

∫∫

Q"2, "2R

y = "2

The procedure for finding limits of integration in rectangular coordinates also works for polar coordinates. We illustrate this using the region R shown in Figure 14.24. To evaluate ∫∫R f ( r , θ ) dA in polar coordinates, integrating first with respect to r and then with respect to θ, take the following steps.

y Leaves at r = 2 L 2 R Enters at r = " 2 csc u

u

x

0 (b) y

"2

∫∫ R

R

Smallest u is p . 4 x

Finding the limits of integration in polar coordinates. Enters at r = 1 Leaves at r = 1 + cos u L

u

u = −p 2

r =2 2 cos θ

f ( r , θ ) r dr d θ.

Solution

FIGURE 14.24

p 2

θ=π 2

∫θ = π 4 ∫r =

EXAMPLE 1 Find the limits of integration for integrating f ( r, θ ) over the region R that lies inside the cardioid r = 1 + cos θ and outside the circle r = 1.

(c)

u=

f ( r , θ ) dA =

y=x

0

y

1. Sketch. Sketch the region and label the bounding curves (Figure 14.24a). 2. Find the r-limits of integration. Imagine a ray L from the origin cutting through R in the direction of increasing r. Mark the r-values where L enters and leaves R. These are the r-limits of integration. They usually depend on the angle θ that L makes with the positive x-axis (Figure 14.24b). 3. Find the θ-limits of integration. Find the smallest and largest θ-values that bound R. These are the θ-limits of integration (Figure 14.24c). The polar iterated integral is

Largest u is p . 2 L

2

f ( r , θ ) r dr d θ.

Finding Limits of Integration

(a)

r sin u = y = " 2 or r = " 2 csc u

r = g 2 (θ ) 1

R

x

0

θ=β

∫θ = α ∫r = g ( θ )

f ( r , θ ) dA =

1

2

x

1. We first sketch the region and label the bounding curves (Figure 14.25). 2. Next we find the r-limits of integration. A typical ray from the origin enters R where r = 1 and leaves where r = 1 + cos θ. 3. Finally, we find the θ-limits of integration. The rays from the origin that intersect R run from θ = −π 2 to θ = π 2. The integral is π 2

1+ cos θ

∫−π 2 ∫1

f ( r , θ )r dr d θ.

If f ( r, θ ) is the constant function whose value is 1, then the integral of f over R is the area of R.

r = 1 + cos u

Area in Polar Coordinates

The area of a closed and bounded region R in the polar coordinate plane is

FIGURE 14.25

Finding the limits of integration in polar coordinates for the region in Example 1.

A =

∫∫

r dr d θ.

R

Area Differential in Polar Coordinates dA = r dr d θ

This formula for area is consistent with all earlier formulas.

14.4  Double Integrals in Polar Form

y p 4

Leaves at r = " 4 cos 2u

–p 4

Find the area enclosed by the lemniscate r 2 = 4 cos 2θ.

EXAMPLE 2

Solution We graph the lemniscate to determine the limits of integration (Figure 14.26) and see from the symmetry of the region that the total area is 4 times the first-quadrant portion.

x Enters at r=0

879

r 2 = 4 cos 2u

A = 4  ∫

FIGURE 14.26

To integrate over the shaded region, we run r from 0 to 4 cos 2θ and θ from 0 to π 4 (Example 2).

= 4  ∫

π 4 0 π 4 0

∫0

4 cos 2 θ

r dr d θ = 4 ∫

π 4 0

r = 4 cos 2 θ

⎡r2 ⎤ ⎢⎣ 2 ⎥⎦ r =0

dθ

⎤π 4 2 cos 2θ d θ = 4 sin 2θ ⎥ = 4. ⎥⎦ 0

Changing Cartesian Integrals into Polar Integrals The procedure for changing a Cartesian integral ∫∫R f ( x , y ) dx dy into a polar integral has two steps. First substitute x = r cos θ and y = r sin θ, and replace dx dy by r dr dθ in the Cartesian integral. Then supply polar limits of integration for the boundary of R. The Cartesian integral then becomes

∫∫

f ( x , y ) dx dy =

R

∫∫

f ( r cos θ, r sin θ ) r dr dθ,

G

where G denotes the same region of integration, but now described in polar coordinates. This is like the substitution method in Chapter 5 except that there are now two variables to substitute for instead of one. Notice that the area differential dx dy is replaced not by dr dθ but by r dr dθ. A more general discussion of changes of variables (substitutions) in multiple integrals is given in Section 14.8. y

1

EXAMPLE 3

Evaluate

y = "1 − x2 r=1

∫∫ e x

2 + y2

dy dx,

R

u=p −1

u=0 0

1

FIGURE 14.27

where R is the semicircular region bounded by the x-axis and the curve y = (Figure 14.27). x

The semicircular region in Example 3 is the region 0 ≤ r ≤ 1,

0 ≤ θ ≤ π.

1 − x2

Solution In Cartesian coordinates, the integral in question is a nonelementary integral and there is no direct way to integrate e x 2 + y 2 with respect to either x or y. Yet this integral and others like it are important in mathematics—in statistics, for example—and we need to evaluate it. Polar coordinates make this possible. Substituting x = r cos θ and y = r sin θ and replacing dy dx by r dr dθ give

∫∫ e x

2 + y2

dy dx =

R

=

π

1

∫0 ∫0 e r π

∫0

2

r dr d θ =

π

∫0

r =1

⎡ 1 er2 ⎤ dθ ⎢⎣ 2 ⎥⎦ r =0

1( π e − 1) d θ = ( e − 1). 2 2

The r in the r dr dθ is what allowed us to integrate e r 2. Without it, we would have been unable to find an antiderivative for the first (innermost) iterated integral.

EXAMPLE 4

Evaluate the integral 1

∫0 ∫0

1− x 2

( x 2 + y 2 ) dy dx.

880

Chapter 14 Multiple Integrals

Solution Integration with respect to y gives 1⎛

∫0 ⎜⎜⎝ x 2  

1 − x2  +  

(1 − x 2 )

3

32

⎞⎟ ⎟⎟ dx , ⎠

which is difficult to evaluate without tables. Things go better if we change the original integral to polar coordinates. The region of integration in Cartesian coordinates is given by the inequalities 0 ≤ y ≤ 1 − x 2 and 0 ≤ x ≤ 1, which correspond to the interior of the unit quarter circle x 2 + y 2 = 1 in the first quadrant. (See Figure 14.27, first quadrant.) Substituting the polar coordinates x = r cos θ,  y = r sin θ, 0 ≤ θ ≤ π 2, and 0 ≤ r ≤ 1, and replacing dy dx by r dr dθ in the double integral, we get 1

∫0 ∫0

1− x 2

( x 2 + y 2 ) dy dx =

=

π 2

1

∫0 ∫0 (r 2 ) r dr d θ π 2

∫0

r =1

⎡r4 ⎤ dθ = ⎢⎣ 4 ⎥⎦ r =0

π 2

∫0

π 1 dθ = . 4 8

The polar coordinate transformation is effective here because x 2 + y 2 simplifies to r 2 and the limits of integration become constants.

z z=9−

x2

−

y2

9

Solution The region of integration R is bounded by the unit circle x 2 + y 2 = 1, which is described in polar coordinates by r = 1, 0 ≤ θ ≤ 2π. The solid region is shown in Figure 14.28. The volume is given by the double integral 2π

1

2π

1

∫∫ ( 9 − x 2 − y 2 ) dA = ∫0 ∫0 ( 9 − r 2 ) r dr d θ

−2

r 2 = x 2 + y 2 , dA = r dr dθ.

R

R

2

x2 + y2 = 1

2

FIGURE 14.28

The solid region in

Example 5. y

y = " 3x (1, " 3) R

1 (" 3, 1) p 2 p 3 x + y2 = 4 6 1 2

FIGURE 14.29

Example 6.

∫0 ∫0 ( 9r − r 3 ) dr d θ

=

∫0

=

17 2 π 17π . dθ = 4 ∫0 2

2π

r =1 ⎡9r2 − 1r4 ⎤ dθ ⎢⎣ 2 ⎥ 4 ⎦ r =0

EXAMPLE 6 Using polar integration, find the area of the region R enclosed by the circle x 2 + y 2 = 4, above the line y = 1, and below the line y = 3x.

2 y = 1, or r = csc u

= y

x

0

EXAMPLE 5 Find the volume of the solid region bounded above by the paraboloid z = 9 − x 2 − y 2 and below by the unit circle in the xy-plane.

x

The region R in

Solution A sketch of the region R is shown in Figure 14.29. First we note that the line y = 3x has slope 3 = tan θ, so θ = π 3. Next we observe that the line y = 1 intersects the circle x 2 + y 2 = 4 when x 2 + 1 = 4, or x = 3. Moreover, the radial line from the origin through the point ( 3, 1) has slope 1 3 = tan θ, giving its angle of inclination as θ = π 6. This information is shown in Figure 14.29. Now, for the region R, as θ varies from π 6 to π 3, the polar coordinate r varies from the horizontal line y = 1 to the circle x 2 + y 2 = 4 . Substituting r sin θ for y in the equation for the horizontal line, we have r sin θ = 1, or r = csc θ, which is the polar equation of the line. The polar equation for the circle is r = 2. So in polar coordinates, for π 6 ≤ θ ≤ π 3, r varies from r = csc θ to r = 2. It follows that the iterated integral for the area is

14.4  Double Integrals in Polar Form

∫∫

dA =

R

EXERCISES

π3

2

∫π 6 ∫csc θ

881

r dr dθ r =2

π3

⎡1r2⎤ dθ ⎢⎣ 2 ⎥⎦ r = csc θ

π3

1 [ 4 − csc 2 θ ] d θ 2

=

∫π 6

=

∫π 6

=

⎤π 3 1⎡ ⎢ 4θ + cot θ ⎥ 2 ⎢⎣ ⎦⎥ π 6

=

1 4π 1 1 4π + − + 2 3 2 6 3

(

)

(

)

3 =

π− 3 . 3

14.4 7. The region enclosed by the circle x 2 + y 2 = 2 x

Regions in Polar Coordinates

In Exercises 1–8, describe the given region in polar coordinates. 1.

8. The region enclosed by the semicircle x 2 + y 2 = 2 y,   y ≥ 1

2. y

y

9

Evaluating Polar Integrals

In Exercises 9–22, change the Cartesian integral into an equivalent polar integral. Then evaluate the polar integral.

4 1

0

9

x

3.

0

4

x

4. y

y

∫−1 ∫0

10.

∫0 ∫0

11.

∫0 ∫0

12.

∫−a ∫−

13.

∫0 ∫0

15.

∫1 ∫1

17.

∫−1 ∫− 1− x

18.

∫−1 ∫− 1− x

1 "3

−1

0

1

x 0

5.

x

1

6. y

y 2

1− x 2

1

9.

dy dx

1

1− y 2

2

4− y 2

a

a 2 −x 2

6

1

ln 2

( x 2 + y 2 ) dx dy

a 2 −x 2

y

dy dx

x dx dy

3

0

( x 2 + y 2 ) dx dy

x

dy dx

0

2

2 (1 + x 2 + y 2 )

( ln 2 )2 − y 2

19.

∫0 ∫0

20.

∫−1 ∫− 1− y

21.

∫0 ∫ x

e

x 2 + y2

2

16.

∫ 2∫

dy dx  

dx dy

2 1 0

0 1

2" 3

1

2

x

1

1− y 2 2

ln ( x 2 + y 2 + 1) dx dy

x −2

1

2− x 2

(x

+ 2 y ) dy dx

x

∫0 ∫0

2 dy dx x 2 + y2

1+

1− x 2 2

2

14.

2

y dy dx

y 4− y 2

dx dy

882

22.

Chapter 14 Multiple Integrals

2

∫1 ∫0

2x−x 2

1 (x 2 + y2 )

2

38. Converting to a polar integral Integrate f ( x , y ) = [ ln ( x 2 + y 2 ) ] ( x 2 + y 2 ) over the region 1 ≤ x 2 + y 2 ≤ e 2 .

dy dx

In Exercises 23–26, sketch the region of integration, and convert each polar integral or sum of integrals into a Cartesian integral or sum of integrals. Do not evaluate the integrals. π2

1

π2

csc θ

π4

2 sec θ

23.

∫0 ∫0 r 3 sin θ cos θ dr d θ

24.

∫π 6 ∫1

25. 26.

arctan

∫0

4 3

40. Volume of noncircular right cylinder The region enclosed by the lemniscate r 2 = 2 cos 2θ is the base of a solid right cylinder whose top is bounded by the sphere z = 2 − r 2 . Find the cylinder’s volume.

r 2 cos θ dr d θ

∫0 ∫0

41. Converting to polar integrals r 5 sin 2 θ dr d θ

3 sec θ

∫0

39. Volume of noncircular right cylinder The region that lies inside the cardioid r = 1 + cos θ and outside the circle r = 1 is the base of a solid right cylinder. The top of the cylinder lies in the plane z = x. Find the cylinder’s volume.

r 7 dr d θ   +

a. The usual way to evaluate the improper integral ∞ I = ∫0 e −x 2 dx is first to calculate its square: π2

4 csc θ

∫arctan 4 ∫0

r 7 dr d θ

I2 =

3

Area in Polar Coordinates

27. Find the area of the region cut from the first quadrant by the curve r = 2( 2 − sin 2θ )1 2. 28. Cardioid overlapping a circle Find the area of the region that lies inside the cardioid r = 1 + cos θ and outside the circle r = 1. 29. One leaf of a rose Find the area enclosed by one leaf of the rose r = 12 cos 3θ. 30. Snail shell Find the area of the region enclosed by the positive x-axis and spiral r = 4θ 3,  0 ≤ θ ≤ 2π. The region looks like a snail shell.

(∫

∞ 0

e −x 2 dx

)( ∫

∞ 0

)

e − y 2 dy =

∞

∞

∫0 ∫0

e −( x

2 + y2 )

dx dy.

Evaluate the last integral using polar coordinates and solve the resulting equation for I. b. Evaluate lim erf( x ) = lim

x →∞

x →∞

x

∫0

2e −t 2 dt. π

42. Converting to a polar integral Evaluate the integral ∞

∞

∫0 ∫0

1 (1 + x 2 + y 2 )

2

dx dy.

31. Cardioid in the first quadrant Find the area of the region cut from the first quadrant by the cardioid r = 1 + sin θ.

43. Existence Integrate the function f ( x , y ) = 1 (1 − x 2 − y 2 ) over the disk x 2 + y 2 ≤ 3 4. Does the integral of f ( x , y ) over the disk x 2 + y 2 ≤ 1 exist? Give reasons for your answer.

32. Overlapping cardioids Find the area of the region common to the interiors of the cardioids r = 1 + cos θ and r = 1 − cos θ.

44. Area formula in polar coordinates Use the double integral in polar coordinates to derive the formula A =

Average Values

In polar coordinates, the average value of a function over a region R (Section 14.3) is given by 1 f ( r , θ ) r dr d θ. Area( R ) ∫∫ R

33. Average height of a hemisphere Find the average height of the hemispherical surface z = a 2 − x 2 − y 2 above the disk x 2 + y 2 ≤ a 2 in the xy-plane. 34. Average height of a cone Find the average height of the (single) cone z = x 2 + y 2 above the disk x 2 + y 2 ≤ a 2 in the xy-plane. 35. Average distance from interior of disk to center Find the average distance from a point P ( x , y ) in the disk x 2 + y 2 ≤ a 2 to the origin. 36. Average distance squared from a point in a disk to a point in its boundary Find the average value of the square of the distance from the point P ( x , y ) in the disk x 2 + y 2 ≤ 1 to the boundary point A(1, 0 ). Theory and Examples

37. Converting

to

[ ln ( x 2 + y 2 ) ]

a

polar

integral

Integrate

f ( x, y ) =

x 2 + y 2 over the region 1 ≤ x 2 + y 2 ≤ e.

β

∫α

1 2 r dθ 2

for the area of the fan-shaped region between the origin and the polar curve r = f (θ),   α ≤ θ ≤ β. 45. Average distance to a given point inside a disk Let P0 be a point inside a circle of radius a and let h denote the distance from P0 to the center of the circle. Let d denote the distance from an arbitrary point P to P0 . Find the average value of d 2 over the region enclosed by the circle. (Hint: Simplify your work by placing the center of the circle at the origin and P0 on the x-axis.) 46. Area Suppose that the area of a region in the polar coordinate plane is A =

3π 4

2 sin θ

∫π 4 ∫csc θ

r dr d θ.

Sketch the region and find its area. 47. Evaluate the integral ∫∫R x 2 + y 2 dA, where R is the region inside the upper semicircle of radius 2 centered at the origin, but outside the circle x 2 + ( y − 1) 2 = 1. 48. Evaluate the integral ∫∫R ( x 2 + y 2 )−2 dA, where R is the region inside the circle x 2 + y 2 = 2 for x ≤ −1.

14.5  Triple Integrals in Rectangular Coordinates

883

d. Change the integrand from Cartesian to polar coordinates. Determine the limits of integration from your plot in part (c) and evaluate the polar integral using the CAS integration utility.

COMPUTER EXPLORATIONS

In Exercises 49–52, use a CAS to change the Cartesian integrals into an equivalent polar integral and evaluate the polar integral. Perform the following steps in each exercise. a. Plot the Cartesian region of integration in the xy-plane. b. Change each boundary curve of the Cartesian region in part (a) to its polar representation by solving its Cartesian equation for r and R. c. Using the results in part (b), plot the polar region of integration in the rR-plane.

1

1

49.

∫0 ∫ x

51.

∫0 ∫−y 3

1

x2

y dy dx + y2

y 3

x2

y dx dy + y2

1

x 2

1

2− y

50.

∫0 ∫0

52.

∫0 ∫ y

x dy dx x 2 + y2 x + y dx dy

14.5 Triple Integrals in Rectangular Coordinates Just as double integrals allow us to deal with more general situations than could be handled by single integrals, triple integrals enable us to solve still more general problems. We use triple integrals to calculate the volumes of three-dimensional shapes and the average value of a function over a three-dimensional region. Triple integrals also arise in the study of vector fields and fluid flow in three dimensions, as we will see in Chapter 15. z

Triple Integrals If F ( x ,  y,  z ) is a function defined on a closed bounded solid region D in space, such as the region occupied by a solid ball or a lump of clay, then the integral of F over D may be defined in the following way. We partition a rectangular boxlike region containing D into rectangular cells by planes parallel to the coordinate axes (Figure 14.30). We number the cells that lie completely inside D from 1 to n in some order, the kth cell having dimensions %x k by %y k by %z k and volume ΔVk = Δx k Δy k Δz k . We choose a point ( x k ,  y k ,  z k ) in each cell and form the sum

(xk, yk, zk) Δzk

D Δyk

Δxk

x

y

Sn =

FIGURE 14.30

Partitioning a solid with rectangular cells of volume %Vk .

n

∑ F ( x k ,  y k ,  z k ) ΔVk .

(1)

k =1

We are interested in what happens as D is partitioned by smaller and smaller cells, so that %x k ,  %y k ,  %z k , and the norm of the partition P , the largest value among %x k ,  %y k ,  %z k , all approach zero. When a single limiting value is attained, no matter how the partitions and points ( x k ,  y k ,  z k ) are chosen, we say that F is integrable over D. As before, it can be shown that when F is continuous and the bounding surface of D is formed from finitely many smooth surfaces joined together along finitely many smooth curves, then F is integrable. In this case, as P l 0 and the number of cells n goes to d, the sums S n approach a limit. We call this limit the triple integral of F over D and write lim S n =

n →∞

∫∫∫

F ( x , y, z ) dV

or

D

lim S n =

P →0

∫∫∫

F ( x , y, z ) dx dy dz.

D

The regions D over which continuous functions are integrable are those having “reasonably smooth” boundaries.

Volume of a Solid Region in Space If F is the constant function whose value is 1, then the sums in Equation (1) reduce to Sn =

n

∑ F ( x k , y k , z k ) ΔV k k =1

=

n

∑ 1 ⋅ ΔV k k =1

=

n

∑ ΔV k . k =1

884

Chapter 14 Multiple Integrals

As Δx k ,  Δy k , and Δz k approach zero, the cells ΔVk become smaller and more numerous and fill up more and more of D. We therefore define the volume of D to be the triple integral n

∑ ΔV k n →∞

=

lim

k =1

DEFINITION

∫∫∫

dV .

D

The volume of a closed and bounded solid region D in space is V =

∫∫∫

dV.

D

This definition is in agreement with our previous definitions of volume, although we omit the verification of this fact. As we will see in a moment, this integral enables us to calculate the volumes of solids enclosed by curved surfaces. These are more general solids than the ones encountered before (Chapter 6 and Section 14.2).

Iterated Integrals We evaluate a triple integral by applying a three-dimensional version of Fubini’s Theorem (Section 14.2) to evaluate it by three repeated single integrations. As with double integrals, there is a geometric procedure for finding the limits of integration for these iterated integrals. To evaluate

∫∫∫

F ( x , y, z ) dV

D

over a solid region D, integrate first with respect to z, then with respect to y, and finally with respect to x. (You might choose a different order of integration, but the procedure is similar, as we illustrate in Example 2.) 1. Sketch. Sketch the solid region D along with its “shadow” R (vertical projection) in the xy-plane. Label the upper and lower bounding surfaces of D and the upper and lower bounding curves of R. z z = f2(x, y)

D z = f1(x, y) y a b x

R

14.5  Triple Integrals in Rectangular Coordinates

885

2. Find the z-limits of integration. Draw a line M passing through a typical point ( x , y ) in R parallel to the z-axis. As z increases, M enters D at z = f1 ( x , y ) and leaves at z = f 2 ( x , y ). These are the z-limits of integration. z M Leaves at z = f2(x, y) D

Enters at z = f1(x, y) y a

x

R

b

(x, y)

3. Find the y-limits of integration. Draw a line L through ( x , y ) parallel to the y-axis. As y increases, L enters R at y = g1 ( x ) and leaves at y = g 2 ( x ). These are the y-limits of integration. z M

D

Enters at y = g1(x) a

y

x b

(x, y)

x

R

L Leaves at y = g2(x)

4. Find the x-limits of integration. Choose x-limits that include all lines through R parallel to the y-axis (x = a and x = b in the preceding figure). These are the x-limits of integration. The integral is x=b

y= g2 ( x )

z = f 2( x ,  y )

∫ x = a ∫ y= g ( x ) ∫z = f 1

1( x ,  y )

F ( x , y, z ) dz dy dx.

Follow similar procedures if you change the order of integration. The “shadow” of the solid region D lies in the plane of the last two variables with respect to which the iterated integration takes place. The limits of an iterated triple integral satisfy these properties: • The limits of the outside integral are constants (they do not depend on any of the three variables of integration), • the limits of the middle integral are functions that may depend on the variable of the outside integral, and • the limits of the inside integral are functions that may depend on two variables: the middle integration variable and the outside integration variable.

886

Chapter 14 Multiple Integrals

The preceding procedure applies whenever a solid region D is bounded above and below by a surface, and when the “shadow” region R is bounded by a lower and upper curve. It does not apply to regions with more complicated shapes (such as regions containing holes); although, sometimes such regions can be subdivided into simpler regions for which the procedure does apply. We illustrate this method of finding the limits of integration in our first example. EXAMPLE 1 Let S be the sphere of radius 5 centered at the origin, and let D be the solid region under the sphere that lies above the plane z = 3. Set up the limits of integration for evaluating the triple integral of a function F ( x ,  y,  z ) over the region D. Solution The solid region under the sphere that lies above the plane z = 3 is enclosed by the surfaces x 2 + y 2 + z 2 = 25 and z = 3. To find the limits of integration, we first sketch the solid region, as shown in Figure 14.31. The “shadow region” R in the xy-plane is a circle of some radius centered at the origin. By considering a side view of the region D, we can determine that the radius of this circle is 4; see Figure 14.32a. If we fix a point ( x , y ) in R and draw a vertical line M above ( x , y ), then we see that this line enters the region D at the height z = 3 and leaves the region at the height z = 25 − x 2 − y 2; see Figure 14.31. This gives us the z-limits of integration. To find the y-limits of integration, we consider a line L that lies in the region R, passes through the point ( x , y ), and is parallel to the y-axis. For clarity we have separately pictured the region R and the line L in Figure 14.32b. The line L enters R when y = − 16 − x 2 and exits when y = 16 − x 2 . This gives us the y-limits of integration. Finally, as L sweeps across R from left to right, the value of x varies from x = −4 to x = 4. This gives us the x-limits of integration. Therefore, the triple integral of F over the region D is given by

∫∫∫

F ( x , y, z ) dz dy dx =

D

16 − x 2

4

∫−4 ∫− 16− x ∫3

25 − x 2 − y 2

2

F ( x , y, z ) dz dy dx.

z Surface

z=3

x2

+

y2

+

z2

= 25

M

Leaves at z = "25 − x2 − y2 Enters at z=3

D

R (x, y)

4

5

y

4 5 x

FIGURE 14.31

Finding the limits of integration for evaluating the triple integral of a function defined over the portion of the sphere of radius 5 that lies above the plane z = 3 (Example 1).

887

14.5  Triple Integrals in Rectangular Coordinates

y L

Leaves at y = "16 − x 2

4 z

R

(x, y)

5 D

−4

0

4

x

3

R −5 −4

0

4

5

−4

y

(a)

Enters at y = −"16 − x 2 (b)

FIGURE 14.32

(a) Side view of the solid region from Example 1, looking down the x-axis. The dashed right triangle has a hypotenuse of length 5 and sides of lengths 3 and 4. In this side view, the shadow region R lies between 4 and 4 on the y-axis. (b) The “shadow region” R shown face-on in the xy-plane.

z z=y−x M

The region D in Example 1 has a great deal of symmetry, which makes visualization easier. Even without symmetry, the steps in finding the limits of integration are the same, as shown in the next example.

C(0, 1, 1)

D x

B(0, 1, 0)

O

L

(x, y)

y=1

y=x

R

1

y

EXAMPLE 2 Set up the limits of integration for evaluating the triple integral of a function F ( x , y, z ) over the tetrahedron D whose vertices are O ( 0, 0, 0 ),  A(1, 1, 0 ), B( 0, 1, 0 ), and  C ( 0, 1, 1). Use the order of integration dz dy dx. Solution The solid region D and its “shadow” R in the xy-plane are shown in Figure 14.33a. The “top” face is contained in the plane through the points O, A, and C. Following the procedure introduced in Example 7 of Section 11.5, we first form a normal vector to that plane:

i

n = OA × OC = 1 0

A(1, 1, 0)

x (a)

j

k

1 1

0 = i − j + k, 1

and then use this vector and the coordinates of O to set up an equation for the plane:

y L

Leaves at y=1

1( x − 0 ) − 1( y − 0 ) + 1( z − 0 ) = 0 x − y + z = 0.

1 R (x, y) y=x Enters at y=x 0

1

x

(b)

FIGURE 14.33

(a) The tetrahedron in Example 2, showing how the limits of integration are found for the order dz dy dx. (b) The “shadow region” R shown face-on in the xy-plane.

The “side” face of D is parallel to the xz-plane, the “back” face lies in the yz-plane, and the “bottom” face is contained in the xy-plane. To find the z-limits of integration, fix a point ( x , y ) in the shadow region R, and consider the vertical line M that passes through ( x,   y ) and is parallel to the z-axis. This line enters D at the height z  0 , and it exits at height z = y − x. To find the y-limits of integration we again fix a point ( x , y ) in R, but now we consider a line L that lies in R, passes through ( x , y ), and is parallel to the y-axis. This line is shown in Figure 14.33a and also in the face-on view of R that is pictured in Figure 14.33b. The line L enters R when y  x and exits when y  1. Finally, as L sweeps across R, the value of x varies from x  0 to x  1. Therefore, the triple integral of F over the region D is given by

∫∫∫ D

F ( x , y, z ) dz dy dx =

1

1

∫0 ∫ x ∫0

y− x

F ( x , y, z ) dz dy dx.

888

Chapter 14 Multiple Integrals

In the next example we project the region D onto the xz-plane instead of the xy-plane, to show how to use a different order of integration. EXAMPLE 3 Find the volume of the tetrahedron D from Example 2 by integrating F ( x , y, z ) = 1 over the region using the order dz dy dx. Then do the same calculation using the order dy dz dx. Solution Using the limits of integration that we found in Example 2, we calculate the volume of the tetrahedron as follows:

V = =

1

1

1

1

∫0 ∫ x ∫0

y− x

dz dy dx

∫0 ∫ x ( y − x ) dy dx 1

∫0

=

∫0 ( 2 − x + 2 x 2 ) dx 1

1

1 1 1 1 = ⎡⎢ x − x 2 + x 3 ⎤⎥ ⎣2 2 6 ⎦0

= z

(0, 1, 1)

1

Line x+z=1

y=x+z

L

D R

y=1 (0, 1, 0)

y

M (x, z)

1

x Enters at y=x+z

Leaves at y=1 (1, 1, 0)

1 . 6

Finding the limits of integration for evaluating the triple integral of a function defined over the tetrahedron D (Example 3).

Integrate over y.

Evaluate.

Integrate over x.

Evaluate.

Now we will compute the volume using the order of integration dy dz dx. The procedure for finding the limits of integration is similar, except that we find the limits for y first, then for z, and then for x. The region D is the same tetrahedron as before, but now the “shadow region” R lies in the xz-plane, as shown in Figure 14.34. To find the y-limits of integration, we fix a point ( x , z ) in the shadow R and consider the line M that passes through ( x , z ) and is parallel to the y-axis. As shown in Figure 14.34, this line enters D when y = x + z, and it leaves when y = 1. Next we find the z-limits of integration. The line L that passes through a point ( x , z ) in R and is parallel to the z-axis enters R when z = 0 and exits when z = 1 − x (see Figure 14.34). Finally, as L sweeps across R, the value of x varies from x = 0 to x = 1. Therefore, the volume of the tetrahedron is V =

x

FIGURE 14.34

Integrate over z and evaluate.

y =1

⎡ 1 y 2 − xy ⎤ ⎢⎣ 2 ⎥⎦ y = x dx

=

1

Integrand is 1 when computing volume.

1

1− x

1

1− x

1

∫0 ∫0 ∫ x + z dy dz dx

=

∫0 ∫0

=

∫0

=

∫0

=

1 1( 1 − x ) 2 dx 2 ∫0

=−

(1

− x − z ) dz dx

1

z =1− x ⎡ (1 − x ) z − 1 z 2 ⎤ dx ⎢⎣ ⎥ 2 ⎦ z=0

1

⎡ (1 − x ) 2 − 1 (1 − x ) 2 ⎤ dx ⎢⎣ ⎥⎦ 2

1 1( 1 1 − x )3 ⎤⎥ = . ⎦0 6 6

Next we set up and evaluate a triple integral over a more complicated region.

14.5  Triple Integrals in Rectangular Coordinates

889

EXAMPLE 4 Find the volume of the solid region D enclosed by the surfaces z = x 2 + 3 y 2 and z =  8 − x 2 − y 2 . Solution The volume is

V =

∫∫∫

dz dy dx,

D

the integral of F ( x , y, z ) = 1 over D. To find the limits of integration for evaluating the integral, we first sketch the region. The surfaces (Figure 14.35) intersect on the elliptical cylinder x 2 + 3 y 2 = 8 − x 2 − y 2 or x 2 + 2 y 2 = 4,  z > 0. The boundary of the region R, the projection of D onto the xy-plane, is an ellipse with the same equation: x 2 + 2 y 2 = 4. The “upper” boundary of R is the curve y = ( 4 − x 2 ) 2. The lower boundary is the curve y = − ( 4 − x 2 ) 2.

M

z

Leaves at z = 8 − x 2 − y2

z = 8 − x2 − y2 The curve of intersection

D (−2, 0, 4) z = x2 + 3y2

(2, 0, 4) Enters at z = x2 + 3y2 Enters at y = −"(4 − x2)/2

(−2, 0, 0)

(2, 0, 0) x

x

(x, y)

x2 + 2y2 = 4

R Leaves at y = "(4 − x2)2

L

y

FIGURE 14.35

The volume of the region enclosed by two paraboloids, calculated in Example 4.

Now we find the z-limits of integration. The line M passing through a typical point ( x , y ) in R parallel to the z-axis enters D at z = x 2 + 3 y 2 and leaves at z = 8 − x 2 − y 2 . Next we find the y-limits of integration. The line L through ( x , y ) that lies parallel to the y-axis enters the region R when y = − ( 4 − x 2 ) 2 and leaves when y = ( 4 − x 2 ) 2. Finally, we find the x-limits of integration. As L sweeps across R, the value of x varies from x = −2 at (−2, 0, 0 ) to x = 2 at ( 2, 0, 0 ). The volume of D is V =

∫∫∫

dz dy dx

Integrand is 1 when computing volume.

D

=

2

( 4− x 2 ) 2

∫−2 ∫− ( 4− x

2) 2

8− x 2 − y 2

∫x

2 +3y 2

dz dy dx

Form an iterated integral.

890

Chapter 14 Multiple Integrals

( 4− x 2 ) 2

2

=

∫−2 ∫− ( 4− x

=

∫−2 ⎢⎣( 8 − 2 x 2 ) y − 3 y 3 ⎥⎦ y=− ( 4− x

=

⎛ 4 − x2 8 4 − x2 ∫−2 ⎜⎜⎜⎝ 2( 8 − 2 x 2 ) 2 − 3 2

=

∫−2 ⎢⎢⎣ 8(

=

4 2 2( 4 − x 2 )3 2 dx 3 ∫−2

2) 2

( 8 − 2 x 2 − 4 y 2 ) dy dx

⎡

2

⎤

4

y= ( 4− x 2 ) 2 2) 2

dx

(

2

⎡

2

= 8π 2.

4 − x2 2

)

32

−

(

8 4 − x2 3 2

)

32

Integrate over z and evaluate.

Integrate over y.

)

3 2⎞

⎟⎟ dx ⎟⎠

Evaluate.

⎤ ⎥ dx ⎦⎥

After integration with the substitution x = 2 sin θ

Average Value of a Function in Space The average value of a function F over a solid region D in space is defined by the formula Average   value  of  F  over  D =

1   F dV . volume of  D ∫∫∫

(2)

D

For example, if F ( x , y, z ) = x 2 + y 2 + z 2 , then the average value of F over D is the average distance of points in D from the origin. If F ( x , y, z ) is the temperature at ( x , y, z ) on a solid that occupies a region D in space, then the average value of F over D is the average temperature of the solid. EXAMPLE 5 Find the average value of F ( x , y, z ) = xyz throughout the cubical region D bounded by the coordinate planes and the planes x = 2,  y = 2, and z = 2 in the first octant. Solution We sketch the cube with enough detail to show the limits of integration (Figure 14.36). We then use Equation (2) to calculate the average value of F over the cube. The volume of the region D is ( 2 )( 2 )( 2 ) = 8. The value of the integral of F over the cube is

z

2

2

2

2

∫0 ∫0 ∫0

xyz dx dy dz = =

D y

2 x

in Example 5.

2

2

∫0

x=2

⎡ x 2 yz ⎤ dy dz = ⎢⎣ 2 ⎥⎦ x=0 y= 2

⎡ 2 ⎤ dz = ⎢ y z⎥ ⎣ ⎦ y= 0

2

∫0

2

2

∫0 ∫0

2 yz dy dz 2

⎡ ⎤ 4 z dz = ⎢ 2 z 2 ⎥ = 8. ⎣ ⎦0

With these values, Equation (2) gives 2

FIGURE 14.36

2

∫0 ∫0

The region of integration

Average value of 1 = volume xyz over the cube

∫∫∫ cube

xyz dV =

( 81 )(8 ) = 1.

In evaluating the integral, we chose the order dx dy dz, but any of the other five possible orders would have done as well.

Properties of Triple Integrals Triple integrals have the same algebraic properties as double and single integrals. Simply replace the double integrals in the four properties given in Section 14.2, page 864, with triple integrals.

14.5  Triple Integrals in Rectangular Coordinates

891

14.5

EXERCISES

Triple Integrals in Different Iteration Orders

Finding Equivalent Iterated Integrals

1. Evaluate the integral in Example 3, taking F ( x , y, z ) = 1 to find the volume of the tetrahedron in the order dz dx dy.

21. Here is the region of integration of the integral 1

dz dy dx.

z Top: y + z = 1 Side: 1 y = x2 −11 (−1, 1, 0)

4. Volume of solid Write six different iterated triple integrals for the volume of the solid region in the first octant enclosed by the cylinder x 2 + z 2 = 4 and the plane y = 3. Evaluate one of the integrals.

6. Volume inside paraboloid beneath a plane Let D be the solid region bounded by the paraboloid z = x 2 + y 2 and the plane z = 2 y. Write triple iterated integrals in the order dz dx dy and dz dy dx that give the volume of D. Do not evaluate either integral.

1− y

2

3. Volume of tetrahedron Write six different iterated triple integrals for the volume of the tetrahedron cut from the first octant by the plane 6 x + 3 y + 2 z = 6. Evaluate one of the integrals.

5. Volume enclosed by paraboloids Let D be the solid region bounded by the paraboloids z = 8 − x 2 − y 2 and z = x 2 + y 2 . Write six different triple iterated integrals for the volume of D. Evaluate one of the integrals.

1

∫−1 ∫ x ∫0

2. Volume of rectangular solid Write six different iterated triple integrals for the volume of the rectangular solid in the first octant bounded by the coordinate planes and the planes x = 1,   y = 2, and z = 3. Evaluate one of the integrals.

1

1 x

y

(1, 1, 0)

Rewrite the integral as an equivalent iterated integral in the order a. dy dz dx

b. dy dx dz

c. dx dy dz

d. dx dz dy

e. dz dx dy. 22. Here is the region of integration of the integral 1

0

∫0 ∫−1 ∫0

y2

dz dy dx

Evaluating Triple Iterated Integrals

Evaluate the integrals in Exercises 7–20. 1

1

7.

∫0 ∫0 ∫0 ( x 2 + y 2 + z 2 ) dz dy dx

8.

∫0 ∫0 ∫ x 1

8− x 2 − y 2

3y

2

3− 3 x

2 +3 y 2

3− 3 x − y

9.

10.

∫0 ∫0

12.

∫−1 ∫0 ∫0 ( x + y + z ) dy dx dz

13.

∫0 ∫0

15.

∫0 ∫0 ∫0

17.

1

1

π

2− x

π

π

e

18.

∫0 ∫1 ∫1

19.

∫0 ∫0

20.

π4

7

∫0

11.

e2

e

∫1 ∫1 ∫1 π6

1

e3

1 dx dy dz xyz

e

∫0 ∫0 ∫0

dz dy dx

dz dy dx

se s ln r  

3

∫0 ∫0 ∫−2 y sin z dx dy dz

2t

∫−∞ e x dx dt d υ

4−q 2

1 x

Rewrite the integral as an equivalent iterated integral in the order

2

14.

∫0 ∫−

16.

∫0 ∫0

( ln t ) 2 dt dr ds t

y

0 (1, −1, 0)

1

4− y 2 4− y 2

1− x 2

2x+ y

∫0

4− x 2 − y

∫3

dz dx dy

x dz dy dx

cos( u + υ + w ) du d υ dw ( uυw-space )

ln sec υ

2

9− x 2

2− x − y

∫0 ∫0 ∫0 1

dz dy dx

1 z = y2

(1, −1, 1)

2

9− x 2

3

1

∫0

dz dx dy

z

(0, −1, 1)

1

a. dy dz dx

b. dy dx dz

c. dx dy dz

d. dx dz dy

e. dz dx dy. Finding Volumes Using Triple Integrals

Find the volumes of the solid regions in Exercises 23–36. 23. The region between the cylinder z = y 2 and the xy-plane that is bounded by the planes x = 0,   x = 1,   y = −1,   y = 1 z

( rst -space )

( t υ x -space )

y

q dp dq dr ( pqr -space ) r +1

x

892

Chapter 14 Multiple Integrals

24. The region in the first octant bounded by the coordinate planes and the planes x + z = 1,   y + 2 z = 2 z

29. The region common to the interiors of the cylinders x 2 + y 2 = 1 and x 2 + z 2 = 1, one-eighth of which is shown in the accompanying figure z

y

x2 + y2 = 1

x

25. The region in the first octant bounded by the coordinate planes, the plane y + z = 2, and the cylinder x = 4 − y 2

x 2 + z2 = 1

z

y

y

x

30. The region in the first octant bounded by the coordinate planes and the surface z = 4 − x 2 − y x

z

26. The wedge cut from the cylinder x 2 + y 2 = 1 with z ≥ 0 by the planes z = − y and z = 0 z

y x

y

31. The region in the first octant bounded by the coordinate planes, the plane x + y = 4, and the cylinder y 2 + 4 z 2 = 16

x

27. The tetrahedron in the first octant bounded by the coordinate planes and the plane passing through (1, 0, 0 ), ( 0, 2, 0 ), and ( 0, 0, 3 )

z

z (0, 0, 3) y

(0, 2, 0) y

(1, 0, 0) x

x

32. The region cut from the cylinder x 2 + y 2 = 4 by the plane z = 0 and the plane x + z = 3 z

28. The region in the first octant bounded by the coordinate planes, the plane y = 1 − x , and the surface z = cos( π x 2 ),  0 ≤ x ≤ 1 z

y y x

x

14.6  Applications

33. The region between the planes 2 x + 2 y +   z = 4 in the first octant 34. The finite region bounded by x + z = 8,   z = y,   y = 8, and z = 0

x + y + 2z = 2 the

planes

and

z = x,

44.

2

4− x 2

∫0 ∫0

x

∫0

893

sin 2 z dy dz dx 4−z

Theory and Examples

45. Finding an upper limit of an iterated integral Solve for a:

35. The region cut from the solid elliptical cylinder x 2 + 4 y 2 ≤ 4 by the xy-plane and the plane z = x + 2

4−a− x 2

1

∫0 ∫0

36. The region bounded in back by the plane x = 0, on the front and sides by the parabolic cylinder x = 1 − y 2 , on the top by the paraboloid z = x 2 + y 2 , and on the bottom by the xy-plane

4− x 2 − y

∫a

dz dy dx =

4 . 15

46. Ellipsoid For what value of c is the volume of the ellipsoid x 2 + ( y 2 ) 2 + ( z c ) 2 = 1 equal to 8π ? 47. Minimizing a triple integral mizes the value of the integral

Average Values

In Exercises 37–40, find the average value of F ( x , y, z ) over the given region.

What domain D in space mini-

∫∫∫ ( 4 x 2 + 4 y 2 + z 2 − 4 ) dV ?

37. F ( x , y, z ) = x 2 + 9 over the cube in the first octant bounded by the coordinate planes and the planes x = 2,   y = 2, and z = 2

D

Give reasons for your answer.

38. F ( x , y, z ) = x + y − z over the rectangular box in the first octant bounded by the coordinate planes and the planes x = 1,   y = 1, and z = 2

48. Maximizing a triple integral What domain D in space maximizes the value of the integral

39. F ( x , y, z ) = x 2 + y 2 + z 2 over the cube in the first octant bounded by the coordinate planes and the planes x = 1,   y = 1, and z = 1

∫∫∫ (1 − x 2 − y 2 − z 2 ) dV ? D

Give reasons for your answer.

40. F ( x , y, z ) = xyz over the cube in the first octant bounded by the coordinate planes and the planes x = 2,   y = 2, and z = 2

COMPUTER EXPLORATIONS

Changing the Order of Integration

In Exercises 49–52, use a CAS integration utility to evaluate the triple integral of the given function over the specified solid region.

Evaluate the integrals in Exercises 41–44 by changing the order of integration in an appropriate way. 4

1

2

1

1

1

41.

∫0 ∫0 ∫2 y

42.

∫0 ∫0 ∫ x

43.

∫0 ∫ z ∫0

1

1

2

4 cos( x 2 ) dx dy dz 2 z

50. F ( x , y, z ) = xyz over the solid bounded below by the paraboloid z = x 2 + y 2 and above by the plane z = 1 z 51. F ( x , y, z ) = 3 2 over the solid bounded below (x 2 + y2 + z 2 ) by the cone z = x 2 + y 2 and above by the plane z = 1

12 xze zy 2 dy dx dz

ln 3

3

49. F ( x , y, z ) = x 2 y 2 z over the solid cylinder bounded by x 2 + y 2 = 1 and the planes z = 0 and z = 1

πe 2 x sin π y 2 dx dy dz y2

52. F ( x , y, z ) = x 4 + y 2 + z 2 over the solid sphere x 2 + y 2 + z2 ≤ 1

14.6 Applications z Δ mk = d(xk , yk , zk ) ΔVk

D

Masses and First Moments

(xk , yk , zk )

x

FIGURE 14.37

This section shows how to calculate the masses and moments of two- and threedimensional objects in Cartesian coordinates. The definitions and ideas are similar to the single-variable case we studied in Section 6.6, but now we can consider more general situations.

If δ ( x , y, z ) is the density (mass per unit volume) of an object occupying a solid region D in space, the integral of δ over D gives the mass of the object. To see why, imagine partitioning the object into n mass elements like the one in Figure 14.37. The object’s mass is the limit y

To define an object’s mass, we first imagine it to be partitioned into a finite number of mass elements Δm k .

n

∑ Δm k n →∞

M = lim

k =1

n

∑ δ ( x k , y k , z k ) ΔV k n →∞

= lim

k =1

=

∫∫∫

δ ( x , y, z ) dV .

D

The first moment of a solid region D about a coordinate plane is defined as the triple integral over D of the (signed) distance from a point ( x , y, z ) in D to the plane multiplied

894

Chapter 14 Multiple Integrals

by the density of the solid at that point. For instance, the first moment about the yz-plane is the integral M yz =

∫∫∫

x δ ( x , y, z ) dV .

D

The center of mass is found from the first moments. For instance, the x-coordinate of the center of mass is x = M yz M . For a two-dimensional object, such as a thin, flat plate, we calculate first moments about the coordinate axes by simply dropping the z-coordinate. So the first moment about the y-axis is the double integral over the region R forming the plate of the (signed) distance from the axis multiplied by the density, or My =

∫∫

x δ ( x , y ) dA.

R

Table 14.1 summarizes the formulas. TABLE 14.1 Mass and first moment formulas

THREE-DIMENSIONAL SOLID Mass:

M =

∫∫∫ δ dV

δ = δ ( x , y, z ) is the density at ( x , y , z ).

D

First moments about the coordinate planes: M yz =

∫∫∫

x δ dV ,

Mxz =

D

∫∫∫

y δ dV ,

Mxy =

D

x =

Center of mass:

M yz , M

y =

∫∫∫

z δ dV

D

Mxz , M

z =

Mxy M

TWO-DIMENSIONAL PLATE Mass:

M =

∫∫

δ dA

δ = δ ( x , y ) is the density at ( x , y ).

R

My =

First   moments:

∫∫ x δ dA,

Mx =

R

z

x =

Center   of   mass:

My , M

y =

∫∫ y δ dA R

Mx M

z = 4 − x2 − y2

EXAMPLE 1 Find the center of mass of a solid of constant density δ bounded below by the disk R: x 2 + y 2 ≤ 4 in the plane z = 0 and above by the paraboloid z = 4 − x 2 − y 2 (Figure 14.38).

c.m. R

Solution By symmetry x = y = 0. To find z , we first calculate

0 y x + 2

x

y2

FIGURE 14.38

=4

Finding the center of mass of a solid (Example 1).

Mxy =

z = 4− x 2 − y 2

∫∫ ∫z = 0

z δ dz dy dx =

R

=

δ 2 ( 4 − x 2 − y 2 ) dy dx 2 ∫∫ R

∫∫ R

z = 4− x 2 − y 2

⎡ z2 ⎤ ⎢ ⎥ ⎣ 2 ⎦ z=0

δ dy dx

14.6  Applications

=

δ 2π 2 ( 4 − r 2 ) 2 r dr d θ 2 ∫0 ∫0

=

δ 2π 2 ∫0

895

Polar coordinates simplify the integration.

r =2 16δ 2 π 32πδ ⎡ − 1   ( 4 − r 2 )3 ⎤ dθ = dθ = . ⎢⎣ 6 ⎥⎦ 3 ∫0 3 r =0

A similar calculation gives the mass: M =

4− x 2 − y 2

∫∫ ∫0

δ dz dy dx = 8πδ.

R

Therefore, z = ( Mxy M ) = 4 3 and the center of mass is ( x , y , z ) = ( 0, 0, 4 3 ). When the density of a solid object or plate is constant (as in Example 1), the center of mass is called the centroid of the object. To find a centroid, we set δ equal to 1 and proceed to find x , y , and z as before, by dividing first moments by masses. These calculations are also valid for two-dimensional objects. EXAMPLE 2 Find the centroid of the region in the first quadrant that is bounded above by the line y = x and below by the parabola y = x 2 .

y 1

Solution We sketch the region and include enough detail to determine the limits of integration (Figure 14.39). We then set δ equal to 1 and evaluate the appropriate formulas from Table 14.1:

(1, 1) y=x y = x2

0

x

1

FIGURE 14.39

The centroid of this region is found in Example 2.

1

∫0 ∫ x

x

1 dy dx = 2

∫0

1

x

∫0

M = Mx =

∫0 ∫ x

= My =

2

1

( x2

1

x

∫0

2

∫0 ∫ x

2

y dy dx = −

1

1

⎡ ⎤ y= x dx = ⎢ y⎥ ⎣ ⎦ y= x 2

1

∫0

1 ⎡ x2 − x3 ⎤ = 1 ⎥ 3 ⎦0 6 ⎣ 2

( x − x 2 ) dx = ⎢

y= x

⎡ y2 ⎤ ⎢ ⎥ 2 dx ⎣ 2 ⎦ y= x

)

x4 x3 x 5 ⎤1 1 dx = ⎡⎢ − ⎥ = 2 10 ⎦ 0 15 ⎣ 6

x dy dx =

1

∫0

⎡ ⎤ y= x dx = ⎢ xy ⎥ ⎣ ⎦ y= x 2

1

∫0

1 ⎡ x3 − x4 ⎤ = 1 . ⎥ 4 ⎦0 12 ⎣ 3

( x 2 − x 3 ) dx = ⎢

From these values of M , M x , and M y , we find x = yk

Δmk

and

y =

1 15 Mx 2 = = . M 16 5

The centroid is the point (1 2, 2 5 ).

rku

rk u

Note that each coordinate of the centroid of a region is equal to the average value of the corresponding variable over the region.

Axis of rotation L

FIGURE 14.40

My 1 12 1 = = M 16 2

To find an integral for the amount of energy stored in a rotating shaft, we first imagine the shaft to be partitioned into small blocks. Each block has its own kinetic energy. We add the contributions of the individual blocks to find the kinetic energy of the shaft.

Moments of Inertia An object’s first moments (Table 14.1) give us information related to balance and to the torque the object experiences about different axes in a gravitational field. If the object is a rotating shaft, we are interested in how much energy is stored in the shaft and how much energy is generated by a shaft rotating at a particular angular velocity. This is captured by the second moment or moment of inertia. Think of partitioning the shaft into small blocks of mass Δm k and let rk denote the distance from the kth block’s center of mass to the axis of rotation (Figure 14.40). If the

896

Chapter 14 Multiple Integrals

shaft rotates at a constant angular velocity of ω = dθ dt radians per second, the block’s center of mass will trace its orbit at a linear speed of d dθ ( r θ ) = rk = rk ω. dt k dt

υk =

The block’s kinetic energy will be approximately 1 1 1 Δ m k υ k 2 = Δ m k ( rk ω ) 2 = ω 2 rk 2 Δ m k . 2 2 2 The kinetic energy of the shaft will be approximately

∑ 12 ω 2rk 2 Δm k . The integral approached by these sums as the shaft is partitioned into smaller and smaller blocks gives the shaft’s kinetic energy: KE shaft =

1

∫ 2 ω 2r 2 dm

1 2 ω r 2 dm.  2 ∫

=

(1)

The factor I =

∫ r 2 dm

is the moment of inertia of the shaft about its axis of rotation, and we see from Equation (1) that the shaft’s kinetic energy is KE shaft =

The moment of inertia of a shaft resembles in some ways the inertial mass of a locomotive. To start a locomotive with mass m moving at a linear velocity υ, we need to provide a kinetic energy of KE = (1 2 ) mυ 2 . To stop the locomotive we have to remove this amount of energy. To start a shaft with moment of inertia I rotating at an angular velocity ω, we need to provide a kinetic energy of KE = (1 2 ) I ω 2 . To stop the shaft we have to take this amount of energy back out. The shaft’s moment of inertia is analogous to the locomotive’s mass. What makes the locomotive hard to start or stop is its mass. What makes the shaft hard to start or stop is its moment of inertia. The moment of inertia depends not only on the mass of the shaft but also on its distribution. Mass that is farther away from the axis of rotation contributes more to the moment of inertia. We now derive a formula for the moment of inertia for a solid in space. If r ( x , y, z ) is the distance from the point ( x , y, z ) in D to a line L, then the moment of inertia of the mass Δm k = δ ( x k , y k , z k ) ΔVk about the line L (as in Figure 14.40) is approximately ΔIk = r 2 ( x k , y k , z k ) Δm k . The moment of inertia about L of the entire object is

z

x "x2 + y2

IL = lim

n →∞

y

dV

"y2 + z2

x

FIGURE 14.41

the axes.

y

x

Distances from dV to

k =1

= lim

n

n →∞

Ix =

∑ r 2 ( x k , y k , z k ) δ ( x k , y k , z k ) ΔV k k =1

=

∫∫∫ r 2δ dV . D

∫∫∫ ( y 2 + z 2 ) δ ( x , y, z ) dV . D

y

z

n

∑ ΔIk

If L is the x-axis, then r 2 = y 2 + z 2 (Figure 14.41) and

"x2 + z2

0 x

1 2 Iω . 2

Similarly, if L is the y-axis or the z-axis, we have Iy =

∫∫∫ ( x 2 + z 2 ) δ ( x , y, z ) dV D

and

Iz =

∫∫∫ ( x 2 + y 2 ) δ ( x , y, z ) dV . D

14.6  Applications

897

Table 14.2 summarizes the formulas for these moments of inertia (second moments because they invoke the squares of the distances). It shows the definition of the polar moment about the origin as well. TABLE 14.2 Moments of inertia (second moments) formulas

THREE-DIMENSIONAL SOLID Ix =

About the x-axis:

∫∫∫ ( y 2 + z 2 ) δ dV

δ = δ ( x , y, z )

D

Iy =

About the y-axis:

∫∫∫ ( x 2 + z 2 ) δ dV D

Iz =

About the z-axis:

∫∫∫ ( x 2 + y 2 ) δ dV D

∫∫∫ r 2 ( x , y, z ) δ dV  

IL =

About a line L:

r ( x , y, z ) = distance from the point  ( x , y, z )  to line  L

D

TWO-DIMENSIONAL PLATE Ix =

About the x-axis:

∫∫ y 2  δ dA

δ = δ ( x, y)

R

Iy =

About the y-axis:

∫∫

x 2  δ dA

R

∫∫ r 2 ( x , y ) δ dA

IL =

About a line L:

r ( x , y ) = distance from  ( x , y )  to  L

R

About the origin (polar moment):

I0 =

∫∫ ( x 2 + y 2 ) δ dA =

Ix + I y

R

Find Ix ,   I y ,   Iz for the rectangular solid of constant density δ shown in

EXAMPLE 3 Figure 14.42.

Solution The formula for I x gives

z

Ix =

c2

b2

a2

∫−c 2 ∫−b 2 ∫−a 2 ( y 2 + z 2 ) δ dx dy dz.

We can avoid some of the work of integration by observing that ( y 2 + z 2 )δ is an even function of x, y, and z since δ is constant. The rectangular solid consists of eight symmetric pieces, one in each octant. We can evaluate the integral on one of these pieces and then multiply by 8 to get the total value:

c

y a x

Center of block

b

FIGURE 14.42

Finding Ix ,   Iy , and Iz for the block shown here. The origin lies at the center of the block (Example 3).

Ix = 8 ∫

c2 0

= 4 aδ ∫ = 4 aδ ∫

b2

a2

∫0 ∫0

( y 2 + z 2 ) δ dx dy dz = 4 aδ

y=b 2 y3 ⎤ dz + z 2y⎥ ⎢ ⎣ 3 ⎦ y=0

c 2⎡ 0 c2 0

b z b dz + ( 24 2 ) 3

2

c2

b2

∫0 ∫0

( y 2 + z 2 ) dy dz

898

Chapter 14 Multiple Integrals

= 4 aδ

δ( b ( b48c + c48b ) = abc 12 3

3

2

+ c2 ) =

M( 2 b + c 2 ). 12

M = abcδ

Similarly,

y

(1, 2)

2

Iy =

M( 2 a + c2 ) 12

Iz =

and

M 2 ( a + b 2 ). 12

y = 2x x=1

0

1

EXAMPLE 4 A thin plate covers the triangular region bounded by the x-axis and the lines x = 1 and y = 2 x in the first quadrant. The plate’s density at the point ( x , y ) is δ ( x , y ) = 6 x + 6 y + 6. Find the plate’s moments of inertia about the coordinate axes and the origin.

x

The triangular region covered by the plate in Example 4.

FIGURE 14.43

Solution We sketch the plate and put in enough detail to determine the limits of integration for the integrals we have to evaluate (Figure 14.43). The moment of inertia about the x-axis is

Ix = = Beam A

1

2x

∫0 ∫0 1

∫0

y 2 δ ( x , y ) dy dx =

1

2x

∫0 ∫0

( 6 xy 2 + 6 y 3 + 6 y 2 ) dy dx

y= 2x ⎡ 2 xy 3 + 3 y 4 + 2 y 3 ⎤ dx = ⎢⎣ ⎥ ⎦ y= 0 2

1

∫0 ( 40 x 4 + 16 x 3 ) dx

1

⎡ ⎤ = ⎢ 8 x 5 + 4 x 4 ⎥ = 12. ⎣ ⎦0 Axis

Similarly, the moment of inertia about the y-axis is Iy =

2x

x 2 δ ( x , y ) dy dx =

39 . 5

Notice that we integrate y 2 times density in calculating Ix , and x 2 times density to find I y . Since we know Ix and I y , we do not need to evaluate an integral to find I 0; we can use the equation I0 = Ix + Iy from Table 14.2 instead:

Beam B

Axis

The greater the polar moment of inertia of the cross-section of a beam about the beam’s longitudinal axis, the stiffer the beam. Beams A and B have the same cross-sectional area, but A is stiffer.

FIGURE 14.44

1

∫0 ∫0

I0 = 12 +

39 60 + 39 99 . = = 5 5 5

The moment of inertia also plays a role in determining how much a horizontal metal beam will bend under a load. The stiffness of the beam is a constant times I, the moment of inertia of a typical cross-section of the beam about the beam’s longitudinal axis. The greater the value of I, the stiffer the beam and the less it will bend under a given load. That is why we use I-beams instead of beams whose cross-sections are square. The flanges at the top and bottom of the beam hold most of the beam’s mass away from the longitudinal axis to increase the value of I (Figure 14.44).

Probability The probability that a continuous random variable X takes values between a and b is found by integrating a probability density function f (Appendix A.8), P(a ≤ X ≤ b) =

b

∫a

f ( x ) dx.

A similar process applies to probabilities involving two continuous random variables. The probability that a pair of random variables ( X , Y ) takes values lying within a particular

14.6  Applications

899

region is determined by a joint probability density function f . Integrating the joint probability density function over a region R in the plane gives the probability that the pair of random variables take values in that region: P (( X , Y ) ∈ R ) =

∫∫

f ( x , y ) dx dy.

R

If the region is a rectangle, then this expression has the simple form P ( a ≤ X ≤ b  and c ≤ Y ≤ d ) =

d

b

∫c ∫ a

f ( x , y ) dx dy.

A joint probability density function f is defined by three basic properties. The first property ensures that there are no negative probabilities, and the second implies that the total probability of all possible outcomes is one. The final property describes the connection of f to probabilities.

DEFINITION A joint probability density function f is a function that satisfies

three conditions: 1. f ( x , y ) ≥ 0 2.

∞

∞

∫−∞ ∫−∞ f ( x , y ) dx dy = 1

3. P (( X, Y ) ∈ R ) =

∫∫

f ( x , y ) dx dy.

R

A pair of random variables has a uniform distribution on a region R with finite area A if f ( x , y ) = 1 A for any ( x , y ) ∈ R, and f ( x , y ) = 0 otherwise. EXAMPLE 5 A random number generator is used to generate two random real numbers X and Y in succession. The first number X is chosen between 0 and 10, and the second number Y is chosen between 0 and 5. The random number generation is done by a process that gives a uniform distribution. Find the joint probability density function f for the pair of numbers ( X , Y ) and use it to compute the probability that X is larger than Y. Solution The joint probability density function f is constant on the rectangle 0 ≤ x ≤ 10, 0 ≤ y ≤ 5, because ( X , Y ) is uniformly distributed. The area of the rectangle is 50, so f takes the value 1 50 inside this rectangle:

⎧⎪ 1 50, f ( x , y ) = ⎪⎨ ⎪⎪ 0, ⎩

y 5

0

5

FIGURE 14.45

10

x

The pair of random variables X and Y take values anywhere in this rectangle with equal probability. In the shaded region we have X > Y .

if 0 ≤ x ≤ 10 and 0 ≤ y ≤ 5, otherwise.

To compute the probability that X > Y , we integrate the joint probability density function f over the region in the rectangle where X > Y . This region is bounded on the left by the line x = y and on the right by the line x = 10. An integral over this region has limits of integration given by y ≤ x ≤ 10, 0 ≤ y ≤ 5 (see Figure 14.45). The probability is given by P( X > Y ) =

5

10

∫0 ∫ y

1 3 dx dy = . 50 4

There is a 75% probability that the first number is larger than the second.

900

Chapter 14 Multiple Integrals

Using the joint probability density function

EXAMPLE 6

⎧⎪ e −( x + y ) , f ( x , y ) = ⎪⎨ ⎪⎪ 0, ⎩

if 0 < x  and 0 < y otherwise

find the probability that 1 < X < 2 and 2 < Y < 3. Solution

P (1 < X < 2,  2 < Y < 3) =

3

2

∫2 ∫1 e − x + y (

)

dx dy = e −5 + e −3 − 2e −4 ≈ 0.01989.

There is slightly less than a 2% probability that X and Y fall within these bounds.

Means and Expected Values The mean, or expected value, of a random variable is (Appendix A.8) μ =

∞

∫−∞ x  f ( x ) dx.

When X and Y have joint probability density function f , the expected value of X and the expected value of Y are μX =

∞

∞

∫−∞ ∫−∞ x  f ( x , y ) dx dy

and

μY =

∞

∞

∫−∞ ∫−∞ y  f ( x , y ) dx dy.

These indicate the average value expected for each of X and Y. The expected values μ X and μY are sometimes called the first moments of the distribution, because their defining formulas have the same form as those seen in Table 14.1 for the moments of a twodimensional plate. The joint probability density function plays the role in computing μ X that the mass density function plays in computing the x-coordinate x of the center of mass, and the same applies to μY and y . One can roughly think of the joint probability density function as measuring the probability concentration per unit area on the plane, just as density measures the mass per unit area for a plate. EXAMPLE 7 Find the expected values μ X and μY for the joint probability density function in Example 5. Solution For the joint probability density function in Example 5, we compute

μX =

5

10

∫0 ∫0

x (1 50 ) dx dy = 5

and μY =

5

10

∫0 ∫0

y (1 50 ) dx dy = 2.5.

The expected value of X is 5 and that of Y is 2.5.

EXERCISES

14.6

Plates of Constant Density

1. Finding a center of mass Find the center of mass of a thin plate of density δ = 3 bounded by the lines x = 0,   y = x , and the parabola y = 2 − x 2 in the first quadrant. 2. Finding moments of inertia Find the moments of inertia about the coordinate axes of a thin rectangular plate of constant density

δ gm cm 2 bounded by the lines x = 3 and y = 3 in the first quadrant. 3. Finding a centroid Find the centroid of the region in the first quadrant bounded by the x-axis, the parabola y 2 = 2 x , and the line x + y = 4.

14.6  Applications

4. Finding a centroid Find the centroid of the triangular region cut from the first quadrant by the line x + y = 3. 5. Finding a centroid Find the centroid of the region cut from the first quadrant by the circle x 2 + y 2 = a 2 . 6. Finding a centroid Find the centroid of the region between the x-axis and the arch y = sin x ,  0 ≤ x ≤ π. 7. Finding moments of inertia Find the moment of inertia about the x-axis of a thin plate of density δ = 1 gm cm 2 bounded by the circle x 2 + y 2 = 4. Then use your result to find I y and I 0 for the plate. 8. Finding a moment of inertia Find the moment of inertia with respect to the y-axis of a thin sheet of constant density δ = 1 gm cm 2 bounded by the curve y = ( sin 2 x ) x 2 and the interval π ≤ x ≤ 2π of the x-axis. 9. The centroid of an infinite region Find the centroid of the infinite region in the second quadrant enclosed by the coordinate axes and the curve y = e x . (Use improper integrals in the massmoment formulas.) 10. The first moment of an infinite plate Find the first moment about the y-axis of a thin plate of density δ ( x , y ) = 1 covering the 2 infinite region under the curve y = e −x 2 in the first quadrant.

Solids with Constant Density

21. Moments of inertia Find the moments of inertia of the rectangular box of constant density δ ( x , y, z ) = 1 shown here with respect to its edges by calculating Ix ,   Iy , and Iz . z c

b y a x

22. Moments of inertia The coordinate axes in the figure run through the centroid of a solid wedge parallel to the labeled edges. Find Ix ,   Iy , and Iz if a = b = 6, c = 4, and the density is δ ( x , y, z ) = 1. z

Plates with Varying Density

11. Finding a moment of inertia Find the moment of inertia about the x-axis of a thin plate bounded by the parabola x = y − y 2 and the line x + y = 0 if δ ( x , y ) = x + y. 12. Finding mass Find the mass of a thin plate occupying the smaller region cut from the ellipse x 2 + 4 y 2 = 12 by the parabola x = 4 y 2 if δ ( x , y ) = 5 x kg m 2 . 13. Finding a center of mass Find the center of mass of a thin triangular plate bounded by the y-axis and the lines y = x and y = 2 − x if δ ( x , y ) = 6 x + 3 y + 3. 14. Finding a center of mass and moment of inertia Find the center of mass and moment of inertia about the x-axis of a thin plate bounded by the curves x = y 2 and x = 2 y − y 2 if the density at the point ( x , y ) is δ ( x , y ) = y + 1. 15. Center of mass, moment of inertia Find the center of mass and the moment of inertia about the y-axis of a thin rectangular plate cut from the first quadrant by the lines x = 6 and y = 1 if δ ( x , y ) = x + y + 1. 16. Center of mass, moment of inertia Find the center of mass and the moment of inertia about the y-axis of a thin plate bounded by the line y = 1 and the parabola y = x 2 if the density is δ ( x , y ) = y + 1. 17. Center of mass, moment of inertia Find the center of mass and the moment of inertia about the y-axis of a thin plate bounded by the x-axis, the lines x = ±1, and the parabola y = x 2 if δ ( x , y ) = 7 y + 1. 18. Center of mass, moments of inertia Find the center of mass and the moments of inertia about the x-axis of a thin rectangular plate bounded by the lines x = 0,   x = 20,   y = −1, and y = 1 if δ ( x , y ) = 1 + ( x 20 ). 19. Center of mass, moments of inertia Find the center of mass, the moment of inertia about the coordinate axes, and the polar moment of inertia of a thin triangular plate bounded by the lines y = x ,   y = −x , and y = 1 if δ ( x , y ) = y + 1 kg m 2. 20. Center of mass, moments of inertia Repeat Exercise 19 for δ ( x , y ) = 3 x 2 + 1 kg m 2.

901

Centroid at (0, 0, 0)

a 2 b 3

c

y

c 3

a b

x

23. Center of mass and moments of inertia A solid “trough” of constant density δ ( x , y, z ) = 1 is bounded below by the surface z = 4 y 2 , above by the plane z = 4, and on the ends by the planes x = 1 and x = −1. Find the center of mass and the moments of inertia with respect to the three axes. 24. Center of mass A solid of constant density is bounded below by the plane z = 0, on the sides by the elliptical cylinder x 2 + 4 y 2 = 4, and above by the plane z = 2 − x (see the accompanying figure). a. Find x and y. b. Evaluate the integral Mxy =

(1 2 ) 4 − x 2

2

2− x

∫−2 ∫−(1 2) 4− x ∫0 2

z dz dy dx,

using integral tables to carry out the final integration with respect to x. Then divide Mxy by M to verify that z = 5 4. z

z=2−x 2

x = −2

1 2 x

y

x 2 + 4y 2 = 4

902

Chapter 14 Multiple Integrals

25. a. Center of mass Find the center of mass of a solid of constant density bounded below by the paraboloid z = x 2 + y 2 and above by the plane z = 4.

31. A solid cube in the first octant is bounded by the coordinate planes and by the planes x = 1,   y = 1, and z = 1. The density of the cube is δ ( x , y, z ) = x + y + z + 1.

b. Find the plane z = c that divides the solid into two parts of equal volume. This plane does not pass through the center of mass.

32. A wedge like the one in Exercise 22 has dimensions a = 2,   b = 6, and c = 3. The density is δ ( x , y, z ) = x + 1. Notice that if the density is constant, the center of mass will be ( 0, 0, 0 ).

26. Moments A solid cube of constant density δ ( x , y, z ) = 1, 2 units on a side, is bounded by the planes x = ±1,   z = ±1, y = 3, and y = 5. Find the center of mass and the moments of inertia about the coordinate axes. 27. Moment of inertia about a line A wedge like the one in Exercise 22 has a = 4,   b = 6, c = 3, and a constant density δ ( x , y, z ) = 1. Make a quick sketch to check for yourself that the square of the distance from a typical point ( x , y, z ) of the wedge to the line L : z = 0,   y = 6 is r 2 = ( y − 6 ) 2 + z 2 . Then calculate the moment of inertia of the wedge about L. 28. Moment of inertia about a line A wedge like the one in Exercise 22 has a = 4,   b = 6, c = 3, and a constant density δ ( x , y, z ) = 1. Make a quick sketch to check for yourself that the square of the distance from a typical point ( x , y, z ) of the wedge to the line L : x = 4,   y = 0 is r 2 = ( x − 4 ) 2 + y 2 . Then calculate the moment of inertia of the wedge about L.

33. Mass Find the mass of the solid bounded by the planes x + z = 1,   x − z = −1,   y = 0, and the surface y = z. The density of the solid is δ ( x , y, z ) = 2 y + 5 kg m 3 . 34. Mass Find the mass of the solid region bounded by the parabolic surfaces z = 16 − 2 x 2 − 2 y 2 and z = 2 x 2 + 2 y 2 if the density of the solid is δ ( x , y, z ) = x 2 + y 2 . Theory and Examples

The Parallel Axis Theorem Let L c.m. be a line through the center of mass of a body of mass m and let L be a parallel line h units away from L c.m. . The Parallel Axis Theorem says that the moments of inertia I c.m. and IL of the body about L c.m. and L satisfy the equation IL = I c.m. + mh 2 .

As in the two-dimensional case, the theorem gives a quick way to calculate one moment when the other moment and the mass are known. 35. Proof of the Parallel Axis Theorem

Solids with Varying Density

In Exercises 29 and 30, find a. the mass of the solid.

b. the center of mass.

29. A solid region in the first octant is bounded by the coordinate planes and the plane x + y + z = 2. The density of the solid is δ ( x , y, z ) = 2 x gm cm 3 .

a. Show that the first moment of a body in space about any plane through the body’s center of mass is zero. (Hint: Place the body’s center of mass at the origin and let the plane be the yz-plane. What does the formula x = Myz M then tell you?) z

30. A solid in the first octant is bounded by the planes y = 0 and z = 0 and by the surfaces z = 4 − x 2 and x = y 2 (see the accompanying figure). Its density function is δ ( x , y, z ) = kxy,   k a constant.

Lc.m. L

v = xi + yj hi

z

(x, y, z)

i v−h

4

y

c.m. D

x

z = 4 − x2

(h, 0, 0)

b. To prove the Parallel Axis Theorem, place the body with its center of mass at the origin, with the line L c.m. along the z-axis and the line L perpendicular to the xy-plane at the point ( h, 0, 0 ). Let D be the region of space occupied by the body. Then, in the notation of the figure,

x = y2 y 2 x

(2)

IL =

∫∫∫

v − h i 2 dm.

D

(2, " 2, 0)

In Exercises 31 and 32, find a. the mass of the solid. b. the center of mass. c. the moments of inertia about the coordinate axes.

Expand the integrand in this integral and complete the proof. 36. The moment of inertia about a diameter of a solid sphere of constant density and radius a is ( 2 5 ) ma 2 , where m is the mass of the sphere. Find the moment of inertia about a line tangent to the sphere. 37. The moment of inertia of the solid in Exercise 21 about the z-axis is Iz = abc ( a 2 + b 2 ) 3. a. Use Equation (2) to find the moment of inertia of the solid about the line parallel to the z-axis through the solid’s center of mass.

14.7  Triple Integrals in Cylindrical and Spherical Coordinates

b. Use Equation (2) and the result in part (a) to find the moment of inertia of the solid about the line x = 0,   y = 2b. 38. If a = b = 6 and c = 4, the moment of inertia of the solid wedge in Exercise 22 about the x-axis is Ix = 208. Find the moment of inertia of the wedge about the line y = 4,   z = −4 3 (the edge of the wedge’s narrow end).

⎪⎧ 6 x 2 y, 41. f ( x , y ) = ⎪⎨ ⎪⎪ 0, ⎪⎩

903

if 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1, otherwise.

⎧⎪ 3 ( x 2 + y 2 ), 2 42. f ( x ,   y ) = ⎪⎨ ⎪⎪ 0, ⎪⎩

if 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1, otherwise.

For Exercises 39–42, verify that f gives a joint probability density function. Then find the expected values μ X and μY .

43. Suppose that f is a uniform joint probability density function on 0 ≤ x < 2,  0 ≤ y < 3. What is the formula for f ? What is the probability that X < Y ?

⎧⎪ x + y, 39. f ( x , y ) = ⎪⎨ ⎪⎪ 0, ⎩

44. The following formula defines a joint probability density function. What is the value of C? What are the expected values μ X and μY ?

Joint Probability Density Functions

⎧⎪ 4 xy, 40. f ( x , y ) = ⎪⎨ ⎪⎪ 0, ⎩

if 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1, otherwise.

⎪⎧ Cxy, f ( x , y ) = ⎪⎨ ⎪⎪ 0, ⎩

if 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1, otherwise.

if 0 ≤ x ≤ 2 and 0 ≤ y ≤ 3, otherwise.

14.7 Triple Integrals in Cylindrical and Spherical Coordinates When a calculation in physics, engineering, or geometry involves a cylinder, cone, or sphere, we can often simplify our work by using cylindrical or spherical coordinates, which are introduced in this section. The procedure for transforming to these coordinates and evaluating the resulting triple integrals is similar to the transformation to polar coordinates in the plane discussed in Section 14.4.

z P(r, u, z)

z

0 x

Integration in Cylindrical Coordinates u

r y y

x

We obtain cylindrical coordinates for space by combining polar coordinates in the xy-plane with the usual z-axis. This assigns to every point in space coordinate triples of the form ( r , θ, z ), as shown in Figure 14.46. Here we require r ≥ 0 .

FIGURE 14.46

The cylindrical coordinates of a point in space are r, θ, and z.

Cylindrical coordinates represent a point P in space by ordered triples ( r , θ, z ) in which

DEFINITION

1. r and θ are polar coordinates for the vertical projection of P on the xy-plane, with r ≥ 0 , and 2. z is the rectangular vertical coordinate.

z u = u0, r and z vary z = z0, r and u vary z0

The values of x, y, r, and θ in rectangular and cylindrical coordinates are related by the usual equations.

0 a

x

FIGURE 14.47

u0

y r = a, u and z vary

Constant-coordinate equations in cylindrical coordinates yield cylinders and planes.

Equations Relating Rectangular ( x, y, z ) and Cylindrical ( r , θ , z ) Coordinates

x = r cos θ, r2

=

x2

y = r sin θ, +

y2,

z = z,

tan θ = y x

In cylindrical coordinates, the equation r = a describes not just a circle in the xy-plane but an entire cylinder about the z-axis (Figure 14.47). The z-axis is given by r = 0. The equation θ = θ 0 describes the half-plane that contains the z-axis and makes an angle θ 0 with the positive x-axis. And, just as in rectangular coordinates, the equation z = z 0 describes a plane perpendicular to the z-axis.

904

Chapter 14 Multiple Integrals

Cylindrical coordinates are good for describing cylinders whose axes run along the z-axis and planes that either contain the z-axis or lie perpendicular to the z-axis. Surfaces like these have equations of constant coordinate value: r = 4 π θ= 3 z = 2. z r Δr Δu

r Δu

Δz Δu Δr

r

FIGURE 14.48

In cylindrical coordinates the volume of the wedge is approximated by the product ΔV =   r Δz Δr Δθ.

Cylinder, radius 4, axis the z-axis Half-plane containing the z-axis Plane perpendicular to the z-axis

When computing triple integrals over a solid region D in cylindrical coordinates, we partition the region into n small cylindrical wedges, rather than into rectangular boxes. In the kth cylindrical wedge, r, θ, and z change by Δrk , Δθ k , and Δz k , and the largest of these numbers among all the cylindrical wedges is called the norm of the partition. We express the triple integral as a limit of Riemann sums using these wedges. The volume of such a cylindrical wedge ΔVk is obtained by taking the area ΔAk of its base in the rθ-plane and multiplying by the height Δz k (Figure 14.48). For a point ( rk , θ k , z k ) in the center of the kth wedge, we calculated in polar coordinates that ΔAk = rk Δrk Δθ k . So ΔVk = Δz k rk Δrk Δθ k =  rk Δz k Δrk Δθ k, and a Riemann sum for f over D has the form Sn =

n

∑ f ( rk , θ k , z k ) rk Δ z k Δ rk Δ θ k . k =1

The triple integral of a function f over D is obtained by taking a limit of such Riemann sums with partitions whose norms approach zero: Volume Differential in Cylindrical Coordinates

lim S n =

n →∞

dV = r dz dr dθ

∫∫∫ D

f dV =

∫∫∫

f r dz dr d θ.

D

Triple integrals in cylindrical coordinates are then evaluated as iterated integrals, as in the following example. Although the definition of cylindrical coordinates makes sense without any restrictions on θ, in most situations when integrating, we will need to restrict θ to an interval of length 2π. So we impose the requirement that α ≤ θ ≤ β , where 0 ≤ β − α ≤ 2π.

z

Top Cartesian: z = x 2 + y 2 Cylindrical: z = r 2

M

EXAMPLE 1 Find the limits of integration in cylindrical coordinates for integrating a function f ( r , θ, z ) over the solid region D bounded below by the plane z = 0, laterally by the circular cylinder x 2 + ( y − 1) 2 = 1, and above by the paraboloid z = x 2 + y 2 . Solution The base of D is also the region’s projection R on the xy-plane. The boundary of R is the circle x 2 + ( y − 1) 2 = 1. Its polar coordinate equation is

D

x 2 + ( y − 1) 2 = 1 y

2 u

x

R

(r, u)

L

Cartesian: x 2 + ( y − 1)2 = 1 Polar: r = 2 sin u

FIGURE 14.49

Finding the limits of integration for evaluating an integral in cylindrical coordinates (Example 1).

x 2 + y 2 − 2y + 1 = 1 r 2 − 2r sin θ = 0 r = 2 sin θ. The region is sketched in Figure 14.49. We find the limits of integration, starting with the z-limits. A line M through a typical point ( r, θ ) in R parallel to the z-axis enters D at z = 0 and leaves at z = x 2 + y 2 = r 2 . Next we find the r-limits of integration. A ray L through ( r, θ ) from the origin enters R at r = 0 and leaves at r = 2 sin θ.

14.7  Triple Integrals in Cylindrical and Spherical Coordinates

905

Finally, we find the θ-limits of integration. As L sweeps across R, the angle θ it makes with the positive x-axis runs from θ = 0 to θ = π. The integral is

∫∫∫

f ( r , θ, z ) dV =

D

π

2 sin θ

∫0 ∫0

r2

∫0

f ( r , θ, z ) r dz dr d θ.

Example 1 illustrates a good procedure for finding limits of integration in cylindrical coordinates. The procedure is summarized as follows.

How to Integrate in Cylindrical Coordinates To evaluate

∫∫∫

f ( r , θ, z ) dV

D

over a solid region D in space in cylindrical coordinates, integrating first with respect to z, then with respect to r, and finally with respect to θ, take the following steps. 1. Sketch. Sketch the solid region D along with its projection R on the xy-plane. Label the surfaces and curves that bound D and R.

z z = g2(r, u)

D z = g1(r, u) y x

R

2. Find the z-limits of integration. Draw a line M through a typical point ( r, θ ) of R parallel to the z-axis. As z increases, M enters D at z = g1 ( r, θ ) and leaves at z = g 2 ( r, θ ). These are the z-limits of integration.

z M

z = g2(r, u)

D z = g1(r, u) y x

R (r, u)

906

Chapter 14 Multiple Integrals

3. Find the r-limits of integration. Draw a ray L through ( r, θ ) from the origin. The ray enters R at r = h1 (θ) and leaves at r = h 2 (θ). These are the r-limits of integration. z M z = g2(r, u)

D z = g1(r, u)

a

b

u

x r = h1(u)

y R (r, u) u=b

u=a

r = h2(u)

L

4. Find the θ-limits of integration. As L sweeps across R, the angle θ it makes with the positive x-axis runs from θ = α to θ = β. These are the θ-limits of integration. The integral is

∫∫∫

f ( r , θ, z ) dV =

4

z = x2 + y2 = r2

M

x2 + y2 = 4 r=2

u x

y

(r, u) L

FIGURE 14.50

Example 2 shows how to find the centroid of this solid.

z = g2(r , θ )

r = h2 (θ ) 1

D

z

θ=β

∫θ = α ∫r = h ( θ ) ∫ z = g ( r , θ )

f ( r , θ, z ) r dz dr d θ.

1

EXAMPLE 2 Find the centroid ( δ = 1) of the solid enclosed by the cylinder x 2 + y 2 = 4, bounded above by the paraboloid z = x 2 + y 2 , and bounded below by the xy-plane. Solution We sketch the solid, bounded above by the paraboloid z = r 2 and below by the plane z = 0 (Figure 14.50). Its base R is the disk 0 ≤ r ≤ 2 in the xy-plane. The solid’s centroid ( x , y , z ) lies on its axis of symmetry, here the z-axis. This makes x = y = 0. To find z , we divide the first moment Mxy by the mass M. To find the limits of integration for the mass and moment integrals, we continue with the four basic steps. We completed our initial sketch. The remaining steps give the limits of integration. The z-limits. A line M through a typical point ( r, θ ) in the base parallel to the z-axis enters the solid at z = 0 and leaves at z = r 2 . The r-limits. A ray L through ( r, θ ) from the origin enters R at r = 0 and leaves at r = 2. The θ-limits. As L sweeps over the base like a clock hand, the angle θ it makes with the positive x-axis runs from θ = 0 to θ = 2π. The value of Mxy is

Mxy = =

2π

2

2π

2

r2

∫0 ∫0 ∫0

z r dz dr d θ =

r5 dr d θ = 2

∫0 ∫0

2π

∫0

2π

2

∫0 ∫0

z =r 2

⎡ z2 ⎤ r dr d θ ⎢ 2⎥ ⎣ ⎦ z=0

r =2

⎡r6 ⎤ dθ = ⎢ 12 ⎥ ⎣ ⎦ r =0

2π

∫0

16 32π . dθ = 3 3

The value of M is M = =

2π

2

2π

2

r2

∫0 ∫0 ∫0 ∫0 ∫0

r dz dr d θ =

r 3 dr d θ =

2π

∫0

2π

2

∫0 ∫0 r =2

z =r 2

⎡z⎤ r dr d θ ⎢⎣ ⎥⎦ z=0

⎡r4 ⎤ dθ = ⎢⎣ 4 ⎥⎦ r =0

2π

∫0

4 d θ = 8π.

907

14.7  Triple Integrals in Cylindrical and Spherical Coordinates

Therefore, Mxy 32π 1 4 = = , M 3 8π 3

z =

and the centroid is ( 0, 0, 4 3 ). Notice that the centroid lies on the z-axis, outside the solid.

Spherical Coordinates and Integration Spherical coordinates locate points in space with two angles and one distance, as shown in Figure 14.51. The first coordinate, ρ = OP , is the point’s  distance from the origin and is never negative. The second coordinate, φ, is the angle OP makes with the positive z-axis. It is required to lie in the interval [ 0, π ]. The third coordinate is the angle θ as measured in cylindrical coordinates.

φ is the Greek letter phi, pronounced “fee.”

Spherical coordinates represent a point P in space by ordered triples ( ρ , φ, θ ) in which

DEFINITION z

1. ρ is the distance from P to the origin ( ρ ≥ 0 ).  2. φ is the angle OP makes with the positive z-axis ( 0 ≤ φ ≤ π ). 3. θ is the angle from cylindrical coordinates.

P(r, f, u) f r

z = r cos f

0 x

u

r

y

y

x

FIGURE 14.51

The spherical coordinates ρ , φ, and θ and their relation to x, y, z, and r.

On maps of Earth, θ is related to the longitude of a point on the planet and φ to its latitude, while ρ is related to elevation above Earth’s surface. The equation ρ = a describes the sphere of radius a centered at the origin (Figure 14.52). The equation φ = φ 0 describes a single cone whose vertex lies at the origin and whose axis lies along the z-axis. (We broaden our interpretation to include the xy-plane as the cone φ = π 2.) If φ 0 is greater than π 2, the cone φ = φ 0 opens downward. The equation θ = θ 0 describes the half-plane that contains the z-axis and makes an angle θ 0 with the positive x-axis. Equations Relating Spherical Coordinates to Cartesian and Cylindrical Coordinates

z f = f0, r and u vary

f0

P(r, f0, u0)

r = ρ sin φ,

x = r cos θ = ρ sin φ cos θ,

z = ρ cos φ,

y = r sin θ = ρ sin φ sin θ,

ρ =

x 2 + y2 + z 2 =

EXAMPLE 3 Find a x 2 + y 2 + ( z − 1) 2 = 1.

spherical

(1)

r 2 + z 2.

coordinate

equation

for

the

sphere

Solution We use Equations (1) to substitute for x, y, and z: u0

y

x 2 + y 2 + ( z − 1) 2 = 1 ρ 2 sin 2 φ cos 2 θ + ρ 2 sin 2 φ sin 2 θ + ( ρ cos φ − 1) 2 = 1

x r = a, f and u vary

FIGURE 14.52

ρ 2 sin 2 u = u0, r and f vary

Constant-coordinate equations in spherical coordinates yield spheres, single cones, and half-planes.

φ ( cos 2 θ

+

sin 2

θ) +

 1

ρ2

cos 2 φ

Eqs. (1)

− 2ρ cos φ + 1 = 1

ρ 2 ( sin 2 φ + cos 2 φ ) = 2ρ cos φ  1

ρ 2 = 2ρ cos φ ρ = 2 cos φ.

Includes ρ = 0

908

Chapter 14 Multiple Integrals

The angle φ varies from 0 at the north pole of the sphere to π 2 at the south pole; the angle θ does not appear in the expression for ρ, reflecting the symmetry about the z-axis (see Figure 14.53).

z x2 + y2 + (z − 1)2 = 1 r = 2 cos f

2

1

EXAMPLE 4 f

y x

Solution 2 Use algebra. If we use Equations (1) to substitute for x, y, and z, we obtain the same result:

The sphere in

Example 3. z

f=p 4

z =

x 2 + y2

ρ cos φ =

ρ 2 sin 2 φ

ρ cos φ = ρ sin φ "x2

f=p 4

+

y2

φ =

FIGURE 14.54

2π 3 π θ = . 3

φ= r sin fΔu

Δr

O

π . 4

ρ =4

The cone in Example 4.

z

f

r rΔf

Δf y Δu

Sn =

x

In spherical coordinates we use the volume of a spherical wedge, which closely approximates that of a rectangular box. Volume Differential in Spherical Coordinates dV = ρ 2 sin φ d ρ dφ d θ

0 ≤φ ≤ π

Sphere, radius 4, center at origin Cone opening down from the origin, making an angle of 2π 3 radians with the positive z-axis Half-plane, hinged along the z-axis, making an angle of π 3 radians with the positive x-axis

When computing triple integrals over a solid region D in spherical coordinates, we partition the region into n spherical wedges. The size of the kth spherical wedge, which contains a point ( ρ k , φ k , θ k ), is given by the changes Δρ k ,  Δφ k , and Δθ k in ρ , φ, and θ. Such a spherical wedge has one edge a circular arc of length ρ k Δφ k , another edge a circular arc of length ρ k sin φ k Δθ k , and thickness Δρ k . The spherical wedge closely approximates a rectangular box of these dimensions when Δρ k ,  Δφ k , and Δθ k are all small (Figure 14.55). It can be shown that the volume of this spherical wedge ΔVk is ΔVk = ρ k 2 sin φ k Δρ k Δφ k Δθ k for ( ρ k , φ k , θ k ), a point chosen inside the wedge. The corresponding Riemann sum for a function f ( ρ , φ, θ ) is

u

FIGURE 14.55

Includes  ρ = 0 ( the origin )

Spherical coordinates are useful for describing spheres centered at the origin, halfplanes hinged along the z-axis, and cones whose vertices lie at the origin and whose axes lie along the z-axis. Surfaces like these have equations of constant coordinate value:

y

x

Example 3 ρ ≥ 0, sin φ ≥ 0

cos φ = sin φ z =

x 2 + y2 .

Solution 1 Use geometry. The cone is symmetric with respect to the z-axis and cuts the first quadrant of the yz-plane along the line z = y. The angle between the cone and the positive z-axis is therefore π 4 radians. The cone consists of the points whose spherical coordinates have φ equal to π 4, so its equation is φ = π 4. (See Figure 14.54.)

r

FIGURE 14.53

Find a spherical coordinate equation for the cone z =

n

∑ f ( ρ k , φ k , θ k ) ρ k 2 sin φ k Δρ k Δφ k Δθ k . k =1

As the norm of a partition approaches zero, and the spherical wedges get smaller, the limit of the Riemann sums is the triple integral:

lim S n =

n →∞

∫∫∫ D

f ( ρ , φ, θ ) dV =

∫∫∫ D

f ( ρ , φ, θ ) ρ 2 sin φ dρ d φ d θ.

14.7  Triple Integrals in Cylindrical and Spherical Coordinates

909

To evaluate integrals in spherical coordinates, we usually integrate first with respect to ρ. The procedure for finding the limits of integration is as follows. As with cylindrical coordinates, we restrict θ in the form α ≤ θ ≤ β and 0 ≤ β − α ≤ 2π.

How to Integrate in Spherical Coordinates To evaluate

∫∫∫

f ( ρ , φ, θ ) dV

D

over a solid region D in space in spherical coordinates, integrating first with respect to ρ, then with respect to φ, and finally with respect to θ, take the following steps. 1. Sketch. Sketch the solid region D along with its projection R on the xy-plane. Label the surfaces that bound D. z

z

r = g2(f, u)

fmax

fmin

f

M r = g2(f, u)

D r = g1(f, u) D r = g1(f, u) R

y

x

u=a x

z M D

Sphere r = 1

f

Cone f = p 3

u

FIGURE 14.56

Example 5.

L

2. Find the ρ -limits of integration. Draw a ray M from the origin through D, making an angle φ with the positive z-axis. Also draw the projection of M on the xy-plane (call the projection L). The ray L makes an angle θ with the positive x-axis. As ρ increases, M enters D at ρ = g1 ( φ, θ ) and leaves at ρ = g 2 ( φ, θ ). These are the ρ-limits of integration shown in the above figure. 3. Find the φ-limits of integration. For any given θ, the angle φ that M makes with the positive z-axis runs from φ = φ min to φ = φ max . The φ-limits of integration may depend on θ, but they are often constant. 4. Find the θ-limits of integration. The ray L sweeps over R as θ runs from α to β. These are the θ-limits of integration. The integral is

∫∫∫ D

f ( ρ , φ, θ ) dV =

θ=β

φ = φ max

∫ θ = α ∫φ = φ

min

ρ = g2(φ , θ )

∫ρ = g ( φ , θ )

f ( ρ , φ, θ ) ρ 2 sin φ dρ d φ d θ.  

1

EXAMPLE 5 Find the volume of the “ice cream cone” D bounded above by the sphere ρ = 1 and bounded below by the cone φ = π 3.

R

x

u=b y

R u

L

y

The ice cream cone in

Solution The volume is V = ∫∫∫ D ρ 2 sin φ dρ dφ dθ, the integral of f ( ρ , φ, θ ) = 1 over D. To find the limits of integration for evaluating the integral, we begin by sketching D and its projection R on the xy-plane (Figure 14.56).

910

Chapter 14 Multiple Integrals

The ρ -limits of integration. We draw a ray M from the origin through D, making an angle φ with the positive z-axis. We also draw L, the projection of M on the xy-plane, along with the angle θ that L makes with the positive x-axis. Ray M enters D at ρ = 0 and leaves at ρ = 1. The φ-limits of integration. The cone φ = π 3 makes an angle of π 3 with the positive z-axis. For any given θ, the angle φ can run from φ = 0 to φ = π 3. The θ-limits of integration. The ray L sweeps over R as θ runs from 0 to 2π. The volume is V =

∫∫∫ ρ 2 sin φ dρ dφ d θ

=

D

π3

π 3 ⎡ ρ 3 ⎤ ρ =1

2π

=

∫0 ∫0

=

∫0

2π

2π

  sin φ d φ d θ = ⎢ ⎥ ⎣ 3 ⎦ ρ=0

φ=π 3 ⎡ − 1 cos φ ⎤ dθ = ⎢⎣ 3 ⎥⎦ φ=0

1

∫0 ∫0 ∫0 ρ 2 sin φ dρ dφ d θ

2π

∫0

2π

π 31

∫0 ∫0

3

sin φ d φ d θ

(− 16 + 13 ) d θ = 16 (2π) = π3 .

EXAMPLE 6 A solid of constant density δ = 1 occupies the solid region D in Example 5. Find the solid’s moment of inertia about the z-axis. Solution In rectangular coordinates, the moment is

Iz =

∫∫∫ ( x 2 + y 2 ) dV. D

In spherical coordinates, x 2 + y 2 = ( ρ sin φ cos θ ) 2 + ( ρ sin φ sin θ ) 2 = ρ 2 sin 2 φ. Hence, Iz =

∫∫∫ ( ρ 2 sin 2 φ )ρ 2 sin φ dρ dφ d θ

=

D

∫∫∫ ρ 4 sin 3 φ dρ dφ d θ. D

For the region D in Example 5, this becomes Iz =

2π

π3

1

∫0 ∫0 ∫0

ρ 4 sin 3 φ dρ d φ d θ =

2π

π3

∫0 ∫0

ρ =1

⎡ρ5 ⎤ sin 3 φ d φ d θ ⎢ ⎥ ⎣ 5 ⎦ ρ=0

=

cos 3 φ ⎤ φ = π 3 1 2π π 3( 1 2π ⎡ dθ 1 − cos 2 φ ) sin φ d φ d θ = ∫ ⎢ − cos φ + ⎥ ∫ ∫ 0 5 0 5 0 ⎣ 3 ⎦ φ=0

=

1 2π 1 1 1 1 2π 5 1 π − + +1− dθ = ∫ dθ = (2π) = . ∫ 5 0 2 24 3 5 0 24 24 12

(

)

Coordinate Conversion Formulas

Cylindrical to Rectangular

Spherical to Rectangular

Spherical to Cylindrical

x = r cos θ

x = ρ sin φ cos θ

r = ρ sin φ

y = r sin θ

y = ρ sin φ sin θ

z = ρ cos φ

z = z

z = ρ cos φ

θ = θ

Corresponding formulas for dV in triple integrals: dV = dx dy dz = r dz dr dθ = ρ 2 sin φ dρ dφ dθ

14.7  Triple Integrals in Cylindrical and Spherical Coordinates

911

In the next section we offer a more general procedure for determining dV in cylindrical and spherical coordinates. The results, of course, will be the same.

14.7

EXERCISES

In Exercises 1–12, sketch the region described by the following cylindrical coordinates in three-dimensional space. π 1. r = 2 2. θ = 4 3. z = −1 4. z = r 5. r = θ 7.

r2

+

6. z = r sin θ

z2

π 8. 1 ≤ r ≤ 2, 0 ≤ θ ≤ 3

= 4

9. r ≤ z ≤

9−

r2

10. 0 ≤ r ≤ 2 sin θ, 1 ≤ z ≤ 3 11. 0 ≤ r ≤ 4 cos θ, 0 ≤ θ ≤

π , 0 ≤ z ≤ 5 2

In Exercises 13–22, sketch the region described by the following spherical coordinates in three-dimensional space. π 6

14. φ =

15. θ =

2 π 3 17. ρ cos φ = 4

16. ρ = csc φ

18. 1 ≤ ρ ≤ 2 sec φ, 0 ≤ φ ≤

π 4

2 −r 2

1

25.

∫0 ∫0 ∫0

2π

2π

θ 2π

1

1

r dz dr d θ

3+ 24 r 2

2 −r 2

24.

r dz dr d θ 26.

2π

3

∫ 0 ∫ 0 ∫r π

θπ

18−r 2 2 3

2

4 −r 2

32.

∫ 0 ∫r − 2 ∫ 0

2π

1+ cos θ

4r dr d θ dz

( r 2 cos 2 θ + z 2 ) r d θ dr dz 2π

( r sin θ + 1)r d θ dz dr

33. Let D be the solid region bounded below by the plane z = 0, above by the sphere x 2 + y 2 + z 2 = 4, and on the sides by the cylinder x 2 + y 2 = 1. Set up the triple integrals in cylindrical coordinates that give the volume of D using the following orders of integration. b. dr dz dθ

c. dθ dz dr

34. Let D be the solid region bounded below by the cone z = x 2 + y 2 and above by the paraboloid z = 2 − x 2 − y 2 . Set up the triple integrals in cylindrical coordinates that give the volume of D using the following orders of integration. a. dz dr dθ

b. dr dz dθ

c. dθ dz dr

f ( r , θ, z ) r dz dr d θ

as an iterated integral over the solid region D that is bounded below by the plane z = 0, on the side by the cylinder r = cos θ, and on top by the paraboloid z = 3r 2 . 36. Convert the integral 1

∫−1 ∫0

28.

∫0 ∫0 ∫−1 2( r 2 sin 2 θ + z 2 ) r dz dr d θ

r dz dr d θ

3 4 −r 2

∫0 ∫0 ∫−

∫ 0 ∫ 0 ∫r 1

2π

1

∫−1 ∫0 ∫0

D

27.

2π

z

∫0 ∫0 ∫0

30.

∫∫∫

Evaluate the cylindrical coordinate integrals in Exercises 23–28. 2π

1

31.

r 3 dr dz d θ

35. Give the limits of integration for evaluating the integral

Evaluating Integrals in Cylindrical Coordinates

∫ 0 ∫ 0 ∫r

z 3

Finding Iterated Integrals in Cylindrical Coordinates

19. 0 ≤ ρ ≤ 3 csc φ π 20. 0 ≤ ρ ≤ 1, ≤ φ ≤ π, 0 ≤ θ ≤ π 2 21. 0 ≤ ρ cos θ sin φ ≤ 2, 0 ≤ ρ sin θ sin φ ≤ 3, 0 ≤ ρ cos φ ≤ 4 π 22. 4 sec φ ≤ ρ ≤ 5, 0 ≤ φ ≤ 2

23.

3

∫0 ∫0 ∫0

a. dz dr dθ

−π π ≤ θ ≤ , 0 ≤ z ≤ r cos θ 12. 0 ≤ r ≤ 3, 2 2

13. ρ = 3

2π

29.

4 −r 2

z r dz dr d θ

1− y 2

x

∫0 ( x 2 + y 2 ) dz dx dy

to an equivalent integral in cylindrical coordinates and evaluate the result. In Exercises 37–42, set up the iterated integral for evaluating ∫∫∫D f ( r , θ, z ) r dz dr dθ over the given solid region D. 37. D is the right circular cylinder whose base is the circle r = 2 sin θ in the xy-plane and whose top lies in the plane z = 4 − y.

3 r dz dr d θ

z z=4−y

12

Changing the Order of Integration in Cylindrical Coordinates

The integrals we have seen so far suggest that there are preferred orders of integration for cylindrical coordinates, but other orders usually work well and are occasionally easier to evaluate. Evaluate the integrals in Exercises 29–32.

y x

r = 2 sin u

912

Chapter 14 Multiple Integrals

38. D is the right circular cylinder whose base is the circle r = 3 cos θ and whose top lies in the plane z = 5 − x.

42. D is the right prism whose base is the triangle in the xy-plane bounded by the y-axis and the lines y = x and y = 1 and whose top lies in the plane z = 2 − x.

z z

z=5−x

z=2−x 2

y r = 3 cos u x 1

39. D is the solid right cylinder whose base is the region in the xy-plane that lies inside the cardioid r = 1 + cos θ and outside the circle r = 1 and whose top lies in the plane z = 4.

y

x y=x

z

Evaluating Integrals in Spherical Coordinates

4

Evaluate the spherical coordinate integrals in Exercises 43–48.

40. D is the solid right cylinder whose base is the region between the circles r = cos θ and r = 2 cos θ and whose top lies in the plane z = 3 − y. z z=3−y

y

x

r = cos u r = 2 cos u

41. D is the right prism whose base is the triangle in the xy-plane bounded by the x-axis and the lines y = x and x = 1 and whose top lies in the plane z = 2 − y.

∫0 ∫0 ∫0 ( ρ cos φ ) ρ 2 sin φ dρ dφ d θ

45.

∫0 ∫0 ∫0

46.

∫0 ∫0 ∫0 5ρ 3 sin 3 φ dρ dφ d θ

47.

∫0 ∫0 ∫sec φ 3ρ 2 sin φ dρ dφ d θ

48.

∫0 ∫0 ∫0

z=2−y

y 1 x

y=x

2π

π4

2π

π

3π 2

ρ 2 sin φ dρ d φ d θ

2

(1− cos φ ) 2

π

1

2π

π3

2

2π

π4

sec φ

ρ 2 sin φ dρ d φ d θ

( ρ cos φ ) ρ 2 sin φ dρ d φ d θ

Changing the Order of Integration in Spherical Coordinates

The previous integrals suggest there are preferred orders of integration for spherical coordinates, but other orders give the same value and are occasionally easier to evaluate. Evaluate the integrals in Exercises 49–52. 2

π2

0

49.

∫0 ∫−π ∫π 4 ρ 3 sin 2φ dφ d θ dρ

50.

∫π 6 ∫csc φ ∫0

51.

∫0 ∫0 ∫0

52.

∫π 6 ∫−π 2 ∫csc φ 5ρ 4 sin 3 φ dρ d θ dφ

z 2

2 sin φ

44. r=1 r = 1 + cos u

π

∫0 ∫0 ∫0

y

x

π

43.

π3

1

π2

2 csc φ

π

π4

π2

2π

ρ 2 sin φ d θ dρ d φ

12ρ sin 3 φ d φ d θ dρ 2

53. Let D be the region in Exercise 33. Set up the triple integrals in spherical coordinates that give the volume of D using the following orders of integration. a. dρ dφ dθ b. dφ dρ dθ

913

14.7  Triple Integrals in Cylindrical and Spherical Coordinates

54. Let D be the solid region bounded below by the cone z = x 2 + y 2 and above by the plane z = 1. Set up the triple integrals in spherical coordinates that give the volume of D using the following orders of integration. a. dρ dφ dθ

b. dφ dρ dθ

Finding Iterated Integrals in Spherical Coordinates

In Exercises 55–60, (a) find the spherical coordinate limits for the integral that calculates the volume of the given solid and then (b) evaluate the integral. 55. The solid between the sphere ρ = cos φ and the hemisphere ρ = 2,   z ≥ 0 z r = cos f

r=2

2

Finding Triple Integrals

61. Set up triple integrals for the volume of the sphere ρ = 2 in (a) spherical, (b) cylindrical, and (c) rectangular coordinates. 62. Let D be the solid region in the first octant that is bounded below by the cone φ = π 4 and above by the sphere ρ = 3. Express the volume of D as an iterated triple integral in (a) cylindrical and (b) spherical coordinates. Then (c) find the volume. 63. Let D be the smaller cap cut from a solid ball of radius 2 units by a plane 1 unit from the center of the sphere. Express the volume of D as an iterated triple integral in (a) spherical, (b) cylindrical, and (c) rectangular coordinates. Then (d) find the volume by evaluating one of the three triple integrals. 64. Let D be the solid hemisphere x 2 + y 2 + z 2 ≤ 1,   z ≥ 0 . If the density is δ ( x , y, z ) = 1, express the moment of intertia I z as an iterated integral in (a) cylindrical and (b) spherical coordinates. Then (c) find I z . Volumes

Find the volumes of the solids in Exercises 65–70. x

65.

2

2

66. z

z

y

z = 4 − 4 (x 2 + y 2)

56. The solid bounded below by the hemisphere ρ = 1,   z ≥ 0, and above by the surface ρ = 1 + cos φ

z=1−r –1

–1

z r = 1 + cos f

x

r=1

1

1

y

x

z = (x 2 + y 2 ) 2 −1

67.

57. The solid enclosed by the surface ρ = 1 − cos φ 58. The upper portion cut from the solid in Exercise 57 by the xy-plane

r = 3 cos u

59. The solid bounded below by the sphere ρ = 2 cos φ and above by the cone z = x 2 + y 2

x

z

z = "x2 + y2

69.

z = −"1 − r 2

68.

z

z = −y

y

x

y

z

z = "x 2 + y 2

y r = −3 cos u y

x

70.

z z = "1 − x 2 − y 2

z

z = 3"1 − x2 − y2

r = 2 cos f x

y

60. The solid bounded below by the xy-plane, on the sides by the sphere ρ = 2, and above by the cone φ = π 3 z f=

r = sin u

y x

y

r = cos u

p 3

71. Ball and cones Find the volume of the portion of the ball ρ ≤ a that lies between the cones φ = π 3 and φ = 2π 3. r=2

x

x

y

72. Ball and half-planes Find the volume of the region cut from the ball ρ ≤ a by the half-planes θ = 0 and θ = π 6 in the first octant.

914

Chapter 14 Multiple Integrals

73. Ball and plane Find the volume of the smaller region cut from the ball ρ ≤ 2 by the plane z = 1.

92. Centroid Find the centroid of the solid bounded above by the sphere ρ = a and below by the cone φ = π 4.

74. Cone and planes Find the volume of the solid enclosed by the cone z = x 2 + y 2 between the planes z = 1 and z = 2.

93. Centroid Find the centroid of the solid region that is bounded above by the surface z = r , on the sides by the cylinder r = 4, and below by the xy-plane.

75. Cylinder and paraboloid Find the volume of the solid region bounded below by the plane z = 0, laterally by the cylinder x 2 + y 2 = 1, and above by the paraboloid z = x 2 + y 2 . 76. Cylinder and paraboloids Find the volume of the solid region bounded below by the paraboloid z = x 2 + y 2 , laterally by the cylinder x 2 + y 2 = 1, and above by the paraboloid z = x 2 + y 2 + 1. 77. Cylinder and cones Find the volume of the solid cut from the thick-walled cylinder 1 ≤ x 2 + y 2 ≤ 2 by the cones z = ± x 2 + y2. 78. Sphere and cylinder Find the volume of the solid region that lies inside the sphere x 2 + y 2 + z 2 = 2 and outside the cylinder x 2 + y 2 = 1. 79. Cylinder and planes Find the volume of the solid region enclosed by the cylinder x 2 + y 2 = 4 and the planes z = 0 and y + z = 4. 80. Cylinder and planes Find the volume of the solid region enclosed by the cylinder x 2 + y 2 = 4 and the planes z = 0 and x + y + z = 4. 81. Region trapped by paraboloids Find the volume of the solid region bounded above by the paraboloid z = 5 − x 2 − y 2 and below by the paraboloid z = 4 x 2 + 4 y 2 . 82. Paraboloid and cylinder Find the volume of the solid region bounded above by the paraboloid z = 9 − x 2 − y 2 , bounded below by the xy-plane, and lying outside the cylinder x 2 + y 2 = 1. 83. Cylinder and sphere Find the volume of the region cut from the solid cylinder x 2 + y 2 ≤ 1 by the sphere x 2 + y 2 + z 2 = 4. 84. Sphere and paraboloid Find the volume of the solid region bounded above by the sphere x 2 + y 2 + z 2 = 2 and below by the paraboloid z = x 2 + y 2 . Average Values

85. Find the average value of the function f ( r , θ, z ) = r over the solid region bounded by the cylinder r = 1 between the planes z = −1 and z = 1. 86. Find the average value of the function f ( r , θ, z ) = r over the solid ball bounded by the sphere r 2 + z 2 = 1. (This is the sphere x 2 + y 2 + z 2 = 1.) 87. Find the average value of the function f ( ρ , φ, θ ) = ρ over the solid ball ρ ≤ 1. 88. Find the average value of the function f ( ρ , φ, θ ) = ρ cos φ over the upper half of the solid ball ρ ≤ 1,  0 ≤ φ ≤ π 2. Masses, Moments, and Centroids

89. Center of mass A solid of constant density is bounded below by the plane z = 0, above by the cone z = r ,   r ≥ 0, and on the sides by the cylinder r = 1. Find the center of mass. 90. Centroid Find the centroid of the solid region in the first octant that is bounded above by the cone z = x 2 + y 2 , below by the plane z = 0, and on the sides by the cylinder x 2 + y 2 = 4 and the planes x = 0 and y = 0. 91. Centroid

Find the centroid of the solid in Exercise 60.

94. Centroid Find the centroid of the region cut from the solid ball r 2 + z 2 ≤ 1 by the half-planes θ = −π 3,   r ≥ 0, and θ = π 3,   r ≥ 0. 95. Moment of inertia of solid cone Find the moment of inertia of a solid right circular cone of base radius 1 and height 1 about an axis through the vertex parallel to the base if the density is δ = 1. 96. Moment of inertia of ball Find the moment of inertia of a ball of radius a about a diameter if the density is δ = 1. 97. Moment of inertia of solid cone Find the moment of inertia of a solid right circular cone of base radius a and height h about its axis if the density is δ = 1. (Hint: Place the cone with its vertex at the origin and its axis along the z-axis.) 98. Variable density A solid is bounded on the top by the paraboloid z = r 2 , on the bottom by the plane z = 0, and on the sides by the cylinder r = 1. Find the center of mass and the moment of inertia about the z-axis if the density is a. δ ( r , θ, z ) = z b. δ ( r , θ, z ) = r. 99. Variable density A solid is bounded below by the cone z = x 2 + y 2 and above by the plane z = 1. Find the center of mass and the moment of inertia about the z-axis if the density is a. δ ( r , θ, z ) = z b. δ ( r , θ, z ) = z 2 . 100. Variable density A solid ball is bounded by the sphere ρ = a. Find the moment of inertia about the z-axis if the density is a. δ ( ρ , φ, θ ) = ρ 2 b. δ ( ρ , φ, θ ) = r = ρ sin φ. 101. Centroid of solid semi-ellipsoid Show that the centroid of the solid semi-ellipsoid of revolution ( r 2 a 2 ) + ( z 2 h 2 ) ≤ 1, z ≥ 0, lies on the z-axis three-eighths of the way from the base to the top. The special case h = a gives a solid hemisphere. Thus, the centroid of a solid hemisphere lies on the axis of symmetry three-eighths of the way from the base to the top. 102. Centroid of solid cone Show that the centroid of a solid right circular cone is one-fourth of the way from the base to the vertex. (In general, the centroid of a solid cone or pyramid is one-fourth of the way from the centroid of the base to the vertex.) 103. Density of center of a planet A planet is in the shape of a sphere of radius R and total mass M with spherically symmetric density distribution that increases linearly as one approaches its center. What is the density at the center of this planet if the density at its edge (surface) is taken to be zero? 104. Mass of planet’s atmosphere A spherical planet of radius R has an atmosphere whose density is μ = μ 0e −ch , where h is the altitude above the surface of the planet, μ 0 is the density at sea level, and c is a positive constant. Find the mass of the planet’s atmosphere.

14.8  Substitutions in Multiple Integrals

915

107. Symmetry What symmetry will you find in a surface that has an equation of the form r  f ( z ) in cylindrical coordinates? Give reasons for your answer.

Theory and Examples

105. Vertical planes in cylindrical coordinates a. Show that planes perpendicular to the x-axis have equations of the form r  a sec R in cylindrical coordinates.

108. Symmetry What symmetry will you find in a surface that has an equation of the form ρ  f (φ) in spherical coordinates? Give reasons for your answer.

b. Show that planes perpendicular to the y-axis have equations of the form r  b csc R. 106. (Continuation of Exercise 105.) Find an equation of the form r  f (R) in cylindrical coordinates for the plane ax + by = c, c v 0.

14.8 Substitutions in Multiple Integrals y

(u, y) G

u

0 Cartesian uy-plane

This section introduces the ideas involved in coordinate transformations to evaluate multiple integrals by substitution. The method replaces complicated integrals by ones that are easier to evaluate. Substitutions accomplish this by simplifying the integrand, the limits of integration, or both. A thorough discussion of multivariable transformations and substitutions is best left to a more advanced course, but our introduction here shows how the substitutions just studied reflect the general idea derived for single integral calculus.

Substitutions in Double Integrals

x = g(u, y) y = h(u, y)

The polar coordinate substitution of Section 14.4 is a special case of a more general substitution method for double integrals, a method that pictures changes in variables as transformations of regions. Suppose that a region G in the uV-plane is transformed into the region R in the xy-plane by equations of the form

y

x = g ( u, V ) ,

(x, y)

R 0

x

Cartesian xy-plane

FIGURE 14.57

The equations x = g ( u, V ) and y = h ( u, V ) allow us to change an integral over a region R in the xy-plane into an integral over a region G in the uV-plane. HISTORICAL BIOGRAPHY Carl Gustav Jacob Jacobi (1804–1851) Jacobi, one of nineteenth-century Germany’s most accomplished scientists, developed the theory of determinants and transformations into a powerful tool for evaluating multiple integrals and solving differential equations. He also applied transformation methods to study integrals like the ones that arise in the calculation of arc length. To know more, visit the companion Website.

y = h ( u, V ) ,

as suggested in Figure 14.57. We assume the transformation is one-to-one on the interior of G. We call R the image of G under the transformation, and G the preimage of R. Any function f ( x , y ) defined on R can be thought of as a function f ( g ( u, V),  h ( u, V)) defined on G as well. How is the integral of f ( x , y ) over R related to the integral of f ( g ( u, V),  h ( u, V)) over G? To gain some insight into the question, we look again at the single variable case. To be consistent with how we are using them now, we interchange the variables x and u used in the substitution method for single integrals in Chapter 5, so the equation is g(b)

∫g(a)

f ( x ) dx =

b

∫a

f ( g(u)) g′(u) du.

x = g(u), dx = g′(u)du

To propose an analogue for substitution in a double integral ∫∫R f ( x , y ) dx dy, we need a derivative factor like ga(u) as a multiplier that transforms the area element du d V in the region G to its corresponding area element dx dy in the region R. We denote this factor by J. In continuing with our analogy, it is reasonable to assume that J is a function of both variables u and V, just as ga is a function of the single variable u. Moreover, J should register instantaneous change, so partial derivatives are going to be involved in its expression. Since four partial derivatives are associated with the transforming equations x = g ( u, V ) and y = h ( u, V ), it is also reasonable to assume that the factor J ( u, V ) we seek includes them all. These features are captured in the following definition, which is constructed from the partial derivatives and is named after the German mathematician Carl Jacobi.

916

Chapter 14 Multiple Integrals

The Jacobian determinant or Jacobian of the coordinate transformation x = g ( u, υ ),  y = h ( u, υ ) is

DEFINITION

∂x ∂u J ( u, υ ) = ∂y ∂u

∂x ∂υ ∂y ∂x ∂x ∂y = − . ∂u ∂ υ ∂u ∂ υ ∂y ∂υ

(1)

The Jacobian can also be denoted by J ( u, υ ) = Differential Area Change Substituting x = g ( u, υ),   y = h( u, υ) dx dy =

∂( x , y ) ∂( u , υ )

du d υ

u p 2 G

∂( x , y ) ∂( u, υ )

to help us remember how the determinant in Equation (1) is constructed from the partial derivatives of x and y. The array of partial derivatives in Equation (1) behaves just like the derivative g′ in the single variable situation. The Jacobian measures how much the transformation is expanding or contracting the area around the point ( u, υ ). Effectively, the factor J converts the area of the differential rectangle du d υ in G to match its corresponding differential area dx dy in R. We note that, in general, the value of the scaling factor J depends on the point ( u, υ ) in G; that is, the scaling changes as the point ( u, υ ) varies through the region G. Our examples to follow will show how it scales the differential area du d υ for specific transformations. Now we can answer our original question concerning the relationship of the integral of f ( x , y ) over the region R to the integral of f ( g ( u, υ),  h ( u, υ)) over G. THEOREM 3—Substitution for Double Integrals

Suppose that f ( x , y ) is continuous over the region R. Let G be the preimage of R under the transformation x = g ( u, υ ),  y = h ( u, υ ), which is assumed to be one-to-one on the interior of G. If the functions g and h have continuous first partial derivatives within the interior of G, then 0

r

1

∫∫

Cartesian r u-plane

f ( x , y) dx dy =

R

x = r cos u y = r sin u

p 2

R

u=0 1

x

Cartesian xy-plane

FIGURE 14.58

G

∂( x , y ) du d υ. ∂( u, υ )

(2)

EXAMPLE 1 Find the Jacobian for the polar coordinate transformation x = r cos θ, y = r sin θ, and use Equation (2) to write the Cartesian integral ∫∫R f ( x , y ) dx dy as a polar integral.

1

0

f ( g ( u, υ) ,   h ( u, υ))

The derivation of Equation (2) is intricate and properly belongs to a course in advanced calculus, so we do not include it here. We now present examples illustrating the substitution method defined by the equation.

y u =

∫∫

The equations x = r cos θ,   y = r sin θ transform G into R. The Jacobian factor r, calculated in Example 1, scales the differential rectangle dr dθ in G to match the differential area element dx dy in R.

Solution Figure 14.58 shows how the equations x = r cos θ,  y = r sin θ transform the rectangle G: 0 ≤ r ≤ 1, 0 ≤ θ ≤ π 2, into the quarter of a circular disk R bounded by x 2 + y 2 = 1 in the first quadrant of the xy-plane. For polar coordinates, we have r and θ in place of u and υ. With x = r cos θ and y = r sin θ, the Jacobian is

∂x ∂r J (r, θ ) = ∂y ∂r

∂x cos θ ∂θ = ∂y sin θ ∂θ

−r sin θ r cos θ

= r ( cos 2 θ + sin 2 θ ) = r.

14.8  Substitutions in Multiple Integrals

917

Since we assume r ≥ 0 when integrating in polar coordinates, J ( r , θ ) = r = r so that Equation (2) gives

∫∫

f ( x , y ) dx dy =

R

∫∫

f ( r cos θ, r sin θ ) r dr d θ.

(3)

G

This is the same formula we derived independently using a geometric argument for polar area in Section 14.4. Here is an example of a substitution in which the image of a rectangle under the coordinate transformation is a trapezoid. Transformations like this one are called linear transformations, and their Jacobians are constant throughout G. EXAMPLE 2

Evaluate 4

x =( y 2 )+1

∫0 ∫ x = y 2

2x − y dx dy 2

by applying the transformation u =

2x − y , 2

υ =

y 2

(4)

and integrating over an appropriate region in the uυ-plane. Solution We sketch the region R of integration in the xy-plane and identify its boundaries (Figure 14.59). y 2

y y=2

u=0

0

1

y = 2x

x=u+y y = 2y

u=1

G

y=0

y=4

4

R

u

0

y = 2x − 2

x

1 y=0

The equations x = u + υ and y = 2υ transform G into R. Reversing the transformation by the equations u = ( 2 x − y ) 2 and υ = y 2 transforms R into G (Example 2). FIGURE 14.59

To apply Equation (2), we need to find the corresponding uυ-region G and the Jacobian of the transformation. To find them, we first solve Equations (4) for x and y in terms of u and υ. From those equations it is easy to find algebraically that x = u + υ,

y = 2υ.

(5)

We then find the boundaries of G by substituting these expressions into the equations for the boundaries of R (Figure 14.59) xy-equations for the boundary of R x = y 2 x = ( y 2) + 1 y = 0 y = 4

Corresponding uυ-equations for the boundary of G

Simplified uυ-equations

u + υ = 2υ 2 = υ

u = 0

u + υ = ( 2υ 2 ) + 1 = υ + 1 2υ = 0 2υ = 4

u =1 υ = 0 υ = 2

918

Chapter 14 Multiple Integrals

From Equations (5) the Jacobian of the transformation is ∂x ∂u J ( u, υ ) = ∂y ∂u

∂x ∂( u + υ) ∂υ ∂u = ∂y ∂ (2υ) ∂u ∂υ

∂( u + υ) 1 ∂υ = 0 ∂ (2υ) ∂υ

1 2

= 2.

We now have everything we need to apply Equation (2): y 1 0

4

x = ( y 2 )+ 1

∫0 ∫ x = y 2

y=u

G 1 u=1

2

y = −2u

EXAMPLE 3

1

∫0 ∫0

=

u

2x − y dx dy = 2

J ( u, υ ) du d υ

u =1

2⎡ ⎤ (u)( 2 ) du d υ =∫ ⎢ u 2 ⎥ dυ = 0 ⎣ ⎦ u=0

1

1− x

∫0 ∫0

x=

2

∫0 d υ

= 2.

x + y ( y − 2 x ) 2 dy dx.

Solution We sketch the region R of integration in the xy-plane and identify its boundaries (Figure 14.60). The integrand suggests the transformation u = x + y and υ = y − 2 x. Routine algebra produces x and y as functions of u and υ:

y 1

x = x+y=1

x=0

u =1

Evaluate

−2 u y − 3 3 2u y y= + 3 3

υ=2

∫ υ = 0 ∫u = 0 u

u υ − , 3 3

2u υ + . 3 3

y =

(6)

From Equations (6), we can find the boundaries of the uυ-region G (Figure 14.60).

R 0

FIGURE 14.60

y=0

1

x

The equations x = ( u 3 ) − ( υ 3 ) and y = ( 2u 3 ) + ( υ 3 ) transform G into R. Reversing the transformation by the equations u = x + y and υ = y − 2 x transforms R into G (Example 3).

xy-equations for the boundary of R

Corresponding uυ-equations for the boundary of G

x+ y =1

( u3 − υ3 ) + ( 23u + υ3 ) = 1

Simplified uυ-equations u =1

u υ − = 0 3 3 2u υ + = 0 3 3

x = 0 y = 0

υ = u υ = −2u

The Jacobian of the transformation in Equations (6) is ∂x ∂u J ( u, υ ) = ∂y ∂u

∂x 1 ∂υ 3 = ∂y 2 3 ∂υ

−

1 3 1 = . 3 1 3

Applying Equation (2), we evaluate the integral: 1

1− x

∫0 ∫0

x + y ( y − 2 x ) 2 dy dx =

=

∫0 ∫−2u u 1 2υ 2 ( 3 ) d υ du

=

1 1 12 3 u ( u + 8u 3 ) du = 9 ∫0

1

u

1

u =1

υ=u

∫u = 0 ∫υ =−2u u 1 2υ 2

=

J ( u, υ ) d υ du

1 1 1 2 ⎡ 1 3 ⎤ υ=u u ⎢ υ ⎥ du ⎣ 3 ⎦ υ =−2u 3 ∫0 1

∫0

u 7 2 du =

2 9 2 ⎤1 2 u ⎥ = . ⎦0 9 9

In the next example we illustrate a nonlinear transformation of coordinates resulting from simplifying the form of the integrand. Like the polar coordinates’ transformation, nonlinear transformations can map a straight-line boundary of a region into a curved

14.8  Substitutions in Multiple Integrals

919

boundary (or vice versa with the inverse transformation). In general, nonlinear transformations are more complex to analyze than linear ones, and a complete treatment is left to a more advanced course. EXAMPLE 4

Evaluate the integral 2

y

y e x

∫1 ∫1 y

xy

dx dy.

Solution The square root terms in the integrand suggest that we might simplify the integration by substituting u = xy and υ = y x . Squaring these equations gives u 2 = xy and υ 2 = y x, which imply that u 2 υ 2 = y 2 and u 2 υ 2 = x 2 . So we obtain the transformation (in the same ordering of the variables as discussed before)

u and y = uυ , υ with u > 0 and υ > 0. Let’s first see what happens to the integrand itself under this transformation. The Jacobian of the transformation is not constant: x =

∂x ∂u J ( u, υ ) = ∂y ∂u y

y=2

∫∫

R

R

1 xy = 1

0

1

x

2

FIGURE 14.61

The region of integration

R in Example 4. y uy = 2 3 y = 2

2 u = 1 3 xy = 1

xy

2

∫∫ υe u

∫∫ υe u

J ( u, υ ) du d υ =

G

G

2u du d υ = υ

∫∫

2ue u du d υ.

G

y

y  e x

xy

2

∫1 ∫1

dx dy =

2u

2ue u d υ du.

Note the order of integration.

We now evaluate the transformed integral on the right-hand side:

y=13y=x

1

dx dy =

The transformed integrand function is easier to integrate than the original one, so we proceed to determine the limits of integration for the transformed integral. The region of integration R of the original integral in the xy-plane is shown in Figure 14.61. From the substitution equations u = xy and υ = y x, we see that the image of the left-hand boundary xy = 1 for R is the vertical line segment u = 1, 2 ≥ υ ≥ 1, in G (see Figure 14.62). Likewise, the right-hand boundary y = x of R maps to the horizontal line segment υ = 1, 1 ≤ u ≤ 2, in G. Finally, the horizontal top boundary y = 2 of R maps to uυ = 2, 1 ≤ υ ≤ 2, in G. As we move counterclockwise around the boundary of the region R, we also move counterclockwise around the boundary of G, as shown in Figure 14.62. Knowing the region of integration G in the uυ-plane, we can now write equivalent iterated integrals: 2

1

FIGURE 14.62

y e x

∫1 ∫1 y

G

0

−u 2u . υ2 = υ u

If G is the region of integration in the uυ-plane, then by Equation (2) the transformed double integral under the substitution is

y=x

2

∂x 1 ∂υ = υ ∂y υ ∂υ

u

The boundaries of the region G correspond to those of region R in Figure 14.61. Notice that as we move counterclockwise around the region R, we move counterclockwise around the region G as well. The inverse transformation equations u = xy ,   υ = y x produce the region G from the region R.

2

∫1 ∫1

2u

2ue u d υ du = 2 ∫

2

1

υ=2 u

⎡ ⎤ du ⎢ υue u ⎥ ⎣ ⎦ υ =1

2

= 2 ∫ ( 2e u − ue u ) du 1

= 2∫

1

2

(2

− u ) e u du u=2

⎡ ⎤ = 2 ⎢( 2 − u )e u + e u ⎥ ⎣ ⎦ u =1

= 2( e 2 − ( e + e )) = 2e ( e − 2 ).

Integrate by parts.

920

Chapter 14 Multiple Integrals

Substitutions in Triple Integrals The cylindrical and spherical coordinate substitutions in Section 14.7 are special cases of a substitution method that pictures changes of variables in triple integrals as transformations of solid regions. The method is like the method for double integrals given by Equation (2) except that now we work in three dimensions instead of two. Suppose that a solid region G in uυw-space is transformed one-to-one into the solid region D in xyz-space by differentiable equations of the form x = g ( u, υ, w ) ,

y = h ( u, υ, w ) ,

z = k ( u, υ, w ) ,

as suggested in Figure 14.63. Then any function F ( x , y,   z ) defined on D can be thought of as a function F ( g ( u, υ, w ),  h ( u, υ, w ),  k ( u, υ, w )) = H ( u, υ, w ) defined on G. If g, h, and k have continuous first partial derivatives, then the integral of F ( x , y, z ) over D is related to the integral of H ( u, υ, w ) over G by the equation

∫∫∫ F ( x , y, z ) dx dy dz

=

D

∫∫∫ H ( u, υ, w ) J ( u, υ, w )

du d υ dw.

(7)

G

w

z x = g(u, y, w) y = h(u, y, w) z = k(u, y, w)

G

D y

u

y x

Cartesian uyw-space

Cartesian xyz-space

The equations x = g ( u, υ, w ),   y = h ( u, υ, w ), and z = k ( u, υ, w ) allow us to change an integral over a region D in Cartesian xyz-space into an integral over a region G in Cartesian uυw-space using Equation (7). FIGURE 14.63

The factor J ( u, υ, w ), whose absolute value appears in this equation, is the Jacobian determinant

Determinants 2 × 2 and 3 × 3 determinants are evaluated as follows: a

b

c

d

a1

a2

a3

b1

b2

b3

c1

c2

c3

b1

b3

c1

c3

− a2

= ad − bc

= a1

+ a3

b2

b3

c2

c3

b1

b2

c1

c2

∂x ∂u ∂y J ( u, υ, w ) = ∂u ∂z ∂u

∂x ∂υ ∂y ∂υ ∂z ∂υ

∂x ∂w ∂( x , y, z ) ∂y = . ∂( u, υ, w ) ∂w ∂z ∂w

This determinant measures how much the volume near a point in G is being expanded or contracted by the transformation from ( u, υ, w ) to ( x , y, z ) coordinates. As in the two-dimensional case, the derivation of the change-of-variable formula in Equation (7) is omitted. For cylindrical coordinates, r, θ, and z take the place of u, υ, and w. The transformation from Cartesian r θ z-space to Cartesian xyz-space is given by the equations x = r cos θ,

y = r sin θ,

z = z

14.8  Substitutions in Multiple Integrals

Rectangular box with sides parallel to the coordinate axes

z

G u r

(Figure 14.64). The Jacobian of the transformation is ∂x ∂r ∂y J ( r , θ, z ) = ∂r ∂z ∂r

∂x ∂θ ∂y ∂θ ∂z ∂θ

∂x ∂z cos θ ∂y = sin θ ∂z 0 ∂z ∂z

−r sin θ

0

r cos θ

0 = r cos 2 θ + r sin 2 θ = r.

0

1

The corresponding version of Equation (7) is

Cartesian ruz-space

∫∫∫ F ( x , y, z ) dx dy dz

x = r cos u y = r sin u z = z

=

D

∫∫∫ H ( r , θ, z ) r

dr d θ dz.

G

We can drop the absolute value signs because r ≥ 0. For spherical coordinates, ρ , φ, and θ take the place of u, υ, and w. The transformation from Cartesian ρφθ-space to Cartesian xyz-space is given by

z z = constant D

u = constant x

921

x = ρ sin φ cos θ,

z = ρ cos φ

(Figure 14.65). The Jacobian of the transformation (see Exercise 23) is ∂x ∂ρ ∂y J ( ρ , φ, θ ) = ∂ρ ∂z ∂ρ

r = constant y

Cartesian xyz-space

FIGURE 14.64

The equations x = r cos θ,   y = r sin θ, and z = z transform the rectangular box G into a cylindrical wedge D.

y = ρ sin φ sin θ,

∂x ∂φ ∂y ∂φ ∂z ∂φ

∂x ∂θ ∂y = ρ 2 sin φ. ∂θ ∂z ∂θ

The corresponding version of Equation (7) is

∫∫∫ F ( x , y, z ) dx dy dz

=

D

u

∫∫∫ H ( ρ, φ, θ ) ρ 2 sin φ

dρ d φ d θ.

G

Rectangular box with sides parallel to the coordinate axes

u = constant

r = constant

z

D

(x, y, z) f

x = r sin f cos u y = r sin f sin u z = r cos f

r

f = constant

G f r

Cartesian rfu-space

u x

y

Cartesian xyz-space

The equations x = ρ sin φ cos θ,   y = ρ sin φ sin θ, and z = ρ cos φ transform the rectangular box G into the spherical wedge D.

FIGURE 14.65

We can drop the absolute value signs because sin φ is never negative for 0 ≤ φ ≤ π. Note that this is the same result we obtained in Section 14.7. Here is an example of another substitution. Although we could evaluate the integral in this example directly, we have chosen it to illustrate the substitution method in a simple (and fairly intuitive) setting.

922

Chapter 14 Multiple Integrals

w

EXAMPLE 5

Evaluate

1

3

G 2

1 u

4

x =( y 2 )+1

∫0 ∫0 ∫ x = y 2 by applying the transformation y

u = ( 2 x − y ) 2,

x=u+y y = 2y z = 3w

υ = y 2,

w = z 3

(8)

and integrating over an appropriate region in uυw-space. Solution We sketch the solid region D of integration in xyz-space and identify its boundaries (Figure 14.66). In this case, the bounding surfaces are planes. To apply Equation (7), we need to find the corresponding uυw-region G and the Jacobian of the transformation. To find them, we first solve Equations (8) for x, y, and z in terms of u, υ, and w. Routine algebra gives

z Rear plane: y x = , or y = 2x 2

3

( 2x 2− y + 3z ) dx dy dz

x = u + υ,

D

y = 2υ,

z = 3w.

(9)

We then find the boundaries of G by substituting these expressions into the equations for the boundaries of D. 1 4

y

x Front plane: y x = + 1, or y = 2x − 2 2

FIGURE 14.66

The equations x = u + υ,   y = 2υ, and z = 3w transform G into D. Reversing the transformation by the equations u = ( 2 x − y ) 2,   υ = y 2, and w = z 3 transforms D into G (Example 5).

Corresponding uυw-equations for the boundary of G

Simplified uυw-equations

u + υ = 2υ 2 = υ

u = 0

u + υ = ( 2υ 2 ) + 1 = υ + 1

u =1

y = 0

2υ = 0

υ = 0

y = 4

2υ = 4

υ = 2

z = 0

3w = 0

w = 0

z = 3

3w = 3

w =1

xyz-equations for the boundary of D x = y 2 x = ( y 2) + 1

The Jacobian of the transformation, again from Equations (9), is ∂x ∂u ∂y J ( u, υ, w ) = ∂u ∂z ∂u

∂x ∂υ ∂y ∂υ ∂z ∂υ

∂x ∂w 1 ∂y = 0 ∂w 0 ∂z ∂w

1

0

2

0 = 6.

0

3

We now have everything we need to apply Equation (7): 3

4

x = ( y 2 )+ 1

∫0 ∫0 ∫ x = y 2 1

2

1

1

2

1

( 2x 2− y + 3z ) dx dy dz

=

∫0 ∫0 ∫0 ( u + w )

=

∫0 ∫0 ∫0 ( u + w )( 6 ) du d υ dw

= 6∫

1

2

0 ∫0

J ( u, υ, w ) du d υ dw

( 12 + w ) d υ dw = 6∫ 1

⎡ ⎤ = 6 ⎢ w + w 2 ⎥ = 6( 2 ) = 12. ⎣ ⎦0

1 0

= 6∫

1

2

0 ∫0

u =1

⎡ u 2 + uw ⎤ d υ dw ⎢ ⎥ ⎣ 2 ⎦ u=0

υ=2

1 ⎡ υ + υw ⎤ dw = 6 ∫ (1 + 2 w ) dw ⎢⎣ 2 ⎥⎦ 0 υ=0

14.8  Substitutions in Multiple Integrals

923

14.8

EXERCISES

Jacobians and Transformed Regions in the Plane

1. a. Solve the system u = x − y,

8. Use the transformation and parallelogram R in Exercise 4 to evaluate the integral

∫∫

υ = 2x + y

for x and y in terms of u and υ. Then find the value of the Jacobian ∂( x , y ) ∂( u, υ ). b. Find the image under the transformation u = x − y, υ = 2 x + y of the triangular region with vertices ( 0, 0 ), (1, 1), and (1, −2 ) in the xy-plane. Sketch the transformed region in the uυ-plane. 2. a. Solve the system

2( x − y ) dx dy.

R

9. Let R be the region in the first quadrant of the xy-plane bounded by the hyperbolas xy = 1,   xy = 9 and the lines y = x ,   y = 4 x. Use the transformation x = u υ,   y = uυ with u > 0 and υ > 0 to rewrite ⎛ y ⎞ ∫∫ ⎜⎜⎝ x + xy ⎟⎟⎟⎠ dx dy R

u = x + 2 y,

υ = x−y

for x and y in terms of u and υ. Then find the value of the Jacobian ∂( x , y ) ∂( u, υ ). b. Find the image under the transformation u = x + 2 y, υ = x − y of the triangular region in the xy-plane bounded by the lines y = 0,   y = x , and x + 2 y = 2. Sketch the transformed region in the uυ-plane. 3. a. Solve the system

as an integral over an appropriate region G in the uυ-plane. Then evaluate the uυ-integral over G. 10. a. Find the Jacobian of the transformation x = u,   y = uυ and sketch the region G :1 ≤ u ≤ 2,  1 ≤ uυ ≤ 2, in the uυ-plane. b. Then use Equation (2) to transform the integral 2

∫1 ∫1

2

y dy dx x

into an integral over G, and evaluate both integrals.

u = 3 x + 2 y,

υ = x + 4y

for x and y in terms of u and υ. Then find the value of the Jacobian ∂( x , y ) ∂( u, υ ). b. Find the image under the transformation u = 3 x + 2 y, υ = x + 4 y of the triangular region in the xy-plane bounded by the x-axis, the y-axis, and the line x + y = 1. Sketch the transformed region in the uυ-plane. 4. a. Solve the system u = 2 x − 3 y,

υ = −x + y

for x and y in terms of u and υ. Then find the value of the Jacobian ∂( x , y ) ∂( u, υ ). b. Find the image under the transformation u = 2 x − 3 y, υ = −x + y of the parallelogram R in the xy-plane with boundaries x = −3,   x = 0,   y = x , and y = x + 1. Sketch the transformed region in the uυ-plane. Substitutions in Double Integrals

5. Evaluate the integral 2x − y dx dy 2 from Example 1 directly by integration with respect to x and y to confirm that its value is 2. 4

11. Polar moment of inertia of an elliptical plate A thin plate of constant density covers the region bounded by the ellipse x 2 a 2 + y 2 b 2 = 1,   a > 0,   b > 0, in the xy-plane. Find the first moment of the plate about the origin. (Hint: Use the transformation x = ar cos θ,   y = br sin θ.) 12. The area of an ellipse The area πab of the ellipse x 2 a 2 + y 2 b 2 = 1 can be found by integrating the function f ( x , y ) = 1 over the region bounded by the ellipse in the xy-plane. Evaluating the integral directly requires a trigonometric substitution. An easier way to evaluate the integral is to use the transformation x = au,   y = bυ and evaluate the transformed integral over the disk G : u 2 + υ 2 ≤ 1 in the uυ-plane. Find the area this way. 13. Use the transformation in Exercise 2 to evaluate the integral + 2 y ) e ( y− x ) dx dy

by first writing it as an integral over a region G in the uυ-plane.

x =( y 2 )+1

∫∫ ( 2 x 2 − xy − y 2 ) dx dy

(x

14. Use the transformation x = u + (1 2 ) υ,   y = υ to evaluate the integral 2

∫0 ∫ x = y 2

6. Use the transformation in Exercise 1 to evaluate the integral

2− 2 y

23

∫0 ∫ y

( y+ 4 )

2

∫0 ∫ y 2

y 3 ( 2 x − y ) e ( 2 x − y ) dx dy 2

by first writing it as an integral over a region G in the uυ-plane. 15. Use the transformation x = u υ,   y = uυ to evaluate the integral sum 2

y

4

4 y

∫1 ∫1 y( x 2 + y 2 ) dx dy + ∫2 ∫ y 4 ( x 2 + y 2 ) dx dy.

R

for the region R in the first quadrant bounded by the lines y = −2 x + 4,   y = −2 x + 7,   y = x − 2, and y = x + 1. 7. Use the transformation in Exercise 3 to evaluate the integral

∫∫

( 3 x 2 + 14 xy + 8 y 2 ) dx dy

R

for the region R bounded by the lines y = −( 3 2 ) x + 1, y = −( 3 2 ) x + 3,   y = −(1 4 ) x , and y = −(1 4 ) x + 1.

16. Use the transformation x = u 2 − υ 2 ,   y = 2uυ to evaluate the integral 1

2 1− x

∫0 ∫0

x 2 + y 2 dy dx.

(Hint: Show that the image of the triangular region G with vertices ( 0, 0 ), (1, 0 ), (1, 1) in the uυ-plane is the region of integration R in the xy-plane defined by the limits of integration.)

924

Chapter 14 Multiple Integrals

Substitutions in Triple Integrals

17. Evaluate the integral in Example 5 by integrating with respect to x, y, and z. 18. Volume of a solid ellipsoid Find the volume of the solid ellipsoid y2 x2 z2 + 2 + 2 ≤ 1. 2 a b c (Hint: Let x = au,   y = bυ, and z = cw. Then find the volume of an appropriate region in uυw-space.) 19. Evaluate

∫∫∫

xyz dx dy dz

D

over the solid ellipsoid D, y2 x2 z2 + 2 + 2 ≤ 1. a2 b c (Hint: Let x = au,   y = bυ, and z = cw. Then integrate over an appropriate region in uυw-space.) 20. Let D be the solid region in xyz-space defined by the inequalities 1 ≤ x ≤ 2, 0 ≤ xy ≤ 2, 0 ≤ z ≤ 1. Evaluate

∫∫∫ ( x 2 y + 3xyz ) dx dy dz D

by applying the transformation u = x , υ = xy, w = 3z and integrating over an appropriate region G in uυw-space. Theory and Examples

21. Find the Jacobian ∂( x , y ) ∂( u, υ ) of the transformation a. x = u cos υ, y = u sin υ b. x = u sin υ, y = u cos υ. 22. Find the Jacobian ∂( x , y, z ) ∂( u, υ, w ) of the transformation a. x = u cos υ, y = u sin υ, z = w b. x = 2u − 1, y = 3υ − 4, z = (1 2 )( w − 4 ).

CHAPTER 14

23. Evaluate the appropriate determinant to show that the Jacobian of the transformation from Cartesian ρφθ-space to Cartesian xyzspace is ρ 2 sin φ. 24. Substitutions in single integrals How can substitutions in single definite integrals be viewed as transformations of regions? What is the Jacobian in such a case? Illustrate with an example. 25. Centroid of a solid semi-ellipsoid Assuming the result that the centroid of a solid hemisphere lies on the axis of symmetry three-eighths of the way from the base toward the top, show, by transforming the appropriate integrals, that the center of mass of a solid semi-ellipsoid ( x 2 a 2 ) + ( y 2 b 2 ) + ( z 2 c 2 ) ≤ 1,   z ≥ 0, lies on the z-axis three-eighths of the way from the base toward the top. (You can do this without evaluating any of the integrals.) 26. Cylindrical shells In Section 6.2, we learned how to find the volume of a solid of revolution using the shell method. Specifically, if the region between the curve y = f ( x ) and the x-axis from a to b ( 0 < a < b ) is revolved about the y-axis, the b volume of the resulting solid is ∫a 2π x f ( x ) dx. Prove that finding volumes by using triple integrals gives the same result. (Hint: Use cylindrical coordinates with the roles of y and z changed.) 27. Inverse transform The equations x = g ( u, υ ),   y = h ( u, υ ) in Figure 14.57 transform the region G in the uυ-plane into the region R in the xy-plane. Since the substitution transformation is one-to-one with continuous first partial derivatives, it has an inverse transformation, and there are equations u = α( x , y ),   υ = β ( x , y ) with continuous first partial derivatives transforming R back into G. Moreover, the Jacobian determinants of the transformations are related reciprocally by ⎛ ∂( u, υ ) ⎞⎟−1 ∂( x , y ) (10) = ⎜⎜ ⎟ . ∂( u, υ ) ⎝⎜ ∂( x , y ) ⎟⎠ Equation (10) is proved in advanced calculus. Use it to find the area of the region R in the first quadrant of the xy-plane bounded by the lines y = 2 x ,  2 y = x , and the curves xy = 2,  2 xy = 1 for u = xy and υ = y x. 28. (Continuation of Exercise 27.) For the region R described in Exercise 27, evaluate the integral ∫∫ R y 2 dA.

Questions to Guide Your Review

1. Define the double integral of a function of two variables over a bounded region in the coordinate plane. 2. How are double integrals evaluated as iterated integrals? Does the order of integration matter? How are the limits of integration determined? Give examples. 3. How are double integrals used to calculate areas and average values. Give examples. 4. How can you change a double integral in rectangular coordinates into a double integral in polar coordinates? Why might it be worthwhile to do so? Give an example. 5. Define the triple integral of a function f ( x , y, z ) over a bounded solid region in space. 6. How are triple integrals in rectangular coordinates evaluated? How are the limits of integration determined? Give an example.

7. How are double and triple integrals in rectangular coordinates used to calculate volumes, average values, masses, moments, and centers of mass? Give examples. 8. How are triple integrals defined in cylindrical and spherical coordinates? Why might one prefer working in one of these coordinate systems to working in rectangular coordinates? 9. How are triple integrals in cylindrical and spherical coordinates evaluated? How are the limits of integration found? Give examples. 10. How are substitutions in double integrals pictured as transformations of regions in the plane? Give a sample calculation. 11. How are substitutions in triple integrals pictured as transformations of solid regions? Give a sample calculation.

Chapter 14  Practice Exercises

CHAPTER 14

Practice Exercises

Evaluating Double Iterated Integrals In Exercises 1–4, sketch the region of integration and evaluate the double integral.

1. 3.

1y

10

∫1 ∫0 32

∫0 ∫−

ye xy dx dy

9− 4 t 2 9− 4 t 2

t ds dt

x3

1

2.

∫0 ∫0

4.

∫0 ∫

1

e y x dy dx

2− y

5. 7.

( y−4 )

∫0 ∫− 32

2

4− y

∫0 ∫−

dx dy

9− 4 y 2 9− 4 y 2

6.

y dx dy

8.

1

∫0 ∫ x 2

x 2

11.

1

2

2

1

∫0 ∫2 y 4 cos ( x 2 ) dx dy

10.

∫0 ∫ y 2 e x

dy dx y4 + 1

12.

∫0 ∫

8

∫0 ∫

2 3

x

1

1 3y

a. Triangular region The triangle with vertices ( 0, 0 ), (1, 0 ), and (1,   3 ). b. First quadrant The first quadrant of the xy-plane. Evaluating Triple Iterated Integrals Evaluate the integrals in Exercises 23–26. π

2

dx dy

2π sin πx 2 dx dy x2

Areas and Volumes Using Double Integrals

π

π

23.

∫0 ∫0 ∫0

24.

∫ln 6 ∫0 ∫ln 4 e x + y+ z

25.

∫0 ∫0 ∫0

26.

∫1 ∫1 ∫0

ln 7

2 x dy dx

Evaluate the integrals in Exercises 9–12. 9.

22. Integrate f ( x , y ) = 1 (1 + x 2 + y 2 ) 2 over

x dy dx

4− x 2

∫0 ∫0

21. Integrating over a lemniscate Integrate the function f ( x , y ) = 1 (1 + x 2 + y 2 ) 2 over the region enclosed by one loop of the lemniscate ( x 2 + y 2 ) 2 − ( x 2 − y 2 ) = 0.

xy dx dy

y

In Exercises 5–8, sketch the region of integration and write an equivalent integral with the order of integration reversed. Then evaluate both integrals. 4

925

ln 2

1

x2

e

x

cos( x + y + z ) dx dy dz ln 5

x+ y

z

(

( 2x

)

dz dy dx

− y − z ) dz dy dx

2y dy dz dx z3

Volumes and Average Values Using Triple Integrals

27. Volume Find the volume of the wedge-shaped solid region enclosed on the side by the cylinder x = − cos y,   −π 2 ≤ y ≤ π 2, on the top by the plane z = −2 x , and below by the xy-plane.

13. Area between line and parabola Find the area of the region enclosed by the line y = 2 x + 4 and the parabola y = 4 − x 2 in the xy-plane. 14. Area bounded by lines and parabola Find the area of the “triangular” region in the xy-plane that is bounded on the right by the parabola y = x 2 , on the left by the line x + y = 2, and above by the line y = 4. 15. Volume of the region under a paraboloid Find the volume under the paraboloid z = x 2 + y 2 above the triangle enclosed by the lines y = x ,   x = 0, and x + y = 2 in the xy-plane. 16. Volume of the region under a parabolic cylinder Find the volume under the parabolic cylinder z = x 2 above the region enclosed by the parabola y = 6 − x 2 and the line y = x in the xy-plane.

z

z = −2x

x = −cos y

−p 2

p 2

x

y

28. Volume Find the volume of the solid that is bounded above by the cylinder z = 4 − x 2 , on the sides by the cylinder x 2 + y 2 = 4, and below by the xy-plane.

Average Values Find the average value of f ( x , y ) = xy over the regions in Exercises 17 and 18.

z z = 4 − x2

17. The square bounded by the lines x = 1,   y = 1 in the first quadrant 18. The quarter circle x 2 + y 2 ≤ 1 in the first quadrant Polar Coordinates Evaluate the integrals in Exercises 19 and 20 by changing to polar coordinates.

19.

20.

1

1− x 2

∫−1 ∫− 1− x 1

2

1− y 2

∫−1 ∫− 1− y

2

2 dy dx

x

(1 + x 2 + y 2 )

ln ( x 2

+

y2

y x2 + y2 = 4

2

+ 1) dx dy

29. Average value Find the average value of f ( x , y, z ) = 30 xz x 2 + y over the rectangular solid in the first octant bounded by the coordinate planes and the planes x = 1, y = 3, z = 1.

926

Chapter 14 Multiple Integrals

30. Average value Find the average value of ρ over the ball ρ ≤ a (spherical coordinates). Cylindrical and Spherical Coordinates

31. Cylindrical to rectangular coordinates Convert 2π

4 −r 2

2

∫ 0 ∫ 0 ∫r

3r dz dr d θ,

r ≥ 0

39. Moment of inertia of a “thick” sphere Find the moment of inertia of a solid of constant density δ bounded by two concentric spheres of radii a and b ( a < b ) about a diameter. 40. Moment of inertia of an apple Find the moment of inertia about the z-axis of a solid of density δ = 1 enclosed by the spherical coordinate surface ρ = 1 − cos φ. The solid is the red curve rotated about the z-axis in the accompanying figure.

to (a) rectangular coordinates with the order of integration dz dx dy and (b) spherical coordinates. Then (c) evaluate one of the integrals.

z

r = 1 − cos f

32. Rectangular to cylindrical coordinates (a) Convert to cylindrical coordinates. Then (b) evaluate the new integral. (x 2 + y2 )

1− x 2

1

∫0 ∫− 1− x ∫−( x 2

2 + y2 )

21xy 2 dz dy dx

33. Rectangular to spherical coordinates (a) Convert to spherical coordinates. Then (b) evaluate the new integral. 1

1− x 2

∫−1 ∫− 1− x ∫ 2

1 x 2 + y2

35. Cylindrical to rectangular coordinates Set up an integral in rectangular coordinates equivalent to the integral 3

∫0 ∫1 ∫1

4 −r 2

r 3 ( sin θ cos θ ) z 2 dz dr dθ.

Arrange the order of integration to be z first, then y, then x. 36. Rectangular to cylindrical coordinates The volume of a solid is 2

∫0 ∫0

2x−x 2

∫−

x

dz dy dx

34. Rectangular, cylindrical, and spherical coordinates Write an iterated triple integral for the integral of f ( x , y, z ) = 6 + 4 y over the region in the first octant bounded by the cone z = x 2 + y 2 , the cylinder x 2 + y 2 = 1, and the coordinate planes in (a) rectangular coordinates, (b) cylindrical coordinates, and (c) spherical coordinates. Then (d) find the integral of f by evaluating one of the triple integrals.

π2

y

4− x 2 − y 2 4− x 2 − y 2

41. Centroid Find the centroid of the “triangular” region bounded by the lines x = 2,   y = 2 and the hyperbola xy = 2 in the xy-plane. 42. Centroid Find the centroid of the region between the parabola x + y 2 − 2 y = 0 and the line x + 2 y = 0 in the xy-plane. 43. Polar moment Find the polar moment of inertia about the origin of a thin triangular plate of constant density δ = 3 bounded by the y-axis and the lines y = 2 x and y = 4 in the xy-plane. 44. Polar moment Find the polar moment of inertia about the center of a thin rectangular sheet of constant density δ = 1 bounded by the lines a. x = ±2, y = ± 1 in the xy-plane b. x = ±a, y = ±b in the xy-plane.

dz dy dx.

a. Describe the solid by giving equations for the surfaces that form its boundary. b. Convert the integral to cylindrical coordinates, but do not evaluate the integral. 37. Spherical versus cylindrical coordinates Triple integrals involving spherical shapes do not always require spherical coordinates for convenient evaluation. Some calculations may be accomplished more easily with cylindrical coordinates. As a case in point, find the volume of the solid region bounded above by the sphere x 2 + y 2 + z 2 = 8 and below by the plane z = 2 by using (a) cylindrical coordinates and (b) spherical coordinates. Masses and Moments

38. Finding Iz in spherical coordinates Find the moment of inertia about the z-axis of a solid of constant density δ = 1 that is bounded above by the sphere ρ = 2 and below by the cone φ = π 3 (spherical coordinates).

(Hint: Find Ix . Then use the formula for Ix to find Iy , and add the two to find I0 .) 45. Inertial moment Find the moment of inertia about the x-axis of a thin plate of constant density δ covering the triangle with vertices ( 0, 0 ), ( 3, 0 ), and ( 3, 2 ) in the xy-plane. 46. Plate with variable density Find the center of mass and the moments of inertia about the coordinate axes of a thin plate bounded by the line y = x and the parabola y = x 2 in the xyplane if the density is δ ( x , y ) = x + 1. 47. Plate with variable density Find the mass and first moments about the coordinate axes of a thin square plate bounded by the lines x = ±1,   y = ±1 in the xy-plane if the density is δ ( x , y ) = x 2 + y 2 + 1 3. 48. Triangles with same inertial moment Find the moment of inertia about the x-axis of a thin triangular plate of constant density δ whose base lies along the interval [ 0, b ] on the x-axis and whose vertex lies on the line y = h above the x-axis. As you will see, it does not matter where on the line this vertex lies. All such triangles have the same moment of inertia about the x-axis.

927

Chapter 14  Additional and Advanced Exercises

49. Centroid Find the centroid of the region in the polar coordinate plane defined by the inequalities 0 ≤ r ≤ 3,   −π 3 ≤ θ ≤ π 3. 50. Centroid Find the centroid of the region in the first quadrant bounded by the rays θ = 0 and θ = π 2 and the circles r = 1 and r = 3. 51. a. Centroid Find the centroid of the region in the polar coordinate plane that lies inside the cardioid r = 1 + cos θ and outside the circle r = 1.

Substitutions

53. Show that if u = x − y and υ = y, then for any continuous f , ∞

∞

e −s( u + υ )   f ( u, υ ) du d υ.

54. What relationship must hold between the constants a, b, and c to make ∞

∞

∫−∞ ∫−∞ e −( ax

b. Sketch the region and show the centroid in your sketch. 52. a. Centroid Find the centroid of the plane region defined by the polar coordinate inequalities 0 ≤ r ≤ a, −α ≤ θ ≤ α (0 < α ≤ π). How does the centroid move as α → π − ?

∞

x

∫0 ∫0 e −sx  f ( x − y, y ) dy dx =∫0 ∫0

2 + 2 bxy + cy 2 )

dx dy = 1?

(Hint: Let s = αx + β y and t = γ x + δ y, where ( αδ − βγ ) 2 = ac − b 2 . Then ax 2 + 2bxy + cy 2 = s 2 + t 2 .)

b. Sketch the region for α = 5π 6 and show the centroid in your sketch.

CHAPTER 14

Additional and Advanced Exercises

Volumes

1. Sand pile: double and triple integrals The base of a sand pile covers the region in the xy-plane that is bounded by the parabola x 2 + y = 6 and the line y = x. The height of the sand above the point ( x , y ) is x 2 . Express the volume of sand as (a) a double integral and (b) a triple integral. Then (c) find the volume. 2. Water in a hemispherical bowl A hemispherical bowl of radius 5 cm is filled with water to within 3 cm of the top. Find the volume of water in the bowl. 3. Solid cylindrical region between two planes Find the volume of the portion of the solid cylinder x 2 + y 2 ≤ 1 that lies between the planes z = 0 and x + y + z = 2. 4. Sphere and paraboloid Find the volume of the solid region bounded above by the sphere x 2 + y 2 + z 2 = 2 and below by the paraboloid z = x 2 + y 2 . 5. Two paraboloids Find the volume of the solid region bounded above by the paraboloid z = 3 − x 2 − y 2 and below by the paraboloid z = 2 x 2 + 2 y 2 .

7. Hole in solid ball A circular cylindrical hole is bored through a ball, the axis of the hole being a diameter of the sphere. The volume of the remaining solid is V = 2  ∫

2π 0

3

∫0 ∫1

4−z 2

r dr dz d θ.

a. Find the radius of the hole and the radius of the sphere. b. Evaluate the integral. 8. Ball and cylinder Find the volume of material cut from the ball r 2 + z 2 ≤ 9 by the cylinder r = 3 sin θ. 9. Two paraboloids Find the volume of the solid region enclosed by the surfaces z = x 2 + y 2 and z = ( x 2 + y 2 + 1) 2. 10. Cylinder and surface z = xy Find the volume of the solid region in the first octant that lies between the cylinders r = 1 and r = 2 and is bounded below by the xy-plane and above by the surface z = xy. Changing the Order of Integration

11. Evaluate the integral

6. Spherical coordinates Find the volume of the solid region enclosed by the spherical coordinate surface ρ = 2 sin φ (see accompanying figure).

∞

∫0

e −ax − e −bx dx. x

(Hint: Use the relation e −ax − e −bx = x

z r = 2 sin f

b

∫a e −xy dy

to form a double integral, and evaluate the integral by changing the order of integration.) 12. a. Polar coordinates Show, by changing to polar coordinates, that y a sin β

∫0

a2−y2

∫ y cot β

(

ln ( x 2 + y 2 ) dx dy = a 2 β ln a −

x

where a > 0 and 0 < β < π 2.

)

1 , 2

928

Chapter 14 Multiple Integrals

b. Rewrite the Cartesian integral with the order of integration reversed. 13. Reducing a double to a single integral By changing the order of integration, show that the following double integral can be reduced to a single integral: x

u

∫0 ∫0 e m x −t  f (t ) dt du (

)

x

∫0 ( x − t )e m x −t  f (t ) dt.

=

(

)

over the rectangle x 0 ≤ x ≤ x 1 ,   y 0 ≤ y ≤ y1 is F ( x 1 , y1 ) − F ( x 0 , y1 ) − F ( x 1 , y 0 ) + F ( x 0 , y 0 ). 21. Suppose that f ( x , y ) can be written as a product f ( x , y ) = F ( x ) G ( y) of a function of x and a function of y. Then the integral of f over the rectangle R: a ≤ x ≤ b,   c ≤ y ≤ d can be evaluated as a product as well, by the formula

∫∫

Similarly, it can be shown that x

υ

u

R

∫0 ∫0 ∫0 e m x −t  f (t ) dt du d υ (

)

=

x (x

∫0

− t ) 2 m( x −t )   f (t ) dt. e 2

∫0

f (x)

(∫

x 0

∫∫

f ( x , y ) dA =

)

=

∫0

f ( y)

(∫

1 y

)

g ( x − y ) f ( x ) dx dy

1 1 1 = ∫ ∫ g ( x − y ) f ( x ) f ( y) dx dy. 2 0 0 Masses and Moments

16. Polar inertia of triangular plate Find the polar moment of inertia about the origin of a thin triangular plate of constant density δ = 3 bounded by the y-axis and the lines y = 2 x and y = 4 in the xy-plane. 17. Mass and polar inertia of a counterweight The counterweight of a flywheel of constant density 1 has the form of the smaller segment cut from a circle of radius a by a chord at a distance b from the center ( b < a ). Find the mass of the counterweight and its polar moment of inertia about the center of the wheel. 18. Centroid of a boomerang Find the centroid of the boomerangshaped region between the parabolas y 2 = −4 ( x − 1) and y 2 = −2( x − 2 ) in the xy-plane. Theory and Examples

a

b

∫0 ∫0 e max( b x

2 2 ,  a 2 y 2 )

dy dx,

where a and b are positive numbers and ⎪⎧ b 2 x 2 max ( b 2 x 2 , a 2 y 2 ) = ⎪⎨ 2 2 ⎪⎪ a y ⎪⎩

if  b 2 x 2 < a 2 y 2 .

20. Show that

∫∫

∂ 2 F ( x, y ) dx dy ∂x ∂y

)

d

G ( y) dy .

c

(1)

∫c ( ∫ a d

b

)

(i)

=

∫c ( G ( y)∫a F ( x ) dx ) dy

(ii)

=

∫c ( ∫a F ( x ) dx ) G( y) dy

(iii)

=

(∫

(iv)

ln 2

π2

∫0 ∫0

d

b

d

b

b a

F ( x ) dx

e x cos y dy dx

)∫

d c

G ( y ) dy.

c.

2

1

x

∫1 ∫−1 y 2 dx dy

22. Let D u f denote the derivative of f ( x , y ) = ( x 2 + y 2 ) 2 in the direction of the unit vector u = u1i + u 2 j. a. Finding average value Find the average value of D u f over the triangular region cut from the first quadrant by the line x + y = 1. b. Average value and centroid Show in general that the average value of D u f over a region in the xy-plane is the value of D u f at the centroid of the region. 23. The value of Γ( 1 2 ) The gamma function, Γ( x ) =

∞

∫0

t x −1   e −t dt ,

extends the factorial function from the nonnegative integers to other real values. Of particular interest in the theory of differential equations is the number

( 12 ) = ∫

∞ 0

t (1 2 )−1 e −t dt =

∞

∫0

e −t dt. t

(2)

a. If you have not yet done Exercise 41 in Section 14.4, do it now to show that I =

if  b 2 x 2 ≥ a 2 y 2

)( ∫

a. Give reasons for Steps (i) through (iv).

Γ

19. Evaluate

F ( x ) dx

When it applies, Equation (1) can be a time-saver. Use it to evaluate the following integrals. b.

15. Minimizing polar inertia A thin plate of constant density is to occupy the triangular region in the first quadrant of the xy-plane having vertices ( 0, 0 ), ( a, 0 ), and ( a, 1 a ). What value of a will minimize the plate’s polar moment of inertia about the origin?

b a

F ( x ) G ( y) dx dy

R

g ( x − y ) f ( y) dy dx 1

(∫

The argument is that

14. Transforming a double integral to obtain constant limits Sometimes a multiple integral with variable limits can be changed into one with constant limits. By changing the order of integration, show that 1

f ( x , y ) dA =

∞

∫0

e − y 2 dy =

π . 2

b. Substitute y = t in Equation (2) to show that Γ (1 2 ) = 2 I = π . 24. Total electrical charge over circular plate The electrical charge distribution on a circular plate of radius R meters is σ ( r , θ ) = kr (1 − sin θ ) coulomb m 2 (k a constant). Integrate σ over the plate to find the total charge Q.

Chapter 14  Additional and Advanced Exercises

suspended 1 unit above the origin and its axis the ray from ( 0, 0, 1) to ∞. Use cylindrical coordinates to evaluate

25. A parabolic rain gauge A bowl is in the shape of the graph of z = x 2 + y 2 from z = 0 to z = 30 cm. You plan to calibrate the bowl to make it into a rain gauge. What height in the bowl would correspond to 3 cm of rain? 9 cm of rain? 26. Water in a satellite dish A parabolic satellite dish is 2 m wide and 1 2 m deep. Its axis of symmetry is tilted 30 degrees from the vertical. a. Set up, but do not evaluate, a triple integral in rectangular coordinates that gives the amount of water the satellite dish will hold. (Hint: Put your coordinate system so that the satellite dish is in “standard position” and the plane of the water level is slanted.) (Caution: The limits of integration are not “nice.”) b. What would be the smallest tilt of the satellite dish so that it holds no water?

∫∫∫ z ( r 2 + z 2 )

−5 2

dV.

D

b

28. Hypervolume We have learned that ∫a 1  dx is the length of the interval [ a, b ] on the number line (one-dimensional space), ∫∫ R 1 dA is the area of region R in the xy-plane (two-dimensional space), and ∫∫∫D 1 dV is the volume of the region D in threedimensional space (xyz-space). We could continue: If Q is a region in 4-space (xyzw-space), then ∫∫∫∫ Q 1 dV is the “hyper-volume” of Q. Use your generalizing abilities and a Cartesian coordinate system of 4-space to find the hypervolume inside the unit fourdimensional sphere x 2 + y 2 + z 2 + w 2 = 1.

27. An infinite half-cylinder Let D be the interior of the infinite right circular half-cylinder of radius 1 with its single-end face

CHAPTER 14

929

Technology Application Projects

Mathematica/Maple Projects

Projects can be found within MyLab Math. • Take Your Chances: Try the Monte Carlo Technique for Numerical Integration in Three Dimensions Use the Monte Carlo technique to integrate numerically in three dimensions. • Means and Moments and Exploring New Plotting Techniques, Part II Use the method of moments in a form that makes use of geometric symmetry as well as multiple integration.

15

Integrals and Vector Fields OVERVIEW In this chapter we extend the theory of integration to functions whose domains are curves and surfaces in space. The resulting line and surface integrals give powerful mathematical tools for science and engineering. Line integrals are used to find the work done by a force in moving an object along a path and to find the mass of a curved wire with variable density. Surface integrals are used to find the rate of flow of a fluid across a surface and to describe the interactions of electric and magnetic forces. We present the fundamental theorems of vector integral calculus and discuss their mathematical consequences and physical applications. The theorems of vector calculus are then shown to be generalized versions of the Fundamental Theorem of Calculus.

15.1 Line Integrals of Scalar Functions z t=b

r(t)

Δs k x

t=a

FIGURE 15.1

(x k , yk , z k )

y

The curve r(t ) partitioned into small arcs from t = a to t = b. The length of a typical subarc is Δs k .

To calculate the total mass of a wire lying along a curve in space, or to find the work done by a variable force acting along such a curve, we need a more general notion of integral than was defined in Chapter 5. We need to integrate over a curve C rather than over an interval [ a, b ]. These more general integrals are called line integrals (although path integrals might be more descriptive). We make our definitions for space curves, with curves in the xy-plane being the special case with z-coordinate identically zero. Suppose that f ( x , y, z ) is a real-valued function we wish to integrate over the curve C lying within the domain of f and parametrized by r (t ) = g(t ) i + h(t ) j + k (t )k,  a ≤ t ≤ b. The values of f along the curve are given by the composite function f ( g(t ), h(t ), k (t ) ). We are going to integrate this composition with respect to arc length from t = a to t = b. To begin, we first partition the curve C into a finite number n of subarcs (Figure 15.1). The typical subarc has length Δs k . In each subarc we choose a point ( x k , y k , z k ) and form the sum Sn =

n

( x k , y k , z k ) Δs k , ∑ f   k =1

value of f at a point on the subarc

length of a small subarc of the curve

which is similar to a Riemann sum. Depending on how we partition the curve C and pick ( x k , y k , z k ) in the kth subarc, we may get different values for S n. If f is continuous and the functions g, h, and k have continuous first derivatives, then these sums approach a limit as n increases and the lengths Δs k approach zero. This leads to the following definition, which is similar to that for a single integral. In the definition, we assume that the norm of the partition approaches zero as n → ∞, so that the length of the longest subarc approaches zero.

930

931

15.1  Line Integrals of Scalar Functions

DEFINITION If f is defined on a curve C given parametrically by r (t ) = g(t ) i + h(t ) j + k (t )k,  a ≤ t ≤ b, then the line integral of f over C is

∫C f ( x , y, z ) ds

n

∑ f ( x k , y k , z k ) Δs k , n →∞

= lim

(1)

k =1

provided this limit exists.

If the curve C is smooth for a ≤ t ≤ b (so v = d r dt is continuous and never 0) and the function f is continuous on C, then the limit in Equation (1) can be shown to exist. We can then apply the Fundamental Theorem of Calculus to differentiate the arc length equation, s (t ) =

t

∫a

v(τ ) d τ ,

  

Eq. (3) of Section 12.3 with t 0 = a

to express ds in Equation (1) as ds = v(t ) dt and evaluate the integral of f over C as ds = v = dt

( dxdt )

2

+

( dydt ) + ( dzdt ) 2

2

∫C f ( x , y, z ) ds

=

b

∫a

f ( g(t ), h(t ), k (t ) ) v(t ) dt.

(2)

The integral on the right side of Equation (2) is just an ordinary definite integral, as defined in Chapter 5, where we are integrating with respect to the parameter t. The formula evaluates the line integral on the left side correctly no matter what smooth parametrization is used. Note that the parameter t defines a direction along the path. The starting point on C is the position r(a), and movement along the path is in the direction of increasing t (see Figure 15.1).

How to Evaluate a Line Integral

To integrate a continuous function f ( x , y, z ) over a curve C: 1. Find a smooth parametrization of C, r (t ) = g(t ) i + h(t ) j + k (t )k ,

f ( r(t ) ) = f ( g(t ), h(t ), k (t ) )

a ≤ t ≤ b.

2. Evaluate the integral as

∫C f ( x , y, z ) ds

=

b

∫a

f ( g(t ), h(t ), k (t ) ) v(t ) dt.

z

If f has the constant value 1, then the integral of f over C gives the length of C from t = a to t = b. We also write f ( r(t ) ) for the evaluation f ( g(t ), h(t ), k (t ) ) along the curve r.

(1, 1, 1) C y

EXAMPLE 1 Integrate f ( x , y, z ) = x − 3 y 2 + z over the line segment C joining the origin to the point (1, 1, 1) (Figure 15.2). x (1, 1, 0)

FIGURE 15.2

Example 1.

The integration path in

Solution Since any choice of parametrization will give the same answer, we choose the simplest parametrization we can think of:

r (t ) = t i + t j + t k,

0 ≤ t ≤ 1.

932

Chapter 15 Integrals and Vector Fields

The components have continuous first derivatives, and v(t ) = i + j + k = 1 2 + 1 2 + 1 2 = 3 is never 0, so the parametrization is smooth. The integral of f over C is 1

∫C f ( x , y, z ) ds = ∫0 =

f ( t , t , t ) 3 dt

Eq. (2),  ds = v(t ) dt =

1

∫0 ( t − 3t 2 + t )

=

3 dt

3 dt 1

⎡ ⎤ 3 ⎢ t 2 − t 3 ⎥ = 0. ⎣ ⎦0

1

3 ∫ ( 2t − 3t 2 ) dt = 0

Additivity Line integrals have the useful property that if a piecewise smooth curve C is made by joining a finite number of smooth curves C 1 , C 2 , … , C n end to end (Section 12.1), then the integral of a function over C is the sum of the integrals over the curves that make it up:

∫C f ds z

C2

∫C

f ds + 1

∫C

y

FIGURE 15.3

v =

C 2 : r (t ) = i + j + tk, 0 ≤ t ≤ 1;

C1

Example 2.

∫C

f ds.

(3)

n

Solution We choose the simplest parametrizations for C 1 and C 2 we can find, calculating the lengths of the velocity vectors as we go along:

C 1: r (t ) = t i + tj, 0 ≤ t ≤ 1;

x

f ds +  + 2

Figure 15.3 shows another path from the origin to (1, 1, 1), formed from EXAMPLE 2 two line segments C 1 and C 2 . Integrate f ( x , y, z ) = x − 3 y 2 + z over C 1 ∪ C 2 .

(1, 1, 1)

(0, 0, 0)

=

(1, 1, 0)

The path of integration in

12 + 12 =

v =

2

0 2 + 0 2 + 1 2 = 1.

With these parametrizations we find that

∫C ∪ C 1

f ( x , y, z ) ds = 2

∫C

f ( x , y, z ) ds + 1

1

=

∫0

=

∫0 ( t − 3t 2 + 0 )

=

1

∫C

f ( t , t , 0 ) 2 dt +

f ( x , y, z ) ds

Eq. (3)

2

1

∫0

f (1, 1, t )(1) dt

2 dt +

Eq. (2)

1

∫0 (1 − 3 + t )(1) dt

1 1 t2 t2 2 3 2 ⎡⎢ − t 3 ⎤⎥ + ⎡⎢ − 2t ⎤⎥ = − − . 2 2 ⎣2 ⎦0 ⎣2 ⎦0

Notice three things about the integrations in Examples 1 and 2. First, as soon as the components of the appropriate curve were substituted into the formula for f , the integration became a standard integration with respect to t. Second, the integral of f over C 1 ∪ C 2 was obtained by integrating f over each section of the path and adding the results. Third, the integrals of f over C and C 1 ∪ C 2 had different values. We investigate this third observation in Section 15.3.

15.1  Line Integrals of Scalar Functions

933

The value of a line integral along a path joining two points can change if you change the path between them.

z

EXAMPLE 3 Find the line integral of f ( x , y, z ) = 2 xy + r (t ) = cos t i + sin tj + tk, 0 ≤ t ≤ π.

1

z over the helix

Solution For the helix (Figure 15.4) we find v(t ) = r ′(t ) = − sin t i + cos tj + k and

v(t ) = (− sin t ) 2 + ( cos t ) 2 + 1 = we obtain

2. Evaluating the function f at the point r(t ),

f ( r(t ) ) = f ( cos t , sin t , t ) = 2 cos t sin t + −11 −

− 1 −1 x

t.

The line integral is given by

1

1

t = sin 2t +

π

y

∫C f ( x , y, z ) ds = ∫0

FIGURE 15.4

A line integral is taken over a curve such as this helix from Example 3.

π 1 2 2 ⎡⎢ − cos 2t + t 3 2 ⎤⎥ ⎣ 2 ⎦0 3

= =

( sin 2t + t ) 2 dt

2 2 32 π ≈ 5.25. 3

Mass and Moment Calculations We treat coil springs and wires as masses distributed along smooth curves in space. The distribution is described by a continuous density function δ ( x , y, z ) representing mass per unit length. When a curve C is parametrized by r (t ) = x (t ) i + y(t ) j + z (t )k,  a ≤ t ≤ b, then x, y, and z are functions of the parameter t, the density is the function δ ( x (t ), y(t ), z (t ) ) , and the arc length differential is given by ds =

( dxdt ) + ( dydt ) + ( dzdt ) 2

2

2

dt.

(See Section 12.3.) The spring’s or wire’s mass, center of mass, and moments are then calculated using the formulas in Table 15.1, with the integrations in terms of the parameter t over the interval [ a, b ]. For example, the formula for mass becomes M=

1 (x, ¯ ¯y, ¯z ) center of mass

x

y2

FIGURE 15.5

+

( dxdt ) + ( dydt ) + ( dzdt ) 2

2

2

dt.

These formulas also apply to thin rods, and their derivations are similar to those in Section 6.6. Notice how similar the formulas are to those in Tables 15.1 and 15.2 for double and triple integrals. The double integrals for planar regions, and the triple integrals for solids, become line integrals for coil springs, wires, and thin rods. Notice that the element of mass dm is equal to δ ds in the table, rather than to δ dV as in Table 15.1, and that the integrals are taken over the curve C.

z

−1

b

∫a δ ( x (t ), y(t ), z(t ) )

z2

1 y = 1, z ≥ 0

Example 4 shows how to find the center of mass of a circular arch of variable density.

EXAMPLE 4 A slender metal arch, denser at the bottom than at the top, lies along the semicircle y 2 + z 2 = 1,  z ≥ 0, in the yz-plane (Figure 15.5). Find the center of the arch’s mass if the density at the point ( x , y, z ) on the arch is δ ( x , y, z ) = 2 − z. Solution We know that x = 0 and y = 0 because the arch lies in the yz-plane with its mass distributed symmetrically about the z-axis. To find z , we parametrize the circle as

r (t ) = ( cos t ) j + ( sin t ) k,

0 ≤ t ≤ π.

934

Chapter 15 Integrals and Vector Fields

TABLE 15.1 Mass and moment formulas for coil springs, wires, and thin rods lying along a smooth curve C in space

M =

Mass:

∫C δ ds

   δ = δ( x , y, z ) is the density at ( x , y, z ).

First moments about the coordinate planes: M yz =

∫C x δ ds,

Mxz =

∫C y δ ds,

Mxy =

∫C z δ ds

Coordinates of the center of mass: x = Myz M ,

y = Mxz M ,

z = Mxy M

Moments of inertia about axes and other lines: Ix =

∫C ( y 2 + z 2 )δ ds,

IL =

∫C r 2δ ds

Iy =

∫C ( x 2 + z 2 )δ ds,

Iz =

∫C ( x 2 + y 2 )δ ds,

r = r ( x , y, z ) is the distance from the point ( x , y, z ) to line L.

For this parametrization, v(t ) =

( dxdt ) + ( dydt ) + ( dzdt ) 2

2

2

=

( 0 )2

+ ( − sin t ) 2 + ( cos t ) 2 = 1,

so ds = v dt = dt. The formulas in Table 15.1 then give M =

∫C δ ds

Mxy =

∫C z δ ds

= z =

z Height f (x, y)

x

=

=

∫C z ( 2 − z ) ds

( 2 sin t − sin 2 t ) dt =

π

∫0 =

( 2 − sin t ) dt = 2π − 2 π

∫0

8−π 2

( sin t )( 2 − sin t ) dt Routine integration

Mxy 8−π 1 8−π = ⋅ = ≈ 0.57. 2 2π − 2 4π − 4 M

With z to the nearest hundredth, the center of mass is ( 0, 0, 0.57 ).

Line Integrals in the Plane Plane curve C

t=b

FIGURE 15.6

∫C

∫C ( 2 − z ) ds

y

t=a (x, y)

Δsk

π

∫0

=

The line integral f ds gives the area of the portion of

the cylindrical surface or “wall” beneath z = f ( x , y ) ≥ 0.

Line integrals for curves in the plane have a natural geometric interpretation. If C is a smooth curve in the xy-plane parametrized by r (t ) = x (t ) i + y(t ) j,  a ≤ t ≤ b, we generate a cylindrical surface by moving a straight line along C perpendicular to the plane, holding the line parallel to the z-axis, as in Figure 15.6. If z = f ( x , y ) is a nonnegative continuous function over a region in the plane containing the curve C, then the graph of f is a surface that lies above the plane. The cylinder cuts through this surface, forming a curve on it that lies above the curve C and follows its winding nature. The part of the cylindrical surface that lies beneath the surface curve and above the xy-plane forms a “curved

935

15.1  Line Integrals of Scalar Functions

wall” or “fence” standing on the curve C and orthogonal to the plane. At any point ( x , y ) along the curve, the height of the wall is f ( x , y ). From the definition

∫C f ds

n

∑ f ( x k , y k ) Δs k , n →∞

= lim

k =1

where Δs k → 0 as n → ∞, we see that the line integral shown in the figure.

EXERCISES

∫C f ds is the area of the wall

15.1

Graphs of Vector Equations

1. r (t ) = ti + (1 − t ) j, 0 ≤ t ≤ 1

Match the vector equations in Exercises 1–8 with the graphs (a)–(h) given here.

2. r (t ) = i + j + tk, −1 ≤ t ≤ 1 3. r (t ) = ( 2 cos t ) i + ( 2 sin t ) j, 0 ≤ t ≤ 2π

b.

a.

4. r (t ) = ti, −1 ≤ t ≤ 1

z

z

5. r (t ) = ti + tj + tk, 0 ≤ t ≤ 2

2

−1

6. r (t ) = tj + ( 2 − 2t ) k, 0 ≤ t ≤ 1 7. r (t ) = ( t 2 − 1) j + 2tk, −1 ≤ t ≤ 1

1

y

8. r (t ) = ( 2 cos t ) i + ( 2 sin t ) k, 0 ≤ t ≤ π

x

Evaluating Line Integrals over Space Curves 1

c.

9. Evaluate ∫C ( x + y ) ds, where C is the straight-line segment x = t ,  y = (1 − t ) ,  z = 0, from ( 0, 1, 0 ) to (1, 0, 0 ).

y

x

10. Evaluate ∫C ( x − y + z − 2 ) ds, where C is the straight-line segment x = t ,  y = (1 − t ) ,  z = 1, from ( 0, 1, 1) to (1, 0, 1).

d. z

z

11. Evaluate ∫C ( xy + y + z ) ds along the curve r (t ) = 2ti + tj + ( 2 − 2t ) k, 0 ≤ t ≤ 1.

(2, 2, 2) 1

2

1

y

y

2

13. Find the line integral of f ( x , y, z ) = x + y + z over the straight-line segment from (1, 2, 3 ) to ( 0, −1, 1).

x

x

e.

f. z

14. Find the line integral of f ( x , y, z ) = 3 ( x 2 + y 2 + z 2 ) over the curve r (t ) = ti + tj + tk, 1 ≤ t < ∞.

z

15. Integrate f ( x , y, z ) = x + y − z 2 over the path C 1 followed by C 2 from ( 0, 0, 0 ) to (1, 1, 1) (see accompanying figure) given by

2

C 1: r (t ) = ti + t 2 j, 0 ≤ t ≤ 1

(1, 1, 1)

−1

1 1

12. Evaluate ∫C x 2 + y 2 ds along the curve r (t ) = ( 4 cos t ) i + ( 4 sin t ) j + 3tk, −2π ≤ t ≤ 2π.

C 2 : r (t ) = i + j + tk, 0 ≤ t ≤ 1.

y

z

y x

x

z −2

(1, 1, −1)

g.

(0, 0, 1)

h. z

(0, 0, 0)

z

2

x 2 x

(1, 1, 1)

y C2

y

x

C3

(0, 0, 0) −2

y

C1 (1, 1, 0) (a)

(0, 1, 1)

C1

(1, 1, 1)

2 2

C2

y x (b)

The paths of integration for Exercises 15 and 15.

936

Chapter 15 Integrals and Vector Fields

16. Integrate f ( x , y, z ) = x + y − z 2 over the path C 1 followed by C 2 followed by C 3 from ( 0, 0, 0 ) to (1, 1, 1) (see accompanying figure) given by C 1 : r (t ) = t k , 0 ≤ t ≤ 1 C 2 : r (t ) = t j + k , 0 ≤ t ≤ 1 C 3 : r (t ) = ti + j + k, 0 ≤ t ≤ 1. 17. Integrate f ( x , y, z ) = ( x + y + z ) ( x 2 + y 2 + z 2 ) over the path r (t ) = t i + tj + tk, 0 < a ≤ t ≤ b. 18. Integrate f ( x , y, z ) = − x 2 + z 2 over the circle r (t ) = ( a cos t ) j + ( a sin t ) k,

0 ≤ t ≤ 2π.

In Exercises 27–30, integrate f over the given curve. 27. f ( x , y ) = x 3 y , C: y = x 2 2, 0 ≤ x ≤ 2 28. f ( x , y ) = ( x + y 2 ) ( 0, 0 )

1 + x 2 , C: y = x 2 2 from (1, 1 2 ) to

29. f ( x , y ) = x + y, C: x 2 + y 2 = 4 in the first quadrant from ( 2, 0 ) to ( 0, 2 ) 30. f ( x , y ) = x 2 − y, C: x 2 + y 2 = 4 in the first quadrant from ( 0, 2 ) to ( 2, 2 ) 31. Find the area of one side of the “winding wall” standing perpendicularly on the curve y = x 2, 0 ≤ x ≤ 2, and beneath the curve on the surface f ( x , y ) = x + y. 32. Find the area of one side of the “wall” standing perpendicularly on the curve 2 x + 3 y = 6, 0 ≤ x ≤ 6, and beneath the curve on the surface f ( x , y ) = 4 + 3 x + 2 y.

Line Integrals over Plane Curves

19. Evaluate ∫C x ds, where C is a. the straight-line segment x = t ,  y = t 2, from ( 0, 0 ) to ( 4, 2 ). b. the parabolic curve x = t ,  y = t 2, from ( 0, 0 ) to ( 2, 4 ). 20. Evaluate ∫C

x + 2 y ds, where C is

a. the straight-line segment x = t ,  y = 4 t , from ( 0, 0 ) to (1, 4 ). b. C 1 ∪ C 2 ; C 1 is the line segment from ( 0, 0 ) to (1, 0 ) and C 2 is the line segment from (1, 0 ) to (1, 2 ). 21. Find the line integral of f ( x , y ) = ye x 2 along the curve r (t ) = 4 ti − 3tj, −1 ≤ t ≤ 2. 22. Find the line integral of f ( x , y ) = x − y + 3 along the curve r (t ) = ( cos t ) i + ( sin t ) j, 0 ≤ t ≤ 2π. 23. Evaluate

x2

∫C y 4 3 ds, where C is the curve x

= t 2,  y = t 3, for

33. Mass of a wire Find the mass of a wire that lies along the curve r (t ) = ( t 2 − 1) j + 2tk, 0 ≤ t ≤ 1, if the density is δ = ( 3 2 ) t. 34. Center of mass of a curved wire A wire of density δ ( x , y, z ) = 15 y + 2 lies along the curve r (t ) = ( t 2 − 1) j + 2tk, −1 ≤ t ≤ 1. Find its center of mass. Then sketch the curve and center of mass together. 35. Mass of wire with variable density Find the mass of a thin wire lying along the curve r (t ) = 2ti + 2tj + ( 4 − t 2 ) k, 0 ≤ t ≤ 1, if the density is (a) δ = 3t and (b) δ = 1.

y x along the curve

36. Center of mass of wire with variable density Find the center of mass of a thin wire lying along the curve r (t ) = ti + 2tj + ( 2 3 ) t 3 2 k, 0 ≤ t ≤ 2, if the density is δ = 3 5 + t .

y ) ds , where C is given in the accompany-

37. Moment of inertia of wire hoop A circular wire hoop of constant density δ lies along the circle x 2 + y 2 = a 2 in the xy-plane. Find the hoop’s moment of inertia about the z-axis.

1 ≤ t ≤ 2.

24. Find the line integral of f ( x , y ) = r (t ) = t 3 i + t 4 j, 1 2 ≤ t ≤ 1. 25. Evaluate ∫C ( x + ing figure.

Masses and Moments

38. Inertia of a slender rod A slender rod of constant density lies along the line segment r (t ) = tj + ( 2 − 2t ) k, 0 ≤ t ≤ 1, in the yz-plane. Find the moments of inertia of the rod about the three coordinate axes.

y

C

(1, 1)

39. Two springs of constant density lies along the helix

y=x y = x2 x

(0, 0)

A spring of constant density δ

r (t ) = ( cos t ) i + ( sin t ) j + tk,

0 ≤ t ≤ 2π.

a. Find Iz . 26. Evaluate

1

∫C x 2 + y 2 + 1 ds, where C is given in the accompa-

nying figure. y

(0, 1)

b. Suppose that you have another spring of constant density δ that is twice as long as the spring in part (a) and lies along the helix for 0 ≤ t ≤ 4 π. Do you expect Iz for the longer spring to be the same as that for the shorter one, or should it be different? Check your prediction by calculating Iz for the longer spring. 40. Wire of constant density along the curve

(1, 1)

A wire of constant density δ = 1 lies

r (t ) = ( t cos t ) i + ( t sin t ) j + ( 2 2 3 ) t 3 2 k, (0, 0)

(1, 0)

0 ≤ t ≤ 1.

x

Find z  and Iz . 41. The arch in Example 4 Find Ix for the arch in Example 4.

937

15.2  Vector Fields and Line Integrals: Work, Circulation, and Flux

42. Center of mass and moments of inertia for wire with variable density Find the center of mass and the moments of inertia about the coordinate axes of a thin wire lying along the curve 2 2 32 t2 r (t ) = t i + t j + k, 3 2

0 ≤ t ≤ 2,

43. f ( x , y, z ) = 0 ≤ t ≤ 2

1 + 30 x 2 + 10 y; r (t ) = ti + t 2 j + 3t 2 k,

44. f ( x , y, z ) = 0 ≤ t ≤ 2

1 + x 3 + 5 y 3 ; r (t ) = t i +

1 2 t j+ 3

t k,

45. f ( x , y, z ) = x y − 3z 2 ; r (t ) = ( cos 2t ) i + ( sin 2t ) j + 5tk, 0 ≤ t ≤ 2π

if the density is δ = 1 ( t + 1).

(

In Exercises 43–46, use a CAS to perform the following steps to evaluate the line integrals.

)

9 13 14 z ; 4 r (t ) = ( cos 2t ) i + ( sin 2t ) j + t 5 2 k, 0 ≤ t ≤ 2π

46. f ( x , y, z ) = 1 +

COMPUTER EXPLORATIONS

a. Find ds = v(t ) dt for the path r (t ) = g(t ) i + h(t ) j + k (t )k. b. Express the integrand f ( g(t ), h(t ), k (t ) ) v(t ) as a function of the parameter t. c. Evaluate ∫C f ds using Equation (2) in the text.

15.2 Vector Fields and Line Integrals: Work, Circulation, and Flux Gravitational and electric forces have both a direction and a magnitude. They are represented by a vector at each point in their domain, producing a vector field. In this section we show how to compute the work done in moving an object through such a field by using a line integral involving the vector field. We also discuss velocity fields, such as the vector field representing the velocity of a flowing fluid in its domain. A line integral can be used to find the rate at which the fluid flows along or across a curve within the domain.

Vector Fields FIGURE 15.7

Velocity vectors of a flow around an airfoil.

Suppose a region in the plane or in space is occupied by a moving fluid, such as air or water. The fluid is made up of a large number of particles, and at any instant of time, a particle has a velocity v. At different points of the region at a given (same) time, these velocities can vary. We can think of a velocity vector being attached to each point of the fluid, representing the velocity of a particle at that point. Such a fluid flow is an example of a vector field. Figure 15.7 shows a velocity vector field obtained from air flowing around an airfoil in a wind tunnel. Figure 15.8 shows a vector field of velocity vectors along the streamlines of water moving through a contracting channel. Vector fields are also associated with forces such as gravitational attraction (Figure 15.9) and with magnetic fields and electric fields. There are purely mathematical fields as well. Generally, a vector field is a function that assigns a vector to each point in its domain. A vector field on a three-dimensional domain in space might have a formula like F ( x , y, z ) = M ( x , y, z ) i + N ( x , y, z ) j + P ( x , y, z ) k. The vector field is continuous if the component functions M, N, and P are continuous; it is differentiable if each of the component functions is differentiable. The formula for a field of two-dimensional vectors could look like

FIGURE 15.8 Streamlines in a contracting channel. The water speeds up as the channel narrows, and the velocity vectors increase in length.

F ( x , y ) = M ( x , y ) i + N ( x , y ) j. We encountered another type of vector field in Chapter 12. The tangent vectors T and normal vectors N for a curve in space both form vector fields along the curve. Along a curve r(t ) they might have a component formula similar to the velocity field expression v(t ) = f (t ) i + g(t ) j + h(t )k. If we attach the gradient vector ∇f of a scalar function f ( x , y, z ) to each point of a level surface of the function, we obtain a three-dimensional field on the surface. If we attach the velocity vector to each point of a flowing fluid, we have a three-dimensional field

938

Chapter 15 Integrals and Vector Fields

z

defined on a region in space. These and other fields are illustrated in Figures 15.7–15.16. To sketch the fields, we picked a representative selection of domain points and drew the vectors attached to them. The arrows are drawn with their tails, not their heads, attached to the points where the vector functions are evaluated. y z

x

FIGURE 15.9 Vectors in a gravitational field point toward the center of mass that gives the source of the field. y x

FIGURE 15.10

A surface might represent a filter (or a net or a parachute) in a vector field representing water or wind flow velocity vectors. The arrows show the direction of fluid flow, and their lengths indicate speed.

y

f (x, y, z) = c

y

FIGURE 15.11

The field of gradient vectors ∇f on a level surface f ( x , y, z ) = c. The function f is constant on the surface, and each vector points in the direction where f is increasing fastest.

x

x

FIGURE 15.12

z

x2 + y2 ≤ a2

The radial field F = xi + yj formed by the position vectors of points in the plane. Notice the convention that an arrow is drawn with its tail, not its head, at the point where F is evaluated.

FIGURE 15.13

A “spin” field of rotat-

ing unit vectors F = (− yi + xj ) ( x 2 + y 2 )1 2 in the plane. The field is not defined at the origin.

z = a2 − r2

Gradient Fields

x

0 y

The gradient vector of a differentiable scalar-valued function at a point gives the direction of greatest increase of the function. An important type of vector field is formed by all the gradient vectors of the function (see Section 13.5). We define the gradient field of a differentiable function f ( x , y, z ) to be the field of gradient vectors ∇f =

FIGURE 15.14

The flow of fluid in a long cylindrical pipe. The vectors v = ( a 2 − r 2 ) k inside the cylinder that have their bases in the xy-plane have their tips on the paraboloid z = a 2 − r 2 .

∂f ∂f ∂f i+ j+ k. ∂x ∂y ∂z

At each point ( x , y, z ), the gradient field gives a vector pointing in the direction of greatest increase of f , with magnitude being the value of the directional derivative in that direction. The gradient field might represent a force field, or a velocity field that gives the motion of a fluid, or the flow of heat through a medium, depending on the application being considered.

15.2  Vector Fields and Line Integrals: Work, Circulation, and Flux

939

y

x

0

FIGURE 15.15 The velocity vectors v(t ) of a projectile’s motion make a vector field along the trajectory.

FIGURE 15.16

Data from NASA’s QuikSCAT satellite were used to create this representation of wind speed and wind direction in Hurricane Irene approximately six hours before it made landfall in North Carolina on August 27, 2011. The arrows show wind direction, and speed is indicated by color (rather than length). The maximum wind speeds (over 130 km hour) occurred over a region too small to see in this illustration. (Source: JPL-Caltech/ISRO/NASA)

In many physical applications, f represents a potential energy, and the gradient vector field indicates the corresponding force. In such situations, f is often taken to be negative, so that the force gives the direction of decreasing potential energy. z

EXAMPLE 1 Suppose that a material is heated, that the resulting temperature T at each point ( x , y, z ) in a region of space is given by T = 100 − x 2 − y 2 − z 2 , and that F ( x , y, z ) is defined to be the gradient of T. Find the vector field F.

x y

FIGURE 15.17

The vectors in a temperature gradient field point in the direction of greatest increase in temperature. In this case they are pointing toward the origin.

Solution The gradient field F is the field F = ∇T = −2 xi − 2 yj − 2 zk. At each point in the region, the vector field F gives the direction for which the increase in temperature is greatest. The vectors point toward the origin, where the temperature is greatest. See Figure 15.17.

Line Integrals of Vector Fields In Section 15.1 we defined the line integral of a scalar function f ( x , y, z ) over a path C. We turn our attention now to the idea of a line integral of a vector field F along the curve C. Such line integrals have important applications in the study of fluid flows, work and energy, and electrical or gravitational fields. Assume that the vector field F = M ( x , y, z ) i + N ( x , y, z ) j + P ( x , y, z ) k has continuous components, and that the curve C has a smooth parametrization r (t ) = g(t ) i + h(t ) j + k (t )k,  a ≤ t ≤ b. As discussed in Section 15.1, the parametrization r(t ) defines a direction (or orientation) along C that we call the forward direction. At each point along the path C, the tangent vector T = d r ds = v v is a unit vector tangent to the path and pointing in this forward direction. (The vector v = d r dt is the velocity vector tangent to C at the point, as discussed in Sections 12.1 and 12.3.) The line integral of the vector field is the line integral of the scalar tangential component of F along C. This tangential component is given by the dot product F⋅T = F⋅ so we are led to the following definition.

dr , ds

940

Chapter 15 Integrals and Vector Fields

Let F be a vector field with continuous components defined along a smooth curve C parametrized by r(t ),  a ≤ t ≤ b. Then the line integral of F along C is DEFINITION

∫C F ⋅ T ds

=

∫C ( F ⋅ ds ) ds dr

=

∫C F ⋅ d r.

(1)

We evaluate line integrals of vector fields in a way similar to the way we evaluate line integrals of scalar functions (Section 15.1). The vector field may also be defined on points not meeting the curve, but only the vectors along the curve play a role in the line integral. See Figure 15.18.

z

Evaluating the Line Integral of F = Mi + Nj + Pk Along C: r( t ) = g( t )i + h( t )j + k( t )k

y x

1. Express the vector field F along the parametrized curve C as F ( r(t ) ) by substituting the components x = g(t ),  y = h(t ),  z = k (t ) of r into the scalar components M ( x , y, z ), N ( x , y, z ), P ( x , y, z ) of F.

FIGURE 15.18

A curve (in red) winds through a vector field as in Example 2. The line integral is determined by the vectors that lie along the curve.

2. Find the derivative (velocity) vector d r dt . 3. Evaluate the line integral with respect to the parameter t ,  a ≤ t ≤ b, to obtain

∫C F ⋅ d r

=

b

dr

∫a F( r(t ) ) ⋅ dt dt.

(2)

EXAMPLE 2 Evaluate ∫ C F ⋅ d r, where F ( x , y, z ) = zi + xyj − y 2 k along the curve C given by r (t ) = t 2 i + tj + t k, 0 ≤ t ≤ 1. Solution We have

F ( r (t ) ) =

t i + t 3 j − t 2k

  

z =

t , xy = t 3, −y 2 = −t 2

and dr 1 = 2t i + j +   k. dt 2 t Thus, 1

dr

∫C F ⋅ d r = ∫0 F( r(t ) ) ⋅ dt dt =

Eq. (2)

∫0 ( 2t 3 2 + t 3 − 2 t 3 2 ) dt 1

1

⎡ ⎤1 3 2 52 1 4⎥ 17 ⎢ . = ⎢ + t ⎥ = t 2 5 4 20 ⎢⎣ ⎥⎦ 0

( )(

)

Line Integrals with Respect to dx, dy, or dz When analyzing forces or flows, it is often useful to consider each component direction separately. For example, when analyzing the effect of a gravitational force, we might want to consider motion and forces in the vertical direction, while ignoring horizontal motions. Or we might be interested only in the force exerted horizontally by water pushing against

941

15.2  Vector Fields and Line Integrals: Work, Circulation, and Flux

the face of a dam or in wind affecting the course of a plane. In such situations we want to evaluate a line integral of a scalar function with respect to only one of the coordinates, such as ∫ C M dx. This type of integral is not the same as the arc length line integral ∫ C M ds we defined in Section 15.1, since it picks out displacement in the direction of only one coordinate. To define the integral ∫ C M dx for the scalar function M ( x , y, z ), we specify a vector field F = M ( x , y, z ) i having a component only in the x-direction, and none in the y- or the z-direction. Then, over the curve C parametrized by r (t ) = g(t ) i + h(t ) j + k (t )k for a ≤ t ≤ b, we have x = g(t ),  dx = g′(t ) dt , and F ⋅ dr = F ⋅

dr dt = M ( x , y, z ) i ⋅ ( g′(t ) i + h′(t ) j + k ′(t )k ) dt dt = M ( x , y, z ) g′(t ) dt = M ( x , y, z ) dx.

As in the definition of the line integral of F along C, we define

∫C M ( x , y, z ) dx

=

∫C F ⋅ d r ,

where F = M ( x , y, z ) i.

In the same way, by defining F = N ( x , y, z ) j with a component only in the y-direction, or F = P ( x , y, z ) k with a component only in the z-direction, we obtain the line integrals ∫ C N dy and ∫ C P dz. Expressing everything in terms of the parameter t along the curve C, we have the following formulas for these three integrals:

Line Integral Notation The commonly occurring expression

b

∫C M ( x , y, z ) dx

=

∫a

∫C N ( x , y, z ) dy

=

∫a

∫C P ( x , y, z ) dz

=

∫a

b

b

M ( g(t ), h(t ), k (t ) ) g′(t ) dt

(3)

N ( g(t ), h(t ), k (t ) ) h′(t ) dt

(4)

P ( g(t ), h(t ), k (t ) ) k ′(t ) dt

(5)

∫C M dx + N dy + P dz is a short way of expressing the sum of three line integrals, one for each coordinate direction:

∫C M ( x , y, z ) dx + ∫C N ( x , y, z ) dy +

It often happens that these line integrals occur in combination, and we abbreviate the notation by writing

∫C M ( x , y, z ) dx + ∫C N ( x , y, z ) dy + ∫C P ( x , y, z ) dz

=

∫C M dx + N dy + P dz.

∫C P ( x , y, z ) dz.

To evaluate these integrals, we parametrize C as g(t ) i + h(t ) j + k (t )k and use Equations (3), (4), and (5).

EXAMPLE 3 Evaluate the line integral ∫ C − y dx + z dy + 2 x dz , where C is the helix r (t ) = ( cos t ) i + ( sin t ) j + tk, 0 ≤ t ≤ 2π. Solution We express everything in terms of the parameter t, so x = cos t , y = sin t , z = t, and dx = − sin t dt ,  dy = cos t dt ,  dz = dt. Then

∫C − y dx + z dy + 2 x dz

2π

=

∫0

=

∫0

2π

[ (− sin t )(− sin t ) + t cos t + 2 cos t ] dt [ 2 cos t + t cos t + sin 2 t ] dt

(

)

sin 2t ⎤ 2 π ⎡ t = ⎢ 2 sin t + ( t sin t + cos t ) + − ⎥ 2 4 ⎦0 ⎣ = [ 0 + ( 0 + 1) + ( π − 0 ) ] − [ 0 + ( 0 + 1) + ( 0 − 0 ) ] = π.

942

Chapter 15 Integrals and Vector Fields

Work Done by a Force over a Curve in Space Suppose that the vector field F = M ( x , y, z ) i + N ( x , y, z ) j + P ( x , y, z ) k represents a force throughout a region in space (it might be the force of gravity or an electromagnetic force) and that r (t ) = g(t ) i + h(t ) j + k (t )k ,

Fk

Pk

(xk , yk , zk ) Pk−1

Fk . Tk

Tk

FIGURE 15.19

The work done along the subarc shown here is approximately Fk ⋅ Tk Δs k , where Fk = F ( x k , y k , z k ) and Tk = T ( x k , y k , z k ).

a ≤ t ≤ b,

represents a smooth curve C in the region. The formula for the work done by the force in moving an object along the curve is motivated by the same kind of reasoning we used in Chapter 6 to derive the ordinary single integral for the work done by a continuous force of magnitude F ( x ) directed along an interval of the x-axis. For the curve C in space, we define the work done by a continuous force field F to move an object along C from a point A to another point B as follows. We divide C into n subarcs Pk −1Pk with lengths Δs k , starting at A and ending at B. We choose any point ( x k , y k , z k ) in the subarc Pk −1Pk and let T ( x k , y k , z k ) be the unit tangent vector at the chosen point. The work W k done to move the object along the subarc Pk −1Pk is approximated by the tangential component of the force F ( x k , y k , z k ) times the arc length Δs k, the distance the object moves along the subarc (see Figure 15.19). The total work done in moving the object from point A to point B is then obtained by summing the work done along each of the subarcs, so W =

n

∑ Wk k =1

≈

n

∑ F ( x k , y k , z k ) ⋅ T ( x k , y k , z k ) Δs k . k =1

For any subdivision of C into n subarcs, and for any choice of the points ( x k , y k , z k ) within each subarc, as n → ∞ and Δs k → 0, these sums approach the line integral

∫C F ⋅ T ds. This is the line integral of F along C, which now defines the total work done.

B t=b

DEFINITION Let C be a smooth curve parametrized by r(t ),  a ≤ t ≤ b, and let F be a continuous force field over a region containing C. Then the work done in moving an object from the point A = r(a) to the point B = r(b) along C is

W =

∫C F ⋅ T ds

=

b

∫a

F ( r (t ) ) ⋅

dr dt. dt

(6)

T F

A t=a

FIGURE 15.20

The work done by a force F is the line integral of the scalar component F ⋅ T over the smooth curve from A to B.

The sign of the number we calculate with this integral depends on the direction in which the curve is traversed. If we reverse the direction of motion, then we reverse the direction of T in Figure 15.20 and change the sign of F ⋅ T and its integral. Using the notations we have presented, we can express the work integral in a variety of ways, depending upon what seems most suitable or convenient for a particular discussion. Table 15.2 shows five ways we can write the work integral in Equation (6). In the table, the field components M, N, and P are functions of the intermediate variables x, y, and z, which in turn are functions of the independent variable t along the curve C in the vector field. So along the curve, x = g(t ),  y = h(t ), and z = k (t ) with dx = g′(t ) dt , dy = h′(t ) dt , and dz = k ′(t ) dt.

15.2  Vector Fields and Line Integrals: Work, Circulation, and Flux

943

TABLE 15.2 Different ways to write the work integral for F = Mi + Nj + P k over

the curve C: r( t ) = g( t )i + h( t ) j + k( t )k, a ≤ t ≤ b

∫C F ⋅ T   ds

The definition

=

∫C F ⋅ d r

Vector differential form

=

∫a

=

∫a

=

∫C M dx + N dy + P dz

W =

z

b

b

F⋅

dr dt dt

Parametric vector evaluation

( M g′(t ) + N h′(t ) + P k ′(t ) ) dt

Parametric scalar evaluation Scalar differential form

EXAMPLE 4 Find the work done by the force field F = ( y − x 2 ) i + ( z − y 2 ) j + 2 ( x − z ) k in moving an object along the curve r (t ) = ti + t 2 j + t 3 k, 0 ≤ t ≤ 1, from ( 0, 0, 0 ) to (1, 1, 1) (Figure 15.21). Solution First we evaluate F on the curve r(t ):

(0, 0, 0)

(1, 1, 1)

F = ( y − x 2 ) i + ( z − y 2 ) j + ( x − z 2 )k = ( t 2 − t 2 ) i + ( t 3 − t 4 ) j + ( t − t 6 ) k. 

Substitute  x = t ,  y = t 2,  z = t 3 .

0

y

Then we find d r dt: x r(t) = ti + t 2j + t 3k

FIGURE 15.21

dr d = ( ti + t 2 j + t 3k ) = i + 2tj + 3t 2 k. dt dt

(1, 1, 0)

The curve in Example 4.

Finally, we find F ⋅ d r dt and integrate from t = 0 to t = 1: F⋅

dr = [ ( t 3 − t 4 ) j + ( t − t 6 ) k ] ⋅ ( i + 2tj + 3t 2 k ) dt = ( t 3 − t 4 ) (2t ) + ( t − t 6 ) (3t 2 ) = 2t 4 − 2t 5 + 3t 3 − 3t 8 .

Evaluate dot product.

Thus Work =

b

∫a

F⋅

dr dt = dt

1

∫0 ( 2t 4 − 2t 5 + 3t 3 − 3t 8 ) dt

⎡ ⎤1 2 5 2 6 3 4 3 9⎥ 29 ⎢ . = ⎢ t − t + t − t ⎥ = 6 4 9 ⎥ 60 ⎢⎣ 5 ⎦0 EXAMPLE 5 Find the work done by the force field F = xi + yj + zk in moving an object along the curve C parametrized by r (t ) = cos(πt ) i + t 2 j + sin (πt ) k, 0 ≤ t ≤ 1. Solution We begin by writing F along C as a function of t:

F ( r (t ) ) = cos(πt ) i + t 2 j + sin (πt ) k. Next we compute d r dt: dr = −π sin (πt ) i + 2tj + π cos(πt ) k. dt

944

Chapter 15 Integrals and Vector Fields

We then calculate the dot product: F ( r (t ) ) ⋅

dr = −π sin (πt ) cos(πt ) + 2t 3 + π sin (πt ) cos(πt ) = 2t 3 . dt

The work done is the line integral b

∫a

1

dr dt = dt

F ( r (t ) ) ⋅

∫0 2t 3 dt

=

t 4 ⎤1 1 = . 2 ⎥⎦ 0 2

Flow Integrals and Circulation for Velocity Fields Suppose that F represents the velocity field of a fluid flowing through a region in space (a tidal basin or the turbine chamber of a hydroelectric generator, for example). Under these circumstances, the integral of F ⋅ T along a curve in the region gives the fluid’s flow along, or circulation around, the curve. For instance, the vector field in Figure 15.12 gives zero circulation around the unit circle in the plane. By contrast, the vector field in Figure 15.13 gives a nonzero circulation around the unit circle.

If r(t ) parametrizes a smooth curve C in the domain of a continuous velocity field F, then the flow along the curve from A = r (a) to B = r (b) is

DEFINITION

Flow =

∫C F ⋅ T ds.

(7)

The integral is called a flow integral. If the curve starts and ends at the same point, so that A = B, the flow is called the circulation around the curve.

The direction we travel along C matters. If we reverse the direction, then T is replaced by −T and the sign of the integral changes. We evaluate flow integrals the same way we evaluate work integrals. EXAMPLE 6 A fluid’s velocity field is F = xi + zj + yk. Find the flow along the helix r (t ) = ( cos t ) i + ( sin t ) j + tk, 0 ≤ t ≤ π 2. Solution We evaluate F on the curve r(t ):

F = xi + zj + yk = ( cos t ) i + tj + ( sin t ) k

Substitute  x = cos t ,  z = t ,  y = sin t.

and then find d r dt: dr = ( − sin t ) i + ( cos t ) j + k. dt The dot product of F with d r dt is F⋅

dr = ( cos t )(− sin t ) + (t )( cos t ) + ( sin t )(1) dt = − sin t cos t + t cos t + sin t.

Finally, we integrate F ⋅ ( d r dt ) from t = 0 to t = Flow =

t=b

∫t = a

F⋅

dr dt = dt

π 2

⎡ cos 2 t ⎤ =⎢ + t sin t ⎥ ⎢⎣ 2 ⎥⎦ 0

π 2

∫0

π : 2

(− sin t cos t + t cos t + sin t ) dt

(

= 0+

) (

)

π π 1 1 − +0 = − . 2 2 2 2

945

15.2  Vector Fields and Line Integrals: Work, Circulation, and Flux

y

EXAMPLE 7 Find the circulation of the field F = ( x − y ) i + xj around the circle r (t ) = ( cos t ) i + ( sin t ) j, 0 ≤ t ≤ 2π (Figure 15.22). Solution On the circle, F = ( x − y ) i + xj = ( cos t − sin t ) i + ( cos t ) j, and

dr = ( − sin t ) i + ( cos t ) j. dt

x

Then F⋅

dr 2 t + cos 2 t = − sin t cos t + sin   dt 1

FIGURE 15.22

The vector field F and curve r(t ) in Example 7.

gives Circulation =

2π

∫0

F⋅

dr dt = dt

2π

∫0

(1 − sin t cos t ) dt

sin 2 t ⎤ 2 π ⎡ = ⎢t − ⎥ = 2π. ⎢⎣ 2 ⎥⎦ 0 As Figure 15.22 suggests, a fluid with this velocity field is circulating counterclockwise around the circle. The circle is also traversed counterclockwise as t increases from 0 to 2π, so the circulation is positive.

Flux Across a Simple Closed Plane Curve

Simple, not closed

Simple, closed

Not simple, not closed

Not simple, closed

A curve in the xy-plane is simple if it does not cross itself (Figure 15.23). When a curve starts and ends at the same point, it is a closed curve or loop. To find the rate at which a fluid is entering or leaving a region enclosed by a smooth simple closed curve C in the xyplane, we calculate the line integral over C of F ⋅ n, the scalar component of the fluid’s velocity field in the direction of the curve’s outward-pointing normal vector. We use only the normal component of F, while ignoring the tangential component, because the normal component leads to the flow across C. The value of this integral is the flux of F across C. Flux is Latin for flow, but many flux calculations involve no motion at all. When F is an electric or magnetic field, for instance, the integral of F ⋅ n is still called the flux of the field across C.

FIGURE 15.23

Distinguishing between curves that are simple and curves that are closed. Closed curves are also called loops.

If C is a smooth simple closed curve in the domain of a continuous vector field F = M ( x , y ) i + N ( x , y ) j in the plane, and if n is the outwardpointing unit normal vector on C, the flux of F across C is DEFINITION

Flux of  F  across  C =

∫C F ⋅ n ds.

(8)

Notice the difference between flux and circulation. The flux of F across C is the line integral with respect to arc length of F ⋅ n, the scalar component of F in the direction of the outward normal. The circulation of F around C is the line integral with respect to arc length of F ⋅ T, the scalar component of F in the direction of the unit tangent vector. Flux is the integral of the normal component of F; circulation is the integral of the tangential component of F. In Section 15.6 we will define flux across a surface. To evaluate the integral for flux in Equation (8), we begin with a smooth parametrization x = g(t ),

y = h(t ),

a ≤ t ≤ b,  

946

Chapter 15 Integrals and Vector Fields

z

For clockwise motion, k × T points outward. y x

k

C

T k×T

that traces the curve C exactly once as t increases from a to b. We can find the outward unit normal vector n by crossing the curve’s unit tangent vector T with the vector k. But which order do we choose, T × k or k × T? Which one points outward? It depends on which way C is traversed as t increases. If the motion is clockwise, k × T points outward; if the motion is counterclockwise, T × k points outward (Figure 15.24). The usual choice is n = T × k, the choice that assumes counterclockwise motion. Thus, even though the value of the integral in Equation (8) does not depend on which way C is traversed, the formulas we are about to derive for computing n and evaluating the integral assume counterclockwise motion. In terms of components,

z

(

For counterclockwise motion, T × k points outward.

k

C

T×k

F ⋅ n = M ( x, y )

k

dy dx − N ( x, y ) . ds ds

Hence,

T

∫C F ⋅ n ds

FIGURE 15.24

To find an outward unit normal vector for a smooth simple curve C in the xy-plane that is traversed counterclockwise as t increases, we take n = T × k. For clockwise motion, we take n = k × T.

)

j

dx dy 0 ds ds 0 0 1

If F = M ( x , y ) i + N ( x , y ) j, then y

x

i

dy dy dx dx n = T×k = i+ j ×k = i− j. ds ds ds ds

=

∫C ( M ds − N ds ) ds dy

dx

=

DC M dy − N dx.

We put a directed circle on the last integral as a reminder that the integration around the closed curve C is to be in the counterclockwise direction. To evaluate this integral, we express M, dy, N, and dx in terms of the parameter t and integrate from t = a to t = b. We do not need to know n or ds explicitly to find the flux.

Calculating Flux Across a Smooth Closed Plane Curve

( Flux of  F = Mi + N j across C ) =

DC M dy − N dx

(9)

The integral can be evaluated from any smooth parametrization x = g(t ), y = h(t ),  a ≤ t ≤ b, that traces C counterclockwise exactly once. EXAMPLE 8 Find the flux of F = ( x − y ) i + xj across the circle x 2 + y 2 = 1 in the xy-plane. (The vector field and curve were shown in Figure 15.22.) Solution The parametrization r (t ) = ( cos t ) i + ( sin t ) j, 0 ≤ t ≤ 2π , traces the circle counterclockwise exactly once. We can therefore use this parametrization in Equation (9). With

M = x − y = cos t − sin t ,

dy = d ( sin t ) = cos t dt ,

N = x = cos t ,

dx = d ( cos t ) = − sin t dt ,

we find Flux = =

DC

2π

∫0

2π

∫0

M dy − N dx = cos 2 t dt =

2π

∫0

( cos 2 t − sin t cos t + cos t sin t ) dt

Eq. (9)

1 + cos 2t sin 2t ⎤ 2 π ⎡t dt = ⎢ + ⎥ = π. 2 4 ⎦0 ⎣2

The flux of F across the circle is π. Since the answer is positive, the net flow across the curve is outward. A net inward flow would have given a negative flux.

15.2  Vector Fields and Line Integrals: Work, Circulation, and Flux

15.2

EXERCISES Vector Fields

Find the gradient fields of the functions in Exercises 1–4.

15.

∫C ( x 2 + y 2 ) dy, where C is given in the accompanying figure

1. f ( x , y, z ) = ( x 2 + y 2 + z 2 )−1 2

y

2. f ( x , y, z ) = ln x 2 + y 2 + z 2 3. g ( x , y, z ) = e z − ln ( x 2 + y 2 )

(3, 3)

4. g ( x , y, z ) = xy + yz + xz

C

5. Give a formula F = M ( x , y ) i + N ( x , y ) j for the vector field in the plane that has the property that F points toward the origin with magnitude inversely proportional to the square of the distance from ( x , y ) to the origin. (The field is not defined at ( 0, 0 ).) 6. Give a formula F = M ( x , y ) i + N ( x , y ) j for the vector field in the plane that has the properties that F = 0 at ( 0, 0 ) and that at any other point ( a, b ), F is tangent to the circle x 2 + y 2 = a 2 + b 2 and points in the clockwise direction with magnitude F = a2 + b2.

(0, 0)

16.

∫C

C

8. F = [ 1 ( x 2 + 1) ] j

7. F = 3 yi + 2 xj + 4 zk z i − 2 xj +

10. F = xyi + yzj + xzk

yk

11. F = ( 3 x 2 − 3 x ) i + 3zj + k

y = 3x

z

C1

x

(0, 0)

17. Along the curve r (t ) = ti − j + t 2 k, 0 ≤ t ≤ 1, evaluate each of the following integrals. a.

∫C ( x + y − z ) dx

c.

∫C ( x + y − z ) dz

b.

∫C ( x + y − z ) dy

18. Along the curve r (t ) = ( cos t ) i + ( sin t ) j − ( cos t ) k, 0 ≤ t ≤ π , evaluate each of the following integrals.

12. F = ( y + z ) i + ( z + x ) j + ( x + y ) k

(0, 0, 0)

(1, 3)

(0, 3)

a. The straight-line path C 1:  r (t ) = ti + tj + tk, 0 ≤ t ≤ 1 c. The path C 3 ∪ C 4 consisting of the line segment from ( 0, 0, 0 ) to (1, 1, 0 ) followed by the segment from (1, 1, 0 ) to (1, 1, 1)

x

y

In Exercises 7–12, find the line integrals of F from ( 0, 0, 0 ) to (1, 1, 1) over each of the following paths in the accompanying figure. b. The curved path C 2 :  r (t ) = ti + t 2 j + t 4 k, 0 ≤ t ≤ 1

(3, 0)

x + y dx, where C is given in the accompanying figure

Line Integrals of Vector Fields

9. F =

a.

∫C xz dx

c.

∫C xyz dz

(1, 1, 1)

b.

∫C xz dy

Work

C2

C4

y

C3

In Exercises 19–22, find the work done by F over the curve in the direction of increasing t. 19. F = xyi + yj − yzk

x (1, 1, 0)

r (t ) = t i + t 2 j + t k , 0 ≤ t ≤ 1 20. F = 2 yi + 3 xj + ( x + y ) k

Line Integrals with Respect to x, y, and z

In Exercises 13–16, find the line integrals along the given path C. 13. 14.

∫C ( x − y ) dx, where C:  x ∫C

947

= t ,  y = 2t + 1, for 0 ≤ t ≤ 3

x dy, where C:  x = t ,  y = t 2 , for 1 ≤ t ≤ 2 y

r (t ) = ( cos t ) i + ( sin t ) j + ( t 6 ) k, 0 ≤ t ≤ 2π 21. F = zi + xj + yk r (t ) = ( sin t ) i + ( cos t ) j + tk, 0 ≤ t ≤ 2π 22. F = 6 zi + y 2 j + 12 xk r (t ) = ( sin t ) i + ( cos t ) j + ( t 6 ) k, 0 ≤ t ≤ 2π

948

Chapter 15 Integrals and Vector Fields

Line Integrals in the Plane

23. Evaluate ∫C xy dx + ( x + y ) dy along the curve y = (−1, 1) to ( 2, 4 ).

x2

from

24. Evaluate ∫C ( x − y ) dx + ( x + y ) dy counterclockwise around the triangle with vertices ( 0, 0 ), (1, 0 ), and ( 0, 1). 25. Evaluate ∫C F ⋅ T ds for the vector field F = x 2 i − yj along the curve x = y 2 from ( 4, 2 ) to (1, −1). 26. Evaluate ∫C F ⋅ d r for the vector field F = yi − xj counterclockwise along the unit circle x 2 + y 2 = 1 from (1, 0 ) to ( 0, 1).

by the vector field F = δ v, where v = xi + y 2 j is a velocity field measured in meters per second. Find the flux of F across the curve r(t ). 38. The flow of a gas with a density of δ = 0.3 kg m 2 over the closed curve r (t ) = ( cos t ) i + ( sin t ) j, 0 ≤ t ≤ 2π , is given by the vector field F = δ v, where v = x 2 i − yj is a velocity field measured in meters per second. Find the flux of F across the curve r(t ). 39. Find the flow of the velocity field F = y 2 i + 2 xyj along each of the following paths from ( 0, 0 ) to ( 2, 4 ). a.

b.

y

Work, Circulation, and Flux in the Plane

(2, 4)

27. Work Find the work done by the force F = xyi + ( y − x ) j over the straight line from (1, 1) to ( 2, 3 ).

29. Circulation and flux fields F1 = xi + yj

y = x2

y )2

Find the circulation and flux of the and

(2, 4)

y = 2x

+ 28. Work Find the work done by the gradient of f ( x , y ) = counterclockwise around the circle x 2 + y 2 = 4 from ( 2, 0 ) to itself. (x

F2 = − yi + xj

around and across each of the following curves. a. The circle r (t ) = ( cos t ) i + ( sin t ) j, 0 ≤ t ≤ 2π

x

2

(0, 0)

(0, 0)

x

40. Find the circulation of the field F = yi + ( x + 2 y ) j around each of the following closed paths. a.

y (−1, 1)

(1, 1)

30. Flux across a circle Find the flux of the fields and

2

c. Use any path from ( 0, 0 ) to ( 2, 4 ) different from parts (a) and (b).

b. The ellipse r (t ) = ( cos t ) i + ( 4 sin t ) j, 0 ≤ t ≤ 2π F1 = 2 xi − 3 yj

y

x

F2 = 2 xi + ( x − y ) j (−1, −1)

across the circle r (t ) = ( a cos t ) i + ( a sin t ) j,

0 ≤ t ≤ 2π.

In Exercises 31–34, find the circulation and flux of the field F around and across the closed semicircular path that consists of the semicircular arch r1 (t ) = ( a cos t ) i + ( a sin t ) j, 0 ≤ t ≤ π , followed by the line segment r2 (t ) = ti, −a ≤ t ≤ a. 31. F = xi + yj

32. F = x 2 i + y 2 j

33. F = − yi + xj

34. F = − y 2 i + x 2 j

35. Flow integrals Find the flow of the velocity field F = ( x + y ) i − ( x 2 + y 2 ) j along each of the following paths from (1, 0 ) to (−1, 0 ) in the xy-plane. a. The upper half of the circle x 2 + y 2 = 1 b. The line segment from (1, 0 ) to (−1, 0 ) c. The line segment from (1, 0 ) to ( 0, −1) followed by the line segment from ( 0, −1) to (−1, 0 ) 36. Flux across a triangle Find the flux of the field F in Exercise 35 outward across the triangle with vertices (1, 0 ), ( 0, 1), (−1, 0 ). 37. The flow of a gas with a density of δ = 0.001 kg m 2 over the closed curve r (t ) = (− sin t ) i + ( cos t ) j, 0 ≤ t ≤ 2π , is given

b.

y

(1, −1) x 2 + y2 = 4

x

c. Use any closed path different from parts (a) and (b). 41. Find the work done by the force F = y 2 i + x 3 j, where force is measured in newtons, in moving an object over the curve r (t ) = 2ti + t 2 j, 0 ≤ t ≤ 2, where distance is measured in meters. 42. Find the work done by the force F = e y i + ( ln x ) j + 3zk, where force is measured in newtons, in moving an object over the curve r (t ) = e t i + ( ln t ) j + t 2 k, 1 ≤ t ≤ e, where distance is measured in meters. y x i+ j, 43. Find the flow of the velocity field F = y+1 x +1 where velocity is measured in meters per second, over the curve r (t ) = t 2 i + tj, 0 ≤ t ≤ 1.

15.2  Vector Fields and Line Integrals: Work, Circulation, and Flux

44. Find the flow of the velocity field F = ( y + z ) i + xj − yk, where velocity is measured in meters per second, over the curve r (t ) = e t i − e 2 t j + e −t k, 0 ≤ t ≤ ln 2. 45. Salt water with a density of δ = 0.25 g cm 2 flows over the curve r (t ) = t i + tj, 0 ≤ t ≤ 4, according to the vector field F = δ v, where v = xyi + ( y − x ) j is a velocity field measured in centimeters per second. Find the flow of F over the curve r(t ). 46. Propyl alcohol with a density of δ = 0.2 g cm 2 flows over the closed curve r (t ) = ( sin t ) i − ( cos t ) j, 0 ≤ t ≤ 2π , according to the vector field F = δ v, where v = ( x − y ) i + x 2 j is a velocity field measured in centimeters per second. Find the circulation of F around the curve r(t ).

949

53. Work and area Suppose that f (t ) is differentiable and positive for a ≤ t ≤ b. Let C be the path r (t ) = ti + f (t ) j,  a ≤ t ≤ b, and F = yi. Is there any relation between the value of the work integral

∫C F ⋅ d r and the area of the region bounded by the t-axis, the graph of f , and the lines t = a and t = b? Give reasons for your answer. 54. Work done by a radial force with constant magnitude A particle moves along the smooth curve y = f ( x ) from ( a, f (a) ) to ( b, f (b) ). The force moving the particle has constant magnitude k and always points away from the origin. Show that the work done by the force is

Vector Fields in the Plane

∫C F ⋅ T ds

47. Spin field Draw the spin field F = −

y i+ x 2 + y2

x j x 2 + y2

(see Figure 15.13) along with its horizontal and vertical components at a representative assortment of points on the circle x 2 + y 2 = 4. 48. Radial field Draw the radial field F = xi + yj (see Figure 15.12) along with its horizontal and vertical components at a representative assortment of points on the circle x 2 + y 2 = 1. 49. A field of tangent vectors a. Find a field G = P ( x , y ) i + Q ( x , y ) j in the xy-plane with the property that at any point ( a, b ) ≠ ( 0, 0 ) , G is a vector of magnitude a 2 + b 2 tangent to the circle x 2 + y 2 = a 2 + b 2 and pointing in the counterclockwise direction. (The field is undefined at ( 0, 0 ).) b. How is G related to the spin field F in Figure 15.13? 50. A field of tangent vectors

12 12 = k ⎡⎢ ( b 2 + ( f (b) ) 2 ) − ( a 2 + ( f (a) ) 2 ) ⎤⎥ . ⎣ ⎦

Flow Integrals in Space

In Exercises 55–58, F is the velocity field of a fluid flowing through a region in space. Find the flow along the given curve in the direction of increasing t. 55. F = −4 xyi + 8 yj + 2k r (t ) = t i + t 2 j + k , 0 ≤ t ≤ 2 56. F = x 2 i + yzj + y 2 k r (t ) = 3tj + 4 tk, 0 ≤ t ≤ 1 57. F = ( x − z ) i + xk r (t ) = ( cos t ) i + ( sin t ) k, 0 ≤ t ≤ π 58. F = −yi + xj + 2k r (t ) = (−2 cos t ) i + ( 2 sin t ) j + 2tk, 0 ≤ t ≤ 2π 59. Circulation Find the circulation of F = 2 xi + 2 zj + 2 yk around the closed path consisting of the following three curves traversed in the direction of increasing t. C 1: r (t ) = ( cos t ) i + ( sin t ) j + tk, 0 ≤ t ≤ π 2 C 2 : r (t ) = j + ( π 2 )(1 − t ) k, 0 ≤ t ≤ 1 C 3 : r (t ) = ti + (1 − t ) j, 0 ≤ t ≤ 1

a. Find a field G = P ( x , y ) i + Q ( x , y ) j in the xy-plane with the property that at any point ( a, b ) ≠ ( 0, 0 ) , G is a unit vector tangent to the circle x 2 + y 2 = a 2 + b 2 and pointing in the clockwise direction.

z

C1

b. How is G related to the spin field F in Figure 15.13? 51. Unit vectors pointing toward the origin Find a field F = M ( x , y ) i + N ( x , y ) j in the xy-plane with the property that at each point ( x , y ) ≠ ( 0, 0 ) , F is a unit vector pointing toward the origin. (The field is undefined at ( 0, 0 ).) 52. Two “central” fields Find a field F = M ( x , y ) i + N ( x , y ) j in the xy-plane with the property that at each point ( x , y ) ≠ ( 0, 0 ) , F points toward the origin and F is (a) the distance from ( x , y ) to the origin, (b) inversely proportional to the distance from ( x , y ) to the origin. (The field is undefined at ( 0, 0 ).)

p a0, 1, b 2

(0, 1, 0)

(1, 0, 0) C3 x

C2

y

60. Zero circulation Let C be the ellipse in which the plane 2 x + 3 y − z = 0 meets the cylinder x 2 + y 2 = 12. Show, without evaluating either line integral directly, that the circulation of the field F = xi + yj + zk around C in either direction is zero.

950

Chapter 15 Integrals and Vector Fields

61. Flow along a curve The field F = xyi + yj − yzk is the velocity field of a flow in space. Find the flow from ( 0, 0, 0 ) to (1, 1, 1) along the curve of intersection of the cylinder y = x 2 and the plane z = x. (Hint: Use t = x as the parameter.)

c. Evaluate ∫ F ⋅ d r. C

63. F = xy 6 i + 3 x ( xy 5 + 2 ) j; r (t ) = ( 2 cos t ) i + ( sin t ) j, 0 ≤ t ≤ 2π 3 2 64. F = i+ j; r (t ) = ( cos t ) i + ( sin t ) j, 1 + x2 1 + y2 0 ≤ t ≤ π

z

z=x

a. Find dr for the path r (t ) = g(t ) i + h(t ) j + k (t )k. b. Evaluate the force F along the path.

(1, 1, 1) y

y = x2

65. F = ( y + yz cos xyz ) i + ( x 2 + xz cos xyz ) j + ( z + xy cos xyz ) k; r (t ) = ( 2 cos t ) i + ( 3 sin t ) j + k, 0 ≤ t ≤ 2π 66. F = 2 xyi − y 2 j + ze x k; r (t ) = −ti + 1≤ t ≤ 4

x

62. Flow of a gradient field Find the flow of the field F = ∇( xy 2 z 3 ): a. Once around the curve C in Exercise 58, clockwise as viewed from above; b. Along the line segment from (1, 1, 1) to ( 2, 1, −1).

t j + 3tk,

67. F = ( 2 y + sin x ) i + ( z 2 + (1 3 ) cos y ) j + x 4 k; r (t ) = ( sin t ) i + ( cos t ) j + ( sin 2t ) k, −π 2 ≤ t ≤ π 2 1 68. F = ( x 2 y ) i + x 3 j + xyk; r (t ) = ( cos t ) i + ( sin t ) j + 3 ( 2 sin 2 t − 1) k, 0 ≤ t ≤ 2π

COMPUTER EXPLORATIONS

In Exercises 63–68, use a CAS to perform the following steps for finding the work done by force F over the given path:

15.3 Path Independence, Conservative Fields, and Potential Functions A gravitational field G is a vector field that represents the effect of gravity at a point in space due to the presence of a massive object. The gravitational force on a body of mass m placed in the field is given by F = mG. Similarly, an electric field E is a vector field in space that represents the effect of electric forces on a charged particle placed within it. The force on a body of charge q placed in the field is given by F = q E. In gravitational and electric fields, the amount of work it takes to move a mass or charge from one point to another depends on the initial and final positions of the object—not on which path is taken between these positions. In this section we study vector fields with this independence-ofpath property and the calculation of work integrals associated with them.

Path Independence If A and B are two points in an open region D in space, the line integral of F along C from A to B for a field F defined on D usually depends on the path C taken, as we saw in Section 15.1. For some special fields, however, the integral’s value is the same for all paths from A to B.

DEFINITIONS Let F be a vector field defined on an open region D in space, and suppose that for any two points A and B in D, the line integral ∫ C F ⋅ d r along a path C from A to B in D is the same over all paths from A to B. Then the integral ∫ C F ⋅ d r is path independent in D and the field F is conservative on D.

The word conservative comes from physics, where it refers to fields in which the principle of conservation of energy holds. When a line integral is independent of the path C from B point A to point B, we sometimes represent the integral by the symbol ∫ A rather than the usual line integral symbol ∫ C . This substitution helps us remember the path-independence

15.3  Path Independence, Conservative Fields, and Potential Functions

951

property by indicating that the integral depends only on the initial and final points, not on the path connecting them. Under reasonable differentiability conditions that we will specify, we will show that a field F is conservative if and only if it is the gradient field of a scalar function f —that is, if and only if F = ∇f for some f . The function f then has a special name.

y

If F is a vector field defined on D and F = ∇f for some scalar function f on D, then f is called a potential function for F.

DEFINITION

Simply connected

A gravitational potential is a scalar function whose gradient field is a gravitational field, an electric potential is a scalar function whose gradient field is an electric field, and so on. As we will see, once we have found a potential function f for a field F, we can evaluate all the line integrals in the domain of F over any path between A and B by

x

(a)

B

z

∫A

F ⋅ dr =

B

∫A

∇f ⋅ d r = f ( B) − f ( A).

(1)

If you think of ∇f for functions of several variables as analogous to the derivative f ′ for functions of a single variable, then you see that Equation (1) is the vector calculus rendition of the Fundamental Theorem of Calculus formula b

∫a Simply connected

y

f ′( x ) dx = f (b) − f (a).

Conservative fields have other important properties. For example, saying that F is conservative on D is equivalent to saying that the integral of F around every closed path in D is zero. Certain conditions on the curves, fields, and domains must be satisfied for Equation (1) to be valid. We discuss these conditions next.

x (b) y

Assumptions on Curves, Vector Fields, and Domains

C1 Not simply connected x

(c) z

C2 Not simply connected

y

x (d)

FIGURE 15.25

Four connected regions. In (a) and (b), the regions are simply connected. In (c) and (d), the regions are not simply connected because the curves C 1 and C 2 cannot be contracted to a point inside the regions containing them.

In order for the computations and results we derive below to be valid, we must assume certain properties for the curves, surfaces, domains, and vector fields we consider. We give these assumptions in the statements of theorems, and they also apply to the examples and exercises unless otherwise stated. The curves we consider are piecewise smooth. Such curves are made up of finitely many smooth pieces connected end to end, as discussed in Section 12.1. For such curves we can compute lengths and, except at finitely many points where the smooth pieces connect, tangent vectors. We consider vector fields F whose components have continuous first partial derivatives. The domains D we consider are connected. For an open region, this means that any two points in D can be joined by a smooth curve that lies in the region. Some results require D to be simply connected, which means that every loop in D can be contracted to a point in D without ever leaving D. The plane with a disk removed is a two-dimensional region that is not simply connected; a loop in the plane that goes around the disk cannot be contracted to a point without going into the “hole” left by the removed disk (see Figure 15.25c). Similarly, if we remove a line from space, the remaining region D is not simply connected. A curve encircling the line cannot be shrunk to a point while remaining inside D. Connectivity and simple connectivity are not the same, and neither property implies the other. Think of connected regions as being in “one piece” and of simply connected regions as not having any “loop-catching holes.” All of space itself is both connected and simply connected. Figure 15.25 illustrates some of these properties. Caution Some of the results in this chapter can fail to hold if applied to situations where the conditions we’ve imposed are not met. In particular, the component test for conservative fields, given later in this section, is not valid on domains that are not simply connected (see Example 5). The condition will be stated when needed.

952

Chapter 15 Integrals and Vector Fields

Line Integrals in Conservative Fields A gradient field F is obtained by differentiating a scalar function f . A theorem analogous to the Fundamental Theorem of Calculus gives a way to evaluate the line integrals of gradient fields. Like the Fundamental Theorem of Calculus, Theorem 1 gives a direct way to evaluate line integrals without having to take limits of Riemann sums and without needing to compute a line integral by the procedure used in Section 15.2. Before proving Theorem 1, we give an example.

THEOREM 1—Fundamental Theorem of Line Integrals

Let C be a smooth curve joining the point A to the point B in the plane or in space and parametrized by r(t ). Let f be a differentiable function with a continuous gradient vector F = ∇f on a domain D containing C. Then

∫C F ⋅ d r

= f ( B) − f ( A).

Suppose the force field F = ∇f is the gradient of the function

EXAMPLE 1

f ( x , y, z ) = −

1 . x 2 + y2 + z 2

Find the work done by F in moving an object along a smooth curve C joining (1, 0, 0 ) to ( 0, 0, 2 ) that does not pass through the origin. Solution An application of Theorem 1 shows that the work done by F along any smooth curve C joining the two points and not passing through the origin is

∫C F ⋅ d r

= f ( 0, 0, 2 ) − f (1, 0, 0 ) = −

1 ( ) 3 − −1 = . 4 4

The gravitational force due to a planet, and the electric force associated with a charged particle, can both be modeled by the field F given in Example 1 up to a constant that depends on the units of measurement. When used to model gravity, the function f in Example 1 represents gravitational potential energy. The sign of f is negative, and f approaches −∞ near the origin. This choice ensures that the gravitational force F, the gradient of f , points toward the origin, so that objects fall down rather than up. Proof of Theorem 1 Suppose that A and B are two points in the region D and that C:  r (t ) = g(t ) i + h(t ) j + k (t )k,  a ≤ t ≤ b, is a smooth curve in D joining A to B. In Section 13.5 we found that the derivative of a scalar function f along a path C is the dot product ∇f ( r (t ) ) ⋅ r ′(t ), so we have

∫C F ⋅ d r = ∫C ∇f t=b

=

∫t = a

=

∫a

b

⋅ dr

∇f ( r (t ) ) ⋅ r ′(t ) dt

d f ( r (t ) ) dt dt

F = ∇f

Eq. (2) of Section 15.2 for computing dr

Eq. (7) of Section 13.5 giving derivative along a path

= f ( r (b) ) − f ( r (a) )

Fundamental Theorem of Calculus

= f ( B) − f ( A).

r (a) = A,   r (b) = B

15.3  Path Independence, Conservative Fields, and Potential Functions

953

We see from Theorem 1 that the line integral of a gradient field F = ∇f is straightforward to compute once we know the function f . Many important vector fields arising in applications are indeed gradient fields. The next result, which follows from Theorem 1, shows that any conservative field is of this type.

THEOREM 2—Conservative Fields Are Gradient Fields

Let F = Mi + N j + Pk be a vector field whose components are continuous throughout an open connected region D in space. Then F is conservative if and only if F is a gradient field ∇f for a differentiable function f . Theorem 2 says that F = ∇f if and only if, for any two points A and B in the region D, the value of the line integral ∫ C F ⋅ d r is independent of the path C joining A to B in D. z D

(x0, y, z) L B0 B (x, y, z) x0

C0 A

y

x x

The function f ( x , y, z ) in the proof of Theorem 2 is computed by a line integral ∫C 0 F ⋅ d r = f ( B 0 ) from A to B 0 , plus a line integral ∫ L F ⋅ d r along a line segment L parallel to the x-axis and joining B 0 to B located at ( x , y, z ). The value of f at A is f ( A) = 0. FIGURE 15.26

Proof of Theorem 2 If F is a gradient field, then F = ∇f for a differentiable function f , and Theorem 1 shows that ∫ C F ⋅ d r = f ( B) − f ( A). The value of the line integral does not depend on C, but only on its endpoints A and B. So the line integral is path independent and F satisfies the definition of a conservative field. On the other hand, suppose that F is a conservative vector field. We want to find a function f on D satisfying ∇f = F. First, pick a point A in D and set f ( A) = 0. For any other point B in D define f ( B) to equal ∫ C F ⋅ d r, where C is any smooth path in D from A to B. The value of f ( B) does not depend on the choice of C, since F is conservative. To show that ∇f = F, we need to demonstrate that ∂ f ∂ x = M , ∂ f ∂ y = N , and ∂ f ∂ z = P. Suppose that B has coordinates ( x , y, z ). By the definition of f , the value of the function f at a nearby point B 0 located at ( x 0 , y, z ) is ∫ C 0 F ⋅ d r, where C 0 is any path from A to B 0 . We take a path C = C 0 ∪ L from A to B formed by first traveling along C 0 to arrive at B 0 and then traveling along the line segment L from B 0 to B (Figure 15.26). When B 0 is close to B, the segment L lies in D and, since the value f ( B) is independent of the path from A to B, f ( x , y, z ) =

∫C

F ⋅ dr + 0

∫ L F ⋅ d r.

Differentiating, we have ∂ ∂ f ( x , y, z ) = ∂x ∂x

(∫

C0

F ⋅ dr +

∫ L F ⋅ d r ).

Only the last term on the right depends on x, so ∂ ∂ f ( x , y, z ) = ∂x ∂x

∫ L F ⋅ d r.

Now we parametrize L as r (t ) = ti + yj + zk,  x 0 ≤ t ≤ x. Then d r dt = i, and since x F = Mi + N j + Pk, it follows that F ⋅ d r dt = M and ∫ L F ⋅ d r = ∫ x M ( t , y, z ) dt. 0 Differentiating then gives ∂ ∂ f ( x , y, z ) = ∂x ∂x

∫x

x

M ( t , y, z ) dt = M ( x , y, z )

0

by the Fundamental Theorem of Calculus. The partial derivatives ∂ f ∂ y = N and ∂ f ∂ z = P follow similarly, showing that F = ∇f .

954

Chapter 15 Integrals and Vector Fields

EXAMPLE 2

Find the work done by the conservative field f ( x , y, z ) = xyz ,

F = yzi + xzj + xyk = ∇f , where

in moving an object along any smooth curve C joining the point A(−1, 3, 9 ) to B (1, 6, − 4 ). Solution With f ( x , y, z ) = xyz , we have B

∫C F ⋅ d r = ∫ A

∇f ⋅ d r

F = ∇f and path independence

= f ( B) − f ( A) = xyz

(1,6,− 4 )

Theorem 1

− xyz

( − 1,3,9 )

= (1)( 6 )( −4 ) − ( −1)( 3 )( 9 ) = −24 + 27 = 3. A very useful property of line integrals in conservative fields comes into play when the path of integration is a closed curve, or loop. We often use the notation DC for integration around a closed path (discussed with more detail in the next section).

THEOREM 3—Loop Property of Conservative Fields

The following statements are equivalent. 1. DC F ⋅ d r = 0 around every loop (that is, closed curve C ) in D. 2. The field F is conservative on D. B C2

B

−C 2

Proof that Part 1 ⇒ Part 2 We want to show that for any two points A and B in D, the integral of F ⋅ d r has the same value over any two paths C 1 and C 2 from A to B. We reverse the direction on C 2 to make a path −C 2 from B to A (Figure 15.27). Together, C 1 and −C 2 make a closed loop C, and by assumption,

C1

C1

A

∫ C F ⋅ d r − ∫C

A

1

FIGURE 15.27

If we have two paths from A to B, one of them can be reversed to make a loop. B

C2

B

−C 2

∫C F ⋅ d r + ∫−C 1

F ⋅ dr = 2

∫C F ⋅ d r

= 0.

Thus, the integrals over C 1 and C 2 give the same value. Note that the definition of F ⋅ d r shows that changing the direction along a curve reverses the sign of the line integral. Proof that Part 2 ⇒ Part 1 We want to show that the integral of F ⋅ d r is zero over any closed loop C. We pick two points A and B on C and use them to break C into two pieces: C 1 from A to B followed by C 2 from B back to A (Figure 15.28). Then

DC F ⋅ dr

C1

F ⋅ dr = 2

=

∫C F ⋅ d r + ∫C 1

F ⋅ dr = 2

B

∫A

F ⋅ dr −

B

∫A

F ⋅ d r = 0.

C1

The following diagram summarizes the results of Theorems 2 and 3. A

A Theorem 2

FIGURE 15.28

If A and B lie on a loop, we can reverse part of the loop to make two paths from A to B.

F = ∇f  on  D

⇔

Theorem 3

F  conservative on  D

⇔

DC F ⋅ dr

= 0

over any loop in D

955

15.3  Path Independence, Conservative Fields, and Potential Functions

Two questions arise: 1. How do we know whether a given vector field F is conservative? 2. If F is in fact conservative, how do we find a potential function f (so that F = ∇f )?

Finding Potentials for Conservative Fields The test for a vector field being conservative involves the equivalence of certain first partial derivatives of the field components.

Component Test for Conservative Fields

Let F = M ( x , y, z ) i + N ( x , y, z ) j + P ( x , y, z ) k be a field on an open simply connected domain whose component functions have continuous first partial derivatives. Then F is conservative if and only if ∂P ∂N , = ∂y ∂z

∂M ∂P , = ∂z ∂x

and

∂N ∂M . = ∂x ∂y

(2)

We can view the component test as saying that on a simply connected region, the vector

(

)

⎛ ∂P ⎛ ∂N ∂ N ⎟⎞ ∂M ∂P ∂ M ⎞⎟ ⎜⎜ − − j + ⎜⎜ − ⎟i + ⎟k ⎝ ∂y ⎝ ∂x ∂ z ⎟⎠ ∂z ∂x ∂ y ⎟⎠

(3)

is zero if and only if F is conservative. This interesting vector curl F is called the curl of F. We study it in Sections 15.4 and 15.7. Proof that Equations (2) hold if F is conservative is a potential function f such that F = Mi + N j + P k = ∇f =

If F is conservative, then there

∂f ∂f ∂f i+ j+ k. ∂x ∂y ∂z

Hence,

( ) = ∂∂y ∂f z

∂P ∂ ∂f = ∂y ∂y ∂z

2

=

∂2 f ∂z ∂y

=

∂ ⎛⎜ ∂ f ⎞⎟ ∂N . ⎟= ∂ z ⎜⎝ ∂ y ⎟⎠ ∂z

Mixed Derivative Theorem, Section 13.3

The others in Equations (2) are proved similarly. The second half of the proof, that Equations (2) imply that F is conservative, is a consequence of Stokes’ Theorem, taken up in Section 15.7, and requires our assumption that the domain of F be simply connected. Once we know that F is conservative, we often want to find a potential function for F. This requires solving the equation ∇f = F or ∂f ∂f ∂f i+ j+ k = Mi + N j + P k ∂x ∂y ∂z

956

Chapter 15 Integrals and Vector Fields

for f . We accomplish this by integrating the three equations ∂f = M, ∂x

∂f = N, ∂y

∂f = P, ∂z

as illustrated in the next example. EXAMPLE 3 Show that F = ( e x cos y + yz ) i + ( xz − e x sin y ) j + ( xy + z ) k is conservative over its natural domain, and find a potential function for it. Solution The natural domain of F is all of space, which is open and simply connected. We apply the test in Equations (2) to

M = e x cos y + yz ,

N = xz − e x sin y,

P = xy + z

and calculate ∂P ∂N , = x = ∂y ∂z

∂M ∂P , = y = ∂z ∂x

∂N ∂M = −e x sin y + z = . ∂x ∂y

The partial derivatives are continuous, so these equalities tell us that F is conservative, so there is a function f with ∇f = F (Theorem 2). We find f by integrating the equations ∂f = e x cos y + yz , ∂x

∂f = xz − e x sin y, ∂y

∂f = xy + z. ∂z

(4)

We integrate the first equation with respect to x, holding y and z fixed, to get f ( x , y, z ) = e x cos y + xyz + g ( y, z ). We write the constant of integration as a function of y and z because its value may depend on y and z, though not on x. We then calculate ∂ f ∂ y from this equation and match it with the expression for ∂ f ∂ y in Equations (4). This gives −e x sin y + xz +

∂g = xz − e x sin y, ∂y

so ∂ g ∂ y = 0. Therefore, g is a function of z alone, and f ( x , y, z ) = e x cos y + xyz + h( z ). We now calculate ∂ f ∂ z from this equation and match it to the formula for ∂ f ∂ z in Equations (4). This gives xy +

dh = xy + z , dz

or

dh = z, dz

so h( z ) =

z2 + C. 2

Hence, f ( x , y, z ) = e x cos y + xyz +

z2 + C. 2

We found infinitely many potential functions of F, one for each value of C.

15.3  Path Independence, Conservative Fields, and Potential Functions

EXAMPLE 4

957

Show that F = ( 2 x − 3 ) i − zj + ( cos z ) k is not conservative.

Solution We apply the Component Test in Equations (2) and find immediately that

∂P ∂ = ( cos z ) = 0, ∂y ∂y

∂N ∂ (−z ) = −1. = ∂z ∂z

The two are unequal, so F is not conservative. No further testing is required.

EXAMPLE 5

Show that the vector field F =

−y x i+ 2 j + 0k x 2 + y2 x + y2

satisfies the equations in the Component Test but is not conservative over its natural domain. Explain why this is possible. Solution We have M = −y ( x 2 + y 2 ) ,  N = x ( x 2 + y 2 ) , and P = 0. If we apply the Component Test, we find

∂P ∂N = 0 = , ∂y ∂z

∂P ∂M = 0 = , ∂x ∂z

and

y2 − x 2 ∂M ∂N = . 2 = 2 2 ∂y ∂x (x + y )

So it may appear that the field F passes the Component Test. However, the test assumes that the domain of F is simply connected, which is not the case here. Since x 2 + y 2 cannot equal zero, the natural domain is the complement of the z-axis and contains loops that cannot be contracted to a point. One such loop is the unit circle C in the xy-plane. The circle is parametrized by r (t ) = ( cos t ) i + ( sin t ) j, 0 ≤ t ≤ 2π. This loop wraps around the z-axis and cannot be contracted to a point while staying within the complement of the z-axis. To show that F is not conservative, we compute the line integral D F ⋅ d r around the C loop C. First we write the field in terms of the parameter t: F =

x2

− sin t cos t −y x i+ 2 j = i+ j = ( − sin t ) i + ( cos t ) j. 2 2 2 2 2 + y x + y sin t + cos t sin t + cos 2 t

Next we find d r dt = ( − sin t ) i + ( cos t ) j and then calculate the line integral as

DC

F ⋅ dr =

DC

F⋅

dr dt = dt

2π

∫0

( sin 2 t + cos 2 t ) dt = 2π.

Since the line integral of F around the loop C is not zero, the field F is not conservative, by Theorem 3. The field F is displayed in Figure 15.31d in the next section.

Example 5 shows that the Component Test does not apply when the domain of the field is not simply connected. However, if we change the domain in the example so that it is restricted to the ball of radius 1 centered at the point ( 2, 2, 2 ), or to any similar ballshaped region that does not contain a piece of the z-axis, then this new domain D is simply connected. Now the partial derivative Equations (2), as well as all the assumptions of the Component Test, are satisfied. In this new situation, the field F in Example 5 is conservative on D. Just as we must be careful with a function when determining whether it satisfies a property throughout its domain (such as continuity, which is required for the Intermediate Value Property), so must we also be careful with a vector field in determining the properties it may or may not have over its assigned domain.

958

Chapter 15 Integrals and Vector Fields

Exact Differential Forms It is often convenient to express work and circulation integrals in the differential form

∫C M dx + N dy + P dz discussed in Section 15.2. Such line integrals are relatively easy to evaluate if M dx + N dy + P dz is the total differential of a function f and if C is any path joining the point A to the point B, for then ∂f

∂f

∫C M dx + N dy + P dz = ∫C ∂ x dx + ∂ y dy + =

B

∫A

∂f dz ∂z

∇f ⋅ d r

∇f is conservative.

= f ( B) − f ( A).

Theorem 1

Thus, B

∫ A df

= f ( B) − f ( A),

just as with differentiable functions of a single variable.

DEFINITIONS Any expression M ( x , y, z ) dx + N ( x , y, z ) dy + P ( x , y, z ) dz

is a differential form. A differential form is exact on a domain D in space if M dx + N dy + P dz =

∂f ∂f ∂f dx + dy + dz = df ∂y ∂z ∂x

for some scalar function f throughout D.

Notice that if M dx + N dy + P dz = df on D, then F = Mi + N j + Pk is the gradient field of f on D. Conversely, if F = ∇f , then the form M dx + N dy + P dz is exact. The test for the form being exact is therefore the same as the test for F being conservative.

Component Test for Exactness of M dx + N dy + P dz

The differential form M dx + N dy + P dz is exact on an open simply connected domain if and only if ∂P ∂N , = ∂y ∂z

∂M ∂P , = ∂z ∂x

and

∂N ∂M = . ∂x ∂y

This is equivalent to saying that the field F = Mi + N j + Pk is conservative.

EXAMPLE 6

Show that y dx + x dy + 4 dz is exact, and evaluate the integral ( 2,3,−1)

∫ 1,1,1 (

)

y dx + x dy + 4 dz

over any path from (1, 1, 1) to ( 2, 3, −1).

15.3  Path Independence, Conservative Fields, and Potential Functions

959

Solution Note that the domain of F fills all of three-dimensional space, so it is simply connected. We let M = y,  N = x ,  P = 4 and apply the Component Test for Exactness:

∂P ∂N , = 0 = ∂y ∂z

∂M ∂P , = 0 = ∂z ∂x

∂N ∂M . =1= ∂x ∂y

These equalities tell us that y dx + x dy + 4 dz is exact, so y dx + x dy + 4 dz = df for some function f , and the integral’s value is f ( 2, 3, −1) − f (1, 1, 1). We find f up to a constant by integrating the equations ∂f = y, ∂x

∂f = x, ∂y

∂f = 4. ∂z

(5)

From the first equation we get f ( x , y, z ) = xy + g ( y, z ). The second equation tells us that ∂f ∂g = x+ = x, ∂y ∂y

or

∂g = 0. ∂y

Hence, g is a function of z alone, and f ( x , y, z ) = xy + h( z ). The third of Equations (5) tells us that ∂f dh = 0+ = 4, dz ∂z

or

h( z ) = 4 z + C.

Therefore, f ( x , y, z ) = xy + 4 z + C. The value of the line integral is independent of the path taken from (1, 1, 1) to ( 2, 3, −1) and equals f ( 2, 3, −1) − f (1, 1, 1) = 2 + C − ( 5 + C ) = −3.

EXERCISES

15.3

Testing for Conservative Fields

Which fields in Exercises 1–6 are conservative, and which are not? 1. F = yzi + xzj + xyk 2. F = ( y sin z ) i + ( x sin z ) j + ( xy cos z ) k 3. F = yi + ( x + z ) j − yk

11. F = ( ln x + sec 2 ( x + y )) i + ⎛ ⎞⎟ y z ⎜⎜ sec 2 ( x + y ) + k ⎟j + 2 ⎝ y 2 + z 2 ⎟⎠ y + z2 12. F =

4. F = − yi + xj

⎛ y x i + ⎜⎜⎜ + ⎝1 + x 2 y 2 1 + x 2y2

5. F = ( z + y ) i + z j + ( y + x ) k 6. F = ( e x cos y ) i − ( e x sin y ) j + zk Finding Potential Functions

In Exercises 7–12, find a potential function f for the field F. 7. F = 2 xi + 3 yj + 4 zk 8. F = ( y + z ) i + ( x + z ) j + ( x + y ) k 9. F = e y + 2 z ( i + xj + 2 xk ) 10. F = ( y sin z ) i + ( x sin z ) j + ( xy cos z ) k

z ⎟⎞⎟ j + 1 − y 2 z 2 ⎟⎠ ⎛ y 1 ⎞⎟ ⎜⎜ ⎜⎝ 1 − y 2 z 2 + z ⎟⎟⎠ k

Exact Differential Forms

In Exercises 13–17, show that the differential forms in the integrals are exact. Then evaluate the integrals. ( 2,3,− 6 )

13.

∫ 0,0,0

14.

∫ 1,1,2

(

)

( 3,5,0 )

(

)

2 x dx + 2 y dy + 2 z dz

yz dx + xz dy + xy dz

960

15. 16. 17.

Chapter 15 Integrals and Vector Fields

(1,2,3 )

∫ 0,0,0 (

( 3,3,1)

∫ 0,0,0 (

(

c. The x-axis from (1, 0, 0 ) to ( 0, 0, 0 ) followed by the parabola z = x 2,  y = 0 from ( 0, 0, 0 ) to (1, 0, 1)

4 dz 2 x dx − y 2 dy − 1 + z2

)

( 0,1,1)

∫ 1,0,0

b. The helix r (t ) = ( cos t ) i + ( sin t ) j + ( t 2π ) k, 0 ≤ t ≤ 2π

2 xy dx + ( x 2 − z 2 ) dy − 2 yz dz

)

z

sin y cos x dx + cos y sin x dy + dz

)

1 (1, 0, 1)

Finding Potential Functions to Evaluate Line Integrals

z = x2

Although they are not defined on all of space R 3, the fields associated with Exercises 18–22 are conservative. Find a potential function for each field, and evaluate the integrals as in Example 6. (1,π 2,2 )

18.

∫ 0,2,1

19.

∫ 1,1,1

20.

∫ 1,2,1

21. 22.

(

)

(1,2,3 )

(

( 2,1,1)

(

)

(

z2 dy + 2 z ln y dz y

⎛x2 ⎞ ( 2 x ln y − yz ) dx + ⎜⎜ − xz ⎟⎟⎟ dy − xy dz ⎝ y ⎠

( 2,2,2 )

∫ 1,1,1

⎛1 ⎞ 1 2 cos y dx + ⎜⎜ − 2 x sin y ⎟⎟⎟ dy + dz ⎝y ⎠ z

3 x 2 dx +

)

)

(0, 0, 0)

x

( 2,2,2 )

(

)

(1, 0, 0)

30. Work along different paths Find the work done by F = e yz i + ( xze yz + z cos y ) j + ( xye yz + sin y ) k over the following paths from (1, 0, 1) to (1, π 2, 0 ). a. The line segment x = 1,  y = πt 2,  z = 1 − t , 0 ≤ t ≤ 1 z

⎛1 y x ⎞ 1 dx + ⎜⎜ − 2 ⎟⎟⎟ dy − 2 dz ⎝z y y ⎠ z

∫ −1,−1,−1

y

1 (1, 0, 1)

2 x dx + 2 y dy + 2 z dz x 2 + y2 + z 2

p 2 y

Applications and Examples

23. Revisiting Example 6

Evaluate the integral

( 2,3,−1)

∫ 1,1,1 (

)

y dx + x dy + 4 dz

from Example 6 by finding parametric equations for the line segment from (1, 1, 1) to ( 2, 3, −1) and evaluating the line integral of F = yi + xj + 4 k along the segment. Since F is conservative, the integral is independent of the path. 24. Evaluate

x

p Q1, 2 , 0R

1

b. The line segment from (1, 0, 1) to the origin followed by the line segment from the origin to (1, π 2, 0 ) z 1 (1, 0, 1)

∫C x 2 dx + yz dy + ( y 2

2 ) dz

p 2

(0, 0, 0)

y

along the line segment C joining ( 0, 0, 0 ) to ( 0, 3, 4 ). Independence of path Show that the values of the integrals in Exercises 25 and 26 do not depend on the path taken from A to B. B

25.

∫ A z 2 dx + 2 y dy + 2 xz dz

26.

∫A

B

x

c. The line segment from (1, 0, 1) to (1, 0, 0 ), followed by the x-axis from (1, 0, 0 ) to the origin, followed by the parabola y = π x 2 2,  z = 0 from there to (1, π 2, 0 )

x dx + y dy + z dz x 2 + y2 + z 2

z

In Exercises 27 and 28, find a potential function for F. ⎛ 1 − x 2 ⎞⎟ 2x 27. F = i + ⎜⎜ j, ⎝ y 2 ⎟⎟⎠ y

p Q1, 2 , 0R

1

{( x , y ):  y > 0 }

1 (1, 0, 1)

⎛ex ⎞ 28. F = ( e x ln y ) i + ⎜⎜ + sin z ⎟⎟⎟ j + ( y cos z ) k ⎝ y ⎠ 29. Work along different paths Find the work done by F = ( x 2 + y ) i + ( y 2 + x ) j + ze z k over the following paths from (1, 0, 0 ) to (1, 0, 1). a. The line segment x = 1,  y = 0, 0 ≤ z ≤ 1

(0, 0, 0)

y

y = p x2 2 x (1, 0, 0)

p Q1, 2 , 0R

15.4  Green’s Theorem in the Plane

31. Evaluating a work integral two ways Let F = ∇( x 3 y 2 ) and let C be the path in the xy-plane from (−1, 1) to (1, 1) that consists of the line segment from (−1, 1) to ( 0, 0 ) followed by the line segment from ( 0, 0 ) to (1, 1). Evaluate ∫C F ⋅ d r in two ways. a. Find parametrizations for the segments that make up C and evaluate the integral. b. Use f ( x , y ) = x 3 y 2 as a potential function for F. 32. Integral along different paths Evaluate the line integral ∫C 2 x cos y dx − x 2 sin y dy along the following paths C in the xy-plane. a. The parabola y = ( x − 1) 2 from (1, 0 ) to ( 0, 1) b. The line segment from (−1, π ) to (1, 0 ) c. The x-axis from (−1, 0 ) to (1, 0 ) d. The astroid r (t ) = ( cos 3 t ) i + ( sin 3 t ) j, 0 ≤ t ≤ 2π, counterclockwise from (1, 0 ) back to (1, 0 ) y (0, 1)

961

34. Gradient of a line integral Suppose that F = ∇f is a conservative vector field and g ( x , y, z ) =

( x , y ,z )

∫ 0,0,0 (

)

F ⋅ d r.

Show that ∇g = F. 35. Path of least work You have been asked to find the path along which a force field F will perform the least work in moving a particle between two locations. A quick calculation on your part shows F to be conservative. How should you respond? Give reasons for your answer. 36. A revealing experiment By experiment, you find that a force field F performs only half as much work in moving an object along path C 1 from A to B as it does in moving the object along path C 2 from A to B. What can you conclude about F? Give reasons for your answer. 37. Work by a constant force Show that the work done by a constant force field F = a i + b j + c k in moving a particle along any path from A to B is W = F ⋅ AB. 38. Gravitational field

(−1, 0)

(1, 0)

a. Find a potential function for the gravitational field x

F = −GmM

x i + y j + zk (x 2 + y2 + z 2 )

32

(G, m, and M are constants).

(0, −1)

33. a. Exact differential form How are the constants a, b, and c related if the following differential form is exact? ( ay 2 + 2czx ) dx + y ( bx + cz ) dy + ( ay 2 + cx 2 ) dz

b. Let P1 and P2 be points at distances s1 and s 2 from the origin. Show that the work done by the gravitational field in part (a) in moving a particle from P1 to P2 is ⎛1 1⎞ GmM ⎜⎜ − ⎟⎟⎟. ⎝ s2 s1 ⎠

b. Gradient field For what values of b and c will F = ( y 2 + 2czx ) i + y ( bx + cz ) j + ( y 2 + cx 2 ) k be a gradient field?

15.4 Green’s Theorem in the Plane If F is a conservative field, then we know F = ∇f for a differentiable function f , and we can calculate the line integral of F over any path C joining point A to point B as ∫ C F ⋅ d r = f ( B) − f ( A). In this section we derive a method for computing a work or flux integral over a closed curve C in the plane. This method, which can be used even when the field F is not conservative, comes from Green’s Theorem. It enables us to convert the line integral into a double integral over the region enclosed by C. The discussion is given in terms of velocity fields of fluid flows (a fluid is a liquid or a gas) because they are easy to visualize. However, Green’s Theorem applies to any vector field, independent of any particular interpretation of the field, provided the assumptions of the theorem are satisfied. We introduce two new ideas for Green’s Theorem: circulation density around an axis perpendicular to the plane and divergence (or flux density).

Spin Around an Axis: The k-Component of Curl Suppose that F ( x , y ) = M ( x , y ) i + N ( x , y ) j is the velocity field of a fluid flowing in the plane and that the first partial derivatives of M and N are continuous at each point of a region R. Let ( x , y ) be a point in R, and let A be a small rectangle with one corner at ( x , y )

962

Chapter 15 Integrals and Vector Fields

that, along with its interior, lies entirely in R. The sides of the rectangle, parallel to the coordinate axes, have lengths of Δx and Δy. Assume that the components M and N do not change sign throughout a small region containing the rectangle A. The first idea we use to convey Green’s Theorem quantifies the rate at which a floating paddle wheel, with axis perpendicular to the plane, spins at a point in a fluid flowing in a plane region. This idea gives some sense of how the fluid is circulating around axes located at different points and perpendicular to the plane. Physicists sometimes refer to this as the circulation density of a vector field F at a point. To obtain it, we consider the velocity field F( x, y ) = M ( x, y ) i + N ( x, y ) j and the rectangle A in Figure 15.29 (where we assume both components of F are positive). (x, y + Δy)

(x + Δ x, y + Δy) F · (−i) < 0

F · (−j) < 0

Δy

A

F(x, y)

F·j>0 F·i>0 (x + Δ x, y)

(x, y) Δx

FIGURE 15.29

The rate at which a fluid flows along the bottom edge of a rectangular region A in the direction i is approximately F ( x , y ) ⋅ i Δx , which is positive for the vector field F shown here. To approximate the rate of circulation at the point ( x , y ), we calculate the (approximate) flow rates along each edge in the directions of the red arrows, sum these rates, and then divide the sum by the area of A. Taking the limit as Δx → 0 and Δy → 0 gives the rate of the circulation per unit area.

The circulation rate of F around the boundary of A is the sum of flow rates along the sides in the tangential direction. For the bottom edge, the flow rate is approximately F ( x , y ) ⋅ i Δ x = M ( x , y ) Δ x. This is the scalar component of the velocity F ( x , y ) in the tangent direction i times the length of the segment. The flow rates may be positive or negative, depending on the components of F. We approximate the net circulation rate around the rectangular boundary of A by summing the flow rates along the four edges as defined by the following dot products. Top:

F ( x , y + Δy ) ⋅ (−i) Δx = −M ( x , y + Δy ) Δx

Bottom:

F ( x , y ) ⋅ i Δx = M ( x , y ) Δx

Right:

F ( x + Δx , y ) ⋅ j Δy = N ( x + Δx , y ) Δy

Left:

F ( x , y ) ⋅ (− j) Δy = −N ( x , y ) Δy

We sum opposite pairs to get Top and bottom:

⎞ ⎛ ∂M −( M ( x , y + Δy ) − M ( x , y )) Δx ≈ −⎜⎜ Δy ⎟⎟⎟ Δx ⎠ ⎝ ∂y

Right and left:

( N ( x + Δ x , y ) − N ( x , y )) Δ y ≈

( ∂∂Nx Δx ) Δy.

963

15.4  Green’s Theorem in the Plane

Vertical axis

Adding these last two equations gives the net circulation rate relative to the counterclockwise orientation,

k

⎛ ∂N ∂ M ⎞⎟ Circulation rate around rectangle ≈ ⎜⎜ − ⎟ Δx Δy. ⎝ ∂x ∂ y ⎟⎠ (x 0 , y 0 )

We now divide by Δx Δy to estimate the circulation rate per unit area, or circulation density, for the rectangle: Circulation around rectangle ∂N ∂M ≈ − . ∂x ∂y rectangle area

curl F (x 0 , y 0 ) . k > 0 Counterclockwise circulation

Vertical axis

We let Δx and Δy approach zero to define the circulation density of F at the point ( x , y ). If we see a counterclockwise rotation looking downward onto the xy-plane from the tip of the unit k vector, then the circulation density is positive (Figure 15.30).

k

The circulation density of a vector field F = Mi + N j at the point ( x , y ) is the scalar expression

DEFINITION

∂N ∂M − . ∂x ∂y

(x 0 , y 0 )

(1)

curl F (x 0 , y 0 ) . k < 0 Clockwise circulation

FIGURE 15.30

In the flow of an incompressible fluid over a plane region, the k-component of the curl measures the rate of the fluid’s rotation at a point. The k-component of the curl is positive at points where the rotation is counterclockwise and negative where the rotation is clockwise.

The expression in Equation (1) is the the k-component of the curl of F, which was introduced in Equation (3) of Section 15.3: ∂N ∂M = ( curl  F ) ⋅ k. − ∂x ∂y If water is moving about a region in the xy-plane in a thin layer, then the k-component of the curl at a point ( x 0 , y 0 ) gives a way to measure how fast and in what direction a small paddle wheel spins if it is put into the water at ( x 0 , y 0 ) with its axis perpendicular to the plane, parallel to k (Figure 15.30). Looking downward onto the xy-plane, it spins counterclockwise when ( curl  F ) ⋅ k is positive and clockwise when the k-component is negative. EXAMPLE 1 The following vector fields represent the velocity of a gas flowing in the xy-plane. Find the circulation density of each vector field and interpret its physical meaning. Figure 15.31 displays the vector fields. (a) Uniform expansion or compression: F ( x , y ) = cxi + cyj (b) Uniform rotation: F ( x , y ) = −cyi + cxj (c) Shearing flow: F ( x , y ) = yi −y x (d) Whirlpool effect: F ( x , y ) = 2 i+ 2 j x + y2 x + y2

c a constant

Solution

∂ ∂ (a) Uniform expansion: ( curl  F ) ⋅ k = (cy) − (cx ) = 0. The gas is not circulating ∂ ∂ x y at very small scales. ∂ ∂ (b) Rotation: ( curl  F ) ⋅ k = (cx ) − (−cy) = 2c. The constant circulation density ∂x ∂y indicates rotation around every point. If c > 0, the rotation is counterclockwise; if c < 0, the rotation is clockwise.

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Chapter 15 Integrals and Vector Fields

y

y

x

x

(a)

(b)

y

y

x

(c)

FIGURE 15.31

y

x

(d)

Velocity fields of a gas flowing in the plane (Example 1).

∂ ( y) = −1. The circulation density is constant and negative, ∂y so a paddle wheel floating in water undergoing such a shearing flow spins clockwise. The rate of rotation is the same at each point. The average rotational effect of the fluid flow is to push fluid clockwise around each of the small circles shown in Figure 15.32.

(c) Shear: ( curl  F ) ⋅ k = −

x

(d) Whirlpool: ( curl  F ) ⋅

FIGURE 15.32

A shearing flow pushes the fluid clockwise around each point (Example 1c).

k=

⎞⎟ x y2 − x 2 y2 − x 2 ∂ ⎛⎜ ∂ ⎛⎜ −y ⎞⎟ − = 0. ⎟⎟ − ⎟⎟ = 2 ⎜⎜ 2 ⎜⎜ 2 2 2 2 ∂x ⎝ x + y ⎠ ∂y⎝ x + y ⎠ ( x + y 2 ) ( x 2 + y 2 )2

The circulation density is 0 at every point away from the origin (where the vector field is undefined and the whirlpool effect blows up), and the gas is not circulating at any point for which the vector field is defined. One form of Green’s Theorem tells us how circulation density can be used to calculate the line integral for flow in the xy-plane. (The flow integral was defined in Section 15.2.) A second form of the theorem tells us how we can calculate the flux integral, which gives the flow across the boundary, from flux density. We define this idea next and then present both versions of the theorem.

Divergence Consider again the velocity field F ( x , y ) = M ( x , y ) i + N ( x , y ) j in a domain containing the rectangle A, as shown in Figure 15.33. As before, we assume the field components do not change sign throughout a small region containing the rectangle A. Our interest now is to determine the rate at which the fluid leaves A by flowing across its boundary.

15.4  Green’s Theorem in the Plane

(x, y + Δy)

965

(x + Δx, y + Δ y) F·j>0

F · (−i) < 0 A

Δy F·i>0

F(x, y) F · (−j) < 0

(x + Δx, y)

(x, y) Δx

FIGURE 15.33

The rate at which the fluid leaves the rectangular region A across the bottom edge in the direction of the outward normal −j is approximately F ( x , y ) ⋅ (− j) Δx , which is negative for the vector field F shown here. To approximate the flow rate at the point ( x , y ), we calculate the (approximate) flow rates across each edge in the directions of the red arrows, sum these rates, and then divide the sum by the area of A. Taking the limit as Δx → 0 and Δy → 0 gives the flow rate per unit area.

The rate at which fluid leaves the rectangle across the bottom edge is approximately (Figure 15.33) F ( x , y ) ⋅ (− j) Δx = −N ( x , y ) Δx. This is the scalar component of the velocity at ( x , y ) in the direction of the outward normal times the length of the segment. If the velocity is in meters per second, for example, the flow rate will be in meters per second times meters, or square meters per second. The rates at which the fluid crosses the other three sides in the directions of their outward normals can be estimated in a similar way. The flow rates may be positive or negative, depending on the signs of the components of F. We approximate the net flow rate across the rectangular boundary of A by summing the flow rates across the four edges as defined by the following dot products. Fluid   Flow   Rates:

Top:

F ( x , y + Δy ) ⋅ j Δx = N ( x , y + Δy ) Δx

Bottom:

F ( x , y ) ⋅ (− j) Δx = −N ( x , y ) Δx

Right:

F ( x + Δx , y ) ⋅ i Δy = M ( x + Δx , y ) Δy

Left:

F ( x , y ) ⋅ (−i) Δy = −M ( x , y ) Δy

Summing opposite pairs gives Top and bottom:

⎞ ⎛ ∂N Δy ⎟⎟⎟ Δx , ( N ( x , y + Δy ) − N ( x , y )) Δx ≈ ⎜⎜ ⎠ ⎝ ∂y

Right and left:

( M ( x + Δ x , y ) − M ( x , y )) Δ y ≈

( ∂∂Mx Δx ) Δy.

Adding these last two equations gives the net effect of the flow rates, or the ⎛ ∂M ∂ N ⎞⎟ Flux across rectangle boundary ≈ ⎜⎜ + ⎟ Δx Δy. ⎝ ∂x ∂ y ⎟⎠ We now divide by Δx Δy to estimate the total flux per unit area, or flux density, for the rectangle: ⎛ ∂M Flux across rectangle boundary ∂ N ⎞⎟ ≈ ⎜⎜ + ⎟. ⎝ ∂x rectangle area ∂ y ⎟⎠

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Chapter 15 Integrals and Vector Fields

div F is the symbol for divergence.

Finally, we let Δx and Δy approach zero to define the flux density of F at the point ( x , y ). The mathematical term for the flux density is the divergence of F. The symbol for it is div F, which is pronounced “divergence of F” or “div F.” The divergence (flux density) of a vector field F = Mi + N j at the point ( x , y ) is

DEFINITION

div F =

Source: div F (x 0 , y0 ) > 0 A gas expanding at the point (x 0 , y0 )

Sink: div F (x 0 , y0 ) < 0 A gas compressing at the point (x 0 , y0 )

If a gas is expanding at a point ( x 0 , y 0 ) , the lines of flow have positive divergence; if the gas is compressing, the divergence is negative.

(2)

A gas is compressible, unlike a liquid, and the divergence of its velocity field measures to what extent it is expanding or compressing at each point. Intuitively, if a gas is expanding at the point ( x 0 , y 0 ) , the lines of flow would diverge there (hence the name), and since the gas would be flowing out of a small rectangle about ( x 0 , y 0 ) , the divergence of F at ( x 0 , y 0 ) would be positive. If the gas were compressing instead of expanding, the divergence would be negative (Figure 15.34). EXAMPLE 2 Find the divergence, and interpret what it means, for each vector field in Example 1 representing the velocity of a gas flowing in the xy-plane. Solution

∂ ∂ (cx ) + (cy) = 2c: If c > 0, the gas is undergoing uniform expansion; ∂x ∂y if c < 0, it is undergoing uniform compression.

(a) div F =

(b) div F =

FIGURE 15.34

∂M ∂N + . ∂x ∂y

∂ ∂ (−cy) + (cx ) = 0: The gas is neither expanding nor compressing. ∂x ∂y

∂ ( y) = 0: The gas is neither expanding nor compressing. ∂x ⎞⎟ 2 xy 2 xy ∂ ⎛⎜ −y ⎞⎟ ∂ ⎛⎜ x − = 0: Again, ⎟⎟ + ⎟= (d) div F = ⎜⎜ 2 ⎜⎜ 2 2 2 ∂ x ⎝ x + y ⎠ ∂ y ⎝ x + y ⎟⎠ ( x 2 + y 2 ) 2 ( x 2 + y 2 )2 the divergence is zero at all points in the domain of the velocity field. (c) div F =

Cases (b), (c), and (d) of Figure 15.31 are plausible models for the two-dimensional flow of a liquid. In fluid dynamics, when the velocity field of a flowing fluid always has divergence equal to zero, as in those cases, the flow is said to be incompressible.

Two Forms for Green’s Theorem A simple closed curve C can be traversed in two possible directions. (Recall that a curve is simple if it does not cross itself.) The curve is traversed counterclockwise, and said to be positively oriented, if the region it encloses is always to the left when moving along the curve. If the curve is traversed clockwise, then the enclosed region is on the right when moving along the curve, and the curve is said to be negatively oriented. The line integral of a vector field F along C reverses sign if we change the orientation. We use the notation

DC F( x , y ) ⋅ dr for the line integral when the simple closed curve C is traversed counterclockwise, with its positive orientation. In one form, Green’s Theorem says that the counterclockwise circulation of a vector field around a simple closed curve is the double integral of the k-component of the curl of the field over the region enclosed by the curve. Recall the defining Equation (5) for circulation in Section 15.2.

967

15.4  Green’s Theorem in the Plane

THEOREM 4—Green’s Theorem (Circulation-Curl or Tangential Form) Circulation around  C = ( curl  F )

⋅k =

DC F ⋅ T ds

∂N ∂M − ∂x ∂y

Let C be a piecewise smooth, simple closed curve enclosing a region R in the plane. Let F = Mi + N j be a vector field with M and N having continuous first partial derivatives in an open region containing R. Then the counterclockwise circulation of F around C equals the double integral of ( curl  F ) ⋅ k over R.

DC

F ⋅ T ds =

DC

M dx + N dy =

⎛ ∂N

∫∫ ⎜⎜⎝ ∂ x

−

R

Counterclockwise circulation

∂ M ⎞⎟ ⎟ dx dy ∂ y ⎟⎠

(3)

Curl integral

A second form of Green’s Theorem says that the outward flux of a vector field across a simple closed curve in the plane equals the double integral of the divergence of the field over the region enclosed by the curve. Recall the formulas for flux in Equations (8) and (9) in Section 15.2.

THEOREM 5—Green’s Theorem (Flux-Divergence or Normal Form) Flux of  F  across  C = div F =

∂M ∂N + ∂x ∂y

DC F ⋅ n ds

Let C be a piecewise smooth, simple closed curve enclosing a region R in the plane. Let F = Mi + N j be a vector field with M and N having continuous first partial derivatives in an open region containing R. Then the outward flux of F across C equals the double integral of div F over the region R enclosed by C.

DC F ⋅ n ds

=

DC M dy − N dx

Outward flux

=

⎛ ∂M

∫∫ ⎜⎜⎝ ∂ x R

+

∂ N ⎞⎟ ⎟ dx dy ∂ y ⎟⎠

(4)

Divergence integral

The two forms of Green’s Theorem are equivalent. Applying Equation (3) to the field G 1 = −N i + Mj gives Equation (4), and applying Equation (4) to G 2 = N i − Mj gives Equation (3). Both forms of Green’s Theorem can be viewed as two-dimensional generalizations of the Fundamental Theorem of Calculus from Section 5.4. The counterclockwise circulation of F around C, defined by the line integral on the left-hand side of Equation (3), is the integral of its rate of change (circulation density) over the region R enclosed by C, which is the double integral on the right-hand side of Equation (3). Likewise, the outward flux of F across C, defined by the line integral on the left-hand side of Equation (4), is the integral of its rate of change (flux density) over the region R enclosed by C, which is the double integral on the right-hand side of Equation (4). EXAMPLE 3

Verify both forms of Green’s Theorem for the vector field F ( x , y ) = ( x − y ) i + xj

and the region R bounded by the unit circle C: r (t ) = ( cos t ) i + ( sin t ) j, 0 ≤ t ≤ 2π. Solution First we evaluate the counterclockwise circulation of F = Mi + N j around C. On the curve C we have x = cos t and y = sin t . Evaluating F ( r(t ) ) and computing the derivatives of the components of r, we have

M = x − y = cos t − sin t ,

dx = d ( cos t ) = − sin t dt ,

N = x = cos t ,

dy = d ( sin t ) = cos t dt.

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Chapter 15 Integrals and Vector Fields

Therefore,

DC F ⋅ T ds = DC M dx + N dy t = 2π

=

∫t = 0

=

∫0

2π

( cos t − sin t )(− sin t ) dt + ( cos t )( cos t ) dt

(−sin t cos t + 1) dt = 2π.

This gives the left side of Equation (3). Next we find the curl integral, the right side of Equation (3). Since M = x − y and N = x, we have ∂M = 1, ∂x

∂M = −1, ∂y

∂N = 1, ∂x

∂ M ⎞⎟ ⎟ dx dy = dy ⎟⎠

∫∫ (1 − (−1)) dx dy

∂N = 0. ∂y

Therefore, ⎛ ∂N

∫∫ ⎜⎜⎝ ∂ x R

−

R

= 2 ∫∫ dx dy = 2( area inside the unit circle ) = 2π. R

Thus, the right and left sides of Equation (3) both equal 2π, as asserted by the circulationcurl version of Green’s Theorem. Figure 15.35 displays the vector field and circulation around C. Now we compute the two sides of Equation (4) in the flux-divergence form of Green’s Theorem, starting with the outward flux:

y

T

t = 2π

x

DC M dy − N dx = ∫t = 0

T

=

2π

∫0

( cos t − sin t )( cos t dt ) − ( cos t )(− sin t dt )

cos 2 t dt = π.

Next we compute the divergence integral: FIGURE 15.35

The vector field in Example 3 has a counterclockwise circulation of 2π around the unit circle.

⎛ ∂M

∫∫ ⎜⎜⎝ ∂ x R

+

∂ N ⎞⎟ ⎟ dx dy = ∂ y ⎟⎠

∫∫ (1 + 0 ) dx dy = ∫∫ R

dx dy = π.

R

Hence the right and left sides of Equation (4) both equal π, as asserted by the flux-divergence version of Green’s Theorem.

Using Green’s Theorem to Evaluate Line Integrals If we construct a closed curve C by piecing together a number of different curves end to end, the process of evaluating a line integral over C can be lengthy because there are so many different integrals to evaluate. If C bounds a region R to which Green’s Theorem applies, however, we can use Green’s Theorem to change the line integral around C into one double integral over R. EXAMPLE 4

Evaluate the line integral

DC xy dy − y

2

dx,

where C is the boundary of the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1.

969

15.4  Green’s Theorem in the Plane

Solution We can use either form of Green’s Theorem to change the line integral into a double integral over the square, where C is the square’s boundary and R is its interior.

1. With the Tangential Form Equation (3): Taking M = −y 2 and N = xy gives the result:

DC

− y 2 dx + xy dy =

⎛ ∂N

∫∫ ⎜⎜⎝ ∂ x

−

R

=

1

∂ M ⎞⎟ ⎟ dx dy = ∂ y ⎟⎠

1

1

∫0 ∫0 3 y dx dy = ∫0

∫∫ ( y − (−2 y )) dx dy R

x =1

⎤ ⎡ dy = ⎢ 3 xy ⎥ ⎣ ⎦ x = 0 

1

3

⎤

1

3

∫0 3 y dy = 2 y 2 ⎥⎦ 0 = 2 .

2. With the Normal Form Equation (4): Taking M = xy,  N = y 2, gives the same result:

DC xy dy − y

2

dx =

⎛ ∂M

∫∫ ⎜⎜⎝ ∂ x R

+

∂ N ⎞⎟ ⎟ dx dy = ∂ y ⎟⎠

∫∫ ( y + 2 y ) dx dy = R

3 . 2

EXAMPLE 5 Calculate the outward flux of the vector field F ( x , y ) = 2e xy i + y 3 j across the boundary of the square −1 ≤ x ≤ 1, −1 ≤ y ≤ 1. Solution Calculating the flux with a line integral would take four integrations, one for each side of the square. With Green’s Theorem, we can change the line integral to one double integral. With M = 2e xy,  N = y 3, C the square’s boundary, and R the square’s interior, we have

Flux = =

DC F ⋅ n ds ⎛ ∂M

∫∫ ⎜⎜⎝ ∂ x R

= =

1

+

=

DC M dy − N dx

∂ N ⎞⎟ ⎟ dx dy ∂ y ⎟⎠

Green’s Theorem, Eq. (4)

1

1

⎡

⎤

x =1

∫−1 ∫−1 ( 2 ye xy + 3 y 2 ) dx dy = ∫−1 ⎢⎣ 2e xy + 3xy 2 ⎥⎦ x =−1 dy 1

∫−1

⎡ ⎣

1

⎤ = 4. ⎦ −1

( 2e y + 6 y 2 − 2e − y ) dy = ⎢ 2e y + 2 y 3 + 2e − y ⎥

Proof of Green’s Theorem for Special Regions

y P2 (x, f2 (x))

C 2: y = f2 (x)

Let C be a smooth simple closed curve in the xy-plane with the property that lines parallel to the axes cut it at no more than two points. Let R be the region enclosed by C and suppose that M, N, and their first partial derivatives are continuous at every point of some open region containing C and R. We want to prove the circulation-curl form of Green’s Theorem, ⎛ ∂N

⎜⎜ ⎝ ∂x DC M dx + N dy = ∫∫ R

R

−

∂ M ⎞⎟ ⎟ dx dy. ∂ y ⎟⎠

(5)

Figure 15.36 shows C made up of two directed parts: P1(x, f1(x))

C1: y = f1(x) 0

a

FIGURE 15.36

x

b

C 1: y = f1 ( x ), a ≤ x ≤ b, x

The boundary curve C is made up of C 1 , the graph of y = f1 ( x ), and C 2 , the graph of y = f 2 ( x ).

C 2 : y = f 2 ( x ), b ≥ x ≥ a.

For any x between a and b, we can integrate ∂ M ∂ y with respect to y from y = f1 ( x ) to y = f 2 ( x ) and obtain f2 ( x)

∫ f (x) 1

⎤ y= f2 ( x) ∂M dy = M ( x , y )⎥ = M ( x , f 2 ( x ) ) − M ( x , f1 ( x ) ). ∂y ⎥⎦ y = f1 ( x )

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Chapter 15 Integrals and Vector Fields

We can then integrate this with respect to x from a to b: b

f2 (x )

∫a ∫ f ( x ) 1

∂M dy dx = ∂y

b

∫a

[ M ( x , f 2 ( x ) ) − M ( x , f1 ( x ) ) ] dx a

= −∫ M ( x , f 2 ( x ) ) dx − b

= −∫ M dx − C2

∫C

b

∫a

M ( x , f1 ( x ) ) dx

M dx 1

=−

DC M dx.

Therefore, reversing the order of the equations, we have

DC y d

C′1: x = g1( y)

y

Q 2(g2( y), y) Q1(g1( y), y)

x

(6)

R

DC N dy = ∫∫ R

C′2 : x = g2( y)

0

⎛ ∂M ⎞

∫∫ ⎜⎜⎝− ∂ y ⎟⎟⎟⎠ dx dy.

Equation (6) is half the result we need for Equation (5). We derive the other half by integrating ∂ N ∂ x first with respect to x and then with respect to y, as suggested by Figure 15.37. This shows the curve C of Figure 15.36 decomposed into the two directed parts C 1′ :  x = g1 ( y), d ≥ y ≥ c and C 2′ :  x = g 2 ( y),  c ≤ y ≤ d. The result of this double integration is

R c

M dx =

∂N dx dy. ∂x

(7)

Summing Equations (6) and (7) gives Equation (5). This concludes the proof.

FIGURE 15.37

The boundary curve C is made up of C 1′ , the graph of x = g1 ( y), and C 2′ , the graph of x = g 2 ( y).

Green’s Theorem also holds for more general regions, such as those shown in Figure 15.38. Notice that the region in Figure 15.38c is not simply connected. The curves C 1 and C h on its boundary are oriented so that the region R is always on the left-hand side as the curves are traversed in the directions shown, and cancelation occurs over common boundary arcs traversed in opposite directions. With this convention, Green’s Theorem is valid for regions that are not simply connected. The proof proceeds by summing the contributions to the integral of a collection of special regions, which overlap along their boundaries. Cancelation occurs along arcs that are traversed twice, once in each direction, as in Figure 15.38c. We do not give the full proof here.

y

y

y C1 C

b

R

x

0 (a)

FIGURE 15.38

R

C

a

R

0

a

Ch 0

b (b)

h

1

x

x (c)

Other regions to which Green’s Theorem applies. In (c) the axes convert the region into four simply connected regions, and we sum the line integrals along the oriented boundaries.

15.4  Green’s Theorem in the Plane

15.4

EXERCISES

Computing the k-Component of Curl(F)

19. F = ( x + e x sin y ) i + ( x + e x cos y ) j

In Exercises 1–6, find the k-component of curl(F) for the following vector fields on the plane.

C: The right-hand loop of the lemniscate r 2 = cos 2θ

(

2. F = ( x 2 − y ) i + ( y 2 ) j 3. F = ( xe y ) i + ( ye x ) j 4. F = ( x 2 y) i + ( xy 2 ) j

21. Find the counterclockwise circulation and outward flux of the field F = xyi + y 2 j around and over the boundary of the region enclosed by the curve y = x 2 and the line y = x .

5. F = ( y sin x ) i + ( x sin y ) j 6. F = ( x y ) i − ( y x ) j Verifying Green’s Theorem

In Exercises 7–10, verify the conclusion of Green’s Theorem by evaluating both sides of Equations (3) and (4) for the field F = Mi + N j. Take the domains of integration in each case to be the disk R:  x 2 + y 2  ≤  a 2 and its bounding circle C:  r = ( a cos t ) i + ( a sin t ) j, 0 ≤ t ≤ 2π.

22. Find the counterclockwise circulation and the outward flux of the field F = (− sin y ) i + ( x cos y ) j around and over the boundary of the square 0 ≤ x ≤ π 2, 0 ≤ y ≤ π 2. 23. Find the outward flux of the field ⎛ x ⎞⎟ F = ⎜⎜ 3 xy − ⎟ i + ( e x + tan −1 y ) j ⎝ 1 + y 2 ⎟⎠

8. F = yi

7. F = − yi + xj

10. F = −x 2 yi + xy 2 j

9. F = 2 xi − 3 yj

across the cardioid r = a (1 + cos θ ) ,  a > 0.

Circulation and Flux

In Exercises 11–20, use Green’s Theorem to find the counterclockwise circulation and outward flux for the field F and the curve C. 11. F = ( x − y ) i + ( y − x ) j C: The square bounded by x = 0,  x = 1,  y = 0, and y = 1 12. F = ( x 2 + 4 y ) i + ( x + y 2 ) j C: The square bounded by x = 0,  x = 1,  y = 0, and y = 1 13. F = ( y 2 − x 2 ) i + ( x 2 + y 2 ) j

24. Find the counterclockwise circulation of F = ( y + e x ln y ) i + ( e x y ) j around the boundary of the region that is bounded above by the curve y = 3 − x 2 and below by the curve y = x 4 + 1. Work

In Exercises 25 and 26, find the work done by F in moving a particle once counterclockwise around the given curve. 25. F = 2 xy 3 i + 4 x 2 y 2 j C: The boundary of the “triangular” region in the first quadrant enclosed by the x-axis, the line x = 1, and the curve y = x 3

C: The triangle bounded by y = 0,   x = 3, and y = x 14. F = ( x + y ) i − ( x 2 + y 2 ) j

26. F = ( 4 x − 2 y ) i + ( 2 x − 4 y ) j

C: The triangle bounded by y = 0,  x = 1, and y = x y 2 )i

Using Green’s Theorem

y x=

Apply Green’s Theorem to evaluate the integrals in Exercises 27–30.

(1, 1) C

C 1

x 2 + 2y 2 = 2

y = x2 x

(0, 0)

−2

2

1 4 x yj 2

18. F =

y

C

x 2 + y2 = 1

y=

dx + x 2 dy )

DC ( 3 y dx + 2 x dy )

29.

DC ( 6 y + x ) dx + ( y + 2 x ) dy C: The circle ( x − 2 ) 2 + ( y − 3 ) 2 = 4

−1 x2

2

C: The boundary of 0 ≤ x ≤ π , 0 ≤ y ≤ sin x

(2, 2)

y=x

(0, 0)

x i + ( tan −1 y ) j 1 + y2 1

DC ( y

C: The boundary of the triangle enclosed by the lines x = 0, x + y = 1, and y = 0 28.

y C

27.

x

−1

17. F = x 3 y 2 i +

C: The circle ( x − 2 ) 2 + ( y − 2 ) 2 = 4

+ ( x − y ) j 16. F = ( x + 3 y ) i + ( 2 x − y ) j

y y2

)

y i + ln ( x 2 + y 2 ) j x C: The boundary of the region defined by the polar coordinate inequalities 1 ≤ r ≤ 2, 0 ≤ θ ≤ π

20. F = tan −1

1. F = ( x + y ) i + (2 xy) j

15. F = ( xy +

971

1

−x x

−1

x

30.

DC ( 2 x + y

2 ) dx

+ ( 2 xy + 3 y ) dy

C: Any simple closed curve in the plane for which Green’s Theorem holds

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Chapter 15 Integrals and Vector Fields

Calculating Area with Green’s Theorem If a simple closed curve C in the plane and the region R it encloses satisfy the hypotheses of Green’s Theorem, the area of R is given by

Green’s Theorem Area Formula Area of  R =

1 2

DC

x dy − y dx

below by the x-axis, above by the graph of f , and on the sides by the lines x = a and x = b. Show that b

∫a

∫∫ dy dx R

=

1 1 x 2 dy = − xy dx = x 2 dy − xy dx = Ax . D 2D 3D C

∫∫ ( 2 + 2 ) dy dx 1

1

R

1 1 = D 2 x dy − 2 y dx. Use the Green’s Theorem area formula given above to find the areas of the regions enclosed by the curves in Exercises 31–34. 31. The circle r (t ) = ( a cos t ) i + ( a sin t ) j, 0 ≤ t ≤ 2π

y = 1 − cos t

35. Let C be the boundary of a region on which Green’s Theorem holds. Use Green’s Theorem to calculate a.

DC f ( x ) dx + g( y) dy

b.

DC ky dx + hx dy

(k and h constants).

36. Integral dependent only on area Show that the value of

DC xy

2

dx + ( x 2 y + 2 x ) dy

around any square depends only on the area of the square and not on its location in the plane. 37. Evaluate the integral

DC 4 x

3y

dx + x 4 dy

C

1 1 x 3 dy = − x 2 y dx = D 3D 4 C

C

DC x

3

dy − x 2 y dx = I y .

43. Green’s Theorem and Laplace’s equation Assuming that all the necessary derivatives exist and are continuous, show that if f ( x , y ) satisfies the Laplace equation ∂2 f ∂2 f + = 0, 2 ∂x ∂y2

32. The ellipse r (t ) = ( a cos t ) i + ( b sin t ) j, 0 ≤ t ≤ 2π 33. The astroid r (t ) = ( cos 3 t ) i + ( sin 3 t ) j, 0 ≤ t ≤ 2π

C

42. Moment of inertia Let I y be the moment of inertia about the y-axis of the region in Exercise 41. Show that

C

34. One arch of the cycloid x = t − sin t ,

DC y dx.

41. Area and the centroid Let x be the x-coordinate of the centroid of a region R that is bounded by a piecewise smooth, simple closed curve C in the xy-plane. If A is the area of R, show that

The reason is that, by Equation (4) run backward, Area of  R =

f ( x ) dx = −

then ∂f

∂f

DC ∂ y dx − ∂ x dy =

0

for all closed curves C to which Green’s Theorem applies. (The converse is also true: If the line integral is always zero, then f satisfies the Laplace equation.) 44. Maximizing work Among all smooth, simple closed curves in the plane, oriented counterclockwise, find the one along which the work done by F =

( 14 x

2y

+

)

1 3 y i + xj 3

is greatest. (Hint: Where is ( curl  F ) ⋅ k positive?) 45. Regions with many holes Green’s Theorem holds for a region R with any finite number of holes as long as the bounding curves are smooth, simple, and closed and we integrate over each component of the boundary in the direction that keeps R on our immediate left as we proceed along the curve (see accompanying figure).

for any closed path C. 38. Evaluate the integral

DC − y

3

dy + x 3 dx

for any closed path C. 39. Area as a line integral Show that if R is a region in the plane bounded by a piecewise smooth, simple closed curve C, then Area of  R =

DC x dy = −DC y dx.

40. Definite integral as a line integral Suppose that a nonnegative function y = f ( x ) has a continuous first derivative on [ a, b ]. Let C be the boundary of the region in the xy-plane that is bounded

a. Let f ( x , y ) = ln ( x 2 + y 2 ) and let C be the circle x 2 + y 2 = a 2 . Evaluate the flux integral

DC ∇f

⋅ n ds.

15.5  Surfaces and Area

b. Let K be an arbitrary smooth, simple closed curve in the plane that does not pass through ( 0, 0 ). Use Green’s Theorem to show that

DK ∇f

⋅ n ds

973

COMPUTER EXPLORATIONS

In Exercises 49–52, use a CAS and Green’s Theorem to find the counterclockwise circulation of the field F around the simple closed curve C. Perform the following CAS steps. a. Plot C in the xy-plane. b. Determine the integrand ( ∂ N ∂ x ) − ( ∂ M ∂ y ) for the tangential form of Green’s Theorem.

has two possible values, depending on whether ( 0, 0 ) lies inside K or outside K. 46. Bendixson’s criterion The streamlines of a planar fluid flow are the smooth curves traced by the fluid’s individual particles. The vectors F = M ( x , y ) i + N ( x , y ) j of the flow’s velocity field are the tangent vectors of the streamlines. Show that if the flow takes place over a simply connected region R (no holes or missing points) and that if Mx + Ny ≠ 0 throughout R, then none of the streamlines in R is closed. In other words, no particle of fluid ever has a closed trajectory in R. The criterion Mx + N y ≠ 0 is called Bendixson’s criterion for the nonexistence of closed trajectories. 47. Establish Equation (7) to finish the proof of the special case of Green’s Theorem.

c. Determine the (double integral) limits of integration from your plot in part (a), and evaluate the curl integral for the circulation. 49. F = ( 2 x − y ) i + ( x + 3 y ) j, C: The ellipse  x 2 + 4 y 2 = 4 50. F = ( 2 x 3 − y 3 ) i + ( x 3 + y 3 ) j, C: The ellipse 

y2 x2 + =1 4 9

51. F = x −1e y i + ( e y ln x + 2 x ) j, C: The boundary of the region defined by y = 1 + x 4 (below) and y = 2 (above) 52. F = xe y i + ( 4 x 2 ln y ) j, C: The triangle with vertices ( 0, 0 ), ( 2, 0 ), and ( 0, 4 )

48. Curl component of conservative fields Can anything be said about the curl component of a conservative two-dimensional vector field? Give reasons for your answer.

15.5 Surfaces and Area We have described curves in the plane in three different ways. y = f (x)

Explicit form: Implicit form:

u = constant

y

Parametric vector form: y = constant (u, y)

We have analogous descriptions of surfaces in space. z = f ( x, y ) F ( x , y, z ) = 0.

Explicit form:

R

Implicit form:

u

0

F ( x, y ) = 0 r (t ) = f (t ) i + g(t ) j, a ≤ t ≤ b.

There is also a parametric form for surfaces that gives the position of a point on the surface as a vector function of two variables. We discuss this new form in this section and apply the form to obtain the area of a surface as a double integral. Double integral formulas for areas of surfaces given in implicit and explicit forms are then obtained as special cases of the more general parametric formula.

Parametrization

z Curve y = constant

Parametrizations of Surfaces S

Suppose

P

r ( u, υ ) = f ( u, υ ) i + g ( u, υ ) j + h ( u, υ ) k Curve u = constant

x

r(u, y) = f (u, y)i + g(u, y)j + h(u, y)k, position vector to surface point y

FIGURE 15.39

A parametrized surface S expressed as a vector function of two variables defined on a region R.

(1)

is a continuous vector function that is defined on a region R in the uυ-plane and is one-to-one on the interior of R (Figure 15.39). We call the range of r the surface S defined or traced by r. Equation (1) together with the domain R constitutes a parametrization of the surface. The variables u and υ are the parameters, and R is the parameter domain. To simplify our discussion, we take R to be a rectangle defined by inequalities of the form a ≤ u ≤ b, c ≤ υ ≤ d. The requirement that r be one-to-one on the interior of R ensures that S does not cross itself. Notice that Equation (1) is the vector equivalent of three parametric equations: x = f ( u, υ ) ,

y = g ( u, υ ) ,

z = h ( u, υ ).

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Chapter 15 Integrals and Vector Fields

Cone: z = "x2 + y2 =r

EXAMPLE 1

z

Find a parametrization of the cone z =

1

x 2 + y2 ,

0 ≤ z ≤ 1.

Solution Here, cylindrical coordinates provide a parametrization. A typical point ( x , y, z ) on the cone (Figure 15.40) has x = r cos θ,  y = r sin θ, and z = x 2 + y 2 = r, with 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π. Taking u = r and υ = θ in Equation (1) gives the parametrization

r(r, u) = (r cos u)i + (r sin u)j + rk

r ( r , θ ) = ( r cos θ ) i + ( r sin θ ) j + rk,

(x, y, z) = (r cos u, r sin u, r)

The parametrization is one-to-one on the interior of the domain, though not on the boundary where r = 0 (mapped to the tip of the cone) or where θ = 0 or θ = 2π (where the cone glues together along a seam above the x-axis).

u r y

x

0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π.

FIGURE 15.40

The cone in Example 1 can be parametrized using cylindrical coordinates. z

(x, y, z) = (a sin f cos u, a sin f sin u, a cos f)

Find a parametrization of the sphere x 2 + y 2 + z 2 = a 2 .

EXAMPLE 2

Solution Spherical coordinates provide what we need. A typical point ( x , y, z ) on the sphere (Figure 15.41) has x = a sin φ cos θ,  y = a sin φ sin θ, and z = a cos φ, 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π. Taking u = φ and υ = θ in Equation (1) gives the parametrization

r ( φ, θ ) = ( a sin φ cos θ ) i + ( a sin φ sin θ ) j + ( a cos φ ) k,

a

0 ≤ φ ≤ π, f

u a

Again, the parametrization is one-to-one on the interior of the domain, though not on its boundary.

r(f, u) a y

x

EXAMPLE 3

Find a parametrization of the cylinder x 2 + ( y − 3 ) 2 = 9,

FIGURE 15.41

The sphere in Example 2 can be parametrized using spherical coordinates. z

0 ≤ θ ≤ 2π.

Cylinder: x2 + (y − 3)2 = 9 or r = 6 sin u

0 ≤ z ≤ 5.

Solution In cylindrical coordinates, a point ( x , y, z ) has x = r cos θ,   y = r sin θ, and z = z. For points on the cylinder x 2 + ( y − 3 ) 2 = 9 (Figure 15.42), the equation is the same as the polar equation for the cylinder’s base in the xy-plane:

x 2 + ( y 2 − 6y + 9) = 9 r 2 − 6r sin θ = 0

x 2 + y 2 = r 2,  y = r sin θ

or r = 6 sin θ,

z

0 ≤ θ ≤ π.

A typical point on the cylinder therefore has

r(u, z)

x = r cos θ = 6 sin θ cos θ = 3 sin 2θ (x, y, z) =(3 sin 2u, 6 sin2u, z)

x

r = 6 sin u

FIGURE 15.42

y

The cylinder in Example 3 can be parametrized using cylindrical coordinates.

y = r sin θ = 6 sin 2 θ z = z. Taking u = θ and υ = z in Equation (1) gives the parametrization r ( θ, z ) = ( 3 sin 2θ ) i + ( 6 sin 2 θ ) j + zk, 0 ≤ θ ≤ π, 0 ≤ z ≤ 5, which is one-to-one on the interior of the domain.

Surface Area Our goal is to find a double integral that gives the area of a curved surface S based on the parametrization r ( u, υ ) = f ( u, υ ) i + g ( u, υ ) j + h ( u, υ ) k , a ≤ u ≤ b, c ≤ υ ≤ d.

975

15.5  Surfaces and Area

We need S to be smooth for the construction we now describe. The definition of smoothness involves the partial derivatives of r with respect to u and υ: ∂f ∂r = i+ ∂u ∂u ∂f ∂r = rυ = i+ ∂υ ∂υ ru =

∂g j+ ∂u ∂g j+ ∂υ

∂h k ∂u ∂h k. ∂υ

DEFINITION A parametrized surface r ( u, υ ) = f ( u, υ ) i + g ( u, υ ) j + h ( u, υ ) k

is smooth if ru and rυ are continuous and if ru × rυ is never zero on the interior of the parameter domain.

The condition that ru × rυ is never the zero vector in the definition of smoothness means that the two vectors ru and rυ are nonzero and never lie along the same line, so they always determine a plane tangent to the surface. We relax this condition on the boundary of the domain, but this does not affect the area computations. Now consider a small rectangle ΔAuυ in R with sides on the lines u = u 0 , u = u 0 + Δu,  υ = υ 0 , and υ = υ 0 + Δυ (Figure 15.43). Each side of ΔAuυ maps onto a curve on the surface S, and together these four curves bound a “curved patch element” Δσ uυ . In the notation of the figure, the side υ = υ 0 maps to curve C 1 , the side u = u 0 maps onto C 2 , and their common vertex ( u 0 , υ 0 ) maps to P0 . z C1: y = y0 y

C2: u = u0 ru × ry

S

d

Δsuy

y0 + Δy

Δ Auy

y0

y = y0 + Δy z

u0

u0 + Δu

b

Δsuy

u

C1: y = y0 y

A rectangular area element ΔAuυ in the uυ-plane maps onto a curved patch element Δσ uυ on S.

FIGURE 15.43

Δyry

P0 z

C2

Δuru C1

x

Δsuy

y

FIGURE 15.45

The area of the parallelogram determined by the vectors Δuru and Δυrυ approximates the area of the surface patch element Δσ uυ .

C2: u = u0

ru

x a

ry

P0

u = u0 + Δu

R

c 0

P0

Parametrization

x

y

FIGURE 15.44

A magnified view of a surface patch element Δσ uυ .

Figure 15.44 shows an enlarged view of Δσ uυ . The partial derivative vector ru ( u 0 , υ 0 ) is tangent to C 1 at P0 . Likewise, rυ ( u 0 , υ 0 ) is tangent to C 2 at P0 . The cross product ru × rυ is normal to the surface at P0 . (Here is where we begin to use the assumption that S is smooth. We want to be sure that ru × rυ ≠ 0.) We next approximate the surface patch element Δσ uυ by the parallelogram on the tangent plane whose sides are determined by the vectors Δuru and Δυrυ (Figure 15.45). The area of this parallelogram is Δu ru × Δυrυ = ru × rυ Δu Δυ.

(2)

A partition of the region R in the uυ-plane by rectangular regions ΔAuυ induces a partition of the surface S into surface patch elements Δσ uυ . We approximate the area of each surface

976

Chapter 15 Integrals and Vector Fields

patch element Δσ uυ by the parallelogram area in Equation (2) and sum these areas together to obtain an approximation of the surface area of S:

∑

ru × rυ Δu Δυ.

(3)

n

As Δu and Δυ approach zero independently, the number of area elements n approaches ∞ and the continuity of ru and rυ guarantees that the sum in Equation (3) approaches the d b double integral ∫c ∫ a ru × rυ du d υ. This double integral over the region R is used to define the area of the surface S.

The area of the smooth surface

DEFINITION

r ( u, υ ) = f ( u, υ ) i + g ( u, υ ) j + h ( u, υ ) k ,

a ≤ u ≤ b, c ≤ υ ≤ d

is A =

∫∫

ru × rυ dA =

R

d

b

∫c ∫ a

ru × rυ du d υ.

(4)

We can abbreviate the integral in Equation (4) by writing dσ for ru × rυ du d υ. The surface area differential dσ is analogous to the arc length differential ds in Section 12.3.

Surface Area Differential for a Parametrized Surface

∫∫

dσ = ru × rυ du d υ

dσ

(5)

S

Surface area differential, also called surface area element

EXAMPLE 4

Differential formula for surface area

Find the surface area of the cone in Example 1 (Figure 15.40).

Solution In Example 1, we found the parametrization

r ( r , θ ) = ( r cos θ ) i + ( r sin θ ) j + rk,

0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π.

To apply Equation (4), we first find rr × rθ : rr × rθ =

i

j

k

cos θ

sin θ

1

−r sin θ

r cos θ 0

= −( r cos θ ) i − ( r sin θ ) j + ( r cos 2 θ + r sin 2 θ ) k.  r

Thus, rr × rθ =

r 2 cos 2 θ + r 2 sin 2 θ + r 2 =

2π

1

2π

1

A=

∫0 ∫0

=

∫0 ∫0

rr × rθ dr dθ 2r dr dθ =

2r 2 =

2r. The area of the cone is

Eq. (4) with  u = r ,   υ = θ 2π

∫0

2 2 dθ = (2π) = π 2 square units. 2 2

15.5  Surfaces and Area

977

Find the surface area of a sphere of radius a.

EXAMPLE 5

Solution We use the parametrization from Example 2:

r ( φ, θ ) = ( a sin φ cos θ ) i + ( a sin φ sin θ ) j + ( a cos φ ) k, 0 ≤ φ ≤ π , 0 ≤ θ ≤ 2π. For rφ × rθ , we get i

j

rφ × rθ = a cos φ cos θ −a sin φ sin θ

k

a cos φ sin θ − a sin φ a sin φ cos θ

0

= ( a 2 sin 2 φ cos θ ) i + ( a 2 sin 2 φ sin θ ) j + ( a 2 sin φ cos φ ) k. Thus, rφ × rθ =

a 4 sin 4 φ cos 2 θ + a 4 sin 4 φ sin 2 θ + a 4 sin 2 φ cos 2 φ a 4 sin 4 φ + a 4 sin 2 φ cos 2 φ =

=

a 4 sin 2 φ ( sin 2 φ + cos 2 φ )

= a 2 sin 2 φ = a 2 sin φ because sin φ ≥ 0 for 0 ≤ φ ≤ π. Therefore, the area of the sphere is A=

z

x = cos z, y = 0

p 2 r = cos z is the radius of a circle at height z.

=

2π

π

∫0 ∫0 2π

∫0

a 2 sin φ dφ dθ

⎡ 2 ⎤ φ=π dθ = ⎢ −a cos φ ⎥ ⎥⎦ φ = 0 ⎣⎢

2π

∫0

2a 2 dθ = 4 πa 2

square units.

This gives the well-known formula for the surface area of a sphere. EXAMPLE 6 Let S be the “football” surface formed by rotating the curve x = cos z , y = 0, −π 2 ≤ z ≤ π 2 around the z-axis (see Figure 15.46). Find a parametrization for S and compute its surface area.

(x, y, z) 1

1

x

−

FIGURE 15.46

y

Solution Example 2 suggests finding a parametrization of S based on its rotation around the z-axis. If we rotate a point ( x , 0, z ) on the curve x = cos z ,  y = 0 about the z-axis, we obtain a circle at height z above the xy-plane that is centered on the z-axis and has radius r = cos z (see Figure 15.46). The point sweeps out the circle through an angle of rotation θ, 0 ≤ θ ≤ 2π. We let ( x , y, z ) be an arbitrary point on this circle, and define the parameters u = z and υ = θ. Then we have x = r cos θ = cos u cos υ,  y = r sin θ = cos u sin υ, and z = u, giving a parametrization for S as

r ( u, υ ) = cos u cos υi + cos u sin υ j + u k, −

p 2

The “football” surface in Example 6 obtained by rotating the curve x = cos z about the z-axis.

π π ≤ u ≤ , 0 ≤ υ ≤ 2π. 2 2

Next we use Equation (5) to find the surface area of S. Differentiation of the parametrization gives ru = − sin u cos υ i − sin u sin υ j + k and rυ = − cos u sin υ i + cos u cos υ j. Computing the cross product, we have i

j

k

ru × rυ = − sin u cos υ

− sin u sin υ

1

− cos u sin υ

cos u cos υ

0

= − cos u cos υ i − cos u sin υ j − ( sin u cos u cos 2 υ + cos u sin u sin 2 υ ) k.

978

Chapter 15 Integrals and Vector Fields

Taking the magnitude of the cross product gives ru × rυ = =

cos 2 u   ( cos 2 υ + sin 2 υ ) + sin 2 u cos 2 u cos 2 u   (1 + sin 2 u )

= cos u   1 + sin 2 u.

cos u ≥ 0 for  −

π π ≤ u ≤ 2 2

From Equation (4) the surface area is given by the integral A =

2π

π 2

∫0 ∫−π 2 cos u

1 + sin 2 u du d υ.

To evaluate the integral, we substitute w = sin u and dw = cos u du, −1 ≤ w ≤ 1. Since the surface S is symmetric across the xy-plane, we need only integrate with respect to w from 0 to 1 and multiply the result by 2. In summary, we have A = 2∫

2π 0

= 2∫ =

2π 0

2π

∫0

1

∫0

1 + w 2 dw d υ w =1

⎡ w 1 + w 2 + 1 ln ( w + ⎢⎣ 2 2

1 1 2 ⎡⎢ 2 + ln (1 + ⎣2 2

= 2π [ 2 + ln (1 +

1 + w 2 )⎤⎥ dυ ⎦ w=0

Integral Table Formula 35

2 )⎤⎥ d υ ⎦

2 ) ].

Implicit Surfaces Surfaces are often presented as level sets of a function, described by an equation such as F ( x , y, z ) = c,

Surface F(x, y, z) = c

S p

R

The vertical projection or “shadow” of S on a coordinate plane

FIGURE 15.47

As we soon see, the area of a surface S in space can be calculated by evaluating a related double integral over the vertical projection or “shadow” of S on a coordinate plane. The unit vector p is normal to the plane.

for some constant c. Such a level surface does not come with an explicit parametrization and is called an implicitly defined surface. Implicit surfaces arise, for example, as equipotential surfaces in electric or gravitational fields. Figure 15.47 shows a piece of such a surface. It may be difficult to find explicit formulas for the functions f , g, and h that describe the surface in the form r ( u, υ ) = f ( u, υ ) i + g ( u, υ ) j + h ( u, υ ) k. We now show how to compute the surface area differential dσ for implicit surfaces. Figure 15.47 shows a piece of an implicit surface S that lies above its “shadow” region R in the plane beneath it. The surface is defined by the equation F ( x , y, z ) = c, and we choose p to be a unit vector normal to the plane region R. We assume that the surface is smooth (F is differentiable and ∇F is nonzero and continuous on S) and that ∇F ⋅ p ≠ 0, so the surface never folds back over itself. Assume that the normal vector p is the unit vector k, so the region R in Figure 15.47 lies in the xy-plane. By assumption, we then have ∇F ⋅ p = ∇F ⋅ k = Fz ≠ 0 on S. The Implicit Function Theorem (see Section 13.4) implies that S is then the graph of a differentiable function z = h ( x , y ) , although the function h ( x , y ) is not explicitly known. Define the parameters u and υ by u = x and υ = y. Then z = h ( u, υ ) and r ( u, υ ) = u i + υ j + h ( u, υ ) k

(6)

gives a parametrization of the surface S. We use Equation (4) to find the area of S. Calculating the partial derivatives of r, we find ru = i +

∂h k ∂u

and

rυ = j +

∂h k. ∂υ

979

15.5  Surfaces and Area

Applying the Chain Rule for implicit differentiation (see Equation (2) in Section 13.4) to F ( x , y, z ) = c, where x = u, y = υ, and z = h ( u, υ ) , we obtain the partial derivatives F ∂h = − x ∂u Fz

F ∂h = − y. ∂υ Fz

and

Fz ≠ 0

Substitution of these derivatives into the derivatives of r gives ru = i −

Fx k Fz

rυ = j −

and

Fy k. Fz

From a routine calculation of the cross product, we find ru × rυ =

F Fx i+ yj+k Fz Fz

i j

k

1 0

−Fx Fz

0 1

−Fy Fz

cross product of ru rυ

1 = ( Fx i +   Fy j +   Fz k ) Fz =

∇F ∇F = Fz ∇F ⋅ k

=

∇F . ∇F ⋅ p

p = k

Therefore, the surface area differential is given by dσ = ru × rυ du d υ =

∇F dx dy. ∇F ⋅ p

u = x  and υ = y

We obtain similar calculations if instead the vector p = j is normal to the xz-plane when Fy ≠ 0 on S, or if p = i is normal to the yz-plane when Fx ≠ 0 on S. Combining these results with Equation (4) then gives the following general formula. Formula for the Surface Area of an Implicit Surface

The area of the surface F ( x , y, z ) = c over a closed and bounded plane region R is Surface area =

∫∫ R

z

∇F dA, ∇F ⋅ p

(7)

where p = i, j, or k is normal to R and ∇F ⋅ p ≠ 0.

4 S z = x2 + y2

Thus, the area is the double integral over R of the magnitude of ∇F divided by the magnitude of the scalar component of ∇F normal to R. We reached Equation (7) under the assumption that ∇F ⋅ p ≠ 0 throughout R and that ∇F is continuous. Whenever the integral exists, however, we define its value to be the area of the portion of the surface F ( x , y, z ) = c that lies over R. (Recall that the projection is assumed to be one-to-one.)

R 0 x2 + y2 = 4

y

x

FIGURE 15.48

The area of this parabolic surface is calculated in Example 7.

EXAMPLE 7 Find the area of the surface cut from the bottom of the paraboloid x 2 + y 2 − z = 0 by the plane z = 4. Solution We sketch the surface S and the region R below it in the xy-plane (Figure 15.48). The surface S is part of the level surface F ( x , y, z ) = x 2 + y 2 − z = 0, and R is the disk x 2 + y 2 ≤ 4 in the xy-plane. To get a unit vector normal to the plane of R, we can take p = k.

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Chapter 15 Integrals and Vector Fields

At any point ( x , y, z ) on the surface, we have F ( x , y, z ) = x 2 + y 2 − z ∇F = 2 x i + 2 y j − k ∇F =

4x 2 + 4 y 2 + 1

(2 x ) 2 + (2 y) 2 + (−1) 2 =

∇F ⋅ p = ∇F ⋅ k = −1 = 1. In the region R,  dA = dx dy. Therefore, Surface area =

∇F dA ∇F ⋅ p

∫∫ R

=

∫∫

Eq. (7)

4 x 2 + 4 y 2 + 1 dx dy

x 2 + y 2 ≤4

=

2π

2

∫0 ∫0

4r 2 + 1 r dr dθ

Polar coordinates

2π

r =2 ⎡ 1 ( 4r 2 + 1)3 2 ⎤ ⎢⎣ 12 ⎥⎦ r = 0 dθ

2π

1 π (17 3 2 − 1) dθ = (17 17 − 1). 12 6

=

∫0

=

∫0

Example 7 illustrates how to find the surface area for a function z = f ( x , y ) over a region R in the xy-plane. Actually, the surface area differential can be obtained in two ways, and we show this in the next example.

EXAMPLE 8 Derive the surface area differential dσ of the surface z = f ( x , y ) over a region R in the xy-plane (a) parametrically using Equation (5), and (b) implicitly, as in Equation (7). Solution

(a) We parametrize the surface by taking x = u,  y = υ, and z = f ( x , y ) over R. This gives the parametrization r ( u, υ ) = u i + υ j + f ( u, υ ) k. Computing the partial derivatives gives ru = i + f u k,  rυ = j + f υ k and ru × rυ = − f u i − f υ j + k.

i

j

k

1

0

fu

0

1

fυ

Then ru × rυ du d υ = f u 2 + f υ 2 + 1 du d υ. Substituting for u and υ. then gives the surface area differential dσ =

f x 2 + f y 2 + 1 dx dy.

(b) We define the implicit function F ( x , y, z ) = f ( x , y ) − z. Since ( x , y ) belongs to the region R, the unit normal to the plane of R is p = k. Then ∇F = f x i + f y j − k so

981

15.5  Surfaces and Area

that ∇F ⋅ p = −1 = 1, ∇F = f x 2 + f y 2 + 1, and ∇F ∇F ⋅ p = ∇F . The surface area differential is again given by dσ =

f x 2 + f y 2 + 1 dx dy.

The surface area differential derived in Example 8 gives the following formula for calculating the surface area of the graph of a function defined explicitly as z = f ( x , y ). Formula for the Surface Area of a Graph z = f ( x, y )

For a graph z = f ( x , y ) over a region R in the xy-plane, the surface area formula is A =

∫∫

f x 2 + f y 2 + 1 dx dy.

(8)

R

EXERCISES

15.5

Finding Parametrizations

a. Inside the cylinder x 2 + z 2 = 3

In Exercises 1–16, find a parametrization of the surface. (There are many correct ways to do these, so your answers may not be the same as those in the back of the text.)

b. Inside the cylinder y 2 + z 2 = 2

1. The paraboloid z =

x2

+

2. The paraboloid z = 9 −

y 2,  z

x2

≤ 4

− y 2,  z ≥ 0

3. Cone frustum The first-octant portion of the cone z = x 2 + y 2 2 between the planes z = 0 and z = 3 4. Cone frustum The portion of the cone z = 2 x 2 + y 2 between the planes z = 2 and z = 4 5. Spherical cap The cap cut from the sphere x 2 + y 2 + z 2 = 9 by the cone z = x 2 + y 2 6. Spherical cap The portion of the sphere x 2 + y 2 + z 2 = 4 in the first octant between the xy-plane and the cone z = x 2 + y 2 7. Spherical band The portion of the sphere + between the planes z = 3 2 and z = − 3 2 x2

y2

+

z2

= 3

8. Spherical cap The upper portion cut from the sphere x 2 + y 2 + z 2 = 8 by the plane z = −2 9. Parabolic cylinder between planes The surface cut from the parabolic cylinder z = 4 − y 2 by the planes x = 0,  x = 2, and z = 0 10. Parabolic cylinder between planes The surface cut from the parabolic cylinder y = x 2 by the planes z = 0,  z = 3, and y = 2 11. Circular cylinder band The portion of the y 2 + z 2 = 9 between the planes x = 0 and x = 3

cylinder

12. Circular cylinder band The portion of the cylinder x 2 + z 2 = 4 above the xy-plane between the planes y = −2 and y = 2 13. Tilted plane inside cylinder The portion of the plane x+ y+z =1 a. Inside the cylinder x 2 + y 2 = 9 b. Inside the cylinder y 2 + z 2 = 9 14. Tilted plane inside cylinder The portion of the plane x − y + 2z = 2

15. Circular cylinder band The portion of the cylinder ( x − 2 ) 2 + z 2 = 4 between the planes y = 0 and y = 3 16. Circular cylinder band The portion of the cylinder y 2 + ( z − 5 ) 2 = 25 between the planes x = 0 and x = 10 Surface Area of Parametrized Surfaces

In Exercises 17–26, use a parametrization to express the area of the surface as a double integral. Then evaluate the integral. (There are many correct ways to set up the integrals, so your integrals may not be the same as those in the back of the text. They should have the same values, however.) 17. Tilted plane inside cylinder The portion of the plane y + 2 z = 2 inside the cylinder x 2 + y 2 = 1 18. Plane inside cylinder The portion of the plane z = −x inside the cylinder x 2 + y 2 = 4 19. Cone frustum The portion of the cone z = 2 x 2 + y 2 between the planes z = 2 and z = 6 20. Cone frustum The portion of the cone z = between the planes z = 1 and z = 4 3

x 2 + y2 3

21. Circular cylinder band The portion of the cylinder x 2 + y 2 = 1 between the planes z = 1 and z = 4 22. Circular cylinder band The portion of the cylinder x 2 + z 2 = 10 between the planes y = −1 and y = 1 23. Parabolic cap The cap cut from the paraboloid z = 2 − x 2 − y 2 by the cone z = x 2 + y 2 24. Parabolic band The portion of the paraboloid z = x 2 + y 2 between the planes z = 1 and z = 4 25. Sawed-off sphere The lower portion cut from the sphere x 2 + y 2 + z 2 = 2 by the cone z = x 2 + y 2 26. Spherical band The portion of the sphere x 2 + y 2 + z 2 = 4 between the planes z = −1 and z = 3

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Chapter 15 Integrals and Vector Fields

Planes Tangent to Parametrized Surfaces

The tangent plane at a point P0 ( f ( u 0 , υ 0 ) ,   g ( u 0 , υ 0 ) ,   h ( u 0 , υ 0 )) on a parametrized surface r ( u, υ ) = f ( u, υ ) i + g ( u, υ ) j + h ( u, υ ) k is the plane through P0 normal to the vector ru ( u 0 , υ 0 ) × rυ ( u 0 , υ 0 ) , the cross product of the tangent vectors ru ( u 0 , υ 0 ) and rυ ( u 0 , υ 0 ) at P0 . In Exercises 27–30, find an equation for the plane tangent to the surface at P0 . Then find a Cartesian equation for the surface, and sketch the surface and tangent plane together.

is a parametrization of the resulting surface of revolution, where 0 ≤ υ ≤ 2π is the angle from the xy-plane to the point r ( u, υ ) on the surface. (See the accompanying figure.) Notice that f (u) measures distance along the axis of revolution and g(u) measures distance from the axis of revolution. y

27. Cone The cone r ( r , θ ) = ( r cos θ ) i + ( r sin θ ) j + rk, r ≥ 0, 0 ≤ θ ≤ 2π at the point P0 ( 2, 2, 2 ) corresponding to ( r, θ ) = ( 2, π 4 )

( f (u), g(u), 0) r(u, y)

28. Hemisphere The hemisphere surface r ( φ, θ ) = ( 4 sin φ cos θ ) i + ( 4 sin φ sin θ ) j + ( 4 cos φ ) k, 0 ≤ φ ≤ π 2, 0 ≤ θ ≤ 2π , at the point P0 ( 2, 2, 2 3 ) corresponding to ( φ, θ ) = ( π 6, π 4 ) 29. Circular cylinder The circular cylinder r ( θ, z ) = ( 3 sin 2θ ) i + ( 6 sin 2 θ ) j + zk, 0 ≤ θ ≤ π , at the point P0 ( 3 3 2, 9 2, 0 ) corresponding to ( θ, z ) = ( π 3, 0 ) (See Example 3.) 30. Parabolic cylinder The parabolic cylinder surface r ( x , y ) = xi + yj − x 2 k, −∞ < x < ∞, −∞ < y < ∞, at the point P0 (1, 2, −1) corresponding to ( x , y ) = (1, 2 ) More Parametrizations of Surfaces

31. a. A torus of revolution (doughnut) is obtained by rotating a circle C in the xz-plane about the z-axis in space. (See the accompanying figure.) If C has radius r > 0 and center ( R, 0, 0 ), show that a parametrization of the torus is r ( u, υ ) = (( R + r cos u ) cos υ ) i + (( R + r cos u ) sin υ ) j + ( r sin u ) k, where 0 ≤ u ≤ 2π and 0 ≤ υ ≤ 2π are the angles in the figure. b. Show that the surface area of the torus is A = 4 π 2 Rr. z

C y g(u) z

f (u) x

b. Find a parametrization for the surface obtained by revolving the curve x = y 2, y ≥ 0, about the x-axis. 33. a. Parametrization of an ellipsoid The parametrization x = a cos θ,  y = b sin θ, 0 ≤ θ ≤ 2π gives the ellipse ( x 2 a 2 ) + ( y 2 b 2 ) = 1. Using the angles θ and φ in spherical coordinates, show that r ( θ, φ ) = ( a cos θ sin φ ) i + ( b sin θ sin φ ) j + ( c cos φ ) k is a parametrization of the ellipsoid

( x 2 a 2 ) + ( y 2 b 2 ) + ( z 2 c 2 ) = 1. b. Write an integral for the surface area of the ellipsoid, but do not evaluate the integral. 34. Hyperboloid of one sheet

C r u 0

a. Find a parametrization for the hyperboloid of one sheet x 2 + y 2 − z 2 = 1 in terms of the angle θ associated with the circle x 2 + y 2 = r 2 and the hyperbolic parameter u associated with the hyperbolic function r 2 − z 2 = 1. (Hint: cosh 2 u − sinh 2 u = 1. )

x

R

z

b. Generalize the result in part (a) to the hyperboloid ( x 2 a 2 ) + ( y 2 b 2 ) − ( z 2 c 2 ) = 1. 35. (Continuation of Exercise 34.) Find a Cartesian equation for the plane tangent to the hyperboloid x 2 + y 2 − z 2 = 25 at the point ( x 0 , y 0 , 0 ) , where x 0 2 + y 0 2 = 25.

y

36. Hyperboloid of two sheets Find a parametrization of the hyperboloid of two sheets ( z 2 c 2 ) − ( x 2 a 2 ) − ( y 2 b 2 ) = 1.

x u

y

32. Parametrization of a surface of revolution Suppose that the parametrized curve C: ( f (u),  g(u) ) is revolved about the x-axis, where g(u) > 0 for a ≤ u ≤ b. a. Show that r ( u, υ ) = f (u) i + ( g(u) cos υ ) j + ( g(u)sin υ ) k

Surface Area for Implicit and Explicit Forms

37. Find the area of the surface cut from the paraboloid x 2 + y 2 − z = 0 by the plane z = 2. 38. Find the area of the band cut from the paraboloid x 2 + y 2 − z = 0 by the planes z = 2 and z = 6. 39. Find the area of the region cut from the plane x + 2 y + 2 z = 5 by the cylinder whose walls are x = y 2 and x = 2 − y 2 .

15.6  Surface Integrals

40. Find the area of the portion of the surface x 2 − 2 z = 0 that lies above the triangle bounded by the lines x = 3,  y = 0, and y = x in the xy-plane. − 2 y − 2 z = 0 that lies above 41. Find the area of the surface the triangle bounded by the lines x = 2,  y = 0, and y = 3 x in the xy-plane. x2

42. Find the area of the cap cut from the sphere x 2 + y 2 + z 2 = 2 by the cone z = x 2 + y 2 . 43. Find the area of the ellipse cut from the plane z = cx (c a constant) by the cylinder x 2 + y 2 = 1. 44. Find the area of the upper portion of the cylinder x 2 + z 2 = 1 that lies between the planes x = ±1 2 and y = ±1 2. 45. Find the area of the portion of the paraboloid x = 4 − − that lies above the ring 1 ≤ y 2 + z 2 ≤ 4 in the yz-plane. 46. Find the area of the surface cut from the paraboloid x 2 + y + z 2 = 2 by the plane y = 0. y2

z2

53. The surface in the first octant cut from the cylinder y = ( 2 3 ) z 3 2 by the planes x = 1 and y = 16 3 54. The portion of the plane y + z = 4 that lies above the region cut from the first quadrant of the xz-plane by the parabola x = 4 − z 2 55. Use the parametrization r ( x , z ) = x i + f ( x , z ) j + zk and Equation (5) to derive a formula for dσ associated with the explicit form y = f ( x , z ). 56. Let S be the surface obtained by rotating the smooth curve y = f ( x ),  a ≤ x ≤ b, about the x-axis, where f ( x ) ≥ 0. a. Show that the vector function r ( x , θ ) = xi + f ( x ) cos θ j + f ( x ) sin θk is a parametrization of S, where θ is the angle of rotation around the x-axis (see the accompanying figure). z

47. Find the area of the surface x 2 − 2 ln x + 15 y − z = 0 above the square R: 1 ≤ x ≤ 2, 0 ≤ y ≤ 1, in the xy-plane. 48. Find the area of the surface 2 x 3 2 + 2 y 3 2 − 3z = 0 above the square R: 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, in the xy-plane. Find the area of the surfaces in Exercises 49–54. 49. The surface cut from the bottom of the paraboloid z = x 2 + y 2 by the plane z = 3 50. The surface cut from the “nose” of the paraboloid x = 1 − y 2 − z 2 by the yz-plane 51. The portion of the cone z = x 2 + y 2 that lies over the region between the circle x 2 + y 2 = 1 and the ellipse 9 x 2 + 4 y 2 = 36 in the xy-plane. (Hint: A formula from geometry states that the area inside the ellipse x 2 a 2 + y 2 b 2 = 1 is πab.) 52. The triangle cut from the plane 2 x + 6 y + 3z = 6 by the bounding planes of the first octant. Calculate the area three ways, using different explicit forms.

983

0

y

(x, y, z)

z

u f (x)

x

b. Use Equation (4) to show that the surface area of this surface of revolution is given by A =

b

∫a

2π f ( x ) 1 + [ f ′( x ) ]2 dx.

15.6 Surface Integrals To compute the mass of a surface, the flow of a liquid across a curved membrane, or the total electrical charge on a surface, we need to integrate a function over a curved surface in space. Such a surface integral is the two-dimensional extension of the line integral concept used to integrate over a one-dimensional curve. Like line integrals, surface integrals arise in two forms. The first occurs when we integrate a scalar function over a surface, such as integrating a mass density function defined on a surface to find its total mass. This form corresponds to line integrals of scalar functions defined in Section 15.1 and can be used to find the mass of a thin wire. The second form involves surface integrals of vector fields, analogous to the line integrals for vector fields defined in Section 15.2. An example occurs when we want to measure the net flow of a fluid across a surface submerged in the fluid (just as we previously defined the flux of F across a curve). In this section we investigate these ideas and their applications.

Surface Integrals Suppose that the function G ( x , y, z ) gives the mass density (mass per unit area) at each point on a surface S. Then we can calculate the total mass of S as an integral in the following way.

984

Chapter 15 Integrals and Vector Fields

Pk

Δyry

z

Assume, as in Section 15.5, that the surface S is defined parametrically on a region R in the uυ-plane, r ( u, υ ) = f ( u, υ ) i + g ( u, υ ) j + h ( u, υ ) k ,

Δuru Δsk = Δsuy x

(xk, yk, zk)

In Figure 15.49, we see how a subdivision of R (considered as a rectangle for simplicity) divides the surface S into corresponding curved surface elements, or patches, of area Δσ uυ ≈ ru × rυ du d υ.

y

FIGURE 15.49 The area of the patch Δσ k is approximated by the area of the tangent parallelogram determined by the vectors Δu ru and Δυ rυ . The point ( x k , y k , z k ) lies on the surface patch, beneath the parallelogram shown here.

( u, υ ) ∈ R.

As we did for the subdivisions when defining double integrals in Section 14.2, we number the surface element patches in some order with their areas given by Δσ1 ,  Δσ 2 , … ,  Δσ n . To form a Riemann sum over S, we choose a point ( x k , y k , z k ) in the kth patch, multiply the value of the function G at that point by the area Δσ k , and add together the products: n

∑ G ( x k , y k , z k ) Δσ k . k =1

Depending on how we pick ( x k , y k , z k ) in the kth patch, we may get different values for this Riemann sum. Then we take the limit as the number of surface patches increases, their areas shrink to zero, and both Δu → 0 and Δυ → 0. This limit, whenever it exists independent of all choices made, defines the surface integral of G over the surface S as

∫∫

G ( x , y, z ) dσ = lim

n →∞

S

n

∑ G ( x k , y k , z k ) Δσ k .

(1)

k =1

Notice the analogy with the definition of the double integral (Section 14.2) and with the line integral (Section 15.1). If S is a piecewise smooth surface, and G is continuous over S, then the surface integral defined by Equation (1) can be shown to exist. The formula for evaluating the surface integral depends on the manner in which S is described—parametrically, implicitly, or explicitly—as discussed in Section 15.5. Formulas for a Surface Integral of a Scalar Function

1. For a smooth surface S defined parametrically as r ( u, υ ) = f ( u, υ ) i + g ( u, υ ) j + h ( u, υ ) k , ( u, υ ) ∈ R , and a continuous function G ( x , y, z ) defined on S, the surface integral of G over S is given by the double integral over R,

∫∫ G ( x , y, z ) dσ

=

S

∫∫ G ( f ( u, υ ),  g( u, υ ),  h( u, υ ))

ru × rυ du d υ. (2)

R

2. For a surface S given implicitly by F ( x , y, z ) = c, where F is a continuously differentiable function, with S lying above its closed and bounded shadow region R in the coordinate plane beneath it, the surface integral of the continuous function G over S is given by the double integral over R,

∫∫ G ( x , y, z ) dσ S

=

∇F

∫∫ G ( x , y, z ) ∇F ⋅ p

dA,

(3)

R

where p is a unit vector normal to R and ∇F ⋅ p ≠ 0. 3. For a surface S given explicitly as the graph of z = f ( x , y ) , where f is a continuously differentiable function over a region R in the xy-plane, the surface integral of the continuous function G over S is given by the double integral over R,

∫∫ G ( x , y, z ) dσ S

=

∫∫ G ( x , y, f ( x , y )) R

f x 2 + f y 2 + 1 dx dy.

(4)

15.6  Surface Integrals

985

The surface integral in Equation (1) takes on different meanings in different applications. If G has the constant value 1, the integral gives the area of S. If G gives the mass density of a thin shell of material modeled by S, the integral gives the mass of the shell. If G gives the charge density of a thin shell, the integral gives the total charge. Integrate G ( x , y, z ) = x 2 over the cone z =

EXAMPLE 1

x 2 + y 2 , 0 ≤ z ≤ 1.

Solution Using Equation (2) and the calculations from Example 4 in Section 15.5, we have rr × rθ = 2r and

∫∫

x 2 dσ =

S

2π

2∫

= =

1

∫0 ∫0 ( r 2 cos 2 θ )(

2 4

2π

x = r cos θ

1

∫0 r 3 cos 2 θ dr dθ

0 2π

∫0

2r ) dr dθ

2π 2 ⎡θ 1 ⎤ = π 2. + sin 2 θ ⎥⎦ 0 4 ⎢⎣ 2 4 4

cos 2 θ dθ =

Surface integrals behave like other double integrals, the integral of the sum of two functions being the sum of their integrals and so on. The domain Additivity Property takes the form

∫∫ G dσ

=

S

∫∫ G dσ + ∫∫ G dσ +  + ∫∫ G dσ. S1

S2

Sn

When S is partitioned by smooth curves into a finite number of smooth patches with nonoverlapping interiors (i.e., if S is piecewise smooth), then the integral over S is the sum of the integrals over the patches. Thus, the integral of a function over the surface of a cube is the sum of the integrals over the faces of the cube. We integrate over a “turtle shell” of welded plates by integrating over one plate at a time and adding the results. EXAMPLE 2 Integrate G ( x , y, z ) = xyz over the surface of the cube cut from the first octant by the planes x = 1,  y = 1, and z = 1 (Figure 15.50).

z

Solution We integrate xyz over each of the six sides and add the results. Since xyz = 0 on the sides that lie in the coordinate planes, the integral over the surface of the cube reduces to

Side A

1

∫∫

xyz dσ =

0

y

1 x

1

∫∫

xyz dσ +

xyz dσ +

Si de B

∫∫

xyz dσ.

Si de C

Side A is the surface f ( x , y, z ) = z = 1 over the square region R xy : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, in the xy-plane. For this surface and region,

Side C

p = k,

Side B

FIGURE 15.50

∫∫

Si de A

Cube surface

∇f = k , dσ =

The cube in Example 2.

∇f = 1,

∇f ⋅ p = k ⋅ k = 1

∇f 1 dA = dx dy = dx dy ∇f ⋅ p 1

Eq. (3)

xyz = xy(1) = xy and

∫∫

Si de A

xyz dσ =

∫∫ xy dx dy

=

R xy

1

1

∫0 ∫0 xy dx dy

=

1

∫0

y 1 dy = . 4 2

Symmetry tells us that the integrals of xyz over sides B and C are also 1 4. Hence,

∫∫

Cube surface

xyz dσ =

1 1 1 3 + + = . 4 4 4 4

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Chapter 15 Integrals and Vector Fields

EXAMPLE 3 Integrate G ( x , y, z ) = 1 − x 2 − y 2 over the “football” surface S formed by rotating the curve x = cos z ,  y = 0, −π 2 ≤ z ≤ π 2, around the z-axis. Solution The surface is displayed in Figure 15.46, and in Example 6 of Section 15.5 we found the parametrization

x = cos u cos υ, y = cos u sin υ, z = u, −

π π ≤ u ≤ 2 2

and 0 ≤ υ ≤ 2π,

where υ represents the angle of rotation from the xz-plane about the z-axis. Substituting this parametrization into the expression for G gives 1 − x 2 − y2 =

1 − ( cos 2 u )( cos 2 υ + sin 2 υ ) =

1 − cos 2 u = sin u .

The surface area differential for the parametrization was found to be (Example 6, Section 15.5) dσ = cos u 1 + sin 2 u du d υ. These calculations give the surface integral

∫∫

1 − x 2 − y 2 dσ =

S

2π

= 2∫ =

2π 0

2π

1

FIGURE 15.51

y

The surface S in

x (1 + 2 z ) =

x 1 + y2.

With z = f ( x , y ) = y 2 2, we use Equation (4) to evaluate the surface integral: dσ =

f x 2 + f y 2 + 1 dx dy =

0 + y 2 + 1 dx dy

and

S

x

w dw d υ

w = 1 + sin 2 u, dw = 2 sin u cos u du When u = 0,  w = 1. When u = π 2,   w = 2.

2 3 2⎤2 4π w ⎥ = ( 2 2 − 1). ⎦1 3 3

G ( x , y, z ) =

∫∫ G ( x , y, z ) dσ = ∫∫ (

1

sin u cos u 1 + sin 2 u du d υ

Solution The function G on the surface S is given by

z = 12 y 2

x+y=1

2

sin u = sin(−u) for −π 2 < u < 0

EXAMPLE 4 Evaluate ∫∫ S x (1 + 2 z ) dσ on the portion of the cylinder z = y 2 2 over the triangular region R: x ≥ 0,  y ≥ 0,  x + y ≤ 1, in the xy-plane (Figure 15.51).

a0, 1, 2b

1

sin u cos u 1 + sin 2 u du d υ

π 2

∫0

∫0 ∫1

= 2π ⋅

z

π 2

∫0 ∫−π 2

x 1 + y 2 ) 1 + y 2 dx dy

R

1

1− x

1

1 x ⎡⎢ (1 − x ) + (1 − x )3 ⎤⎥ dx ⎣ ⎦ 3

=

∫0 ∫0

=

∫0

=

∫0 ( 3 x 1 2 − 2 x 3 2 + x 5 2 − 3 x 7 2 ) dx

Example 4.

1

4

x (1 + y 2 ) dy dx

1

8 4 2 2 9 2 ⎤1 x ⎥ = ⎡⎢ x 3 2 − x 5 2 + x 7 2 − ⎣9 ⎦0 5 7 27 =

8 4 2 2 284 − + − = ≈ 0.30. 9 5 7 27 945

Integrate and evaluate.

Routine algebra

987

15.6  Surface Integrals

Orientation of a Surface

(a)

(b)

FIGURE 15.52

(a) An outward-pointing vector field and (b) an inward-pointing vector field give the two possible orientations of a sphere. d

c

a

b Start Finish db ac

A curve C with a parametrization r(t ) has a natural orientation, or direction, that comes from the direction of increasing t. The unit tangent vector T along C points in this forward direction at each point on the curve. There are two possible orientations for a curve, corresponding to whether we follow the direction of the tangent vector T at each point, or the direction of −T. To specify an orientation on a surface in space S, we do something similar, but this time we specify a normal vector at each point on the surface. A parametrization of a surface r ( u, υ ) gives a vector ru × rυ that is normal to the surface, and so gives an orientation wherever the parametrization applies. A second choice of orientation is found by taking −( ru × rυ ), giving a vector that points to the opposite side of the surface at each point. In essence, an orientation is a way of consistently choosing one of the two sides of a surface. Not all surfaces have orientations, but a surface that does have one also has a second, opposite orientation. Each point on the sphere in Figure 15.52 has one normal vector pointing inward, toward the center of the sphere, and another opposite normal vector pointing outward. We specify one of two possible orientations for the sphere by choosing either the inward vector at each point, or alternatively the outward vector at each point. When we can choose a continuous field of unit normal vectors n on a smooth surface S, we say that S is orientable (or two-sided). Spheres and other smooth surfaces that are the boundaries of solid regions in space are orientable since we can choose an outward-pointing unit vector n at each point to specify an orientation. A surface together with its normal field n, or, equivalently, a surface with a consistent choice of sides, is called an oriented surface. The vector n at any point gives the positive direction or positively oriented side at that point (Figure 15.52). Not all surfaces can be oriented. The Möbius band in Figure 15.53 is an example of a surface that is not orientable. No matter how you try to construct a continuous unit normal vector field (shown as the shafts of thumbtacks in the figure), starting at one point and moving the vector continuously around the surface in the manner shown will return it to the starting point, but pointing in the opposite direction. No choice of a vectors can give a continuous normal vector field on the Möbius band, so the Möbius band is not orientable.

FIGURE 15.53

To make a Möbius band, take a rectangular strip of paper abcd, give the end bc a single twist, and paste the ends of the strip together to match a with c and b with d. The Möbius band is a nonorientable, or one-sided, surface.

Surface Integrals of Vector Fields In Section 15.2 we defined the line integral of a vector field along a path C as ∫ F ⋅ T ds, C where T is the unit tangent vector to the path pointing in the forward-oriented direction. We have a similar definition for surface integrals.

Let F be a vector field in three-dimensional space with continuous components defined over a smooth surface S having a chosen field of normal unit vectors n orienting S. Then the surface integral of F over S is

DEFINITION

∫∫

F ⋅ n dσ.

(5)

S

This integral is also called the flux of the vector field F across S.

If F is the velocity field of a three-dimensional fluid flow, then the flux of F across S is the net rate at which fluid is crossing S per unit time in the chosen positive direction n defined by the orientation of S. Fluid flows are discussed in more detail in Section 15.7.

988

Chapter 15 Integrals and Vector Fields

Computing a Surface Integral for a Parametrized Surface EXAMPLE 5 Find the flux of F = yzi + xj − z 2 k through the parabolic cylinder y = x 2, 0 ≤ x ≤ 1, 0 ≤ z ≤ 4, in the direction n indicated in Figure 15.54.

z 4 (1, 0, 4)

Solution On the surface we have x = x ,  y = x 2, and z = z, so we have the parametrization r ( x , z ) = xi + x 2 j + zk, 0 ≤ x ≤ 1, 0 ≤ z ≤ 4. The cross product of tangent vectors,

y = x2

i

n

j

r x × rz = 1 2 x 1

0

y

0

k 0 = 2 xi − j,

rx = i + 2 xj rz = k

1

can be used to find unit normal vectors to the surface,

1 x

n =

FIGURE 15.54

Finding the flux through the surface of a parabolic cylinder (Example 5).

r x × rz = r x × rz

2 xi − j . 4x 2 + 1

We can equally well choose the unit normal vectors −n that point in the opposite direction. The first choice is shown in Figure 15.54. On the surface we have y = x 2, so the vector field there is F = yzi + xj − z 2 k = x 2 zi + xj − z 2 k. Thus, F⋅n =

1

( ( x 2 z )(2 x ) + ( x )(−1) + (−z 2 )( 0 )) =

4x 2 + 1 The flux of F outward through the surface is

∫∫

F ⋅ n dσ =

S

4

1

∫0 ∫0

2x 3z − x r x × rz dx dz 4x 2 + 1

2x 3z − x . 4x 2 + 1

dσ = rx × rz dx dz

2x 3z − x 4 x 2 + 1 dx dz 4x 2 + 1 x =1 4 1 4 1 1 = ∫ ∫ ( 2 x 3 z − x ) dx dz = ∫ ⎡⎢ x 4 z − x 2 ⎤⎥ dz 0 ⎣2 0 0 2 ⎦ x=0 =

4

1

∫0 ∫0

4 1( 1 ) dz = ( z − 1) 2 ⎤⎥ z 1 − ∫0 2 ⎦0 4 1( ) 1( ) = 9 − 1 = 2. 4 4

=

4

There is a simple formula for the flux of F across a parametrized surface r ( u, υ ). Since dσ = ru × rυ du d υ and n =

ru × rυ , ru × rυ

it follows that Flux Across a Parametrized Surface Flux = ± ∫∫ F ⋅ ( ru × rυ ) du d υ R

∫∫ S

F ⋅ n dσ =

∫∫ R

F⋅

ru × rυ r × rυ du d υ = ru × rυ u

∫∫

F ⋅ ( ru × rυ ) du d υ.

R

The other choice of unit normal vector, −n, would add a negative sign to this formula. The choice of n or −n depends on the direction in which we choose to measure the flux across the surface. This integral for flux simplifies the computation in Example 5 by eliminating the need to compute the canceled factor ru × rυ . Since F ⋅ ( r x × rz ) = ( yzi + xj − z 2 k = x 2 zi + xj − z 2 k ) ⋅ ( 2 xi − j ) = ( x 2 z )(2 x ) + ( x )(−1) = 2 x 3 z − x ,

15.6  Surface Integrals

989

we obtain directly Flux =

∫∫

F ⋅ n dσ =

S

4

1

∫0 ∫0 ( 2 x 3 z − x ) dx dz

= 2

in Example 5.

Computing a Surface Integral for a Level Surface If S is part of a level surface g ( x , y, z ) = c, then n may be taken to be one of the two fields n = ±

∇g , ∇g

(6)

depending on which one gives the preferred direction. The corresponding flux is Flux =

∫∫

F ⋅ n dσ

S

=

∫∫ ⎜⎜⎝ F ⋅

⎛

±∇g ⎞⎟ ∇g dA ⎟ ∇g ⎟⎠ ∇g ⋅ p

∫∫

±∇g dA. ∇g ⋅ p

R

=

F⋅

R

Eqs. (6) and (3)

(7)

EXAMPLE 6 Find the flux of F = yzj + z 2 k through the surface S cut from the cylinder y 2 + z 2 = 1, z ≥ 0, by the planes x = 0 and x = 1, in the direction away from the x-axis. Solution The normal field on S (Figure 15.55) in the specified direction may be calculated from the gradient of g ( x , y, z ) = y 2 + z 2 to be

n = + z y2

+

z2

∇g = ∇g

With p = k, we also have

=1

dσ =

n

Rxy

1

FIGURE 15.55

Eq. (3)

F ⋅ n = ( yzj + z 2 k ) ⋅ ( yj + zk )

y

(1, −1, 0)

∇g 2 1 dA = dA = dA. 2z z ∇g ⋅ k

We can drop the absolute value bars because z ≥ 0 on S. The value of F ⋅ n on the surface is

S

x

2 y j + 2 zk 2 y j + 2 zk = = yj + zk. 2 2 2 1 4 y + 4z

= y 2z + z 3 = z( y 2 + z 2 ) = z.

(1, 1, 0)

Calculating the flux of a vector field through the surface S. The area of the shadow region R xy is 2 (Example 6).

y 2 + z 2 = 1 on  S.

The surface projects onto the shadow region R xy , which is the rectangle in the xy-plane shown in Figure 15.55. Therefore, the flux of F through S in the direction away from the x-axis is

∫∫ S

F ⋅ n dσ =

∫∫ (z)( z dA ) = ∫∫ 1

R xy

dA = area ( R xy ) = 2.

R xy

Moments and Masses of Thin Shells Thin shells of material like bowls, metal drums, and domes are modeled with surfaces. Their moments and masses are calculated with the formulas in Table 15.3. The derivations are similar to those in Section 6.6. The formulas resemble those for line integrals in Table 15.1, Section 15.1.

990

Chapter 15 Integrals and Vector Fields

TABLE 15.3 Mass and moment formulas for very thin shells

Mass: M =

∫∫ δ dσ

δ = δ( x , y, z ) = density at  ( x , y, z )  is mass per unit area.

S

First moments about the coordinate planes:

∫∫

M yz =

x δ dσ ,

Mxz =

S

∫∫

y δ dσ ,

Mxy =

S

∫∫

z δ dσ

S

Coordinates of center of mass: x = M yz M ,

y = Mxz M ,

z = Mxy M

Moments of inertia about coordinate axes: Ix =

∫∫ ( y 2 + z 2 ) δ dσ,

Iy =

S

IL =

∫∫

r 2δ

dσ

∫∫ ( x 2 + z 2 ) δ dσ,

Iz =

S

∫∫ ( x 2 + y 2 ) δ dσ, S

r ( x , y, z ) = distance from point  ( x , y, z )  to line  L

S

z a a0, 0, 2b

EXAMPLE 7 Find the center of mass of a thin hemispherical shell of radius a and constant density δ.

x2 + y2 + z2 = a2

Solution We model the shell with the hemisphere

f ( x , y, z ) = x 2 + y 2 + z 2 = a 2 ,

S

R

a x

a

y

z ≥ 0

(Figure 15.56). The symmetry of the surface about the z-axis tells us that x = y = 0. It remains only to find z from the formula z = Mxy M . The mass of the shell is

x2 + y2 = a2

M =

FIGURE 15.56

The center of mass of a thin hemispherical shell of constant density lies on the axis of symmetry halfway from the base to the top (Example 7).

∫∫ δ dσ S

= δ ∫∫ d σ = (δ )( area of  S ) = 2πa 2δ.

δ = constant

S

To evaluate the integral for Mxy , we take p = k and calculate ∇f = 2 x i + 2 y j + 2 z k = 2 x 2 + y 2 + z 2 = 2a ∇f ⋅ p = ∇f ⋅ k = 2 z = 2 z dσ =

∇f a dA = dA. ∇f ⋅ p z

Eq. (3)

Then Mxy =

∫∫ S

a zδ dσ = δ ∫∫ z dA = δ a ∫∫ dA = δ a (πa 2 ) = δπa 3 z R

M a π a 3δ = . z = xy = M 2πa 2δ 2 The shell’s center of mass is the point ( 0, 0, a 2 ).

R

15.6  Surface Integrals

z

EXAMPLE 8 Find the center of mass of a thin shell of density δ = 1 z 2 cut from the 2 cone z = x + y 2 by the planes z = 1 and z = 2 (Figure 15.57).

2 z = "x 2 + y 2 1

Solution Since the surface and the density function δ are symmetric about the z-axis, we have x = y = 0. We now proceed to find z = Mxy M . Working as in Example 4 of Section 15.5, we have

r ( r , θ ) = ( r cos θ ) i + ( r sin θ ) j + rk,

1 ≤ r ≤ 2, 0 ≤ θ ≤ 2π ,

and

y

x

991

FIGURE 15.57

The cone frustum formed when the cone z = x 2 + y 2 is cut by the planes z = 1 and z = 2 (Example 8).

rr × rθ =

2r.

Therefore, M =

∫∫ δ dσ

=

S

2∫

=

2π 0

2π

∫0 ∫1

2

1 r2

⎡ ⎤2 ⎢ ln r ⎥ dθ = ⎥⎦ 1 ⎣⎢

2r dr dθ 2∫

2π 0

ln 2 dθ

= 2π 2 ln 2, Mxy =

∫∫ δ z dσ

=

S

=

2∫

=

2∫

z =

2π 0 2π 0

2π

∫0 ∫1

2

1 r 2r dr dθ r2

2

∫1 dr dθ d θ = 2π 2,

Mxy 2π 2 1 . = = ln 2 M 2π 2 ln 2

The shell’s center of mass is the point ( 0, 0, 1 ln 2 ).

EXERCISES

15.6

Surface Integrals of Scalar Functions

In Exercises 1–8, integrate the given function over the given surface.

8. Spherical cap H ( x , y, z ) = yz , over the part of the sphere x 2 + y 2 + z 2 = 4 that lies above the cone z = x 2 + y 2

1. Parabolic cylinder G ( x , y, z ) = x , over the parabolic cylinder y = x 2, 0 ≤ x ≤ 2, 0 ≤ z ≤ 3

9. Integrate G ( x , y, z ) = x + y + z over the surface of the cube cut from the first octant by the planes x = a,  y = a,  z = a.

2. Circular cylinder G ( x , y, z ) = z , over the cylindrical surface y 2 + z 2 = 4, z ≥ 0, 1 ≤ x ≤ 4 3. Sphere G ( x , y, z ) = x 2, over the unit sphere x 2 + y 2 + z 2 = 1

10. Integrate G ( x , y, z ) = y + z over the surface of the wedge in the first octant bounded by the coordinate planes and the planes x = 2 and y + z = 1.

4. Hemisphere G ( x , y, z ) = z 2 , over the hemisphere x 2 + y 2 + z 2 = a 2,  z ≥ 0

11. Integrate G ( x , y, z ) = xyz over the surface of the rectangular solid cut from the first octant by the planes x = a,  y = b, and z = c.

5. Portion of plane F ( x , y, z ) = z , over the portion of the plane x + y + z = 4 that lies above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, in the xy-plane

12. Integrate G ( x , y, z ) = xyz over the surface of the rectangular solid bounded by the planes x = ±a,  y = ±b, and z = ±c.

6. Cone F ( x , y, z ) = z − x , over the cone z = 0 ≤ z ≤1 7. Parabolic dome H ( x , y, z ) = dome z = 1 − x 2 − y 2 ,  z ≥ 0

x2

x2

+

y 2,

5 − 4 z , over the parabolic

13. Integrate G ( x , y, z ) = x + y + z over the portion of the plane 2 x + 2 y + z = 2 that lies in the first octant. 14. Integrate G ( x , y, z ) = x y 2 + 4 over the surface cut from the parabolic cylinder y 2 + 4 z = 16 by the planes x = 0,  x = 1, and z = 0.

992

Chapter 15 Integrals and Vector Fields

15. Integrate G ( x , y, z ) = z − x over the portion of the graph of z = x + y 2 above the triangle in the xy-plane having vertices ( 0, 0, 0 ) ,   (1, 1, 0 ), and ( 0, 1, 0 ). (See accompanying figure.)

19. Parabolic cylinder F = z 2 i + xj − 3zk through the surface cut from the parabolic cylinder z = 4 − y 2 by the planes x = 0,  x = 1, and z = 0 in the direction away from the x-axis 20. Parabolic cylinder F = x 2 j − xzk through the surface cut from the parabolic cylinder y = x 2, −1 ≤ x ≤ 1, by the planes z = 0 and z = 2 in the direction away from the yz-plane

z (1, 1, 2)

21. Sphere F = zk across the portion of the sphere x 2 + y 2 + z 2 = a 2 in the first octant in the direction away from the origin

1 z = x + y2

22. Sphere F = xi + yj + zk across the sphere x 2 + y 2 + z 2 = a 2 in the direction away from the origin

(0, 1, 1)

23. Plane F = 2 xyi + 2 yzj + 2 xzk upward across the portion of the plane x + y + z = 2a that lies above the square 0 ≤ x ≤ a, 0 ≤ y ≤ a, in the xy-plane

(0, 0, 0)

1

24. Cylinder F = xi + yj + zk through the portion of the cylinder x 2 + y 2 = 1 cut by the planes z = 0 and z = a in the direction away from the z-axis

(0, 1, 0) y

25. Cone F = xyi − zk through the cone z = 0 ≤ z ≤ 1, in the direction away from the z-axis

1

x 2 + y 2,

26. Cone F = y 2 i + xzj − k through the cone z = 2 x 2 + y 2 , 0 ≤ z ≤ 2, in the direction away from the z-axis

x (1, 1, 0)

27. Cone frustum F = −xi − yj + z 2 k through the portion of the cone z = x 2 + y 2 between the planes z = 1 and z = 2 in the direction away from the z-axis

16. Integrate G ( x , y, z ) = x over the surface given by z = x 2 + y for 0 ≤ x ≤ 1, −1 ≤ y ≤ 1. 17. Integrate G ( x , y, z ) = xyz over the triangular surface with vertices (1, 0, 0 ) , ( 0, 2, 0 ), and ( 0, 1, 1).

28. Paraboloid F = 4 xi + 4 yj + 2k through the surface cut from the bottom of the paraboloid z = x 2 + y 2 by the plane z = 1 in the direction away from the z-axis In Exercises 29 and 30, find the surface integral of the field F over the portion of the given surface in the specified direction.

z (0, 1, 1)

1

29. F ( x , y, z ) = −i + 2 j + 3k S: rectangular surface z = 0, 0 ≤ x ≤ 2, 0 ≤ y ≤ 3, direction k (0, 2, 0)

x

y

(1, 0, 0)

18. Integrate G ( x , y, z ) = x − y − z over the portion of the plane x + y = 1 in the first octant between z = 0 and z = 1 (see the figure below). z 1

S: rectangular surface y = 0, −1 ≤ x ≤ 2, 2 ≤ z ≤ 7, direction −j In Exercises 31–36, use Equation (7) to find the surface integral of the field F over the portion of the sphere x 2 + y 2 + z 2 = a 2 in the first octant in the direction away from the origin. 31. F ( x , y, z ) = zk 32. F ( x , y, z ) = − yi + xj 33. F ( x , y, z ) = yi − xj + k

(0, 1, 1)

(1, 0, 1)

1 x

30. F ( x , y, z ) = yx 2 i − 2 j + xzk

y

1

Finding Flux or Surface Integrals of Vector Fields

In Exercises 19–28, use a parametrization to find the flux ∫∫ S F ⋅ n dσ across the surface in the specified direction.

34. F ( x , y, z ) = zxi + zyj + z 2 k 35. F ( x , y, z ) = xi + yj + zk x i + y j + zk 36. F ( x , y , z ) = x 2 + y2 + z 2 37. Find the flux of the field F ( x , y, z ) = z 2 i + xj − 3zk through the surface cut from the parabolic cylinder z = 4 − y 2 by the planes x = 0,  x = 1, and z = 0 in the direction away from the x-axis. 38. Find the flux of the field F ( x , y, z ) = 4 xi + 4 yj + 2k through the surface cut from the bottom of the paraboloid z = x 2 + y 2 by the plane z = 1 in the direction away from the z-axis.

15.7  Stokes’ Theorem

39. Let S be the portion of the cylinder y = e x in the first octant that projects parallel to the x-axis onto the rectangle R yz : 1 ≤ y ≤ 2, 0 ≤ z ≤ 1, in the yz-plane (see the accompanying figure). Let n be the unit vector normal to S that points away from the yz-plane. Find the flux of the field F ( x , y, z ) = −2 i + 2 yj + zk across S in the direction of n.

993

46. Conical surface of constant density Find the moment of inertia about the z-axis of a thin shell of constant density δ cut from the cone 4 x 2 + 4 y 2 − z 2 = 0,  z ≥ 0, by the circular cylinder x 2 + y 2 = 2 x (see the accompanying figure). z

z 1

Ry z

4x2 + 4y2 − z2 = 0

1 x y = ex

S

2

y

z ≥0 y

40. Let S be the portion of the cylinder y = ln x in the first octant whose projection parallel to the y-axis onto the xz-plane is the rectangle R xz : 1 ≤ x ≤ e, 0 ≤ z ≤ 1. Let n be the unit vector normal to S that points away from the xz-plane. Find the flux of F = 2 yj + zk through S in the direction of n. 41. Find the outward flux of the field F = 2 xyi + 2 yzj + 2 xzk across the surface of the cube cut from the first octant by the planes x = a,  y = a, and z = a. 42. Find the outward flux of the field F = xzi + yzj + k across the surface of the upper cap cut from the ball x 2 + y 2 + z 2 ≤ 25 by the plane z = 3. Moments and Masses

43. Centroid Find the centroid of the portion of the sphere x 2 + y 2 + z 2 = a 2 that lies in the first octant. 44. Centroid Find the centroid of the surface cut from the cylinder y 2 + z 2 = 9,  z ≥ 0, by the planes x = 0 and x = 3 (resembles the surface in Example 6).

x

2

x2

+

y2

= 2x or r = 2 cos u

47. Spherical shells Find the moment of inertia about a diameter of a thin spherical shell of radius a and constant density δ. (Work with a hemispherical shell and double the result.) 48. Conical Surface Find the centroid of the lateral surface of a solid cone of base radius a and height h (cone surface minus the base). 49. A surface S lies on the plane 2 x + 3 y + 6 z = 12 directly above the rectangle in the xy-plane with vertices ( 0, 0 ) , (1, 0 ) , ( 0, 2 ), and (1, 2 ). If the density at a point ( x , y, z ) on S is given by δ ( x , y, z ) = 4 xy + 6 z mg cm 2, find the total mass of S. 1 1 50. A surface S lies on the paraboloid z = x 2 + y 2 directly 2 2 above the triangle in the xy-plane with vertices ( 0, 0 ) , ( 2, 0 ), and ( 2, 4 ). If the density at a point ( x , y, z ) on S is given by δ ( x , y, z ) = 9 xy g cm 2, find the total mass of S.

45. Thin shell of constant density Find the center of mass and the moment of inertia about the z-axis of a thin shell of constant density δ cut from the cone x 2 + y 2 − z 2 = 0 by the planes z = 1 and z = 2.

15.7 Stokes’ Theorem To calculate the counterclockwise circulation of a two-dimensional vector field F = Mi + N j around a simple closed curve in the plane, Green’s Theorem says we can compute the double integral over the region enclosed by the curve of the scalar quantity ( ∂ N ∂ x − ∂ M ∂ y ). This expression is the k-component of a curl vector field, and it measures the rate of rotation of F at each point in the region around an axis parallel to k. For a vector field in three-dimensional space, the rotation at each point is around an axis that is parallel to the curl vector at that point. When a closed curve C in space is the boundary of an oriented surface, we will see that the circulation of F around C is equal to the surface integral of the curl vector field. This result extends Green’s Theorem from regions in the plane to general surfaces in space having a smooth boundary curve.

994

Chapter 15 Integrals and Vector Fields

Curl F (x, y, z)

FIGURE 15.58

The circulation vector at a point ( x , y, z ) in a plane in a threedimensional fluid flow. Notice its righthand relation to the rotating particles in the fluid.

The Curl Vector Field Suppose that F is the velocity field of a fluid flowing in space. Particles near the point ( x , y, z ) in the fluid tend to rotate around an axis through ( x , y, z ) that is parallel to a certain vector we are about to identify. This vector points in the direction for which the rotation is counterclockwise when viewed looking down onto the plane of the circulation from the tip of the arrow representing the vector. This is the direction your right-hand thumb points when your fingers curl around the axis of rotation in the way consistent with the rotating motion of the particles in the fluid (see Figure 15.58). The length of the vector measures the rate of rotation. The vector, introduced in Equation (3) of Section 15.3, is called the curl vector for the vector field F = Mi + N j + Pk, and it is given by

(

)

⎛ ∂P ⎛ ∂N ∂ N ⎞⎟ ∂M ∂P ∂ M ⎞⎟ curl  F = ⎜⎜ − − j + ⎜⎜ − ⎟⎟⎠ i + ⎟ k. ⎝ ∂y ⎝ ∂z ∂z ∂x ∂x ∂ y ⎟⎠

(1)

This information is a consequence of Stokes’ Theorem, the generalization to space of the circulation-curl form of Green’s Theorem. Notice that ( curl  F ) ⋅ k = ( ∂ N ∂ x − ∂ M ∂ y ) , which is consistent with our discussion in Section 15.4 when F = M ( x , y ) i + N ( x , y ) j. The formula for curl F in Equation (1) is often expressed using the symbol ∇ = i ∇ is the symbol “del.”

∂ ∂ ∂ + j +k . ∂y ∂z ∂x

(2)

The symbol ∇ is pronounced “del,” and we can use this symbol to express the curl of F with the formula i ∂ ∇×F = ∂x M

j

k

∂ ∂y N

∂ ∂z P

(

)

⎛ ∂N ⎛ ∂P ∂M ∂P ∂ M ⎞⎟ ∂ N ⎞⎟ j + ⎜⎜ − − = ⎜⎜ − ⎟ k. ⎟⎟⎠ i + ⎝ ⎝ ∂y ∂ y ⎟⎠ ∂z ∂x ∂x ∂z We often use this cross product notation to write the curl symbolically as “del cross F.” curl  F = ∇ × F EXAMPLE 1

(3)

Find the curl of F = ( x 2 − z ) i + xe z j + xyk.

Solution We use Equation (3) and the determinant form for the cross product, which gives,

curl  F = ∇ × F i =

j

k

∂ ∂ ∂ ∂x ∂ y ∂z x 2 − z xe z xy

(

)

⎞ ⎛∂ ∂ ∂ 2 ∂ (x − z) j = ⎜⎜ ( xy) − ( xe z ) ⎟⎟⎟ i − ( xy) − ⎠ ⎝∂y ∂x ∂z ∂z ⎞ ⎛∂ ∂ 2 ( x − z )⎟⎟⎟ k + ⎜⎜ ( xe z ) − ⎠ ⎝ ∂x ∂y

Curl F is a vector, not a scalar.

= ( x − xe z ) i − ( y + 1) j + ( e z − 0 ) k = x (1 − e z ) i − ( y + 1) j + e z k.

995

15.7  Stokes’ Theorem

As we will see, the operator ∇ has a number of other applications. For instance, when applied to a scalar function f ( x , y, z ), it gives the gradient of f : S

n

∇f =

∂f ∂f ∂f i+ j+ k. ∂x ∂y ∂z

In this setting it is read sometimes as “del f ” and sometimes as “grad f .” C

FIGURE 15.59

The orientation of the bounding curve C gives it a right-hand relation to the normal field n. If the thumb of a right hand points along n, the fingers curl in the direction of C.

Stokes’ Theorem Stokes’ Theorem generalizes Green’s Theorem to three dimensions. The circulation-curl form of Green’s Theorem relates the counterclockwise circulation of a vector field around a simple closed curve C in the xy-plane to a double integral over the plane region R enclosed by C. Stokes’ Theorem relates the circulation of a vector field around the boundary C of an oriented surface S in space (Figure 15.59) to a surface integral over the surface S. We require that the surface be piecewise smooth, which means that it is a finite union of smooth surfaces joining along smooth curves.

THEOREM 6—Stokes’ Theorem

Let S be a piecewise smooth oriented surface having a piecewise smooth boundary curve C. Let F = Mi + N j + Pk be a vector field whose components have continuous first partial derivatives on an open region containing S. Then the circulation of F around C in the direction counterclockwise with respect to the surface’s unit normal vector n equals the integral of the curl vector field ∇ × F over S:

DC F ⋅ dr

=

∫∫ ( ∇ × F ) ⋅ n dσ

(4)

S

Counterclockwise circulation

Curl integral

Notice from Equation (4) that if two different oriented surfaces S 1 and S 2 have the same boundary C, their curl integrals are equal:

∫∫ ( ∇ × F ) ⋅ n 1 dσ S1

=

∫∫ ( ∇ × F ) ⋅ n 2 dσ. S2

Both curl integrals equal the counterclockwise circulation integral on the left side of Equation (4) as long as the unit normal vectors n 1 and n 2 correctly orient the surfaces. So the curl integral is independent of the surface and depends only on circulation along the boundary curve. This independence of surface resembles the path independence for the flow integral of a conservative velocity field along a curve, where the value of the flow integral depends only on the endpoints (that is, the boundary points) of the path. In that sense, the curl field ∇ × F is analogous to the gradient field ∇f of a scalar function f . If C is a curve in the xy-plane, oriented counterclockwise, and R is the region in the xy-plane bounded by C, then dσ = dx dy and (∇

⎛ ∂N ∂ M ⎞⎟ × F ) ⋅ n = ( ∇ × F ) ⋅ k = ⎜⎜ − ⎟. ⎝ ∂x ∂ y ⎟⎠

Under these conditions, Stokes’ equation becomes

DC F ⋅ dr

=

⎛ ∂N

∫∫ ⎜⎜⎝ ∂ x R

−

∂ M ⎞⎟ ⎟ dx dy, ∂ y ⎟⎠

which is the circulation-curl form of the equation in Green’s Theorem. Conversely, by reversing these steps we can rewrite the circulation-curl form of Green’s Theorem for twodimensional fields in del notation as

996

Chapter 15 Integrals and Vector Fields

Green:

DC F ⋅ dr

k Curl

R

Stokes: n S

Circulation

(5)

R

See Figure 15.60.

Solution The hemisphere looks much like the surface in Figure 15.59 with the bounding circle C in the xy-plane (see Figure 15.61). We calculate the counterclockwise circulation around C (as viewed from above) using the parametrization r (θ) = ( 3 cos θ ) i + ( 3 sin θ ) j, 0 ≤ θ ≤ 2π:

d r = ( −3 sin θ dθ ) i + ( 3 cos θ dθ ) j

FIGURE 15.60

When applied to curves and surfaces in the plane, Stokes’ Theorem gives the circulation-curl version of Green’s Theorem. But Stokes’ Theorem also applies more generally, to curves and surfaces not lying in the plane.

F = yi − xj = ( 3 sin θ ) i − ( 3 cos θ ) j F ⋅ d r = −9 sin 2 θ dθ − 9 cos 2 θ dθ = −9 dθ

DC

F ⋅ dr =

2π

∫0

−9 dθ = −18π.

When evaluating the right side of Equation (4), we choose the orientation of the unit normal vector so that it points away from the origin, giving it a right-hand relation to the prescribed orientation of the curve C (see Figure 15.61). We have

z x2 + y2 + z2 = 9

∫∫ ( ∇ × F ) ⋅ k dA.

EXAMPLE 2 Evaluate both sides of Equation (4) for the hemisphere S:  x 2 + y 2 + z 2 = 9,  z ≥ 0; its bounding circle C:  x 2 + y 2 = 9,  z = 0, traversed counterclockwise (when viewed from above); and the field F = yi − xj.

Circulation

Curl

=

n

(

)

⎛ ∂P ⎛ ∂N ∂ N ⎞⎟ ∂M ∂P ∂ M ⎞⎟ ∇ × F = ⎜⎜ − − j + ⎜⎜ − ⎟i + ⎟k ⎝ ∂y ⎝ ∂x ∂ z ⎟⎠ ∂z ∂x ∂ y ⎟⎠

k y C: x 2

+

y2

= ( 0 − 0 ) i + ( 0 − 0 ) j + (−1 − 1) k = −2k

=9

x i + y j + zk x i + y j + zk = 3 x 2 + y2 + z 2

n=

x

FIGURE 15.61

A hemisphere and a disk, each with boundary C (Examples 2 and 3).

dσ = (∇

Section 15.6, Example 7, with a = 3

3 dA z

× F ) ⋅ n dσ = (−2k) ×

Unit normal vector

( xi + y3j + zk ) dσ = − 23z 3z dA = −2 dA

and

∫∫ ( ∇ × F ) ⋅ n dσ S

=

∫∫

−2 dA = −18π.

x 2 + y 2 ≤9

The circulation around the circle equals the integral of the curl over the hemisphere, as it should from Stokes’ Theorem. The surface integral in Stokes’ Theorem can be computed using any surface having boundary curve C, provided the surface is properly oriented and lies within the domain of the field F. The next example illustrates this fact for the circulation around the curve C in Example 2. EXAMPLE 3 Calculate the circulation around the bounding circle C in Example 2, using the disk of radius 3 centered at the origin in the xy-plane as the surface S (instead of the hemisphere). See Figure 15.61. Solution As in Example 2, ∇ × F = −2k. When the surface is the described disk in the xy-plane, we have the normal vector n = k, chosen to give a counterclockwise direction for C as required by Stokes’ Theorem, so that (∇

× F ) ⋅ n dσ = −2k ⋅ k dA = −2 dA

15.7  Stokes’ Theorem

z

997

and C:

x2

+

y2

= 4, z = 2

∫∫ ( ∇ × F ) ⋅ n dσ

=

S

n

∫∫

−2 dA = −18π ,

x 2 + y 2 ≤9

a simpler calculation than before.

S: r(r, u) = (r cos u)i + (r sin u)j + rk x

FIGURE 15.62

in Example 4.

y

The curve C and cone S

EXAMPLE 4 Find the circulation of the field F = ( x 2 − y ) i + 4 zj + x 2 k around the curve C in which the plane z = 2 meets the cone z = x 2 + y 2 , counterclockwise as viewed from above (Figure 15.62). Solution Stokes’ Theorem enables us to find the circulation by integrating over the surface of the cone. Traversing C in the counterclockwise direction viewed from above corresponds to taking the inner normal n to the cone, the normal with a positive k-component. We parametrize the cone as

r ( r , θ ) = ( r cos θ ) i + ( r sin θ ) j + rk,

0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π.

We then have n= =

−( r cos θ ) i − ( r sin θ ) j + rk rr × rθ = rr × rθ r 2

Section 15.5, Example 4

1 (−( cos θ ) i − ( sin θ ) j + k ) 2

dσ = r 2 dr dθ

Section 15.5, Example 4

∇ × F = −4 i − 2 xj + k

Computation of curl

= −4 i − 2r cos θ j + k.

x = r cos θ

Accordingly, (∇

1 ( 4 cos θ + 2r cos θ sin θ + 1) 2 1 = ( 4 cos θ + r sin 2θ + 1) , 2

× F) ⋅ n =

and the circulation is ( ∇ × F ) ⋅ n dσ DC F ⋅ dr = ∫∫ S

=

2π

2

∫0 ∫0

Stokes’ Theorem, Eq. (4)

1 ( 4 cos θ + r sin 2θ + 1)( r 2 dr dθ ) = 4 π. 2

EXAMPLE 5 The cone used in Example 4 is not the easiest surface to use for calculating the circulation around the bounding circle C lying in the plane z = 2. If instead we use the flat disk of radius 2 centered on the z-axis and lying in the plane z = 2, then the normal vector to the surface S is n = k (chosen to give a counterclockwise direction for the curve C). Just as in the computation for Example 4, we still have ∇ × F = −4 i − 2 xj + k. However, now we get ( ∇ × F ) ⋅ n = 1, so that

∫∫ ( ∇ × F ) ⋅ n dσ S

=

∫∫

1 dA = 4 π.

The shadow is the disk of radius 2 in the xy-plane.

x 2 + y 2 ≤4

This result agrees with the circulation value found in Example 4.

998

Chapter 15 Integrals and Vector Fields

z

EXAMPLE 6 Find a parametrization for the surface S formed by the part of the hyperbolic paraboloid z = y 2 − x 2 lying inside the cylinder of radius one around the z-axis and for the boundary curve C of S. (See Figure 15.63.) Then verify Stokes’ Theorem for S using the normal having positive k-component and the vector field F = yi − xj + x 2 k.

1 n

S

−1

1

1

Solution As the unit circle is traversed in the xy-plane, the z-coordinate of the surface with the curve C as boundary is given by y 2 − x 2. We choose the orientation for the curve C to be counterclockwise when viewed from above (see Figure 15.63). A parametrization of C is given by

y

x C

r (t ) = ( cos t ) i + ( sin t ) j + ( sin 2 t − cos 2 t ) k, 0 ≤ t ≤ 2π z

with dr = ( − sin t ) i + ( cos t ) j + ( 4 sin t cos t ) k, 0 ≤ t ≤ 2π. dt

1 S

n

Along the curve r(t ) the formula for the vector field F is F = ( sin t ) i − ( cos t ) j + ( cos 2 t ) k.

y

The counterclockwise circulation along C is the value of the line integral x C

2π

∫0

F⋅

FIGURE 15.63

2π

dr dt = dt

∫0

=

∫0

The surface and vector field for Example 6.

(− sin 2 t − cos 2 t + 4 sin t cos 3 t ) dt

2π

( 4 sin t cos 3 t − 1) dt 2π

⎤ ⎡ = ⎢ − cos 4 t − t ⎥ = −2π. ⎣ ⎦0 We now compute the same quantity by integrating ( ∇ × F ) ⋅ n over the surface S. We use polar coordinates and parametrize S by noting that above the point ( r, θ ) in the plane, the z–coordinate of S is y 2 − x 2 = r 2 sin 2 θ − r 2 cos 2 θ. A parametrization of S is r ( r , θ ) = ( r cos θ ) i + ( r sin θ ) j + r 2 ( sin 2 θ − cos 2 θ ) k, 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π. We next compute ( ∇ × F ) ⋅ n dσ. We have

∇×F =

i

j

∂ ∂x y

∂ ∂y −x

k ∂ = −2 xj − 2k = −( 2r cos θ ) j − 2k ∂z x2

and rr = ( cos θ ) i + ( sin θ ) j + 2r ( sin 2 θ − cos 2 θ ) k rθ = ( −r sin θ ) i + ( r cos θ ) j + 4r 2 ( sin θ cos θ ) k i rr × rθ =

j

cos θ

sin θ

−r sin θ

r cos θ

k 2r ( sin 2 θ

− cos 2 θ )

4r 2 ( sin θ cos θ )

= 2r 2 ( 2 sin 2 θ cos θ − sin 2 θ cos θ + cos 3 θ ) i − 2r 2 ( 2 sin θ cos 2 θ + sin 3 θ + sin θ cos 2 θ ) j + rk. Note that the k-component is always nonnegative. Therefore, we take n = +

( rr × rθ )

rr × rθ

.

15.7  Stokes’ Theorem

999

We now obtain 2π

1

2π

1

2π

1

∫∫ ( ∇ × F ) ⋅ n dσ = ∫0 ∫0 ( ∇ × F ) ⋅ S

rr × rθ r × rθ dr dθ rr × rθ r

=

∫0 ∫0 ( ∇ × F ) ⋅ ( rr × rθ ) dr dθ

=

∫0 ∫0 [ 4r 3 ( 2 sin θ cos 3 θ + sin 3 θ   cos θ + sin θ cos 3 θ ) − 2r ] dr dθ

=

∫0

=

∫0

2π

2π

⎤ r =1 ⎡ 4 dθ ⎢ r ( 3 sin θ cos 3 θ + sin 3 θ cos θ ) − r 2 ⎥ ⎢⎣ ⎥⎦ r = 0

Integrate.

( 3 sin θ cos 3 θ + sin 3 θ cos θ − 1) dθ

Evaluate.

2π 3 1 = ⎡⎢ − cos 4 θ + sin 4 θ − θ ⎤⎥ ⎣ 4 ⎦0 4

(

= −

)

3 3 + 0 − 2π + − 0 + 0 = −2π. 4 4

So the surface integral of ( ∇ × F ) ⋅ n over S equals the counterclockwise circulation of F along C, as asserted by Stokes’ Theorem. Sphere x 2 + y2 + z2 = 1

z

Circle C in the 1 plane z =

EXAMPLE 7

"2

F = ( x 2 + z ) i + ( y 2 + 2x ) j + ( z 2 − y )k

1

1

along the curve of intersection of the sphere x 2 + y 2 + z 2 = 1 with the cone z = x 2 + y 2 traversed in the counterclockwise direction around the z-axis when viewed from above. 1

y

x Cone z = "x 2 + y 2

FIGURE 15.64

Example 7.

Calculate the circulation of the vector field

Circulation curve C in

Solution The sphere and cone intersect when 1 = ( x 2 + y 2 ) + z 2 = z 2 + z 2 = 2 z 2, or z = 1 2 (see Figure 15.64). We apply Stokes’ Theorem to the curve of intersection x 2 + y 2 = 1 2 considered as the boundary of the enclosed disk in the plane z = 1 2. The normal vector to the surface that gives a counterclockwise orientation to the boundary curve is then n = k. We calculate the curl vector as

∇×F =

i

j

∂ ∂x x2 + z

∂ ∂y y 2 + 2x

k ∂ = −i + j + 2k, ∂z z2 − y

so that ( ∇ × F ) ⋅ k = 2. The circulation around the disk is ( ∇ × F ) ⋅ k dσ DC F ⋅ dr = ∫∫ S

=

∫∫

2 dσ = 2 ⋅ area of disk = 2 ⋅ π

S

( 12 )

2

= π.

Paddle Wheel Interpretation of ∇ × F Suppose that F is the velocity field of a fluid moving in a region R in space containing the closed curve C. Then

DC F ⋅ dr

1000

Chapter 15 Integrals and Vector Fields

is the circulation of the fluid around C. By Stokes’ Theorem, the circulation is equal to the flux of ∇ × F through any suitably oriented surface S with boundary C:

DC F ⋅ dr

Curl F

=

∫∫ ( ∇ × F ) ⋅ n dσ. S

Suppose we fix a point Q in the region R and a direction u at Q. Take C to be a circle of radius ρ, with center at Q, whose plane is normal to u. If ∇ × F is continuous at Q, the average value of the u-component of ∇ × F over the circular disk S bounded by C approaches the u-component of ∇ × F at Q as the radius ρ → 0:

Q

(( ∇ × F ) ⋅ u )

= lim

ρ→ 0

Q

1 πρ 2

∫∫ (∇ × F) ⋅ u dσ. S

If we apply Stokes’ Theorem and replace the surface integral by a line integral over C, we get

FIGURE 15.65

A small paddle wheel in a fluid spins fastest at point Q when its axle points in the direction of curl F.

(( ∇ × F ) ⋅ u )

= lim

ρ→0

Q

z

1 πρ 2

DC F ⋅ dr.

(6)

The left-hand side of Equation (6) has its maximum value when u is the direction of ∇ × F. When ρ is small, the limit on the right-hand side of Equation (6) is approximately

v

1 πρ 2

P(x, y, z)

DC F ⋅ dr,

which is the circulation around C divided by the area of the disk (circulation density). Suppose that a small paddle wheel of radius ρ is introduced into the fluid at Q, with its axle directed along u (Figure 15.65). The circulation of the fluid around C affects the rate of spin of the paddle wheel. The wheel spins fastest when the circulation integral is maximized; therefore, it spins fastest when the axle of the paddle wheel points in the direction of ∇ × F. F = v(−yi + xj)

0 r

y

P(x, y, 0) x

FIGURE 15.66

A steady rotational flow parallel to the xy-plane, with constant angular velocity ω in the positive (counterclockwise) direction (Example 8). z

2x + y + z = 2 R

= ( 0 − 0 ) i + ( 0 − 0 ) j + ( ω − (−ω ) ) k = 2ωk,

=

∫∫ ( ∇ × F ) ⋅ n dσ

=

S

∫∫

2ωk ⋅ k dx dy = (2ω )(πρ 2 ).

S

y = 2 − 2x

The planar surface in

(∇

× F ) ⋅ k = 2ω =

1 πρ 2

DC F ⋅ dr,

which is consistent with Equation (6) when u = k.

(0, 2, 0)

Example 9.

)

Solving this last equation for 2ω , we see that

C

FIGURE 15.67

(

⎛ ∂N ⎛ ∂P ∂M ∂P ∂ M ⎞⎟ ∂ N ⎞⎟ i+ − − ∇ × F = ⎜⎜ − j + ⎜⎜ ⎟k ⎟ ⎟ ⎝ ∂x ⎝ ∂y ∂z ∂x ∂ y ⎟⎠ ∂z ⎠

DC F ⋅ dr

n

x

Solution With F = −ω yi + ω xj, we find the curl

and therefore ( ∇ × F ) ⋅ k = 2ω . By Stokes’ Theorem, the circulation of F around a circle C of radius ρ (traversed counterclockwise when viewed from above) bounding a disk S in a plane normal to ∇ × F, say the xy-plane, is (0, 0, 2)

(1, 0, 0)

EXAMPLE 8 A fluid of constant density rotates around the z-axis with velocity F = ω (−yi + xj ) , where ω is a positive constant called the angular velocity of the rotation (Figure 15.66). Find ∇ × F and relate it to the circulation density.

y

EXAMPLE 9 Use Stokes’ Theorem to evaluate ∫ C F ⋅ d r, if F = xzi + xyj + 3 xzk and C is the boundary of the portion of the plane 2 x + y + z = 2 in the first octant, traversed counterclockwise as viewed from above (Figure 15.67).

15.7  Stokes’ Theorem

f ( x , y, z ) = 2

Solution The plane is the level surface f ( x , y, z ) = 2 x + y + z. The unit normal vector

n =

of

the

1001

function

( 2i + j + k ) ∇f 1 ( = = 2i + j + k ) 2i + j + k ∇f 6

is consistent with the counterclockwise motion around C. To apply Stokes’ Theorem, we find i curl  F = ∇ × F =

∂ ∂x xz

j

k

∂ ∂ = ( x − 3z ) j + yk. ∂ y ∂z xy 3 xz

On the plane, z equals 2 − 2 x − y, so ∇ × F = ( x − 3( 2 − 2 x − y )) j + y k = ( 7 x + 3 y − 6 ) j + y k and (∇

× F) ⋅ n =

1 ( 1 ( 7x + 3y − 6 + y ) = 7 x + 4 y − 6 ). 6 6

The surface area differential is dσ =

∇f 6 dA = dx dy. 1 ∇f ⋅ k

Formula (7) in Section 15.5

The circulation is ( ∇ × F ) ⋅ n dσ DC F ⋅ dr = ∫∫ S

z

C: x2 + 4y2 = 1, z = 1

z = x2 + 4y2

FIGURE 15.68

2− 2 x

1

2− 2 x

∫0 ∫0

=

∫0 ∫0

1 ( 7 x + 4 y − 6 ) 6 dy dx 6 ( 7x

+ 4 y − 6 ) dy dx = −1.

EXAMPLE 10 Let the surface S be the elliptic paraboloid z = x 2 + 4 y 2 lying beneath the plane z = 1 (Figure 15.68). We define the orientation of S by taking the inner normal vector n to the surface, which is the normal having a positive k-component. Find the flux of ∇ × F across S in the direction n for the vector field F = yi − xzj + xz 2 k.

n

x

1

=

Stokes’ Theorem, Eq. (4)

y

The portion of the elliptic paraboloid in Example 10, showing its curve of intersection C with the plane z = 1 and its inner normal orientation by n.

Solution We use Stokes’ Theorem to calculate the curl integral by finding the equivalent counterclockwise circulation of F around the curve of intersection C of the paraboloid z = x 2 + 4 y 2 and the plane z = 1, as shown in Figure 15.68. Note that the orientation of S is consistent with traversing C in a counterclockwise direction around the z-axis. The curve C is the ellipse x 2 + 4 y 2 = 1 in the plane z = 1. We can parametrize the ellipse by x = cos t ,  y = 12 sin t ,  z = 1 for 0 ≤ t ≤ 2π , so C is given by

1 r (t ) = ( cos t ) i + ( sin t ) j + k, 2

0 ≤ t ≤ 2π.

To compute the circulation integral DC F ⋅ d r, we evaluate F along C and find the velocity vector d r dt: F ( r (t ) ) =

1 ( sin t ) i − ( cos t ) j + ( cos t ) k 2

1002

Chapter 15 Integrals and Vector Fields

and dr 1 = −( sin t ) i + ( cos t ) j. dt 2 Then

DC

2π

F ⋅ dr =

∫0

=

∫0

=−

2π

F ( r (t ) ) ⋅

dr dt dt

(− 12 sin

t−

2

)

1 cos 2 t dt 2

1 2π dt = −π. 2 ∫0

Therefore, by Stokes’ Theorem the flux of the curl across S in the direction n for the field F is

∫∫ ( ∇ × F ) ⋅ n dσ

= −π.

S

E

Proof Outline of Stokes’ Theorem for Polyhedral Surfaces

A

D

Let S be a polyhedral surface consisting of a finite number of plane regions or faces. (See Figure 15.69 for examples.) We apply Green’s Theorem to each separate face of S. There are two types of faces: 1. Those that are surrounded on all sides by other faces. 2. Those that have one or more edges that are not adjacent to other faces.

B

C (a)

The boundary of S consists of those edges of the type 2 faces that are not adjacent to other faces. In Figure 15.69a, the triangles EAB, BCE, and CDE represent a part of S, with ABCD part of the boundary of the surface, boundary(S). Although Green’s Theorem was stated for curves in the xy-plane, a generalized form applies to curves that lie in a plane in space. In the generalized form, the theorem asserts that the line integral of F around the curve enclosing the plane region R normal to n equals the double integral of ( curl  F ) ⋅ n over R. Applying this generalized form to the three triangles of Figure 15.69a in turn, and adding the results, gives ⎛ ⎞⎟ ⎛ ⎜⎜ ⎜ ⎟ ⎜⎜ D + D + D ⎟⎟⎟ F ⋅ d r = ⎜⎜⎜ ∫∫ + ∫∫ + ∫∫ ⎝ EAB ⎝ EAB BCE BCE CDE CDE ⎠

⎞⎟ ⎟⎟ ( ∇ × F ) ⋅ n dσ. ⎟⎠

(7)

(b)

FIGURE 15.69

(a) Part of a polyhedral surface. (b) Other polyhedral surfaces.

The three line integrals on the left-hand side of Equation (7) combine into a single line integral taken around the periphery ABCDE because the integrals along interior segments cancel in pairs. For example, the integral along segment BE in triangle ABE is opposite in sign to the integral along the same segment in triangle EBC. The same holds for segment CE. Hence, Equation (7) reduces to

D

F ⋅ dr =

ABCDE

∫∫

(∇

× F ) ⋅ n dσ.

ABCDE

When we apply Green’s Theorem to all the faces and add the results, we get F ⋅ dr D boundary( S )

=

∫∫ ( ∇ × F ) ⋅ n dσ. S

15.7  Stokes’ Theorem

1003

This is Stokes’ Theorem for the polyhedral surface S in Figure 15.69a. More general polyhedral surfaces are shown in Figure 15.69b, and the proof can be extended to them. General smooth surfaces can be obtained as limits of polyhedral surfaces.

Stokes’ Theorem for Surfaces with Holes n

Stokes’ Theorem holds for an oriented surface S that has one or more holes (Figure 15.70). The surface integral over S of the normal component of ∇ × F equals the sum of the line integrals around all the boundary curves of the tangential component of F, where the curves are to be traced in the direction induced by the orientation of S. For such surfaces the theorem is unchanged, but C is considered as a union of simple closed curves.

An Important Identity S

The following identity arises frequently in mathematics and the physical sciences.

FIGURE 15.70

Stokes’ Theorem also holds for oriented surfaces with holes. Consistent with the orientation of S, the outer curve is traversed counterclockwise around n, and the inner curves surrounding the holes are traversed clockwise.

curl grad  f = 0

∇ × ∇f = 0

(8)

Forces arising in the study of electromagnetism and gravity are often associated with a potential function f . The identity (8) says that these forces have curl equal to zero. The identity (8) holds for any function f ( x , y, z ) whose second partial derivatives are continuous. The proof goes like this:

∇ × ∇f =

C S

or

i

j

∂ ∂x ∂f ∂x

∂ ∂y ∂f ∂y

k ∂ = ( f zy − f yz ) i − ( f zx − f xz ) j + ( f yx − f xy ) k. ∂z ∂f ∂z

If the second partial derivatives are continuous, the mixed second derivatives in parentheses are equal (Theorem 2, Section 13.3) and the vector is zero. (a)

Conservative Fields and Stokes’ Theorem In Section 15.3, we found that a field F being conservative in an open region D in space is equivalent to the integral of F around every closed loop in D being zero. This, in turn, is equivalent in simply connected open regions to saying that ∇ × F = 0 (which gives a test for determining whether F is conservative for such regions).

THEOREM 7—Curl F = 0 Related to the Closed-Loop Property

If ∇ × F = 0 at every point of a simply connected open region D in space, then on any piecewise-smooth closed path C in D, (b)

FIGURE 15.71

(a) In a simply connected open region in space, a simple closed curve C is the boundary of a smooth surface S. (b) Smooth curves that cross themselves can be divided into loops to which Stokes’ Theorem applies.

DC F ⋅ dr

= 0.

Sketch of a Proof Theorem 7 can be proved in two steps. The first step is for simple closed curves (loops that do not cross themselves), like the one in Figure 15.71a. A theorem from topology, a branch of advanced mathematics, states that every smooth simple closed

1004

Chapter 15 Integrals and Vector Fields

curve C in a simply connected open region D is the boundary of a smooth two-sided surface S that also lies in D. Hence, by Stokes’ Theorem,

DC F ⋅ dr

=

∫∫ ( ∇ × F ) ⋅ n dσ

= 0.

S

The second step is for curves that cross themselves, like the one in Figure 15.71b. The idea is to break these into simple loops spanned by orientable surfaces, apply Stokes’ Theorem one loop at a time, and add the results. The following diagram summarizes the results for conservative fields defined on connected, simply connected open regions. For such regions, the four statements are equivalent to each other. Theorem 2, Section 15.3

F conservative on D

F = ∇f on D Vector identity (Eq. 8) (continuous second partial derivatives)

Theorem 3, Section 15.3

FC

over any closed path in D

EXERCISES

1. F = ( x + y − z ) i + ( 2 x − y + 3z ) j + ( 3 x + 2 y + z ) k 2. F =

− y)i +

Theorem 7 Domain’s simple connectivity and Stokes’ Theorem

15.7

In Exercises 1–6, find the curl of each vector field F. (x2

∇ × F = 0 throughout D

F ∙ dr = 0

( y2

− z)j +

(z2

− x )k

3. F = ( xy + z ) i + ( yz + x ) j + ( xz + y ) k 4. F = ye z i + ze x j − xe y k 5. F = x 2 yzi + xy 2 z j + xyz 2 k y x z 6. F = i− j+ k yz xz xy

11. F = ( y 2 + z 2 ) i + ( x 2 + y 2 ) j + ( x 2 + y 2 ) k C: The square bounded by the lines x = ±1 and y = ±1 in the xy-plane, counterclockwise when viewed from above 12. F = x 2 y 3 i + j + zk C: The intersection of the cylinder x 2 + y 2 = 4 and the hemisphere x 2 + y 2 + z 2 = 16, z ≥ 0, counterclockwise when viewed from above Integral of the Curl Vector Field

Using Stokes’ Theorem to Find Line Integrals

In Exercises 7–12, use the surface integral in Stokes’ Theorem to calculate the circulation of the field F around the curve C in the indicated direction. 7. F = x 2 i + 2 xj + z 2 k C: The ellipse + = 4 in the xy-plane, counterclockwise when viewed from above 4x 2

y2

8. F = 2 yi + 3 xj − z 2 k C: The circle x 2 + y 2 = 9 in the xy-plane, counterclockwise when viewed from above 9. F = yi + xzj + x 2 k C: The boundary of the triangle cut from the plane x + y + z = 1 by the first octant, counterclockwise when viewed from above 10. F = ( y 2 + z 2 ) i + ( x 2 + z 2 ) j + ( x 2 + y 2 ) k C: The boundary of the triangle cut from the plane x + y + z = 1 by the first octant, counterclockwise when viewed from above

13. Let n be the unit normal in the direction away from the origin of the elliptic shell S: 4 x 2 + 9 y 2 + 36 z 2 = 36,

z ≥ 0,

and let F = yi + x 2 j + ( x 2 + y 4 )3 2 sin e

xyz

k.

Find the value of

∫∫ ( ∇ × F ) ⋅ n dσ. S

(Hint: One parametrization of the ellipse at the base of the shell is x = 3 cos t , y = 2 sin t , 0 ≤ t ≤ 2π.) 14. Let n be the unit normal in the direction away from the origin of the parabolic shell S: 4 x 2 + y + z 2 = 4,

y ≥ 0,

15.7  Stokes’ Theorem

and let

1005

Theory and Examples

(

F = −z +

)

(

)

1 1 i + ( tan −1 y ) j + x + k. 2+ x 4+z

Find the value of

∫∫ ( ∇ × F ) ⋅ n dσ. S

15. Let S be the cylinder x 2 + y 2 = a 2, 0 ≤ z ≤ h, together with its top, x 2 + y 2 ≤ a 2,  z = h. Let F = − yi + xj + x 2 k. Use Stokes’ Theorem to find the flux of ∇ × F through S in the direction away from the origin.

25. Let C be the smooth curve r (t ) = ( 2 cos t ) i + ( 2 sin t ) j  + ( 3 − 2 cos 3 t ) k, oriented to be traversed counterclockwise around the z-axis when viewed from above. Let S be the piecewise smooth cylindrical surface x 2 + y 2 = 4, below the curve for z ≥ 0, together with the base disk in the xy-plane. Note that C lies on the cylinder S and above the xy-plane (see the accompanying figure). Verify Equation (4) in Stokes’ Theorem for the vector field F = yi − xj + x 2 k. z

C

16. Evaluate

∫∫ ( ∇ × ( yi )) ⋅ n dσ, S

where S is the hemisphere x 2 + y 2 + z 2 = 1,  z ≥ 0, in the direction away from the origin.

y

2

2

17. Suppose F = ∇ × A, where A = ( y + z 2 ) i + e xyz j + cos( xz ) k. Determine the flux of F through the hemisphere x 2 + y 2 + z 2 = 1,  z ≥ 0, in the direction away from the origin. 18. Repeat Exercise 17 for the flux of F across the entire unit sphere. Stokes’ Theorem for Parametrized Surfaces

In Exercises 19–24, use the surface integral in Stokes’ Theorem to calculate the flux of the curl of the field F across the surface S. 19. F = 2 zi + 3 xj + 5 yk S: r ( r , θ ) = ( r cos θ ) i + ( r sin θ ) j + ( 4 − r 2 ) k, 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π , in the direction away from the origin. 20. F = ( y − z ) i + ( z − x ) j + ( x + z ) k S: r ( r , θ ) = ( r cos θ ) i + ( r sin θ ) j + ( 9 − r 2 ) k, 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π , in the direction away from the origin.

x

26. Verify Stokes’ Theorem for the vector field F = 2 xyi + xj  + ( y + z ) k and surface z = 4 − x 2 − y 2 ,  z ≥ 0, oriented with unit normal n pointing upward. 27. Zero circulation Use Equation (8) and Stokes’ Theorem to show that the circulations of the following fields around the boundary of any smooth orientable surface in space are zero. a. F = 2 xi + 2 yj + 2 zk

b. F = ∇( xy 2 z 3 )

c. F = ∇ × ( xi + yj + zk )

d. F = ∇f

28. Zero circulation Let f ( x , y, z ) = ( x 2 + y 2 + z 2 )−1 2. Show that the clockwise circulation of the field F = ∇f around the circle x 2 + y 2 = a 2 in the xy-plane is zero a. by taking r = ( a cos t ) i + ( a sin t ) j, 0 ≤ t ≤ 2π , and integrating F ⋅ d r over the circle. b. by applying Stokes’ Theorem. 29. Let C be a simple closed smooth curve in the plane 2 x + 2 y + z = 2, oriented as shown here. Show that

DC 2 y dx + 3z dy − x dz

21. F = x 2 yi + 2 y 3 zj + 3zk S: r ( r , θ ) = ( r cos θ ) i + ( r sin θ ) j + rk,

z

0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π , in the direction away from the z-axis.

2x + 2y + z = 2

2 C

22. F = ( x − y ) i + ( y − z ) j + ( z − x ) k S: r ( r , θ ) = ( r cos θ ) i + ( r sin θ ) j + ( 5 − r ) k, 0 ≤ r ≤ 5, 0 ≤ θ ≤ 2π , in the direction away from the z-axis. 23. F = 3 yi + ( 5 − 2 x ) j + ( z 2 − 2 ) k S: r ( φ, θ ) = ( 3 sin φ cos θ ) i + ( 3 sin φ sin θ ) j +

( 3 cos φ ) k, 0 ≤ φ ≤ π 2, 0 ≤ θ ≤ 2π, in the direction away from the origin. 24. F = y 2 i + z 2 j + xk

O

a1

y

1 x

depends only on the area of the region enclosed by C and not on the position or shape of C. 30. Show that if F = xi + yj + zk, then ∇ × F = 0.

S: r ( φ, θ ) = ( 2 sin φ cos θ ) i + ( 2 sin φ sin θ ) j + ( 2 cos φ ) k,

31. Find a vector field with twice-differentiable components whose curl is xi + yj + zk, or prove that no such field exists.

0 ≤ φ ≤ π 2, 0 ≤ θ ≤ 2π , in the direction away from the origin.

32. Does Stokes’ Theorem say anything special about circulation in a field whose curl is zero? Give reasons for your answer.

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Chapter 15 Integrals and Vector Fields

33. Let R be a region in the xy-plane that is bounded by a piecewise smooth simple closed curve C, and suppose that the density is δ = 1 and the moments of inertia of R about the x- and y-axes are known to be I x and I y . Evaluate the integral

DC where r =

∇(r 4 ) ⋅ n ds,

x 2 + y 2 , in terms of I x and I y .

34. Zero curl, yet the field is not conservative Show that the curl of F =

is zero but that

DC F ⋅ dr is not zero if C is the circle x 2 + y 2 = 1 in the xy-plane. (Theorem 7 does not apply here because the domain of F is not simply connected. The field F is not defined along the z-axis, so there is no way to contract C to a point without leaving the domain of F.)

−y x i+ 2 j + zk x 2 + y2 x + y2

15.8 The Divergence Theorem and a Unified Theory The divergence form of Green’s Theorem in the plane states that the net outward flux of a vector field across a simple closed curve can be calculated by integrating the divergence of the field over the region enclosed by the curve. The corresponding theorem in three dimensions, called the Divergence Theorem, states that the net outward flux of a vector field across a closed surface in space can be calculated by integrating the divergence of the field over the solid region enclosed by the surface. In this section we prove the Divergence Theorem and show how it simplifies the calculation of flux, which is the integral of the field over the closed oriented surface. We also derive Gauss’s law for flux in an electric field and the continuity equation of hydrodynamics. Finally, we summarize the chapter’s vector integral theorems in a single unifying principle generalizing the Fundamental Theorem of Calculus.

Divergence in Three Dimensions The divergence of a vector field F = M ( x , y, z ) i + N ( x , y, z ) j + P ( x , y, z ) k is the scalar function div F = ∇ ⋅ F =

∂M ∂N ∂P + + . ∂x ∂y ∂z

(1)

The symbol “div F” is read as “divergence of F” or “div F.” The notation ∇ ⋅ F is read “del dot F.” Div F has the same physical interpretation in three dimensions as it has in two. If F is the velocity field of a flowing gas, the value of div F at a point ( x , y, z ) is the rate at which the gas is compressing or expanding at ( x , y, z ). The gas is expanding if div F is positive and compressing if div F is negative. The divergence is the flux per unit volume, or flux density, at the point. EXAMPLE 1 The following vector fields represent the velocity of a gas flowing in space. Find the divergence of each vector field and interpret its physical meaning. Figure 15.72 displays the vector fields. (a) Expansion: F ( x , y, z ) = xi + yj + zk (b) Compression: F ( x , y, z ) = −xi − yj − zk (c) Rotation about the z-axis: F ( x , y, z ) = −yi + xj (d) Shearing along parallel horizontal planes: F ( x , y, z ) = zj Solution

∂ ∂( ) ∂ (x) + y + ( z ) = 3: The gas is undergoing constant uniform ∂x ∂y ∂z expansion at all points.

(a) div F =

15.8  The Divergence Theorem and a Unified Theory

1007

∂ ∂ ∂ (−x ) + (−y) + (−z ) = −3: The gas is undergoing constant uni∂x ∂y ∂z form compression at all points.

(b) div F =

z

(c) div F = y

∂ ∂ (−y) + ( x ) = 0: The gas is neither expanding nor compressing at any ∂x ∂y

point. ∂ ( z ) = 0: Again, the divergence is zero at all points in the domain of the ∂y velocity field, so the gas is neither expanding nor compressing at any point.

x

(d) div F = (a)

Divergence Theorem

z

The Divergence Theorem says that under suitable conditions, the outward flux of a vector field across a closed surface equals the triple integral of the divergence of the field over the three-dimensional region enclosed by the surface. y x

THEOREM 8—Divergence Theorem

Let F be a vector field whose components have continuous first partial derivatives, and let S be a piecewise smooth oriented closed surface. The flux of F across S in the direction of the surface’s outward unit normal field n equals the triple integral of the divergence ∇ ⋅ F over the solid region D enclosed by the surface:

(b) z

∫∫

F ⋅ n dσ =

S

∫∫∫ ∇ ⋅ F dV .

(2)

D

Outward flux

Divergence integral

y x

EXAMPLE 2 Evaluate both sides of Equation (2) for the expanding vector field F = xi + yj + zk over the sphere x 2 + y 2 + z 2 = a 2 (Figure 15.73). (c)

Solution The outer unit normal to S, calculated from the gradient of f ( x , y, z ) = x 2 + y 2 + z 2 − a 2, is

z

n = y

2 ( x i + y j + zk ) x i + y j + zk . = 2 2 2 a 4( x + y + z )

x 2 + y 2 + z 2 = a 2  on  S

It follows that F ⋅ n dσ =

x

x 2 + y2 + z 2 a2 dσ = a dσ. dσ = a a

Therefore, the outward flux is (d)

FIGURE 15.72

Velocity fields of a gas flowing in space (Example 1).

∫∫

F ⋅ n dσ =

S

∫∫ a dσ S

= a ∫∫ d σ = a(4 πa 2 ) = 4 πa 3.

Area of S is 4 πa 2.

S

For the right-hand side of Equation (2), the divergence of F is ∇⋅F =

∂ ∂( ) ∂ (x) + y + ( z ) = 3, ∂x ∂y ∂z

so we obtain the divergence integral,

∫∫∫ D

∇ ⋅ F dV =

∫∫∫ D

3 dV = 3

( 43 πa ) = 4πa . 3

3

1008

Chapter 15 Integrals and Vector Fields

z

Many vector fields of interest in applied science have zero divergence at each point. A common example is the velocity field of a circulating incompressible liquid, since it is neither expanding nor contracting. Other examples include constant vector fields F = a i + b j + ck, and velocity fields for shearing action along a fixed plane (see Example 1d). If F is a vector field whose divergence is zero at each point in the region D, then the integral on the right-hand side of Equation (2) equals 0. So if S is any closed surface for which the Divergence Theorem applies, then the outward flux of F across S is zero. We state this important application of the Divergence Theorem.

y

x

FIGURE 15.73

A uniformly expanding vector field and a sphere (Example 2).

COROLLARY The outward flux across a piecewise smooth oriented closed surface S is zero for any vector field F having zero divergence at every point of the region enclosed by the surface.

EXAMPLE 3 Find the flux of F = xyi + yzj + xzk outward through the surface of the cube cut from the first octant by the planes x = 1,  y = 1, and z = 1. Solution Instead of calculating the flux as a sum of six separate integrals, one for each face of the cube, we can calculate the flux by integrating the divergence

∂ ∂ ∂ ( xy) + ( yz ) + ( xz ) = y + z + x ∂x ∂y ∂z

∇⋅F = over the cube’s interior: Flux =

∫∫

F ⋅ n dσ =

Cube surface

=

z

1

∫∫∫

∇ ⋅ F dV

The Divergence Theorem

Cube interior 1

1

∫0 ∫0 ∫0 ( x + y + z ) dx dy dz

=

3 . 2

Routine integration

EXAMPLE 4 1

(a) Calculate the flux of the vector field F = x 2 i + 4 xyz j + ze x k 2

y

3 x

FIGURE 15.74

The integral of div F over this region equals the total flux across the six sides (Example 4).

out of the box-shaped region D: 0 ≤ x ≤ 3, 0 ≤ y ≤ 2, 0 ≤ z ≤ 1. (See Figure 15.74.) (b) Integrate div F over this region and show that the result is the same value as in part (a), as asserted by the Divergence Theorem. Solution

(a) The region D has six sides. We calculate the flux across each side in turn. Consider the top side in the plane z = 1, having outward normal n = k. The flux across this side is given by F ⋅ n = ze x. Since z = 1 on this side, the flux at a point ( x , y, z ) on the top is e x . The total outward flux across this side is given by the surface integral 2

3

∫0 ∫0 e x dx dy

= 2e 3 − 2.

Routine integration

The outward flux across the other sides is computed similarly, and the results are summarized in the following table.

15.8  The Divergence Theorem and a Unified Theory

Side

Unit normal n

x = 0

−i

x = 3

i

y = 0 y = 2

−j j

z = 0

F⋅n

  0

= 9

18

−4 xyz = 0 4 xyz = 8 xz

−k

−ze x

k

ze x

z =1

Flux across side

−x 2 = 0 x2

  0 18

= 0 =

1009

  0

ex

2e 3

−2

The total outward flux is obtained by adding the terms for each of the six sides: 18 + 18 + 2e 3 − 2 = 34 + 2e 3. (b) We first compute the divergence of F, obtaining div F = ∇ ⋅ F = 2 x + 4 xz + e x. The integral of the divergence of F over D is

∫∫∫

div F dV =

D

1

2

1

2

3

∫0 ∫0 ∫0 ( 2 x + 4 xz + e x ) dx dy dz

=

∫0 ∫0 ( 8 + 18z + e 3 ) dy dz

=

∫0 (16 + 36z + 2e 3 ) dz

1

= 34 + 2e 3 . As asserted by the Divergence Theorem, the integral of the divergence over D equals the outward flux across the boundary surface of D.

Divergence and the Curl If F is a vector field on three-dimensional space, then the curl ∇ × F is also a vector field on three-dimensional space. So we can calculate the divergence of ∇ × F using Equation (1). The result of this calculation is always 0.

If F = Mi + N j + Pk is a vector field with continuous second partial derivatives, then

THEOREM 9

div ( curl  F ) = ∇ ⋅ ( ∇ × F ) = 0.

Proof

From the definitions of the divergence and curl, we have div ( curl  F ) = ∇ ⋅ ( ∇ × F )

(

)

=

∂ ∂M ∂P ∂ ⎛⎜ ∂ N ∂ M ⎞⎟ ∂ ⎛⎜ ∂ P ∂ N ⎞⎟ − + − − ⎟ ⎟+ ∂x ∂ z ⎜⎝ ∂ x ∂ y ⎟⎠ ∂ x ⎜⎝ ∂ y ∂ z ⎟⎠ ∂ y ∂ z

=

∂2P ∂2N ∂2M ∂2P ∂2N ∂2M − + − + − ∂x ∂y ∂x ∂z ∂ y ∂z ∂y ∂x ∂z ∂x ∂z ∂ y

= 0,

1010

Chapter 15 Integrals and Vector Fields

z

because the mixed second partial derivatives cancel by the Mixed Derivative Theorem in Section 13.3. Ryz

D

Rxz

S2 S1 y x

Rxy

FIGURE 15.75

We prove the Divergence Theorem for the kind of threedimensional region shown here. z

n3

Proof of the Divergence Theorem for Special Regions To prove the Divergence Theorem, we take the components of F = Mi + N j + Pk to have continuous first partial derivatives. We first assume that D is a convex region with no holes or bubbles, such as a solid ball, cube, or ellipsoid, and that S is a piecewise smooth surface. In addition, we assume that any line perpendicular to the xy-plane at an interior point of the region R xy that is the projection of D on the xy-plane intersects the surface S in exactly two points, producing surfaces

(n1, n 2, n 3)

k g a

S 1:

n

n1

n2

j

y

x

FIGURE 15.76

z = f1 ( x , y ) ,

( x , y )  in R xy

S 2 : z = f 2 ( x , y ),

b i

Theorem 9 has some interesting applications. If a vector field G = curl  F, then the field G must have divergence 0. Saying this another way, if div G ≠ 0, then G cannot be the curl of any vector field F having continuous second partial derivatives. Moreover, if G = curl  F, then the outward flux of G across any closed surface S is zero by the corollary to the Divergence Theorem, provided the conditions of the theorem are satisfied. So if there is a closed surface for which the surface integral of the vector field G is nonzero, we can conclude that G is not the curl of some vector field F.

The components of n are the cosines of the angles α,  β , and γ that it makes with i, j, and k.

( x , y )  in R xy ,

with f1 ≤ f 2 . We make similar assumptions about the projection of D onto the other coordinate planes. See Figure 15.75, which illustrates these assumptions. The components of the unit normal vector n = n1i + n 2 j + n 3k are the cosines of the angles α,  β , and γ that n makes with i, j, and k (Figure 15.76). This is true because all the vectors involved are unit vectors, giving the direction cosines n1 = n ⋅ i = n i cos α = cos α n 2 = n ⋅ j = n j cos β = cos β n 3 = n ⋅ k = n k cos γ = cos γ. Thus the unit normal vector is given by n = ( cos α ) i + ( cos β ) j + ( cos γ ) k, and F ⋅ n = M cos α + N cos β + P cos γ. In component form, the Divergence Theorem states that

∫∫ ( M cos α + N cos β + P cos γ ) dσ S



=

⎛ ∂M

∂N ∂ P ⎞⎟ + + ⎟ dx dy dz. ∂y ∂ z ⎟⎠ 

∫∫∫ ⎜⎜⎝ ∂ x D

F⋅n

div F

We prove the theorem by establishing the following three equations:

∫∫

M cos α dσ =

∫∫

N cos β dσ =

∫∫

P cos γ dσ =

S

∂M dx dy dz ∂x

(3)

∫∫∫

∂N dx dy dz ∂y

(4)

∫∫∫

∂P dx dy dz ∂z

(5)

D

S

S

∫∫∫ D

D

15.8  The Divergence Theorem and a Unified Theory

z

n

D S2

O

Proof of Equation (5) We prove Equation (5) by converting the surface integral on the left to a double integral over the projection R xy of D on the xy-plane (Figure 15.77). The surface S consists of an upper part S 2 whose equation is z = f 2 ( x , y ) and a lower part S 1 whose equation is z = f1 ( x , y ). On S 2 , the outer normal n has a positive k-component and

z = f2(x, y)

ds

cos γ dσ = dx dy

S1 n

ds

z = f1(x, y)

cos γ dσ = −dx dy. Therefore,

dA = dx dy

∫∫

P cos γ dσ =

S

FIGURE 15.77

The region D enclosed by the surfaces S 1 and S 2 projects vertically onto R xy in the xy-plane.

=

∫∫

∫∫

P ( x , y, f 2 ( x , y )) dx dy −

∫∫

[ P ( x , y, f 2 ( x , y )) − P ( x , y, f1 ( x , y )) ] dx dy

P cos γ dσ

S1

∫∫

P ( x , y, f1 ( x , y )) dx dy

R xy

R xy

=

f 2( x , y )

1

∂P ⎤ dz ⎥ dx dy = ∂z ⎦

Here g is obtuse, so dx dy ds = − cos g .

D

∂P dz dx dy.  ∂z

Divergence Theorem for Other Regions

dx

FIGURE 15.78

An enlarged view of the area patches in Figure 15.77. The relations dσ = ±dx dy cos γ come from Eq. (7) in Section 15.5 with F = F ⋅ n.

The Divergence Theorem can be extended to regions that can be partitioned into a finite number of simple regions of the type just discussed and to regions that can be defined as limits of simpler regions in certain ways. For an example of one step in such a splitting process, suppose that D is the region between two concentric spheres and that F has continuously differentiable components throughout D and on the bounding surfaces. Split D by an equatorial plane and apply the Divergence Theorem to each half separately. The bottom half, D1 , is shown in Figure 15.79. The surface S 1 that bounds D1 consists of an outer hemisphere, a plane washer-shaped base, and an inner hemisphere. The Divergence Theorem says that

∫∫

z

F ⋅ n 1 dσ =

S1

O y n1

x

The lower half of the solid region between two concentric spheres.

∫∫∫ ∇ ⋅ F dV .

(6)

D1

The unit normal n 1 that points outward from D1 points away from the origin along the outer surface, equals k along the flat base, and points toward the origin along the inner surface. Next apply the Divergence Theorem to D 2 and its surface S 2 (Figure 15.80):

k

D1

∫∫∫

This proves Equation (5). The proofs for Equations (3) and (4) follow the same pattern; just permute x , y, z;  M , N , P;  α, β , γ , in order, and get those results from Equation (5). This proves the Divergence Theorem for these special regions.

k

FIGURE 15.79

⎡

∫∫ ⎢⎣ ∫ f ( x , y ) R xy

Here g is acute, so dx dy ds = cos g .

dy

P cos γ dσ +

R xy

n

n

∫∫ S2

=

g

dA dx dy . = cos γ cos γ

See Figure 15.78. On S 1 , the outer normal n has a negative k-component and

Rxy

k g

dσ =

because

y x

1011

∫∫ S2

F ⋅ n 2 dσ =

∫∫∫ ∇ ⋅ F dV .

(7)

D2

As we follow n 2 over S 2 , pointing outward from D 2 , we see that n 2 equals −k along the washer-shaped base in the xy-plane, points away from the origin on the outer sphere, and points toward the origin on the inner sphere. When we add Equations (6) and (7), the integrals over the flat base cancel because of the opposite signs of n 1 and n 2 . We thus arrive at the result

1012

Chapter 15 Integrals and Vector Fields

z

∫∫

F ⋅ n dσ =

S

∫∫∫ ∇ ⋅ F dV , D

with D the region between the spheres, S the boundary of D consisting of two spheres, and n the unit normal to S directed outward from D.

D2 n2

EXAMPLE 5

Find the net outward flux of the field

y

−k

F =

x

FIGURE 15.80

The upper half of the solid region between two concentric spheres.

x i + y j + zk , ρ3

ρ =

x 2 + y2 + z 2

(8)

across the boundary of the region D: 0 < b 2 ≤ x 2 + y 2 + z 2 ≤ a 2 (Figure 15.81). Solution The flux can be calculated by integrating ∇ ⋅ F over D. Note that ρ ≠ 0 in D. We have

∂ρ 1 x = ( x 2 + y 2 + z 2 )−1 2 (2 x ) = ∂x ρ 2

z

and ∂ρ ∂ 1 3x 2 ∂M = ( xρ −3 ) = ρ −3 − 3 xρ −4 = 3 − 5 . ∂x ∂x ∂x ρ ρ

Sa

Similarly, Sb

3y 2 1 ∂N = 3 − 5 ∂y ρ ρ

y

x

Hence, div F =

FIGURE 15.81

Two concentric spheres in an expanding vector field. The outer sphere S a surrounds the inner sphere S b .

∂P 1 3z 2 = 3 − 5 . ∂z ρ ρ

and

3ρ 2 ∂M ∂N ∂P 3 3 3 + + = 3 − 5 ( x 2 + y 2 + z 2 ) = 3 − 5 = 0. ∂x ∂y ∂z ρ ρ ρ ρ

So the net outward flux of F across the boundary of D is zero by the corollary to the Divergence Theorem. There is more to learn about this vector field F, though. The flux leaving D across the inner sphere S b is the negative of the flux leaving D across the outer sphere S a (because the sum of these fluxes is zero). Hence, the flux of F across S b in the direction away from the origin equals the flux of F across S a in the direction away from the origin. Thus, the flux of F across a sphere centered at the origin is independent of the radius of the sphere. What is this flux? To find it, we evaluate the flux integral directly for an arbitrary sphere S a . The outward unit normal on the sphere of radius a is n =

x i + y j + zk x i + y j + zk = . 2 2 2 a x + y +z

Hence, on the sphere, F⋅n =

x i + y j + zk x i + y j + zk x 2 + y2 + z 2 a2 1 ⋅ = = 4 = 2 3 4 a a a a a

and

∫∫ Sa

F ⋅ n dσ =

1 a2

∫∫ d σ Sa

=

1 (4 πa 2 ) = 4 π. a2

The outward flux of F in Equation (8) across any sphere centered at the origin is 4 π. This result does not contradict the Divergence Theorem because F is not continuous at the origin.

15.8  The Divergence Theorem and a Unified Theory

1013

Gauss’s Law: One of the Four Great Laws of Electromagnetic Theory In electromagnetic theory, the electric field created by a point charge q located at the origin is E ( x , y, z ) =

( ) = 4πεq

1 q r 4 πε 0 r 2 r

0

q x i + y j + zk r = , r3 4 πε 0 ρ3

where ε 0 is a physical constant, r is the position vector of the point ( x , y, z ), and ρ = r = x 2 + y 2 + z 2 . From Equation (8), q F. 4 πε 0

E = z

The calculations in Example 5 show that the outward flux of E across any sphere centered at the origin is q ε 0 , but this result is not confined to spheres. The outward flux of E across any closed surface S that encloses the origin (and to which the Divergence Theorem applies) is also q ε 0 . To see why, we have only to imagine a large sphere S a centered at the origin and enclosing the surface S (see Figure 15.82). Because

S

∇⋅E = ∇⋅

y Sphere Sa

x

q q F = ∇⋅F = 0 4 πε 0 4 πε 0

when ρ > 0, the triple integral of ∇ ⋅ E over the region D between S and S a is zero. Hence, by the Divergence Theorem,

∫∫

E ⋅ n dσ = 0.

Boundary of  D

FIGURE 15.82

A sphere S a surrounding another surface S. The tops of the surfaces are removed for visualization.

So the flux of E across S in the direction away from the origin must be the same as the flux of E across S a in the direction away from the origin, which is q ε 0 . This statement, called Gauss’s law, also applies to charge distributions that are more general than the one assumed here, as shown in most physics texts. For any closed surface that encloses the origin, we have Gauss’s law:

∫∫

E ⋅ n dσ =

S

q . ε0

Continuity Equation of Hydrodynamics Let D be a region in space bounded by a closed oriented surface S. If v( x ,  y,  z ) is the velocity field of a fluid flowing smoothly through D,  δ = δ ( t ,  x ,  y,  z ) is the fluid’s density at (x, y, z) at time t, and F = δ v, then the continuity equation of hydrodynamics states that ∂δ = 0. ∂t If the functions involved have continuous first partial derivatives, the equation evolves naturally from the Divergence Theorem, as we now demonstrate. First, the integral ∇⋅F+

n

v Δt

h = (v Δt) . n

∫∫

Δs

F ⋅ n dσ

S

S

FIGURE 15.83

The fluid that flows upward through the patch Δσ in a short time Δt fills a “cylinder” whose volume is approximately base × height = v ⋅ n Δσ Δt.

is the rate at which mass leaves D across S (leaves because n is the outer normal). To see why, consider a patch of area Δσ on the surface (Figure 15.83). In a short time interval Δt, the volume ΔV of fluid that flows across the patch is approximately equal to the volume of a cylinder with base area Δσ and height ( v Δt ) ⋅ n, where v is a velocity vector rooted at a point of the patch: ΔV ≈ v ⋅ n Δ σ Δt.

1014

Chapter 15 Integrals and Vector Fields

The mass of this volume of fluid is about Δm ≈ δ v ⋅ n Δσ Δt , so the rate at which mass is flowing out of D across the patch is about Δm ≈ δ v ⋅ n Δσ. Δt This leads to the approximation

∑ Δm

∑ δ v ⋅ n Δσ

≈

Δt

as an estimate of the average rate at which mass flows across S. Finally, letting Δσ → 0 and Δt → 0 gives the instantaneous rate at which mass leaves D across S as dm = dt

∫∫ δ v ⋅ n d σ,

dm = dt

∫∫

S

which for our particular flow is F ⋅ n d σ.

S

Now let B be a solid sphere centered at a point Q in the flow. The average value of ∇ ⋅ F over B is 1 volume of  B

∫∫∫

∇ ⋅ F dV .

B

It is a consequence of the continuity of the divergence that ∇ ⋅ F actually takes on this value at some point P in B. Thus, by the Divergence Theorem Equation (2),

1 ( ∇ ⋅ F )( P ) = volume of  B =

∫∫∫

∇ ⋅ F dV =

B

∫∫

F ⋅ n dσ

S

volume of  B

rate at which mass leaves B  across its surface  S . volume of  B

(9)

The last term of the equation describes decrease in mass per unit volume. Now let the radius of B approach zero while the center Q stays fixed. The left side of Equation (9) converges to ( ∇ ⋅ F )Q , and the right side converges to (−∂δ ∂ t )Q , since δ = m V. The equality of these two limits is the continuity equation ∇⋅F = −

∂δ . ∂t

The continuity equation “explains” ∇ ⋅ F: The divergence of F at a point is the rate at which the density of the fluid is decreasing there. The Divergence Theorem

∫∫ S

F ⋅ n dσ =

∫∫∫

∇ ⋅ F dV

D

now says that the net decrease in density of the fluid in region D (divergence integral) is accounted for by the mass transported across the surface S (outward flux integral). So, the theorem is a statement about conservation of mass (Exercise 35).

15.8  The Divergence Theorem and a Unified Theory

1015

Unifying the Integral Theorems If we think of a two-dimensional field F = M ( x , y ) i + N ( x , y ) j as a three-dimensional field whose k-component is zero, then ∇ ⋅ F = ( ∂ M ∂ x ) + ( ∂ N ∂ y ), and the normal form of Green’s Theorem can be written as

DC F ⋅ n ds

=

⎛ ∂M

∫∫ ⎜⎜⎝ ∂ x

+

R

∂ N ⎞⎟ ⎟ dx dy = ∂ y ⎟⎠

∫∫ ∇ ⋅ F dA. R

Similarly, ( ∇ × F ) ⋅ k = ( ∂ N ∂ x ) − ( ∂ M ∂ y ) , so the tangential form of Green’s Theorem can be written as

DC F ⋅ T ds

=

⎛ ∂N

∫∫ ⎜⎜⎝ ∂ x R

−

∂ M ⎞⎟ ⎟ dx dy = ∂ y ⎟⎠

∫∫ ( ∇ × F ) ⋅ k dA. R

With the equations of Green’s Theorem expressed in del notation, we can see their relationships to the equations in Stokes’ Theorem and the Divergence Theorem, all summarized here.

Green’s Theorem and Its Generalization to Three Dimensions

Tangential form of Green’s Theorem: Stokes’ Theorem: Normal form of Green’s Theorem:

DC F ⋅ T ds

=

DC F ⋅ T ds

=

DC F ⋅ n ds

=

∫∫

Divergence Theorem:

∫∫ ( ∇ × F ) ⋅ k dA R

∫∫ ( ∇ × F ) ⋅ n dσ S

∫∫ ∇ ⋅ F dA

F ⋅ n dσ =

S

R

∫∫∫ ∇ ⋅ F dV D

Notice how Stokes’ Theorem generalizes the tangential (curl) form of Green’s Theorem from a flat surface in the plane to a surface in three-dimensional space. In each case, the surface integral of curl F over the interior of the oriented surface equals the circulation of F around the boundary. Likewise, the Divergence Theorem generalizes the normal (flux) form of Green’s Theorem from a two-dimensional region in the plane to a three-dimensional region in space. In each case, the integral of ∇ ⋅ F over the interior of the region equals the total flux of the field across the boundary enclosing the region. All these results can be thought of as forms of a single fundamental theorem. The Fundamental Theorem of Calculus in Section 5.4 says that if f (x) is differentiable on ( a, b ) and continuous on [ a, b ], then b

∫a n = −i a

FIGURE 15.84

n=i b

The outward unit normals at the boundary of [ a, b ] in one-dimensional space.

x

df dx = f (b) − f (a). dx

If we let F = f ( x ) i throughout [ a, b ], then df dx = ∇ ⋅ F. If we define the unit vector field n normal to the boundary of [ a, b ] to be i at b and −i at a (Figure 15.84), then f (b) − f (a) = f (b) i ⋅ ( i) + f (a) i ⋅ (−i) = F( b ) ⋅ n + F( a ) ⋅ n = total outward flux of  F  across the boundary of [a, b].

1016

Chapter 15 Integrals and Vector Fields

The Fundamental Theorem now says that F( b ) ⋅ n + F( a ) ⋅ n =

∫

∇ ⋅ F dx.

[ a, b ]

The Fundamental Theorem of Calculus, the normal form of Green’s Theorem, and the Divergence Theorem all say that the integral of the differential operator ∇ ⋅ operating on a field F over a region equals the sum of the normal field components over the boundary enclosing the region. (Here we are interpreting the line integral in Green’s Theorem and the surface integral in the Divergence Theorem as “sums” over the boundary.) Stokes’ Theorem and the tangential form of Green’s Theorem say that, when things are properly oriented, the surface integral of the differential operator ∇ × operating on a field equals the sum of the tangential field components over the boundary of the surface. The beauty of these interpretations is the observance of a single unifying principle, which we can state as follows.

A Unifying Fundamental Theorem of Vector Integral Calculus

The integral of a differential operator acting on a field over a region equals the sum of the field components appropriate to the operator over the boundary of the region.

EXERCISES

15.8 12. Ball F = x 2 i + xzj + 3zk

Calculating Divergence

In Exercises 1–8, find the divergence of the field. 1. F = ( x − y + z ) i + ( 2 x + y − z ) j + ( 3 x + 2 y − 2 z ) k 2. F = ( x ln y ) i + ( y ln z ) j + ( z ln x ) k 3. F = ye xyz i + ze xyz j + xe xyz k 4. F = sin ( xy) i + cos( yz ) j + tan ( xz ) k 5. The spin field in Figure 15.13 6. The radial field in Figure 15.12 7. The gravitational field in Figure 15.9 and Exercise 38a in Section 15.3 8. The velocity field v ( x , y, z ) = ( a 2 − x 2 − y 2 ) k in Figure 15.14 Calculating Flux Using the Divergence Theorem

In Exercises 9–20, use the Divergence Theorem to find the outward flux of F across the boundary of the region D. 9. Cube F = ( y − x ) i + ( z − y ) j + ( y − x ) k D: The cube bounded by the planes x = ±1,  y = ±1, and z = ±1

13. Portion of ball F = x 2 i − 2 xyj + 3 xzk D: The region cut from the first octant by the ball x 2 + y 2 + z2 = 4 14. Cylindrical can F = ( 6 x 2 + 2 xy ) i + ( 2 y + x 2 z ) j + 4 x 2 y 3 k D: The region cut from the first octant by the cylinder x 2 + y 2 = 4 and the plane z = 3 15. Wedge F = 2 xzi − xyj − z 2 k D: The wedge cut from the first octant by the plane y + z = 4 and the elliptic cylinder 4 x 2 + y 2 = 16 16. Ball F = x 3 i + y 3 j + z 3 k D: The ball x 2 + y 2 + z 2 ≤ a 2 17. Thick sphere F =

x 2 + y 2 + z 2 ( x i + y j + zk )

D: The region 1 ≤ x 2 + y 2 + z 2 ≤ 2 18. Thick sphere F = ( xi + yj + zk )

10. F = x 2 i + y 2 j + z 2 k a. Cube

D:

The cube cut from the first octant by the planes x = 1,  y = 1, and z = 1

b. Cube

D:

The cube bounded by the planes x = ±1,  y = ±1, and z = ±1

c. Cylindrical can

D: The ball x 2 + y 2 + z 2 ≤ 4

D: The region cut from the solid cylinder x 2 + y 2 ≤ 4 by the planes z = 0 and z = 1

11. Cylinder and paraboloid F = yi + xyj − zk D: The region inside the solid cylinder x 2 + y 2 ≤ 4 between the plane z = 0 and the paraboloid z = x 2 + y 2

x 2 + y2 + z 2

D: The region 1 ≤ x 2 + y 2 + z 2 ≤ 4 19. Thick sphere F = ( 5 x 3 + 12 xy 2 ) i + ( y 3 + e y sin z ) j + ( 5 z 3 + e y cos z ) k D: The solid region between the spheres x 2 + y 2 + z 2 = 1 and x 2 + y2 + z 2 = 2 y 2z arctan j + 20. Thick cylinder F = ln ( x 2 + y 2 ) i − x x z x 2 + y2 k

(

)

D: The thick-walled cylinder 1 ≤ x 2 + y 2 ≤ 2, −1 ≤ z ≤ 2

15.8  The Divergence Theorem and a Unified Theory

Theory and Examples

21. a. Show that the outward flux of the position vector field F = xi + yj + zk through a smooth closed surface S is three times the volume of the region enclosed by the surface. b. Let n be the outward unit normal vector field on S. Show that it is not possible for F to be orthogonal to n at every point of S. 22. The base of the closed cubelike surface shown here is the unit square in the xy-plane. The four sides lie in the planes x = 0, x = 1,  y = 0, and y = 1. The top is an arbitrary smooth surface whose identity is unknown. Let F = xi − 2 yj + ( z + 3 ) k, and suppose the outward flux of F through Side A is 1 and through Side B is −3. Can you conclude anything about the outward flux through the top? Give reasons for your answer. z

1017

28. Compute the net outward flux of the vector field F = ( xi + yj + zk ) ( x 2 + y 2 + z 2 )3 2 across the ellipsoid 9 x 2 + 4 y 2 + 6 z 2 = 36. 29. Let F be a differentiable vector field, and let g ( x , y, z ) be a differentiable scalar function. Verify the following identities. a. ∇ ⋅ ( g F ) = g∇ ⋅ F + ∇g ⋅ F b. ∇ × ( g F ) = g∇ × F + ∇g × F 30. Let F1 and F2 be differentiable vector fields, and let a and b be arbitrary real constants. Verify the following identities. a. ∇ ⋅ ( a F1 + b F2 ) = a∇ ⋅ F1 + b∇ ⋅ F2 b. ∇ × ( a F1 + b F2 ) = a∇ × F1 + b∇ × F2 c. ∇ ⋅ ( F1 × F2 ) = F2 ⋅ ∇ × F1 − F1 ⋅ ∇ × F2 31. If F = Mi + N j + Pk is a differentiable vector field, we define the notation F ⋅ ∇ to mean M

Top

∂ ∂ ∂ + N +P . ∂x ∂y ∂z

For differentiable vector fields F1 and F2 , verify the following identities. a. ∇ × ( F1 × F2 ) = ( F2 ⋅ ∇ ) F1 − ( F1 ⋅ ∇ ) F2 + ( ∇ ⋅ F2 ) F1 − ( ∇ ⋅ F1 ) F2 1 y

1 Side B x

Side A

(1, 1, 0)

23. Let F = ( y cos 2 x ) i + ( y 2 sin 2 x ) j + ( x 2 y + z ) k. Is there a vector field A such that F = ∇ × A? Explain your answer. 24. Outward flux of a gradient field Let S be the surface of the portion of the ball x 2 + y 2 + z 2 ≤ a 2 that lies in the first octant, and let f ( x , y, z ) = ln x 2 + y 2 + z 2 . Calculate

∫∫ ∇f

⋅ n dσ.

S

b. ∇( F1 ⋅ F2 ) = ( F1 ⋅ ∇ ) F2 + ( F2 ⋅ ∇ ) F1 + F1 × ( ∇ × F2 ) + F2 × ( ∇ × F1 ) 32. Harmonic functions A function f ( x , y, z ) is said to be harmonic in a region D in space if it satisfies the Laplace equation ∇ 2 f = ∇ ⋅ ∇f = throughout D. a. Suppose that f is harmonic throughout a bounded region D enclosed by a smooth surface S and that n is the chosen unit normal vector on S. Show that the integral over S of ∇f ⋅ n, the derivative of f in the direction of n, is zero. b. Show that if f is harmonic on D, then

∫∫

(∇f ⋅ n is the derivative of f in the direction of outward normal n.) 25. Let F be a field whose components have continuous first partial derivatives throughout a portion of space containing a region D bounded by a smooth closed surface S. If F ≤ 1, can any bound be placed on the size of

∫∫∫

∫∫ S

Give reasons for your answer.

F = xyi + ( sin xz + y 2 ) j + ( e xy 2 + x ) k over the surface S surrounding the region D bounded by the planes y = 0,  z = 0,  z = 2 − y and the parabolic cylinder z = 1 − x 2.

∫∫∫

∇f

2

dV .

D

33. Green’s first formula Suppose that f and g are scalar functions with continuous first- and second-order partial derivatives throughout a region D that is bounded by a closed piecewise smooth surface S. Show that

∇ ⋅ F dV ?

27. Calculate the net outward flux of the vector field

f ∇f ⋅ n d σ =

S

D

26. Maximum flux Among all rectangular boxes defined by the inequalities 0 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ 1, find the one for which the total flux of F = (−x 2 − 4 xy ) i − 6 yzj + 12 zk outward through the six sides is greatest. What is the greatest flux?

∂2 f ∂2 f ∂2 f + + = 0 ∂x 2 ∂y2 ∂z 2

f ∇g ⋅ n d σ =

∫∫∫ ( f ∇ 2 g + ∇f

⋅ ∇g ) dV .

(10)

D

Equation (10) is Green’s first formula. (Hint: Apply the Divergence Theorem to the field F = f ∇g.) 34. Green’s second formula (Continuation of Exercise 33.) Interchange f and g in Equation (10) to obtain a similar formula. Then subtract this formula from Equation (10) to show that

∫∫ ( f ∇g − g∇f ) ⋅ n dσ = ∫∫∫ ( f ∇ 2 g − g∇ 2 f ) dV . S

D

This equation is Green’s second formula.

(11)

1018

Chapter 15 Integrals and Vector Fields

35. Conservation of mass Let v ( t , x , y, z ) be a continuously differentiable vector field over the region D in space, and let p( t , x , y, z ) be a continuously differentiable scalar function. The variable t represents the time domain. The Law of Conservation of Mass asserts that d dt

∫∫∫ D

p( t , x , y, z ) dV = −∫∫ p v ⋅ n dσ,

36. The heat diffusion equation Let T ( t , x , y, z ) be a function with continuous second derivatives giving the temperature at time t at the point ( x , y, z ) of a solid occupying a region D in space. If the solid’s heat capacity and mass density are denoted by the constants c and ρ, respectively, the quantity cρT is called the solid’s heat energy per unit volume. Let −k∇T denote the energy flux vector. (Here the constant k is called the conductivity.) Assuming the Law of Conservation of Mass with −k∇T = v and cρT = p in Exercise 35, derive the diffusion (heat) equation

S

where S is the surface enclosing D. a. Give a physical interpretation of the conservation of mass law if v is a velocity flow field and p represents the density of the fluid at point ( x , y, z ) at time t.

∂T = K ∇ 2T , ∂t

b. Use the Divergence Theorem and Leibniz’s Rule, d dt

∫∫∫

p( t , x , y, z ) dV =

D

∫∫∫ D

where K = k (cρ ) > 0 is the diffusivity constant. (Notice that if T ( t , x ) represents the temperature at time t at position x in a uniform conducting rod with perfectly insulated sides, then ∇ 2T = ∂ 2 T ∂ x 2 and the diffusion equation reduces to the onedimensional heat equation in Chapter 13’s Additional Exercises.)

∂p dV , ∂t

to show that the Law of Conservation of Mass is equivalent to the continuity equation, ∇ ⋅ pv +

∂p = 0. ∂t

(In the first term ∇ ⋅ pv, the variable t is held fixed, and in the second term ∂ p ∂ t, it is assumed that the point ( x , y, z ) in D is held fixed.)

CHAPTER 15

Questions to Guide Your Review

1. What are line integrals of scalar functions? How are they evaluated? Give examples. 2. How can you use line integrals to find the centers of mass of springs or wires? Explain.

11. How do you calculate the area of a parametrized surface in space? Of an implicitly defined surface F ( x , y, z ) = 0? Of the surface that is the graph of z = f ( x , y )? Give examples.

3. What is a vector field? What is the line integral of a vector field? What is a gradient field? Give examples.

12. How do you integrate a scalar function over a parametrized surface? Over surfaces that are defined implicitly or in explicit form? Give examples.

4. What is the flow of a vector field along a curve? What is the work done by a vector field moving an object along a curve? How do you calculate the work done? Give examples.

13. What is an oriented surface? What is the surface integral of a vector field in three-dimensional space over an oriented surface? How is it related to the net outward flux of the field? Give examples.

5. What is the Fundamental Theorem of line integrals? Explain how it is related to the Fundamental Theorem of Calculus.

14. What is the curl of a vector field? How can you interpret it?

6. Specify three properties that are special about conservative fields. How can you tell when a field is conservative? 7. What is special about path independent fields? 8. What is a potential function? Show by example how to find a potential function for a conservative field. 9. What is a differential form? What does it mean for such a form to be exact? How do you test for exactness? Give examples. 10. What is Green’s Theorem? Discuss how the two forms of Green’s Theorem extend the Net Change Theorem in Chapter 5.

15. What is Stokes’ Theorem? Explain how it generalizes Green’s Theorem to three dimensions. 16. What is the divergence of a vector field? How can you interpret it? 17. What is the Divergence Theorem? Explain how it generalizes Green’s Theorem to three dimensions. 18. How are Green’s Theorem, Stokes’ Theorem, and the Divergence Theorem related to the Fundamental Theorem of Calculus for ordinary single integrals?

Chapter 15  Practice Exercises

CHAPTER 15

Practice Exercises

Evaluating Line Integrals

Finding and Evaluating Surface Integrals

1. The accompanying figure shows two polygonal paths in space joining the origin to the point (1, 1, 1). Integrate f ( x , y, z ) = 2 x − 3 y 2 − 2 z + 3 over each path. z

z

(0, 0, 0)

(0, 0, 0)

(1, 1, 1)

(1, 1, 0)

x

Path 1

Path 2

2. The accompanying figure shows three polygonal paths joining the origin to the point (1, 1, 1). Integrate f ( x , y, z ) = x 2 + y − z over each path. z

12. Area of a parabolic cap Find the area of the cap cut from the paraboloid y 2 + z 2 = 3 x by the plane x = 1.

14. a. Hemisphere cut by cylinder Find the area of the surface cut from the hemisphere x 2 + y 2 + z 2 = 4,  z ≥ 0, by the cylinder x 2 + y 2 = 2 x.

y

(1, 1, 0)

11. Area of an elliptic region Find the area of the elliptic region cut from the plane x + y + z = 1 by the cylinder x 2 + y 2 = 1.

13. Area of a spherical cap Find the area of the cap cut from the top of the sphere x 2 + y 2 + z 2 = 1 by the plane z = 2 2.

(1, 1, 1)

y x

z

b. Find the area of the portion of the cylinder that lies inside the hemisphere. (Hint: Project onto the xz-plane. Or evaluate the integral ∫ h ds, where h is the altitude of the cylinder and ds is the element of arc length on the circle x 2 + y 2 = 2 x in the xy-plane.) z

(0, 0, 0)

1019

(0, 0, 0)

(1, 1, 1)

Hemisphere z = "4 − r 2

(1, 1, 1)

(1, 0, 0) C1 x

C3

y

C4

(1, 1, 0)

C2

C3

y

(1, 1, 0)

x z

(0, 0, 1) (0, 0, 0)

C6 C5

(0, 1, 1)

C7 (1, 1, 1)

x

y

Cylinder r = 2 cos u

y

x

3. Integrate f ( x , y, z ) =

r (t ) = ( a cos t ) j + ( a sin t ) k, 4. Integrate f ( x , y, z ) =

15. Area of a triangle Find the area of the triangle in which the plane ( x a ) + ( y b ) + ( z c ) = 1  ( a, b, c > 0 ) intersects the first octant. Check your answer with an appropriate vector calculation.

x 2 + z 2 over the circle 0 ≤ t ≤ 2π.

x 2 + y 2 over the involute curve

r (t ) = ( cos t + t sin t ) i + ( sin t − t cos t ) j,

0 ≤ t ≤

3.

Evaluate the integrals in Exercises 5 and 6. 5.

( 4,− 3,0 )

∫(−1,1,1)

dx + dy + dz x+ y+z

6.

(10,3,3 )

∫ 1,1,1 (

)

dx −

z dy − y

y dz z

7. Integrate F = −( y sin z ) i + ( x sin z ) j + ( xy cos z ) k around the circle cut from the sphere x 2 + y 2 + z 2 = 5 by the plane z = −1, clockwise as viewed from above. 8. Integrate F = 3 x 2 yi + ( x 3 + 1) j + 9 z 2 k around the circle cut from the sphere x 2 + y 2 + z 2 = 9 by the plane x = 2. Evaluate the integrals in Exercises 9 and 10. 9.

∫C 8 x sin y dx − 8 y cos x dy C is the square cut from the first quadrant by the lines x = π 2 and y = π 2.

10.

∫C y 2 dx + x 2 dy C is the circle

x2

+

y2

= 4.

16. Parabolic cylinder cut by planes Integrate yz z a. g ( x , y, z ) = b. g ( x , y, z ) = 4y2 + 1 4y2 + 1 over the surface cut from the parabolic cylinder y 2 − z = 1 by the planes x = 0,  x = 3, and z = 0. 17. Circular cylinder cut by planes Integrate g ( x , y, z ) = x 4 y ( y 2 + z 2 ) over the portion of the cylinder y 2 + z 2 = 25 that lies in the first octant between the planes x = 0 and x = 1 and above the plane z = 3. 18. Area of Wyoming The state of Wyoming is bounded by the meridians 111 ° 3 ′ and 104 ° 3 ′ west longitude and by the circles 41 ° and 45 ° north latitude. Assuming that Earth is a sphere of radius R = 6370 km, find the area of Wyoming. Parametrized Surfaces Find parametrizations for the surfaces in Exercises 19–24. (There are many ways to do these, so your answers may not be the same as those in the back of the text.)

19. Spherical band The portion of the sphere x 2 + y 2 + z 2 = 36 between the planes z = −3 and z = 3 3

1020

Chapter 15 Integrals and Vector Fields

20. Parabolic cap The portion of the paraboloid z = −( x 2 + y 2 ) 2 above the plane z = −2 21. Cone

The cone z = 1 +

x 2 + y 2 ,  z ≤ 3

a. By using the parametrization of the curve to evaluate the work integral. b. By evaluating a potential function for F.

22. Plane above square The portion of the plane 4 x + 2 y + 4 z = 12 that lies above the square 0 ≤ x ≤ 2, 0 ≤ y ≤ 2, in the first quadrant

38. Flow along different paths ∇ ( x 2 ze y )

Find the flow of the field F =

23. Portion of paraboloid The portion of the paraboloid y = 2( x 2 + z 2 ) ,  y ≤ 2, that lies above the xy-plane

a. once around the ellipse C in which the plane x + y + z = 1 intersects the cylinder x 2 + z 2 = 25, clockwise as viewed from the positive y-axis.

24. Portion of hemisphere The portion of the hemisphere x 2 + y 2 + z 2 = 10,  y ≥ 0, in the first octant

b. along the curved boundary of the helicoid in Exercise 27 from (1, 0, 0 ) to (1, 0, 2π ).

25. Surface area Find the area of the surface r ( u , υ ) = ( u + υ ) i + ( u − υ ) j + υk ,   0 ≤ u ≤ 1, 0 ≤ υ ≤ 1. 26. Surface integral Integrate f ( x , y, z ) = xy − z 2 over the surface in Exercise 25. 27. Area of a helicoid Find the surface area of the helicoid r ( r , θ ) = ( r cos θ ) i + ( r sin θ ) j + θk, 0 ≤ θ ≤ 2π , 0 ≤ r ≤ 1, in the accompanying figure. z 2p

In Exercises 39 and 40, use the curl integral in Stokes’ Theorem to find the circulation of the field F around the curve C in the indicated direction. 39. Circulation around an ellipse F = y 2 i − yj + 3z 2 k C: The ellipse in which the plane 2 x + 6 y − 3z = 6 meets the cylinder x 2 + y 2 = 1, counterclockwise as viewed from above 40. Circulation around a circle F = ( x 2 + y ) i + ( x + y ) j + ( 4 y 2 − z )k C: The circle in which the plane z = − y meets the sphere x 2 + y 2 + z 2 = 4, counterclockwise as viewed from above Masses and Moments

41. Wire with different densities Find the mass of a thin wire lying along the curve r (t ) = 2t i + 2t j + ( 4 − t 2 ) k, 0 ≤ t ≤ 1, if the density at t is (a) δ = 3t and (b) δ = 1.

(1, 0, 2p)

42. Wire with variable density Find the center of mass of a thin wire lying along the curve r (t ) = ti + 2tj + ( 2 3 ) t 3 2 k, 0 ≤ t ≤ 2, if the density at t is δ = 3 5 + t .

(1, 0, 0) y

x

28. Surface integral Evaluate the integral ∫ ∫ S where S is the helicoid in Exercise 27.

x 2 + y 2 + 1 dσ,

Conservative Fields Which of the fields in Exercises 29–32 are conservative, and which are not?

29. F = xi + yj + zk 30. F = ( xi + yj + zk ) ( x 2 + y 2 + z 2 )3 2 31. F = xe y i + ye z j + ze x k 32. F = ( i + zj + yk ) ( x + yz ) Find potential functions for the fields in Exercises 33 and 34. 33. F = 2 i + ( 2 y + z ) j + ( y + 1) k 34. F = ( z cos xz ) i + e y j + ( x cos xz ) k Work and Circulation In Exercises 35 and 36, find the work done by each field along the paths from ( 0, 0, 0 ) to (1, 1, 1) in Exercise 1.

35. F = 2 xyi + j + x 2 k 36. F = 2 xyi + x 2 j + k 37. Finding work in two ways Find the work done by xi + yj F = 32 (x 2 + y2 ) over the plane curve r (t ) = ( e t cos t ) i + ( e t sin t ) j from the point (1, 0 ) to the point ( e 2 π , 0 ) in two ways:

43. Wire with variable density Find the center of mass and the moments of inertia about the coordinate axes of a thin wire lying along the curve r (t ) = t i +

2 2 32 t2 t j + k, 3 2

0 ≤ t ≤ 2,

if the density at t is δ = 1 ( t + 1). 44. Center of mass of an arch A slender metal arch lies along the semicircle y = a 2 − x 2 in the xy-plane. The density at the point ( x , y ) on the arch is δ ( x , y ) = 2a − y. Find the center of mass. 45. Wire with constant density A wire of constant density δ = 1 lies along the curve r (t ) = ( e t cos t ) i + ( e t sin t ) j + e t k, 0 ≤ t ≤ ln 2. Find z and I z . 46. Helical wire with constant density Find the mass and center of mass of a wire of constant density δ that lies along the helix r (t ) = ( 2 sin t ) i + ( 2 cos t ) j + 3tk, 0 ≤ t ≤ 2π. 47. Inertia and center of mass of a shell Find I z and the center of mass of a thin shell of density δ ( x , y, z ) = z cut from the upper portion of the sphere x 2 + y 2 + z 2 = 25 by the plane z = 3. 48. Moment of inertia of a cube Find the moment of inertia about the z-axis of the surface of the cube cut from the first octant by the planes x = 1,  y = 1, and z = 1 if the density is δ = 1. Flux Across a Plane Curve or Surface Use Green’s Theorem to find the counterclockwise circulation and outward flux for the fields and curves in Exercises 49 and 50.

49. Square F = ( 2 xy + x ) i + ( xy − y ) j C: The square bounded by x = 0, x = 1, y = 0, y = 1

Chapter 15  Additional and Advanced Exercises

50. Triangle F = ( y − 6 x 2 ) i + ( x + y 2 ) j

55. Spherical cap F = −2 xi − 3 yj + zk

C: The triangle made by the lines y = 0,  y = x , and x = 1 51. Zero line integral Show that

DC

ln x sin y dy −

cos y dx = 0 x

for any closed curve C to which Green’s Theorem applies. 52. a. Outward flux and area Show that the outward flux of the position vector field F = xi + yj across any closed curve to which Green’s Theorem applies is twice the area of the region enclosed by the curve. b. Let n be the outward unit normal vector to a closed curve to which Green’s Theorem applies. Show that it is not possible for F = xi + yj to be orthogonal to n at every point of C. In Exercises 53–56, find the outward flux of F across the boundary of D. 53. Cube F = 2 xyi + 2 yzj + 2 xzk D: The cube cut from the first octant by the planes x = 1,  y = 1, and  z = 1 54. Spherical cap F = xzi + yzj + k D: The entire surface of the upper cap cut from the ball x 2 + y 2 + z 2 ≤ 25 by the plane z = 3

CHAPTER 15

1021

D: The upper region cut from the ball x 2 + y 2 + z 2 ≤ 2 by the paraboloid z = x 2 + y 2 56. Cone and cylinder F = ( 6 x + y ) i − ( x + z ) j + 4 yzk D: The region in the first octant bounded by the cone z = x 2 + y 2 , the cylinder x 2 + y 2 = 1, and the coordinate planes 57. Hemisphere, cylinder, and plane Let S be the surface that is bounded on the left by the hemisphere x 2 + y 2 + z 2 = a 2, y ≤ 0, in the middle by the cylinder x 2 + z 2 = a 2, 0 ≤ y ≤ a, and on the right by the plane y = a. Find the flux of F = yi + zj + xk outward across S. 58. Cylinder and planes Find the outward flux of the field F = 3 xz 2 i + yj − z 3 k across the surface of the solid in the first octant that is bounded by the cylinder x 2 + 4 y 2 = 16 and the planes y = 2 z ,  x = 0, and z = 0. 59. Cylindrical can Use the Divergence Theorem to find the flux of F = xy 2 i + x 2 yj + yk outward through the surface of the region enclosed by the cylinder x 2 + y 2 = 1 and the planes z = 1 and z = −1. 60. Hemisphere Find the flux of F = ( 3z + 1) k upward across the hemisphere x 2 + y 2 + z 2 = a 2,  z ≥ 0 , (a) with the Divergence Theorem and (b) by evaluating the flux integral directly.

Additional and Advanced Exercises

Finding Areas with Green’s Theorem

Use the Green’s Theorem area formula in Exercises 15.4 to find the areas of the regions enclosed by the curves in Exercises 1–4. 1. The limaçon x = 2 cos t − cos 2t ,  y = 2 sin t , 0 ≤ t ≤ 2π

3. The eight curve x = (1 2 ) sin 2t ,  y = sin t , 0 ≤ t ≤ π (one loop) y 1

y

x 0

x

1

−1

2. The deltoid 0 ≤ t ≤ 2π

x = 2 cos t + cos 2t ,  y = 2 sin t − sin 2t , y

0

4. The teardrop x = 2a cos t − a sin 2t ,  y = b sin t , 0 ≤ t ≤ 2π y b

3

x

0

2a

x

1022

Chapter 15 Integrals and Vector Fields

Theory and Applications

5. a. Give an example of a vector field F ( x , y, z ) that has value 0 at only one point and such that curl F is nonzero everywhere. Be sure to identify the point and compute the curl. b. Give an example of a vector field F ( x , y, z ) that has value 0 on precisely one line and such that curl F is nonzero everywhere. Be sure to identify the line and compute the curl. c. Give an example of a vector field F ( x , y, z ) that has value 0 on a surface and such that curl F is nonzero everywhere. Be sure to identify the surface and compute the curl. 6. Find all points ( a, b, c ) on the sphere x 2 + y 2 + z 2 = R 2 where the vector field F = yz 2 i + xz 2 j + 2 xyzk is normal to the surface and F ( a, b, c ) ≠ 0 . 7. Find the mass of a spherical shell of radius R such that at each point ( x , y, z ) on the surface, the mass density δ ( x , y, z ) is its distance to some fixed point ( a, b, c ) of the surface. 8. Find the mass of a helicoid r ( r , θ ) = ( r cos θ ) i + ( r sin θ ) j + θk, 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π, if the density function is δ ( x , y, z ) = 2 x 2 + y 2 . See Practice Exercise 27 for a figure. 9. Among all rectangular regions 0 ≤ x ≤ a, 0 ≤ y ≤ b, find the one for which the total outward flux of F = ( x 2 + 4 xy ) i − 6 yj across the four sides is least. What is the least flux? 10. Find an equation for the plane through the origin such that the circulation of the flow field F = zi + xj + yk around the circle of intersection of the plane with the sphere x 2 + y 2 + z 2 = 4 is a maximum. + = 4 from ( 2, 0 ) to ( 0, 2 ) in 11. A string lies along the circle the first quadrant. The density of the string is ρ ( x , y ) = xy. x2

a. Show that the surface integral giving the magnitude of the total force on the ball due to the fluid’s pressure is

n

∑ g x k y k 2 Δs k n →∞

=

k =1

∫C g xy 2 ds,

12. A thin sheet lies along the portion of the plane x + y + z = 1 in the first octant. The density of the sheet is δ ( x , y, z ) = xy. a. Partition the sheet into a finite number of subpieces to show that the work done by gravity to move the sheet straight down to the xy-plane is given by n →∞

n

∑ g x k y k z k Δσ k k =1

=

∫∫

g xyz dσ ,

S

where g is the gravitational constant. b. Find the total work done by evaluating the surface integral in part (a). c. Show that the total work done equals the work required to move the sheet’s center of mass ( x , y , z ) straight down to the xy-plane.

∫∫

w ( 4 − z ) dσ.

S

b. Since the ball is not moving, it is being held up by the buoyant force of the liquid. Show that the magnitude of the buoyant force on the sphere is Buoyant force =

∫∫

w ( z − 4 ) k ⋅ n dσ ,

S

where n is the outer unit normal at ( x , y, z ). This illustrates Archimedes’ principle that the magnitude of the buoyant force on a submerged solid equals the weight of the displaced fluid. c. Use the Divergence Theorem to find the magnitude of the buoyant force in part (b). 14. Fluid force on a curved surface A cone in the shape of the surface z = x 2 + y 2 , 0 ≤ z ≤ 2, is filled with a liquid of constant weight density w. Assuming the xy-plane is “ground level,” show that the total force on the portion of the cone from z = 1 to z = 2 due to liquid pressure is the surface integral F =

∫∫

w ( 2 − z ) dσ.

S

Evaluate the integral. 15. Faraday’s law If E ( t , x , y, z ) and B( t , x , y, z ) represent the electric and magnetic fields at point ( x , y, z ) at time t, a basic principle of electromagnetic theory says that ∇ × E = −∂ B ∂t . In this expression ∇ × E is computed with t held fixed and ∂ B ∂t is calculated with ( x , y, z ) fixed. Use Stokes’ Theorem to derive Faraday’s law,

where g is the gravitational constant. c. Show that the total work done equals the work required to move the string’s center of mass ( x , y ) straight down to the x-axis.

=

k =1

DC E ⋅ dr

b. Find the total work done by evaluating the line integral in part (a).

Work = lim

n

∑ w ( 4 − z k ) Δσ k n →∞

Force = lim

y2

a. Partition the string into a finite number of subarcs to show that the work done by gravity to move the string straight down to the x-axis is given by Work = lim

13. Archimedes’ principle If an object such as a ball is placed in a liquid, it will either sink to the bottom, float, or sink a certain distance and remain suspended in the liquid. Suppose a fluid has constant weight density w and that the fluid’s surface coincides with the plane z = 4. A spherical ball remains suspended in the fluid and occupies the region x 2 + y 2 + ( z − 2 ) 2 ≤ 1.

= −

∂ ∂t

∫∫

B ⋅ n dσ ,

S

where C represents a wire loop through which current flows counterclockwise with respect to the surface’s unit normal n, giving rise to the voltage

DC E ⋅ dr around C. The surface integral on the right side of the equation is called the magnetic flux, and S is any oriented surface with boundary C. 15. Let F = −

GmM r r3

be the gravitational force field defined for r ≠ 0. Use Gauss’s law in Section 15.8 to show that there is no continuously differentiable vector field H satisfying F = ∇ × H.

Chapter 15  Technology Application Projects

17. If f ( x , y, z ) and g ( x , y, z ) are continuously differentiable scalar functions defined over the oriented surface S with boundary curve C, prove that

∫∫ ( ∇f × ∇g ) ⋅ n dσ

=

S

surface. Also, the magnitude dσ = dV is the element of surface area (by Equation 5 in Section 15.5). Derive the identity dσ = ( EG − F 2 )1 2 du d υ,

DC f ∇g ⋅ dr.

18. Suppose that ∇ ⋅ F1 = ∇ ⋅ F2 and ∇ × F1 = ∇ × F2 over a region D enclosed by the oriented surface S with outward unit normal n and that F1 ⋅ n = F2 ⋅ n on S. Prove that F1 = F2 throughout D.

where E = ru 2, F = ru ⋅ rυ , and G = ru 2. 21. Show that the volume V of a region D in space enclosed by the oriented surface S with outward normal n satisfies the identity

19. Prove or disprove that if ∇ ⋅ F = 0 and ∇ × F = 0, then F = 0. 20. Let S be an oriented surface parametrized by r ( u, υ ). Define the notation dV = ru du × rυ d υ so that dV is a vector normal to the

CHAPTER 15

1023

V =

1 r ⋅ n dσ , 3 ∫∫ S

where r is the position vector of the point ( x , y, z ) in D.

Technology Application Projects

Mathematica/Maple Projects

Projects can be found within MyLab Math. • Work in Conservative and Nonconservative Force Fields Explore integration over vector fields and experiment with conservative and nonconservative force functions along different paths in the field. • How Can You Visualize Green’s Theorem? Explore integration over vector fields and use parametrizations to compute line integrals. Both forms of Green’s Theorem are explored. • Visualizing and Interpreting the Divergence Theorem Verify the Divergence Theorem by formulating and evaluating certain divergence and surface integrals.

16

First-Order Differential Equations

Chapter 17 is available online. To access this chapter, visit the companion Website.

OVERVIEW Many real-world problems, when formulated mathematically, lead to differential equations. We encountered a number of these equations in previous chapters when studying phenomena such as the motion of an object along a straight line, the decay of a radioactive material, the growth of a population, and the cooling of a heated object placed within a medium of lower temperature. Section 4.8 introduced differential equations of the form dy dx = f ( x ), where f is given and y is an unknown function of x. We learned that when f is continuous over some interval, the general solution y( x ) is found directly by integration, y = ∫ f ( x ) dx. In Section 7.2 we investigated differential equations of the form dy dx = f ( x , y), where f is a function of both the independent variable x and the dependent variable y. There we learned how to find the general solution for the special case when the differential equation is separable. In this chapter we further extend our study to include other commonly occurring first-order differential equations. These differential equations involve only first derivatives of the unknown function y( x ), and they model phenomena varying from simple electrical circuits to the concentration of a chemical in a container. Differential equations involving second derivatives are examined in Chapter 17.

16.1 Solutions, Slope Fields, and Euler’s Method We begin this section by defining general differential equations involving first derivatives. We then look at slope fields, which give a geometric picture of the solutions to such equations. Many differential equations cannot be solved by obtaining an explicit formula for the solution. However, we can often find numerical approximations to solutions. We present one such method here, called Euler’s method, which is the basis for many other numerical methods as well.

General First-Order Differential Equations and Solutions A first-order differential equation is an equation dy = f ( x , y) dx

(1)

in which f ( x , y) is a function of two variables defined on a region in the xy-plane. The equation is of first order because it involves only the first derivative dy dx and not higher-order derivatives. In a typical situation y represents an unknown function of x, and f ( x , y) is a known function. Some examples of first-order differential equations are

1024

16.1  Solutions, Slope Fields, and Euler’s Method

1025

y′ = x + y,  y′ = y x , and y′ = 3 xy . In all cases we should think of y as an unknown function of x whose derivative is given by f ( x , y). The equations y′ = f ( x , y)

d y = f ( x , y) dx

and

are equivalent to Equation (1), and all three forms will be used interchangeably in the text. A solution of Equation (1) is a differentiable function y = y( x ) defined on an interval I of x-values (possibly an infinite interval) such that d y( x ) = f ( x , y( x ) ) dx on that interval. That is, when y( x ) and its derivative y′( x ) are substituted into Equation (1), the resulting equation is true for all x over the interval I. EXAMPLE 1

Show that every member of the family of functions y =

C +2 x

is a solution of the first-order differential equation dy 1 = (2 − y) dx x on the interval ( 0, ∞ ), where C is any constant. Solution Differentiating y = C x + 2 gives

( )

dy d 1 C = C + 0 = − 2. dx x x dx We need to show that the differential equation is satisfied when we substitute into it the expressions ( C x ) + 2 for y, and −C x 2 for dy dx . That is, we need to verify that for all x ∈ ( 0, ∞ ) , C 1 C − 2 = ⎢⎡ 2 − + 2 ⎤⎥ . ⎦ x x⎣ x

(

This last equation follows side: 1 x

)

immediately by expanding the expression on the right-hand

(

)

( )

⎡ 2 − C + 2 ⎤ = 1 −C = − C . ⎢⎣ ⎥⎦ x x x x2

Therefore, for every value of C, the function y = C x + 2 is a solution of the differential equation. The differential equation in Example 1 has a whole family of solutions, one for each value of C. A natural question to consider is whether this family contains all the solutions to the differential equation, or whether there are others that can somehow arise. The general solution to a first-order differential equation is the name given to an expression that contains all possible solutions. The general solution always contains an arbitrary constant, but a solution may contain an arbitrary constant without being the general solution. Establishing when a solution is the general solution is left to a more extensive development of the theory of differential equations. As with antiderivatives, we often need a particular rather than the general solution to a first-order differential equation y′ = f ( x , y). A common way to pick out one of the collection of possible solutions is to specify the value of y at a point x = x 0 . The particular solution satisfying the initial condition y( x 0 ) = y 0 is the solution y = y( x ) whose value is y 0 when x = x 0 . Thus the graph of the particular solution passes through the point ( x 0 , y 0 ) in the xy-plane. A first-order initial value problem is a differential equation y′ = f ( x , y) whose solution must satisfy an initial condition y( x 0 ) = y 0 .

1026

Chapter 16 First-Order Differential Equations

EXAMPLE 2

Show that the function 1 y = ( x + 1) − e x 3

is a solution to the first-order initial value problem dy = y − x, dx

y(0) =

2 . 3

Solution The equation

dy = y−x dx is a first-order differential equation with f ( x , y) = y − x. On the left side of the equation: y 2 a0, b 3 −4

−2

2 1 −1

(

2

4

x

On the right side of the equation: 1 1 y − x = ( x + 1) − e x − x = 1 − e x . 3 3

−2 −3

The function satisfies the initial condition because

−4

1 1 2 y(0) = ⎡⎢ ( x + 1) − e x ⎤⎥ = 1− = . ⎣ 3 ⎦ x=0 3 3

FIGURE 16.1

Graph of the solution to the initial value problem in Example 2.

)

dy 1 1 d = x + 1 − e x = 1 − e x. 3 3 dx dx

y = (x + 1) − 1e x 3

The graph of the function is shown in Figure 16.1.

Slope Fields: Viewing Solution Curves Each time we specify an initial condition y( x 0 ) = y 0 for the solution of a differential equation y′ = f ( x , y), the solution curve (graph of the solution) is required to pass through the point ( x 0 , y 0 ) and to have slope f ( x 0 , y 0 ) there. We can picture these slopes graphically by drawing short line segments of slope f ( x , y) at selected points ( x , y) in the region of the xy-plane that constitutes the domain of f. Each segment has the same slope as the solution curve through ( x , y) and so is tangent to the curve there. The resulting picture is called a slope field (or direction field) and gives a visualization of the general shape of the solution curves. Figure 16.2a shows a slope field, with a particular solution sketched into it in Figure 16.2b. We see how these line segments indicate the direction the solution curve takes at each point it passes through. y

−4

−2

y

4

4

2

2

0

2

4

x

−4

−2

0

−2

−2

−4

−4

(a)

FIGURE 16.2

2 a0, b 3

2

4

(b)

(a) Slope field for

dy = y − x. (b) The particular soludx

( 23 ) (Example 2).

tion curve through the point 0,

x

16.1  Solutions, Slope Fields, and Euler’s Method

y

EXAMPLE 3

1027

For the differential equation y′ = 2 y − x ,

2

draw line segments representing the slope field at the points (1, 2 ), (1, 1), ( 2, 2 ), and ( 2, 1). 1

0

1

2

Solution At (1, 2 ) the slope is 2( 2 ) − (1) = 3. Similarly, the slope at ( 2, 2 ) is 2, at ( 2, 1) is 0, and at (1, 1) is 1. We indicate this on the graph by drawing short line segments of the given slope through each point, as in Figure 16.3.

x

Figure 16.4 shows three slope fields, and we see how the solution curves behave by following the tangent line segments in these fields. Slope fields are useful because they display the overall behavior of the family of solution curves for a given differential equation. For instance, the slope field in Figure 16.4b reveals that every solution y( x ) to the differential equation specified in the figure satisfies lim y( x ) = 0. We will see that

(a) y 2

x →±∞

knowing the overall behavior of the solution curves is often critical to understanding and predicting outcomes in a real-world system modeled by a differential equation. Constructing a slope field with pencil and paper can be quite tedious. Our examples were generated by computer software.

1

0

1

2

x

(b)

FIGURE 16.3

(a) The slope field for y′ = 2 y − x is shown at four points. (b) The slope field at several hundred additional points in the plane. (a) y′ = y − x 2

(b) y′ = −

2 xy 1 + x2

(c) y′ = (1 − x)y + x 2

FIGURE 16.4

Slope fields (top row) and selected solution curves (bottom row). In computer renditions, slope segments are sometimes portrayed with arrows, as they are here, but they should be considered as just tangent line segments.

Euler’s Method

y

y = L(x) = y 0 + f (x 0, y 0)(x − x 0 ) y = y (x)

y0

0

(x 0, y 0) x0

FIGURE 16.5

x

The linearization L ( x ) of y = y( x ) at x = x 0 .

If we do not require or cannot find an exact solution that gives an explicit formula for an initial value problem y′ = f ( x , y),  y( x 0 ) = y 0 , we can often use a computer to generate a table of approximate numerical values of y for values of x in an appropriate interval. Such a table is called a numerical solution of the problem, and the process by which we generate the table is called a numerical method. Given a differential equation dy dx = f ( x , y) and an initial condition y( x 0 ) = y 0 , we can approximate the solution y = y( x ) by its linearization L ( x ) = y( x 0 ) + y′( x 0 )( x − x 0 ) or

L ( x ) = y 0 + f ( x 0 , y 0 )( x − x 0 ).

The function L ( x ) gives a good approximation to the solution y( x ) in a short interval about x 0 (Figure 16.5). The basis of Euler’s method is to patch together a string of linearizations to approximate the curve over a longer stretch. Here is how the method works. We know the point ( x 0 , y 0 ) lies on the solution curve. Suppose that we specify a new value for the independent variable to be x 1 = x 0 + dx. (Recall that dx = Δx in the definition of differentials.) If the increment dx is small, then y1 = L ( x 1 ) = y 0 + f ( x 0 , y 0 ) dx

1028

Chapter 16 First-Order Differential Equations

y (x1, y1)

is a good approximation to the exact solution value y = y( x 1 ). So from the point ( x 0 , y 0 ) , which lies exactly on the solution curve, we have obtained the point ( x 1 , y1 ) , which lies very close to the point ( x 1 , y( x 1 ) ) on the solution curve (Figure 16.6). Using the point ( x 1 , y1 ) and the slope f ( x 1 , y1 ) of the solution curve through ( x 1 , y1 ) , we take a second step. Setting x 2 = x 1 + dx, we use the linearization of the solution curve through ( x 1 , y1 ) to calculate

y = y(x) (x1, y(x1))

(x 0, y 0)

0

x0

dx

y 2 = y1 + f ( x 1 , y1 ) dx.

x

x1 = x 0 + dx

This gives the next approximation ( x 2 , y 2 ) to values along the solution curve y = y( x ) (Figure 16.7). Continuing in this fashion, we take a third step from the point ( x 2 , y 2 ) with slope f ( x 2 , y 2 ) to obtain the third approximation

FIGURE 16.6 The first Euler step approximates y( x 1 ) with y1 = L ( x 1 ).

y 3 = y 2 + f ( x 2 , y 2 ) dx ,

y (x 3, y 3)

Euler approximation (x , y ) 2 2 (x1, y1)

x0

EXAMPLE 4 Find the first three approximations y1 , y 2 , y 3 using Euler’s method for the initial value problem y′ = 1 + y, y(0) = 1,

True solution curve y = y(x)

(x 0, y 0)

0

Error

dx

x1

dx

x2

dx

and so on. We are building an approximation to a solution by following the direction of the slope field of the differential equation. The steps in Figure 16.7 are drawn large to illustrate the construction process, so the approximation looks crude. In practice, dx would be chosen small enough to make the red curve hug the blue one and give a better approximation.

x3

x

FIGURE 16.7 Three steps in the Euler approximation to the solution of the initial value problem y′ = f ( x , y), y( x 0 ) = y 0 . As we take more steps, the errors involved usually accumulate, but not in the exaggerated way shown here.

starting at x 0 = 0 with dx = 0.1. Solution We already have the starting values x 0 = 0 and y 0 = 1. Next we determine the values of x at which the Euler approximations will take place: x 1 = x 0 + dx = 0.1, x 2 = x 0 + 2 dx = 0.2, and x 3 = x 0 + 3 dx = 0.3. Then we find

First :

y1 = y 0 + f ( x 0 , y 0 ) dx = y 0 + (1 + y 0 ) dx = 1 + (1 + 1)( 0.1) = 1.2

Second :

f ( x , y) = 1 + y y 0 = 1, dx = 0.1

y 2 = y1 + f ( x 1 , y1 ) dx = y1 + (1 + y1 ) dx = 1.2 + (1 + 1.2 )( 0.1) = 1.42

Third :

y1 = 1.2, dx = 0.1

y 3 = y 2 + f ( x 2 , y 2 ) dx = y 2 + (1 + y 2 ) dx = 1.42 + (1 + 1.42 )( 0.1) = 1.662

y 2 = 1.42, dx = 0.1

The step-by-step process used in Example 4 can be continued with more points. Using equally spaced values for the independent variable in the table for the numerical solution, and generating n of them, set x 1 = x 0 + dx x 2 = x 1 + dx  x n = x n −1 + dx. Then calculate the approximations to the solution, y1 = y 0 + f ( x 0 , y 0 ) dx y 2 = y1 + f ( x 1 , y1 ) dx  y n = y n −1 + f ( x n −1 , y n −1 ) dx.

16.1  Solutions, Slope Fields, and Euler’s Method

HISTORICAL BIOGRAPHY Leonhard Euler (1707–1783) Born in Basel, Switzerland, Leonhard Euler was the dominant mathematical figure of his century and the most prolific mathematician who ever lived. He was also an astronomer, physicist, engineer, and chemist. He was the first scientist to give the function concept prominence in his work, thereby setting a strong foundation for the development of calculus and other areas of mathematics. To know more, visit the companion Website.

1029

The number of steps n can be as large as we like, but errors involving the representation of numbers in software can accumulate if n is too large. Euler’s method is easy to implement on a computer or calculator. A typical software program takes as input x 0 and y 0 , the number of steps n, and the step size dx. It then calculates the approximate solution values y1 , y 2 , …, y n in iterative fashion, as just described. Solving the separable equation in Example 4, we find that the exact solution to the initial value problem is y = 2e x − 1. We use this information in Example 5. EXAMPLE 5

Use Euler’s method to solve y′ = 1 + y,

y(0) = 1,

on the interval 0 ≤ x ≤ 1, starting at x 0 = 0 and taking (a) dx = 0.1 and (b) dx = 0.05. Compare the approximations with the values of the exact solution y = 2e x − 1. Solution

(a) We used a computer to generate the approximate values in Table 16.1. The “error” column is obtained by subtracting the unrounded Euler values from the unrounded values found using the exact solution. All entries are then rounded to four decimal places. TABLE 16.1 Euler solution of y′ = 1 + y, y(0) = 1, step size dx = 0.1

y 4 3 2

y = 2e x − 1

1

0

1

x

x

y (Euler)

y (exact)

Error

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

1 1.2 1.42 1.662 1.9282 2.2210 2.5431 2.8974 3.2872 3.7159 4.1875

1 1.2103 1.4428 1.6997 1.9836 2.2974 2.6442 3.0275 3.4511 3.9192 4.4366

0 0.0103 0.0228 0.0377 0.0554 0.0764 0.1011 0.1301 0.1639 0.2033 0.2491

FIGURE 16.8

The graph of y = 2e x − 1 superimposed on a scatterplot of the Euler approximations shown in Table 16.1 (Example 5).

By the time we reach x = 1 (after 10 steps), the error is about 5.6% of the exact solution. A plot of the exact solution curve with the scatterplot of Euler solution points from Table 16.1 is shown in Figure 16.8. (b) One way to try to reduce the error is to decrease the step size. Table 16.2 shows the results and their comparisons with the exact solutions when we decrease the step size to 0.05, doubling the number of steps to 20. As in Table 16.1, all computations are performed before rounding. This time when we reach x = 1, the relative error is only about 2.9%. It might be tempting to reduce the step size even further in Example 5 to obtain greater accuracy. Each additional calculation, however, not only requires additional computer time but, more importantly, adds to the buildup of round-off errors due to the approximate representations of numbers inside the computer.

1030

Chapter 16 First-Order Differential Equations

The analysis of error and the investigation of methods to reduce it when making numerical calculations are important. An area of mathematics called numerical analysis studies advanced numerical methods that are more accurate than Euler’s method. TABLE 16.2 Euler solution of y′ = 1 + y, y(0) = 1, step size dx = 0.05

x

y (Euler)

y (exact)

Error

0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00

1 1.1 1.205 1.3153 1.4310 1.5526 1.6802 1.8142 1.9549 2.1027 2.2578 2.4207 2.5917 2.7713 2.9599 3.1579 3.3657 3.5840 3.8132 4.0539 4.3066

1 1.1025 1.2103 1.3237 1.4428 1.5681 1.6997 1.8381 1.9836 2.1366 2.2974 2.4665 2.6442 2.8311 3.0275 3.2340 3.4511 3.6793 3.9192 4.1714 4.4366

0 0.0025 0.0053 0.0084 0.0118 0.0155 0.0195 0.0239 0.0287 0.0340 0.0397 0.0458 0.0525 0.0598 0.0676 0.0761 0.0853 0.0953 0.1060 0.1175 0.1300

16.1

EXERCISES

In Exercises 1–4, match the differential equations with their slope fields, graphed here. y

y

4 2 −2

2

2

4

x

−4

−2

2

−2

−2

−4

−4

(a)

4

4

2

2

4 −4

−4

y

y

Slope Fields

(b)

4

x

−2

2

4

x

−4

−2

2

−2

−2

−4

−4

(c)

1. y′ = x + y 3. y′ = −

x y

(d)

2. y′ = y + 1 4. y′ = y 2 − x 2

4

x

16.1  Solutions, Slope Fields, and Euler’s Method

In Exercises 5 and 6, copy the slope fields, and sketch in some of the solution curves. 5. y′ = ( y + 2 )( y − 2 ) y 4

In Exercises 15–20, use Euler’s method to calculate the first three approximations to the given initial value problem for the specified increment size. Calculate the exact solution and investigate the accuracy of your approximations. Round your results to four decimal places. 2y , y(1) = −1, dx = 0.5 x 16. y′ = x (1 − y ) , y(1) = 0, dx = 0.2

−2

2

4

17. y′ = 2 xy + 2 y,

x

18. y′ =

−2

y 2 (1

T 20. y′ = ye x ,

2

24. Use Euler’s method with dx = 1 3 to estimate y(2) if y′ = x sin y and y(0) = 1. What is the exact value of y(2)?

−2

2

4

25. Show that the solution of the initial value problem

x

y′ = x + y,

y = −1 − x + (1 + x 0 + y 0 )  e x − x 0 . 26. What integral equation is equivalent to the initial value problem y′ = f ( x ), y( x 0 ) = y 0?

In Exercises 7–12, write an equivalent first-order differential equation and initial condition for y. x x 1 7. y = −1 + ∫ ( t − y(t ) ) dt 8. y = ∫ dt 1 1 t

e

∫x

t2

∫−2 te y(t ) dt

a. ( 0, 1)

+ ( y(t ) ) dt

a. ( 0, 1)

14. f (y)

2

2

1

1 2

3

y

c. ( 0, −1)

b. ( 0, 4 )

c. ( 0, 5 )

29. y′ = y ( x + y ) with

f (y)

1

b. ( 0, 2 )

28. y′ = 2( y − 4 ) with

x

2

13.

In Exercises 27–32, obtain a slope field and add to it graphs of the solution curves passing through the given points. a. ( 0, 1)

11. y = x + 4 +

y(t ) dt

COMPUTER EXPLORATIONS

27. y′ = y with

x

∫0 (1 + y(t ) ) sin t dt

In Exercises 13 and 14, consider the differential equation y′ = f ( y) and the given graph of f. Make a rough sketch of a direction field for each differential equation.

−3 −2 −1 0

y( x 0 ) = y 0

is

Integral Equations

12. y = ln x +

dx = 0.5

23. Use Euler’s method with dx = 0.5 to estimate y(5) if y′ = y 2 x and y(1) = −1. What is the exact value of y(5)?

−4

∫0

y(0) = 2,

dx = 0.5

dx = 0.1

4

−2

10. y = 1 +

y(−1) = 1,

y(0) = 2,

22. Use Euler’s method with dx = 0.2 to estimate y(2) if y′ = y x and y(1) = 2. What is the exact value of y(2)? y

x

dx = 0.2

21. Use Euler’s method with dx = 0.2 to estimate y(1) if y′ = y and y(0) = 1. What is the exact value of y(1)?

6. y′ = y ( y + 1)( y − 1)

9. y = 2 −

y(0) = 3,

+ 2 x ),

2 T 19. y′ = 2 xe x ,

−4

−4

Using Euler’s Method

15. y′ =

2 −4

1031

−3 −2 −1

−1

−1

−2

−2

b. ( 0, − 2 )

c. ( 0, 1 4 )

d. (−1, −1)

b. ( 0, 2 )

c. ( 0, −1)

d. ( 0, 0 )

c. ( 0, 3 )

d. (1, −1)

30. y′ = y 2 with a. ( 0, 1)

31. y′ = ( y − 1)( x + 2 ) with b. ( 0, 1) a. ( 0, −1) xy 32. y′ = 2 with x +4 a. ( 0, 2 )

1

2

3

y

b. ( 0, − 6 )

c. (−2 3, − 4 )

In Exercises 33 and 34, obtain a slope field, and graph the particular solution over the specified interval. Use your CAS DE solver to find the general solution of the differential equation. 33. A logistic equation 0 ≤ y ≤ 3

y′ = y ( 2 − y ) , y(0) = 1 2; 0 ≤ x ≤ 4,

34. y′ = ( sin x )( sin y ), y(0) = 2; − 6 ≤ x ≤ 6, − 6 ≤ y ≤ 6

1032

Chapter 16 First-Order Differential Equations

Exercises 35 and 36 have no explicit solution in terms of elementary functions. Use a CAS to explore graphically each of the differential equations. 35. y′ = cos ( 2 x − y ) ,

y(0) = 2;

36. A Gompertz equation 0 ≤ x ≤ 4, 0 ≤ y ≤ 3

0 ≤ x ≤ 5,

Use a CAS to explore graphically each of the differential equations in Exercises 43–46. Perform the following steps to help with your explorations.

0 ≤ y ≤ 5

y′ = y (1 2 − ln y ) ,

a. Plot a slope field for the differential equation in the given xy-window.

y(0) = 1 3;

b. Find the general solution of the differential equation using your CAS DE solver.

37. Use a CAS to find the solutions of y′ + y = f ( x ), subject to the initial condition y(0) = 0, if f ( x ) is b. sin 2 x

a. 2x

c. 3e

x 2

d. 2e

−x 2

c. Graph the solutions for the values of the arbitrary constant C = −2, −1, 0, 1, 2 superimposed on your slope field plot.

cos 2 x.

Graph all four solutions over the interval −2 ≤ x ≤ 6 to compare the results.

d. Find and graph the solution that satisfies the specified initial condition over the interval [ 0, b ]. e. Find the Euler numerical approximation to the solution of the initial value problem with 4 subintervals of the x-interval, and plot the Euler approximation superimposed on the graph produced in part (d).

38. a. Use a CAS to plot the slope field of the differential equation y′ =

+ 4x + 2 2( y − 1)

3x 2

over the region −3 ≤ x ≤ 3 and −3 ≤ y ≤ 3.

f. Repeat part (e) for 8, 16, and 32 subintervals. Plot these three Euler approximations superimposed on the graph from part (e).

b. Separate the variables and use a CAS integrator to find the general solution in implicit form.

g. Find the error ( y ( exact ) − y ( Euler )) at the specified point x = b for each of your four Euler approximations. Discuss the improvement in the percentage error.

c. Using a CAS implicit function grapher, plot solution curves for the arbitrary constant values C = −6, − 4, − 2, 0, 2, 4, 6. d. Find and graph the solution that satisfies the initial condition y(0) = −1. In Exercises 39–42, use Euler’s method with the specified step size to estimate the value of the solution at the given point x*. Find the value of the exact solution at x*. 39. y′ =

2 xe x 2 ,

40. y′ =

2y 2 ( x

41. y′ =

x y,

42. y′ = 1 +

y(0) = 2, dx = 0.1, x* = 1 − 1) ,

y2,

y(2) = −1 2, dx = 0.1, x* = 3

y > 0,

y(0) = 1, dx = 0.1, x* = 1

43. y′ = x + y, b =1

y(0) = −7 10; − 4 ≤ x ≤ 4, − 4 ≤ y ≤ 4;

44. y′ = −x y , y(0) = 2; − 3 ≤ x ≤ 3, − 3 ≤ y ≤ 3; b = 2 45. y′ = y ( 2 − y ) , b = 3

y(0) = 1 2; 0 ≤ x ≤ 4, 0 ≤ y ≤ 3;

46. y′ = ( sin x )( sin y ) , y(0) = 2; − 6 ≤ x ≤ 6, − 6 ≤ y ≤ 6; b = 3π 2

y(0) = 0, dx = 0.1, x* = 1

16.2 First-Order Linear Equations A first-order linear differential equation is one that can be written in the form dy + P ( x ) y = Q( x ), dx

(1)

where P and Q are continuous functions of x. Equation (1) is the linear equation’s standard form. Since the exponential growth decay equation dy dx = ky (Section 7.2) can be put in the standard form dy − ky = 0, dx we see it is a linear equation with P ( x ) = −k and Q( x ) = 0. Equation (1) is linear (in y) because y and its derivative dy dx occur only to the first power, they are not multiplied together, nor do they appear as the argument of a function (such as sin y, e y, or dy dx).

EXAMPLE 1

Put the following equation in standard form: x

dy = x 2 + 3 y, dx

x > 0.

16.2  First-Order Linear Equations

1033

Solution

dy = x 2 + 3y dx dy 3 = x+ y dx x dy 3 − y = x dx x x

Divide by x. Standard form with P ( x ) = −3 x and  Q( x ) = x

Notice that P ( x ) is −3 x , not +3 x . The standard form is y′ + P ( x ) y = Q( x ), so the minus sign is part of the formula for P ( x ).

Solving Linear Equations We solve the equation dy + P ( x ) y = Q( x ) dx by multiplying both sides by a positive function υ( x ) that transforms the left-hand side into the derivative of the product υ( x ) ⋅ y. We will show how to find υ in a moment, but first we want to show how, once found, it provides the solution we seek. Here is why multiplying by υ( x ) works: dy + P ( x ) y = Q( x ) dx υ( x )

Original equation is in standard form.

dy + P ( x )υ( x ) y = υ( x )Q( x ) dx

Multiply by positive υ( x ). υ( x ) is chosen to make dy d( υ + Pυ y = υ ⋅ y ). dx dx

d( υ( x ) ⋅ y ) = υ( x )Q( x ) dx υ( x ) ⋅ y = y =

∫ υ( x )Q( x ) dx

Integrate with respect to x.

1 υ( x )Q( x ) dx υ( x ) ∫

Divide by υ( x ).

(2)

Equation (2) expresses the solution of Equation (1) in terms of the functions υ( x ) and Q( x ). We call υ( x ) an integrating factor for Equation (1) because its presence makes the equation integrable. Why doesn’t the formula for P ( x ) appear in the solution as well? It does, but indirectly, in the construction of the positive function υ( x ) . We have dy d + Pυ y ( υ y) = υ dx dx dy dy dυ υ + y = υ + Pυ y dx dx dx dυ = P υ y. y dx

Condition imposed on υ Derivative Product Rule The terms υ

This last equation will hold if dυ = Pυ dx

∫

dυ = P dx υ

Variables separated, υ > 0

dυ = υ

Integrate both sides.

∫ P dx

dy cancel. dx

1034

Chapter 16 First-Order Differential Equations

ln υ =

Since υ > 0, we do not need absolute value signs in ln υ.

∫ P dx

e ln υ = e ∫ P dx υ = e

∫ P dx

Exponentiate both sides to solve for υ.

(3)

.

Thus a formula for the general solution to Equation (1) is given by Equation (2), where υ( x ) is given by Equation (3). However, rather than memorizing the formula, just remember how to find the integrating factor once you have the standard form so P ( x ) is correctly identified. Any antiderivative of P works for Equation (3).

Integrating Factors

To solve the linear equation y′ + P ( x ) y = Q( x ), multiply both sides by the integrating factor υ( x ) = e ∫ P ( x ) dx and integrate both sides.

When you integrate the product on the left-hand side in this procedure, you always obtain the product υ( x ) y of the integrating factor and solution function y because of the way υ is defined. EXAMPLE 2

Solve the equation x

HISTORICAL BIOGRAPHY Adrien-Marie Legendre (1752–1833) Legendre, a French mathematician, taught at the École polytechnique and won a research prize from the Berlin Academy in 1782. He encountered and developed polynomials, today named for him, in his research on the gravitational attraction of ellipsoids. He devoted 40 years to this research to elliptic integrals.

dy = x 2 + 3 y, dx

x > 0.

Solution First we put the equation in standard form (Example 1):

dy 3 − y = x, dx x so P ( x ) = −3 x is identified. The integrating factor is υ( x ) = e ∫ P ( x ) dx = e ∫ (−3 x ) dx = e −3 ln

Constant of integration is 0, so υ is as simple as possible.

x

= e −3 ln x

To know more, visit the companion Website.

= e

ln x −3

x > 0

1 = 3. x

Next we multiply both sides of the standard form by υ( x ) and integrate:

(

) = x1

dy 1 3 ⋅ − y x 3 dx x 1 dy 3 − 4y 3 x dx x d 1 y dx x 3 1 y x3 1 y x3

(

)

3

⋅x

1 x2 1 = 2 x =

1 dx x2 1 = − + C. x =

∫

Left-hand side is

d( υ ⋅ y ). dx

Integrate both sides.

Solving this last equation for y gives the general solution:

(

y = x3 −

)

1 + C = −x 2 + Cx 3, x

x > 0.

16.2  First-Order Linear Equations

EXAMPLE 3

1035

Find the particular solution of 3 xy′ − y = ln x + 1,

x > 0,

that satisfying y(1) = −2. Solution With x > 0, we write the equation in standard form:

y′ −

ln x + 1 1 . y = 3x 3x

Then the integrating factor is given by υ = e ∫ −1 ( 3 x ) dx = e (−1 3) ln x = x −1 3.

  x > 0

Thus 1 ( ln x + 1) x −4 3 dx. 3∫ Integration by parts of the right-hand side gives x −1 3 y =

x −1 3 y = −x −1 3 ( ln x + 1) +

  

  

 

Left-hand side is υ y.

∫ x −4 3 dx + C.

Therefore, x −1 3 y = −x −1 3 ( ln x + 1) − 3 x −1 3 + C or, solving for y, y = −( ln x + 4 ) + Cx 1 3. When x = 1 and y = −2, this last equation becomes −2 = −( 0 + 4 ) + C , so C = 2. Substitution into the equation for y gives the particular solution y = 2 x 1 3 − ln x − 4.

In solving the linear equation in Example 2, we integrated both sides of the equation after multiplying each side by the integrating factor. However, we can shorten the amount of work, as in Example 3, by remembering that the left-hand side always integrates into the product υ( x ) ⋅ y of the integrating factor times the solution function. From Equation (2) this means that υ( x ) y =

∫ υ( x )Q( x ) dx.

(4)

We need only integrate the product of the integrating factor υ( x ) with Q( x ) on the righthand side of Equation (1) and then equate the result with υ( x ) y to obtain the general solution. Nevertheless, to emphasize the role of υ( x ) in the solution process, we sometimes follow the complete procedure as illustrated in Example 2. Observe that if the function Q( x ) is identically zero in the standard form given by Equation (1), the linear equation is separable and can be solved by the method of Section 7.2: dy + P ( x ) y = Q( x ) dx dy + P(x) y = 0 dx dy = −P ( x ) dx. y

Q( x ) = 0 Separating the variables

1036

Chapter 16 First-Order Differential Equations

V

+

RL Circuits

−

The diagram in Figure 16.9 represents an electrical circuit whose total resistance is a constant R ohms and whose self-inductance, shown as a coil, is L henries, also a constant. There is a switch whose terminals at a and b can be closed to connect a constant electrical source of V volts. Ohm’s Law, V = RI, has to be augmented for such a circuit. The correct equation accounting for both resistance and inductance is

Switch a

b

i R

L

FIGURE 16.9

The RL circuit

L

in Example 4.

di + Ri = V, dt

(5)

where i is the current in amperes and t is the time in seconds. By solving this equation, we can predict how the current will flow after the switch is closed. EXAMPLE 4 The switch in the RL circuit in Figure 16.9 is closed at time t = 0. How will the current flow as a function of time? Solution Equation (5) is a first-order linear differential equation for i as a function of t. Its standard form is

di R V + i = , dt L L

(6)

and the corresponding solution, given that i = 0 when t = 0, is i =

i I= V R

I e

L R

2

L R

3

L R

4

L R

t →∞

t

t →∞

( VR − VR e

The growth of the current in the RL circuit in Example 4. I is the current’s steady-state value. The number t = L R is the time constant of the circuit. The current gets to within 5% of its steady-state value in 3 time constants (Exercise 27).

L

) = VR − VR ⋅ 0 = VR .

di + Ri = V, dt

I = V R is the current that will flow in the circuit if either L = 0 (no inductance) or di dt = 0 (steady current, i = constant) (Figure 16.10). Equation (7) expresses the solution of Equation (6) as the sum of two terms: a steady-state solution V R and a transient solution −( V R ) e −( R L )t that tends to zero as t → ∞.

16.2

First-Order Linear Equations

Solve the differential equations in Exercises 1–14. dy + y = e x, x > 0 1. x dx dy + 2e x y = 1 dx sin x 3. xy′ + 3 y = , x > 0 x2

2. e x

−( R L )t

At any given time, the current is less than V R, but as time passes, the current approaches the steady-state value V R. According to the equation

FIGURE 16.10

EXERCISES

(7)

(We leave the calculation of the solution to Exercise 28.) Since R and L are positive, −( R L ) is negative and e −( R L )t → 0 as t → ∞. Thus,

i = V (1 − e−RtL) R

lim i = lim

0

V V − e −( R L )t. R R

4. y′ + ( tan x ) y = cos 2 x , −π 2 < x < π 2 dy 1 + 2y = 1 − , x > 0 x dx 6. (1 + x )  y′ + y = x 7. 2 y′ = e x 2 + y

5. x

8. e 2 x y′ + 2e 2 x y = 2 x 9. xy′ − y = 2 x ln x cos x dy 10. x = − 2 y, x > 0 dx x

16.2  First-Order Linear Equations

11. ( t − 1)3 12. ( t + 1)

ds + 4 ( t − 1) 2 s = t + 1, t > 1 dt

1037

value I = V R , the decaying current (see accompanying figure) obeys the equation

ds 1 + 2 s = 3( t + 1 ) + , t > −1 ( t + 1) 2 dt

L

di + Ri = 0, dt

13. sin θ

dr + ( cos θ )r = tan θ, 0 < θ < π 2 dθ

which is Equation (5) with V = 0.

14. tan θ

dr + r = sin 2 θ, 0 < θ < π 2 dθ

b. How long after the switch is thrown will it take the current to fall to half its original value?

Solving Initial Value Problems

Solve the initial value problems in Exercises 15–20. dy + 2 y = 3, y(0) = 1 15. dt 16. t

dy + 2 y = t 3, t > 0, y(2) = 1 dt

17. θ

dy + y = sin θ, θ > 0, y ( π 2 ) = 1 dθ

18. θ

dy − 2 y = θ 3 sec θ tan θ, θ > 0, y ( π 3 ) = 2 dθ

a. Solve the equation to express i as a function of t.

c. Show that the value of the current when t = L R is I e. (The significance of this time is explained in the next exercise.) i V R

I e 0

dy e x2 19. ( x + 1) − 2( x 2 + x ) y = , x > −1, y(0) = 5 x +1 dx 20.

dy + xy = x , y(0) = −6 dx

21. Solve the exponential growth/decay initial value problem for y as a function of t by thinking of the differential equation as a firstorder linear equation with P ( x ) = −k and Q( x ) = 0: dy = ky ( k constant), y(0) = y 0 dt 22. Solve the following initial value problem for u as a function of t: du k + u = 0 ( k  and m  positive constants), u(0) = u 0 dt m a. as a first-order linear equation. b. as a separable equation.

L R

2

L R

3

L R

t

27. Time constants Engineers call the number L R the time constant of the RL circuit in Figure 16.10. The significance of the time constant is that the current will reach 95% of its final value within 3 time constants of the time the switch is closed (Figure 16.10). Thus, the time constant gives a built-in measure of how rapidly an individual circuit will reach equilibrium. a. Find the value of i in Equation (7) that corresponds to t = 3 L R , and show that it is about 95% of the steady-state value I = V R. b. Approximately what percentage of the steady-state current will be flowing in the circuit 2 time constants after the switch is closed (i.e., when t = 2 L R )? 28. Derivation of Equation (7) in Example 4 a. Show that the solution of the equation

Theory and Examples

di R V + i = dt L L

23. Is either of the following equations correct? Give reasons for your answers. a. x ∫

1 dx = x ln x + C x

1 dx = x ln x + Cx x 24. Is either of the following equations correct? Give reasons for your answers. b. x ∫

1 a. cos x 1 b. cos x

∫ cos x dx ∫

= tan x + C

C cos x dx = tan x + cos x

25. Current in a closed RL circuit How many seconds after the switch in an RL circuit is closed will it take the current i to reach half of its steady-state value? Notice that the time depends on R and L, not on how much voltage is applied. 26. Current in an open RL circuit If the switch is thrown open after the current in an RL circuit has built up to its steady-state

is i =

V + Ce −( R L )t. R

b. Then use the initial condition i(0) = 0 to determine the value of C. This will complete the derivation of Equation (7). c. Show that i = V R is a solution of Equation (6) and that i = Ce −( R L )t satisfies the equation di R + i = 0. dt L A Bernoulli differential equation is of the form dy + P ( x ) y = Q( x ) y n . dx Observe that, if n = 0 or 1, the Bernoulli equation is linear. For other values of n, the substitution u = y 1−n transforms the Bernoulli equation into the linear equation du ( + 1 − n ) P ( x )u = (1 − n ) Q( x ). dx

1038

Chapter 16 First-Order Differential Equations

For example, in the equation dy − y = e −x y 2, dx we have n = 2, so that u = y 1−2 = y −1 and du dx = − y −2 dy dx . Then dy dx = − y 2 du dx = −u −2 du dx . Substitution into the original equation gives du −u −2 − u −1 = e − x u −2 , dx or, equivalently, du + u = −e −x. dx This last equation is linear in the (unknown) dependent variable u.

HISTORICAL BIOGRAPHY Jakob Bernoulli (1654–1705) Jakob Bernoulli was born in Switzerland and received his degree in 1671 after studying philosophy and theology at the request of his father and mathematics and astronomy against the will of his father. Working on problems in optics and mechanics, Bernoulli contributed to important developments in infinitesimal geometry and calculus. To know more, visit the companion Website.

Solve the Bernoulli equations in Exercises 29–32. 29. y′ − y = − y 2 31. xy′ + y = y −2

30. y′ − y = xy 2 32. x 2 y′ + 2 xy = y 3

16.3 Applications We now look at four applications of first-order differential equations. The first application analyzes an object moving along a straight line while subject to a force opposing its motion. The second is a model of population growth. The third application considers a curve or curves intersecting each curve in a second family of curves orthogonally (that is, at right angles). The final application analyzes chemical concentrations entering and leaving a container. The various models involve separable or linear first-order equations.

Motion with Resistance Proportional to Velocity In some cases it is reasonable to assume that the resistance encountered by a moving object, such as a car coasting to a stop, is proportional to the object’s velocity. The faster the object moves, the more its forward progress is resisted by the air through which it passes. Picture the object as a mass m moving along a coordinate line with position function s and velocity υ at time t. From Newton’s second law of motion, the resisting force opposing the motion is dυ Force = mass × acceleration = m . dt If the resisting force is proportional to velocity, we have dυ dυ k ( k > 0 ). = −k υ or = − υ dt dt m This is a separable differential equation representing exponential change. The solution to the equation with initial condition υ = υ 0 at t = 0 is (Section 7.2) m

υ = υ 0 e −( k m )t.

(1)

What can we learn from Equation (1)? For one thing, we can see that if m is something large, like the mass of a 20,000-ton ore boat in Lake Erie, it will take a long time for the velocity to approach zero (because t must be large in the exponent of the equation in order to make kt m large enough for υ to be small). We can learn even more if we integrate Equation (1) to find the position s as a function of time t. Suppose that an object is coasting to a stop and the only force acting on it is a resistance proportional to its speed. How far will it coast? To find out, we start with Equation (1) and solve the initial value problem ds = υ 0 e −( k m )t , dt

s(0) = 0.

16.3  Applications

1039

Integrating with respect to t gives s = −

υ 0 m −( k m )t e + C. k

Substituting s = 0 when t = 0 gives 0 = −

υ0m +C k

C =

and

υ0m . k

The body’s position at time t is therefore s (t ) = −

υ 0 m −( k m )t υ m υ m e + 0 = 0 (1 − e −( k m )t ). k k k

(2)

To find how far the body will coast, we find the limit of s(t ) as t → ∞. Since −( k m ) < 0, we know that e −( k m )t → 0 as t → ∞, so that lim s(t ) = lim

t →∞

t →∞

υ0m υ m υ m (1 − e −( k m )t ) = 0 (1 − 0 ) = 0 . k k k

Thus, Distance coasted =

υ0m . k

(3)

The number υ 0 m k is only an upper bound (albeit a useful one). It is true to life in one respect, at least: If m is large, the body will coast a long way.

EXAMPLE 1 For a 90-kg ice skater, the k in Equation (1) is about 5 kg/s. How long will it take the skater to coast from 3.3 m/s (11.88 km/h) to 0.3 m/s? How far will the skater coast before coming to a complete stop? Solution We answer the first question by solving Equation (1) for t:

33e −t 18 = 0.3 e

−t 18

Eq. (1) with k = 5,

= 1 11

m = 90, υ 0 = 3.3, υ = 0.3

−t 18 = ln (1 11) = − ln 11 t = 18 ln 11 ≈ 43 s. We answer the second question with Equation (3): Distance coasted =

υ0m 3.3 ⋅ 90 = = 59.4 m. 5 k

Inaccuracy of the Exponential Population Growth Model In Section 7.2 we modeled population growth with the Law of Exponential Change: dP = kP , dt

P (0) = P0 ,

where P is the population at time t, k > 0 is a constant growth rate, and P0 is the size of the population at time t = 0. In Section 7.2 we found the solution P = P0 e kt to this model. To assess the model, notice that the exponential growth differential equation says that dP dt = k P

(4)

is constant. This rate is called the relative growth rate. We can use this to predict total future world population based on historical data. Table 16.3 gives the world population at

1040 P

Chapter 16 First-Order Differential Equations

TABLE 16.3 World population (midyear)

World population (1980–2020)

9000

8000 P=

4454e 0.017t

5000

4000

0

10

40

FIGURE 16.11

t

The value of the solution P = 4454e 0.017 t is 8792 when t = 40, which is nearly 13% more than the actual population in 2020.

Year

Population (millions)

ΔP P

1980

4454

76 4454 ≈ 0.0171

1981

4530

80 4530 ≈ 0.0177

1982

4610

80 4610 ≈ 0.0174

1983

4690

80 4690 ≈ 0.0171

1984

4770

81 4770 ≈ 0.0170

1985

4851

82 4851 ≈ 0.0169

1986

4933

85 4933 ≈ 0.0172

1987

5018

87 5018 ≈ 0.0173

1988

5105

85 5105 ≈ 0.0167

1989

5190

midyear for the years 1980 to 19816. Taking dt = 1 and dP ≈ ΔP, we see from the table that the relative growth rate in Equation (4) is approximately equal to 0.017. Thus, based on the tabled data with t = 0 representing 1980, t = 1 representing 1981, and so forth, the world population could be modeled by the initial value problem

Orthogonal trajectory

dP = 0.017 P , dt FIGURE 16.12

An orthogonal trajectory intersects the family of curves at right angles, or orthogonally. y

P (0) = 4454.

The solution to this initial value problem gives the population function P = 4454e 0.017 t. In year 2008 (so t = 28 ), the solution predicts the world population in midyear to be about 7169 million, or 7.2 billion (Figure 16.11), which is more than the actual population of 6707 million, an error of about 7%. The error grows as the number of years increases. For 2020 ( t = 40 ) the model predicts a population of 8792 million. The reported population for 2020 is 7795 million, an overprediction error of about 13%. A more realistic model would consider environmental, economic, and other factors affecting the growth rate, which has been steadily declining. We consider one such model in Section 16.4.

Orthogonal Trajectories x

FIGURE 16.13

Every straight line through the origin is orthogonal to the family of circles centered at the origin.

An orthogonal trajectory of a family of curves is a curve that intersects each curve of the family at right angles, or orthogonally (Figure 16.12). For instance, each straight line through the origin is an orthogonal trajectory of the family of circles x 2 + y 2 = a 2, centered at the origin (Figure 16.13). Such mutually orthogonal systems of curves are of particular importance in physical problems related to electrical potential, where the curves in one family correspond to strength of an electric field, and those in the other family correspond to constant electric potential. They also occur in hydrodynamics and heat-flow problems. EXAMPLE 2 Find the orthogonal trajectories of the family of curves xy = a, where a ≠ 0 is an arbitrary constant. Solution The curves xy = a form a family of hyperbolas having the coordinate axes as asymptotes. First we find the slopes of each curve in this family, or their dy dx values. Differentiating xy = a implicitly gives

x

dy + y = 0 dx

or

dy y = − . dx x

16.3  Applications

y

x2 − y2 = b b≠0 0

Thus the slope of the tangent line at any point ( x , y ) on one of the hyperbolas xy = a is y′ = −y x . On an orthogonal trajectory the slope of the tangent line at this same point must be the negative reciprocal, or x y. Therefore, the orthogonal trajectories must satisfy the differential equation dy x = . y dx

x x y = a, a≠0

1041

This differential equation is separable, and we solve it as in Section 7.2: y dy = x dx

∫ y dy

=

∫ x dx

Separate variables. Integrate both sides.

1 2 1 y = x2 + C 2 2 y 2 − x 2 = b,

FIGURE 16.14

Each curve is orthogonal to every curve it meets in the other family (Example 2).

(5)

where b = 2C is an arbitrary constant. The orthogonal trajectories are the family of hyperbolas given by Equation (5) and sketched in Figure 16.14.

Mixture Problems Suppose a chemical in a liquid solution (or dispersed in a gas) runs into a container holding the liquid (or the gas) with, possibly, a specified amount of the chemical dissolved as well. The mixture is kept uniform by stirring and flows out of the container at a known rate. In this process, it is often important to know the concentration of the chemical in the container at any given time. The differential equation describing the process is based on the formula ⎛ rate at which ⎞⎟ ⎛ rate at which ⎞⎟ Rate of change ⎜⎜ ⎟ ⎟ ⎜⎜ of amount = ⎜⎜ chemical ⎟⎟⎟ − ⎜⎜ chemical ⎟⎟⎟. ⎟ ⎜⎜ ⎟⎟ ⎜⎜ in container arrives ⎠ ⎝ departs. ⎟⎟⎠ ⎝

(6)

If y(t ) is the amount of chemical in the container at time t, and V (t ) is the total volume of liquid in the container at time t, then the departure rate of the chemical at time t is y( t ) ⋅ ( outflow rate ) V (t ) ⎛ concentration in ⎞⎟ ⎟ ⋅ ( outflow rate ). = ⎜⎜⎜ ⎜⎝ container at time t ⎟⎟⎠

Departure rate =

(7)

Accordingly, Equation (6) becomes dy y( t ) = ( chemical’s arrival rate ) − ⋅ ( outflow rate ). dt V (t )

(8)

If, say, y is measured in kilograms, V in liters, and t in minutes, the units in Equation (8) are kilograms kilograms kilograms liters . = − ⋅ minutes minutes liters minutes

EXAMPLE 3 In an oil refinery, a storage tank contains 10,000 L of gasoline that initially has 50 kg of an additive dissolved in it. In preparation for winter weather, gasoline containing 0.2 kg of additive per liter is pumped into the tank at a rate of 200 L/min.

1042

Chapter 16 First-Order Differential Equations

The well-mixed solution is pumped out at a rate of 220 L/min. How much of the additive is in the tank 20 min after the pumping process begins (Figure 16.15)? 200 L/min containing 0.2 kg/L

220 L/min containing y kg/L V

FIGURE 16.15

The storage tank in Example 3 mixes input liquid with stored liquid to produce an output liquid.

Solution Let y be the amount (in kilograms) of additive in the tank at time t. We know that y = 50 when t = 0. The number of liters of gasoline and additive in solution in the tank at any time t is

(

V (t ) = 10, 000 L + 200

)

L L ( − 220 t min ) = (10, 000 − 20t ) L. min min

Therefore, y (t ) ⋅ outflow rate V (t ) y = 200 10, 000 − 20t

Rate out =

(

=

Eq. (7)

)

Outflow rate is 220 L min and V = 10,000 − 20t.

220 y kg . 10, 000 − 20t min

Also,

(

Rate in = 0.2

)(

)

kg kg L 200 = 40 . L min min

The differential equation modeling the mixture process is dy 220 y = 40 − dt 10, 000 − 20t

  

Eq. (8)

in kilograms per minute. To solve this differential equation, we first write it in standard linear form: dy 220 + y = 40. dt 10, 000 − 20t Thus, P(t ) = 220 (10, 000 − 20t ) and Q(t ) = 40. The integrating factor is 220

υ(t ) = e ∫ P dt = e ∫ 10,000 − 20 t

dt

= e −11 ln (10,000 − 20 t )

   10,000 − 20t > 0 −11

= (10, 000 − 20t )

.

16.3  Applications

1043

Multiplying both sides of the standard equation by υ(t ) and integrating both sides gives (10, 000 (10, 000

− 20t )−11 ⋅

− 20t )−11

( dydt + 10, 000220− 20t y ) = 40 10, 000 − 20t (

)−11

dy + 220 (10, 000 − 20t )−12 y = 40 (10, 000 − 20t )−11 dt d ( [ 10, 000 − 20t )−11 y ] = 40 (10, 000 − 20t )−11 dt

∫ 40 (10, 000 − 20t )−11 dt

(10, 000

− 20t )−11 y =

(10, 000

− 20t )−11 y = 40 ⋅

− 20t )−10 + C. ( − 10 )( − 20 )

(10, 000

The general solution is y = 0.2 (10, 000 − 20t ) + C (10, 000 − 20t )11. Because y = 50 when t = 0, we can determine the value of C: 50 = 0.2 (10, 000 − 0 ) + C (10, 000 − 0 )11 C = −

1950 (10, 000 )11

.

The particular solution of the initial value problem is y = 0.2 (10, 000 − 20t ) −

1950 (10, 000 )11

(10, 000

− 20t )11.

The amount of additive in the tank 20 min after the pumping begins is

y (20) = 0.2[ 10, 000 − 20 ( 20 ) ] −

EXERCISES

1950 (10, 000 )11

[ 10, 000

− 20 ( 20 ) ]11 ≈ 675 kg.

16.3

Motion Along a Line

1. Coasting bicycle A 66-kg cyclist on a 7-kg bicycle starts coasting on level ground at 9 m s. The k in Equation (1) is about 3.9 kg s.

TABLE 16.4 Ashley Sharritts skating data

t (s)

s (m)

t (s)

s (m)

t (s)

s (m)

0

0

2.24

3.05

4.48

4.77

0.16

0.31

2.40

3.22

4.64

4.82

0.32

0.57

2.56

3.38

4.80

4.84

0.48

0.80

2.72

3.52

4.96

4.86

0.64

1.05

2.88

3.67

5.12

4.88

0.80

1.28

3.04

3.82

5.28

4.89

0.96

1.50

3.20

3.96

5.44

4.90

a. About how far will the ship coast before it is dead in the water?

1.12

1.72

3.36

4.08

5.60

4.90

b. About how long will it take the ship’s speed to drop to 1 m s?

1.28

1.93

3.52

4.18

5.76

4.91

1.44

2.09

3.68

4.31

5.92

4.90

1.60

2.30

3.84

4.41

6.08

4.91

1.76

2.53

4.00

4.52

6.24

4.90

1.92

2.73

4.16

4.63

6.40

4.91

2.08

2.89

4.32

4.69

6.56

4.91

a. About how far will the cyclist coast before reaching a complete stop? b. How long will it take the cyclist’s speed to drop to 1 m s? 2. Coasting battleship An Iowa class battleship has mass around 51,000 metric tons (51,000,000 kg) and a k value in Equation (1) of about 59, 000 kg s. Assume that the ship loses power when it is moving at a speed of 9 m s.

3. The data in Table 16.4 were collected with a motion detector and a CBL™ by Valerie Sharritts, then a mathematics teacher at St. Francis DeSales High School in Columbus, Ohio. The table shows the distance s (meters) coasted on inline skates in t s by her daughter Ashley when she was 10 years old. Find a model for Ashley’s position given by the data in Table 16.4 in the form of

1044

Chapter 16 First-Order Differential Equations

Equation (2). Her initial velocity was υ 0 = 2.75 m s, her mass m = 39.92 kg , and her total coasting distance was 4.91 m. 4. Coasting to a stop Table 16.5 shows the distance s (meters) coasted on inline skates in terms of time t (seconds) by Kelly Schmitzer. Find a model for her position in the form of Equation (2). Her initial velocity was υ 0 = 0.80 m s, her mass m = 49.90 kg, and her total coasting distance was 1.32 m.

Mixture Problems

13. Salt mixture A tank initially contains 400 L of brine in which 20 kg/L of salt are dissolved. A brine containing 0.2 kg/L of salt runs into the tank at the rate of 20 L/min. The mixture is kept uniform by stirring and flows out of the tank at the rate of 16 L/min. a. At what rate (kilograms per minute) does salt enter the tank at time t? b. What is the volume of brine in the tank at time t? c. At what rate (kilograms per minute) does salt leave the tank at time t?

TABLE 16.5 Kelly Schmitzer skating data

t (s)

s (m)

t (s)

s (m)

t (s)

s (m)

d. Write down and solve the initial value problem describing the mixing process.

0

0

1.5

0.89

3.1

1.30

0.1

0.07

1.7

0.97

3.3

1.31

e. Find the concentration of salt in the tank 25 min after the process starts.

0.3

0.22

1.9

1.05

3.5

1.32

0.5

0.36

2.1

1.11

3.7

1.32

0.7

0.49

2.3

1.17

3.9

1.32

0.9

0.60

2.5

1.22

4.1

1.32

1.1

0.71

2.7

1.25

4.3

1.32

1.3

0.81

2.9

1.28

4.5

1.32

Orthogonal Trajectories

In Exercises 5–10, find the orthogonal trajectories of the family of curves. Sketch several members of each family. 6. y = cx 2

5. y = mx 8.

2x 2

+

y2

=

c2

9. y =

ce −x

7. kx 2 + y 2 = 1 10. y = e kx

11. Show that the curves 2 x 2 + 3 y 2 = 5 and y 2 = x 3 are orthogonal. 12. Find the family of solutions of the given differential equation and the family of orthogonal trajectories. Sketch both families. a. x dx + y dy = 0 b. x dy − 2 y dx = 0

14. Mixture problem A 800-L tank is half full of distilled water. At time t = 0, a solution containing 50 grams/L of concentrate enters the tank at the rate of 20 L/min, and the well-stirred mixture is withdrawn at the rate of 12 L/min. a. At what time will the tank be full? b. At the time the tank is full, how many kilograms of concentrate will it contain? 15. Fertilizer mixture A tank contains 400 L of fresh water. A solution containing 0.1 kg/L of soluble lawn fertilizer runs into the tank at the rate of 4 L/min, and the mixture is pumped out of the tank at the rate of 12 L/min. Find the maximum amount of fertilizer in the tank and the time required to reach the maximum. 16. Carbon monoxide pollution An executive conference room of a corporation contains 120 m 3 of air initially free of carbon monoxide. Starting at time t = 0, cigarette smoke containing 4% carbon monoxide is blown into the room at the rate of 0.008 m 3 min. A ceiling fan keeps the air in the room well circulated and the air leaves the room at the same rate of 0.008 m 3 min. Find the time when the concentration of carbon monoxide in the room reaches 0.01%.

16.4 Graphical Solutions of Autonomous Equations In Chapter 4 we learned that the sign of the first derivative tells where the graph of a function is increasing and where it is decreasing. The sign of the second derivative tells the concavity of the graph. We can build on our knowledge of how derivatives determine the shape of a graph to solve differential equations graphically. We will see that the ability to discern physical behavior from graphs is a powerful tool in understanding real-world systems. The starting ideas for a graphical solution are the notions of phase line and equilibrium value. We arrive at these notions by investigating, from a point of view quite different from that studied in Chapter 4, what happens when the derivative of a differentiable function is zero.

Equilibrium Values and Phase Lines When we differentiate implicitly the equation 1 ( ln 5 y − 15 ) = x + 1, 5

16.4  Graphical Solutions of Autonomous Equations

1045

we obtain ⎞⎟ dy 1 ⎛⎜ 5 = 1. ⎟ 5 ⎜⎝ 5 y − 15 ⎟⎠ dx Solving for y′ = dy dx , we find y′ = 5 y − 15 = 5( y − 3 ). In this case the derivative y′ is a function of y only (the dependent variable) and is zero when y = 3. A differential equation for which dy dx is a function of y only is called an autonomous differential equation. Let’s investigate what happens when the derivative in an autonomous equation equals zero. We assume any derivatives are continuous.

If dy dx = g( y) is an autonomous differential equation, then the values of y for which dy dx = 0 are called equilibrium values or rest points.

DEFINITION

Thus, equilibrium values are those at which no change occurs in the dependent variable, so y is at rest. The emphasis is on the value of y where dy dx = 0, not the value of x, as we studied in Chapter 4. For example, the equilibrium values for the autonomous differential equation dy = ( y + 1)( y − 2 ) dx are y = −1 and y = 2. To construct a graphical solution to an autonomous differential equation, we first make a phase line for the equation, a plot on the y-axis that shows the equation’s equilibrium values along with the intervals where dy dx and d 2 y dx 2 are positive and negative. Then we know where the solutions are increasing and decreasing, and the concavity of the solution curves. These are the essential features we found in Section 4.4, so we can determine the shapes of the solution curves without having to find formulas for them.

EXAMPLE 1

Draw a phase line for the equation dy = ( y + 1)( y − 2 ) , dx

and use it to sketch solutions to the equation. Solution

1. Draw a number line for y and mark the equilibrium values y = −1 and y = 2, where dy dx = 0. y

−1

2

2. Identify and label the intervals where y′ > 0 and y′ < 0. This step resembles what we did in Section 4.3, only now we are marking the y-axis instead of the x-axis.

y′ > 0

y′ < 0 –1

y′ > 0

y

2

We can encapsulate the information about the sign of y′ on the phase line itself. Since y′ > 0 on the interval to the left of y = −1, a solution of the differential equation

1046

Chapter 16 First-Order Differential Equations

with a y-value less than −1 will increase from there toward y = −1. We display this information by drawing an arrow on the interval pointing to −1. y

−1

2

Similarly, y′ < 0 between y = −1 and y = 2, so any solution with a value in this interval will decrease toward y = −1. For y > 2, we have y′ > 0, so a solution with a y-value greater than 2 will increase from there without bound. In short, solution curves below the horizontal line y = −1 in the xy-plane rise toward y = −1. Solution curves between the lines y = −1 and y = 2 fall away from y = 2 toward y = −1. Solution curves above y = 2 rise away from y = 2 and keep going up. 3. Calculate y ″ and mark the intervals where y ″ > 0 and y′′ < 0. To find y ″, we differentiate y′ with respect to x, using implicit differentiation. y′ = ( y + 1)( y − 2 ) = y 2 − y − 2 d( ) d( 2 y′ = y − y − 2) dx dx = 2 yy′ − y′

y″ =

= ( 2 y − 1) y′ = ( 2 y − 1)( y + 1)( y − 2 ).

y

1 2 0 −1

y′ > 0 y″ < 0

y′ < 0 y″ < 0 y′ < 0 y″ > 0

x

y′ > 0 y″ < 0

FIGURE 16.16

Substitute for y′. Differentiate implicitly with respect to x.

From this formula, we see that y ″ changes sign at y = −1, y = 1 2, and y = 2. We add the sign information to the phase line.

y′ > 0 y″ > 0 2

Formula for y′

Graphical solutions from Example 1 include the horizontal lines y = −1 and y = 2 through the equilibrium values. No two solution curves can ever cross or touch each other.

y′ < 0 y″ > 0 −1

y′ < 0 y″ < 0 1 2

y′ > 0 y″ > 0 y 2

4. Sketch an assortment of solution curves in the xy-plane. The horizontal lines y = −1, y = 1 2, and y = 2 partition the plane into horizontal bands in which we know the signs of y′ and y ″. In each band, this information tells us whether the solution curves rise or fall and how they bend as x increases (Figure 16.16). The “equilibrium lines” y = −1 and y = 2 are also solution curves. (The constant functions y = −1 and y = 2 satisfy the differential equation.) Solution curves that cross the line y = 1 2 have an inflection point there. The concavity changes from concave down (above the line) to concave up (below the line). As predicted in Step 2, solutions in the middle and lower bands approach the equilibrium value y = −1 as x increases. Solutions in the upper band rise steadily away from the value y = 2.

Stable and Unstable Equilibria Look at Figure 16.16 once more, in particular at the behavior of the solution curves near the equilibrium values. Once a solution curve has a value near y = −1, it tends steadily toward that value; y = −1 is a stable equilibrium. The behavior near y = 2 is just the opposite: All solutions except the equilibrium solution y = 2 itself move away from it as x increases. We call y = 2 an unstable equilibrium. If the solution is at that value, it stays, but if it is off by any amount, no matter how small, it moves away. (Sometimes an equilibrium value is unstable because a solution moves away from it only on one side of the point.) Now that we know what to look for, we can already see this behavior on the initial phase line (the second diagram in Step 2 of Example 1). The arrows lead away from y = 2 and, once to the left of y = 2, toward y = −1.

16.4  Graphical Solutions of Autonomous Equations

1047

We now present several applied examples for which we can sketch a family of solution curves to the differential equation models using the method in Example 1.

Newton’s Law of Cooling In Section 7.2 we solved analytically the differential equation

dH 0 dt

H

15

FIGURE 16.17

First step in constructing the phase line for Newton’s Law of Cooling. The temperature tends toward the equilibrium (surrounding-medium) value in the long run. dH >0 dt

dH 0

modeling Newton’s Law of Cooling. Here H is the temperature of an object at time t, and H S is the constant temperature of the surrounding medium. Suppose that the surrounding medium (say, a room in a house) has a constant Celsius temperature of 15 °C. We can then express the difference in temperature as H (t ) − 15. Assuming H is a differentiable function of time t, by Newton’s Law of Cooling, there is a constant of proportionality k > 0 such that dH = −k ( H − 15 ) dt

H

15

FIGURE 16.18

The complete phase line for Newton’s Law of Cooling. H Initial temperature

dH = −k ( H − H S ) , dt

Temperature of surrounding medium

15

(1)

(minus k to give a negative derivative when H > 15). Since dH dt = 0 at H = 15, the temperature 15 °C is an equilibrium value. If H > 15, Equation (1) tells us that ( H − 15 ) > 0 and dH dt < 0. If the object is hotter than the room, it will get cooler. Similarly, if H < 15, then ( H − 15 ) < 0 and dH dt > 0. An object cooler than the room will warm up. Thus, the behavior described by Equation (1) agrees with our intuition of how temperature should behave. These observations are captured in the initial phase line diagram in Figure 16.17. The value H = 15 is a stable equilibrium. We determine the concavity of the solution curves by differentiating both sides of Equation (1) with respect to t:

( )

Initial temperature

FIGURE 16.19

t

Temperature versus time. Regardless of initial temperature, the object’s temperature H (t ) tends toward 15 °C, the temperature of the surrounding medium.

d dH d = (−k ( H − 15 )) dt dt dt d 2H dH . = −k dt 2 dt Since −k is negative, we see that d 2 H dt 2 is positive when dH dt < 0 and negative when dH dt > 0. Figure 16.18 adds this information to the phase line. The completed phase line shows that if the temperature of the object is above the equilibrium value of 15 °C, the graph of H (t ) will be decreasing and concave upward. If the temperature is below 15 °C (the temperature of the surrounding medium), the graph of H (t ) will be increasing and concave downward. We use this information to sketch typical solution curves (Figure 16.19). From the upper solution curve in Figure 16.19, we see that as the object cools down, the rate at which it cools slows down because dH dt approaches zero. This observation is implicit in Newton’s Law of Cooling and contained in the differential equation, but the flattening of the graph as time advances gives an immediate visual representation of the phenomenon.

A Falling Body Encountering Resistance Newton observed that the rate of change of the momentum of a moving object is equal to the net force applied to it. In mathematical terms, F =

d ( mυ), dt

(2)

1048

Chapter 16 First-Order Differential Equations

where F is the net force acting on the object, and m and υ are the object’s mass and velocity. If m varies with time, as it will if the object is a rocket burning fuel, the right-hand side of Equation (2) expands to m

dυ dm +υ dt dt

using the Derivative Product Rule. In many situations, however, m is constant, dm dt = 0, and Equation (2) takes the simpler form

Fr = ky

F = m m

dυ dt

F p = mg,

Fp = mg

FIGURE 16.20

An object falling under the propulsion due to gravity, with a resistive force assumed to be proportional to the velocity. dy 0 dt

y

mg k

the force due to gravity. If, however, we think of a real body falling through the air—say, a penny from a great height or a parachutist from an even greater height—we know that at some point air resistance is a factor in the speed of the fall. A more realistic model of free fall would include air resistance, shown as a force Fr in the schematic diagram in Figure 16.20. For low speeds well below the speed of sound, physical experiments have shown that Fr is approximately proportional to the body’s velocity. The net force on the falling body is therefore F = F p − Fr ,

FIGURE 16.21

Initial phase line for the falling body encountering resistance. dy >0 dt

dy 0 and the body speeds up. These observations are captured in the initial phase line diagram in Figure 16.21. We determine the concavity of the solution curves by differentiating both sides of Equation (4) with respect to t: t

Typical velocity curves for a falling body encountering resistance. The value υ = mg k is the terminal velocity.

(

)

d 2υ d k k dυ . = g− υ = − dt 2 dt m m dt We see that d 2 υ dt 2 < 0 when υ < mg k and that d 2 υ dt 2 > 0 when υ > mg k. Figure 16.22 adds this information to the phase line. Notice the similarity to the phase line for Newton’s Law of Cooling (Figure 16.18). The solution curves are similar as well (Figure 16.23).

16.4  Graphical Solutions of Autonomous Equations

1049

Figure 16.23 shows two typical solution curves. Regardless of the initial velocity, we see the body’s velocity tending toward the limiting value υ = mg k. This value, a stable equilibrium point, is called the body’s terminal velocity. Skydivers can vary their terminal velocity from 153 km/h to 290 km/h by changing the amount of body area opposing the fall, which affects the value of k.

The Logistic Model for Population Growth In Section 16.3 we examined population growth using the model of exponential change. That is, if P represents the number of individuals and we neglect departures and arrivals, then dP = kP, dt

(5)

where k > 0 is the birth rate minus the death rate per individual per unit time. Because the natural environment has only a limited number of resources to sustain life, it is reasonable to assume that only a maximum population M can be accommodated. As the population approaches this limiting population or carrying capacity, resources become less abundant and the growth rate k decreases. A simple relationship exhibiting this behavior is k = r ( M − P ), where r > 0 is a constant. Notice that k decreases as P increases toward M and that k is negative if P is greater than M. Substituting r ( M − P ) for k in Equation (5) gives the differential equation dP = r ( M − P ) P = rMP − rP 2. dt dP 0 dt

P M

0

FIGURE 16.24

The initial phase line for logistic growth (Equation 6).

d 2P >0 dt 2

d 2P 0 if 0 < P < M and dP dt < 0 if P > M. These observations are recorded on the phase line in Figure 16.24. We determine the concavity of the population curves by differentiating both sides of Equation (6) with respect to t: d 2P d = ( rMP − rP 2 ) 2 dt dt dP dP = rM − 2rP dt dt dP = r ( M − 2P ) . dt

dP 0 dt

d 2P >0 dt 2

(6)

(7)

P 0

M 2

FIGURE 16.25

M

The completed phase line for logistic growth (Equation 6).

If P = M 2, then d 2 P dt 2 = 0. If P < M 2, then ( M − 2 P ) and dP dt are positive and d 2 P dt 2 > 0. If M 2 < P < M , then ( M − 2 P ) < 0, dP dt > 0, and d 2 P dt 2 < 0. If P > M, then ( M − 2 P ) and dP dt are both negative and d 2 P dt 2 > 0. We add this information to the phase line (Figure 16.25). The lines P = M 2 and P = M divide the first quadrant of the tP-plane into horizontal bands in which we know the signs of both dP dt and d 2 P dt 2. In each band, we know how the solution curves rise and fall, and how they bend as time passes. The equilibrium lines P = 0 and P = M are both population curves. Population curves crossing the line

Chapter 16 First-Order Differential Equations

P = M 2 have an inflection point there, giving them a sigmoid shape (curved in two directions like a letter S). Figure 16.26 displays typical population curves. Notice that each population curve approaches the limiting population M as t → ∞. P

Population

1050

Limiting population

M

M 2

t

Time

FIGURE 16.26

Population curves for logistic growth.

The Logistic Equation in Neural Networks and Machine Learning While Figure 16.26 gives a general idea of the behavior of solutions to the Logistic Equation (6), we have not yet found explicit solutions. Exact formulas for solutions of first order differential equations cannot always be found, but they can be derived for the case of the Logistic Equation, where the solutions are called logistic functions. In Example 2 we find the solutions lying between y = 0 and y = 1 for the Logistic Equation in the case where M = 1 and r is an arbitrary positive constant. dy = ry − ry 2 that satisfy dx 0 < y < 1. Where does a solution cross the horizontal line y = 1 2, and what is the slope of its graph at this point? EXAMPLE 2

Find the solutions to the Logistic Equation

Solution To solve the differential equation we use the method of separation of variables introduced in Section 7.2.

dy = ry − ry 2 = ry (1 − y ) dx 1 dy = r dx y (1 − y ) 1 dy = y (1 − y )

∫ r dx

1 1 + dy = 1− y y

∫ r dx

∫ ∫

ln y + C 1 − ln (1 − y ) + C 2 = rx + C 3 ln y − ln (1 − y ) = rx + C ln

Partial fractions y = y and 1 − y = 1 − y, since 0 < y < 1. Combine constants, C = C 3 − C1 − C 2 .

y = rx + C 1− y y = e rx +C 1− y e rx +C 1 + e rx +C 1 . y = 1 + e −rx −C

y =

Exponentiate.

Use algebra to solve for y. Divide through by e rx + C.

16.4  Graphical Solutions of Autonomous Equations

1051

This gives an explicit formula for all the solutions whose graphs lie strictly between y = 0 and y = 1. When a solution crosses the line y = 1 2 we have 1 1 = − rx − C 1+e 2 1 + e −rx −C = 2 e −rx −C = 1 rx + C = 0 x = −C r . So the solution graph crosses at the point (−C r, 1 2 ). The slope is found by evaluating dy dx = ry − ry 2 at this point, dy r (−C r ) = ry − ry 2 = r (1 2 ) − r (1 2 ) 2 = . 4 dx Figure 16.27 shows the graph of a logistic function with r = 3 and C = −6. y

1 Slope r /4 1

-2

Centered at x = -C /r 1

2

3

4

x

FIGURE 16.27

The constants r and C determine the steepness and horizontal displacement of a solution to the logistic equation. In this example r = 3 and C = −6.

Logistic functions have applications in many areas beyond the study of population growth. A field of Computer Science called Machine Learning develops methods to use a large collection of experimental data, called a training set, to construct a predictor function. The training set might consist of thousands of images of signs, for example, and the predictor function might decide whether a newly obtained image represents a Stop sign. One highly successful approach to Machine Learning is the method of Neural Networks, which creates predictor functions based on a model of interacting neurons. Neural network models are built by taking repeated compositions of linear and logistic functions. Linear functions, such as L ( x ) = ax + b can give accurate approximations of a function f nearby to a point where f is differentiable, as seen in Chapter 3. The optimal choices for the constants a and b in L ( x ) are found by minimizing an error function that is calculated using the training set in a process called linear regression. Logistic functions have several features that make them a useful complement to linear functions in constructing predictor functions. They have values lying between 0 and 1 and are well suited to modeling probabilities. They are differentiable and specified by a small number of constants, such as the constants r and C in Example 2. These constants can be adjusted, or tuned, to minimize the error of a prediction. Logistic functions are nonlinear, and taking compositions of linear and logistic functions allows for the approximation of much more complicated functions than linear functions alone. A more complete discussion of the utility of logistic functions involves multivariable functions and their derivatives, which are introduced in Chapter 13.

1052

Chapter 16 First-Order Differential Equations

16.4

EXERCISES

Phase Lines and Solution Curves

c. Show that if P > M for all t, then lim P (t ) = M .

In Exercises 1–8,

d. What happens if P < m for all t?

a. Identify the equilibrium values. Which are stable and which are unstable? b. Construct a phase line. Identify the signs of y′ and y ″. c. Sketch several solution curves. dy = y2 − 4 dx

dy 1. = ( y + 2 )( y − 3 ) dx

2.

dy 3. = y3 − y dx

dy 4. = y 2 − 2y dx

5. y′ =

6. y′ = y −

y,

y > 0

7. y′ = ( y − 1)( y − 2 )( y − 3 ) 8. y′ =

y3

−

y,

y > 0

y2

The autonomous differential equations in Exercises 16–12 represent models for population growth. For each exercise, use a phase line analysis to sketch solution curves for P (t ), selecting different starting values P(0). Which equilibria are stable, and which are unstable?

11.

e. Discuss the solutions to the differential equation. What are the equilibrium points of the model? Explain the dependence of the steady-state value of P on the initial values of P. About how many permits should be issued? Applications and Examples

Models of Population Growth

9.

t →∞

dP = 1 − 2P dt

10.

dP = P (1 − 2 P ) dt

dP = 2P ( P − 3) dt

12.

dP 1 = 3 P (1 − P ) P − dt 2

(

)

13. Catastrophic change in logistic growth Suppose that a healthy population of some species is growing in a limited environment and that the current population P0 is fairly close to the carrying capacity M 0 . You might imagine a population of fish living in a freshwater lake in a wilderness area. Suddenly a catastrophe such as the Mount St. Helens volcanic eruption contaminates the lake and destroys a significant part of the food and oxygen on which the fish depend. The result is a new environment with a carrying capacity M1 considerably less than M 0 and, in fact, less than the current population P0 . Starting at some time before the catastrophe, sketch a “before-and-after” curve that shows how the fish population responds to the change in environment. 14. Controlling a population The fish and game department in a certain state is planning to issue hunting permits to control the deer population (one deer per permit). It is known that if the deer population falls below a certain level m, the deer will become extinct. It is also known that if the deer population rises above the carrying capacity M, the population will decrease back to M through disease and malnutrition. a. Discuss the reasonableness of the following model for the growth rate of the deer population as a function of time: dP = rP ( M − P )( P − m ) , dt where P is the population of the deer and r is a positive constant of proportionality. Include a phase line. b. Explain how this model differs from the logistic model dP dt = rP ( M − P ). Is it more or less reasonable than the logistic model?

15. Skydiving If a body of mass m falling from rest under the action of gravity encounters an air resistance proportional to the square of velocity, then the body’s velocity t seconds into the fall satisfies the equation dυ = mg − k υ 2, m k > 0, dt where k is a constant that depends on the body’s aerodynamic properties and the density of the air. (We assume that the fall is too short to be affected by changes in the air’s density.) a. Draw a phase line for the equation. b. Sketch a typical velocity curve. c. For a 45-kg skydiver ( mg = 441) and with time in seconds and distance in meter, a typical value of k is 0.15. What is the diver’s terminal velocity? Repeat for an 80-kg skydiver. 16. Resistance proportional to X A body of mass m is projected vertically downward with initial velocity υ 0 . Assume that the resisting force is proportional to the square root of the velocity, and find the terminal velocity from a graphical analysis. 17. Sailing A sailboat is running along a straight course with the wind providing a constant forward force of 200 N. The only other force acting on the boat is resistance as the boat moves through the water. The resisting force is numerically equal to five times the boat’s speed, and the initial velocity is 1 m s. What is the maximum velocity in meters per second of the boat under this wind? 18. The spread of information Sociologists recognize a phenomenon called social diffusion, which is the spreading of a piece of information, technological innovation, or cultural fad among a population. The members of the population can be divided into two classes: those who have the information and those who do not. In a fixed population whose size is known, it is reasonable to assume that the rate of diffusion is proportional to the number who have the information times the number yet to receive it. If X denotes the number of individuals who have the information in a population of N people, then a mathematical model for social diffusion is given by dX = kX ( N − X ) , dt where t represents time in days and k is a positive constant. a. Discuss the reasonableness of the model. b. Construct a phase line identifying the signs of X ′ and X ′′. c. Sketch representative solution curves. d. Predict the value of X for which the information is spreading most rapidly. How many people eventually receive the information?

16.5  Systems of Equations and Phase Planes

1053

19. Current in an RL circuit The accompanying diagram represents an electrical circuit whose total resistance is a constant R ohms and whose self-inductance, shown as a coil, is L henries, also a constant. There is a switch whose terminals at a and b can be closed to connect a constant electrical source of V volts. From Section 16.2, we have di L + Ri = V, dt

proportional to its velocity. Suppose that there is also a resistive buoyant force exerted by the shampoo. According to Archimedes’ principle, the buoyant force equals the weight of the fluid displaced by the pearl. Using m for the mass of the pearl and P for the mass of the shampoo displaced by the pearl as it descends, complete the following steps.

where i is the current in amperes and t is the time in seconds. −

b. Using υ(t ) for the pearl’s velocity as a function of time t, write a differential equation modeling the velocity of the pearl as a falling body.

Switch

c. Construct a phase line displaying the signs of υ′ and υ′′.

V

+

a

a. Draw a schematic diagram showing the forces acting on the pearl as it sinks, as in Figure 16.20.

d. Sketch typical solution curves.

b

e. What is the terminal velocity of the pearl?

i

Logistic Functions R

L

Use a phase line analysis to sketch the solution curve assuming that the switch in the RL circuit is closed at time t = 0. What happens to the current as t → ∞ ? This value is called the steadystate solution. 20. A pearl in shampoo Suppose that a pearl is sinking in a thick fluid, like shampoo, subject to a frictional force opposing its fall and

21. Write the formula for a logistic function that has values between y = 0 and y = 1, crosses the line y = 1 2 at x = 0, and has slope 5 at this point. 22. Write the formula for a logistic function that has values between y = 0 and y = 1, crosses the line y = 1 2 at x = 0, and has slope 1 5 at this point.

16.5 Systems of Equations and Phase Planes In some situations we are led to consider not one, but several, first-order differential equations. Such a collection is called a system of differential equations. In this section we present an approach to understanding systems through a graphical procedure known as a phase-plane analysis. We present this analysis in the context of modeling the populations of trout and bass living in a common pond.

Phase Planes A general system of two first-order differential equations may take the form dx = F ( x , y ), dt dy = G ( x , y ). dt In this system we often think of t as representing time and take x (t ) and y(t ) to be two functions of t. Such a system of equations is called autonomous because dx dt and dy dt do not depend on the independent variable time t, but only on the dependent variables x and y. A solution of such a system consists of a pair of functions x (t ) and y(t ) that satisfies both of the differential equations simultaneously for every t over some time interval (finite or infinite). We cannot look at just one of these equations in isolation to find solutions x (t ) or y(t ) since each derivative depends on both x and y. To gain insight into the solutions, we look at both dependent variables together by plotting the points ( x (t ), y(t ) ) in the xy-plane starting at some specified point. Therefore the solution functions define a solution curve through the specified point, called a trajectory of the system. The xy-plane itself, in which these trajectories reside, is referred to as the phase plane. Thus we consider both solutions together and study the behavior of all the solution trajectories in the phase plane. It can be

1054

Chapter 16 First-Order Differential Equations

proved that two trajectories can never cross or touch each other. (Solution trajectories are examples of parametric curves, which will be examined in detail in Chapter 9.)

A Competitive-Hunter Model Imagine two species of fish, say trout and bass, competing for the same limited resources (such as food and oxygen) in a certain pond. We let x (t ) represent the number of trout and y(t ) the number of bass living in the pond at time t. In reality, x (t ) and y(t ) are always integer valued, but we will approximate them with real-valued differentiable functions. This allows us to apply the methods of differential equations. Several factors affect the rates of change of these populations. As time passes, each species breeds, so we assume its population increases proportionally to its size. Taken by itself, this would lead to exponential growth in each of the two populations. However, there is a countervailing effect from the fact that the two species are in competition. A large number of bass tends to cause a decrease in the number of trout, and vice versa. Our model takes the size of this effect to be proportional to the frequency with which the two species interact, which in turn is proportional to xy, the product of the two populations. These considerations lead to the following model for the growth of the trout and bass in the pond: dx = ( a − by ) x, dt dy = ( m − nx ) y. dt

(1a) (1b)

Here x (t ) represents the trout population, y(t ) the bass population, and a, b, m, n are positive constants. A solution of this system then consists of a pair of functions x (t ) and y(t ) that give the population of each fish species at time t. Each equation in (1) contains both of the unknown functions x and y, so we are unable to solve them individually. Instead, we will use a graphical analysis to study the solution trajectories of this competitive-hunter model. We now examine the nature of the phase plane in the trout-bass population model. We will be interested in the 1st quadrant of the xy-plane, where x ≥ 0 and y ≥ 0, since populations cannot be negative. First, we determine where the bass and trout populations are both constant. Noting that the ( x (t ), y(t ) ) values remain unchanged when dx dt = 0 and dy dt = 0, we see that Equations (1a and 1b) then become (a

− by ) x = 0,

(m

− nx ) y = 0.

This pair of simultaneous equations has two solutions: ( x , y) = ( 0, 0 ) and ( x , y) = ( m n , a b ). At these ( x , y) values, called equilibrium or rest points, the two populations remain at constant values over all time. The point ( 0, 0 ) represents a pond containing no members of either fish species; the point ( m n , a b ) corresponds to a pond with an unchanging number of each fish species. Next, we note that if y = a b, then Equation (1a) implies dx dt = 0, so the trout population x (t ) is constant. Similarly, if x = m n, then Equation (1b) implies dy dt = 0, and the bass population y(t ) is constant. This information is recorded in Figure 16.28. y Bass

a b

y Bass

y Bass

dx = 0 dt

a b

dy =0 dt Trout (a)

FIGURE 16.28

x

m n (b)

Trout

x

m a a , b n b

m n (c)

Trout

x

Rest points in the competitive-hunter model given by Equations (1a) and (1b).

16.5  Systems of Equations and Phase Planes

y Bass

m n

Trout

x

FIGURE 16.29

To the left of the line x = m n the trajectories move upward, and to the right they move downward. y Bass

a b

Trout

x

FIGURE 16.30 Above the line y = a b the trajectories move to the left, and below it they move to the right.

In setting up our competitive-hunter model, we do not generally know precise values of the constants a, b, m, n. Nonetheless, we can analyze the system of Equations (1) to learn the nature of its solution trajectories. We begin by determining the signs of dx dt and dy dt throughout the phase plane. Although x (t ) represents the number of trout and y(t ) the number of bass at time t, we are thinking of the pair of values ( x (t ), y(t ) ) as a point tracing out a trajectory curve in the phase plane. When dx dt is positive, x (t ) is increasing and the point is moving to the right in the phase plane. If dx dt is negative, the point is moving to the left. Likewise, the point is moving upward where dy dt is positive and downward where dy dt is negative. We saw that dy dt = 0 along the vertical line x = m n. To the left of this line, dy dt is positive since dy dt = ( m − nx ) y and x < m n. So the trajectories on this side of the line are directed upward. To the right of this line, dy dt is negative and the trajectories point downward. The directions of the associated trajectories are indicated in Figure 16.29. Similarly, above the horizontal line y = a b, we have dx dt < 0 and the trajectories head leftward; below this line they head rightward, as shown in Figure 16.30. Combining this information gives four distinct regions in the plane A, B, C, D, with their respective trajectory directions shown in Figure 16.31. Next, we examine what happens near the two equilibrium points. The trajectories near ( 0, 0 ) point away from it, upward and to the right. The behavior near the equilibrium point ( m n , a b ) depends on the region in which a trajectory begins. If it starts in region B, for instance, then it will move downward and leftward toward the equilibrium point. Depending on where the trajectory begins, it may move downward into region D, leftward into region A, or perhaps straight into the equilibrium point. If it enters into regions A or D, then it will continue to move away from the rest point. We say that both rest points are unstable, meaning (in this setting) there are trajectories near each point that head away from them. These features are indicated in Figure 16.32. It turns out that in each of the half-planes above and below the line y = a b, there is exactly one trajectory approaching the equilibrium point ( m n , a b ) (see Exercise 7). Above these two trajectories the bass population increases, and below them it decreases. The two trajectories approaching the equilibrium point are suggested in Figure 16.33.

y Bass

y Bass A A

y Bass

B

B

a b

Bass win

a b C

(0, 0)

1055

D m n

Trout

m a a , b n b C

x (0, 0)

FIGURE 16.31

Composite graphical analysis of the trajectory directions in the four regions determined by x = m n and y = a b.

FIGURE 16.32

D m n

Trout win Trout

x

Motion along the trajectories near the rest points ( 0, 0 ) and ( m n , a b ).

(0, 0)

Trout

x

FIGURE 16.33

Qualitative results of analyzing the competitive-hunter model. There are exactly two trajectories approaching the point ( m n , a b ).

Our graphical analysis leads us to conclude that, under the assumptions of the competitivehunter model, it is unlikely that both species will reach equilibrium levels. This is because it would be almost impossible for the fish populations to move exactly along one of the two approaching trajectories for all time. Furthermore, the initial populations point ( x 0 , y 0 ) determines which of the two species is likely to survive over time, and mutual coexistence of the species is highly improbable.

1056

Chapter 16 First-Order Differential Equations

Limitations of the Phase-Plane Analysis Method

FIGURE 16.34

Trajectory direction near the rest point ( 0, 0 ).

Unlike the situation for the competitive-hunter model, it is not always possible to determine the behavior of trajectories near a rest point. For example, suppose we know that the trajectories near a rest point, chosen here to be the origin ( 0, 0 ), behave as in Figure 16.34. The information provided by Figure 16.34 is not sufficient to distinguish among the three possible trajectories shown in Figure 16.35. Even if we could determine that a trajectory near an equilibrium point resembles that of Figure 16.35c, we would still not know how the other trajectories behave. It could happen that a trajectory closer to the origin behaves like the motions displayed in Figure 16.35a or 16.35b. The spiraling trajectory in Figure 16.35c can never actually reach the rest point in a finite time period.

y

y

(x0, y0)

y

(x0, y0) x

(a)

(x0, y0) x

(b)

x

(c)

FIGURE 16.35

Three possible trajectory motions: (a) periodic motion, (b) motion toward an asymptotically stable rest point, and (c) motion near an unstable rest point.

y x2

+

y2

=1

Another Type of Behavior

(x1, y1)

The system dx = y + x − x ( x 2 + y 2 ), dt dy = −x + y − y ( x 2 + y 2 ) dt

x

(x0, y0)

FIGURE 16.36

The solution x 2 + y 2 = 1 is a limit cycle.

EXERCISES

(2a) (2b)

can be shown to have only one equilibrium point at ( 0, 0 ). Yet any trajectory starting on the unit circle traverses it clockwise because, when x 2 + y 2 = 1, we have dy dx = −x y (see Exercise 2). If a trajectory starts inside the unit circle, it spirals outward, asymptotically approaching the circle as t → ∞. If a trajectory starts outside the unit circle, it spirals inward, again asymptotically approaching the circle as t → ∞. The circle x 2 + y 2 = 1 is called a limit cycle of the system (Figure 16.36). In this system, the values of x and y eventually become periodic.

16.5

1. List three of the important considerations that are ignored in the competitive-hunter model as presented in the text. 2. For the system (2a) and (2b), show that any trajectory starting on the unit circle x 2 + y 2 = 1 will traverse the unit circle in a periodic solution. First introduce polar coordinates and rewrite the system as dr dt = r (1 − r 2 ) and −dθ dt = −1.

3. Develop a model for the growth of trout and bass, assuming that in isolation trout demonstrate exponential decay [so that a < 0 in Equations (1a) and (1b)] and that the bass population grows logistically with a population limit M. Analyze graphically the motion in the vicinity of the rest points in your model. Is coexistence possible?

16.5  Systems of Equations and Phase Planes

1057

4. How might the competitive-hunter model be validated? Include a discussion of how the various constants a, b, m, and n might be estimated. How could state conservation authorities use the model to ensure the survival of both species?

b. Separate the variables, integrate, and exponentiate to obtain

5. Consider another competitive-hunter model defined by

c. Let f ( y) = y a e by and g( x ) = x m e nx. Show that f ( y) has a unique maximum of M y = ( a eb ) a when y = a b as shown in Figure 16.37. Similarly, show that g( x ) has a unique maximum Mx = ( m en ) m when x = m n , also shown in Figure 16.37.

⎛ dx x⎞ = a ⎜⎜1 − ⎟⎟⎟ x − bxy, ⎝ dt k1 ⎠ ⎛ dy y⎞ = m ⎜⎜1 − ⎟⎟⎟ y − nxy, ⎝ dt k2 ⎠

y a e −by = Kx m e −nx , where K is a constant of integration.

f (y)

where x and y represent trout and bass populations, respectively. a. What assumptions are implicitly being made about the growth of trout and bass in the absence of competition?

My

y ae−by

b. Interpret the constants a, b, m, n, k1 , k 2, a b , and m n in terms of the physical problem.

y

a b

c. Perform a graphical analysis: i) Find the possible equilibrium levels. ii) Determine whether coexistence is possible. iii) Pick several typical starting points, and sketch typical trajectories in the phase plane.

g(x)

Mx x me−nx

iv) Interpret the outcomes predicted by your graphical analysis in terms of the constants a, b, m, n, k1 , and k 2 . Note: When you get to part (iii), you should realize that five cases exist. You will need to analyze all five cases. 6. An economic model Consider the following economic model. Let P be the price of a single item on the market. Let Q be the quantity of the item available on the market. Both P and Q are functions of time. If one considers price and quantity as two interacting species, the following model might be proposed:

(

)

dP b = aP −P , dt Q dQ = cQ ( fP − Q ) , dt where a, b, c, and f are positive constants. Justify and discuss the adequacy of the model. a. If a = 1, b = 20,000, c = 1, and f = 30, find the equilibrium points of this system. If possible, classify each equilibrium point with respect to its stability. If a point cannot be readily classified, give some explanation. b. Perform a graphical stability analysis to determine what will happen to the levels of P and Q as time increases. c. Give an economic interpretation of the curves that determine the equilibrium points. 7. Two trajectories approach equilibrium Show that the two trajectories leading to ( m n , a b ) shown in Figure 16.33 are unique by carrying out the following steps. a. From system (1a) and (1b) apply the Chain Rule to derive the following equation: ( m − nx ) y dy . = ( a − by ) x dx

x

m n

FIGURE 16.37

Graphs of the functions f ( y) = y a e by and g( x ) = x m e nx .

d. Consider what happens as ( x , y) approaches ( m n , a b ). Take limits in part (b) as x → m n and y → a b to show that either

( )( ex )⎤⎥⎦ = K

ya lim ⎡⎢ by ⎣ e

x→m n y→ a b

nx m

or M y Mx = K. Thus any solution trajectory that approaches ( m n , a b ) must satisfy ⎛ M ⎞ xm ya = ⎜⎜⎜ y ⎟⎟⎟ nx . ⎝ Mx ⎠ e e by

( )

e. Show that only one trajectory can approach ( m n , a b ) from below the line y = a b. Pick y 0 < a b. From Figure 16.37 you can see that f ( y 0 ) < M y , which implies that

( )= y

My x m M x e nx

a 0

e by 0 < M y .

This in turn implies that xm < Mx . e nx

1058

Chapter 16 First-Order Differential Equations

Figure 16.37 tells you that for g( x ) there is a unique value x 0 < m n satisfying this last inequality. That is, for each y < a b there is a unique value of x satisfying the equation in part (d). Thus there can exist only one trajectory solution approaching ( m n , a b ) from below, as shown in Figure 16.38. f. Use a similar argument to show that the solution trajectory leading to ( m n , a b ) is unique if y 0 > a b. y Bass

9. What happens to the rabbit population if there are no foxes present? 10. What happens to the fox population if there are no rabbits present? 11. Show that ( 0, 0 ) and ( c d , a b ) are equilibrium points. Explain the meaning of each of these points. 12. Show, by differentiating, that the function C (t ) = a ln y(t ) − by(t ) − dx (t ) + c ln x (t ) is constant when x (t ) and y(t ) are positive and satisfy the predatorprey equations.

a b y0

Unique x 0

m n

Trout

x

FIGURE 16.38

For any y < a b , only one solution trajectory leads to the rest point ( m n , a b ). 8. Show that the second-order differential equation y ″ = F ( x , y, y′ ) can be reduced to a system of two first-order differential equations dy = z, dx dz = F ( x , y, z ). dx

While x and y may change over time, C (t ) does not. Thus, C is a conserved quantity and its existence gives a conservation law. A trajectory that begins at a point ( x , y) at time t = 0 gives a value of C that remains unchanged at future times. Each value of the constant C gives a trajectory for the autonomous system, and these trajectories close up, rather than spiraling inward or outward. The rabbit and fox populations oscillate through repeated cycles along a fixed trajectory. Figure 16.39 shows several trajectories for the predator-prey system. y

Fox population

a b

Can something similar be done to the nth-order differential equation y ( n ) = F ( x , y, y′, y ″, … , y ( n−1) )? 0

c d

x Rabbit population

Lotka-Volterra Equations for a Predator-Prey Model

In 1925 Lotka and Volterra introduced the predator-prey equations, a system of equations that models the populations of two species, one of which preys on the other. Let x (t ) represent the number of rabbits living in a region at time t, and y(t ) the number of foxes in the same region. As time passes, the number of rabbits increases at a rate proportional to their population, and decreases at a rate proportional to the number of encounters between rabbits and foxes. The foxes, which compete for food, increase in number at a rate proportional to the number of encounters with rabbits but decrease at a rate proportional to the number of foxes. The number of encounters between rabbits and foxes is assumed to be proportional to the product of the two populations. These assumptions lead to the autonomous system dx = ( a − by ) x , dt dy = (−c + dx ) y, dt where a, b, c, d are positive constants. The values of these constants vary according to the specific situation being modeled. We can study the nature of the population changes without setting these constants to specific values.

FIGURE 16.39

Some trajectories along which C is

conserved. 13. Using a procedure similar to that in the text for the competitivehunter model, show that each trajectory is traversed in a counterclockwise direction as time t increases. Along each trajectory, both the rabbit and fox populations fluctuate between their maximum and minimum levels. The maximum and minimum levels for the rabbit population occur where the trajectory intersects the horizontal line y = a b. For the fox population, they occur where the trajectory intersects the vertical line x = c d. When the rabbit population is at its maximum, the fox population is below its maximum value. As the rabbit population declines from this point in time, we move counterclockwise around the trajectory, and the fox population grows until it reaches its maximum value. At this point the rabbit population has declined to x = c d and is no longer at its peak value. We see that the fox population reaches its maximum value at a later time than the rabbits. The predator population lags behind that of the prey in achieving its maximum values. This lag effect is shown in Figure 16.40, which graphs both x (t ) and y(t ).

Chapter 16  Practice Exercises

14. At some time during a trajectory cycle, a wolf invades the rabbit– fox territory, eats some rabbits, and then leaves. Does this mean that the fox population will from then on have a lower maximum value? Explain your answer.

Rabbits x(t)

(Different scale used for foxes and rabbits)

Populations

x, y

1059

Foxes y(t)

t Lag time

FIGURE 16.40

The fox and rabbit populations oscillate periodically, with the maximum fox population lagging the maximum rabbit population.

CHAPTER 16

Questions to Guide Your Review

1. What is a first-order differential equation? When is a function a solution of such an equation? 2. What is a general solution? What is a particular solution? 3. What is the slope field of a differential equation y′ = f ( x , y)? What can we learn from such fields? 4. Describe Euler’s method for solving the initial value problem y′ = f ( x , y), y( x 0 ) = y 0 numerically. Give an example. Comment on the method’s accuracy. Why might you want to solve an initial value problem numerically? 5. How do you solve linear first-order differential equations? 6. What is an orthogonal trajectory of a family of curves? Describe how one is found for a given family of curves.

CHAPTER 16

8. How do you construct the phase line for an autonomous differential equation? How does the phase line help you produce a graph that qualitatively depicts a solution to the differential equation? 9. Why is the exponential model unrealistic for predicting long-term population growth? How does the logistic model correct for the deficiency in the exponential model for population growth? What is the logistic differential equation? What is the form of its solution? Describe the graph of the logistic solution. 10. What is an autonomous system of differential equations? What is a solution to such a system? What is a trajectory of the system?

Practice Exercises

In Exercises 1–22, solve the differential equation. 1. y′ = xe y x − 2

2. y′ = xy e x 2

3. sec x dy + x cos 2 y dx = 0

4. 2 x 2 dx − 3 y csc x dy = 0

5. y′ =

7. What is an autonomous differential equation? What are its equilibrium values? How do they differ from critical points? What is a stable equilibrium value? Unstable?

ey

6. y′ = xe x − y csc y

xy

7. x ( x − 1) dy − y dx = 0 9. 2 y′ − y = xe x 2

8. y′ = ( y 2 − 1) x −1 y′ + y = e −x sin x 2 12. xy′ − y = 2 x ln x 10.

11. xy′ + 2 y = 1 − x −1

Initial Value Problems In Exercises 23–28, solve the initial value problem. dy 23. ( x + 1) + 2 y = x , x > −1, y(0) = 1 dx

dy + 2 y = x 2 + 1, x > 0, y(1) = 1 dx dy + 3 x 2 y = x 2, y(0) = −1 25. dx π = 0 26. x dy + ( y − cos x ) dx = 0, y 2 24. x

( )

13. (1 + e x ) dy + ( y e x + e −x ) dx = 0

27. xy′ + ( x − 2 ) y = 3 x 3e −x , y(1) = 0

14. e −x dy + ( e −x y − 4 x ) dx = 0

28. y dx + ( 3 x − xy + 2 ) dy = 0, y(2) = −1, y < 0

15. ( x +

3 y 2 ) dy

+ y dx = 0 (Hint: d ( xy ) = y dx + x dy)

16. x dy + ( 3 y − x −2 cos x ) dx = 0, x > 0 17. y′ = sin 3 x cos 2 y 19. dy + x ( 2 y − e

x−x 2

21. y′ = xy ln x ln y

18. x dy − ( x 4 − y ) dx = 0

) dx = 0 20. y′ + 3 x 2 y = 7 x 2 22. xy′ + 2 y ln x = ln x

Euler’s Method In Exercises 29 and 30, use Euler’s method to solve the initial value problem on the given interval starting at x 0 with dx = 0.1.

T 29. y′ = y + cos x ,

y(0) = 0; 0 ≤ x ≤ 2; x 0 = 0

T 30. y′ = ( 2 − y )( 2 x + 3 ) , y (−3 ) = 1; − 3 ≤ x ≤ −1; x 0 = −3

1060

Chapter 16 First-Order Differential Equations

In Exercises 31 and 32, use Euler’s method with dx = 0.05 to estimate y(c), where y is the solution to the given initial value problem. dy x − 2y = , y(0) = 1 T 31. c = 3; dx x +1 dy x 2 − 2y + 1 = , y(1) = 1 T 32. c = 4; dx x

Thus, the velocity remains positive as long as υ 0 ≥ 2 gR . The velocity υ 0 = 2 gR is the moon’s escape velocity. A body projected upward with this velocity or a greater one will escape from the moon’s gravitational pull. b. Show that if υ 0 =

(

s = R 1+

In Exercises 33 and 34, use Euler’s method to solve the initial value problem graphically, starting at x 0 = 0 with b. dx = −0.1.

a. dx = 0.1. dy 1 = x + y + 2 , y(0) = −2 T 33. dx e dy x2 + y = − y , y(0) = 0 T 34. dx e + x

Slope Fields In Exercises 35–38, sketch part of the equation’s slope field. Then add to your sketch the solution curve that passes through the point P (1, −1). Use Euler’s method with x 0 = 1 and dx = 0.2 to estimate y(2). Round your answers to four decimal places. Find the exact value of y(2) for comparison.

35. y′ = x

36. y′ = 1 x

37. y′ = xy

38. y′ = 1 y

Autonomous Differential Equations and Phase Lines In Exercises 39 and 40:

a. Identify the equilibrium values. Which are stable and which are unstable?

2 gR , then 3υ 0 t 2R

)

23

.

42. Coasting to a stop Table 16.6 shows the distance s (meters) coasted on inline skates in t s by Johnathon Krueger. Find a model for his position in the form of Equation (2) of Section 16.3. His initial velocity was υ 0 = 0.86 m s, his mass m = 30.84 kg , and his total coasting distance was 0.97 m. TABLE 16.6 Johnathon Krueger skating data

t (s)

s(m)

t (s)

s(m)

t (s)

s(m)

0

0

0.93

0.61

1.86

0.93

0.13

0.08

1.06

0.68

2.00

0.94

0.27

0.19

1.20

0.74

2.13

0.95

0.40

0.28

1.33

0.79

2.26

0.96

0.53

0.36

1.46

0.83

2.39

0.96

0.67

0.45

1.60

0.87

2.53

0.97

0.80

0.53

1.73

0.90

2.66

0.97

b. Construct a phase line. Identify the signs of y′ and y ″. c. Sketch a representative selection of solution curves. dy dy = y2 − 1 40. = y − y2 39. dx dx Applications

41. Escape velocity The gravitational attraction F exerted by an airless moon on a body of mass m at a distance s from the moon’s center is given by the equation F = −mg R 2 s −2, where g is the acceleration of gravity at the moon’s surface and R is the moon’s radius (see accompanying figure). The force F is negative because it acts in the direction of decreasing s. F=−

mgR 2

Mass m

Mixture Problems In Exercises 43 and 44, let S represent the kilograms of salt in a tank at time t minutes. Set up a differential equation representing the given information and the rate at which S changes. Then solve for S and answer the particular questions. 1 43. A mixture containing kg of salt per liter flows into a tank at the 4 rate of 24 L/min, and the well-stirred mixture flows out of the tank at the rate of 16 L/min. The tank initially holds 600 liters of solution containing 6 kilograms of salt.

a. How many liters of solution are in the tank after 1 minute? after 10 minutes? after 1 hour? b. How many kilograms of salt are in the tank after 1 minute? after 10 minutes? after 1 hour?

s2

s R

44. Pure water flows into a tank at the rate of 16 L/min, and the wellstirred mixture flows out of the tank at the rate of 20 L/min. The tank initially holds 800 liters of solution containing 25 kilograms of salt. a. How many liters of solution are in the tank after 1 minute? after 10 minutes? after 200 minutes?

Moon’s center

a. If the body is projected vertically upward from the moon’s surface with an initial velocity υ 0 at time t = 0, use Newton’s second law, F = ma, to show that the body’s velocity at position s is given by the equation υ2 =

2 gR 2 + υ 0 2 − 2 gR. s

b. How many kilograms of salt are in the tank after 1 minute? after 30 minutes? c. When will the tank have exactly 5 kilograms of salt, and how many liters of solution will be in the tank?

Chapter 16  Technology Application Projects

CHAPTER 16

Additional and Advanced Exercises

Theory and Applications

1. Transport through a cell membrane Under some conditions, the result of the movement of a dissolved substance across a cell’s membrane is described by the equation

4. (Continuation of Exercise 3.) Assume the hypotheses of Exercise 3, and assume that y1 ( x ) and y 2 ( x ) are both solutions to the first-order linear equation satisfying the initial condition y( x 0 ) = y 0 . a. Verify that y( x ) = y1 ( x ) − y 2 ( x ) satisfies the initial value problem

dy A = k ( c − y ). V dt In this equation, y is the concentration of the substance inside the cell, and dy dt is the rate at which y changes over time. The letters k, A, V, and c stand for constants, k being the permeability coefficient (a property of the membrane), A the surface area of the membrane, V the cell’s volume, and c the concentration of the substance outside the cell. The equation says that the rate at which the concentration within the cell changes is proportional to the difference between it and the outside concentration.

y′ + P ( x ) y = 0,

d( υ( x )[ y1 ( x ) − y 2 ( x ) ]) = 0. dx Conclude that υ( x )[ y1 ( x ) − y 2 ( x ) ] ≡ constant. c. From part (a), we have y1 ( x 0 ) − y 2 ( x 0 ) = 0. Since υ( x ) > 0 for a < x < b, use part (b) to establish that y1 ( x ) − y 2 ( x ) ≡ 0 on the interval ( a, b ). Thus y1 ( x ) = y 2 ( x ) for all a < x < b.

b. Find the steady-state concentration, lim y(t ). t →∞

2. Height of a rocket If an external force F acts upon a system whose mass varies with time, Newton’s law of motion is

Homogeneous Equations A first-order differential equation of the form

( )

dy y = F dx x

d ( mυ ) dm = F + (υ + u) . dt dt In this equation, m is the mass of the system at time t, υ is its velocity, and υ + u is the velocity of the mass that is entering (or leaving) the system at the rate dm dt . Suppose that a rocket of initial mass m 0 starts from rest, but is driven upward by firing some of its mass directly backward at the constant rate of dm dt = −b units per second and at constant speed relative to the rocket u = −c. The only external force acting on the rocket is F = −mg due to gravity. Under these assumptions, show that the height of the rocket above the ground at the end of t seconds (t small compared to m 0 b) is ⎡ m − bt m 0 − bt ⎤ 1 y = c ⎢t + 0 ln ⎥ − gt 2. 2 b m 0 ⎥⎦ ⎢⎣ 3. a. Assume that P ( x ) and Q( x ) are continuous over the interval [ a, b ]. Use the Fundamental Theorem of Calculus, Part 1, to show that any function y satisfying the equation υ( x ) y =

∫ υ( x ) Q( x ) dx + C

for υ( x ) = e ∫ P ( x ) dx is a solution to the first-order linear equation dy + P ( x ) y = Q( x ). dx x

b. If C = y 0 υ( x 0 ) − ∫x 0 υ(t )Q(t ) dt , then show that any solution y in part (a) satisfies the initial condition y( x 0 ) = y 0 .

y( x 0 ) = 0.

b. For the integrating factor υ( x ) = e ∫ P ( x ) dx , show that

a. Solve the equation for y(t ), using y 0 to denote y(0).

CHAPTER 16

1061

is called homogeneous. It can be transformed into an equation whose variables are separable by defining the new variable υ = y x . Then y = υ x and dy dυ = υ+ x . dx dx Substituting into the original differential equation and collecting terms with like variables then give the separable equation dx dυ + = 0. x υ − F(υ) After solving this separable equation, we obtain the solution of the original equation when we replace υ by y x . Solve the homogeneous equations in Exercises 5–10. First put the equation in the form of a homogeneous equation. 5. ( x 2 + y 2 ) dx + xy dy = 0 6. x 2 dy + ( y 2 − xy ) dx = 0 7. ( xe y

x

+ y ) dx − x dy = 0

8. ( x + y ) dy + ( x − y ) dx = 0 y y−x 9. y′ = + cos x x y y y 10. x sin − y cos dx + x cos dy = 0 x x x

(

)

Technology Application Projects

Mathematica/Maple Projects

Projects can be found within MyLab Math. • Drug Dosages: Are They Effective? Are They Safe? Formulate and solve an initial value model for the absorption of a drug in the bloodstream. • First-Order Differential Equations and Slope Fields Plot slope fields and solution curves for various initial conditions to selected first-order differential equations.

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17

Second-Order Differential Equations Denis Kalinichenko/Shutterstock

OVERVIEW  In this chapter we extend our study of differential equations to those of second order, equations that involve second derivatives of a function. Second-order differential equations arise in many applications in the sciences and engineering. For instance, they can be applied to the study of vibrating springs and electric circuits. You will learn how to solve such differential equations by several methods in this chapter.

17.1 Second-Order Linear Equations An equation of the form

P ( x ) y′′( x ) + Q( x ) y′( x ) + R( x ) y( x ) = G ( x ), (1)

which is linear in y and its derivatives, is called a second-order linear differential equation. We assume that the functions P, Q, R, and G are continuous throughout some open interval I. If G ( x ) is identically zero on I, the equation is said to be homogeneous; otherwise it is called nonhomogeneous. Therefore, the form of a second-order linear homogeneous differential equation is

P ( x ) y′′ + Q( x ) y′ + R ( x ) y = 0. (2)

We also assume that P ( x ) is never zero for any x ∈ I . Two fundamental results are important to solving Equation (2). The first of these says that if we know two solutions y1 and y 2 of the linear homogeneous equation, then any linear combination y = c1 y1 + c 2 y 2 is also a solution for any constants c1 and c 2.

THEOREM 1—The Superposition Principle

If y1 ( x ) and y 2 ( x ) are two solutions to the linear homogeneous equation (2), then for any constants c1 and c 2, the function y( x ) = c1 y1 ( x ) + c 2 y 2 ( x ) is also a solution to Equation (2).

17-1

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17-2

Chapter 17  Second-Order Differential Equations

Proof  Substituting y into Equation (2), we have P ( x ) y ′′ + Q( x ) y ′ + R( x ) y = P( x )( c1 y 1 + c 2 y 2 )″ + Q( x )( c1 y 1 + c 2 y 2 )′ + R( x )( c1 y 1 + c 2 y 2 ) = P ( x )( c1 y 1″ + c 2 y 2 ″ ) + Q( x )( c1 y 1′ + c 2 y 2 ′ ) + R( x )( c1 y1 + c 2 y 2 ) P( x ) y1″ + Q( x ) y1′ + R( x ) y1) + c 2 ( P ( x ) y 2 ″ + Q( x ) y 2 ′ + R( x ) y 2 ) = c1 ( = 0,  y1  is a solution

= 0,  y 2  is a solution

= c1 (0) + c 2 (0) = 0. Therefore, y = c1 y1 + c 2 y 2 is a solution of Equation (2). Theorem 1 immediately establishes the following facts concerning solutions to the linear homogeneous equation. 1. A sum of two solutions y1 + y 2 to Equation (2) is also a solution. (Choose c1 = c 2 = 1.) 2. A constant multiple ky1 of any solution y1 to Equation (2) is also a solution. (Choose c1 = k and c 2 = 0.) 3. The trivial solution y( x ) ≡ 0 is always a solution to the linear homogeneous equation. (Choose c1 = c 2 = 0 .) The second fundamental result about solutions to the linear homogeneous equation concerns its general solution, or solution containing all solutions. This result says that there are two solutions y1 and y 2 such that any solution is some linear combination of them for suitable values of the constants c1 and c 2. However, not just any pair of solutions will do. The solutions must be linearly independent, which means that neither y1 nor y 2 is a constant multiple of the other. For example, the functions f ( x ) = e x and g( x ) = xe x are linearly independent, whereas f ( x ) = x 2 and g( x ) = 7 x 2 are not (they are linearly dependent). These results on linear independence and the following theorem are proved in more advanced courses.

THEOREM 2 If P, Q, and R are continuous over the open interval I and P ( x ) is never zero on I, then the linear homogeneous equation (2) has two linearly independent solutions y1 and y 2 on I. Moreover, if y1 and y 2 are any two linearly independent solutions of Equation (2), then the general solution is given by

y( x ) = c1 y1 ( x ) + c 2 y 2 ( x ), where c1 and c 2 are arbitrary constants.

We now turn our attention to finding two linearly independent solutions to the special case of Equation (2) where P, Q, and R are constant functions.

Constant-Coefficient Homogeneous Equations Suppose we wish to solve the second-order homogeneous differential equation

ay′′ + by′ + cy = 0, (3)

where a, b, and c are constants. To solve Equation (3), we seek a function that, when multiplied by a constant and added to a constant times its first derivative plus a constant times its second derivative, sums identically to zero. One function that behaves this way is the exponential function y = e rx , when r is a constant. Two differentiations of this exponential

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17.1  Second-Order Linear Equations

17-3

function give y′ = re rx and y′′ = r 2e rx, which are just constant multiples of the original exponential. If we substitute y = e rx into Equation (3), we obtain ar 2e rx + bre rx + ce rx = 0. Since the exponential function is never zero, we can divide this last equation through by e rx . Thus, y = e rx is a solution to Equation (3) if and only if r is a solution to the algebraic equation

ar 2 + br + c = 0. (4)



Equation (4) is called the auxiliary equation (or characteristic equation) of the differential equation ay′′ + by′ + cy = 0. The auxiliary equation is a quadratic equation with roots r1 =

−b +

b 2 − 4 ac 2a

and

r2 =

−b −

b 2 − 4 ac . 2a

There are three cases to consider, which depend on the value of the discriminant b 2 − 4 ac. Case 1: b 2 − 4 ac > 0.  In this case the auxiliary equation has two real and unequal roots r1 and r2 . Then y 1 = e r1x and y 2 = e r2 x are two linearly independent solutions to Equation (3) because e r2 x is not a constant multiple of e r1x (see Exercise 61). From Theorem 2 we conclude the following result.

THEOREM 3 If r1 and r2 are two real and unequal roots of the auxiliary equation ar 2 + br + c = 0, then

y = c1e r1x + c 2 e r2 x is the general solution to ay ′′ + by ′ + cy = 0.

EXAMPLE 1   Find the general solution of the differential equation y ′′ − y ′ − 6 y = 0. Solution  Substitution of y = e rx into the differential equation yields the auxiliary equation

r 2 − r − 6 = 0, which factors as (r

− 3 )( r + 2 ) = 0.

The roots are r1 = 3 and r2 = −2. Thus, the general solution is y = c1e 3 x + c 2e −2 x . Case 2: b 2 − 4 ac = 0.  In this case r1 = r2 = −b 2a. To simplify the notation, let r = −b 2a. Then we have one solution y1 = e rx with 2ar + b = 0. Since multiplication of e rx by a constant fails to produce a second linearly independent solution, suppose we try

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Chapter 17  Second-Order Differential Equations

multiplying by a function instead. The simplest such function would be u( x ) = x , so let’s see where y 2 = xe rx is also a solution. Substituting y 2 into the differential equation gives ay 2 ″ + by 2 ′ + cy 2 = a ( 2re rx + r 2 xe rx ) + b ( e rx + rxe rx ) + cxe rx = ( 2ar + b )e rx + ( ar 2 + br + c ) xe rx = 0 ( e rx ) + ( 0 ) xe rx = 0. The first term is zero because r = −b 2a; the second term is zero because r solves the auxiliary equation. The functions y 1 = e rx and y 2 = xe rx are linearly independent (see Exercise 62). From Theorem 2 we conclude the following result. THEOREM 4 If r is the only (repeated) real root of the auxiliary equation ar 2 + br + c = 0, then

y = c1e rx + c 2 xe rx is the general solution to ay ′′ + by ′ + cy = 0. EXAMPLE 2   Find the general solution to y ′′ + 4 y ′ + 4 y = 0. Solution  The auxiliary equation is

r 2 + 4r + 4 = 0, which factors into (r

+ 2 ) 2 = 0.

Thus, r = −2 is a double root. Therefore, the general solution is y = c1e −2 x + c 2 xe −2 x . Case 3: b 2 − 4 ac < 0.  In this case the auxiliary equation has two complex roots: r1 = α + iβ and r2 = α − iβ , where α and β are real numbers and i 2 = −1. (These real numbers are α = −b 2a and β = 4 ac − b 2 2a. ) These two complex roots then give rise to two linearly independent solutions:

y1 = e ( α + iβ )x = e αx ( cos β x + i sin β x ) and

y 2 = e ( α−iβ )x = e αx ( cos β x − i sin β x ).

(The expressions involving the sine and cosine terms follow from Euler’s identity, as seen in the discussion of Taylor series.) However, the solutions y1 and y 2 are complex valued rather than real valued. Nevertheless, because of the superposition principle (Theorem 1), we can obtain from them the two real-valued solutions y3 =

1 1 y + y 2 = e αx cos β x 2 1 2

and

y4 =

1 1 y − y 2 = e αx sin β x. 2i 1 2i

The functions y 3 and y 4 are linearly independent (see Exercise 63). From Theorem 2 we conclude the following result. THEOREM 5 If r1 = α + iβ and r2 = α − iβ are two complex roots of the

auxiliary equation ar 2 + br + c = 0 , then

y = e αx ( c1 cos β x + c 2 sin β x ) is the general solution to ay′′ + by′ + cy = 0.

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17.1  Second-Order Linear Equations

17-5

EXAMPLE 3   Find the general solution to the differential equation y′′ − 4 y′ + 5 y = 0. Solution  The auxiliary equation is

r 2 − 4r + 5 = 0. The roots are the complex pair r = ( 4 ± 16 − 20 ) 2, or r1 = 2 + i and r2 = 2 − i. Thus, α = 2 and β = 1 give the general solution y = e 2 x ( c1 cos x + c 2 sin x ).

Initial Value and Boundary Value Problems To determine a unique solution to a first-order linear differential equation, it was sufficient to specify the value of the solution at a single point. Since the general solution to a secondorder equation contains two arbitrary constants, it is necessary to specify two conditions. One way of doing this is to specify the value of the solution function and the value of its derivative at a single point: y( x 0 ) = y 0 and y′( x 0 ) = y1. These conditions are called initial conditions. The following result is proved in more advanced texts and guarantees the existence of a unique solution for both homogeneous and nonhomogeneous second-order linear initial value problems.

THEOREM 6 If P,   Q,   R, and G are continuous throughout an open interval I,

then there exists one and only one function y ( x ) satisfying both the differential equation P ( x ) y′′( x ) + Q( x ) y′( x ) + R( x ) y( x ) = G ( x ) on the interval I, and the initial conditions y( x 0 ) = y 0

and

y′( x 0 ) = y1

at the specified point x 0 ∈ I .

It is important to realize that any real values can be assigned to y 0 and y1, and Theorem 6 applies. Here is an example of an initial value problem for a homogeneous equation. EXAMPLE 4   Find the particular solution to the initial value problem y′′ − 2 y′ + y = 0,

y(0) = 1, y′(0) = −1.

Solution  The auxiliary equation is

r 2 − 2r + 1 = ( r − 1) 2 = 0. The repeated real root is r = 1, giving the general solution y = c1e x + c 2 xe x . Then y′ = c1e x + c 2 ( x + 1)e x . From the initial conditions we have 1 = c1 + c 2 ⋅ 0

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and

−1 = c1 + c 2 ⋅ 1.

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Chapter 17  Second-Order Differential Equations

y

–4 –3

–2

–1

0

Thus, c1 = 1 and c 2 = −2. The unique solution satisfying the initial conditions is 1

x

y = e x − 2 xe x . The solution curve is shown in Figure 17.1.

–2

Another approach to determine the values of the two arbitrary constants in the general solution to a second-order differential equation is to specify the values of the solution function at two different points in the interval I. That is, we solve the differential equation subject to the boundary values

y = e x – 2xe x –4 –6

y( x 1 ) = y1 and

–8

FIGURE 17.1  Particular solution curve

for Example 4.

y( x 2 ) = y 2 ,

where x 1 and x 2 both belong to I. Here again the values for y1 and y 2 can be any real numbers. The differential equation together with specified boundary values is called a boundary value problem. Unlike the result stated in Theorem 6, boundary value problems do not always possess a solution, or more than one solution may exist (see Exercise 65). These problems are studied in more advanced texts, but here is an example for which there is a unique solution. EXAMPLE 5   Solve the boundary value problem y(0) = 0, y

y′′ + 4 y = 0,

( 12π ) = 1.

Solution  The auxiliary equation is r 2 + 4 = 0, which has the complex roots r = ±2i. The general solution to the differential equation is

y = c1 cos 2 x + c 2 sin 2 x. The boundary conditions are satisfied if y(0) = c1 ⋅ 1 + c 2 ⋅ 0 = 0 π π π y = c1 cos + c 2 sin = 1. 12 6 6

( )

( )

( )

It follows that c1 = 0 and c 2 = 2. The solution to the boundary value problem is y = 2 sin 2 x.

EXERCISES 

17.1

In Exercises 1–30, find the general solution of the given equation.

23.

d 2y dy +4 + 4y = 0 dx 2 dx

24.

d 2y dy −6 + 9y = 0 dx 2 dx

25.

d 2y dy +6 + 9y = 0 dx 2 dx

26. 4

d 2y dy − 12 + 9y = 0 dx 2 dx

27. 4

d 2y dy +4 + y = 0 dx 2 dx

28. 4

d 2y dy −4 + y = 0 dx 2 dx

29. 9

d 2y dy +6 + y = 0 dx 2 dx

30. 9

d 2y dy − 12 + 4y = 0 dx 2 dx

1. y′′ − y′ − 12 y = 0

2. 3 y′′ − y′ = 0

3. y′′ + 3 y′ − 4 y = 0

4. y′′ − 9 y = 0

5. y′′ − 4 y = 0

6. y′′ − 64 y = 0

7. 2 y′′ − y′ − 3 y = 0

8. 9 y′′ − y = 0

9. 8 y′′ − 10 y′ − 3 y = 0

10. 3 y′′ − 20 y′ + 12 y = 0

11. y′′ + 9 y = 0

12. y′′ + 4 y′ + 5 y = 0

13. y′′ + 25 y = 0

14. y′′ + y = 0

15. y′′ − 2 y′ + 5 y = 0

16. y′′ + 16 y = 0

In Exercises 31–40, find the unique solution of the second-order initial value problem.

17. y′′ + 2 y′ + 4 y = 0

18. y′′ − 2 y′ + 3 y = 0

31. y′′ + 6 y′ + 5 y = 0, y(0) = 0,   y′(0) = 3

19. y′′ + 4 y′ + 9 y = 0

20. 4 y′′ − 4 y′ + 13 y = 0

32. y′′ + 16 y = 0, y(0) = 2,   y′(0) = −2

21. y′′ = 0

22. y′′ + 8 y′ + 16 y = 0

33. y′′ + 12 y = 0, y(0) = 0,   y′(0) = 1

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17.2  Nonhomogeneous Linear Equations

17-7

34. 12 y′′ + 5 y′ − 2 y = 0, y(0) = 1,   y′(0) = −1

58. 4 y′′ − 4 y′ + y = 0, y(0) = −1, y′(0) = 2

35. y′′ + 8 y = 0, y(0) = −1,   y′(0) = 2

59. 3 y′′ + y′ − 14 y = 0, y(0) = 2, y′(0) = −1

36. y′′ + 4 y′ + 4 y = 0, y(0) = 0,   y′(0) = 1

60. 4 y′′ + 4 y′ + 5 y = 0, y(π) = 1, y′(π) = 0

37. y′′ − 4 y′ + 4 y = 0, y(0) = 1,   y′(0) = 0

61. Prove that the two solution functions in Theorem 3 are linearly independent.

38. 4 y′′ − 4 y′ + y = 0, y(0) = 4,   y′(0) = 4

62. Prove that the two solution functions in Theorem 4 are linearly independent.

d 2y dy dy + 12 + 9 y = 0, y (0) = 2,   (0) = 1 dx 2 dx dx dy d 2y dy 40. 9 2 − 12 + 4 y = 0, y(0) = −1,   (0) = 1 dx dx dx 39. 4

63. Prove that the two solution functions in Theorem 5 are linearly independent.

41. y ′′ − 2 y ′ − 3 y = 0

42. 6 y ′′ − y ′ − y = 0

64. Prove that if y1 and y 2 are linearly independent solutions to the homogeneous equation (2), then the functions y 3 = y1 + y 2 and y 4 = y1 − y 2 are also linearly independent solutions.

43. 4 y ′′ + 4 y ′ + y = 0

44. 9 y ′′ + 12 y ′ + 4 y = 0

65. a.  Show that there is no solution to the boundary value problem

45. 4 y ′′ + 20 y = 0

46. y ′′ + 2 y ′ + 2 y = 0

47. 25 y ′′ + 10 y ′ + y = 0

48. 6 y ′′ + 13 y ′ − 5 y = 0

49. 4 y ′′ + 4 y ′ + 5 y = 0

50. y ′′ + 4 y ′ + 6 y = 0

In Exercises 41–55, find the general solution.

51. 16 y ′′ − 24 y ′ + 9 y = 0

52. 6 y ′′ − 5 y ′ − 6 y = 0

53. 9 y ′′ + 24 y ′ + 16 y = 0

54. 4 y ′′ + 16 y ′ + 52 y = 0

55. 6 y ′′ − 5 y ′ − 4 y = 0

y′′ + 4 y = 0, y (0) = 0, y (π) = 1. b. Show that there are infinitely many solutions to the boundary value problem y′′ + 4 y = 0, y(0) = 0, y(π ) = 0. 66. Show that if a, b, and c are positive constants, then all solutions of the homogeneous differential equation

In Exercises 56–60, solve the initial value problem.

ay′′ + by′ + cy = 0

56. y′′ − 2 y′ + 2 y = 0, y (0) = 0, y′(0) = 2

approach zero as x → ∞.

57. y′′ + 2 y′ + y = 0, y(0) = 1, y′(0) = 1

17.2 Nonhomogeneous Linear Equations In this section we study two methods for solving second-order linear nonhomogeneous differential equations with constant coefficients. These are the methods of undetermined coefficients and variation of parameters. We begin by considering the form of the general solution.

Form of the General Solution Suppose we wish to solve the nonhomogeneous equation

ay′′ + by′ + cy = G ( x ), (1)

where a, b, and c are constants and G is continuous over some open interval I. Let y c = c1 y1 + c 2 y 2 be the general solution to the associated complementary equation

ay′′ + by′ + cy = 0. (2)

(We learned how to find y c in Section 17.1.) Now suppose we could somehow come up with a particular function y p that solves the nonhomogeneous equation (1). Then the sum

y = y c + y p (3)

also solves the nonhomogeneous equation (1) because a ( y c + y p )″ + b ( y c + y p )′ + c ( y c + y p ) = ( ay c″ + by c′ + cy c ) + ( ay p″ + by p′ + cy p ) = 0 + G(x) = G ( x ).

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y c solves Eq. (2) and y p solves Eq. (1)

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17-8

Chapter 17  Second-Order Differential Equations

Moreover, if y = y ( x ) is the general solution to the nonhomogeneous equation (1), it must have the form of Equation (3). The reason for this last statement follows from the observation that for any function y p satisfying Equation (1), we have a ( y − y p )″ + b ( y − y p )′ + c ( y − y p ) = ( ay′′ + by′ + cy ) − ( ay p ″ + by p ′ + cy p ) = G ( x ) − G ( x ) = 0. Thus, y c = y − y p is the general solution to the homogeneous equation (2). We have established the following result.

THEOREM 7  The general solution y = y ( x ) to the nonhomogeneous differential equation (1) has the form

y = yc + yp, where the complementary solution y c is the general solution to the associated homogeneous equation (2), and y p is any particular solution to the nonhomogeneous equation (1).

The Method of Undetermined Coefficients This method for finding a particular solution y p to the nonhomogeneous equation (1) applies to special cases for which G ( x ) is a sum of terms of various polynomials p( x ) multiplying an exponential with possibly sine or cosine factors. That is, G ( x ) is a sum of terms of the following forms: p1 ( x )e rx ,

p 2 ( x )e αx cos β x ,

p 3 ( x )e αx sin β x.

For instance, 1 − x ,  e 2 x,  xe x,   cos x , and 5e x − sin 2 x represent functions in this category. (Essentially these are functions solving homogeneous linear differential equations with constant coefficients, but the equations may be of order higher than two.) We now present several examples illustrating the method. EXAMPLE 1   Solve the nonhomogeneous equation y ′′ − 2 y ′ − 3 y = 1 − x 2 . Solution  The auxiliary equation for the complementary equation y ′′ − 2 y ′ − 3 y = 0 is

r 2 − 2r − 3 = ( r + 1)( r − 3 ) = 0. It has the roots r = −1 and r = 3, giving the complementary solution y c = c1e − x + c 2 e 3 x . Now G ( x ) = 1 − x 2 is a polynomial of degree 2. It would be reasonable to assume that a particular solution to the given nonhomogeneous equation is also a polynomial of degree 2 because if y is a polynomial of degree 2, then y′′ − 2 y′ − 3 y is also a polynomial of degree 2. So we seek a particular solution of the form y p = Ax 2 + Bx + C. We need to determine the unknown coefficients A, B, and C. When we substitute the polynomial y p and its derivatives into the given nonhomogeneous equation, we obtain 2 A − 2( 2 Ax + B ) − 3( Ax 2 + Bx + C ) = 1 − x 2 , or, collecting terms with like powers of x, −3 Ax 2 + (−4 A − 3B ) x + ( 2 A − 2 B − 3C ) = 1 − x 2 .

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17.2  Nonhomogeneous Linear Equations

17-9

This last equation holds for all values of x if its two sides are identical polynomials of degree 2. Thus, we equate corresponding powers of x to get −3 A = −1,

−4 A − 3B = 0,

2 A − 2 B − 3C = 1.

and

These equations imply in turn that A = 1 3,  B = −4 9, and C = 5 27. Substituting these values into the quadratic expression for our particular solution gives yp =

1 2 4 5 x − x+ . 3 9 27

By Theorem 7, the general solution to the nonhomogeneous equation is y = y c + y p = c1e − x + c 2e 3 x +

1 2 4 5 x − x+ . 3 9 27

EXAMPLE 2   Find a particular solution of y′′ − y′ = 2 sin x. Solution  If we try to find a particular solution of the form

y p = A sin x and substitute the derivatives of y p in the given equation, we find that A must satisfy the equation − A sin x + A cos x = 2 sin x for all values of x. Since this requires A to equal both −2 and 0 at the same time, we conclude that the nonhomogeneous differential equation has no solution of the form A sin x . It turns out that the required form is the sum y p = A sin x + B cos x. The result of substituting the derivatives of this new trial solution into the differential equation is − A sin x − B cos x − ( A cos x − B sin x ) = 2 sin x , or (B

− A )sin x − ( A + B ) cos x = 2 sin x.

This last equation must be an identity. Equating the coefficients for like terms on each side then gives B− A = 2

and

A + B = 0.

Simultaneous solution of these two equations gives A = −1 and B = 1. Our particular solution is y p = cos x − sin x. EXAMPLE 3   Find a particular solution of y ′′ − 3 y ′ + 2 y = 5e x . Solution  If we substitute

y p = Ae x and its derivatives into the differential equation, we find that Ae x − 3 Ae x + 2 Ae x = 5e x , or 0 = 5e x .

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17-10

Chapter 17  Second-Order Differential Equations

However, the exponential function is never zero. The trouble can be traced to the fact that y = e x is already a solution of the related homogeneous equation y ′′ − 3 y ′ + 2 y = 0. The auxiliary equation is r 2 − 3r + 2 = ( r − 1)( r − 2 ) = 0, which has r = 1 as a root. So we would expect Ae x to become zero when substituted into the left-hand side of the differential equation. The appropriate way to modify the trial solution in this case is to multiply Ae x by x. Thus, our new trial solution is y p = Axe x . The result of substituting the derivatives of this new candidate into the differential equation is ( Axe x + 2 Ae x ) − 3( Axe x + Ae x ) + 2 Axe x = 5e x ,

or − Ae x = 5e x . Thus, A = −5 gives our sought-after particular solution y p = −5 xe x . EXAMPLE 4   Find a particular solution of y ′′ − 6 y ′ + 9 y = e 3 x . Solution  The auxiliary equation for the complementary equation

r 2 − 6r + 9 = ( r − 3 ) 2 = 0 has r = 3 as a repeated root. The appropriate choice for y p in this case is neither Ae 3 x nor Axe 3 x because the complementary solution contains both of those terms already. Thus, we choose a term containing the next higher power of x as a factor. When we substitute y p = Ax 2 e 3 x and its derivatives into the given differential equation, we get ( 9 Ax 2e 3 x + 12 Axe 3 x + 2 Ae 3 x ) − 6( 3 Ax 2e 3 x + 2 Axe 3 x ) + 9 Ax 2e 3 x = e 3 x ,

or 2 Ae 3 x = e 3 x . Thus, A = 1 2, and the particular solution is yp =

1 2 3x x e . 2

When we wish to find a particular solution of Equation (1) and the function G ( x ) is the sum of two or more terms, we choose a trial function for each term in G ( x ) and add them. EXAMPLE 5   Find the general solution to y′′ − y′ = 5e x − sin 2 x. Solution  We first check the auxiliary equation

r 2 − r = 0. Its roots are r = 1 and r = 0. Therefore, the complementary solution to the associated homogeneous equation is y c = c1e x + c 2 .

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17.2  Nonhomogeneous Linear Equations

17-11

We now seek a particular solution y p . That is, we seek a function that will produce 5e x − sin 2 x when substituted into the left-hand side of the given differential equation. One part of y p is to produce 5e x , the other − sin 2 x. Since any function of the form c1e x is a solution of the associated homogeneous equation, we choose our trial solution y p to be the sum y p = Axe x + B cos 2 x + C sin 2 x , including xe x where we might otherwise have included only e x . When the derivatives of y p are substituted into the differential equation, the resulting equation is ( Axe x + 2 Ae x − 4 B cos 2 x − 4C sin 2 x ) − ( Axe x + Ae x − 2 B sin 2 x + 2C cos 2 x ) = 5e x − sin 2 x , or Ae x − ( 4 B + 2C ) cos 2 x + ( 2 B − 4C )sin 2 x = 5e x − sin 2 x. This equation will hold if A = 5,

4 B + 2C = 0,

2 B − 4C = −1,

or A = 5,  B = −1 10, and C = 1 5. Our particular solution is y p = 5 xe x −

1 1 cos 2 x + sin 2 x. 10 5

The general solution to the differential equation is y = y c + y p = c1e x + c 2 + 5 xe x −

1 1 cos 2 x + sin 2 x. 10 5

You may find the following table helpful in solving the problems at the end of this section.

TABLE 17.1  The method of undetermined coefficients for selected equations

of the form ay′′ + by′ + cy = G ( x).

If G ( x ) has a term that is a constant multiple of . . . e rx

sin kx ,   cos kx px 2 + qx + m

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And if . . . r is not a root of the auxiliary equation r is a single root of the auxiliary equation r is a double root of the auxiliary equation ki is not a root of the auxiliary equation 0 is not a root of the auxiliary equation 0 is a single root of the auxiliary equation 0 is a double root of the auxiliary equation

Then include this expression in the trial function for yp Ae rx Axe rx Ax 2 e rx B cos kx + C sin kx Dx 2 + Ex + F Dx 3 + Ex 2 + Fx Dx 4 + Ex 3 + Fx 2

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17-12

Chapter 17  Second-Order Differential Equations

The Method of Variation of Parameters This is a general method for finding a particular solution of the nonhomogeneous equation (1) once the general solution of the associated homogeneous equation is known. The method consists of replacing the constants c1 and c 2 in the complementary solution by functions υ1 = υ1 ( x ) and υ 2 = υ 2 ( x ) and requiring (in a way to be explained) that the resulting expression satisfy the nonhomogeneous equation (1). There are two functions to be determined, and requiring that Equation (1) be satisfied is only one condition. As a second condition, we also require that υ1 ′ y1 + υ 2 ′ y 2 = 0. (4)

Then we have

y = υ1 y1 + υ 2 y 2 , y′ = υ1 y1 ′ + υ 2 y 2 ′, y′′ = υ1 y1 ″ + υ 2 y 2 ″ + υ1 ′ y1 ′ + υ 2 ′ y 2 ′. If we substitute these expressions into the left-hand side of equation (1), we obtain υ1 ( ay1 ″ + by1 ′ + cy1 ) + υ 2 ( ay 2 ″ + by 2 ′ + cy 2 ) + a ( υ1 ′ y1 ′ + υ 2 ′ y 2 ′ ) = G ( x ). The first two parenthetical terms are zero since y 1 and y 2 are solutions of the associated homogeneous equation (2). So the nonhomogeneous equation (1) is satisfied if, in addition to equation (4), we require that

a ( υ1 ′ y1 ′ + υ 2 ′ y 2 ′ ) = G ( x ). (5)

Equations (4) and (5) can be solved together as a pair υ1 ′ y1 + υ 2 ′ y 2 = 0, υ1 ′ y1 ′ + υ 2 ′ y 2 ′ =

G(x) a

for the unknown functions υ1 ′ and υ 2 ′. The usual procedure for solving this simple system is to use the method of determinants (also known as Cramer’s Rule), which will be demonstrated in the examples to follow. Once the derivative functions υ1 ′ and υ 2 ′ are known, the two functions υ1 = υ1 ( x ) and υ 2 = υ 2 ( x ) can be found by integration. Here is a summary of the method. Variation of Parameters Procedure

To use the method of variation of parameters to find a particular solution to the nonhomogeneous equation ay′′ + by′ + cy = G ( x ), we can work directly with Equations (4) and (5). It is not necessary to rederive them. The steps are as follows. 1. Solve the associated homogeneous equation ay′′ + by′ + cy = 0 to find the functions y1 and y 2. 2. Solve the equations υ1 ′ y1 + υ 2 ′ y 2 = 0, υ1 ′ y1 ′ + υ 2 ′ y 2 ′ =

G(x) a

simultaneously for the derivative functions υ1 ′ and υ 2 ′. 3. Integrate υ1 ′ and υ 2 ′ to find the functions υ1 = υ1 ( x ) and υ 2 = υ 2 ( x ). 4. Write down the particular solution to the nonhomogeneous equation (1) as y p = υ1 y1 + υ 2 y 2 .

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17.2  Nonhomogeneous Linear Equations

17-13

EXAMPLE 6   Find the general solution to the equation y′′ + y = tan x. Solution  The solution of the homogeneous equation

y′′ + y = 0 is given by y c = c1 cos x + c 2 sin x. Since y1 ( x ) = cos x and y 2 ( x ) = sin x , the conditions to be satisfied in Equations (4) and (5) are υ1 ′ cos x + υ 2 ′ sin x = 0, −υ1 ′ sin x + υ 2 ′ cos x = tan x.

a =1

Solving this system gives

υ1 ′ =

0

sin x

tan x

cos x

cos x

sin x

−sin x

cos x

=

− tan x   sin x −sin 2 x = . 2 2 cos x + sin x cos x

Likewise,

υ2 ′ =

cos x

0

− sin x

tan x

cos x

sin x

−sin x

cos x

= sin x.

After integrating υ1 ′ and υ 2 ′, we have υ1 ( x ) =

∫

− sin 2 x dx cos x

= −∫ ( sec x − cos x ) dx = − ln sec x + tan x + sin x , and υ2 (x) =

∫ sin x dx

= − cos x.

Note that we have omitted the constants of integration in determining υ1 and υ 2 . They would merely be absorbed into the arbitrary constants in the complementary solution. Substituting υ1 and υ 2 into the expression for y p in Step 4 gives y p = [ − ln sec x + tan x + sin x ]cos x + (− cos x )sin x = ( − cos x ) ln sec x + tan x . The general solution is y = c1 cos x + c 2 sin x − ( cos x ) ln sec x + tan x . EXAMPLE 7   Solve the nonhomogeneous equation y′′ + y′ − 2 y = xe x . Solution  The auxiliary equation is

r 2 + r − 2 = ( r + 2 )( r − 1) = 0,

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Chapter 17  Second-Order Differential Equations

giving the complementary solution y c = c1e −2 x + c 2e x . The conditions to be satisfied in Equations (4) and (5) are   υ1 ′e −2 x + υ 2 ′e x = 0, −2υ1 ′e −2 x + υ 2 ′e x = xe x .

   a =1

Solving the above system for υ1 ′ and υ 2 ′ gives

υ1 ′ =

0

ex

xe x

ex

e −2 x

ex

−2e −2 x

ex

=

−xe 2 x 1 = − xe 3 x . 3e − x 3

Likewise,

υ2′ =

e −2 x

0

−2e −2 x

xe x

=

3e − x

xe − x x = . 3e − x 3

Integrating to obtain the parameter functions, we have 1 − xe 3 x dx 3 1 xe 3 x e 3x =− −∫ dx 3 3 3 1( = 1 − 3x ) e 3 x 27

υ1 ( x ) =

∫

(

)

and υ2 (x) =

∫

x x2 dx = . 3 6

Therefore,

( )

(1 − 3 x ) e 3 x  − 2 x x2 x y p =  e + e  27 6   1 x 1 1 = e − xe x + x 2e x . 27 9 6

The general solution to the differential equation is y = c1e −2 x + c 2 e x −

1 x 1 xe + x 2e x , 9 6

where the term (1 27 )e x in y p has been absorbed into the term c 2 e x in the complementary solution.

EXERCISES 

17.2

Solve the equations in Exercises 1–16 by the method of undetermined coefficients.

7. 7′′ − y′ − 2 y = 20 cos x

8. y ′′ + y = 2 x + 3e x

1. y ′′ − 3 y ′ − 10 y = −3

2. y ′′ − 3 y ′ − 10 y = 2 x − 3

9. y ′′ − y = e x + x 2

10. y′′ + 2 y′ + y = 6 sin 2 x

3. y′′ − y′ = sin x

4. y ′′ + 2 y ′ + y = x 2

11. y′′ − y′ − 6 y =

5. y′′ + y = cos 3 x

6. y ′′ + y = e 2 x

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e −x

− 7 cos x

12. y ′′ + 3 y ′ + 2 y = e −x + e −2 x − x

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17.3  Applications

13.

d 2y dy +5 = 15 x 2 dx 2 dx

14.

d 2y dy − = −8 x + 3 dx 2 dx

15.

d 2y dy −3 = e 3 x − 12 x dx 2 dx

16.

d 2y dy +7 = 42 x 2 + 5 x + 1 dx 2 dx

Solve the equations in Exercises 17–28 by variation of parameters. 17. y ′′ + y ′ = x 18. y′′ + y = tan x ,

−

19. y′′ + y = sin x

π π < x < 2 2 20. y ′′ + 2 y ′ + y = e x

21. y ′′ + 2 y ′ + y = e −x

22. y ′′ − y = x

23. y ′′ − y = e x

24. y′′ − y = sin x

25. y′′ + 4 y′ + 5 y = 10

26. y′′ − y′ = 2 x

27. 28.

d 2y dx 2 d 2y dx 2

+ y = sec x , −

−

π π < x < 2 2

dy = e x cos x ,   x > 0 dx

In each of Exercises 29–32, the given differential equation has a particular solution y p of the form given. Determine the coefficients in y p . Then solve the differential equation. 29. y ′′ − 5 y ′ = xe 5 x , y p = Ax 2 e 5 x + Bxe 5 x 30. y′′ − y′ = cos x + sin x , y p = A cos x + B sin x 31. y′′ + y = 2 cos x + sin x , y p = Ax cos x + Bx sin x 32. y ′′ + y ′ − 2 y = xe x , y p = Ax 2 e x + Bxe x In Exercises 33–36, solve the given differential equations (a) by variation of parameters and (b) by the method of undetermined coefficients. 33.

d 2y dy − = e x + e −x dx 2 dx

35.

d 2y dy d 2y dy −4 − 5 y = e x + 4 36. 2 − 9 = 9e 9 x dx 2 dx dx dx

34.

d 2y dy −4 + 4 y = 2e 2 x dx 2 dx

Solve the differential equations in Exercises 37–46. Some of the equations can be solved by the method of undetermined coefficients, but others cannot. 37. y′′ + y = cot x , 0 < x < π 38. y′′ + y = csc x , 0 < x < π

43. y ′′ + 2 y ′ = x 2 − e x

44. y′′ + 9 y = 9 x − cos x

π π 45. y′′ + y = sec x tan x , − < x < 2 2 46. y ′′ − 3 y ′ + 2 y = e x − e 2 x The method of undetermined coefficients can sometimes be used to solve first-order ordinary differential equations. Use the method to solve the equations in Exercises 47–50. 47. y ′ − 3 y = e x

48. y ′ + 4 y = x

49. y ′ − 3 y =

50. y′ + y = sin x

5e 3 x

Solve the differential equations in Exercises 51 and 52 subject to the given initial conditions. 51.

d 2y π π + y = sec 2 x , − < x < ; y (0) = y′ (0) = 1 2 2 dx 2

52.

d 2y 2 + y = e 2 x ; y (0) = 0, y′ (0) = 5 dx 2

In Exercises 53–58, verify that the given function is a particular solution to the specified nonhomogeneous equation. Find the general solution, and evaluate its arbitrary constants to find the unique solution satisfying the equation and the given initial conditions. 53. y′′ + y′ = x , y p =

x2 − x , y (0) = 0, y′ (0) = 0 2

54. y′′ + y = x , y p = 2 sin x + x , y (0) = 0, y′ (0) = 0 1 55. y′′ + y′ + y = 4e x ( cos x − sin x ), 2 y p = 2e x cos x , y(0) = 0, y′(0) = 1 56. y′′ − y′ − 2 y = 1 − 2 x , y p = x − 1, y(0) = 0, y′(0) = 1 57. y′′ − 2 y′ + y = 2e x , y p = x 2e x , y(0) = 1, y′(0) = 0 58. y′′ − 2 y′ + y = x −1e x ,   x > 0, y p = xe x ln x , y(1) = e, y′(1) = 0 In Exercises 59 and 60, two linearly independent solutions y1 and y 2 are given to the associated homogeneous equation of the variablecoefficient nonhomogeneous equation. Use the method of variation of parameters to find a particular solution to the nonhomogeneous equation. Assume x > 0 in each exercise.

39. y ′′ − 8 y ′ = e 8 x

40. y′′ + 4 y = sin x

59. x 2 y′′ + 2 xy′ − 2 y = x 2 , y1 = x −2 ,   y 2 = x

41. y ′′ − y ′ =

42. y ′′ + 4 y ′ + 5 y = x + 2

60. x 2 y′′ + xy′ − y = x , y1 = x −1 ,   y 2 = x

x3

17-15

17.3 Applications In this section we apply second-order differential equations to the study of vibrating springs and electric circuits.

Vibrations A spring has its upper end fastened to a rigid support, as shown in Figure 17.2. An object of mass m is suspended from the spring and stretches it a length s when the spring comes to rest in an equilibrium position. According to Hooke’s Law (Section 6.5), the tension

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17-16

Chapter 17  Second-Order Differential Equations

force in the spring is ks, where k is the spring constant. The force due to gravity pulling down on the spring is mg, and equilibrium requires that ks = mg. (1)



Suppose that the object is pulled down an additional amount y 0 beyond the equilibrium position and then released. We want to study the object’s motion, that is, the vertical position of its center of mass at any future time. Let y, with positive direction downward, denote the displacement position of the object away from the equilibrium position y = 0 at any time t after the motion has started. Then the forces acting on the object are (see Figure 17.3)

s y50 Mass m at equilibrium

y

the propulsion force due to gravity,

Fs = k ( s + y ),

the restoring force of the spring’s tension,

Fr = δ

FIGURE 17.2 Mass m stretches a spring

by length s to the equilibrium position at y = 0.

Fp = mg, dy , dt

a frictional force assumed proportional to velocity.

The frictional force tends to slow the motion of the object. The resultant of these forces is F = Fp − Fs − Fr , and by Newton’s second law F = ma, we must then have m

d 2y dy = mg − ks − ky − δ . dt 2 dt

By Equation (1), mg − ks = 0, so this last equation becomes

s y50 Fs A position after release

y

y0

Start position

Fr

Fp

y

FIGURE 17.3  The propulsion force

(weight) Fp pulls the mass downward, but the spring restoring force Fs and frictional force Fr pull the mass upward. The motion starts at y = y 0 with the mass vibrating up and down.

m

d 2y dy +δ + ky = 0, (2) dt 2 dt

subject to the initial conditions y (0) = y 0 and y′ (0) = 0. (Here we use the prime notation to denote differentiation with respect to time t.) You might expect that the motion predicted by Equation (2) will be oscillatory about the equilibrium position y = 0 and eventually damp to zero because of the frictional force. This is indeed the case, and we will show how the constants m, δ, and k determine the nature of the damping. You will also see that if there is no friction (so δ = 0), then the object will simply oscillate indefinitely.

Simple Harmonic Motion Suppose first that there is no frictional force. Then δ = 0 and there is no damping. If we substitute ω = k m to simplify our calculations, then the second-order equation (2) becomes y′′ + ω 2 y = 0,

y (0) = y 0

with

and

y′ (0) = 0.

The auxiliary equation is r 2 + ω 2 = 0, which has the imaginary roots r = ±ωi. The general solution to the differential equation in (2) is

y = c1 cos ωt + c 2 sin ωt. (3)

To fit the initial conditions, we compute y′ = −c1ω sin ωt + c 2 ω cos ωt and then substitute the conditions. This yields c1 = y 0 and c 2 = 0. The particular solution

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y = y 0 cos ωt (4)

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17.3  Applications

C=

1 c12 + c22 c1 f

describes the motion of the object. Equation (4) represents simple harmonic motion of amplitude y 0 and period T = 2π ω. The general solution given by Equation (3) can be combined into a single term by using the trigonometric identity sin ( ωt + φ ) = cos ωt sin φ + sin ωt cos φ.

c2

FIGURE 17.4  c1 = C sin φ and

17-17

To apply the identity, we take (see Figure 17.4)

c 2 = C cos φ.

c1 = C sin φ

and

c 2 = C cos φ,

where C =

c12 + c 22

and

φ = tan −1

c1 . c2

Then the general solution in Equation (3) can be written in the alternative form y = C sin ( ωt + φ ). (5)



Here C and φ may be taken as two new arbitrary constants, replacing the two constants c1 and c 2. Equation (5) represents simple harmonic motion of amplitude C and period T = 2π ω. The angle ωt + φ is called the phase angle, and φ may be interpreted as its initial value. A graph of the simple harmonic motion represented by Equation (5) is given in Figure 17.5. y

C C sin f

Period T = 2p v

t

0 –C y = C sin(vt + f)

FIGURE 17.5  Simple harmonic motion of amplitude C and period T with initial phase angle φ (Equation 5).

Damped Motion Assume now that there is friction in the spring system, so δ ≠ 0. If we substitute ω = k m and 2b = δ m, then the differential equation (2) is

y′′ + 2by′ + ω 2 y = 0. (6)

The auxiliary equation is r 2 + 2br + ω 2 = 0, with roots r = −b ± b 2 − ω 2 . Three cases now present themselves, depending on the relative sizes of b and ω. Case 1: b = ω.  The double root of the auxiliary equation is real and equals r = ω. The general solution to Equation (6) is

y = ( c 1 + c 2 t ) e −ω t . This situation of motion is called critical damping and is not oscillatory. Figure 17.6a shows an example of this kind of damped motion.

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17-18

Chapter 17  Second-Order Differential Equations

Case 2: b > ω.  The roots of the auxiliary equation are real and unequal, and they are given by r1 = −b + b 2 − ω 2 and r2 = −b − b 2 − ω 2 . The general solution to Equation (6) is given by

y = c1e (−b +

b 2 − ω 2 )t

+ c 2 e ( −b −

b 2 − ω 2 )t

.

Here again the motion is not oscillatory and both r1 and r2 are negative. Thus y approaches zero as time goes on. This motion is referred to as overdamping (see Figure 17.6b). Case 3: b < ω.  The roots to the auxiliary equation are complex and are given by r = −b ± i ω 2 − b 2 . The general solution to Equation (6) is given by

y = e −bt ( c1 cos ω 2 − b 2 t + c 2 sin ω 2 − b 2 t ). This situation, called underdamping, represents damped oscillatory motion. It is analogous to simple harmonic motion of period T = 2π ω 2 − b 2 except that the amplitude is not constant but damped by the factor e −bt . Therefore, the motion tends to zero as t increases, so the vibrations tend to die out as time goes on. Notice that the period T = 2π ω 2 − b 2 is larger than the period T0 = 2π ω in the friction-free system. Moreover, the larger the value of b = δ ( 2 m ) in the exponential damping factor, the more quickly the vibrations tend to become unnoticeable. A curve illustrating underdamped motion is shown in Figure 17.6c. y

0

y

t

0

y = (1 + t)e–t (a) Critical damping

y

t y = 2e –2t – e –t (b) Overdamping

0

t y = e –t sin (5t + p/4) (c) Underdamping

FIGURE 17.6  Three examples of damped vibratory motion for a spring system

with friction, so δ ≠ 0.

An external force F (t ) can also be added to the spring system modeled by Equation (2). The forcing function may represent an external disturbance on the system. For instance, if the equation models an automobile suspension system, the forcing function might represent periodic bumps or potholes in the road affecting the performance of the suspension system. Or it might represent the effects of winds when modeling the vertical motion of a suspension bridge. Inclusion of a forcing function results in the second-order nonhomogeneous equation

m

d 2y dy +δ + ky = F (t ). (7) 2 dt dt

Such equations are studied in the theory of Differential Equations.

Electric Circuits The basic quantity in electricity is the charge q (analogous to the idea of mass). In an electric field we use the flow of charge, or current I = dq dt, as we might use velocity in a gravitational field. There are many similarities between motion in a gravitational field and the flow of electrons (the carriers of charge) in an electric field. Consider the electric circuit shown in Figure 17.7. It consists of four components: voltage source, resistor, inductor, and capacitor. Think of electrical flow as being like a fluid flow, where the voltage source is the pump and the resistor, inductor, and capacitor tend to block the flow. A battery or generator is an example of a source, producing a

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17.3  Applications

17-19

R, Resistor Voltage E source

L, Inductor

C, Capacitor

FIGURE 17.7  An electric circuit.

voltage that causes the current to flow through the circuit when the switch is closed. An electric light bulb or appliance would provide resistance. The inductance is due to a magnetic field that opposes any change in the current as it flows through a coil. The capacitance is normally created by two metal plates that alternate charges and thus reverse the current flow. The following symbols specify the quantities relevant to the circuit. q:  charge at a cross section of a conductor, measured in coulombs (abbreviated c) I  :   current or rate of change of charge dq dt (flow of electrons) at a cross section of a conductor, measured in amperes (abbreviated A) E:  electric (potential) source, measured in volts (abbreviated V) V:  difference in potential between two points along the conductor, measured in volts (V) Ohm observed that the current I flowing through a resistor, caused by a potential difference across it, is (approximately) proportional to the potential difference (voltage drop). He named his constant of proportionality 1 R and called R the resistance. So Ohm’s law is I =

1 V. R

Similarly, it is known from physics that the voltage drops across an inductor and a capacitor are, respectively, L

dI dt

and

q , C

where L is the inductance and C is the capacitance (with q the charge on the capacitor). The German physicist Gustav R. Kirchhoff (1824–1887) formulated the law that the sum of the voltage drops in a closed circuit is equal to the supplied voltage E (t ). Symbolically, this says that RI + L

q dI + = E (t ). dt C

Since I = dq dt, Kirchhoff’s law becomes

L

d 2q dq 1 + R + q = E (t ). (8) dt 2 dt C

The second-order differential equation (8), which models an electric circuit, has exactly the same form as Equation (7) modeling vibratory motion. Both models can be solved using the methods developed in Section 17.2.

Summary The following chart summarizes our analogies between the physics of motion of an object in a spring system and the flow of charged particles in an electric circuit.

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17-20

Chapter 17  Second-Order Differential Equations

Linear Second-Order Constant-Coefficient Models

Mechanical System my′′ + δ y′ + ky = F (t ) y y′ y′′ m δ k F (t )

EXERCISES 

displacement velocity acceleration mass damping constant spring constant forcing function

Electrical System 1 Lq′′ + Rq′ + q = E (t ) C q charge current q′ change in current q′′ L inductance R resistance 1C where C is the capacitance E (t ) voltage source

17.3

1. A 70-N weight is attached to the lower end of a coil spring suspended from the ceiling and having a spring constant of 15 N/m. The resistance in the spring–mass system is numerically equal to 15 times the instantaneous velocity. At t = 0, the weight is set in motion from a position 0.6 m below its equilibrium position by giving it a downward velocity of 0.6 m/s. Write an initial value problem that models the given situation. 2. A 36-N weight stretches a spring 1.2 m. The spring–mass system resides in a medium offering a resistance to the motion that is numerically equal to 20 times the instantaneous velocity. If the weight is released at a position 0.6 m above its equilibrium position with a downward velocity of 0.9 m/s, write an initial value problem modeling the given situation. 3. A 90-N weight is hung on a 0.4-m spring and stretches it 0.15 m. The weight is pulled down 0.1 m and 30 N are added to the weight. If the weight is now released with a downward velocity of υ 0 m/s, write an initial value problem modeling the vertical displacement. 4. A 49-N weight is suspended by a spring that is stretched 0.05 m by the weight. Assume a resistance whose magnitude is 300 g  N times the instantaneous velocity υ in meters per second. If the weight is pulled down 0.08 m below its equilibrium position and released, formulate an initial value problem modeling the behavior of the spring–mass system. 5. An (open) electric circuit consists of an inductor, a resistor, and a capacitor. There is an initial charge of 2 coulombs on the capacitor. At the instant the circuit is closed, a current of 3 amperes is present and a voltage of E (t ) = 20 cos t is applied. In this circuit the voltage drop across the resistor is 4 times the instantaneous change in the charge, the voltage drop across the capacitor is 10 times the charge, and the voltage drop across the inductor is 2 times the instantaneous change in the current. Write an initial value problem to model the circuit. 6. An inductor of 2 henrys is connected in series with a resistor of 12 ohms, a capacitor of 1 16 farad, and a 300-volt battery.

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Initially, the charge on the capacitor is zero and the current is zero. Formulate an initial value problem modeling this electric circuit. 7. A 49-N weight is attached to the lower end of a coil spring suspended from the ceiling and having a spring constant of 10 N/m. The resistance in the spring–mass system is numerically equal to 10 times the instantaneous velocity. At t = 0, the weight is set in motion from a position 0.6 m below its equilibrium position by giving it a downward velocity of 0.6 m/s. At the end of π s, determine whether the mass is above or below the equilibrium position and by what distance. 8. A 29.4-N weight stretches a spring 1.225 m. The spring–mass system resides in a medium offering a resistance to the motion equal to 18 times the instantaneous velocity. If the weight is released at a position 0.6 m above its equilibrium position with a downward velocity of 0.9 m/s, find its position relative to the equilibrium position 2 s later. 9. A 98-N weight is hung on a 0.6 m spring stretching it 0.2 m. The weight is pulled down 0.15 m and 49 N are added to the weight. If the weight is now released with a downward velocity of υ 0 m/s, find the position of mass relative to the equilibrium in terms of υ 0 and valid for any time t ≥ 0. 10. A mass of 15 kg is attached to a spring whose constant is 375/4  N/m. Initially the mass is released 1 m above the equilibrium position with a downward velocity of 3 m/s, and the subsequent motion takes place in a medium that offers a damping force numerically equal to 45 times the instantaneous velocity. An external force f (t ) is driving the system, but assume that initially f (t ) ≡ 0. Formulate and solve an initial value problem that models the given system. Interpret your results. 11. A 50-N weight is suspended by a spring that is stretched 0.05 m by the weight. Assume a resistance whose magnitude is 100 N times the instantaneous velocity in meters per second. If the weight is pulled down 0.1 m below its equilibrium position and released, find the time required to reach the equilibrium position for the first time.

23/03/23 11:42 AM



12. A weight stretches a spring 0.2 m. It is set in motion at a point 0.05 m below its equilibrium position with a downward velocity of 0.05 m/s. a. When does the weight return to its equilibrium position? b. When does it reach its highest point? c. Show that the maximum velocity is 0.05 10 g  m s. 13. A weight of 50 N stretches a spring 0.25 m. The weight is drawn down 0.05 m below its equilibrium position and given an initial velocity of 0.1 m/s. An identical spring has a different weight attached to it. This second weight is drawn down from its equilibrium position a distance equal to the amplitude of the first motion and then given an initial velocity of 0.6 m/s. If the amplitude of the second motion is twice that of the first, what weight is attached to the second spring? 14. A weight stretches one spring 0.05 m and a second weight stretches another spring 0.15 m. If both weights are simultaneously pulled down 0.02 m below their respective equilibrium positions and then released, find the first time after t = 0 when their velocities are equal. 15. A weight of 80 N stretches a spring 1 m. The weight is pulled down 1.5 m below the equilibrium position and then released. What initial velocity υ 0 given to the weight would have the effect of doubling the amplitude of the vibration? 16. A mass weighing 40 N stretches a spring 0.1 m. The spring– mass system resides in a medium with a damping constant of 32 N-s/m. If the mass is released from its equilibrium position with a velocity of 0.1 m/s in the downward direction, find the time required for the mass to return to its equilibrium position for the first time. 17. A weight suspended from a spring executes damped vibrations with a period of 2 s. If the damping factor decreases by 90% in 10 s, find the acceleration of the weight when it is 0.1 m below its equilibrium position and is moving upward with a speed of 0.8 m/s. 18. A 50-N weight stretches a spring 0.6 m. If the weight is pulled down 0.15 m below its equilibrium position and released, find the highest point reached by the weight. Assume the spring–mass system resides in a medium offering a resistance of 30 N times the instantaneous velocity in meters per second. 19. An LRC circuit is set up with an inductance of 1 5 henry, a resistance of 1 ohm, and a capacitance of 5 6 farad. Assuming the initial charge is 2 coulombs and the initial current is 4 amperes, find the solution function describing the charge on the capacitor at any time. What is the charge on the capacitor after a long period of time? 20. An (open) electric circuit consists of an inductor, a resistor, and a capacitor. There is an initial charge of 2 coulombs on the capacitor. At the instant the circuit is closed, a current of 3 amperes is present but no external voltage is being applied. In this circuit the voltage drops at three points are numerically related as follows: across

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17.3  Applications

17-21

the capacitor, 10 times the charge; across the resistor, 4 times the instantaneous change in the charge; and across the inductor, 2 times the instantaneous change in the current. Find the charge on the capacitor as a function of time. 21. A 78.4-N weight stretches a spring 1.225 m. This spring–mass system is in a medium with a damping constant of 72 N-s/m, and an external force given by f (t ) = 25.6 + 6.4e −2 t (in newtons) is being applied. What is the solution function describing the position of the mass at any time if the mass is released from 0.6 m below the equilibrium position with an initial velocity of 1.2 m/s downward? 22. A 10-kg mass is attached to a spring having a spring constant of 140 N m. The mass is started in motion from the equilibrium position with an initial velocity of 1 m s in the upward direction and with an applied external force given by f (t ) = 5 sin t (in newtons). The mass is in a viscous medium with a coefficient of resistance equal to 90 N-s m. Formulate an initial value problem that models the given system; solve the model and interpret the results. 23. A 2-kg mass is attached to the lower end of a coil spring suspended from the ceiling. The mass comes to rest in its equilibrium position thereby stretching the spring 1.96 m. The mass is in a viscous medium that offers a resistance in newtons numerically equal to 4 times the instantaneous velocity measured in meters per second. The mass is then pulled down 2 m below its equilibrium position and released with a downward velocity of 3 m s. At this same instant an external force given by f (t ) = 20 cos t (in newtons) is applied to the system. At the end of π s determine if the mass is above or below its equilibrium position and by how much. 24. A 39.2-N weight stretches a spring 1.225 m. The spring–mass system resides in a medium offering a resistance to the motion equal to 24 times the instantaneous velocity, and an external force given by f (t ) = 28.8 + 19.2e −t (in newtons) is being applied. If the weight is released at a position 0.6 m above its equilibrium position with downward velocity of 0.9 m/s, find its position relative to the equilibrium after 2 s have elapsed. 25. Suppose L = 10 henrys, R = 10 ohms, C = 1 500 farads, E = 100 volts, q (0) = 10 coulombs, and q′ (0) = i (0) = 0. Formulate and solve an initial value problem that models the given LRC circuit. Interpret your results. 26. A series circuit consisting of an inductor, a resistor, and a capacitor is open. There is an initial charge of 2 coulombs on the capacitor, and 3 amperes of current is present in the circuit at the instant the circuit is closed. A voltage given by E (t ) = 20 cos t is applied. In this circuit the voltage drops are numerically equal to the following: across the resistor, to 4 times the instantaneous change in the charge; across the capacitor, to 10 times the charge; and across the inductor, to 2 times the instantaneous change in the current. Find the charge on the capacitor as a function of time. Determine the charge on the capacitor and the current at time t = 10.

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17-22

Chapter 17  Second-Order Differential Equations

17.4 Euler Equations In Section 17.1 we introduced the second-order linear homogeneous differential equation P ( x ) y′′ ( x ) + Q ( x ) y′ ( x ) + R ( x ) y ( x ) = 0 and showed how to solve this equation when the coefficients P, Q, and R are constants. If the coefficients are not constant, we cannot generally solve this differential equation in terms of elementary functions we have studied in calculus. In this section you will learn how to solve the equation when the coefficients have the special forms P ( x ) = ax 2 ,

Q ( x ) = bx ,

R ( x ) = c,

and

where a, b, and c are constants. These special types of equations are called Euler equations in honor of Leonhard Euler, who studied them and showed how to solve them. Such equations arise in the study of mechanical vibrations.

The General Solution of Euler Equations Consider the Euler equation ax 2 y′′ + bxy′ + cy = 0,



x > 0. (1)

To solve Equation (1), we first make the change of variables z = ln x

and

y ( x ) = Y ( z ).

We next use the chain rule to find the derivatives y′ ( x ) and y′′ ( x ): y′ ( x ) =

d d dz 1 Y (z) = Y (z) = Y ′ (z) dx dz dx x

and y′′ ( x ) =

d d 1 1 1 dz 1 1 y′ ( x ) = Y ′ ( z ) = − 2 Y ′ ( z ) + Y ′′ ( z ) = − 2 Y ′ ( z ) + 2 Y ′′ ( z ). dx dx dx x x x x x

Substituting these two derivatives into the left-hand side of Equation (1), we find

(

)

(

)

1 1 1 Y ′( z ) + 2 Y ′′ ( z ) + bx Y ′ ( z ) + cY ( z ) x2 x x = aY ′′ ( z ) + ( b − a )Y ′ ( z ) + cY ( z ).

ax 2 y′′ + bxy′ + cy = ax 2 −

Therefore, the substitutions give us the second-order linear differential equation with constant coefficients

aY ′′ ( z ) + ( b − a )Y ′ ( z ) + cY ( z ) = 0. (2)

We can solve Equation (2) using the method of Section 17.1. That is, we find the roots of the associated auxiliary equation

ar 2 + ( b − a )r + c = 0 (3)

to find the general solution for Y ( z ). After finding Y ( z ), we can determine y ( x ) from the substitution z = ln x. EXAMPLE 1   Find the general solution of the equation x 2 y ′′ + 2 xy ′ − 2 y = 0. Solution  This is an Euler equation with a = 1, b = 2, and c = −2. The auxiliary equation (3) for Y ( z ) is

r 2 + ( 2 − 1) r − 2 = ( r − 1)( r + 2 ) = 0,

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17.4  Euler Equations

17-23

with roots r = −2 and r = 1. The solution for Y ( z ) is given by Y ( z ) = c1e −2 z + c 2 e z . Substituting z = ln x gives the general solution for y ( x ): y ( x ) = c1e −2 ln x + c 2e ln x = c1 x −2 + c 2 x.

EXAMPLE 2   Solve the Euler equation x 2 y ′′ − 5 xy ′ + 9 y = 0. Solution  Since a = 1, b = −5, and c = 9, the auxiliary equation (3) for Y ( z ) is

r 2 + (−5 − 1) r + 9 = ( r − 3 ) 2 = 0. The auxiliary equation has the double root r = 3, giving Y ( z ) = c1e 3 z + c 2 ze 3 z . Substituting z = ln x into this expression gives the general solution y ( x ) = c1e 3 ln x + c 2 ln x e 3 ln x = c1 x 3 + c 2 x 3 ln x.

EXAMPLE 3   Find the particular solution to x 2 y′′ − 3 xy′ + 68 y = 0 that satisfies the initial conditions y(1) = 0 and y′(1) = 1. Solution  Here a = 1, b = −3, and c = 68 substituted into the auxiliary equation (3) give

r 2 − 4r + 68 = 0. The roots are r = 2 + 8i and r = 2 − 8i, giving the solution Y ( z ) = e 2 z ( c1 cos 8 z + c 2 sin 8 z ). Substituting z = ln x into this expression gives y( x ) = e 2 ln x ( c1 cos( 8 ln x ) + c 2 sin ( 8 ln x )). From the initial condition y(1) = 0, we see that c1 = 0 and y( x ) = c 2 x 2 sin ( 8 ln x ). y

To fit the second initial condition, we need the derivative

10

y′( x ) = c 2 ( 8 x cos( 8 ln x ) + 2 x sin ( 8 ln x )).

5

Since y′(1) = 1, we immediately obtain c 2 = 1 8. Therefore, the particular solution satisfying both initial conditions is

0 –5

2

4

6

8

10

2 y = x sin (8 ln x) 8

–10

FIGURE 17.8  Graph of the solution

to Example 3.

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x

y (x) =

1 2 x sin ( 8 ln x ). 8

Since −1 ≤ sin ( 8 ln x ) ≤ 1, the solution satisfies −

x2 x2 ≤ y (x) ≤ . 8 8

A graph of the solution is shown in Figure 17.8.

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17-24

Chapter 17  Second-Order Differential Equations

EXERCISES 

17.4

In Exercises 1–24, find the general solution to the given Euler equation. Assume x > 0 throughout.

21. 9 x 2 y′′ + 15 xy′ + y = 0

1. x 2 y′′ + 2 xy′ − 2 y = 0

2. x 2 y′′ + xy′ − 4 y = 0

3. x 2 y′′ − 6 y = 0

4. x 2 y′′ + xy′ − y = 0

23. 16 x 2 y′′ + 56 xy′ + 25 y = 0

5. x 2 y′′ − 5 xy′ + 8 y = 0

6. 2 x 2 y′′ + 7 xy′ + 2 y = 0

7. 3 x 2 y′′ + 4 xy′ = 0

8. x 2 y′′ + 6 xy′ + 4 y = 0

In Exercises 25–30, solve the given initial value problem.

9. x 2 y′′ − xy′ + y = 0

10. x 2 y′′ − xy′ + 2 y = 0

25. x 2 y′′ + 3 xy′ − 3 y = 0, y(1) = 1, y′(1) = −1

11.

x 2 y′′

− xy′ + 5 y = 0

13. x 2 y′′ + 3 xy′ + 10 y = 0 15.

4 x 2 y′′

+ 8 xy′ + 5 y = 0

17. x 2 y′′ + 3 xy′ + y = 0 19.

x 2 y′′

+ xy′ = 0

12.

x 2 y′′

22. 16 x 2 y′′ − 8 xy′ + 9 y = 0 24. 4 x 2 y′′ − 16 xy′ + 25 y = 0

+ 7 xy′ + 13 y = 0

26. 6 x 2 y′′ + 7 xy′ − 2 y = 0, y(1) = 0, y′(1) = 1

14. x 2 y′′ − 5 xy′ + 10 y = 0 16.

4 x 2 y′′

27. x 2 y′′ − xy′ + y = 0, y(1) = 1, y′(1) = 1

− 4 xy′ + 5 y = 0

28. x 2 y′′ + 7 xy′ + 9 y = 0, y(1) = 1, y′(1) = 0

18. x 2 y′′ − 3 xy′ + 9 y = 0 20.

4 x 2 y′′

29. x 2 y′′ − xy′ + 2 y = 0, y(1) = −1, y′(1) = 1

+ y = 0

30. x 2 y′′ + 3 xy′ + 5 y = 0, y(1) = 1, y′(1) = 0

17.5 Power-Series Solutions In this section we extend our study of second-order linear homogeneous equations with variable coefficients. With the Euler equations in Section 17.4, the power of the variable x in the nonconstant coefficient had to match the order of the derivative with which it was paired: x 2 with y′′,  x 1 with y′, and x 0 ( =1) with y. Here we drop that requirement so we can solve more general equations.

Method of Solution The power-series method for solving a second-order homogeneous differential equation consists of finding the coefficients of a power series y (x) =



∞

∑ cn x n

= c 0 + c1 x + c 2 x 2 +  (1)

n=0

which solves the equation. To apply the method we substitute the series and its derivatives into the differential equation to determine the coefficients c 0 ,   c1 ,   c 2 , … . The technique for finding the coefficients is similar to that used in the method of undetermined coefficients presented in Section 17.2. In our first example we demonstrate the method in the setting of a simple equation whose general solution we already know. This is to help you become more comfortable with solutions expressed in series form. EXAMPLE 1   Solve the equation y′′ + y = 0 by the power-series method. Solution  We assume the series solution takes the form of

y =

∞

∑ cn x n n=0

and calculate the derivatives y′ =

∞

∑ nc n x n−1 n =1

M17_HASS5901_15_GE_C17.indd 24

and

y′′ =

∞

∑ n( n − 1)c n x n−2 . n=2

08/03/2023 16:53



17.5  Power-Series Solutions

17-25

Substitution of these forms into the second-order equation gives us ∞

∞

n=2

n=0

∑ n ( n − 1) c n x n − 2 + ∑ c n x n

= 0.

Next, we equate the coefficients of each power of x to zero as summarized in the following table. Power of x

Coefficient equation

x0

2 ( 1) c 2 + c 0 = 0

or

x1

3( 2 )c 3 + c1 = 0

or

x2

4 ( 3 )c 4 + c 2 = 0

or

x3

5( 4 ) c 5 + c 3 = 0

or

x4

6( 5 ) c 6 + c 4 = 0

or



 n ( n − 1) c n + c n − 2 = 0

x n−2

or

1 c2 = − c0 2 1 c3 = − c 3⋅2 1 1 c4 = − c 4 ⋅3 2 1 c c5 = − 5⋅4 3 1 c c6 = − 6⋅5 4  cn = −

1 c n ( n − 1) n − 2

From the table we notice that the coefficients with even indices ( n = 2k,   k = 1, 2, 3, …) are related to each other and the coefficients with odd indices ( n = 2 k + 1) are also interrelated. We treat each group in turn. Even indices: Here n = 2 k , so the power is x 2 k −2 . From the last line of the table, we have 2 k ( 2 k − 1)c 2 k + c 2 k −2 = 0 or c2k = −

1 . c 2 k ( 2 k − 1) 2 k −2

From this recursive relation we find     1 1 1   1 c2k = −  −   −  −  c0  2 k ( 2 k − 1)   ( 2 k − 2 )( 2 k − 3 )   4 ( 3 )   2  =

( − 1) k

( 2 k )!

c0.

Odd indices: Here n = 2 k + 1, so the power is x 2 k −1. Substituting this into the last line of the table yields ( 2k

+ 1)(2 k ) c 2 k +1 + c 2 k −1 = 0

or c 2 k +1 = −

( 2k

1 . c + 1)(2 k ) 2 k −1

Thus,     1 1 1  1  c 2 k +1 =  −  −   −  −  c1  ( 2 k + 1)(2 k )   ( 2 k − 1)( 2 k − 2 )   5( 4 )   3( 2 )  =

M17_HASS5901_15_GE_C17.indd 25

( − 1) k

( 2k

+ 1)!

c1 .

08/03/2023 16:55

17-26

Chapter 17  Second-Order Differential Equations

Writing the power series by grouping its even and odd powers together and substituting for the coefficients yields y=

∞

∑ cn x n n=0

=

∞

∑ c2k x 2k

+

k=0

= c0

∞

∑

k=0

( − 1)

∞

∑ c 2 k +1 x 2 k +1 k=0

k

(2 k )!

∞

x 2 k + c1 ∑

k=0

( − 1) k

+ 1)!

( 2k

x 2 k +1 .

From our study of Taylor series, we see that the first series on the right-hand side of the last equation represents the cosine function, and the second series represents the sine. Thus, the general solution to y′′ + y = 0 is y = c 0 cos x + c1 sin x. EXAMPLE 2   Find the general solution to y′′ + xy′ + y = 0. Solution  We assume the series solution form

y =

∞

∑ cn x n n=0

and calculate the derivatives ∞

∑ nc n x n−1

y′ =

y′′ =

and

n =1

∞

∑ n(n − 1) c n x n−2 . n=2

Substitution of these forms into the second-order equation yields ∞

∞

∞

n=2

n =1

n=0

∑ n( n − 1)c n x n−2 + ∑ nc n x n + ∑ c n x n

= 0.

We equate the coefficients of each power of x to zero as summarized in the following table. Power of x

Coefficient equation 2(1)c 2 + c 0 = 0

x0

or

x1

3( 2 )c 3 +

c1 + c1 = 0

or

x2

4 ( 3 )c 4 + 2c 2 + c 2 = 0

or

x3

5( 4 )c 5 + 3c 3 + c 3 = 0

or

x4

6( 5 ) c 6 + 4c 4 + c 4 = 0

or

 xn

 (n

+ 2 )( n + 1)c n + 2 + ( n + 1)c n = 0

or

1 c2 = − c0 2 1 c 3 = − c1 3 1 c4 = − c2 4 1 c5 = − c3 5 1 c6 = − c4 6  c n+2 = −

1 c n+2 n

From the table notice that the coefficients with even indices are interrelated and the coefficients with odd indices are also interrelated. Even indices: Here n = 2 k − 2, so the power is x 2 k −2 . From the last line in the table, we have 1 c 2 k = − c 2 k −2 . 2k

M17_HASS5901_15_GE_C17.indd 26

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17.5  Power-Series Solutions

17-27

From this recurrence relation we obtain

(

)(

) ( )( )( )

1 1 1 1 1 −  − − − c0 2k 2k − 2 6 4 2 ( − 1) k = c0 . ( 2 )( 4 )( 6 )  (2k )

c 2k = −

Odd indices: Here n = 2 k − 1, so the power is x 2 k −1. From the last line in the table, we have c 2 k +1 = −

1 c . 2 k + 1 2 k −1

From this recurrence relation we obtain

(

c 2 k +1 = − =

)(

) ( )( )

1 1 1 1 −  − − c1 2k + 1 2k − 1 5 3 ( − 1) k

( 3 )( 5 )  ( 2k

+ 1)

c1 .

Writing the power series by grouping its even and odd powers and substituting for the coefficients yields y=

∞

∑ c 2k x 2k

+

k=0

= c0

∞

∞

∑ c 2k +1x 2k +1 k=0

)k

∑ ( 2 )( 4−)1 (2k ) x 2k (

k=0

∞

+ c1 ∑

k =0

( − 1) k

( 3 )( 5 )  ( 2k

+ 1)

x 2 k +1 .

EXAMPLE 3   Find the general solution to (1 − x 2 ) y′′ − 6 xy′ − 4 y = 0,

x < 1.

Solution  Notice that the leading coefficient is zero when x = ±1. Thus, we assume the solution interval I : −1 < x < 1. Substitution of the series form

y =

∞

∑ cn x n n=0

and its derivatives gives us

(1 − x 2 ) ∞

∞

∞

∞

n=2

n =1

n=0

∑ n( n − 1)c n x n−2 − 6 ∑ nc n x n − 4 ∑ c n x n

∑ n ( n − 1) c n x n − 2 − n=2

M17_HASS5901_15_GE_C17.indd 27

∞

∞

∞

n=2

n =1

n=0

= 0,

∑ n( n − 1)c n x n − 6 ∑ nc n x n − 4 ∑ c n x n = 0.

08/03/2023 16:56

17-28

Chapter 17  Second-Order Differential Equations

Next, we equate the coefficients of each power of x to zero as summarized in the following table. Power of x

Coefficient equation

x0

2(1)c 2

− 4c 0 = 0

or

c2 =

x1

3( 2 )c 3

− 6(1)c1 − 4c1 = 0

or

c 3 = 53 c1

x2

4 ( 3 ) c 4 − 2 ( 1) c 2 − 6 ( 2 ) c 2 − 4 c 2 = 0

or

c4 =

x3

5( 4 )c 5 − 3( 2 )c 3 − 6( 3 )c 3 − 4c 3 = 0

or

c 5 = 75 c 3

 xn



4 2 c0

6 4 c2



( n + 2 )( n + 1)c n + 2 − [ n ( n − 1) + 6n + 4 ]c n = 0 ( n + 2 )( n + 1)c n + 2 − ( n + 4 )( n + 1)c n = 0

or

c n+2 =

n+4 c n+2 n

Again we notice that the coefficients with even indices are interrelated and those with odd indices are interrelated. Even indices: Here n = 2 k − 2, so the power is x 2 k . From the right-hand column and the last line of the table, we get 2k + 2 c 2 k −2 2k 2k + 2 2k 2k − 2 6 4 = c  2k 2k − 2 2k − 4 4 2 0 = ( k + 1) c 0 .

c 2k =

(

)(

)(

)

( )

Odd indices: Here n = 2 k − 1, so the power is x 2 k +1. The right-hand column and the last line of the table give us 2k + 3 c 2k + 1 2 k −1 2k + 3 2k + 1 2k − 1 7 5 = c  2k + 1 2k − 1 2k − 3 5 3 1 2k + 3 = c1 . 3

c 2 k +1 =

(

)(

)(

)

( )

The general solution is y=

∞

∑ cn x n n=0

=

∞

∑ c2k x 2k k=0

= c0

+

∞

∑ c 2 k +1 x 2 k +1 k=0

∞

∑ ( k + 1) x 2 k k=0

∞

2 k + 3 2 k +1 x . 3 k=0

+ c1 ∑

EXAMPLE 4   Find the general solution to y′′ − 2 xy′ + y = 0. Solution  Assuming that

y =

∞

∑ cn x n , n=0

M17_HASS5901_15_GE_C17.indd 28

08/03/2023 16:56



17.5  Power-Series Solutions

17-29

substitution into the differential equation gives us ∞

∞

∞

n=2

n =1

n=0

∑ n( n − 1)c n x n−2 − 2 ∑ nc n x n + ∑ c n x n

= 0.

We next determine the coefficients, listing them in the following table. Power of x

Coefficient equation

x0

2 ( 1) c 2

x1

3( 2 ) c 3 − 2c1

x2

4 ( 3 ) c 4 − 4c 2

x3

5( 4 )c 5 − 6c 3

x4

6( 5 )c 6 − 8c 4

 xn

(n

1 + c 0 = 0 or c 2 = − c 0 2 1 + c 1 = 0 or c 3 = c 3⋅2 1 3 + c 2 = 0 or c 4 = c 4⋅3 2 5 + c 3 = 0 or c 5 = c 5⋅4 3 7 + c 4 = 0 or c 6 = c 6⋅5 4  

+ 2 )( n + 1)c n + 2 − ( 2n − 1)c n = 0 or c n + 2 =

(n

2n − 1 c + 2 )( n + 1) n

From the recursive relation c n+2 =

(n

2n − 1 c , + 2 )( n + 1) n

we write out the first few terms of each series for the general solution:

(

)

1 2 3 21 6 x − x4 − x − 2 4! 6! 1 5 45 7 + c1 x + x 3 + x 5 + x + . 3! 5! 7!

y = c0 1 −

(

EXERCISES 

17.5

In Exercises 1–18, use power series to find the general solution of the differential equation.

9. ( x 2 − 1) y′′ + 2 xy′ − 2 y = 0

1. y ′′ + 2 y ′ = 0

11. ( x 2 − 1) y′′ − 6 y = 0

2. y ′′ + 2 y ′ + y = 0 3. y ′′ + 4 y = 0 4. y ′′ − 3 y ′ + 2 y = 0 5. x 2 y ′′ − 2 xy ′ + 2 y = 0 6. y′′ − xy′ + y = 0 7. (1 + x ) y′′ − y = 0 8. (1 − x 2 ) y′′ − 4 xy′ + 6 y = 0

M17_HASS5901_15_GE_C17.indd 29

)

10. y′′ + y′ − x 2 y = 0 12. xy′′ − ( x + 2 ) y′ + 2 y = 0 13. ( x 2 − 1) y′′ + 4 xy′ + 2 y = 0 14. y′′ − 2 xy′ + 4 y = 0 15. y′′ − 2 xy′ + 3 y = 0 16. (1 − x 2 ) y′′ − xy′ + 4 y = 0 17. y′′ − xy′ + 3 y = 0 18. x 2 y′′ − 4 xy′ + 6 y = 0

08/03/2023 16:56

ANSWERS TO ODD-NUMBERED EXERCISES

Chapter 17 SECTION 17.1, pp. 17-6–17-7

1. y 5. y 9. y 13. y 17. y 19. y 21. y 25. y 29. y 33. y 35. y 37. y 41. y

= c1e −3 x + c 2e 4 x   3. y = c1e −4 x + c 2e x = c1e −2 x + c 2e 2 x   7.  y = c1e −x + c 2e 3 x 2 = c1e −x 4 + c 2e 3 x 2   11. y = c1 cos 3 x + c 2 sin 3 x = c1 cos 5 x + c 2 sin 5 x  15.  y = e x ( c1 cos 2 x + c 2 sin 2 x ) − x = e ( c1 cos 3x + c 2 sin 3x ) = e −2 x ( c1 cos 5 x + c 2 sin 5 x ) = c1 + c 2 x   23. y = c1e −2 x + c 2 xe −2 x = c1e −3 x + c 2 xe −3 x   27. y = c1e −x 2 + c 2 xe −x 2 3 3 = c1e −x 3 + c 2 xe −x 3  31. y = − e −5 x + e −x 4 4 1 = sin 2 3x 2 3 1 = − cos 2 2 x + sin 2 2 x 2 = ( 1 − 2 x )e 2 x  39. y = 2( 1 + 2 x )e −3 x 2 = c1e −x + c 2e 3 x   43. y = c1e −x 2 + c 2 xe −x 2

35. y = 37. y = 39. y = 41. y = 43. y = 45. y = =

49. 51. 53. 55. 57.

49. y = e −x 2 ( c1 cos x + c 2 sin x )  51. y = c1e 3 x 4 + c 2 xe 3 x 4

59.

53. y = c1e

57. y = ( 1 +

+ c 2 xe

−4 x 3

−x 2

  55. y = c1e + c 2e 15 −7 x 3 11 2 x = e + e 13 13

2 x )e −x   59. y

SECTION 17.2, pp. 17-14–17-15

1. y = c1e 5 x + c 2e −2 x +

1 1 cos x − sin x 2 2 1 c1 cos x + c 2 sin x − cos 3 x 8 c1e 2 x + c 2e −x − 6 cos x − 2 sin x 1 c1e x + c 2e −x − x 2 − 2 + xe x 2 1 −x 49 7 3 x − 2 x c1e + c 2e − e + cos x + sin x 4 50 50 3 6 c1 + c 2 e − 5 x + x 3 + x 2 − x 5 25 4 1 c1 + c 2e 3 x + 2 x 2 + x + xe 3 x 3 3 1 2 − x c1 + c 2 e + x − x 2 1 c1 cos x + c 2 sin x − x cos x 2 1 − x 2 ( c1 + c 2 x ) e + x e −x 2 1 c1e x + c 2e −x + xe x 2 e −2 x ( c1 cos x + c 2 sin x ) + 2 A cos x + B sin x + x sin x + cos x ln ( cos x ) 1 2 5x 1 c1 + c 2 e 5 x + x e − xe 5 x 10 25 1 c1 cos x + c 2 sin x − x cos x + x sin x 2

3. y = c1 + c 2e x + 5. y = 7. y = 9. y = 11. y = 13. y = 15. y = 17. y = 19. y = 21. y = 23. y = 25. y = 27. y = 29. y = 31. y =

3 10

1 x e 2 y = ce 3 x + 5 xe 3 x y = 2 cos x + sin x − 1 + sin x ln ( sec x + tan x ) 1 y = −e −x + 1 + x 2 − x 2 y = 2( e x − e −x ) cos x − 3e −x sin x y = (1 − x + x 2 )e x 1 yp = x 2 4

47. y = ce 3 x −

45. y = c1 cos 5 x + c 2 sin 5 x   47. y = c1e −x 5 + c 2 xe −x 5 −4 x 3

1 −x e + xe x 2 1 4 c1e 5 x + c 2e −x − e x − 8 5 c1 cos x + c 2 sin x − ( sin x )[ ln ( csc x + cot x ) ] 1 c1 + c 2e 8 x + xe 8 x 8 c1 + c 2 e x − x 4 4 − x 3 − 3 x 2 − 6 x 1 c1 + c 2 e − 2 x − e x + x 3 6 − x 2 4 + x 4 3 c1 cos x + c 2 sin x + ( x − tan x ) cos x − sin x ln ( cos x ) c1 cos x + c 2 ′ sin x + x cos x − ( sin x ) ln ( cos x )

33. y = c1 + c 2e x +

4x 3

SeCtion 17.3, pp. 17-20–17-21

1 y″ + y′ + y = 0, y(0) = 0.6, y′(0) = 0.6 2 120 3. y″ + 600y = 0, y(0) = 0.05, y′(0) = y0 9.8 5. 2q″ + 4q′ + 10q = 20 cos t, q(0) = 2, q′(0) = 3

1.

7. 0.0259 m (above equilibrium) y0 9. y(t) = 0.05 cos (5.715t) + sin (5.715t) 5.715 (in meters) 11. 1.806 s 13. 45 N 15. 8.13 m > s 17. 0.6238 m > s2 (acceleration upward) 19. q(t) = - 8e-3t + 10e-2t, lim q(t) = 0 tSq

21. y(t) = 0.3 + 0.6e-t - 0.1e-2t - 0.2e-8t 23. y(p) = - 2 m (above equilibrium)

25. q(t) =

49 1199 1199 1199 49 1 cos t + t ≤e-t>2 + ¢ sin 2 2 5 5 995

SECTION 17.4, p. 17-24

c1 c + c 2 x   3. y = 12 + c 2 x 3 x2 x 5. y = c1 x 2 + c 2 x 4   7. y = c1 x −1 3 + c 2 1. y =

9. y = x ( c1 + c 2 ln x ) 11. y = x [ c1 cos( 2 ln x ) + c 2 sin ( 2 ln x ) ] 1 13. y = [ c1 cos( 3 ln x ) + c 2 sin ( 3 ln x ) ] x 1 15. y = [ c cos( ln x ) + c 2 sin ( ln x ) ] x 1

17-30

M17B_HASS5582_15_SE_ANS.indd 30

22/02/22 7:57 PM



17-31

Chapter 17: Answers to Odd-Numbered Exercises

1 ( c + c 2 ln x )  19. y = c1 + c 2 ln x x 1 1 21. y = 3 ( c1 + c 2 ln x )  23. y = x −5 4 ( c1 + c 2 ln x ) x 1 x 25. y = +   27. y = x 2x 3 2 29. y = x [ − cos( ln x ) + 2 sin ( ln x ) ]

5. y = c1 x + c 2 x 2

SECTION 17.5, p. 17-29

13. y = c 0 1 + x 2 +

17. y =

(

)

c 2 1. y = c 0 + c1 x − + x 3 −  = c 0 − 1 e −2 x 3 2 2 2 3 3. y = c 0 ( 1 − 2 x + ) + c1 x − x +  3 = c 0 cos 2 x + c1 sin 2 x x2

(

M17B_HASS5582_15_SE_ANS.indd 31

)

( 12 x − 16 x 5 9. y = c ( 1 − x + x 12 7. y = c 0 1 + 0

2

2

3

4

) ( − ) + c x

+  + c1 x +

1 3 x + 6

)

1

11. y = c 0 ( 1 − 3 x 2 + ) + c1 ( x − x 3 )

( 3 15. y = c ( 1 − x 2 3 17. y = c ( 1 − x 2 0

0

2

2

) (

(

2 4 3 x +  + c1 x + x 3 + x 5 +  3 5 1 3 +  + c1 x − x +  2 1 4 1 + x +  + c1 x − x 3 8 3

)

)

)

)

(

)

22/02/22 7:57 PM

18

Complex Functions

INTRODUCTION  The study of derivatives, integrals, and series can be extended from the real to the complex numbers, where it is called complex analysis. The study of complex functions is far from a straightforward extension of the study of real variable functions. Differentiable functions of a complex variable have many properties not found in their real relatives. For example, while we can find real functions that are once but not twice differentiable, this is not true for complex functions. If a complex function is differentiable once on an open region, then its second, third, and all higher order derivatives also exist. In Chapter 9 we saw differentiable functions that are not equal to any power series. In contrast, a complex differentiable function can always be expressed as a convergent power series. The study of complex functions gives profound new insights into the properties of the real-valued functions we have already encountered and opens the way to many new applications. The study of complex functions, particularly in higher dimensions, remains an active area of mathematical research. In this chapter we introduce the complex numbers and then study some of the many beautiful aspects of complex differentiable functions.

18.1 Complex Numbers Complex numbers are expressed in the form a + ib, or a + bi, where a and b are real numbers and i is a symbol for a new type of number whose square equals −1; that is, i 2 = −1. We refer to a as the real part of the complex number and to b as the imaginary part. The words “real” and “imaginary” have connotations that tend to place −1 in a less favorable position in our minds than a real number such as 2. In fact, a good deal of imagination, or inventiveness, was required to construct the real number system, which forms the basis of calculus (see Appendix A.7). We begin this chapter with a review of the successive stages of invented number systems.

The Hierarchy of Numbers The first stage of number development was the recognition of the counting numbers 1, 2, 3, … , which we now call the natural numbers or the positive integers. Certain arithmetical operations on the positive integers, such as addition and multiplication, keep us entirely within this system. That is, if m and n are any positive integers, then their sum m + n and product mn are also positive integers. Some equations can be solved entirely within the system of positive integers. For example, we can solve 3 + x = 7 using only positive integers. But other simple equations, such as 7 + x = 3, cannot be solved if only positive integers are at our disposal. The invention of the number zero and the negative numbers allowed the solution of equations such as 7 + x = 3. Using the integers … , −3, −2, −1, 0, 1, 2, 3, …

18-1

M18_HASS5901_15_GE_C18.indd 1

23/03/23 10:10 AM

18-2

Chapter 18  Complex Functions

"2

1

1

FIGURE 18.1  The diagonal of the unit square has irrational length.

we can always find the missing integer x that solves the equation m + x = p when we are given the other two integers m and p in the equation. Addition and multiplication of integers always keep us within the system of integers. However, division does not, and so fractions m n , where m and n are integers with n nonzero, were invented. The resulting system is called the rational numbers and is rich enough to perform all of the rational operations of arithmetic, including addition, subtraction, multiplication, and division (although division by zero is excluded since it is meaningless). Yet there are still simple polynomial equations that cannot be solved within the system of rational numbers. The ancient Greeks realized that there is no rational number that solves the equation x 2 = 2, even though the Pythagorean Theorem implies that the length x of the diagonal of the unit square satisfies this equation. (See Figure 18.1.) To see why x 2 = 2 has no rational solution, consider the following argument. Suppose that there did exist some positive integers p and q with no common factor other than 1 such that the fraction x = p q satisfies x 2 = 2. Writing this out, we see that p 2 q 2 = 2, and therefore p 2 = 2q 2 . Then p 2 is also an even integer. Since the square of an odd number is odd, we conclude that p itself must be an even number (for if p were odd, then p 2 would also be odd). So p is divisible by 2, and p = 2 k for some integer k. Hence p 2 = 4 k 2 . Since we already saw that p 2 = 2q 2 , it follows that 2q 2 = p 2 = 4 k 2 , and therefore q 2 = 2 k 2 . Consequently, q2 is an even number and this requires in turn that q itself be even. Therefore, both p and q are divisible by 2, which is contrary to our assumption that they contain no common factors other than 1. Since we have arrived at a contradiction, there cannot exist any such integers p and q, and therefore there is no rational number that solves the equation x 2 = 2. The invention of real numbers addressed this issue (and others). Using real numbers, we can represent every possible physical length. Each real number can be represented as an infinite decimal a.d1d 2 d 3d 4 … , where a is an integer followed by a sequence of decimal digits each between 0 and 9 (Appendix A.7). If the sequence stops or repeats in a periodic pattern, then the decimal represents a rational number. An irrational number is represented by a nonterminating and nonrepeating decimal. The rational and irrational numbers together make up the real number system. Unlike the rational numbers, the real numbers have the completeness property, meaning that there are no “holes” or “gaps” in the real line. This property is essential in calculus, implying for example that continuous functions on a closed interval have an absolute maximum value and an absolute minimum value (Extreme Value Theorem, Section 4.1), and that an increasing and bounded sequence converges (Monotonic Sequence Theorem, Section 9.1). Yet for all of its utility, there are simple equations that cannot be solved when working only within the real number system. As an example, the polynomial equation x 2 + 1 = 0 has no real solutions.

The Complex Numbers We have discussed three systems of numbers that form a hierarchy, with each system containing the previous system. Each system is richer than its predecessor in that it permits additional operations to be performed without going outside the system. 1. Using the integer system, we can solve all equations of the form

x + a = 0,(1)

where a is an integer. 2. Using the rational numbers, we can solve all equations of the form

ax + b = 0,(2) provided that a and b are rational numbers and a ≠ 0.

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18.1  Complex Numbers

18-3

3. Using the real numbers, we can solve all of Equations (1) and (2) and, in addition, all quadratic equations ax 2 + bx + c = 0,(3)



provided that a ≠ 0 and b 2 − 4 ac ≥ 0. The quadratic formula x =



−b ±

b 2 − 4 ac (4) 2a

gives the solutions to Equation (3). When b 2 − 4 ac is negative, there are no real number solutions to the equation ax 2 + bx + c = 0. In particular, the simple quadratic equation x 2 + 1 = 0 cannot be solved using any of the three invented systems of numbers that we have discussed. Thus we are led to the fourth system of numbers, the set of complex numbers a + ib. The symbol i represents a new number whose square equals −1. We call a the real part and b the imaginary part of the complex number a + ib. Real and Imaginary parts

For a complex number z = a + ib, we write Re z = a and Im z = b. Sometimes it is convenient to write a + bi instead of a + ib; both notations describe the same complex number. We define equality and addition for complex numbers in the following way. Equality  a + ib = c + id   if and only if   a = c and b = d .

Two complex numbers a + ib and c + id are equal if and only if their real parts are equal and their imaginary parts are equal. Addition  ( a + ib ) + ( c + id ) = ( a + c ) + i ( b + d )

We sum the real parts and separately sum the imaginary parts. To multiply two complex numbers, we multiply using the distributive rule and then simplify using i 2 = −1. Multiplication  ( a + ib )( c + id )

= ac + iad + ibc + i 2 bd = ( ac − bd ) + i ( ad + bc )   i 2

= −1

EXAMPLE 1   Basic arithmetic operations with complex numbers a. The set of all complex numbers a + i0, where the second number b is zero, has all of the properties of the set of real numbers. For example, addition and multiplication as complex numbers give (a

+ i 0 ) + ( c + i 0 ) = ( a + c ) + i0,

(a

+ i 0 )( c + i0 ) = ac + i 0,

which are numbers of the same type with imaginary part zero. We usually just write a instead of a + i0, and in this sense the complex number system contains the real number system. b. If we multiply a real number a = a + i0 by a complex number c + id, we get a ( c + id ) = ( a + i0 )( c + id ) = ac + iad.

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18-4

Chapter 18  Complex Functions

c. The number 0 = 0 + i 0 plays the role of zero in the complex number system, and the complex number 1 = 1 + i 0 plays the role of unity, or one. d. The complex number i = 0 + i1, which has real part zero and imaginary part one, has the property that its square is i 2 = ( 0 + i1) 2 = ( 0 + i1)( 0 + i1) = ( − 1) + i 0 = − 1. So x = i is a solution to the quadratic equation x 2 + 1 = 0. In the complex number system there are exactly two solutions to this equation, the other solution being x = −i = 0 + i (−1). e. We can divide any two complex numbers, as long as we do not divide by the number 0 = 0 + i 0. When a + ib ≠ 0 (meaning that either a ≠ 0 or b ≠ 0 or both a ≠ 0 and b ≠ 0), we carry out division as follows: ( c + id )( a − ib ) ( ac + bd ) + i ( ad − bc ) c + id ac + bd ad − bc . = = = 2 +i 2 ( a + ib )( a − ib ) a + ib a2 + b2 a + b2 a + b2

In this way we express the quotient in the standard form x + iy, where x = ( ac + bd )( a 2 + b 2 ) and y = ( ad − bc )( a 2 + b 2 ) are real numbers. Note that a 2 + b 2 ≠ 0 since we stipulated that a and b cannot both be zero. f. The number a − ib that was used in part (e) as the multiplier to clear the i from the denominator is called the complex conjugate of a + ib. If we denote the original complex number by z = a + ib, then it is customary to write z (read “z bar”) to denote its complex conjugate: z = a + ib,

z = a − ib.

Multiplying the numerator and denominator of a fraction ( c + id ) ( a + ib ) by the complex conjugate of the denominator will always replace the denominator by a real number. EXAMPLE 2   Computations with complex numbers a. ( 2 + 3i ) + ( 6 − 2i ) = ( 2 + 6 ) + ( 3 − 2 ) i = 8 + i b. ( 2 + 3i ) − ( 6 − 2i ) = ( 2 − 6 ) + ( 3 − ( −2 )) i = −4 + 5i c. ( 2 + 3i )( 6 − 2i ) = ( 2 )( 6 ) + ( 2 )(−2i) + (3i)( 6 ) + (3i)(−2i)         = 12 − 4i + 18i − 6i 2 = 12 + 14i + 6 = 18 − 14i ( 2 + 3i )( 6 + 2i ) 2 + 3i 12 + 4i + 18i + 6i 2 6 + 22i 3 11 d. = = = = + i ( 6 − 2i )( 6 + 2i ) 6 − 2i 36 + 12i − 12i − 4i 2 40 20 20

The Complex Plane We can use the usual xy-coordinate system to represent complex numbers as points or vectors in the plane. In this context we call it the complex plane. There are now two geometric representations of a complex number z = x + iy: 1. as a point P ( x , y ) in the xy-plane,  2. as a vector OP from the origin to P.

y P(x, y)

r

In each representation, the x-axis is called the real axis and the y-axis is the imaginary axis, as in Figure 18.2. Geometric representations of complex numbers are sometimes called Argand Diagrams. In polar coordinates we have

y

u O

x

x = r cos θ,

x

y = r sin θ,

and FIGURE 18.2  The complex number

z = x + iy can be represented both as a point P ( x , y ) and as a vector.

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z = x + iy = r ( cos θ + i sin θ ).(5)

The number 0, or 0 + i 0 is a special case, with r = 0 but no specified θ.

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18.1  Complex Numbers

18-5

We define the absolute value (or magnitude) of a complex number x + iy to be the  length r of a vector OP from the origin to P ( x , y ). We denote the absolute value by vertical bars; thus, x + iy =

x 2 + y2.

We always choose the polar coordinates r and θ so that r is nonnegative, and then r = x + iy . The polar angle θ is called the argument of z and is written θ = arg z. There is more than one choice for the argument. Any integer multiple of 2π may be added to θ to produce another appropriate angle. The following equation gives a useful formula connecting a complex number z, its conjugate z , and its absolute value z : z ⋅ z = z 2.

Euler’s Formula The identity e iθ = cos θ + i sin θ (6)



is called Euler’s Formula. The origins of this formula will be seen in Section 18.6. We show a vector representing e iθ in Figure 18.3. Using Equation (6), we can write Equation (5) as z = re iθ. y

y

eiu = cos u + i sin u

r=1 u = arg z O

eiu = cos u + i sin u (cos u, sin u)

x

(a)

u O

x

(b)

FIGURE 18.3  The complex number z = e iθ as a vector and as a point.

The notation exp( A) is also used for e A, so we could write z = r exp(iθ). Euler’s formula leads to the following rules for calculating products, quotients, powers, and roots of complex numbers in polar form.

Products To multiply two complex numbers in polar form, we multiply their absolute values and add their arguments. To see why, let

z 1 = r1e iθ1 and

z 2 = r2e iθ2 ,(7)

so that z 1 = r1 , arg z 1 = θ1;

z 2 = r2 , arg z 2 = θ 2 .

In the Cartesian a + bi form we have z 1 = r1 cos θ1 + ir1 sin θ1 and z 2 = r2 cos θ 2 + ir2 sin θ 2 .

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18-6

Chapter 18  Complex Functions

Their product is then z 1z 2 = r1e iθ1 r2e iθ2 = r1r2 ( cos θ1 + i sin θ1 )( cos θ 2 + i sin θ 2 ) = r1r2 ([ cos θ1 cos θ 2 − sin θ1 sin θ 2 ] + i[ cos θ1  sin θ 2 + sin θ1  cos θ 2 ])

y z 1z 2

= r1r2 ( cos ( θ1 + θ 2 ) + i sin ( θ1 + θ 2 ))

u1

= r1r2e

r2

r1

u2

z1



u1

FIGURE 18.4 When z 1 and z 2 are multiplied, z 1z 2 = r1 ⋅ r2 and arg( z 1z 2 ) = θ1 + θ 2 .

y z1 = 1 + i "2

0

−p 6 2

−1

z1 ⋅ z 2 =

"3 − 1

p 12

1 + "3

r1e iθ1   r2e iθ2 = r1r2e i( θ1 + θ2 ).(8)

EXAMPLE 3   We compute the product of z 1 = 1 + i and z 2 = 3 − i. We first plot these complex numbers in Figure 18.5, and we can then read off their polar representations

z 1z 2

2"2

p 4

.

The absolute value of the product is r1r2 and the argument of the product is the sum of the arguments of z 1 and z 2 . We have shown that the product of two complex numbers is represented by a vector whose length is the product of the lengths of the two factors and whose angle with the x-axis is the sum of the corresponding angles of the factors. (Figure 18.4.) In particular, Equation (8) implies that multiplying a vector by e iθ causes its vector representation to be rotated counterclockwise through an angle θ. Multiplication by i rotates 90 °, by −1 rotates 180 °, by −i rotates 270°, and so on.

x

O

1

Angle sum formulas

Therefore

z2

r1r 2

i( θ1 + θ 2 )

=2

x

1

=2

z 2 = "3 − i

( i4π ) ⋅ 2 exp(− i6π ) iπ iπ 2 exp( − ) 4 6 iπ 2 exp( ) 12 π π 2 ( cos + i sin ) 12 12

2 exp

=2

≈ 2.73 + 0.73i.

FIGURE 18.5  To multiply two complex numbers, multiply their absolute values and add their arguments.



EXAMPLE 4  Computing Reciprocals Show that if z = re iθ and r > 0, then 1 1 = e −iθ . z r Solution  Applying the formula for products, Equation (8), we see that

( 1r e ) = 1 ⋅ e

( re iθ )

−iθ

i0

= 1.

Dividing both sides by z = re iθ gives 1 −iθ 1 1 e = iθ = . r re z

Quotients We now look at how to divide z 1 = r1e iθ1 by z 2 = r2e iθ2 in polar form. We assume that r2 > 0 so that we are not dividing by zero. To compute the quotient z 1 z 2 we multiply z 1 by 1 z 2 . This gives 1 1  z1 r = z 1   = r1e iθ1  e −iθ2  = 1 e i( θ1 −θ2 ) ,  Equation  ( 8 )  z2   r2  z2 r2

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18.1  Complex Numbers

18-7

so that z1 z = 1 z2 z2 and z  arg 1  = θ1 − θ 2 = arg z 1 − arg z 2 .  z2  That is, we divide lengths and subtract angles to compute the quotient of complex numbers. EXAMPLE 5  Let z 1 = 1 + i and z 2 =

3 − i, as in Example 3. Then

(

)

2e iπ 4 2 5 πi 12 1+i 5π 5π = = e ≈ 0.707 cos + i   sin ≈ 0.183 + 0.683i. −iπ 6 2e 2 12 12 3−i

Powers If n is a positive integer, we can apply the product formulas in Equation (8) to find the polar form of a power z n = z ⋅ z ⋅  ⋅ z.   n factors With z = re iθ , we obtain z n = (re iθ ) n = r n e i( θ +  + θ )  = r n e inθ. n summands



(9)

The length r = z is raised to the nth power, and the angle θ = arg z is multiplied by n. If we take r = 1 in Equation (9), we obtain De Moivre’s Theorem.

De Moivre’s Theorem



( cos θ + i sin θ ) n = cos nθ + i sin nθ (10)

If we expand the left side of the equation in De Moivre’s Theorem using the Binomial Theorem and reduce it to the form a + ib, we obtain trigonometric identities, namely formulas for cos nθ and sin nθ as polynomials of degree n in cos θ and sin θ. EXAMPLE 6  If n = 3 in Equation (10), we have

( cos θ + i sin θ )3 = cos(3θ) + i sin (3θ). The left side of this equation expands to cos 3 θ + 3i cos 2 θ sin θ − 3 cos θ sin 2 θ − i sin 3 θ. The real part of this must equal cos(3θ) and the imaginary part must equal sin (3θ). Therefore, cos(3θ) = cos 3 θ − 3 cos θ sin 2 θ, sin(3θ) = 3 cos 2 θ sin θ − sin 3 θ. 

Roots If x is a positive real number, then it has exactly one real nth root if n is an odd integer and exactly two nth roots if n is even. If x is a negative real number, then x has exactly one real nth root if n is an odd integer and no real nth root if n is even. In contrast, if z = re iθ is a

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18-8

Chapter 18  Complex Functions

complex number different from zero, then there are precisely n different complex numbers, w 0 , w1 , … , w n −1 , that are nth roots of z. To see why, let w = ρe iα be an nth root of z = re iθ. Then wn = z or ρ n e inα = re iθ. Since both r and ρ are positive real numbers, this implies that ρ n = r, and so ρ =

n

r

is the real, positive nth root of r. Although we cannot say that nα and θ must be equal, since the argument of a complex number is only defined up to multiples of 2π, we can say that they may differ only by an integer multiple of 2π. That is, nα = θ + 2 k π , k = 0, ±1, ± 2, … . Therefore, α =

θ 2π +k . n n

This gives the following collection of n distinct complex numbers,



n

re iθ =

n

(

θ 2π r exp  i + k  n n

) ,

k = 0, ±1, ± 2, … .(11)

Each of these has the property that its nth power equals z, so these n complex numbers give all of the nth roots of z = re iθ. The expression n re iθ on the left of Equation (11) defines a new concept, the nth root of a complex number, while the term n r on the right refers to the real nth root. There is exactly one positive real number whose nth power equals the positive real number r, and that number gives the value of n r. On the other hand, the expression n re iθ refers to a collection of n distinct numbers. In practice we often want to pick out a particular nth root, the principal nth root. We can do that by requiring that z = re iθ have −π < θ ≤ π, and then taking the principal nth root to be n

( ) .

θ r exp  i  n

There might appear to be infinitely many different answers in Equation (11), corresponding to the infinitely many possible values of k, but k = m + n gives the same complex number as k = m. This is true because the complex number

(

z = reiu

w0

u O

2p 3

u 3

x

2p 3 w2

FIGURE 18.6  The three cube roots of

z = re iθ

M18_HASS5901_15_GE_C18.indd 8

(

)

)

θ m exp  i + 2π  ,  n n 

r r 13

(

and the complex number

2p 3

w1

)

θ m+n  m  θ  exp  i + 2π  = exp  i n + 2π n + 2π   n n

y

are equal, since they have arguments that differ by 2π. Therefore we need only take n consecutive values for k to obtain all the different complex numbers that are nth roots of z. For convenience, we take k = 0, 1, 2, … , n − 1. All the nth roots of re iθ lie on a circle centered at the origin and having radius equal to the real, positive nth root of r. One of them has argument w 0 = θ n. The others are uniformly spaced around the circle, each being separated from its neighbors by an angle equal to 2π n . (Figure 18.6 illustrates the placement of the three cube roots, w 0 , w1 , w 2 , of the complex number z = re iθ.)

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18-9

18.1  Complex Numbers

y

p 2

w1 p 2

EXAMPLE 7   Find the four fourth roots of −16.

2

w0 p 4

−16

x p 2

w2

Solution  As our first step, we plot the number −16 in the complex plane (Figure 18.7) and determine its polar representation re iθ. Here, z = −16, r = +16, and θ = π. One of the fourth roots of 16e iπ is 2e iπ 4. We obtain others by successive additions of 2π 4 = π 2 to the argument of this first one. Hence,

and the four roots are

16e iπ = 2e ikπ 4 ,

k = 1, 3, 5, 7,(12)

π π w 0 = 2  cos + i sin  =  4 4

w3

p 2

4

3π 3π w1 = 2  cos + i sin  =  4 4 

FIGURE 18.7  The four fourth roots

5π 5π w 2 = 2  cos + i sin  =  4 4  π π 7 7 w 3 = 2  cos + i sin  =  4 4 

of −16

2 (1 + i ) , 2 ( −1 + i ), 2 ( −1 − i ), 2 (1 − i ).



The Fundamental Theorem of Algebra One might say that the invention of 2 is all well and good and leads to a number system that is richer than the rational number system alone, and the complex number system is richer still, but where will this process end? Will we need to develop still more systems so as to obtain 4 −1 (a solution of x 4 = −1), 6 −1 (a solution of x 6 = −1), and so on? It turns out this is not necessary. These numbers are already expressible in the complex number system as a + ib for appropriate a and b. In fact, the Fundamental Theorem of Algebra says that with the introduction of the complex numbers, we now have enough numbers to factor every polynomial into a product of linear factors. We have enough numbers to solve every possible polynomial equation.

The Fundamental Theorem of Algebra

Every polynomial equation of the form a n z n + a n −1z n −1 +  + a1z + a 0 = 0, in which the coefficients a 0 , a1 , … , a n −1 , a n are any complex numbers, whose degree n is greater than or equal to one, and whose leading coefficient a n is not zero, has exactly n roots in the complex number system, provided each multiple root of multiplicity m is counted as m roots.

A proof of this theorem can be found in most texts on the theory of functions of a complex variable.

EXERCISES 

18.1

Operations with Complex Numbers

2. Solve the following equations for the real numbers x and y.

1. Express the following products of complex numbers in the form a + ib where a and b are real numbers.

( 3 + 4i ) 2 − 2 ( x − iy ) = x + iy a.

(

)

2

( 2 + 3i )( 4 − 2i ) a.

1+i b. 1−i

( 2 − i )( − 2 − 3i ) b.

( 3 − 2i )( x + iy ) = 2 ( x − 2iy ) + 2i − 1 c.

+

1 = 1+i x + iy

c. (−1 − 2i )( 2 + i )

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18-10

Chapter 18  Complex Functions

Graphing and Geometry

23. Find all solutions of the equation x 4 + 4 x 2 + 16 = 0.

3. How may the following complex numbers be obtained from z = x + iy geometrically? Sketch.

24. Solve the equation x 4 + 1 = 0.

z b. −z c. −z d. 1z a. 4. Show that the distance between the two points z 1 and z 2 in the complex plane is z 1 − z 2 . In Exercises 5–10, graph the points z = x + iy that satisfy the given conditions. z < 2 c. z > 2 5. a. z = 2 b. 6. z − 1 = 2 7. z +1 = 1 8. z + 1 = z − 1 9. z + i = z −1 10. z + 1 ≥ z − 1 Express the complex numbers in Exercises 11–14 in the form re iθ , with r ≥ 0 and −π < θ ≤ π , and for each of them draw a diagram in the complex plane. 2 1+i 11. (1 + −3 ) 12. 1−i 13.

1+ 1−

−3 ( 2 + 3i )(1 − 2i ) 14. −3

Theory and Examples

25. Complex numbers and vectors in the plane  Show with a diagram in the complex plane that the law for adding complex numbers is the same as the parallelogram law for adding vectors. 26. Complex arithmetic with conjugates  Show that the conjugate of the sum (product, or quotient) of two complex numbers, z 1 and z 2 , is the same as the sum (product, or quotient) of their conjugates. 27. Complex roots of polynomials with real coefficients come in complex-conjugate pairs  a. Extend the results of Exercise 26 to show that f ( z ) = f ( z ) when f ( z ) = a n z n + a n−1z n−1 +  + a1z + a 0 is a polynomial with real coefficients a 0 , … , a n . b. If z is a root of the equation f ( z ) = 0, where f ( z ) is a polynomial with real coefficients as in part (a), show that the conjugate z is also a root of the equation. (Hint: Let f ( z ) = u + iυ = 0; then both u and υ are zero. Use the fact that f ( z ) = f ( z ) = u − iυ.)

Powers and Roots

28. Absolute value of a conjugate  Show that z = z .

Use De Moivre’s Theorem to express the trigonometric functions in Exercises 15 and 16 in terms of cos θ and sin θ.

29. When z = z   If z and z are equal, what can you say about the location of the point z in the complex plane?

15. cos 4θ 16. sin 4θ 17. Find the three cube roots of 1.

30. Real and imaginary parts  Show that the following relations hold for any complex numbers z , z 1 , and z 2 .

18. Find the two square roots of i.

a. z + z = 2Re( z )

19. Find the three cube roots of −8i.

b. z − z = 2Im( z ) c. 2Re( z ) ≤ z

20. Find the six sixth roots of 64. 21. Find the four solutions of the equation

z4

−

2z 2

+ 4 = 0.

22. Find the six solutions of the equation z 6 + 2 z 3 + 2 = 0.

d. z1 + z 2

2

= z1

2

+ z2

2

+ 2 Re ( z 1 z 2 )

e. z1 + z 2 ≤ z1 + z 2

18.2 Functions of a Complex Variable Complex Domains The domains of the real variable functions discussed in Chapter 1 consisted of subsets of the real line, such as [ 0, 1 ] or ( 0, ∞ ). The domain of a complex function is a subset of the complex numbers, which we can visualize as a region in the complex plane. EXAMPLE 1   Examples of complex domains a. The set of complex numbers z for which z ≤ 1 The corresponding region in the complex plane is the set of points lying inside or on a circle of radius 1 centered at the origin O, shown in Figure 18.8(a). b. The set of complex numbers with z − z 0 ≤ 1 This corresponds to the region inside or on a circle of radius 1 centered at a complex number z 0 = a 0 + b 0 i, as in Figure 18.8b. c. A complex region can have a more general shape. Three more examples of complex domains are shown in Figure 18.8c.

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18.2  Functions of a Complex Variable

y

y

18-11

y

|z | = 1 |z − z 0 | ≤ 1 |z | < 1 0

1

x

z0

b

0 (a) The set of complex numbers z for which z ≤ 1

a

x

0

z

x

(b) The set of complex numbers z for which z − z0 ≤ 1

(c) Another three complex domains.

FIGURE 18.8  Some complex domains.

We usually use the letter z for a variable that describes a point in the domain of a complex function, just as we used x for a real function. We write w = f ( z ) to describe a function that takes as input a complex number z and gives as output a complex number w.

Complex Limits To understand complex limits, we take advantage of the fact that a complex number z = x + iy can be identified with a point ( x , y ) in the xy-plane. This identification allows us to extend our knowledge of real limits to complex numbers. We first consider the case where a complex number z = z (t ) is a function of a real variable t, which we often consider to be time. So we think of a vector whose position in the complex plane changes with time. We can write z (t ) = x (t ) + iy(t ), and then each t describes a vector ( x (t ), y(t ) ) in the complex plane. By describing z (t ) as a vector-valued function we are able to understand limits of z (t ) similarly to the limits of vector functions seen in Section 12.1. The meaning of an expression such as lim z (t ) = z 0 , t →1

which states that z (t ) is approaching z 0 = x 0 + iy 0 as t approaches 1 is that lim z (t ) − z 0 = 0. t →1

More specifically, we make the following definition.

DEFINITION  We say that the complex-valued function z (t ) = x (t ) + iy(t ) has

limit z 0 as t approaches t 0 , and write

lim z (t ) = z 0

t →t0

if, for every number ε > 0, there exists a corresponding number δ > 0 such that

z (t ) − z 0 < ε  whenever 0 < t − t 0 < δ.(13)

One way of interpreting Equation (13) is that z (t ) − z 0 is small when t − t 0 is small. In other words, if we draw a circle around the complex number z 0 in the complex plane, as small as we like, then the point z (t ) = x (t ) + iy(t ) will be inside that circle if we choose t sufficiently close to t 0 . Since z (t ) − z 0 = ( x (t ) + iy(t ) ) − ( x 0 + iy 0 ) ≤ x (t ) − x 0 + y(t ) − y 0 ,  Figure 18.9

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18-12

Chapter 18  Complex Functions

z

| z − z 0|

z0

| y − y0 |

| x − x0 |

FIGURE 18.9 z − z 0 2 = ( x − x 0 ) 2 + ( y − y0 ) 2

we see that if x (t ) − x 0 → 0 and y(t ) − y 0 → 0, then also z (t ) − z 0 → 0. So z (t ) converges to z 0 when its real part converges to the real part of z 0 and its imaginary part converges to the imaginary part of z 0 . On the other hand, since z − z 0 ≥ x − x 0 and z − z 0 ≥ y − y 0 , it follows that if z (t ) → z 0 , then also x (t ) → x 0 and y(t ) → y 0 . In summary, as t → t 0 , z (t ) approaches z 0 if and only if both x (t ) approaches x 0 and y(t ) approaches y 0 , z (t ) → z 0 ⇔ x (t ) → x 0  and  y(t ) → y 0 .(14)



This gives a practical formula that can be used to compute the complex limit of a function z (t ) = x (t ) + iy(t ): lim z (t ) = lim x (t ) + i lim y(t ).

t →t0

t →t0

t →t0

EXAMPLE 2   For the function z (t ) = cos t + i sin t , find lim z (t ). t→π

Solution 

lim z (t ) = lim cos t + i lim sin t = −1. t→π

t→π

t→π

Functions of a Complex Variable We say that f is a complex function on a domain S in the complex plane if to each z in S, the function f assigns a single complex number f ( z ). EXAMPLE 3   The function taking z to z 2 For this function, the domain S is the set of all complex numbers, or the complex plane C, and the function sends a complex number z to the complex number z 2 . As before, we write f ( z ) = z 2 to give an explicit name f to the function that inputs a complex number z and outputs a new complex number f ( z ). For each point z = x + iy, this function f outputs the complex number u + iυ where

u + iυ = ( x + iy ) 2 = x 2 + i(2 xy) + i 2 y 2 = ( x 2 − y 2 ) + i(2 xy).(15)

The real part of the complex variable u + iυ is given by a real function of x and y, u ( x , y ) = x 2 − y 2, and the imaginary part by a second real function υ ( x , y ) = 2 xy. We cannot draw the graph of such a function, as we did with real functions whose graphs were curves or surfaces, since both the domain and the range in this case are the complex plane. We would need four dimensions, or four perpendicular directions, to graph a complex function, but we only have three dimensions to work with. Instead, we can set w = f ( z ) and get a feeling for what this function is doing by drawing some lines or curves in the z-plane and looking at their images in the w-plane. We can visualize the behavior of a complex function by thinking of the domain as a rubber sheet that has a grid of lines drawn on it, parallel to the x and y axes. By drawing the image of this grid under the map, we can get an understanding of how the complex function maps one domain to another. EXAMPLE 4  Compute where f ( z ) = z 2 maps the vertical line x = 1. More generally, compute where it maps the vertical line x = a and the horizontal line y = b. Solution  Looking at u and υ in Equation (15), and setting the value of x to be x = a, we see that u ( a, y ) = a 2 − y 2  and  υ ( a, y ) = 2ay.

When a ≠ 0 we obtain y = υ (2a) in Equation (15); we have that the vertical line x = a is mapped to the curve υ2 u = a2 − .(16) 4a 2

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18.2  Functions of a Complex Variable

18-13

For each nonzero choice of constant a, this equation describes a parabola. For a = 1 the parabola u = 1 − υ 2 4 in the complex w-plane is the image of the line x = 1 in the complex z-plane. If we chose a = −1 in Equation (16), the image of the line x = −1 is the same parabola. This is to be expected since the squares of −z and z coincide under the map z → z 2. In the case a = 0, corresponding to the y-axis, we have u ( 0, y ) = −y 2 and υ ( a, y ) = 0. Thus the point ( 0, y ) on the y-axis is mapped to the point (−y 2, 0 ). The y-axis is folded at 0 so that the positive y-axis and the negative y-axis each map onto the negative real axis. We can get further insight into the behavior of the map f ( z ) = z 2 by examining the images of some horizontal lines. The line y = b in the complex z-plane has image points satisfying u ( x , b ) = x 2 − b 2  and  υ ( x , b ) = 2 xb. We can solve for x, and when b ≠ 0, we see that x = υ (2b) and υ2 − b 2. (17) 4b 2

u =



Again this gives us a parabola for each nonzero choice of b. As before, the lines y = ±b map to the same parabola. The x-axis, corresponding to b = 0, is folded over at the origin. The positive real axis is mapped to itself, while the negative real axis is mapped not to itself, but also to the positive real axis. Four parabolas obtained as the images of horizontal and vertical lines are shown in Figure 18.10. The images of two squares in the first quadrant are seen in Figure 18.11 and Figure 18.13. Green stripes are taken to green, and red to red. These figures give us a feeling for the behavior of the complex function f ( z ) = z 2.  Limits of a function of a complex variable f ( z ) = u ( x , y ) + iυ ( x , y ) are defined by considering the limits of the real and imaginary parts. Each of u and υ is a real-valued function of two variables, and we extend the definition of limits developed for such functions in Section 13.2. y 2 y=1

2 1

1

y = 1/2 −2

y

x = 1/2 x=1

−1

0

1

2

x

−2

−1

0

−1

−1

−2

−2

1

2

x

FIGURE 18.10  The images of the horizontal lines y = 1 2 and

y = 1, and of the vertical lines x = 1 2 and x = 1, under the complex map f ( z ) = z 2.

DEFINITION  We say that the complex-valued function f ( z ) = u ( x , y ) + iυ ( x , y )

has limit w 0 = u 0 + iυ 0 as z approaches z 0 = x 0 + iy 0 , and write lim f ( z ) = w 0

z→z0

if, for every number ε > 0, there exists a corresponding number δ > 0 such that

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18-14

Chapter 18  Complex Functions

y

y 2i 1+i

i

0

x

1

−1

0

x

1

FIGURE 18.11  The image of a square with vertices at 0, 1, i, 1 + i under the

complex map f ( z ) = z 2 y

y 3 + 4i

4i i

1+i

3i

2+i

2i i 0

x

2

1

0

1

2

3

x

4

FIGURE 18.12  The image of a square with vertices at 1, 2, 1 + i, 2 + i under

the same map y

y

i

0

i

0.5

x

0.8 1

−1

0

0.25

0.64

1

x

FIGURE 18.13  A quarter of the unit disk in the complex plane and its image

under the complex map f ( z ) = z 2. Radial lines at angle θ are taken to new radial lines at angle 2θ, and circular arcs of radius r to circular arcs of radius r 2.

If we express f in the form f ( x + iy ) = u ( x , y ) + iυ ( x , y ) , then we have f (z) − w 0 =

( u( x, y ) − u 0 )2 + ( υ( x, y ) − υ 0 )2 .

For a point in the complex plane w 0 = u 0 + iυ 0 , the definition implies (Exercise 7) that lim f ( z ) = w 0 exactly when both z→z0

and

lim

u( x, y ) = u 0

lim

υ( x, y ) = υ0 .

( x , y )→ ( x 0 , y 0 )

( x , y )→ ( x 0 , y 0 )

We often use this formulation in computations. Namely the limit of a function f ( x + iy ) = u ( x , y ) + iυ ( x , y ) as x + iy → x 0 + iy 0 exists if and only if the limit of u ( x , y ) and of υ ( x , y ) each exists as ( x , y ) → ( x 0 , y 0 ).

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18.2  Functions of a Complex Variable

18-15

EXAMPLE 5  Show that a. lim z = z 0 , b. lim z 2 = z 0 2, c. lim z z does not exist. z→z0

z→z0

z→0

Solution 

a. We express f ( z ) = z in the form f ( z ) = u ( x , y ) + iυ ( x , y ) , where u ( x , y ) = x and υ ( x , y ) = y, and compute the limit as z approaches z 0 = x 0 + iy 0 . Then since the real-valued functions u ( x , y ) = x and υ ( x , y ) = y are continuous, lim

u( x, y ) = x 0

lim

υ( x, y ) = y0 .

( x , y )→ ( x 0 , y 0 )

and ( x , y )→ ( x 0 , y 0 )

So lim z = x 0 + iy 0 = z 0 as claimed. z→z0

b. We express f ( z ) = z as f ( z ) = u ( x , y ) + iυ ( x , y ) , where u ( x , y ) = x 2 − y 2 and υ ( x , y ) = 2 xy (Equation 15). Then since the real-valued function u ( x , y ) = x 2 − y 2 is continuous, lim

( x , y )→ ( x 0 , y 0 )

u( x, y ) = x 0 2 − y0 2

and similarly, since the real-valued function υ ( x , y ) = 2 xy is continuous, lim

( x , y )→ ( x 0 , y 0 )

υ( x, y ) = 2x 0 y0 .

So lim z 2 = x 0 2 − y 0 2 + i 2 x 0 y 0 = z 0 2 as claimed. z→z0

c. We express f ( z ) = z z in the form f ( x + iy ) = u ( x , y ) + iυ ( x , y ). f ( x + iy ) =

( x − iy )( x − iy ) x − iy x 2 − y2 −2 xy = = 2 +i 2 . 2 ( x + iy )( x − iy ) x + y2 x + iy x + y

The function u( x , y ) =

x 2 − y2 x 2 + y2

is not continuous at ( 0, 0 ) since it has limit equal to 1 if ( x , y ) → ( 0, 0 ) along the x-axis (with y = 0), and it has limit equal to −1 if ( x , y ) → ( 0, 0 ) along the y-axis (with x = 0). Since different limits are obtained as ( x , y ) → ( 0, 0 ) from different directions, no limit exists for u ( x , y ) and therefore no limit exists for f ( z ).

Continuity A function w = f ( z ) that is defined throughout an open set containing a point z = z 0 is said to be continuous at z 0 if

lim f ( z ) − f ( z 0 ) = 0.(19)

z→z0

Intuitively, this means that as the point z approaches z 0 , its image f ( z ) approaches f ( z 0 ), as shown in Figure 18.14. The notion of continuity for real-valued functions of two variables u and υ was defined in Section 13.2. If we write f as a sum of its real and imaginary parts, f ( z ) = u ( x , y ) + iυ ( x , y ) , then f is continuous at z 0 if and only if u ( x , y ) and υ ( x , y ) are each continuous at ( x 0 , y 0 ) (Exercise 8).

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18-16

Chapter 18  Complex Functions

y

v C2 z

C1

f (z 0 )

z0 w = f (z ) x

0

u

0

z -plane

w -plane

FIGURE 18.14  Choose any circle C 2 around f ( z 0 ) in the

w-plane. A continuous function takes points in the z-plane that lie inside a small enough circle C 1 around z 0 to points inside C 2 .

As with functions of a real variable, polynomial functions are continuous. The proof is similar to the real case, and while we do not give it in full, we illustrate the idea with two examples. EXAMPLE 6   Show that at any point z 0 in the complex plane, the function f ( z ) = z is continuous. Solution  The definition of continuity given in Equation (19) requires us to show that the limit

lim f ( z ) − f ( z 0 ) = lim z − z 0 = 0.

z→z0

z→z0

Using the definition of limit in Equation (18), our challenge is to find a δ so that z − z 0 < ε  whenever 0 < z − z 0 < δ. For this very simple function we can choose δ = ε. So the requirement for continuity is satisfied, and we conclude that f is continuous at any point z 0 . By similar arguments we can show that sums and products of continuous functions are continuous, as are quotients of continuous functions wherever the denominator is nonzero. These arguments combine to show that polynomial functions and rational functions are continuous on their domains. We illustrate such an argument for the function f (z) = z 2 . EXAMPLE 7   Show that at any point z 0 in the complex plane, the function f ( z ) = z 2 is continuous. Solution  For the function f ( z ) = z 2 we have

lim f ( z ) − f ( z 0 ) = lim z 2 − z 02

z→z0

z→z0

= lim z − z 0 z + z 0 z→z0

Factor.

= lim z − z 0 lim z + z 0 z→z0

= 0 ⋅ 2z 0

z→z0

Example 1 and Exercise 9

= 0. The requirement for continuity given by Equation (19) is satisfied, and we conclude that f is continuous at any point z 0 .

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18.3  Derivatives

18-17

Although we will not prove it here, the arguments of Examples 1 and 2 can be extended to show that all complex polynomials are continuous, as are all quotients of complex polynomials where the denominator is nonzero.

EXERCISES 

18.2

1. For each of the following regions in the complex plane, describe the image of the region under the map w = z 2 . (Hint: Use polar coordinates.) z = 1, 0 ≤ arg z ≤ π b. z = 2,  π 2 ≤ arg z ≤ π a.

b. Give parameterizations of the four curves to which the sides of the square are mapped. 7. Use Formula (18) with f ( z ) = u ( x , y ) + iυ ( x , y ), z 0 = x 0 + iy 0 , and w 0 = u 0 + iυ 0 to show that

c. arg z = π 4 d. z < 1, −π ≤ arg z ≤ 0 e. The x-axis

a. Determine the four complex numbers to which the vertices of the square are mapped.

f . The y-axis

2. Show that the two parabolas in the w-plane in Figure 18.10 intersect orthogonally (at right angles) if neither a nor b is zero. 3. Show that the function w = z 3 maps the wedge 0 < arg z < π 3 in the z-plane onto the upper half-plane of the w-plane, the halfplane whose imaginary part satisfies Im( w) > 0. Use polar coordinates and sketch. 4. Show that f ( z ) = z 3 is continuous at z = z 0 for any z 0 .

lim f ( z ) = w 0

z→z0

exactly when both

6. The function w = z 3 maps the square with vertices at 0, 1, i, and 1 + i to the region shown in Figure 18.15.

lim

υ( x, y ) = υ 0 .

( x , y )→( x 0 , y 0 )

8. Suppose that f ( z ) = u ( x , y ) + iυ ( x , y ) is defined on an open set containing the point z 0 = x 0 + iy 0 . Show that f ( z ) is continuous at z 0 if and only if u ( x , y ) and υ ( x , y ) are each continuous at ( x 0 , y 0 ).

y 1+i

i

u( x, y ) = u 0

and

5. Show that f ( z ) = 1 z is continuous at z = z 0 if z 0 ≠ 0.

y

lim

( x , y )→( x 0 , y 0 )

9. a. Show that the constant function f ( z ) = z 0 is continuous.

0

1

x

x

b. Show that if f ( z ) and g( z ) are continuous, then so is ( f + g ) ( z ).

FIGURE 18.15  The image under the complex map f ( z ) = z 3

of a square with vertices at 0, 1, i, and 1 + i

18.3 Derivatives The derivative of a complex function w = f ( z ) at z 0 is defined in the same way as for a real-valued function of a real variable.

DEFINITION  The derivative of a complex-valued function f at z = z 0 is



f ′( z 0 ) = lim

z→z0

f (z) − f (z 0 ) ,(20) z − z0

provided that this limit exists. When this limit exists we say that f is complex differentiable at z 0 .

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18-18

Chapter 18  Complex Functions

As usual, we also use the equivalent notation df (z ) dz 0 to denote the derivative of f at z 0 . A function that has a complex derivative at every point in its domain D is said to be complex differentiable or holomorphic on D. Due to a property of complex differentiable functions, that they are equal to their Taylor series where the Taylor series converges, they are also often called analytic functions. The derivative of a real-valued function exists at a point x 0 if the derivative from the left and from the right exist and agree at x 0 . For a complex differentiable function the derivative from any direction must exist and agree with the derivative computed from any other direction. There are many ways in which z can approach z 0. It can approach z 0 on a horizontal path from the left, from the right, or along a vertical path coming from above or from below. And there are many other possibilities. It is possible for z to approach z 0 along a line coming from any direction, or even along a curved path or spiral that ends at z 0 . The existence of a single limiting value for f (z) − f (z 0 ) z − z0 says that no matter how z 0 is approached, the same value is obtained. Despite this very restrictive condition, there is still a large collection of complex differentiable functions. The very special implications of differentiability and the large number of such functions make the use of complex differentiable functions an extremely powerful tool.

EXAMPLE 1   The complex conjugation function f ( z ) = z takes a point z = x + iy to its complex conjugate z = x − iy. Show that at any point z 0 in the complex plane, this function is not complex differentiable.

y

x = x0

z0

Solution  We compute the limit defining the derivative as we approach the point z 0 = x 0 + iy 0 from two different directions:

y = y0

0

x

FIGURE 18.16  The limit defining a

differentiable function has the same value as z approaches z 0 from any direction, including the horizontal and vertical.

a. along the vertical line x = x 0 and b. along the horizontal line y = y 0 . See Figure 18.16. In case (a), we fix x = x 0 and consider what happens as y → y 0 . On the vertical line x = x 0 we have that z approaches z 0 only from above or below. On this line z → z 0 exactly when y → y 0 . The limit defining a derivative, Equation (20), gives lim

z→z0

f (z) − f (z 0 ) z − z0 = lim z→ z0 z − z 0 z − z0 = lim

( x 0 + iy ) − ( x 0 + iy 0 ) ( x 0 + iy ) − ( x 0 + iy 0 )

= lim

( x 0 − iy ) − ( x 0 − iy 0 ) i( y − y 0 )

y → y0

y → y0

−i ( y − y 0 ) y → y0 i( y − y 0 ) −i = lim y → y0 i = −1. = lim

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18.3  Derivatives

18-19

In case (b), we fix y = y 0 and consider what happens along this horizontal line. With y fixed at y 0 , we have z → z 0 exactly when x → x 0 , and Equation (20) becomes lim

z→z0

f (z) − f (z 0 ) z − z0 = lim z→ z0 z − z 0 z − z0 = lim

x→ x0

= lim

( x + iy 0 ) − ( x 0 + iy 0 ) ( x + iy 0 ) − ( x 0 + iy 0 ) ( x − iy 0 ) − ( x 0 − iy 0 )

x − x0

x→ x0

= lim

x→ x0

x − x0 x − x0

= 1. Two different paths along which z approaches z 0 lead to two different answers. So, there is no single complex number that is equal to the limit, no matter which path is chosen. There is no number that we can call f ′( z 0 ), and we conclude that the function w = z does not have a derivative at any point z 0 in the complex plane.

Rules for Finding Complex Derivatives The formulas for differentiating sums, products, quotients, and powers of the complex variable z are the same as those for a real variable x. For example, if c is any complex number, and if f ( z ) and g( z ) are functions that have derivatives at z = z 0 , then at z = z 0 we have: (1)  (c)′ = 0

Derivative of a constant  c

(2)  (cf )′( z ) = cf ′( z )

Constant Multiple Rule

(3)  ( f + g )′ ( z ) = f ′( z ) + g′( z )

Sum Rule

(4)  ( fg)′( z ) = f ′( z ) g( z ) + f ( z ) g′( z )

Product Rule

(5)  ( f g )′( z ) = [ g( z ) f ′( z ) − f ( z ) g′( z ) ] g( z ) 2, if  g( z ) ≠ 0

Quotient Rule

(6)  ( z n ) ′ = nz n −1, where n is any integer (7)  ( g  f )′( z ) = g′ ( f ( z ) ) f ′( z ),   for f ( z ) in domain of g

Power Rule Chain Rule

These formulas are derived in the same manner as the formulas for functions of a real variable. We give an idea of the techniques used to prove them by establishing one case of the power rule. EXAMPLE 2   Show from the definition that f ( z ) = z 3 has derivative f ′( z ) = 3z 2 . Solution  We compute the limit defining a derivative,

lim

z→z0

f (z) − f (z 0 ) z 3 − z 03 = lim z→ z0 z − z 0 z − z0 = lim

z→ z0

Factor numerator.

( z − z 0 )( z 2 + zz 0 + z 02 )

z − z0

= lim z 2 + zz 0 + z 02 z→ z0

Each of the three terms approaches  z 02 .

= 3z 02 . Note that no matter how z approaches z 0 , the terms z 2 and zz 0 each approaches z 02 , and the limit is 3z 02 . So f ′( z 0 ) = 3z 02 .

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18-20

Chapter 18  Complex Functions

EXERCISES 

18.3 Theory and Examples

In Exercises 1–3, find the derivative with respect to z of the given function at the point z 0 .

5. The directional derivative of a real-valued function f in the complex plane in the direction of a unit vector w is given by f ( z 0 + tw ) − f ( z 0 ) D w f ( z 0 ) = lim .(21) t→0 t Compare this to the derivative of a complex function f ( z ) by computing the complex derivative of f as z approaches z 0 along the line z = z 0 + tw, where w now represents a complex number with magnitude 1. What are the differences between these two formulas?

z +1 1. f ( z ) = , z0 = 1 + i z −1 2. f ( z ) = z 3 + 3z 2 + 3z + 2, z 0 = −1 + 2i 1+i 2 (Here we get two answers depending on our choice of the square root. Give the answer using the principal square root, as defined in Section 18.1.)

3. f ( z ) =

z 2 + 1, z 0 =

4. Use the definition of a complex derivative to find f ′( z 0 ) if f ( z ) = 1 z and z ≠ 0.

18.4 The Cauchy-Riemann Equations When we write z = x + iy and the complex function w = f ( z ) = u + iυ is differentiable, the real part of f , u ( x , y ) , and the imaginary part of f , υ ( x , y ) , are real-valued differentiable functions of the real variables x and y. We can compute the derivative of f at a point z 0 = x 0 + iy 0 by approaching z 0 from any direction we choose. By comparing the derivative f ′( z 0 ) obtained from approaching z 0 along horizontal and vertical directions, we obtain equations relating the partial derivatives of u and υ. If we approach along the horizontal line y = y 0 , then z − z 0 = ( x + iy 0 ) − ( x 0 + iy 0 ) = x − x 0 and therefore f ′( z 0 ) = lim

z→z0

f (z) − f (z 0 ) z − z0

= lim

f ( x + iy 0 ) − f ( x 0 + iy 0 ) x − x0

= lim

u ( x , y 0 ) + iυ ( x , y 0 ) − ( u ( x 0 , y 0 ) + iυ ( x 0 , y 0 )) x − x0

= lim

υ( x, y0 ) − υ( x 0 , y0 ) u( x, y0 ) − u( x 0 , y0 ) + i lim x→ x0 x − x0 x − x0

x→ x0

x→ x0

x→ x0

∂u ∂υ ( x , y ) + i ( x 0 , y 0 ). ∂x 0 0 ∂x Though we have approached z 0 in only one of the many possible ways, we obtained a useful formula for the derivative, equal to what we would get if we approached z 0 from any other direction. For f ( z ) = u ( x + iy ) + iυ ( x + iy ) , we obtain the following explicit formula for the complex derivative of a function that has a complex derivative at z 0 . =

Formula for the Complex Derivative



f ′( z 0 ) =

∂u ∂υ ( x , y ) + i ( x 0 , y 0 )(22) ∂x 0 0 ∂x

The formula in Equation (22) does not depend on which direction is taken to approach z 0 . Instead of approaching z 0 along the x-axis, we can approach along the vertical line x = x 0 , on which z − z 0 = ( x 0 + iy ) − ( x 0 + iy 0 ) = i ( y − y 0 ).

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18.4  The Cauchy-Riemann Equations

18-21

In this case we have f ′( z 0 ) = lim

z→z0

f (z) − f (z 0 ) z − z0

= lim

f ( x 0 + iy ) − f ( x 0 + iy 0 ) i( y − y 0 )

= lim

u ( x 0 , y ) + iυ ( x 0 , y ) − ( u ( x 0 , y 0 ) + iυ ( x 0 , y 0 )) i( y − y 0 )

= lim

u( x 0 , y ) − u( x 0 , y0 ) i[ υ ( x 0 , y ) − υ ( x 0 , y 0 ) ] + lim y → y0 i( y − y 0 ) i( y − y 0 )

y → y0

y → y0

y → y0

=

∂υ ∂u ( x , y ) − i ( x 0 , y 0 ). ∂y 0 0 ∂y

1 i = −i

Since f is complex differentiable, these two computations must give the same value for the derivative. The two real parts in the two equations must be equal, and the imaginary parts in the two equations must be the same. This gives relations between the partial derivatives of u and υ that hold at any point where f ( z ) = u ( x , y ) + iυ ( x , y ) has a complex derivative.

The Cauchy-Riemann Equations

∂u ∂υ ∂u ∂υ = and = − ∂x ∂y ∂y ∂x

EXAMPLE 1   Show that the function f ( z ) = 1 z satisfies the Cauchy-Riemann Equations at any point in the complex plane where z ≠ 0. Solution  Let z = x + iy be a nonzero point in the complex plane, and let

w = f (z) = 1 z . Then w = u + iυ =

x − iy x − iy 1 1 = 2 = ⋅ , x + iy x + iy x − iy x + y2

and therefore u =

x2

−y x ,υ = 2 .  x 2 2 + y x + y2

+ y 2 ≠ 0 since  z ≠ 0.

After computing partial derivatives four times and simplifying, we find that y2 − x 2 ∂u ∂υ = 2 = ∂x ∂y (x 2 + y2 ) and −2 xy ∂u ∂υ = . 2 = − ∂y ∂x (x 2 + y2 ) This shows that the Cauchy-Riemann Equations are satisfied at all points where x 2 + y 2 ≠ 0. The Cauchy-Riemann Equations give relationships between u and υ that must be satisfied by a complex differentiable function. We cannot independently specify functions

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Chapter 18  Complex Functions

u ( x , y ) and υ ( x , y ) and expect the function w = u + iυ to be complex differentiable. However if we take functions u and υ, which have continuous partial derivatives ∂u ∂u ∂ υ ∂υ , , , and ∂x ∂y ∂x ∂y on a region R and satisfy the Cauchy-Riemann Equations on R, then a theorem that we will not prove here states that the function w = u + iυ is complex differentiable on R. This is a very powerful conclusion, since it means that by checking agreement of the computation of the limit lim

z→z0

f (z) − f (z 0 ) z − z0

only as z → z 0 in the horizontal and vertical directions, we know the derivative is the same as z → z 0 in every direction.

EXERCISES 

18.4

For the following functions, find the real and imaginary parts of the function w = f ( z ), w = u + iυ, z = x + iy, and show that they satisfy the Cauchy-Riemann equations.

6. Verify that Laplace’s Equation in Exercise 5 is satisfied by the real and imaginary parts of the functions

z3 1. z 2 2.

z 2, b.

3. z 4 4. 1 z 2, z ≠ 0

c. z 3.

5. A real-valued function with continuous second partial derivatives g ( x , y ) is called harmonic if it satisfies Laplace’s Equation,

7. Show that the following functions are complex differentiable on the complex plane by verifying that their real and imaginary parts have continuous partial derivatives and satisfy the CauchyRiemann Equations.

∂2g ∂2g + = 0. 2 ∂x ∂y2 Show that both the real part u = u ( x , y ) and the imaginary part υ = υ ( x , y ) of a complex differentiable function w = f ( z ) satisfy Laplace’s Equation.

a. z,

a. f ( z ) = e x cos y + ie x sin y b. f ( z ) = sin x cosh y + i cos x sinh y

18.5 Complex Power Series Applying the rules for finding the derivatives of sums, products, and quotients of complex functions, we can see that polynomials and rational functions (quotients of polynomials) are complex differentiable on their domains. We have extended these functions from differentiable functions on the real line to differentiable functions on the complex plane C. We now investigate how to extend other elementary functions such as sin z , cos z , e z , cosh z , and so on that are defined on the real line to functions that are differentiable on C. To do this we will need to make sense of expressions such as sin ( 2 + 3i ) and e 3−i . When evaluating sin ( 2 + 3i ) it is not helpful to try to think of 2 + 3i as an angle whose sine is evaluated, as we did when evaluating the sine of a real number. Rather we define the value of the sine function evaluated on a complex number by extending the expression of the sine function as a power series from real to complex numbers. In Chapter 9 we derived the Taylor series expression sin x = x −

x3 x5 x7 + − + …. 3! 5! 7!

When we use this series to calculate a value such as sin 2, we don’t think of x = 2 as a radian measure of an angle, but just consider it as a number that is input into the series. This leads us to define the function f ( z ) = sin z for a complex variable z by the formula

M18_HASS5901_15_GE_C18.indd 22

sin z = z −

z7 z3 z5 + − += 3! 5! 7!

∞

z 2 n +1

∑ (−1)n ( 2n + 1)!.(23)

n=0

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18.5  Complex Power Series

18-23

For real z, for which z = x + iy with y = 0, this gives the familiar series for sin x. So this new definition of sin z for complex numbers z is consistent with our previous understanding of the sine function. We now consider what it means for such a series to converge.

Sequences As we did with real numbers, we base the concept of convergence of a complex series on the more basic concept of the convergence of complex sequences. A complex sequence is defined to be a function from the positive integers to the complex numbers. These sequences are often presented as an infinite list of complex numbers, { z n } = z1 , z 2 , … , z n , … .

We say that a sequence { z n } converges to a complex number L if for every positive number ε there corresponds an integer N such that z n − L < ε  whenever  n > N . If no such number exists we say that { z n } diverges. While each number z n in the sequence and the limiting number L is complex, the conditions for convergence involve the real numbers z n − L , ε, and δ. Because of this, many of the arguments for convergence of sequences given in Section 9.1 carry over to complex sequences.

Series A complex power series ∞

∑ anz n



= a 0 + a1z + a 2 z 2 + (24)

n=0

converges at a point z if the sequence of partial sums s n = a 0 + a1z + a 2 z 2 +  + a n z n has a limit S = u + iυ as n → ∞. If we separate s n into its real and imaginary parts, by writing s n = u n ( x , y ) + iυ n ( x , y ), then s n → S   if and only if u n → u and υ n → υ.  

As in Equation (14)

Many properties of real series proved in Chapter 9 carry over to complex series. For example, sums of convergent series converge. If ∞

∑ cn n=0

= L and

∞

∑ dn

= M

n=0

∞

are two convergent complex series, then the series ∑ ( c n + d n ) also converges, and n=0

∞

∑ (cn + d n ) =

L + M.

n=0

Proofs of these properties are very similar to the proofs for real series. ∞

The series ∑ a n z n is said to converge absolutely if the corresponding series of abson=0 lute values

∞

∑

a n z n = a 0 + a1z + a 2 z 2 + (25)

n=0

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Chapter 18  Complex Functions

converges. The sum in Equation (25) is a series of real numbers, so we can apply the tests developed in Chapter 9, such as the Ratio Test, Comparison Test, or Integral Test, to verify its convergence. If the series of absolute values does converge, then the following theorem tells us that the original series also converges. THEOREM 1—The Absolute Convergence Test for Complex Series

If a series converges absolutely, then it converges. Proof   Separate each term of the series into its real and imaginary part by writing a k z k = c k + id k , k = 0, 1, 2, … where c k and d k are real numbers. Suppose that the series of absolute values in Equation (25) converges. Then, since ck =

c k2 ≤

c k2 + d k2 = a k z k

dk =

and

d k2 ≤

c k2 + d k2 = a k z k ,

the Direct Comparison Test tells us that the series ∞

∞

∑ ck

and   ∑ d k

k=0

k=0

both converge (Theorem 10 in Chapter 9). From this we can conclude that each of the series ∞

∑ ck

∞

∑ dk ,

and

k=0

k=0

computed without taking absolute values, also converge. Therefore the series ∞

∑ (ck

∞

∑ ck

+ id k ) =

k=0

k=0

∞

+ i ∑ dk k=0

converges. EXAMPLE 1   Show that the series for sin z in Equation (23) converges for all complex numbers z. Solution  We test the series of real numbers ∞

∑



n=0

( − 1) n

z 2 n +1 (26) ( 2n + 1)!

for absolute convergence. For this series we apply the Ratio Test. Set wn =

( − 1) n

z 2 n +1 . ( 2n + 1)!

Then we calculate w n +1 z2 = , ( 2n + 3 )( 2n + 2 ) wn and lim

n →∞

w n +1 z2 = lim = 0.  As n n →∞ ( 2n + 3 )( 2n + 2 ) wn

→ ∞,  denominator  → ∞.

Since this limit is less than one, the Ratio Test implies that the series in Equation (26) converges absolutely. By Theorem 1, the series in Equation (26) itself converges, without the absolute value signs, and we have shown that the complex series for sin z , the series in Equation (23), converges at every point z in the complex plane.

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18.6  Some Complex Functions

EXERCISES 

18.5

In Exercises 1–5, find the regions in which each of the series converges. Sketch the region of absolute convergence.

∞

6. Show that the series ∑

k=0

1. 1 + z + z 2 + z 3 +  2. 1 − 2 z + 3z 2 − 4 z 3 +  ( z − 1) 3 ( z − 1) 4 − 1) 2 + −  +  2 3 4 ( z + i )2 ( z + i )3 ( z + i )4 4. ( z + i ) + + +  +  2 3 4

3. ( z − 1) −

5. 1 −

(z

18-25

(z

zk converges absolutely for all z. k! ∞

7. Show that the series all z.

∑ (−1) k (2z k )!

8. Show that the series z < 1.

∑ (−1) k −1 zk

2k

converges absolutely for

k=0 ∞

k

converges absolutely for

k =1

T 9. Use the series in Equation (23) for sin z to compute sin ( 0.2 + 0.3i ) to three decimals.

( z + 1) 3 ( z + 1) 4 + 1) ( z + 1) 2 + − +  −  2 4 8 16

18.6 Some Complex Functions We have seen how to define the sine function as a complex series by extending the series expansion for real numbers to complex numbers. We can similarly define other basic complex functions by extending the series formulas developed for a real variable x to complex numbers. ez = 1 + z + cos z = 1 −

z2

Arctan z = z −

z3

2! 3

z2 z3 + + = 2! 3!

+

z4

+

z5

4! 5

−

z6

−

z7

6! 7

+=

∞

∑ zn! n

n=0 ∞

∑ (−1)n (2z n)! 2n

n=0

+=

∞

2 n +1

∑ (−1)n 2zn + 1

n=0

Using the Ratio Test as in Example 1 of Section 18.5, we can show that the series for e z and cos z converge absolutely for all z in the complex plane. To define Arctan z as an inverse function for the function tan z , we need to restrict the domain of tan z to a region on which it is one-to-one. The formula for Arctan z in the above power series corresponds to one choice for such a restricted domain and is called the principal value of arctan z. The Ratio Test shows that this series for Arctan z converges when z < 1, and diverges when z > 1. We will explore this idea in more detail when we look at the complex logarithmic function later in this section. If there is an open region of the complex plane containing a point z in which all points except for z are in the domain of a complex differentiable function, then z is called an isolated singularity of f . For example, the point z = 0 is not in the domain of f ( z ) = 1 z and is an isolated singularity of this complex differentiable function. ∞ A basic theorem in the theory of complex functions says that a power series ∑ a n z n n=0 either (1) converges for all z, or (2) converges inside a circle z < R and diverges for z > R, or (3) converges only at z = 0. This theorem closely resembles Theorem 18 in Chapter 9, and its proof, based on the Ratio Test, is similar. THEOREM 2—The Convergence Theorem for Complex Power Series ∞

If a series ∑ a n z n converges at a point z = z 0 , then it converges absolutely for n=0

all z with z < z 0 . If the series diverges at z = z 0 , then it diverges for all z with z > z 0 .

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Chapter 18  Complex Functions

EXAMPLE 1   Show that the largest radius circle centered at 0 inside which the series 1 − z2 + z4 − z6 +  converges is the circle of radius R = 1. Solution  The absolute value of the nth term a n = z 2 n of this series is z 2 n . By Theorem 1, the Absolute Convergence Test, this converges for all z with z < 1. For z = i the series becomes

1 − i2 + i4 − i6 +  = 1 + 1 + 1 + 1 + , which diverges. By Theorem 2 the series diverges for all z with z > 1. So the largest radius circle centered at zero inside which the series converges is the circle of radius one. The largest radius circle around a point z 0 inside of which a series converges is called the circle of convergence for z 0 . The series converges for all points inside this circle and diverges at any point outside it. Understanding the convergence for points lying on the circle of convergence is a more difficult problem that we will not discuss. The circle z = 1 is the circle of convergence centered at zero for the function in Example 1. Many of the basic properties of trigonometric and exponential functions extend to the complex versions of these functions that are defined using series. EXAMPLE 2   Using the series definition of the exponential function z2 z3 + += 2! 3!

ez = 1 + z +

∞

∑ zn! , n

n=0

show that e z e w = e z + w. (27)



Solution  We multiply the series for e z and the series for e w , and gather terms to get a series for the product:

(1 + z + z2! + z3! + )(1 + w + w2! + w3! + ) z w = 1+ z + w +( +z⋅w+ +  . 2! 2! ) 2

(

3

2

2

)

3

2

The general form for the terms whose degree is n is given by zn z n −1 z n−2 w2 z n−k wk wn , ⋅1+ ⋅w+ ⋅ ++ ⋅ ++1⋅ ( n − 1)! ( n − 2 )! ( n − k )! k ! n! n! 2! which in summation form equals n

n−k

∑ ( nz− k )! wk ! .(28)



k

k=0

On the other hand, the series expression for e z + w is e z+w = 1 + z +

(z

( z + w )3 + w )2 + += 2! 3!

∞

∑

n=0

(z

+ w )n , n!

and the term of degree n is (z

+ w )n . n!

By noticing that n ( n − 1)  ( n − k + 1) ( n − k )! n! ⋅ = ( n − k )! k! k !( n − k )!

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18.6  Some Complex Functions

18-27

we may also write this in summation form as (z

+ w )n 1 = n! n! = =

1 n!

 z n + nz n −1w + n ( n − 1) z n −2 w 2 + n ( n − 1)( n − 2 ) z n −3 w 3 +  + w n    2 3! n

∑ ( n −n!k )! k ! z n−k w k k=0

n

n− k

∑ ( nz− k )! wk ! . k

k=0

This last expression is identical to the one that appeared in Equation (28). We have shown that e z e w and e z + w have power series whose terms of degree n are equal for each value of n. It follows that the power series themselves are equal, and so e z e w = e z + w.  In a similar way, we can multiply and add power series term by term to show that sin ( z + w ) = sin( z ) cos( w) + cos( z ) sin( w) and cos ( z + w ) = cos( z ) cos( w) − sin( z ) sin( w). When we substitute iz for z in the series expansion for e z , we obtain (iz ) 2 (iz ) 3 (iz ) 4 + + +  2! 3! 4! z2 z3 z4 = 1 + iz − −i + +  2! 3! 4! z2 z4 z6 z3 z5 z7 = 1− + −  +  + i z − + − +  2! 4! 6! 3! 5! 7! = cos z + i sin z.

e iz = 1 + iz +

(

) (

)

i 2 = −1, i 3 = −i, i 4 = 1, … Series expansions for cos z  and  sin z

This establishes Euler’s Formula, which we stated in Section 18.1, e iz = cos z + i sin z.(29)



Euler’s formula tells us how to make sense of a complex exponent. Namely if we write z = x + iy, then e z = e x + iy = e x e iy = e x ( cos y + i sin y ).(30) Thus the value of e z is expressed in terms of sines and cosines of the real numbers x and y. With this formula, we can begin to understand the behavior of the exponential function on the complex plane. The image of a rectangle in the complex plane under the exponential map, computed using Euler’s formula, is shown in Figure 18.17. The rectangle is wrapped around the origin, which is not in the range of e z. If the rectangle is translated upwards by or downwards by a multiple of 2πi, then the formula for e z yields the same value, as can be seen in Equation (30). The height of the rectangle in Figure 18.17 is 6, just y −1 + 3i

y 1 + 3i

1 e

x

−1 − 3i

−e

e

x

1 − 3i

FIGURE 18.17  The image under the map f ( z ) = e z of a rectangle

with vertices at −1 − 3i, 1 − 3i, −1 + 3i, and 1 + 3i

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Chapter 18  Complex Functions

short of 2π. The top and bottom edges are wrapped around by the exponential function and almost coincide. A slightly taller rectangle with height 2π would have its top and bottom edges mapped by w = e z to coinciding radial segments. If we replace z by −iz in Euler’s formula, we get the companion equation e −iz = cos z − i sin z.(31)



By adding these two equations and dividing by two, we obtain a formula for cos z , cos z =



e iz + e −iz .(32) 2

If we subtract Equation (31) from Equation (29), then we obtain a formula for sin z , sin z =



e iz − e −iz . (33) 2i

The complex versions of the other trigonometric functions are derived from sin z and cos z in the usual way. For example tan z =



sin z 1 e iz − e −iz = .(34) cos z i e iz + e −iz

The usual trigonometric identities can be established by expressing the trigonometric functions as exponentials and then making use of the properties of the exponential function. EXAMPLE 3  Show that cos 2 z + sin 2 z = 1. Solution  We square both sides of Equation (32) and Equation (33) and add:

e 2iz + 2e 0 + e −2iz e 2iz − 2e 0 + e −2iz − 4 4 4e 0 = = 1. 4

cos 2 z + sin 2 z =

Equations (32) and (33) also reveal the close relationship between the trigonometric and hyperbolic functions. We define the complex hyperbolic functions by extending the series formulas for the real case, so that e z + e −z = 1+ 2 e z − e −z sinh z = = z+ 2 cosh z =

z2 z4 z6 + + +  , 2! 4! 6! z3 z5 z7 + + +  . 3! 5! 7!

Then Equations (32) and (33) say that

cos z = cosh iz

and

i sin z = sinh iz.(35)

These relationships explain the similarities between trigonometric identities such as cos 2 z + sin 2 z = 1, and the corresponding identities for hyperbolic functions, such as cosh 2 z − sinh 2 z = 1.

The Logarithmic Function We now define the logarithm of a complex number, extending the familiar notion of the logarithm of a real positive number. It is convenient to work with the extension of the natural logarithm, ln x , and we will assume this choice of base. However we will reserve usage of the notation ln x to the case where the x is a real number.

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18.6  Some Complex Functions

18-29

If x is a positive real number, then the logarithm arises as the inverse of the exponential function, so that ln x = y if and only if x = e y . We wish to have the complex logarithmic function be the inverse of the exponential function in a similar manner. However, when we replace x with an arbitrary complex number z, we are faced with a problem because e z + 2 πi = e z e 2 πi = e z ( cos 2π + i sin 2π ) = e z . The function w = e z is not one-to-one, and therefore does not have an inverse if the domain of the exponential function is the entire complex plane. We ran into this type of problem when we defined inverse functions, such as arcsin x , because sin x is not oneto-one. We handled the issue by restricting the domain of sin x to an interval on which it is one-to-one, namely [ −π 2, π 2 ]. We do the same now, restricting the domain of e z to a subset on which it is one-to-one. This allows for defining an unambiguous inverse for e z , which we call Log z, written with a capital “L.” While there are many possibilities for such a subset of the complex plane, we make a choice that gives what is called the principal branch of the logarithm function. Other choices of notation, such as Ln z , are sometimes used for this function. The idea behind the definition of Log z is most easily understood if we represent a complex number using its polar coordinate form, z = re iθ , where r = z and θ = arg( z ). Now recall the property ln ab = ln a + ln b that is satisfied by the real logarithmic function, whose domain is the positive real numbers. If we retain this property, we are led to define the complex logarithm of a nonzero complex number z to be

log z = ln r + iθ.(36)

There is still an ambiguity to resolve, because θ = arg( z ) is only defined up to multiples of 2π. The description of the complex number z could also have been given as z = re i( θ + 2 π ), or z = re i( θ + 4 π ), or in general z = re i( θ + 2 nπ ),

n = 0, ±1, ± 2, … .

The definition in Equation (36) could be applied to any of these choices that describe z, leading to infinitely many possible values for the logarithm of z, with any pair differing by a multiple of 2πi. To pin down one choice, we require that the angle θ used to specify z should be chosen to satisfy −π < θ ≤ π. (We always take r > 0, which implies that 0 is not in the domain of the logarithmic function.) We call the resulting complex number ln r + iθ, −π < θ ≤ π , the principal value of the complex logarithmic function. The definition of the complex logarithmic function now becomes unambiguous, Log z = ln r + iθ,

If we do not require that θ satisfy −π < θ ≤ π , then we get multiple values for ln r + iθ. We use log z to denote this entire collection of values. So log z denotes all complex numbers w satisfying e w = z. A value of log z other than the principal value differs from it by a multiple of 2πi. Thus all possible values w for the logarithm of z are related to the principal value by

y

w = Log z + 2nπi, 1+i "2

1

Solution  The complex number 1 + i is described with polar coordinates r = θ = π 4 (Figure 18.18). Since −π < π 4 ≤ π , the principal value Log (1 + i ) is x

FIGURE 18.18  The polar description

of 1 + i

M18_HASS5901_15_GE_C18.indd 29

n = 0, ±1, ± 2, … .

EXAMPLE 4   Find all possible values for the logarithm of 1 + i.

1

p 4 0

−π < θ ≤ π.

2,

Log(1 + i ) = ln 2 + ( π 4 ) i. So the possible values w for the logarithm of 1 + i are w = Log(1 + i ) + (2nπ )i = ln 2 + ( π 4 ) i + (2nπ )i,

n = 0, ±1, ± 2, … .

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Chapter 18  Complex Functions

EXERCISES  T 1. Estimate e (1+ i ) sion for e z.

2

18.6 by using the first four terms of the series expan-

2. Express the sines and cosines in the following identity in terms of exponentials and thereby show that sin ( A + B ) = sin( A) cos( B) + cos( A) sin( B) is a consequence of the exponential law e z1 e z 2 = e z1 + z 2 .

11. Show that the real and imaginary parts of w = sin z in Exercise 10 satisfy the Cauchy-Riemann equations. 12. Show that sin ( x + iy ) =

13. Let z = x + iy, w = u + iυ, and w = sin z. Show that for any nonzero real constant b, the line segment {−π < x ≤ π , y = b } in the z-plane maps to the ellipse u2 υ2 + =1 cosh 2 b sinh 2 b

3. Establish the following differentiation formulas by differentiating the appropriate series term-by-term. d sin z de z a. = e z b. = cos z dz dz d Arctan z d cos z 1 c. = − sin z d. = dz 1 + z2 dz 4. The derivative of a complex-valued function f ( x ) = u( x ) + iυ( x ) whose input is a real variable x is defined to be f ′( x ) = u ′( x ) + iυ ′( x ). a. Show that y = e iω x and y = e −iω x are solutions of the differential equation d 2y dx 2

+ w 2 y = 0.

b. Show that y = e ( a + ib )x and y = e ( a−ib ) x are solutions of the differential equation d 2y dy + 2a + ( a 2 + b 2 ) y = 0. dx 2 dx c. Assuming that a and b are real in part 4(b), show that both

sin 2 x + sinh 2 y if x and y are real.

in the w-plane. 14. Show that the only complex roots z = x + iy (x and y real) of the equation sin z = 0 are at points on the real axis ( y = 0 ) at which sin x = 0. 15. Show that cosh ( x + iy ) = cosh x cos y + i sinh x sin y. 16. Show that cosh ( x + iy ) =

cos 2 y + sinh 2 x if x and y are real.

17. If x and y are real and cosh ( x + iy ) = 0, show that x = 0 and cos y = 0. 18. What is the image in the w-plane of a line x = constant in the z-plane if w = e z , b. w = sin z ? Sketch. a. 19. The integral of a complex-valued function of a real variable x, f ( x ) = u( x ) + iυ( x ), is defined to be

∫

f ( x ) dx =

∫ u( x ) dx + i ∫ υ( x ) dx.

Evaluate

∫ e a+ib x dx (

)

and by equating real and imaginary parts, obtain formulas for

∫ e ax cos bx dx and ∫ e ax sin bx dx.

e ax cos bx = Re ( e ( a + ib )x )

Logarithms

e ax sin bx = Im ( e ( a + ib )x )

20. Find the principal value of log z for each of the following complex numbers z:

and are solutions of the given differential equation. 5. Find values of a such that y = equation

e ax

is a solution of the differential

d 2y dy a. 2 + 2 + 5 y = 0, dx dx d4y

b. 4 + 4 y = 0, dx

2 − 2i, b. 3 + i, a. −4, d. 4, c. 1+i . 2i, f. e. 1−i 21. Find all values of log z for each of the following complex numbers z: 2, b. −2, a.

d 3y c. 3 − 8 y = 0. dx

2i, d. −2i, c.

6. Show that the point z (t ) traverses a unit circle with angular velocity ω if z (t ) = e iωt and ω is a real constant.

22. Express w = tan z in terms of exponentials; then solve for z in terms of w and thereby show that

i− e.

3.

7. If x and y are real, show that e x + iy = e x. 8. Show that the real and imaginary parts of w = e z satisfy the Cauchy-Riemann Equations. 9. Show using the relevant series that sin(iz ) = i sinh z , b. cos(iz ) = cosh z. a. 10. Show using the results of Exercises 2 and 9 that sin ( x + iy ) = sin x cosh y + i cos x sinh y, and find a value of z such that sin z = 2.

M18_HASS5901_15_GE_C18.indd 30

arctan w =

1 1 + iw log . 2i 1 − iw

23. a. Show that arcsin z = −i log( iz +

1 − z 2 ).

b. Use part (a) to find a complex number z satisfying sin z = 3. Note that since sin z = sin ( z + 2π ) there are infinitely many choices for such a z.

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18.7  Conformal Maps

18-31

y

24. Sketch the images in the w-plane of the following sets in the z-plane, under the mapping function w = ln z: a. z = constant, −π < arg z < π , b. arg z = constant, 0 < z < ∞. 25. Figure 18.19 shows the image under the complex map f ( z ) = Log z of a square with vertices at 1, e, 1 + ( e − 1)i, and e + ( e − 1)i. a. Find the complex number to which each of the four vertices is mapped. b. Find parameterizations of the four curves forming the boundary of the image.

x

0

FIGURE 18.19  The map w = Log z maps a square to this region.

18.7 Conformal Maps In this section we will introduce the concept of a conformal map. Informally, these maps are functions on open regions in the plane that preserve the angles between curves. There is a close connection between this idea and properties of complex functions. It turns out that complex differentiable functions with nonzero derivative are conformal. This geometric concept plays an important role in many applications. To measure the angle between two differentiable curves that meet at a point z 0 , we take the angle between their tangent vectors, as we did in Chapter 12. We now look at what a map f ( x , y ) = ( u ( x , y ) , f ( x , y )) does to such angles. We can compare the angle between two curves a(t ) and b(t ) intersecting at z 0 with the angle between the two curves c(t ) = f (a(t )) and d (t ) = f (b(t )) that they are sent to by f , as in Figure 18.20. When these angles are unchanged for all choices of curves, we say that the map f is conformal. y

v c(t) = f(a(t)) a(t)

u1

f

b(t)

u2

d(t) = f(b(t)) x z -plane

u w -plane

FIGURE 18.20  A conformal map preserves angles, meaning that θ1 = θ 2

for any pair of curves in the domain of f .

Intuitively, conformal maps act at infinitesimal scales by stretching or contracting equally in all directions, and by rotating. Since angles are preserved, a conformal map f rotates the tangent vector of all curves through a point z 0 by the same angle. The length of a tangent vector can be changed by a conformal map, but any two tangent vectors are stretched or compressed by the same factor. It is a rather amazing fact, shown in Theorem 3, that all complex differentiable maps have this powerful and useful geometric property wherever their derivative is nonzero. Figure 18.21 shows the square domain A = {( x , y ): 0 < x < 1, 0 < y < 1} , which is mapped by f ( z ) = z 2 to the domain B in the complex w-plane. Blue stripes in A are sent by f to blue stripes in B, and red stripes to red. Notice that the edges of the red and blue stripes in A meet at right angles, as do their images in B. The angle between these curves is preserved. Even though the region A itself is stretched and distorted when mapped to B, the angle between any two curves in A is the same as the angle between the images of these curves in B.

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18-32

Chapter 18  Complex Functions

y

y 2i 1+i

i

B A

0

1

x

−1

0

1

x

FIGURE 18.21  The map f ( z ) = z 2 takes the perpendicular hori-

zontal and vertical lines in A to perpendicular curves in B. It preserves the angle between any pair of curves, at all points other than z = 0.

Cartographers making maps of the earth for navigational purposes made use of conformal maps. The most commonly used method for displaying a map of the earth, the Mercator Projection, shown in Figure 18.22, is conformal. It maps the surface of the earth, cut open from the north to the south pole along a longitude, to a rectangle in the plane. To set a course to a distant port, a navigator could draw a straight line on the Mercator map between their location and the destination. This would give a heading to follow, such as south 10 degrees west, or due east. If they then sailed their ship in that direction, they would follow the line drawn on the map till they reached their goal. This was very useful before the days of GPS. Note that while it preserves angles, the Mercator conformal map distorts lengths, and can give a distorted impression of the size of two regions. Antarctica and Greenland appear large relative to regions near the equator in this representation.

Qimono/Pixabay

NASA

FIGURE 18.22  The Mercator projection preserves angles, but distorts

area. Regions near the poles are highly stretched.

Since they preserve angles, conformal maps give an accurate representation of shapes at a small scale. This makes them useful for computer graphics, shape recognition, alignment algorithms, and many other applications. Conformal maps also have many uses in solving problems in electromagnetism and heat distribution. To compute angles between two curves, we use their tangent vectors. If a(t ) = ( x (t ), y(t ) ) is a curve with a(0) = ( x 0 , y 0 ) and a′(0) ≠ 0, the vector a′(0) = x ′(0), y′(0) gives the tangent vector. A second curve b(t ) has tangent vector b′(0) at the point ( x 0 , y 0 ). The angle between these two vectors at z 0 gives the angle between the two curves. A complex-valued map f whose domain includes a disk around z 0 = x 0 + iy 0 maps a(t ) to a new curve, c(t ) = f (a(t )), with tangent vector c′(0) at f ( z 0 ). Similarly d (t ) = f (b(t )) has tangent vector d ′(0) at f ( z 0 ). We can compare the angle between the vectors c′(0) and d ′(0) at f ( z 0 ) with the angle between a′(0) and b′(0) at z 0 . In general there is no reason to expect these angles to be the same. However when f is a complex differentiable map and f ′( z 0 ) ≠ 0, then it turns out that these angles are indeed equal, no matter which point z 0 and which curves a and b are considered. Theorem 3 explains why this is true.

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18.7  Conformal Maps

18-33

THEOREM 3  Suppose that a map f = u + iυ is complex differentiable and

f ′( z ) ≠ 0 on a domain A. Then f is conformal.

Proof   We will show that f preserves the angle between the tangent vectors of any pair of differentiable curves. Suppose that a(t ) and b(t ) are two differentiable curves with a(0) = b(0) = z 0 and that their tangent vectors meet at an angle α. Take a r to be a point on a(t ), t > 0, that has distance r from z 0 , and pick br to be a point on b(t ) at the same distance r from z 0 . Then we can write a r − z 0 = re iθ1 and br − z 0 = re iθ2 . The quotient satisfies ar − z 0 = e i( θ1 −θ2 ). br − z 0 As r → 0, the angle θ1 − θ 2 approaches the angle α between the tangent vectors to the two curves,  a − z 0  α = lim arg  r . r→0  br − z 0  Set c(t ) and d (t ) to be the images of a(t ) and b(t ) under f so that this pair of curves intersects at f ( z 0 ). The angle β at which c(t ) and d (t ) intersect at f ( z 0 ) is  f (a r ) − f ( z 0 )  β = lim arg   r→0  f (br ) − f ( z 0 )  as this limit gives the angle between tangent vectors to the two image curves. We can rewrite this limit as  f (a r ) − f ( z 0 )  β = lim arg   r→0  f (br ) − f ( z 0 )    f (a r ) −    ar − = lim arg    f (br ) − r→0     br −

f ( z 0 )        a r − z 0  z0  .  f ( z 0 )    br − z 0     z0

As r → 0 we have lim

r→0

f (a r ) − f ( z 0 ) = f ′( z 0 ) and ar − z 0

f (b r ) − f ( z 0 ) = f ′( z 0 ) br − z 0

since the complex derivative can be computed as we approach z 0 from any direction. We assumed f ′( z 0 ) ≠ 0, so we can cancel these terms. Then  a − z 0  β = lim arg  r . r→0  br − z 0  This is the same as the formula giving α, so the two angles are equal.

EXAMPLE 1   Conformal maps of a square We consider the square { 0 < x < 1, 0 < y < 1} , shown in Figure 18.23a, and its image under three conformal maps: The image under the map f ( z ) = sin z is shown in Figure 18.23b. The image under the map f ( z ) = cos z is shown in Figure 18.23c. The image under the map f ( z ) = Arctan z is shown in Figure 18.23d.

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18-34

Chapter 18  Complex Functions

y

y

i i

0

1

x

0

1

(a)

x

(b)

y

y 2i 0.5

1

0

1.5

2

x i

−0.5i −i

0 y

(c)

0.5

1

x

(d)

FIGURE 18.23  Conformal images of (a) a square region divided into ten

10i

vertical stripes under the conformal maps (b) sin z, (c) cos z, and (d) Arctan z. 5i −2

−1

0

1

2

x

−5i −10i (a)

Only part of the image of the vertical green stripe in the square is shown in Fig­ ure 18.23d. As z → i the values of Arctan z diverge, and Arctan z is not defined at z = i. The image of the green stripe is not bounded. This illustrates that conformal maps can stretch bounded domains such as the square to unbounded domains. Since these maps are conformal, the images of perpendicular, vertical, and horizontal lines in the square remain perpendicular in the image. In fact, all angles between pairs of curves are preserved by these conformal maps.

y 4i 2i −4

−2

0

2

4

x

In Example 1 we see three regions in the plane that are the images of a square under a conformal map. It turns out that these types of region are not so special, and any simply connected open region in the plane is the image of an open square under a conformal map, with one exception. The exception is the entire complex plane. The existence of a conformal map between any other pair of simply connected open regions is given by the Riemann Mapping Theorem, a deep result from the theory of conformal maps.

−2i −4i (b)

FIGURE 18.24.  (a) A rectangle divided

into 20 vertical stripes and (b) its image under the conformal map f ( z ) = sin z

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EXAMPLE 2   The sine function f ( z ) = sin( z ) has nonzero derivative when restricted to the rectangle {−1.5 < x < 1.5, −10 < y < 10 } , shown in Figure 18.24a. The image of this rectangle lying within the square {−2.5 < x < 2.5, −2.5 < y < 2.5 } is shown in Figure 18.24b. If the rectangle is enlarged to the infinite rectangle {−π 2 < x < π 2, −∞ < y < ∞ } , then its image under the sine function fills up the entire complex plane except for two “slits”, the intervals (−∞, −1 ] and [ 1, ∞ ) on the x-axis.

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Chapter 18  Questions to Guide Your Review

EXERCISES 

18-35

18.7

1. Does the map f ( z ) = z 2 preserve the angles between the intersecting lines y = 0 and y = x ? How does this fit into Theorem 3? 2. For which z is the map f ( z ) = e z conformal? 3. For which z is the map f ( z ) = z 3 conformal? What happens to angles at points where it is not conformal? 4. For the map f ( x , y ) = ( 2 x , y ) , compute the angle of intersection of the images under f of the horizontal line { x = 1} and the vertical line { y = 1}. Is this map conformal?

7. Find the angle by which a curve through the point z 0 is rotated by the map f ( z ) = z 2 if a. z 0 = 1, b. z 0 = i, c. z 0 = 1 + i, d. z 0 = −i, e. z 0 = 2i, and f. z 0 = −1. 8. The conformal factor of f at z 0 is the amount by which it stretches tangent vectors. Where is the conformal factor of f ( z ) = z 2 smaller than one? 9. Given two complex numbers a and b, the map f ( z ) = az + b rotates all vectors by the same angle. What is this angle?

5. Is the map f ( x , y ) = ( x , x + y ) conformal? If not, find two intersecting curves whose angles are not preserved by f .

10. Given two complex numbers a and b, the map f ( z ) = az + b, stretches all vectors by the same factor. What is this factor?

6. Show that the map f ( z ) = ( z − i ) ( z + i ) is a conformal map from the upper half-plane {( x , y ):  y > 0 } to the open unit disk {( x , y ):  x 2 + y 2 < 0 }. A deep result, the Riemann Mapping Theorem, shows that this is not unusual. Any simply connected region in the plane can be conformally mapped to the open unit disk, with the exception of the entire complex plane.

11. The most basic conformal maps have the form f ( z ) = az + b. Show that such a map is conformal when a ≠ 0. When a is not real show that f has a fixed point z 0 satisfying f ( z 0 ) = z 0 . If a = 1 and b ≠ 0, show that f does not have a fixed point.

CHAPTER 18

12. What is the fixed point in Exercise 11 when a. f ( z ) = 3z + i, b. f ( z ) = (1 + i ) z + 4 − 2i ?

Questions to Guide Your Review

1. Define the set of complex numbers. 2. Define, for complex numbers, the concepts of equality, addition, multiplication, and division. 3. A set of numbers is said to be closed under an operation ⊗ if a ⊗ b is in the set whenever a and b are. Is the set of complex numbers closed under the operations of addition, subtraction, multiplication, division, and raising to a power (including complex exponents)? (Be careful to consider all complex numbers in formulating your answer, including 0.) 4. How may the complex number a + ib be represented graphically in the complex plane? 5. Illustrate in the complex plane how the absolute values and arguments of the product and quotient of two complex numbers z 1 and z 2 are related to the absolute values and the arguments of z 1 and z 2 . 6. State De Moivre’s theorem and explain how to find the n distinct complex nth roots of a nonzero complex number a + ib. 7. Prove that if n is an even positive integer, then cos nθ may be expressed as a polynomial with integer coefficients, in cos 2 θ. (Example: cos 2θ = 2 cos 2 θ − 1.)

10. If z is a complex number such that z = z , what else can you conclude about z? 11. If z is a complex number such that z = −z , what else can you conclude about z? 12. Describe the sets of complex numbers z satisfying the following equalities or inequalities, where z 0 = a + ib is a given complex number and k is a positive real constant. z − z 0 = k, a. b. z − z 0 < k, c. z − z 0 > k. 13. Define the concept of continuity for a function of a complex variable z. Define the derivative of f at z = a + ib. 14. What are the Cauchy-Riemann Equations and when are they known to be satisfied? 15. Define the convergence of a series of complex numbers. 16. How do we define e z , sin z , cos z , and tan z for a complex number z = a + ib? 17. What is a conformal map and what is its relationship to a complex differentiable map?

8. Using vectors in the complex plane, illustrate geometrically how to find the n complex nth roots of any complex number a + ib. 9. Using the complex plane, illustrate geometrically how the conjugate and the reciprocal of a complex number a + ib are related to that number.

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18-36

Chapter 18  Complex Functions

Additional and Advanced Exercises

CHAPTER 18 1. Let z = 2 − 2i.

11. Let f ( z ) = z (the conjugate of z).

a. Plot the points z , z , and z 2.

a. Study the behavior of the quotient

b. Plot the three complex cube roots of z 2 (that is, z 2 3 ). 2. Express each of the following complex numbers in the form re iθ with r ≥ 0 and −π < θ ≤ π. Sketch 2

(1 + i )(1 − i 3 ) , a. 3 b. 2 − 2i (three answers).

3. Express the four 4th roots of −16i in the form a + bi. 4. a. Solve the equation

z4

+ 16 = 0, obtaining four distinct roots.

b. Express the five roots of the equation z 5 + 32 = 0 in polar form. 5. a. Find all complex numbers z such that z 4 + 1 + i 3 = 0. b. Express the number e 2+ πi 4 in the form a + bi. 6. Plot the complex number 2 − 2 3i in the complex plane and find a. its two square roots, b. the principal value of its logarithm. 7. Find the values of r and θ such that 3 + 4i = re iθ. 8. If z = 3 − 3i, find all values of log z. 9. Find a complex number a + ib that satisfies the equation e a + ib = 1 − i 3. 10. Express the following in the form a + bi: ( − 1 − i )1 3 (write down all the roots). a.

log( 3 + i 3 ). (There are infinitely many possible answers. b. Give an answer that expresses all of these.) c. e 2+ πi.

M18_HASS5901_15_GE_C18.indd 36

f (z) − f (z1 ) z − z1 when z → z 1 along straight lines of slope m. b. From the results of part (a) what can you conclude about the derivative of f ( z )? ∞

∑ a n z n and f ( z ) =

12. If f ( z ) = a n are real. 13. If f ( z ) =

f ( z ), for z < R, show that the

n=0 ∞

∑ anz n

and f ( z ) = f ( z ), for z < R, show that

n=0

f ( z ) is a constant. 14. Show that

dn ( cos θ + i sin θ ) = i n ( cos θ + i sin θ ). dθ n

15. Draw the set of points in the complex plane that satisfy Re( z ) > 0, b. Im( z ) = 2, ( z − i ) ≤ 0, a. z−i c. < 1, d. e z ≥ 1, z+i e. sin z ≤ 1. 16. If u ( x , y ) and υ ( x , y ) are the real and imaginary parts of a complex differentiable function of z = x + iy, show that the family of curves u = constant is orthogonal (perpendicular) to the family υ = constant at every point of intersection where f ′( z ) ≠ 0. 17. Given that z = e Log z and ( e z1 ) z 2 = e z1z 2 , how would you compute z w for complex numbers z and w? Compute z w for z = 1 + i, w = 2i.

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ANSWERS TO ODD-NUMBERED EXERCISES

Chapter 18

SECTION 18.5, p. 18-25

SECTION 18.1, pp. 18-9–18-10

1. (a) 14 + 8i  (b)  −7 − 4i (c)  −5i 3. (a) By reflecting z across the real axis (b) By reflecting z across the imaginary axis (c) By reflecting z through the origin, or equivalently, rotating it by 180 degrees. (d) By reflecting z across the real axis and then multiplying the length of the vector by 1 z 2 5. (a) Points on the circle x 2 + y 2 = 4 (b) Points inside the circle x 2 + y 2 = 4 (c) Points outside the circle x 2 + y 2 = 4 7. Points on a circle of radius 1, center ( −1, 0 ) 9. Points on the line y = −x   11.  4e 2 πi 3   13.  1e 2 πi 3 15. cos 4 θ − 6 cos 2 θ sin 2 θ + sin 4 θ 1 3 17. 1, − ± i   19.  2i, − 3 − i, 3 − i 2 2 6 2 6 2 ± i, − ± i  23.  1 ± 3i, −1 ± 3i 21. 2 2 2 2 SECTION 18.2, p. 18-17

1. (a) w = 1, 0 ≤ arg w ≤ 2π (b) w = 4, π ≤ arg w ≤ 2π (c) arg w = π 2 (d) w < 1, −2π < arg w ≤ 0 (e) The u-axis, u ≥ 0 (f) The u-axis, u ≤ 0 SECTION 18.3, p. 18-20

1.

−2

( z 0 − 1) 2

= +2

z0 , which gives two values; 2 z0 + 1 1 9π 1 π exp i and 4 exp i . 4 8 8 2 2

3. f ′( z 0 ) =

( )

( )

SECTION 18.4, p. 18-22

1. u = x 2 − y 2, υ = 2 xy 3. u = x 4 − 6 x 2 y 2 + y 4 , υ = 4 x 3 y − 4 xy 3

1. z < 1  3. z − 1 < 1  5. z + 1 < 2 9. 0.208 + 0.298i SECTION 18.6, pp. 18-30–18-31

1. 1.59 + 1.32i 5. (a) −1 ± 2i  (b)  1 ± i, −1 ± i (c)  2, −1 ± i 3 ax e 19. ∫ e az cos bx dx = 2 ( a cos bx + b sin bx ) + C a + b2 e ax ∫ e ax sin bx dx = a 2 + b 2 ( a sin bx − b cos bx ) + C 21. (a) ln 2 + 2nπi (b) ln 2 + ( 2n + 1 ) πi (c) ln 2 + ( 2n + 12 ) πi (d) ln 2 + ( 2n − 12 ) πi (e) ln 2 + ( 2n + 56 ) πi, ( n = 0, ±1, ± 2, …) π 23. (b) + 2nπ − i ln ( 3 + 2 2 ) 2

(

)

SECTION 18.7, p. 18-35

1. No. The angle between the lines is 90 °, but the angle between their images is 180 ° . This is the case in Theorem 3 where f ′( z 0 ) = 0. 3. All z ≠ 0 since f ′( z ) ≠ 0 except at z = 0 and f is analytic. At z = 0 angles multiply by 3. 5. No. The angle between y = 0 and x = 0 is π 2. These two lines are taken by f to y = x and x = 0, which intersect at an angle of π 4. 7. (a) 0 (b)  π 2 (c)  π 4 (d) π 3 2 (e)  π 2 (f)  π 9. arg a  11. z 0 = b ( 1 − a ) ADDITIONAL AND ADVANCED EXERCISES, p. 18-36

3. (a) ±

(

2+

2 −i 2−

) (

2 ,±

2−

(b) sin −1 5 = 2nπ + π 2 − i ln ( 5 + ( n = 0, ±1, ± 2, …)

5. (a) 2 1 4 e iπ( 2+ 3n ) 6 ( n = 0, 1, 2, 3 ) (b)  e2 7. r = 5, θ =

tan −1 43   9.  ln

2 +i 2+

2

24 ) , 2 + i( e 2

)

2)

2 − iπ 3

11. (b) Derivative doesn’t exist.

18-37

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19

Fourier Series and Wavelets

BIOGRAPHY Jean-Baptiste Joseph Fourier (1768–1830) Fourier was a French mathematician, who as a young boy aspired to become an army officer. He was denied that opportunity for military service and, therefore, switched his passion to mathematics. To know more, visit the companion Website.

In Section 9.8 we saw that if a function f has derivatives f ( k ) of order k = 1, 2, …,  n, then we can create a polynomial function Pn that approximates f near a chosen point. This Taylor polynomial gives a close fit to f near that particular point, but the error in the approximation can be large at points that are far away. In this chapter we will introduce Fourier series, which provide a way to approximate a function over an interval, rather than near a single point. Instead of involving polynomials, a Fourier series approximation is a sum of multiples of the functions sin kx and cos kx. In a sense that we will describe, a Fourier series approximation minimizes an error that is distributed over an interval. Additionally, while Taylor polynomials cannot be used to approximate discontinuous functions, Fourier series approximations will often be applicable to such functions. Fourier series, introduced by Joseph Fourier, are well-suited to analyzing periodic functions, such as radio signals and alternating currents. They can also be used to solve heat transfer problems; indeed, this was the problem that motivated Fourier. A large field of mathematics called harmonic analysis began with Fourier series. Fourier series and harmonic analysis have applications to electrical engineering, acoustics, optics, signal processing, image processing, quantum mechanics, econometrics, and many other areas in science and engineering.

19.1 Periodic Functions

For some periodic functions, such as the constant function f ( x ) = 1, there is no smallest P > 0 such that f ( x + P ) = f ( x ) for every x.

Let f be a function whose domain is the real line. We say that f is periodic if there is a positive real number P such that f ( x + P ) = f ( x ) for every real number x. In Section 1.3 we defined the period of f to be the smallest such number P. In this chapter we will call the smallest P the fundamental period of f , and say that any other positive number P that satisfies f ( x + P ) = f ( x ) for every x is a period for f . We will focus in this chapter on functions with period P = 2π , but it is not difficult to adapt to other periods. EXAMPLE 1 (a) The function f ( x ) = sin x is periodic with period 2π because for every real number x we have f ( x + 2π ) = sin ( x + 2π ) = sin x = f ( x ). Likewise, cos x is periodic with period 2π , and 2π is the fundamental period for both functions.

19-1

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19-2

Chapter 19  Fourier Series and Wavelets

(b) Let k be any positive integer. Then the function g( x ) = sin kx has period 2π k because if x is any real number, then

(

g x+

)

(

)

2π 2π  = sin ( kx + 2π ) = sin kx = g( x ). = sin  k x +  k k 

The number 2π k is the fundamental period of g, but it is not the only period. Any positive integer multiple of 2π k is also a period. For example, 2π is a period for g because for every x we have g ( x + 2π ) = sin[ k ( x + 2π ) ] = sin ( kx + 2π k ) = sin kx = g( x ). Similarly, g( x ) = cos kx has fundamental period 2π k , and it is also periodic with period 2π. ( c) We will let 1 denote the constant function that takes the value 1 at every point: 1( x ) = 1

for every real number  x.

The function 1 is periodic, and every positive real number P is a period. There is no smallest period P, so 1 does not have a fundamental period. We will use the functions sin kx and cos kx , along with the constant function 1, to define the Fourier series approximation of a function. Each of these functions is periodic with period 2π. The oscillatory nature of these functions is one reason that they are well-suited for forming approximations. We say that sin kx and cos kx have frequency k, because their graphs show k complete oscillations within the interval [ 0, 2π ] (Figure 19.1). Since cos 0 x = 1 = 1( x ) for every x, we say that the constant function 1 has frequency zero. We introduce a name for this set of functions. y 1 0.5 p 2

−0.5

p

3p 2

2p

x

−1 (a) Graph of cos 2x over the interval [0, 2p] y 1 0.5

−0.5

p 2

p

3p 2

2p

x

−1 (b) Graph of sin 7x over the interval [0, 2p]

FIGURE 19.1  Two functions in the trigonometric

system.

DEFINITION  The collection of functions

1, cos x , sin x , cos 2 x , sin 2 x  cos 3 x , sin 3 x , … is called the trigonometric system.

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19.2  Summing Sines and Cosines

EXERCISES 

19-3

19.1

Periodic Functions

In Exercises 1–6, determine whether the given function is periodic. If it is, find its fundamental period P. 1. 1 + sin x

2. x

3. x

4. e cos x

5. sin 2 x cos x

6. sin x 2

Theory and Examples

7. Show that if f ( x ) and g( x ) are both periodic with the same period P, then the following functions are also periodic with period P.

c. f ( x ) g( x )

d.  e f ( x )

8. Periodic extension  Suppose that the values of f ( x ) are specified for 0 ≤ x < 2π. Then for each 0 ≤ x < 2π and every nonzero integer k, define f ( x + 2 k π ) = f ( x ). Show that this extends the definition of f to the entire real line, and that the resulting function is periodic with period 2π.

f ( x ) + g( x ) a.

b.  cf ( x ), where c is a real number

19.2 Summing Sines and Cosines We discussed Taylor series and Taylor polynomials in Section 9.8. If a function f has derivatives of order k for k = 1, 2, …,  n then the Taylor polynomial of order n at x = 0 is Pn ( x ) = a 0 + a1 x + a 2 x 2 +  + a n x n

 where  a k =

f ( k ) (0) . k!

By choosing the coefficients a k in this way we obtain a polynomial Pn that approximates f near the point x = 0. Taylor series are based on the functions 1,  x ,  x 2 ,  x 3 , …, while Fourier series are instead based on sines and cosines. Just as a Taylor polynomial is a combination of finitely many of the functions 1,  x ,  x 2 ,  x 3 , …, we seek a combination of finitely many elements of the trigonometric system that will approximate a function. We refer to such a finite combination of sines and cosines as a trigonometric polynomial. EXAMPLE 1   Define a periodic function f by letting its values on the interval [ 0, 2π ] be  x − π , f ( x ) =   −π , 



0 ≤ x < 2π , x = 2π ,

(1)

and then let those values repeat every 2π units (Figure 19.2). This gives us a function f that is periodic on the real line with period 2π. Observe that f has a jump discontinuity at x = 2π k for every integer k. y p 1 −2p

−p

−1

p

2p

3p

4p

x

−p

FIGURE 19.2  Graph of the periodic function f defined in

Equation (1) over the interval [ −2π ,  4 π ]

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Chapter 19  Fourier Series and Wavelets

We would like to approximate f by a trigonometric polynomial, which is a sum of finitely many multiples of the functions in the trigonometric system (the real number that we multiply 1 or sin kx or cos kx by is called the amplitude of that term). We cannot do a very good job using only one function from the trigonometric system (see Figure 19.3a, which compares the graphs of f and −2 sin x ). However, we can make a better approximation by using more elements of the trigonometric system. Figure 19.3b shows a trigonometric polynomial that is a sum of two scaled sines, while Figure 19.3c shows that we can do a better approximation using a sum of three scaled sines. As seen in Figure 19.3d, a certain combination of 9 terms gives us a trigonometric polynomial that appears to be a good approximation over most of the domain, though still not close at points near integer multiples of 2π. If we use yet more terms, we can improve the approximation. In particular, the trigonometric polynomial 2 2 sin 3 x −  − sin nx (2) 3 n

p n ( x ) = −2 sin x − sin 2 x −



is a good approximation to f using a sum of n sines. We will explain below how we knew what amplitudes to choose for the sines in this sum, and why no cosine terms are used. y

y p

p 1 −2p

−p

p

−1

2p

3p

4p

x

1 −2p

−p

−1

4p

x

y

y p

p 1 −1

3p

(b) f (x) and p2(x) = −2 sin x − sin 2x

(a) f (x) and p1(x) = −2 sin x

−p

2p

−p

−p

−2p

p

p

2p

3p

4p

x

1 −2p

−p

−1

p

2p

3p

4p

x

−p

−p 2

(c) f (x) and p3(x) = −2 sin x − sin 2x − −sin 3x 3

2

2

(d) f (x) and p9(x) = −2 sin x − sin 2x − −sin 3x − . . . − −sin 9x 3 9

FIGURE 19.3  Trigonometric polynomial approximations to the function f defined in Equation (1) over the interval [ −2π , 4 π ], using sums of scaled sines.

We see in Example 1 that we can closely approximate a particular function over most of the interval [ 0, 2π ] by using sines and cosines from the trigonometric system (for that function we only needed sines). We would like to approximate other functions in a similar way. Given a function f that has period 2π , we seek a function

p n ( x ) = a 0 + a1 cos x + a 2 cos 2 x +  + a n cos nx + b1 sin x + b 2 sin 2 x +  + b n sin nx (3) that approximates f across the interval [ 0, 2π ]. We refer to p n as a trigonometric polynomial (Exercises 5–18 will explain why we use the word “polynomial” in this context). We say that a k is the amplitude of the term a k cos kx , and similarly b k is the amplitude of the term b k sin kx. Since a 0 = a 0  cos 0 x , this number is the amplitude of the zero frequency term. Recalling the constant function 1, which satisfies 1( x ) = 1 for every x, we have a 0 = a 0 1( x ). Thus we also say that a 0 is the amplitude of the constant term in the sum in Equation (3).

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We can write Equation (3) more succinctly using sigma notation: pn (x) = a 0 +



n

n

k =1

k =1

∑ a k cos kx + ∑ b k sin kx. (4)

We will show how to choose the amplitudes a k and b k so that p n is the “best approximation” to f that is a sum of constant multiples of the constant function, the n cosines cos x , cos 2 x , …, cos nx and the n sines sin x , sin 2 x , …, sin nx. To do this, we will demonstrate how to approximate vectors in three and higher dimensions in Section 19.3, and then show in Section 19.4 that ideas that work for approximating vectors by other vectors also work for functions.

EXERCISES 

19.2

Theory and Examples

1. Show that a function of the form p n ( x ) =

n

∑ b k sin kx

is odd,

k =1

that is, p n (−x ) = − p n ( x ) for every x.

2. Show that a function of the form p n ( x ) = a 0 + even, that is, p n (−x ) = p n ( x ) for every x.

which are derived in Chapter 18, and we will use the fact that ( z m ) n = z mn . These exercises will show that trigonometric polynomials equal “generalized” polynomials a n z n +  + a 1 z + a 0 + a − 1 z − 1 +  a −n z −n

n

∑ a k cos kx is k =1

Sums of Sines and Cosines

In Exercises 3 and 4, the values of a 0 ,  a k , and b k that define a trigonon n metric polynomial p n ( x ) = a 0 + ∑ k =1 a k cos kx + ∑ k =1 b k sin kx are given. Use a CAS to plot p n for various values of n over the domain [ −2π , 4 π ]. Does it appear that p n ( x ) is converging to some limit function f ( x ) as n increases? If so, what function? Is the convergence “fast” (the limit is apparent even for small values of n) or “slow” (you must use a large value of n before the graph of p n appears to be close to the graph of f )? π 4 ,  a 2 k = 0,  b k = 0 T 3. a 0 = ,  a 2 k −1 = 2 ( 2 k − 1) 2 π 2 1 ,  a 2 k = 0,  b k = 0 T 4. a 0 = ,  a 2 k −1 = (−1) k ( 2 k − 1) π 2

that use both positive and negative powers of z = e ix with complex numbers a k . Prove the identities stated in Exercises 5–18. We assume that z = e ix where x is a real number. 5. z 0 = 1 6. z + 1 + i = (1 + cos x ) + i (1 + sin x ) 7. 2 z 2 + 3z − 5 = ( 2 cos 2 x + 3 cos x − 5 ) + i( 2 sin 2 x + 3 sin x ) n

8. ∑ a k z k = a 0 + k=0

n

n

k=0

k=0

∑ a k cos kx + i ∑ a k sin kx

9. z −1 = e −ix = cos x − i sin x 10. z −n = e −inx = cos nx − i sin nx , n ≥ 0 11. z + z −1 = 2 cos x 12. z − z −1 = 2i sin x 13. z n + z −n = 2 cos nx , n ≥ 0

Complex Numbers and Trigonometric Polynomials

14. z n − z −n = 2i sin nx , n ≥ 0

The exercises below illustrate the connection between trigonometric polynomials and complex polynomials in the variable z = e ix . Some knowledge of complex numbers is required. In particular, we will use Euler’s Formula e ix = cos x + i sin x and De Moivre’s Theorem,

16. 4iz 2 − 2iz + 2iz −1 − 4iz −2 = −8 sin 2 x + 4 sin x

e inx = ( cos x + i sin x ) n = cos nx + i sin nx , n ≥ 0,

15. 2 z 2 + 3z − 5 + 3z −1 + 2 z −2 = 4 cos 2 x + 6 cos x − 5 17. cos x + 2 sin x − 3 = ( 0.5 − i ) z − 3 + ( 0.5 + i ) z −1 18. 2 cos 3 x − 6 cos 2 x + 8 sin 3 x − 5 = (1 − 4 i ) z 3 − 3 z 2 − 5 − 3 z − 2 + (1 + 4 i ) z − 3

19.3 Vectors and Approximation in Three and More Dimensions Vectors in two and three dimensions were introduced in Chapter 11. If v is a threedimensional vector equal to the vector with initial point at the origin and terminal point ( υ1 ,  υ 2 ,  υ 3 ) , then the component form of v is v = 〈υ1 ,  υ 2 ,  υ 3 〉.

Vectors in Three Dimensions We review some definitions and facts about vectors in three dimensions.

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Chapter 19  Fourier Series and Wavelets

DEFINITION

1. The length of a vector v = 〈υ1 ,  υ 2 ,  υ 3 〉 is v =

υ12 + υ 22 + υ 32 .

If v = 1, then we say that v is a unit vector. 2. The distance between two vectors u = 〈u1 ,  u 2 ,  u 3 〉 and v = 〈υ1 ,  υ 2 ,  υ 3 〉 is ( u 1 − υ1 ) 2 + ( u 2 − υ 2 ) 2 + ( u 3 − υ 3 ) 2 .

u−v =

Observe that u − v = v − u . DEFINITION The dot product of u = 〈u1 ,  u 2 ,  u 3 〉 and v = 〈υ1 ,  υ 2 ,  υ 3 〉 is the

number

u ⋅ v = u1υ1 + u 2 υ 2 + u 3υ 3 . An important fact is that the length squared of a vector equals the dot product of the vector with itself: v

2

= υ12 + υ 22 + υ 32 = v ⋅ v.

Some other properties of dot products and lengths are laid out in Exercises 21–30. Theorem 1 in Section 11.3 shows that the dot product of two nonzero vectors satisfies u ⋅ v = u v cos θ, where θ is the angle between u and v. If u and v are perpendicular then θ = π 2 and therefore u ⋅ v = 0. The converse is also true. That is, if u and v are nonzero vectors with u ⋅ v = u v cos θ = 0, then cos θ = 0 and therefore θ = arccos 0 = π 2. The following definition also allows for one or both of the vectors to be the zero vector. DEFINITION Vectors u and v are orthogonal if u ⋅ v = 0. They are ortho-

normal if u ⋅ v = 0 and additionally u = v = 1.

Often we deal with more than two vectors. We say that three vectors u, v, w are orthogonal if the dot product of any two of them is zero: u ⋅ v = 0,   u ⋅ w = 0, and v ⋅ w = 0. If in addition each of u, v, and w is a unit vector, then we say that the vectors u, v, w are orthonormal. Sometimes we will use the longer but more precise phrasing that { u,  v,  w } is an orthogonal set of vectors (or orthonormal set of vectors). In threedimensional space, it is not possible to have a set of more than three orthonormal vectors, but we will see below that it is possible to have larger collections in higher-dimensional space. We will derive some formulas that relate lengths, distances, and dot products. Statement 3 of the next result is the Pythagorean Theorem for vectors. Exercise 25 shows that the converse of the Pythagorean Theorem holds as well. THEOREM 1 Let u and v be three-dimensional vectors.

1. u − v 2 = u 2 − 2( u ⋅ v ) + v 2 . 2. u + v 2 = u 2 + 2( u ⋅ v ) + v 2 . 3. If u and v are orthogonal, then u + v

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2

= u

2

+ v 2.

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19-7

Proof    Statement 1 follows because u−v

2

= ( u 1 − υ1 ) 2 + ( u 2 − υ 2 ) 2 + ( u 3 − υ 3 ) 2 = u12 − 2u1υ1 + υ12 + u 22 − 2u 2 υ 2 + υ 22 + u 32 − 2u 3υ 3 + υ 32 = ( u12 + u 22 + u 32 ) − 2( u1υ1 + u 2 υ 2 + u 3υ 3 ) + ( υ12 + υ 22 + υ 32 ) = u

2

− 2( u ⋅ v ) + v 2 ,

and the proof of Statement 2 is similar. If u and v are orthogonal then u ⋅ v = 0, so Statement 3 follows from Statement 2. We need one additional definition.

DEFINITION A linear combination of finitely many vectors u 1 ,  u 2 , …,  u n is a sum where we scale each vector u k by a real number a k and add those terms together, giving the vector

a1u 1 + a 2 u 2 +  + a n u n .

EXAMPLE 1  The standard unit vectors in three-dimensional space were introduced in Section 11.2. They are i = 〈1, 0, 0〉,

j = 〈0, 1, 0〉,

k = 〈0, 0, 1〉.

We can form linear combinations of one, two, or three of the standard unit vectors by multiplying the vectors by real numbers and adding them together. For example, using two of the vectors we can create linear combinations such as 2 i + 4 j = 〈2, 0, 0〉 + 〈0, 4, 0〉 = 〈2, 4, 0〉, 5 i − 7k = 〈5, 0, 0〉 − 〈0, 0, 7〉

= 〈5, 0, −7〉,

−3 j − 2k = −〈0, 3, 0〉 − 〈0, 0, 2〉 = 〈0, −3, −2〉. Some linear combinations of all three of the standard unit vectors are 5 i + 2 j − 7k = 〈5, 0, 0〉 + 〈0, 2, 0〉 − 〈0, 0, 7〉 = 〈5, 2, −7〉, π i + 6 j − k = 〈π , 0, 0〉 + 〈0, 6, 0〉 − 〈0, 0, 1〉 = 〈π , 6, −1〉, i−

2 j − 2k = 〈1, 0, 0〉 − 〈0,  2, 0〉 − 〈0, 0, 2〉 = 〈1, − 2, −2〉.

To create a linear combination of one vector, we simply multiply that vector by a real number: 3i = 〈3, 0, 0〉, 5 j = 〈0,  5, 0〉, −9k = 〈0, 0, −9〉. We are not restricted to using just three vectors. For example, one linear combination of u = 〈1, −1, 0〉,   v = 〈2, 0, 2〉,   w = 〈3, 1, 2〉, and y = 〈0, 1, −3〉 is 2u + 3v − w − 3y = 〈2, −2, 0〉 + 〈6, 0, 6〉 − 〈3, 1, 2〉 − 〈0, 3, −9〉 = 〈5, −6, 13〉.

Orthonormal Bases in Three Dimensions Any vector v = 〈υ1 ,  υ 2 ,  υ 3 〉 can be written as a linear combination of the standard unit vectors: v = 〈υ1 ,  υ 2 ,  υ 3 〉 = υ1 〈1, 0, 0〉 + υ 2 〈0, 1, 0〉 + υ 3 〈0, 0, 1〉 = υ1i + υ 2 j + υ 3k.

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Chapter 19  Fourier Series and Wavelets

The dot product between any two of the standard basis vectors is zero: i ⋅ j = 0,   i ⋅ k = 0, and j ⋅ k = 0. Further, each of i, j, and k is a unit vector. Therefore, i, j, k are orthonormal vectors. Now choose any other orthonormal vectors r, s, t. For example, we might take the standard unit vectors i, j, k and rotate them in space around some axis that passes through the origin to obtain new vectors r, s, t. The lengths of the vectors and the angles between them remain unchanged under a rotation, so this process gives us orthonormal vectors. However we obtain them, if we have three orthonormal vectors r, s, t, then we can write any vector v as a linear combination of them (this is proved in texts on linear algebra). That is, if we choose a vector v, then there must be real numbers c1 , c 2 , and c 3 so that v = c1r + c 2 s + c 3t. We now determine these coefficients. Taking the dot product of v with r gives us v ⋅ r = ( c1r + c 2 s + c 3t ) ⋅ r = c1 ( r ⋅ r ) + c 2 ( s ⋅ r ) + c 3 ( t ⋅ r ) = c1 r = c1 .

2

+0+0

v = c1r + c 2 s + c 3t distributive property of the dot product r ⋅ r = r 2, s ⋅ r = 0, t ⋅ r = 0 r has unit length.

A similar calculation shows that c 2 = v ⋅ s and c 3 = v ⋅ t. We state these facts as a theorem. THEOREM 2 Let r, s, t be orthonormal vectors in three-dimensional space. Then every vector v = 〈υ1 ,  υ 2 ,  υ 3 〉 can be written as

v = ( v ⋅ r ) r + ( v ⋅ s ) s + ( v ⋅ t ) t. Using the standard unit vectors, we can write v = 〈υ1 ,  υ 2 ,  υ 3 〉 as v = υ1i + υ 2 j + υ 3k. Theorem 2 tells us that if r, s, t are three orthonormal vectors, then v is also a linear combination of r, s, and t, namely v = c1r + c 2 s + c 3t where c1 = v ⋅ r ,  c 2 = v ⋅ s and c 3 = v ⋅ t. Because every vector v can be written as a unique linear combination of r, s, and t, this collection of vectors is called a basis. Since the vectors are also orthonormal, we use the following name. DEFINITION  A set of three orthonormal vectors { r ,  s,  t } in three-dimensional

space is called an orthonormal basis for three-dimensional space.

z

Approximation Using an Orthonormal Basis in Three Dimensions

y3 k

k v i

j y2 j

y1 i x

y p

FIGURE 19.4  The vector in the xy-plane that is closest to v = υ1i + υ 2 j + υ 3k is p = υ1i + υ 2 j.

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Suppose that v = 〈υ1 ,  υ 2 ,  υ 3 〉 is a vector. Out of all the vectors p = 〈 p1 ,  p 2 , 0〉 in the xy-plane, which one is closest to v? It is the vector p that we obtain by dropping v straight down onto the xy-plane. That is, p = 〈υ1 ,  υ 2 , 0〉 is the vector in the xy-plane that is closest to v (Figure 19.4). Expressing this in terms of linear combinations, if we choose any vector v and write it as v = υ1i + υ 2 j + υ 3k, then the vector in the xy-plane that is closest to v is p = υ1i + υ 2 j. The distance from v to the xy-plane is the length of v − p = υ 3k, which is υ 3 . We saw in Section 5 of Chapter 11 how to find the distance from a point to a generic plane in space. Now we formulate that discussion in terms of orthonormal vectors and best approximation. Suppose that r, s, t are three orthonormal vectors in three-dimensional space. If we fix a vector v, then Theorem 2 tells us that v = ( v ⋅ r ) r + ( v ⋅ s ) s + ( v ⋅ t ) t. How closely can we approximate v by a linear combination of just r and s? The set of all possible linear combinations a1r + a 2 s forms a plane in three-dimensional space, which

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19.3  Vectors and Approximation in Three and More Dimensions

(v · t) t

(v · r) r

t r s

p

v

19-9

we call the plane determined by or spanned by r and s. We seek to find the vector p in this plane that minimizes the error between v and p. This error is the distance between v and p, so we want to find the vector p in the plane that is closest to v. We saw above how to solve this optimization problem for the case where our orthonormal vectors are the standard unit vectors i, j, and k. A similar approach works here. As shown in Figure 19.5, we obtain the closest point by dropping v down perpendicularly onto the plane determined by r and s, landing on the point p = ( v ⋅ r ) r + ( v ⋅ s ) s. If the vector v happens to actually be in the plane determined by r and s, then we will have p = v.

(v · s) s

THEOREM 3 If { r ,   s,   t } is an orthonormal basis for three-dimensional space

and v is a vector, then the linear combination of r and s that is closest to v is

FIGURE 19.5  The vector in the

p = ( v ⋅ r ) r + ( v ⋅ s ) s. (5)



plane determined by orthonormal vectors r and s that is closest to v is p = ( v ⋅ r ) r + ( v ⋅ s ) s.

The vector v − p is orthogonal to both r and s. Proof  Let v be a vector and let p be given by Equation (5). Consider any linear combination w = a1r + a 2 s of r and s. We want to show that the distance from v to p is less than or equal to the distance from v to w (less than or equal to because w could possibly equal p). That is, we must prove that v − p ≤ v − w . Observe that v − w = ( v − p ) + ( p − w ). Since p and w are each linear combinations of r and s, their difference p − w is also a linear combination of r and s. Therefore, since r and s are each orthogonal to t, the vector p − w is orthogonal to t, and hence it is also orthogonal to v − p = ( v ⋅ t ) t. Consequently, v−w

2

= ( v − p) + ( p − w ) 2 = v−p

2

+ p−w

2

Add and subtract. Pythagorean Theorem

2

≥ v−p . Taking square roots gives v − p ≤ v − w . This shows that p is the linear combination of r and s that is closest to v. Finally, v − p = ( v ⋅ t ) t is a multiple of t, which is orthogonal to both r and s.

Vectors in Higher Dimensions Written in component form, a two-dimensional vector v = 〈υ1 ,  υ 2 〉 is an ordered pair of real numbers, while a three-dimensional vector v = 〈υ1 ,  υ 2 ,  υ 3 〉 is an ordered triple of real numbers. Although we cannot easily visualize vectors in higher dimensions, we can still define and understand them. A vector in four dimensions is an ordered quadruple v = 〈υ1 ,  υ 2 ,  υ 3 ,  υ 4 〉 of real numbers. The length of such a vector is v =

υ12 + υ 22 + υ 32 + υ 42 .

There are four axis directions, and the number υ k is the component of v in the direction of the kth axis. Higher-dimensional vectors are similar. If d is a positive integer, then a d-dimensional vector is an ordered list v = 〈υ1 ,  υ 2 , …,  υ d 〉 of d real numbers (also called a d-tuple of numbers). Its length is

v =

υ12 + υ 22 +  + υ d2 =

d

∑ υ k2 . (6) k =1

The distance between two d-dimensional vectors u and v is u − v , the length of the difference of u and v. The dot product of u = 〈u1 ,  u 2 , …,  u d 〉 with v = 〈υ1 ,  υ 2 , …,  υ d 〉 is u ⋅ v = u1υ1 + u 2 υ 2 +  + u d υ d =

d

∑ ukυk. k =1

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Chapter 19  Fourier Series and Wavelets

The length of a vector is related to the dot product of the vector with itself by the formula v ⋅ v = υ12 + υ 22 +  + υ d2 = v 2 . Vectors u and v are orthogonal if u ⋅ v = 0. In this case, the Pythagorean Theorem holds (Exercise 24): If  u ⋅ v = 0, then  u + v

2

= u

2

+ v 2.

The converse of the Pythagorean Theorem also holds (Exercise 25). As a reminder, dot products are defined in any dimension, but cross products are special to dimension 3.

Orthonormal Bases and Approximation in Higher Dimensions The standard unit vectors in d dimensions are the d vectors e 1 = 〈1, 0, 0, … , 0〉, e 2 = 〈0, 1, 0, … , 0〉,  e d = 〈0, 0, 0, … , 1〉. The kth vector e k has a 1 as its kth component, while all other components are zero. The set of unit vectors { e 1 ,  e 2 , …,  e d } is orthonormal, and additionally every vector v in d-dimensional space is a linear combination of these vectors. Therefore, we say that { e 1 ,  e 2 , … ,  e d } forms an orthonormal basis for d-dimensional space. Likewise, any collection of d orthonormal vectors is an orthonormal basis for d-dimensional space, and we have the following explicit formula for writing any vector as a linear combination of the basis vectors (this theorem is proved in texts on linear algebra). The coefficients in this representation are dot products of v with the vectors u k .

THEOREM 4 If u 1 ,  u 2 , …,  u d are d orthonormal vectors in d-dimensional space (so { u 1 ,  u 2 , … ,  u d } is an orthonormal basis for d-dimensional space), then every vector v = 〈υ1 ,  υ 2 , …,  υ d 〉 can be written as

v = ( v ⋅ u1 ) u1 + ( v ⋅ u 2 ) u 2 +  + ( v ⋅ u d ) u d =

d

∑ (v ⋅ u k ) uk. k =1

Now fix an integer 1 ≤ n < d. If  { u 1 ,  u 2 , …,  u d } is an orthonormal basis and v is a vector in d dimensions, how close can we get to v using a linear combination of only the first n vectors u 1 ,  u 2 , …,  u n? This type of optimization problem occurs widely throughout mathematics, science, and engineering. The answer (which is often called the least squares solution) is given in the next theorem. The proof is similar to the proof of Theorem 3 (Exercise 31).

THEOREM 5  Assume that { u 1 ,  u 2 , …,  u d } is an orthonormal basis for d-dimensional space. If v is a vector and 1 ≤ n < d , then the linear combination of u 1 ,  u 2 , …,  u n that is closest to v is



p = ( v ⋅ u1 ) u1 + ( v ⋅ u 2 ) u 2 +  + ( v ⋅ u n ) u n =

n

∑ ( v ⋅ u k ) u k . (7) k =1

The vector v − p is orthogonal to u 1 ,  u 2 , … ,  u n .

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The set of all possible linear combinations a1u 1 + a 2 u 2 +  + a n u n of vectors u 1 ,  u 2 , …,  u n is called the subspace spanned by u 1 ,  u 2 , …,  u n . The vector p given in Equation (7) is the vector in this subspace that is closest to v. It is the vector in that subspace that minimizes the distance between v and the vectors in the subspace, and it is called the orthogonal projection of v onto the span of u 1 ,  u 2 , …,  u n . Geometrically, Theorem 5 tells us that to find the closest vector p, we should simply drop down perpendicularly from v onto the subspace spanned by u 1 ,  u 2 , …,  u n .

v−p v

EXAMPLE 2   Assume that u 1 is a unit vector in d-dimensional space. A linear combination of the single vector u 1 is simply a multiple au 1 , where a is a real number. Therefore, the subspace spanned by u 1 is the set of all multiples au 1 of u 1. We also call this the line through u 1. If v is a vector in d-dimensional space, what is the vector p on the line through u 1 that is closest to v?

p u1

FIGURE 19.6  The vector on the line through u 1 that is closest to v is p = ( v ⋅ u 1 ) u 1. The vector v − p is orthogonal to u 1.

Solution  We will apply Theorem 5 with n = 1. By results from linear algebra, starting with a unit vector u 1 we can find vectors u 2 , …,  u d so that u 1 ,  u 2 , …,  u d forms an orthonormal basis for d-dimensional space. Then, by Equation (7) with n = 1, the vector on the line through u 1 that is closest to v is

p = ( v ⋅ u 1 ) u 1. This agrees with the formula that we found in Section 11.3 for vectors in two- or threedimensional space, where we denoted the vector p by proj u1 v (compare Figure 19.6 to Figure 11.27).

19.3

EXERCISES 

Orthogonal Vectors in Three Dimensions

7. r = 〈1, 0, 0〉,   s = 〈0, −1, 0〉,   t = 〈0, 0,  −1〉,   v = 〈1, 2, 3〉.

In Exercises 1–6, find



a. the dot products r ⋅ s,   r ⋅ t , and s ⋅ t, b. the lengths r , s , t .

9. r = 〈2 3, −2 3, 1 3〉,   s = 〈1

Also determine whether { r ,   s,   t } is

1. r = 〈1, 1, 0〉,   s = 〈1, −1, 0〉,   t = 〈0, 0, 1〉. 2, 0〉,   s = 〈1

2, −1

2, 0〉,   t = 〈0, 0, 1〉.

3. r = 〈1, 1, 1〉,   s = 〈1, 1, −2〉,   t = 〈0, −1, 1〉.

4. r = 〈1 3, 1 3, 1 3〉,   s = 〈1 t = 〈0, −1 2, 1 2 〉.

6, 1



5. r = 〈1 t = 〈1

6, 1



6. r = 〈2 3, −2 3, 1 3〉,   s = 〈1 2, 1 t = 〈−1 ( 3 2 ) , 1 ( 3 2 ) , 2 2 3〉.

3, 1 2, −1

3, 1

3〉,   s = 〈1

6, − 2

3〉,

6, − 2

3〉,

In Exercises 11–14, let u = 〈2, 0, −1, 3, 2〉,   v = 〈−1, 1, 1, −1, 0〉, and w = 〈3, −1, 2, −1, 1〉. 11. Find the linear combinations 2u + v − 3w and −u + 5v − w. 13. Find the dot products u ⋅ v,   u ⋅ w, and v ⋅ w. 14. Find the linear combination ( 2w ⋅ u ) u − ( w ⋅ 3v ) v + ( u ⋅ v ) 4 w.

2, 0〉,

Closest Point in a Plane

In Exercises 7–10, a. verify that { r ,   s,   t } is an orthonormal basis for threedimensional space, b. find the linear combination p of r and s that is closest to v,

M19_HASS5901_15_GE_C19.indd 11

Vectors in Five Dimensions

12. Find the lengths u ,   v , and w .

2, 0〉.

c. find the distance from v to p.

2, 0〉,

10. r = 〈3 5, 3 5,  7 5〉,   s = 〈− 7 4, 0, 3 4〉,   t = 〈9 20,  −4 5, 3 7 20〉,   v = 〈0, 1, 0〉.

d. an orthonormal set of vectors. 2, 1

2, 1

t = 〈−1 ( 3 2 ) , 1 ( 3 2 ) , 2 2 3〉, and v = 〈3, 0, 3〉.

c. an orthogonal set of vectors,

2. r = 〈1

8. r = 〈2 3, −2 3, −1 3〉,   s = 〈1 3, 2 3, −2 3〉, t = 〈2 3, 1 3, 2 3〉,   v = 〈3, 6, 3〉.

In Exercises 15–20, let u 1 = 〈1 2, 1 2, 0, 0, 0〉, u 2 = 〈1 2,   −1 2 ,  0,  0,  0〉, u 3 = 〈0,  0,  1 2,  1 2,  0〉, u 4 = 〈0, 0, 1 2, −1 2, 0〉, u 5 = 〈0, 0, 0, 0, 1〉, and v = 〈2, −4, 4, 2, 1〉. 15. Verify that { u 1 ,  u 2 ,  u 3 ,  u 4 ,  u 5 } is an orthonormal basis for fivedimensional space. 16. Find the vector p on the line through u 1 that is closest to v, and find the distance from v to p. Verify that v − p is orthogonal to u 1.

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Chapter 19  Fourier Series and Wavelets

17. Find the linear combination p of u 1 and u 2 that is closest to v, and find the distance from v to p. Verify that v − p is orthogonal to both u 1 and u 2 . 18. Find the linear combination p of u 1 ,  u 2 , and u 3 that is closest to v, and find the distance from v to p. Verify that v − p is orthogonal to u 1 , u 2 , and u 3 . 19. Find the linear combination p of u 1 ,  u 2 ,  u 3 , and u 4 that is closest to v, and find the distance from v to p. Verify that v − p is orthogonal to u 1 ,  u 2 ,  u 3 , and u 4 . 20. Verify that

( v ⋅ u 1 )u 1 + ( v ⋅ u 2 )u 2 + ( v ⋅ u 3 )u 3 + ( v ⋅ u 4 )u 4 + ( v ⋅ u 5 )u 5 equals v. Theory and Examples

In Exercises 21–30, u = 〈u1 ,  u 2 , …,  u d 〉,   v = 〈υ1 ,  υ 2 , …,  υ d 〉, and w = 〈w1 ,  w 2 , …,  w d 〉 are d-dimensional vectors. 21. Symmetry 

25. Converse of the Pythagorean Theorem Prove that if u + v 2 = u 2 + v 2 , then u ⋅ v = 0. 26. Parallelogram 2 u 2 + 2 v 2.

Law Prove

2

= u

2

that

if

u ⋅ v = 0,

2

=

c. Prove that u ⋅ v ≤ u v . 28. Triangle Inequality  Use Exercises 23 and 27 to prove that u+ v ≤ u + v. 29. Reverse Triangle Inequality  Prove that u = u − v + v ≤ u − v + v and v = v − u + u ≤ v − u + u . Use these facts to prove that u − v ≤ u−v. 30. Parseval Equality  Assume that { u 1 ,  u 2 , …,  u d } is an orthonormal basis for d-dimensional space. Use Theorem 4 to prove that v

2

= ( v ⋅ u 1 )2 + ( v ⋅ u 2 )2 +  + ( v ⋅ u d )2 =

then

+ u−v

b. Prove that b 2 a = w + cv 2, and use the Pythagorean Theorem to show that w + cv 2 = w 2 + c 2 v 2 ≥ c 2 b.

+ 2( u ⋅ v ) + v 2 .

24. Pythagorean Theorem Prove u + v 2 = u 2 + v 2.

2

a. Prove that w and cv are orthogonal.

( a u + b v ) ⋅ w = a ( u ⋅ w ) + b ( v ⋅ w ) and u ⋅ ( a v + b w ) = a ( u ⋅ v ) + b ( u ⋅ w ). 23. Prove that u + v

u+v

27. Cauchy–Schwarz Inequality  Let a = u 2 ,   b = v 2, and c = u ⋅ v.  Set w = b u − cv,   so b u = w + cv.

Prove that u ⋅ v = v ⋅ u.

22. Distributive properties  Prove that if a and b are real numbers, then

that

d

∑ ( v ⋅ u k )2 . k =1

31. Prove Theorem 5.

19.4 Approximation of Functions The Norm of a Function The main goal of Fourier series is to find the best approximation of a function by a trigonometric polynomial. We have seen that for vectors, it is easy to find best approximations when we have an orthonormal collection u 1 ,  u 2 , …, u d of vectors. If we had an orthonormal collection of functions, then we could use similar ideas to find the best approximation. We will show below that the trigonometric system is, after being appropriately rescaled, an analogue of such an orthonormal basis. Before doing this, we must explain what we mean by the norm of a function (because we need unit functions) and the inner product of two functions (because we need orthogonal functions). We will assume that the functions we are working with are periodic with period 2π. Any such function simply repeats the values that it takes in the interval [ 0, 2π ] in any other interval [ 2nπ , 2( n + 1) π ] with n an integer. Vectors and functions are more similar than might be apparent at first glance. A vector v = 〈υ1 ,  υ 2 , …,  υ d 〉 in d dimensions is an ordered list of numbers υ1 ,  υ 2 , …,  υ d . For each integer k = 1, 2, …,  d we have an associated number υ k . If f is a function on [ 0, 2π ], then for each real number x in [ 0, 2π ] we have an associated number f ( x ). Many of the things that we have done with vectors can also be done with functions if we use the analogy that the function values f ( x ) are like the components υ k of a vector. There are infinitely many function values instead of finitely many components, so where we used sums before, we now use “continuous sums,” or integrals. There is one technical issue here that does not arise with vectors. While we can always sum up finitely many numbers, there are functions whose integrals do not exist. We cannot define the norm of such functions. Instead, we restrict our attention to integrable functions, those whose integral does exist. Fortunately, most of the functions that we encounter in real world problems are integrable. All of the functions that we will see in later examples or exercises will be integrable. In particular, any function that is continuous on a finite interval [ a,  b ] is integrable. More generally, any function that is bounded and piecewise continuous on [ a,  b ] (continuous at all but finitely many points) is integrable.

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19-13

The norm of a function is analogous to the length of a vector. The length of a vector v = 〈υ1 ,  υ 2 , …,  υ d 〉 is d

∑ υ k2 .

v =

k =1

We define the norm of a function using a similar formula, replacing components υ k with function values f ( x ), and a finite sum with an integral.

DEFINITION

1. The norm of an integrable function f on the domain [ 0, 2π ] is 2π

∫0

f =

f ( x ) 2 dx.

We say that f is a unit function if f = 1. 2. The distance between two integrable functions f and g on the domain [ 0, 2π ] is 2π

( f ( x ) − g( x ) ) 2 dx.

∫0

f −g =

Our definition of the norm of a function is suited to the applications that we consider in this chapter, but it is only one of many ways to extend the idea of length from vectors to functions. The notion of a norm is explored in the mathematical subject called Real Analysis. We compute the norms of some functions, including the functions that belong to the trigonometric system. EXAMPLE 1 (a) Let f be the function defined in Equation (1). Observe that f ( x ) = x − π for all but one point in the interval [ 0, 2π ]. Since changing the value of a function at a single point does not change the value of its integral, we can ignore that exceptional point and compute that f

2

=

2π

∫0

f ( x ) 2 dx =

Taking square roots, f =

2π

∫0

(x

− π ) 2 dx =

2π 3 . 3

2 3  π 3 2 ≈ 4.55.

(b) Let k be any positive integer. The norm of the function sin kx is sin kx =

2π

∫0

sin 2 kx dx.

We can use trigonometric identities to evaluate this integral, but an easier way is to note that the graph of cos kx equals the graph of sin kx translated π ( 2 k ) units to the left, and therefore the area under the graph of cos 2 kx over [ 0, 2π ] is the same as the area under the graph of sin 2 kx over [ 0, 2π ]; see Figure 19.7. Therefore, using the identity sin 2 x + cos 2 x = 1, we see that 2 sin kx

M19_HASS5901_15_GE_C19.indd 13

2

2π

=

∫0

=

∫0

=

∫0

=

∫0

2π

2π

2π

2π

sin 2 kx dx +

∫0

sin 2 kx dx +

∫0

2π

sin 2 kx dx cos 2 kx dx

( sin 2 kx + cos 2 kx ) dx

1 dx = 2π.

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Chapter 19  Fourier Series and Wavelets

y 1

2p 3

p 3

4p 3

p

5p 3

2p

x

(a) Graph of cos 2 3x over the interval [0, 2p] y 1

2p 3

p 3

4p 3

p

5p 3

2p

x

(b) Graph of sin 2 3x over the interval [0, 2p]

FIGURE 19.7  The area under the graph of cos 2 3 x between 0 and 2π is the same as the area under the graph of sin 2 3 x between 0 and 2π.

The norm of the constant function 1 is not 1 because the length of the interval [ 0, 2π ] is 2π , not 1.

Therefore, sin kx 2 = π for each positive integer k. A similar computation applies to cos kx , so by taking square roots we conclude that sin kx = π = cos kx . (c) The norm of the constant function 1 on the domain [ 0, 2π ] is 1 =

2π

∫0

1( x ) 2 dx =

2π

∫0

1 2 dx =

2π .

The Inner Product of Functions The inner product of functions is analogous to the dot product of vectors. The dot product of a vector u = 〈u1 ,  u 2 , …,  u d 〉 with v = 〈υ1 ,  υ 2 , …,  υ d 〉 is the number u ⋅ v = u1υ1 + u 2 υ 2 +  + u d υ d =

d

∑ ukυk. k =1

The inner product of two functions is a number that is defined by a similar formula. However, because the notation f ⋅ g resembles the definition of the product function f ( x ) g( x ) rather than a number, we denote the inner product of f and g by 〈 f ,  g〉.

The notation u = 〈u1 ,  u 2 〉 denotes the components of a two-dimensional vector, while 〈 f ,  g〉 denotes the inner product of functions f and g. Despite the similar notation, we can tell from context which meaning is intended.

M19_HASS5901_15_GE_C19.indd 14

DEFINITION The inner product of two integrable functions f and g on [ 0, 2π ] is

〈 f ,  g〉 =

2π

∫0

f ( x ) g( x ) dx.

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19.4  Approximation of Functions

19-15

The properties of the inner product of functions are similar to the properties of the dot product of vectors. For example, just as the dot product of a vector with itself is the square of its length, the inner product of a function with itself is the square of its norm: 〈f, f〉 =



2π

∫0

f ( x ) 2 dx =

2

f

. (8)

Hence f is a unit function if and only if  〈 f ,  f 〉 = 1. The definition of orthogonality for functions is similar to the definition for vectors. DEFINITION Let f and g be integrable functions on [ 0, 2π ]. Then f and g are orthogonal if  〈 f ,  g〉 = 0. They are orthonormal if they are orthogonal and, additionally, f = g = 1.

Often we deal with more than two functions. We say that a collection of functions is orthogonal if any two distinct functions in the collection are orthogonal. If in addition each function in the collection is a unit function, then we say that the collection is orthonormal. The next theorem is the analogue for functions of Theorem 1 in Section 19.3. Statement 3 of this theorem is the Pythagorean Theorem for functions. THEOREM 6 Let f and g be integrable functions on [ 0, 2π ].

1. f − g

2

=

f

2

− 2 〈 f ,  g〉 + g

2

2. f + g

2

=

f

2

+ 2 〈 f ,  g〉 + g

2

3. If f and g are orthogonal, then f + g

2

=

f

2

+ g 2.

Proof   Statement 1 follows because 2

f −g

2π

=

∫0

=

∫0

= f

2π

2

(

f ( x ) − g( x ) ) 2 dx

f ( x ) 2 dx − 2 ∫

2π 0

− 2〈 f ,  g〉 + g

f ( x ) g( x ) dx + 2

2π

∫0

g( x ) 2 dx

,

and the proof of Statement 2 is similar. If f and g are orthogonal, then 〈 f ,  g〉 = 0, so Statement 3 follows from Statement 2.

Orthogonality of the Trigonometric System We will show that the trigonometric system is an orthogonal collection of functions. It is not an orthonormal collection because the functions 1, cos kx and sin kx do not have norm 1. EXAMPLE 2 (a) If k is any positive integer, then the inner product of the constant function 1 with sin kx is 〈1, sin kx〉 =

2π

∫0

1 ⋅ sin kx dx =

2π

∫0

sin kx dx.

We see by looking at the graph of sin kx on the interval [ 0, 2π ] that this integral is zero since it has as much area below as above the x-axis, and therefore the signed areas cancel. We demonstrate this explicitly by computing that 〈1, sin kx〉 =

M19_HASS5901_15_GE_C19.indd 15

2π

∫0

sin kx dx = −

cos kx  2 π − cos 2 k π cos 0 1 −1 + = + = 0.  = k  0 k k k k

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Chapter 19  Fourier Series and Wavelets

Consequently, the constant function is orthogonal to sin kx for every integer k > 0. A similar calculation shows that 1 is orthogonal to cos kx for every positive k. (b) We show next that sin x and cos x are orthogonal. Using the u-substitution u = sin x , their inner product is sin 2 2π − sin 2 0 sin 2 x  2 π = = 0.  2 0 2 Looking at their graphs, there is no obvious reason why we should expect that sin x and cos x should be orthogonal. We can understand visually what it means for two vectors u and v in three-dimensional space to be perpendicular (there is a 90-degree angle between them), but there is no easy way to see from their graphs why two functions are orthogonal. 〈sin x , cos x〉 =

2π

∫0

sin x cos x dx =

(c) We compute another special case. The inner product of sin x and sin 2 x is zero because 〈sin x , sin 2 x〉 =

2π

∫0

sin x ( 2 sin x cos x ) dx = 2

sin 3 t  2 π sin 3 2π − sin 3 0 = 2 = 0.  3 0 3

Therefore, sin x and sin 2 x are orthogonal. (d) We have shown that the constant function 1 is orthogonal to sin kx and cos kx for every positive integer k. To complete the proof that the trigonometric system is a collection of orthogonal functions, it remains to prove that for all positive integers m and n we have: 2π

〈sin mx , sin nx〉 =

∫0

〈cos mx , cos nx〉 =

∫0

〈sin mx , cos nx〉 =

∫0

2π

2π

sin mx sin nx dx = 0

whenever  m ≠ n,

cos mx cos nx dx = 0

whenever  m ≠ n,

sin mx cos nx dx = 0

for all  m  and  n.

This can be done through the use of trigonometric identities (Exercises 31–33). The functions in the trigonometric system do not have unit norm. However, if we divide each function by its norm, then we obtain functions whose norm is 1. We set u 0 (x) =

1 , 2π

and for each integer k > 0 let u k (x) =

cos kx π

and

υ k (x) =

sin kx . π

Each function u 0 ,  u k ,  and υ k is a unit function, and the set of these functions is orthonormal. DEFINITION The normalized trigonometric system is the collection of ortho-

normal functions u 0 ,  u1 ,  υ1 ,  u 2 ,  υ 2 , ….  Explicitly, these are cos x sin x cos 2 x sin 2 x cos 3 x sin 3 x 1 ,  ,  ,  ,  ,  ,  , …. π π π π π π 2π

Approximation of Functions Now we find the best approximation to a function using the normalized constant function u 0 , the first n normalized cosine functions u1 ,  u 2 , …,  u n , and the first n normalized sine functions υ1 ,  υ 2 , …,  υ n . That is, out of all of the linear combinations

w = a 0u 0 +

n

∑ a ku k k =1

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+

n

∑ b k υ k , (9) k =1

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19.4  Approximation of Functions

19-17

we identify the one whose distance from f is the smallest. We let p n denote this closest function. THEOREM 7 If f is an integrable function on [ 0, 2π ] and n is a positive integer,

then the linear combination of u 0 and u1 ,  u 2 , …,  u n and υ1 ,  υ 2 , …,  υ n that is closest to f is

p n = 〈 f ,  u 0 〉 u 0 +

n

∑ 〈 f , u k 〉 u k

+

k =1

n

∑ 〈 f ,  υ k 〉 υ k . (10) k =1

Proof  Let p n be the function in Equation (10), and let w be any linear combination of the form given in Equation (9). We must prove that   f − p n ≤ f − w . To simplify the notation, let q = f − pn

z = p n − w.

and

Also let c 0 = 〈 f ,  u 0 〉, and for each positive integer k set c k = 〈 f ,  u k 〉 and d k = 〈 f ,  υ k 〉, so n

∑ c ku k

p n = c 0u 0 +

+

k =1

n

∑ dkυk. k =1

The inner product satisfies a distributive law, just as the dot product of vectors does (Exercise 24). Using this and the fact that u 0 is orthogonal to all of the other functions u k and υ k with k > 0, the inner product of p n with u 0 is 〈 p n ,  u 0 〉 = 〈c 0u 0 +

n

∑ c ku k + k =1

= c 0 〈u 0 ,  u 0 〉 +

n

∑ d k υ k , u 0 〉

n

n

∑ c k 〈u k , u 0 〉 + k =1

= c0 u0

2

+

n

∑ ck

Substitute

k =1

⋅0+

k =1

∑ d k 〈υ k , u 0 〉

Distributive laws

⋅0

Orthogonality

k =1

n

∑ dk k =1

= c0 ⋅ 1 + 0 + 0 = c0.

u 0 is a unit function

It follows from this that q is orthogonal to u 0 because q = f − p n , and therefore 〈q,  u 0 〉 = 〈 f − p n ,  u 0 〉 = 〈 f ,  u 0 〉 − 〈 p n ,  u 0 〉 = c 0 − c 0 = 0. A similar calculation holds for inner products of q with each of the functions u1 ,  u 2 , …,  u n and each of υ1 ,  υ 2 , …,  υ n . That is, q is orthogonal to every one of these functions. Using the distributive law again, it follows that q is orthogonal to every linear combination of u 0 and u1 ,  u 2 , …,  u n and υ1 ,  υ 2 , …,  υ n . Now, p n is a linear combination of those functions, and so is w, so q is orthogonal to both p n and w. Hence 〈q,  z〉 = 〈q,  p n − w〉 = 〈q,  p n 〉 − 〈q,  w〉 = 0 − 0 = 0, and so we see that q and z are orthogonal. Therefore, we can apply the Pythagorean Theorem for functions: f −w

2

= q+z

2

q = f − p n ,  z = p n − w

= q

2

+ z

≥ q

2

+0

=

f − pn

2

Pythagorean Theorem z

2

.

2

≥ 0

q = f − pn

Taking square roots, it follows that f − p n ≤ f − w .

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Chapter 19  Fourier Series and Wavelets

We now show how to write the function p n that is closest to f directly in terms of sines and cosines. We call the function p n given in Equation (11) a trigonometric polynomial of degree n. The numbers a 0 , a k , and b k defined in this theorem are called the Fourier coefficients of f . THEOREM 8 Let f be an integrable function on [ 0, 2π ] and let n be a positive

integer. Define real numbers

1 1 〈 f , 1〉 = 2π 2π 1 a k = 〈 f , cos kx〉 = π 1 b k = 〈 f , sin kx〉 = π a0 =

2π

∫0

f ( x ) dx ,

1 2π f ( x ) cos kx dx , π ∫0 1 2π f ( x ) sin kx dx , π ∫0

where k is a positive integer. Then the best approximation to f by a linear combination of 1, cos x , …, cos nx , sin x , …, sin nx is pn (x) = a 0 +



n

n

k =1

k =1

∑ a k cos kx + ∑ b k sin kx. (11)

Proof   By Theorem 7, the best approximation is the function p n given in Equation (10). Expanding and simplifying shows that this function satisfies Equation (11): pn (x) = 〈 f , u 0 〉 u 0 (x) +

n

n

∑ 〈 f , u k 〉 u k (x) + ∑ 〈 f , υ k 〉 υ k (x) k =1

= =

1 2π

f,

1 + 2π

1 〈 f , 1〉 1 + 2π

= a0 +

k =1

n

∑

f,

k =1

cos kx π

cos kx + π

n

n

∑

k =1

f,

sin kx sin kx π π

n

∑ π1 〈 f , cos kx〉 cos kx + ∑ π1 〈 f , sin kx〉 sin kx k =1

k =1

n

n

k =1

k =1

∑ a k cos kx + ∑ b k sin kx.

EXAMPLE 3  Let f be the function defined in Equation (1). We will verify that the function p n given in Equation (2) is the best approximation to f by a trigonometric polynomial of degree n. We must compute the numbers a 0 , a k , and b k given in Theorem 8. If k is a positive integer, then we use Equation 86 from the Table of Integrals to compute that 1 2π ( x − π ) sin kx dx π ∫0 2π 1 2π = ∫ x sin kx dx − ∫ sin kx dx 0 π 0

bk =

=

1 π

x cos kx  2 π  sin kx 1 − 2π 2 = − .  2 −  −0 =  k  0 k π k k

(

)

A similar calculation shows that a 0 = 0 and a k = 0 for every k > 0. Therefore, the best approximation uses amplitude zero for the constant term and the cosine terms, and amplitude −2 k for the sine term of frequency k, so

(

p n ( x ) = −2 sin x +

)

n sin 2 x sin nx sin kx . ++ = −2 ∑ 2 n k k =1

This is the trigonometric polynomial given in Equation (2).

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19.4  Approximation of Functions

19-19

EXAMPLE 4  Let f be the step function  1, f ( x ) =   2, 

0 ≤ x < π  or  x = 2π , π ≤ x < 2π.

Find the best approximation to f by a trigonometric polynomial of degree n. Solution  We compute the Fourier coefficients for f . If k is a positive integer, then 2π

bk =

1 π

∫0

=

1 π

∫0

=

1 π

2π   − cos kx  π  −2 cos kx     +       0   π  k k

π

f ( x ) sin kx dx

sin kx dx +

2π

1 π

∫π

2 sin kx dx

 2  − , cos k π − 1 ( − 1) k − 1 = = =  k π  kπ kπ  0, 

k  odd, k  even.

A similar calculation shows that a 0 = 3 2, but a k = 0 for every k > 0. Therefore, the best approximation to f uses only a constant term and sine terms with odd frequencies. Specifically, if n is odd then the linear combination of 1, sin x , …, sin nx that is closest to f is pn (x) =

(

)

sin 3 x sin 5 x sin nx 3 2 . − sin x + + ++ 2 3 5 n π

At the point x = π , where f ( x ) jumps from 1 to 2, all of the sine terms vanish, so p n (π ) = 3 2 for every n. This is not the value of f (π ), nor does it approach closer to f (π) as n increases. This does not contradict the fact that p n is the best approximation to f because best does not mean that p n ( x ) is close to f ( x ) at every point. Instead, p n is the best approximation because the error is being measured here by an integral. The error in the approximation is the distance between f and p n , which is 2π

∫0

f − pn =

( f ( x ) − p n ( x ) ) 2 dx .

Out of all linear combinations using frequencies of at most n, the trigonometric polynomial p n minimizes this error, which is defined by an integral that takes all of the values of f ( x ) and p n ( x ) across the entire domain [ 0, 2π ] into account (see Figure 19.8).

Convergence of Fourier Series If f is an integrable function on [ 0, 2π ], then Theorem 8 gives us the best approximation p n to f by a linear combination of   1, cos x , …, cos nx , sin x , …, sin nx. The only information that we needed to prove this theorem is that the functions in the trigonometric system are orthogonal. Looking at Figures 19.3 and 19.7, it appears that the approximation p n becomes closer to f in some sense as n increases. The next theorem makes this precise. Specifically, it is the distance f − p n between f and p n that converges to zero as n → ∞. THEOREM 9 Let f be an integrable function on [ 0, 2π ] and for each positive

integer n let p n be the function defined in Equation (11). Then the distance between f and p n converges to zero as n → ∞: lim f − p n = lim

n →∞

M19_HASS5901_15_GE_C19.indd 19

n →∞

2π

∫0

( f ( x ) − p n ( x ) ) 2 dx = 0.

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Chapter 19  Fourier Series and Wavelets

y

y f

2 1.5

1.5

p1

1

1

0.5

0.5 p 2

p

3p 2

f

2

2p

p3

x

p 2

y

p

3p 2

2p

x

y f

2 1.5

1.5

p5

1

1

0.5

0.5 p 2

p

3p 2

f

2

2p

p15

x

p 2

y

p

3p 2

2p

x

y f

2 1.5

1.5

p25

1

1

0.5

0.5 p 2

p

3p 2

f

2

2p

x

p75

p 2

p

3p 2

2p

x

FIGURE 19.8  The trigonometric polynomials p1 , p 3 , p 5 , p15 , p 25 , and p 75 approximating the

function f from Example 4

The proof of Theorem 9 is harder than the proof of Theorem 8. It can be found in advanced texts on real analysis. We can get some insight into why the proof is difficult by recalling our discussion of orthonormal bases in three dimensions. Whenever we have three orthonormal vectors r, s, t in three-dimensional space, we know that these three vectors form an orthonormal basis, and therefore every vector v can be written as a linear combination of those three vectors, as shown in Theorem 2. A similar fact holds in higherdimensional space. That is, a set of d orthonormal vectors in d-dimensional space will be an orthonormal basis for that space (Theorem 4). However, the normalized trigonometric system is a set of infinitely many orthonormal functions. Just having infinitely many orthonormal functions is not enough to ensure that they form the correct analogue of an orthonormal basis. For example, the cosine functions cos x , cos 2 x , cos 3 x , … are infinitely many orthogonal functions, but they are only “half” of the trigonometric system. The difficulty in the proof of Theorem 9 is in showing that the trigonometric system contains “enough” functions.

M19_HASS5901_15_GE_C19.indd 20

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19.4  Approximation of Functions

19-21

Let a 0 ,  a k , and b k be the Fourier coefficients of f as given in Theorem 8, and let pn (x) = a 0 +

n

n

k =1

k =1

∑ a k cos kx + ∑ b k sin kx

be the trigonometric polynomial defined in Equation (11). Theorem 9 says that if f is an integrable function, then p n converges to f in the sense that the distance 2π

f − p n = ∫ 0 ( f ( x ) − p n ( x ) ) 2 dx between p n and f converges to zero as n → ∞. This does not imply that p n ( x ) must converge to f ( x ) at each point x. We often call

a0 +

∞

∞

k =1

k =1

∑ a k cos kx + ∑ b k sin kx (12)

the Fourier series for f , but it is important to note that this series need not converge. Under the right hypothesis, it will be true that p n ( x ) converges to f ( x ), and hence f ( x ) will equal its Fourier series at x. Loosely speaking, the smoother that f is, the “better” the convergence of p n ( x ) to f ( x ) will be. The following result, whose proof we will omit, states that if f is smooth enough as a periodic function on the real line, then convergence will hold at every point, and the maximum difference between f ( x ) and p n ( x ) shrinks as n increases. THEOREM 10 Let f be a function on [ 0, 2π ] such that f (0) = f (2π ), and

extend f to be a periodic function on the real line by repeating values every 2π units. If f is differentiable at every point and f ′ is continuous, then we have for every x that f ( x ) = lim p n ( x ) = a 0 + n →∞

∞

∞

k =1

k =1

∑ a k cos kx + ∑ b k sin kx.

Further, if for each n we let Mn be the maximum value of f ( x ) − p n ( x ) over all x, then lim M n = 0.

n →∞

The function f pictured in Figure 19.8 does not satisfy the hypotheses of Theorem 10, and it can be shown that for that function the maximum Mn defined in Theorem 10 does not decrease to zero as n increases (this is known as the Gibbs phenomenon).

The Parseval Equality for Fourier Series By Exercise 30 of Section 19.3, if { u 1 ,  u 2 , …,  u d } is an orthonormal basis for d-dimensional space, then the Parseval Equality holds for every vector v in d-dimensional space: v

2

=

d

∑ ( v ⋅ u k )2 . k =1

We will prove an analogous result for Fourier series. THEOREM 11—The Parseval Equality

If f is an integrable function on [ 0, 2π ], then f

M19_HASS5901_15_GE_C19.indd 21

2

∞

∞

k =1

k =1

= 2πa 02 + π ∑ a k2 + π ∑ b k2 .

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19-22

Chapter 19  Fourier Series and Wavelets

Proof  Let p n be the trigonometric polynomial of degree n given in Equation (11). According to Exercise 34, we have 〈 p n , 1〉 = 2πa 0 , while 〈 p n , cos kx〉 = πa k and 〈 p n , sin kx〉 = πb k for k = 1, …,  n.  Therefore, pn

2

= 〈 p n ,  p n 〉 =

Equation (8)

p n ,  a 0 +

n

n

k =1

k =1

∑ a k cos kx + ∑ b k sin kx n

n

k =1

k =1

Equation (11)

∑ a k 〈 p n , cos kx〉 + ∑ b k 〈 p n , sin kx〉

= 〈 p n ,  a 0 1〉 + = a 0 〈 p n , 1〉 +

n

∑ a k πa k

+

k =1

Distributive property

n

∑ b k πb k .

Exercise 34

k =1

n

n

k =1

k =1

= 2πa 02 + π ∑ a k2 + π ∑ b k2 .

Exercise 34 and simplify

Since Theorem 9 tells us that f − p n → 0, it follows that f

2

2

= lim p n

Exercise 28

n →∞

n n   = lim  2πa 02 + π ∑ a k2 + π ∑ b k2  n →∞    k =1 k =1 ∞

∞

k =1

k =1

= 2πa 02 + π ∑ a k2 + π ∑ b k2 .

Calculated above Monotonic Sequence Theorem (Theorem 6 in Section 9.1)

EXAMPLE 5  Let f be the function from Example 4. The square of the norm of f is 2

f

π

∫0

=

2π

∫π

1 2 dx +

2 2 dx = 5π.

We saw in Example 4 that a 0 = 3 2,  a k = 0 for all k > 0,  b k = −2 ( k π) if k is odd, and b k = 0 if k is even. Theorem 10 therefore implies that 5π =

f

2

∞

∞

k =1

k =1

= 2π a 02 + π ∑ a k2 + π ∑ b k2 = 2π =

( )

9 −2 +0+π ∑ 4 k  odd k π

18π 4 +0+ 4 π

∑

k  odd

2

1 . k2

Simplifying, we obtain ∞

∑ ( 2k −1 1)2

=

k =1

∑

k  odd

π2 1 = . 2 k 8

Therefore, ∞

∑ k12 = k =1

∞

∑ ( 2k −1 1)2 + k =1

∞

∞

2 ∑ (21k ) 2 = π8 + 14 ∑ k12 . k =1 k =1

This gives us ∞

π2 3 1 = , ∑ 2 4 k =1 k 8 and hence 

M19_HASS5901_15_GE_C19.indd 22

∞

∑ k12 k =1

=

π2 . (13) 6

23/03/23 9:23 AM



19.4  Approximation of Functions

19.4

EXERCISES  Norms of Functions

In Exercises 1–6, a periodic function with period 2π is defined by specifying its values on the interval [ 0, 2π ]. Find the norm of each function.  x , 1. f ( x ) =   2π − x ,   1, 2. g( x ) =   0, 

19-23

0 ≤ x ≤ π π < x ≤ 2π

21. Apply the Parseval Equality to the function h from Exercise 3 to give another proof of the equalities specified in Exercise 20. 22. Define a function f on [ 0, 2π ] by  x ( π − x ), 0 ≤ x ≤ π f ( x ) =   ( x − π )( x − 2π ), π < x ≤ 2π  Apply the Parseval Equality to this function f to prove that

π 2 < x < 3π 2

∞

∑ ( 2k −1 1)6

0 ≤ x ≤ π 2  or  3π 2 ≤ x ≤ 2π

k =1

Use that equality to prove that

3. h( x ) = ( x − π )

2

 x 2 , 4. j ( x ) =   0, 

0 ≤ x ≤ π

 e x , 5. k ( x ) =   1, 

0 ≤ x < 2π

∞

∑ k16

=

k =1

π < x ≤ 2π

π6 . 945

Theory and Examples

In Exercises 23–28, f ,  g, and h denote integrable functions on [ 0, 2π ]. 23. Symmetry  Prove that 〈 f ,  g〉 = 〈g, f 〉.

x = 2π

 cos x , 6. ( x ) =   1, 

π6 . 960

=

24. Distributive properties  Prove that if a and b are real numbers, then 〈af + bg,  h〉 = a 〈 f ,  h〉 + b 〈g,  h〉 and 〈 f ,  ag + bh〉 = a 〈 f ,  g〉 + b 〈 f ,  h〉.

0 ≤ x ≤ π π < x ≤ 2π

Inner Product of Functions

In Exercises 7–12, use the functions defined in Exercises 1–6 to compute the given inner product. Are the two functions orthogonal?

25. Cauchy–Schwarz Inequality  Prove that 〈 f ,  g〉 ≤ f g . (Hint: Follow the approach given in Exercise 27 of Section 19.3.)

8. 〈k , cos x 〉

26. Triangle Inequality  Prove that f + g ≤ f + g . (Hint: Compare Exercise 28 in Section 19.3.)

9. 〈h, 1〉

10. 〈, 1〉

27. Reverse Triangle Inequality  Prove that 

11. 〈 f , sin 3 x 〉

12. 〈, cos x 〉

7. 〈g,  j〉

f − g

≤ f −g .

(Hint: Compare Exercise 29 in Section 19.3.)

Finding Fourier Series

28. Convergence of norms  Use Theorem 9 and the Reverse Triangle Inequality (Exercise 27) to prove that lim p n = f .

In Exercises 13–18, find the Fourier coefficients a 0 ,  a k , and b k and the trigonometric polynomial p n of degree n for the given function. If available, use a CAS to plot p n for for various values of n over the domain [ −2π , 4 π ].

29. Let f ( x ) = 1 + cos x + sin x. Find the Fourier coefficients a 0 ,  a k , and b k and the trigonometric polynomial p n of degree n corresponding to f .

13. The function f from Exercise 1.

30. Suppose that f ( x ) = c 0 +

n →∞

∑ j =1 c j cos jx + ∑ j =1 d j sin jx. m

m

14. The function g from Exercise 2.

What are the Fourier coefficients a 0 ,  a k , and b k for f ? Prove that

15. The function h from Exercise 3.

p n is equal to f for all n ≥ m.

16. The function j from Exercise 4.

Establish the identities stated in Exercises 31–33, where m and n are positive integers.

17. The function k from Exercise 5. 18. The function  from Exercise 6.

31. If m ≠ n, then 〈sin mx , sin nx 〉 =

The Parseval Equality

19. Apply the Parseval Equality to the function f defined in ∞

Equation (11) to give another proof that ∑ k =1 1

k2

=

π2

6.

20. Apply the Parseval Equality to the function f from Exercise 1 to prove that ∞

∑ ( 2k −1 1)4

=

k =1

Use that equality to prove that ∞

∑

k =1

M19_HASS5901_15_GE_C19.indd 23

π4 1 = . k4 90

π4 . 96

2π

∫0

sin mx sin nx dx = 0.

(Hint: sin A sin B = ( cos ( A − B ) − cos ( A + B )) 2. ) 32. If m ≠ n, then 〈cos mx , cos nx 〉 =

2π

∫0

cos mx cos nx dx = 0.

(Hint: cos A cos B = ( cos( A + B ) + cos( A − B )) 2.) 33. 〈sin mx , cos nx 〉 =

2π

∫0

sin mx cos nx dx = 0.

(Hint: sin A cos B = 1 sin ( A + B ) + sin ( A − B ) 2 2.) 34. Let f be an integrable function on [ 0, 2π ], and let p n be the trigonometric polynomial defined in Equation (11). Prove that: 〈 p n , 1〉 = 2π a 0 . a. b. 〈 p n , cos jx 〉 = πa k for j = 1, …,  n. c. 〈 p n , sin jx 〉 = πb k for j = 1,  …,  n.

07/03/2023 17:52

19-24

Chapter 19  Fourier Series and Wavelets

19.5 Advanced Topic: The Haar System and Wavelets Fourier series approximations are based on the trigonometric system, which is an orthogonal collection of functions. Today many sets of orthonormal functions are used in mathematics, science, and engineering. We will discuss a family of systems collectively known as wavelets, which are especially useful in signal and image processing, data analysis, and related areas. The JPEG 2000 image compression standard uses wavelets. In contrast to Fourier series approximations, which are best suited to applications involving periodic functions, wavelets can be used to approximate functions that are not periodic. The domain of the functions in the wavelet systems that we will discuss will be the entire real line, but each individual wavelet function that we will consider is identically zero outside of some finite interval, and is integrable on that interval. We say that a function that is integrable on an interval [ a,  b ] and identically zero outside of that interval is supported in [ a,  b ]. The function may be zero at points within [ a,  b ]. If f is supported in [ a,  b ], then its norm is b

∫a

f =

f ( x ) 2 dx.

Since f is zero outside of [ a,  b ], we often write this integral as ∞

∫−∞

f =

f ( x ) 2 dx.

We say that f is a unit function if f = 1. If g is another integrable function that is supported in an interval [ c,  d ] (which need not equal [ a,  b ]), then the inner product of f and g is

〈 f ,  g〉 =

∞

∫−∞

f ( x ) g( x ) dx. (14)

The product f ( x ) g( x ) is identically zero outside of the intersection of the intervals [ a,  b ] and [ c,  d ]. We say that two functions f and g are orthogonal if 〈 f ,  g〉 = 0. Functions f and g are orthonormal if they are orthogonal and both are unit functions A f = g = 1B. A collection of functions is orthogonal if any two different functions from the collection are orthogonal, and the collection is orthonormal if it is orthogonal and every function in the collection is a unit function.

The Haar System The first wavelet system was introduced by Alfréd Haar (1885–1933). The construction is based on two simple functions that are each supported in the interval [ 0, 1 ]. The first is the box function or Haar scaling function,  1, ϕ( x ) =   0, 

0 ≤ x < 1, x < 0 or  x ≥ 1,

and the second is the square wave or Haar wavelet,   1,  ψ ( x ) =  −1,   0,  

0 ≤ x

FIGURE A.29

The maximum value of y = x − x 3 on [ 0, 1 ] occurs at the irrational number x = 1 3.

x

O1

For any a and b, either a ≤ b or b ≤ a or both.

O2

If a ≤ b and b ≤ a then a = b.

O3

If a ≤ b and b ≤ c then a ≤ c.

O4

If a ≤ b then a + c ≤ b + c.

O5

If a ≤ b and 0 ≤ c then ac ≤ bc.

O3 is the transitivity law, and O4 and O5 relate ordering to addition and multiplication. We can order the reals, the integers, and the rational numbers, but we cannot order the complex numbers (there is no reasonable way to decide whether a number like i = −1 is bigger or smaller than zero). A field in which the size of any two elements can be compared as above is called an ordered field. Both the rational numbers and the real numbers are ordered fields, and there are many others. We can think of real numbers geometrically, lining them up as points on a line. The completeness property says that the real numbers correspond to all points on the line, with no “holes” or “gaps.” The rationals, in contrast, omit points such as 2 and π, and the integers even leave out fractions like 1 2. The reals, having the completeness property, omit no points. What exactly do we mean by this vague idea of missing holes? To answer this we must give a more precise description of completeness. A number M is an upper bound for a set of numbers if all numbers in the set are smaller than or equal to M . M is a least upper bound if it is the smallest upper bound. For example, M = 2 is an upper bound for the negative numbers. So is M = 1, showing that 2 is not a least upper bound. The least upper bound for the set of negative numbers is M = 0. We define a complete ordered field to be one in which every nonempty set bounded above has a least upper bound. If we work with just the rational numbers, the set of numbers less than 2 is bounded, but it does not have a rational least upper bound, since any rational upper bound M must be larger than 2 and can be replaced by a slightly smaller rational number that is still larger than 2. So the rationals are not complete. In the real numbers, a set that is bounded above always has a least upper bound. The reals are a complete ordered field. The completeness property is at the heart of many results in calculus. One example occurs when we search for a maximum value for a function on a closed interval [ a, b ], as in Section 4.1. The function y = x − x 3 has a maximum value on [ 0, 1 ] at the point x satisfying 1 − 3 x 2 = 0, or x = 1 3. If we limited our consideration to functions defined only on rational numbers, we would have to conclude that the function has no maximum, since 1 3 is irrational (Figure A.29). The Extreme Value Theorem (Section 4.1), which implies that continuous functions on closed intervals [ a, b ] have a maximum value, is not true for functions defined only on the rationals. The Intermediate Value Theorem implies that a continuous function f on an interval [ a, b ] with f (a) < 0 and f (b) > 0 must be zero somewhere in [ a, b ]. The function values cannot jump from negative to positive without there being some point x in [ a, b ] where f ( x ) = 0. The Intermediate Value Theorem also relies on the completeness of the real numbers and is false for continuous functions defined only on the rationals. The function f ( x ) = 3 x 2 − 1 has f (0) = −1 and f (1) = 2, but if we consider f only on the rational numbers, it never equals zero. The only value of x for which f ( x ) = 0 is x = 1 3, an irrational number. We have captured the desired properties of the reals by saying that the real numbers are a complete ordered field. But we’re not quite finished. Greek mathematicians in the

A.7  Theory of the Real Numbers

AP-29

school of Pythagoras tried to impose another property on the numbers of the real line, the condition that all numbers are ratios of integers. They learned that their effort was doomed when they discovered irrational numbers such as 2. How do we know that our efforts to specify the real numbers are not also flawed, for some unseen reason? The artist Escher drew optical illusions of spiral staircases that went up and up until they rejoined themselves at the bottom. An engineer trying to build such a staircase would find that no structure realized the plans the architect had drawn. Could it be that our design for the reals contains some subtle contradiction, and that no construction of such a number system can be made? We resolve this issue by giving a specific description of the real numbers and verifying that the algebraic, order, and completeness properties are satisfied in this model. This is called a construction of the reals, and just as stairs can be built with wood, stone, or steel, there are several approaches to constructing the reals. One construction treats the reals as all the infinite decimals, a.d1d 2 d 3d 4  . In this approach a real number is an integer a followed by a sequence of decimal digits d1 ,  d 2 ,  d 3 , … , each between 0 and 9. This sequence may stop, or repeat in a periodic pattern, or keep going forever with no pattern. In this form, 2.00, 0.3333333 . . .  and 3.1415926535898 . . . represent three familiar real numbers. The real meaning of the dots “. . .” following these digits requires development of the theory of sequences and series, as in Chapter 9. Each real number is constructed as the limit of a sequence of rational numbers given by its finite decimal approximations. An infinite decimal is then the same as a series a+

d1 d + 2 + . 10 100

This decimal construction of the real numbers is not entirely straightforward. It’s easy enough to check that it gives numbers that satisfy the completeness and order properties, but verifying the algebraic properties is rather involved. Even adding or multiplying two numbers requires an infinite number of operations. Making sense of division requires a careful argument involving limits of rational approximations to infinite decimals. A different approach was taken by Richard Dedekind (1831–1916), a German mathematician, who gave the first rigorous construction of the real numbers in 1872. Given any real number x, we can divide the rational numbers into two sets: those less than or equal to x and those greater. Dedekind cleverly reversed this reasoning and defined a real number to be a division of the rational numbers into two such sets. This seems like a strange approach, but such indirect methods of constructing new structures from old are powerful tools in theoretical mathematics. These and other approaches can be used to construct a system of numbers having the desired algebraic, order, and completeness properties. A final issue that arises is whether all the constructions give the same thing. Is it possible that different constructions result in different number systems satisfying all the required properties? If yes, which of these comprises the real numbers? Fortunately, the answer turns out to be no. The reals are the only number system satisfying the algebraic, order, and completeness properties. Confusion about the nature of the numbers and about limits caused considerable controversy in the early development of calculus. Calculus pioneers such as Newton, Leibniz, and their successors, when looking at what happens to the difference quotient Δy f ( x + Δx ) − f ( x ) = Δx Δx as each of Δy and Δx approach zero, talked about the resulting derivative being a quotient of two infinitely small quantities. These “infinitesimals,” written dx and dy, were thought to be some new kind of number, smaller than any fixed number but not zero. Similarly, a definite integral was thought of as a sum of an infinite number of infinitesimals f ( x ) ⋅ dx

AP-30

Appendix A

as x varied over a closed interval. The approximating difference quotients Δy Δx were understood much as today, but it was the quotient of infinitesimal quantities, rather than a limit, that was thought to encapsulate the meaning of the derivative. This way of thinking led to logical difficulties, as attempted definitions and manipulations of infinitesimals ran into contradictions and inconsistencies. The more concrete and computable difference quotients did not cause such trouble, but they were thought of merely as useful calculation tools. Difference quotients were used to work out the numerical value of the derivative and to derive general formulas for calculation, but they were not considered to be at the heart of what the derivative actually was. Today we realize that the logical problems associated with infinitesimals can be avoided by defining the derivative to be the limit of its approximating difference quotients. The ambiguities of the old approach are no longer present, and in the standard theory of calculus, infinitesimals are neither needed nor used.

A.8 Probability The outcome of some events, such as a heavy rock falling from a great height, can be modeled so that we can predict with high accuracy what will happen. On the other hand, many events have more than one possible outcome and which one of them will occur is uncertain. If we toss a coin, a head or a tail will result with each outcome being equally likely, but we do not know in advance which one it will be. If we randomly select and then weigh a person from a large population, there are many possible weights the person might have, and it is not certain whether the weight will be between 80 and 85 kg. We are told it is highly likely, but not known for sure, that an earthquake of magnitude 6.0 or greater on the Richter scale will occur near a major population area in California within the next one hundred years. Events having more than one possible outcome are probabilistic in nature, and when modeling them we assign a probability to the likelihood that a particular outcome may occur. In this section we show how calculus plays a central role in making predictions with probabilistic models.

Random Variables We begin our discussion with some familiar examples of uncertain events for which the collection of all possible outcomes is finite. EXAMPLE 1 (a) If we toss a coin once, there are two possible outcomes { H, T }, where H represents the coin landing head face up and T a tail landing face up. If we toss a coin three times, there are eight possible outcomes, taking into account the order in which a head or tail occurs. The set of outcomes is { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT }. (b) If we roll a six-sided die once, the set of possible outcomes is {1, 2, 3, 4, 5, 6 } representing the six faces of the die. (c) If we select at random two cards from a 52-card deck, there are 52 possible outcomes for the first card drawn and then 51 possibilities for the second card. Since the order of the cards does not matter, there are ( 52 ⋅ 51) 2 = 1,326 possible outcomes altogether. It is customary to refer to the set of all possible outcomes as the sample space for an event. With an uncertain event we are usually interested in which outcomes, if any, are more likely to occur than others, and to how large an extent. In tossing a coin three times, is it more likely that two heads or that one head will result? To answer such questions, we need a way to quantify the outcomes. DEFINITION A random variable is a function X that assigns a numerical value

to each outcome in a sample space.

A.8  Probability

AP-31

Random variables that have only finitely many values are called discrete random variables. A continuous random variable can take on values in an entire interval, and it is associated with a distribution function, which we explain later. EXAMPLE 2 (a) Suppose we toss a coin three times giving the possible outcomes { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT }. Define the random variable X to be the number of heads that appear. So X ( HHT ) = 2, X ( THT ) = 1, and so forth. Since X can only assume the values 0, 1, 2, or 3, it is a discrete random variable. (b) We spin an arrow anchored by a pin located at the origin. The arrow can wind up pointing in any possible direction, and we define the random variable X as the radian angle the arrow makes with the positive x-axis, measured counterclockwise. In this case, X is a continuous random variable that can take on any value in the interval [ 0, 2π ). (c) The weight of a randomly selected person in a given population is a continuous random variable W. The cholesterol level of a randomly chosen person, and the waiting time for service of a person in a queue at a bank, are also continuous random variables. (d) The scores on the national ACT Examination for college admissions in a particular year are described by a discrete random variable S taking on integer values between 1 and 36. If the number of outcomes is large, or for reasons involving statistical analysis, discrete random variables such as test scores are often modeled as continuous random variables (Example 13). (e) We roll a pair of dice and define the random variable X to be the sum of the numbers on the top faces. This sum can only assume the integer values from 2 through 12, so X is a discrete random variable. (f) A tire company produces tires for mid-sized sedans. The tires are guaranteed to last for 50,000 kilometers, but some will fail sooner, and some will last many more kilometers beyond 50,000 kilometers. The lifetime in kilometers of a tire is described by a continuous random variable L.

Probability Distributions A probability distribution describes the probabilistic behavior of a random variable. Our chief interest is in probability distributions associated with continuous random variables, but to gain some perspective we first consider a distribution for a discrete random variable. Suppose we toss a coin three times, with each side H or T equally likely to occur on a given toss. We define the discrete random variable X that assigns the number of heads appearing in each outcome, giving the following. X

{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ 3 2 2 1 2 1 1 0

Next we count the frequency or number of times a specific value of X occurs. Because each of the eight outcomes is equally likely to occur, we can calculate the probability of the random variable X by dividing the frequency of each value by the total number of outcomes. We summarize our results as follows:

Value of X Frequency P( X )

0 1

1 3

18

38

2 3 38

3 1 18

AP-32

Appendix A

P 3 8 1 4 1 8

0

1

2

X

3

FIGURE A.30

Probability bar graph for the random variable X when tossing a fair coin three times.

We display this information in a probability bar graph of the discrete random variable X, as shown in Figure A.30. The values of X are portrayed by intervals of length 1 on the x-axis so the area of each bar in the graph is the probability of the corresponding outcome. For instance, the probability that exactly two heads occurs in the three tosses of the coin is the area of the bar associated with the value X = 2, which is 3 8. Similarly, the probability that two or more heads occurs is the sum of areas of the bars associated with the values X = 2 and X = 3, or 4 8. The probability that either zero or three heads occurs is 1 1 1 8 + 8 = 4 , and so forth. Note that the total area of all the bars in the graph is 1, which is the sum of all the probabilities for X. With a continuous random variable, even when the outcomes are equally likely, we cannot simply count the number of outcomes in the sample space or the frequencies of outcomes that lead to a specific value of X. In fact, the probability that X takes on any particular one of its values is zero. What is meaningful to ask is how probable it is that the random variable takes on a value within some specified interval of values. We capture the information we need about the probabilities of X in a function whose graph behaves much like the bar graph in Figure A.30. That is, we take a nonnegative function f defined over the range of the random variable with the property that the total area beneath the graph of f is 1. The probability that a value of the random variable X lies within some specified interval [ c, d ] is then the area under the graph of f over that interval. The following definition assumes the range of the continuous random variable X is any real value, but the definition is general enough to account for random variables having a range of finite length. DEFINITIONS A probability density function for a continuous random variable is a function f defined over (−∞, ∞ ) and having the following properties:

1. f is continuous, except possibly at a finite number of points. 2. f is nonnegative, so f ≥ 0. 3.

y

∞

∫−∞ f ( x ) dx

= 1.

If X is a continuous random variable with probability density function f , the probability that X assumes a value in the interval between X = c and X = d is given by the integral

y = f (x)

P( c ≤ X ≤ d ) =

∫c

d

f ( x ) dx.

x ∞

L −∞

The probability that a continuous random variable X assumes a particular real value c is c P( X = c ) = ∫c f ( x ) dx = 0 , consistent with our previous assertion. Since the area under the graph of f over the interval [ c, d ] is only a portion of the total area beneath the graph, the probability P( c ≤ X ≤ d ) is always a number between zero and one. Figure A.31 illustrates a probability density function. A probability density function for a random variable X resembles the density function for a wire of varying density. To obtain the mass of a segment of the wire, we integrate the density of the wire over an interval. To obtain the probability that a random variable has values in a particular interval, we integrate the probability density function over that interval.

f(x) dx = 1

y

c

d

x

d

P(c ≤ X ≤ d ) =

c L

FIGURE A.31

f(x) dx

A probability density function for the continuous random variable X.

EXAMPLE 3 of x.

Let f ( x ) = 2e −2 x if 0 ≤ x < ∞ and f ( x ) = 0 for all negative values

(a) Verify that f is a probability density function. (b) The time T in hours until a car passes a spot on a remote road is described by the probability density function f . Find the probability P( T ≤ 1) that a hitchhiker at that spot will see a car within one hour. (c) Find the probability P( T = 1) that a car passes by the spot after precisely one hour.

A.8  Probability

AP-33

Solution

(a) The function f is continuous except at x = 0, and is everywhere nonnegative. Moreover, ∞

∫−∞ f ( x ) dx

=

∞

∫0

2e −2 x dx = lim

b →∞

b

∫0

2e −2 x dx = lim (1 − e −2 b ) = 1. b →∞

So all of the conditions are satisfied and we have shown that f is a probability density function. (b) The probability that a car comes after a time lapse between zero and one hour is given by integrating the probability density function over the interval [ 0, 1 ]. So P( T ≤ 1) =

1

∫0

1

2e −2 t dt = −e −2 t ⎤⎥ = 1 − e −2 ≈ 0.865. ⎦0

This result can be interpreted to mean that if 100 people were to hitchhike at that spot, about 87 of them can expect to see a car within one hour. 1 (c) This probability is the integral ∫1 f (t ) dt, which equals zero. We interpret this to mean that a sufficiently accurate measurement of the time until a car comes by the spot would have no possibility of being precisely equal to one hour. It might be very close, perhaps, but it would not be exactly one hour. We can extend the definition to finite intervals. If f is a nonnegative function with at most finitely many discontinuities over the interval [ a, b ], and its extension F to (−∞, ∞ ), obtained by defining F to be 0 outside of [ a, b ], satisfies the definition for a probability density function, then f is a probability density function for [ a, b ]. This means that b ∫a f ( x ) dx = 1. Similar definitions can be made for the intervals ( a, b ), ( a, b ], and [ a, b ). 4 2( EXAMPLE 4 Show that f ( x ) = x 3 − x ) is a probability density function over 27 the interval [ 0, 3 ]. Solution The function f is continuous and nonnegative over [ 0, 3 ]. Also, 3

∫0

(

)

3 4 2( 4 ⎡ 3 1 4 81 27 − x 3 − x ) dx = x − x 4 ⎤⎥ = = 1. ⎢ 27 27 ⎣ 4 ⎦0 27 4

We conclude that f is a probability density function over [ 0, 3 ].

Exponentially Decreasing Distributions The distribution in Example 3 is called an exponentially decreasing probability density function. These probability density functions always take on the form y

0.10

f (x) =

0

if x < 0

0.1e−0.1x

if x ≥ 0

0.08 0.06 0.04 0.02 0

Area = 1

4

FIGURE A.32

8

12

16

x

An exponentially decreasing probability density function.

⎪⎧ 0 if  x < 0 f ( x ) = ⎪⎨ −cx ⎪⎪⎩ ce if  x ≥ 0 (see Exercise 23). Exponential density functions can provide models for describing random variables such as the lifetimes of light bulbs, radioactive particles, tooth crowns, and many kinds of electronic components. They also model the amount of time until some specific event occurs, such as the time until a pollinator arrives at a flower, the arrival times of a bus at a stop, the time between individuals joining a queue, the waiting time between phone calls at a help desk, and even the lengths of the phone calls themselves. A graph of an exponential density function is shown in Figure A.32. Random variables with exponential distributions are memoryless. If we think of X as describing the lifetime of some object, then the probability that the object survives for at least s + t hours, given that it has survived t hours, is the same as the initial probability that it survives for at least s hours. For instance, the current age t of a radioactive particle does not change the probability that it will survive for at least another time period of length s. Sometimes the exponential distribution is used as a model when the memoryless

AP-34

Appendix A

principle is violated because it provides reasonable approximations that are good enough for their intended use. For instance, this might be the case when predicting the lifetime of an artificial hip replacement or heart valve for a particular patient. Here is an application illustrating the exponential distribution. EXAMPLE 5 An electronics company models the lifetime T in years of a chip they manufacture with the exponential density function ⎧⎪ 0 f (t ) = ⎪⎨ ⎪⎪ 0.1e −0.1t ⎩

if  t < 0 if  t ≥ 0.

Using this model, (a) Find the probability P( T > 2 ) that a chip will last for more than two years. (b) Find the probability P( 4 ≤ T ≤ 5 ) that a chip will fail in the fifth year. (c) If 1000 chips are shipped to a customer, how many can be expected to fail within three years? Solution

(a) The probability that a chip lasts at least two years is P( T > 2 ) =

∞

∫2

b

b →∞ ∫ 2

0.1e −0.1t dt = lim

0.1e −0.1t dt

= lim [ e −0.2 − e −0.1b ] = e −0.2 ≈ 0.819. b →∞

That is, about 82% of the chips last more than two years. (b) The probability is P( 4 ≤ T ≤ 5 ) =

5

∫4

5

0.1e −0.1t dt = −e −0.1t ⎥⎤ = e −0.4 − e −0.5 ≈ 0.064 ⎦4

which means that about 6% of the chips fail during the fifth year. (c) We want the probability 3

0.1e −0.1t dt = −e −0.1t ⎤⎥ = 1 − e −0.3 ≈ 0.259. ⎦0 We can expect that about 259 of the 1000 chips will fail within three years. P( 0 ≤ T ≤ 3 ) =

3

∫0

Expected Values, Means, and Medians Suppose the weight in lbs of a steer raised on a cattle ranch is described by a continuous random variable W with probability density function f ( w) and that the rancher can sell a steer of weight w for g( w) dollars. How much can the rancher expect to earn for a randomly chosen steer on the ranch? To answer this question, we consider a small interval [ w i ,  w i +1 ] of width Δw i and note that the probability a steer has weight in this interval is wi +1

∫w

f ( w) dw ≈ f ( w i ) Δw i .

i

The earning on a steer in this interval is approximately g( w i ). The Riemann sum

∑ g(w i ) f (w i ) Δw i then approximates the amount the rancher would receive for a steer. We assume that steers have a maximum weight, so f is zero outside some finite interval [0, b]. Then taking the limit of the Riemann sum as the width of each interval approaches zero gives the integral ∞

∫−∞ g(w) f (w) dw.

A.8  Probability

AP-35

This integral estimates how much the rancher can expect to earn for a typical steer on the ranch and is the expected value of the function g. The expected values of certain functions of a random variable X have particular importance in probability and statistics. One of the most important of these functions is the expected value of the function g( x ) = x .

The expected value or mean of a continuous random variable X with probability density function f is the number

DEFINITION

μ = E (X ) =

∞

∫−∞ x f ( x ) dx.

The expected value E( X ) can be thought of as a weighted average of the random variable X, where each value of X is weighted by f ( X ). The mean can also be interpreted as the long-run average value of the random variable X, and it is one measure of the centrality of the random variable X. EXAMPLE 6 density function

Find the mean of the random variable X with exponential probability ⎪⎧ 0 f ( x ) = ⎪⎨ −cx ⎪⎪ ce ⎩

if  x < 0 if  x ≥ 0.

Solution From the definition we have

μ=

∞

∫−∞ x f ( x ) dx b

b →∞ ∫ 0

= lim

=

∞

∫0

x ce −cx dx

⎛ ⎤b x ce −cx dx = lim ⎜⎜−x e −cx ⎥ + b →∞ ⎜ ⎥⎦ 0 ⎝

(

⎞

∫0 e −cx dx ⎟⎟⎟⎟⎠ b

)

1 1 1 = lim −be −cb − e −cb + = . b →∞ c c c

l’Hôpital’s Rule on first term

Therefore, the mean is μ = 1 c. From the result in Example 6, knowing the mean or expected value μ of a random variable X having an exponential density function allows us to write its entire formula.

Exponential Density Function for a Random Variable X with Mean μ

⎧⎪ 0 if  x < 0 f ( x ) = ⎪⎨ −1 − x μ ⎪⎪⎩ μ e if  x ≥ 0

EXAMPLE 7 Suppose the time T before a chip fails in Example 5 is modeled instead by the exponential density function with a mean of eight years. Find the probability that a chip will fail within five years. Solution The exponential density function with mean μ = 8 is

⎧⎪ ⎪ f (t ) = ⎪⎨ ⎪⎪ ⎪⎩

0 1 −t 8 e 8

if  t < 0 if  t ≥ 0

AP-36

Appendix A

Then the probability a chip will fail within five years is the definite integral 5

0.125e −0.125 t dt = −e −0.125 t ⎥⎤ = 1 − e −0.625 ≈ 0.465 ⎦0 so about 47% of the chips can be expected to fail within five years. P( 0 ≤ T ≤ 5 ) =

y 0.6

f(x) =

4 2 27 x (3

EXAMPLE 8 Find the expected value for the random variable X with probability density function given by Example 4.

− x)

Solution The expected value is

0.4

3

∫0

μ = E (X ) =

0.2 0

5

∫0

0.5

1

1.5

2

2.5

3

x

=

FIGURE A.33

The expected value of a random variable with this probability density function is μ = 1.8 (Example 8).

3 4 3( 4 ⎡3 4 1 x 3 − x ) dx = x − x 5 ⎤⎥ ⎢ 27 27 ⎣ 4 5 ⎦0

(

)

4 243 243 − = 1.8 27 4 5

From Figure A.33, you can see that this expected value is reasonable because the region beneath the probability density function appears to be balanced about the vertical line x = 1.8. That is, the horizontal coordinate of the centroid of a plate described by the region is x = 1.8. There are other ways to measure the centrality of a random variable with a given probability density function. The median of a continuous random variable X with probability density function f is the number m for which

DEFINITION

m

∫−∞ f ( x ) dx

=

1 2

∞

∫m

and

f ( x ) dx =

1 . 2

The definition of the median means that there is an equal likelihood that the random variable X will be smaller than m or larger than m. EXAMPLE 9 density function

Find the median of a random variable X with exponential probability ⎪⎧ 0 f ( x ) = ⎪⎨ −cx ⎪⎪ ce ⎩

if  x < 0 if  x ≥ 0.

Solution The median m must satisfy

1 = 2

m

∫0

m

ce −cx dx = −e −cx ⎤⎥ = 1 − e −cm . ⎦0

It follows that e −cm =

1 2

m =

or

1 ln 2. c

Also, b

b 1 ⎡ ⎤ = lim ∫ ce −cx dx = lim ⎢ −e −cx ⎥ = lim ( e −cm − e −cb ) = e −cm b →∞ m b →∞ ⎣ b →∞ 2 ⎦m

giving the same value for m. Since 1 c is the mean μ of X with an exponential distribution, we conclude that the median is m = μ ln 2. The mean and median differ because the probability density function is skewed and spreads toward the right.

A.8  Probability

y = f (x)

m

AP-37

Variance and Standard Deviation

x

Random variables with exactly the same mean μ but different distributions can behave very differently (see Figure A.34). The variance of a random variable X measures how spread out the values of X are in relation to the mean, and we measure this dispersion by the expected value of ( X − μ ) 2 . Since the variance measures the expected square of the difference from the mean, we often work instead with its square root.

FIGURE A.34

Probability density functions with the same mean can have different spreads in relation to the mean. The blue and red regions under the curves have equal area.

The variance of a random variable X with probability density function f is the expected value of ( X − μ ) 2 :

DEFINITIONS

∞

∫−∞ ( x − μ )2 f ( x ) dx

Var ( X ) = The standard deviation of X is σX =

Var( X ) =

∞

∫−∞ ( x − μ )2 f ( x ) dx .

EXAMPLE 10 Find the standard deviation of the random variable T in Example 5, and find the probability that T lies within one standard deviation of the mean. Solution The probability density function is the exponential density function with mean μ = 10 by Example 6. To find the standard deviation we first calculate the variance integral: ∞

∞

∫−∞ ( t − μ )2 f (t ) dt = ∫0

(t

− 10 ) 2 ( 0.1e −0.1t ) dt b

( t − 10 ) b →∞ ∫ 0

= lim

2

( 0.1e −0.1t ) dt b

⎡ ⎤ = lim ⎢ (−( t − 10 ) 2 − 20 ( t − 10 )) e −0.1t ⎥ b →∞ ⎢⎣ ⎥⎦ 0 b

20e −0.1t dt b →∞ ∫ 0

+ lim

Integrating by parts

⎤b = [ 0 + (−10 ) 2 + 20 (−10 ) ] − 20 lim (10e −0.1t )⎥ b →∞ ⎥⎦ 0 = −100 − 200 lim ( e −0.1b − 1) = 100. b →∞

The standard deviation is the square root of the variance, so σ = 10.0. To find the probability that T lies within one standard deviation of the mean, we find the probability P( μ − σ ≤ T ≤ μ + σ ). For this example, we have ⎤ 20 −0.1t dt = −e −0.1t 0.1 e ⎥ = 1 − e −2 ≈ 0.865 ∫0 ⎥⎦ 0 This means that about 87% of the chips will fail within twenty years. P(10 − 10 ≤ T ≤ 10 + 10 ) =

20

Uniform Distributions The uniform distribution is very simple, but it occurs commonly in applications. The probability density function for this distribution on the interval [ a, b ] is f (x) =

1 , a ≤ x ≤ b. b−a

If each outcome in the sample space is equally likely to occur, then the random variable X has a uniform distribution. Since f is constant on [ a, b ], a random variable with a uniform

AP-38

Appendix A

distribution is just as likely to be in one subinterval of a fixed length as in any other of the same length. The probability that X assumes a value in a subinterval of [ a, b ] is the length of that subinterval divided by ( b − a ). EXAMPLE 11 An anchored arrow is spun around the origin, and the random variable X is the radian angle the arrow makes with the positive x-axis, measured within the interval [ 0, 2π ). Assuming there is equal probability for the arrow pointing in any direction, find the probability density function and the probability that the arrow ends up pointing between North and East. Solution We model the probability density function with the uniform distribution f ( x ) = 1 2π , 0 ≤ x < 2π , and f ( x ) = 0 elsewhere. The probability that the arrow ends up pointing between North and East is given by

(

P 0 ≤ X ≤

) ∫

π = 2

π 2 0

1 1 dx = . 2π 4

Normal Distributions 2 2 1 f(x) = e−(x−m) 2s s "2p

Numerous applications use the normal distribution, which is defined by the probability density function

x

m−s m m+s

FIGURE A.35

The normal probability density function with mean μ and standard deviation σ.

99.7% within 3 standard deviations of the mean 95% within 2 standard deviations of the mean

f (x) =

2 1 e −( x −μ ) σ 2π

2σ 2

.

It can be shown that the mean of a random variable X with this probability density function is μ and its standard deviation is σ. In applications the values of μ and σ are often estimated using large sets of data. The function is graphed in Figure A.35, and the graph is sometimes called a bell curve because of its shape. Since the curve is symmetric about the mean, the median for X is the same as its mean. It is often observed in practice that many random variables have approximately a normal distribution. Some examples illustrating this phenomenon are the height of a man, the annual rainfall in a certain region, an individual’s blood pressure, the serum cholesterol level in the blood, the brain weights in a certain population of adults, and the amount of growth in a given period for a population of sunflower seeds. The normal probability density function does not have an antiderivative expressible in terms of familiar functions. Once μ and σ are fixed, however, an integral involving the normal probability density function can be computed using numerical integration methods. Usually we use the numerical integration capability of a computer or calculator to estimate the values of these integrals. Such computations show that for any normal distribution, we get the following values for the probability that the random variable X lies within k = 1, 2, 3, or 4 standard deviations of the mean: P ( μ − σ < X < μ + σ ) ≈ 0.68269 P ( μ − 2σ < X < μ + 2σ ) ≈ 0.95450

68% within 1 standard deviation of the mean

P ( μ − 3σ < X < μ + 3σ ) ≈ 0.99730 P ( μ − 4σ < X < μ + 4σ ) ≈ 0.99994

34% 2.14%

34%

13.6%

m − 3s m − 2s m − s

13.6% m

2.14%

m + s m + 2s m + 3s

This means, for instance, that the random variable X will take on a value within two standard deviations of the mean about 95% of the time. About 68% of the time, X will lie within one standard deviation of the mean (see Figure A.36).

FIGURE A.36

Probabilities of the normal distribution within its standard deviation bands.

EXAMPLE 12 An individual’s blood pressure is an important indicator of overall health. A medical study of healthy individuals between 14 and 70 years of age modeled

A.8  Probability

AP-39

their systolic blood pressure using a normal distribution with mean 119.7 mm Hg and standard deviation 10.9 mm Hg. (a) Using this model, what percentage of the population has a systolic blood pressure between 140 and 160 mm Hg, the levels set by the American Heart Association for Stage 1 hypertension? (b) What percentage has a blood pressure between 160 and 180 mm Hg, the levels set by the American Heart Association for Stage 2 hypertension? (c) What percentage has a blood pressure in the normal range of 90–120, as set by the American Heart Association? Solution

(a) Since we cannot find an antiderivative, we use a computer to evaluate the probability integral of the normal probability density function with μ = 119.7 and σ = 10.9: P(140 ≤ X ≤ 160 ) =

160

∫140

2 1 e −( x −119.7 ) 10.9 2π

2(10.9 ) 2

dx ≈ 0.03117.

This means that about 3% of the population in the studied age range have Stage 1 hypertension. (b) Again we use a computer to calculate the probability that the blood pressure is between 160 and 180 mm Hg: P(160 ≤ X ≤ 180 ) =

180

∫160

2 1 e −( x −119.7 ) 10.9 2π

2(10.9 ) 2

dx ≈ 0.00011.

We conclude that about 0.011% of the population has Stage 2 hypertension. (c) The probability that the blood pressure falls in the normal range is P( 90 ≤ X ≤ 120 ) =

120

∫90

2 1 e −( x −119.7 ) 10.9 2π

2(10.9 ) 2

dx ≈ 0.50776.

That is, about 51% of the population has a normal systolic blood pressure. Many national tests are standardized using the normal distribution. The following example illustrates modeling the discrete random variable for scores on a test using the normal distribution function for a continuous random variable. EXAMPLE 13 The ACT is a standardized test taken by high school students seeking admission to many colleges and universities. The test measures knowledge, skills, and proficiency in the areas of English, math, and science, with scores ranging over the interval [ 1, 36 ]. Nearly 1.5 million high school students took the test in 2009, and the composite mean score across the academic areas was μ = 21.1 with standard deviation σ = 5.1. (a) What percentage of the population had an ACT score between 18 and 24? (b) What is the ranking of a student who scored 27 on the test? (c) What is the minimal integer score a student needed to get in order to be in the top 8% of the scoring population? Solution

(a) We use a computer to evaluate the probability integral of the normal probability density function with μ = 21.1 and σ = 5.1: P(18 ≤ X ≤ 24 ) =

24

∫18

2 1 e −( x −21.1) 5.1 2π

2( 5.1) 2

dx ≈ 0.44355.

This means that about 44% of the students had an ACT score between 18 and 24.

AP-40

Appendix A

(b) Again we use a computer to calculate the probability of a student getting a score lower than 27 on the test: P(1 ≤ X < 27 ) =

∫1

27

2 1 e −( x −21.1) 5.1 2π

2( 5.1) 2

dx ≈ 0.87630.

We conclude that about 88% of the students scored below a score of 27, so the student ranked in the top 12% of the population. (c) We look at how many students had a mark higher than 28: P( 28 < X ≤ 36 ) =

36

∫28

1 5.1 2π

e −( x −21.1)

2 2( 5.1) 2

dx ≈ 0.0863.

Since this number gives more than 8% of the students, we look at the next higher integer score: P( 29 < X ≤ 36 ) =

36

∫29

2 1 e −( x −21.1) 5.1 2π

2( 5.1) 2

dx ≈ 0.0595.

Therefore, 29 is the lowest integer score a student could get in order to score in the top 8% of the population (and actually scoring here in the top 6%). The simplest form for a normal distribution of X occurs when its mean is zero and its standard deviation is one. The standard normal probability density function f giving mean μ = 0 and standard deviation σ = 1 is f (x) =

1 −x 2 2 e . 2π

Note that the substitution z = ( x − μ ) σ gives the equivalent integrals b

∫a

2 1 e −(( x −μ ) σ ) 2 dx = σ 2π

β

∫α

1 −z 2 2 e dz , 2π

where α = ( a − μ ) σ and β = ( b − μ ) σ. So we can convert random variable values to the “z-values” to standardize a normal distribution, and then use the integral on the righthand side of the last equation to calculate probabilities for the original random variable normal distribution with mean μ and standard deviation σ. In a normal distribution, we know that 95.5% of the population lies within two standard deviations of the mean, so a random variable X converted to a z-value has more than a 95% chance of occurring in the interval [ −2, 2 ].

EXERCISES

A.8

Probability Density Functions

In Exercises 1–8, determine which are probability density functions and justify your answer. 1. f ( x ) =

1 x over [ 4, 8 ] 18

2. f ( x ) =

1( 2 − x ) over [ 0, 2 ] 2

3. f ( x ) =

2x

⎪⎧⎪ 1 5. f ( x ) = ⎪⎨ x 2 ⎪⎪ ⎪⎪⎩ 0

3]

8. f ( x ) =

x < 0

1 x

over ( 0,   e ]

9. Let f be the probability density function for the random variable L in Example 2f. Explain the meaning of each integral. 32,000

a.

∫25,000

c.

∫0

x ≥1 x X ) 2 sin 2 π x πx 2

AP-41

26. f ( x ) =

1 2 x 9

⎧⎪ 1 ⎪ 28. f ( x ) = ⎪⎨ x ⎪⎪ ⎪⎪⎩ 0

over [ 0, 3 ] 1≤ x ≤ e Otherwise

d. If 200 customers come to the bakery in a day, how many are likely to be served within three minutes? 36. Airport waiting time According to the U.S. Customs and Border Protection Agency, the average airport wait time at Chicago’s O’Hare International airport is 16 minutes for a traveler arriving during the hours 7–8 a.m., and 32 minutes for arrival during the hours 4–5 p.m. The wait time is defined as the total processing time from arrival at the airport until the completion of a passenger’s security screening. Assume the wait time is exponentially distributed. a. What is the probability of waiting between 10 and 30 minutes for a traveler arriving during the 7–8 a.m. hour? b. What is the probability of waiting more than 25 minutes for a traveler arriving during the 7–8 p.m. hour?

AP-42

Appendix A

c. What is the probability of waiting between 35 and 50 minutes for a traveler arriving during the 4–5 p.m. hour? d. What is the probability of waiting less than 20 minutes for a traveler arriving during the 4–5 p.m. hour? 37. Printer lifetime The lifetime of a $200 printer is exponentially distributed with a mean of 2 years. The manufacturer agrees to pay a full refund to a buyer if the printer fails during the first year following its purchase, and a one-half refund if it fails during the second year. If the manufacturer sells 100 printers, how much should it expect to pay in refunds? 38. Failure time The time between failures of a photocopier is exponentially distributed. Half of the copiers at a university require service during the first 2 years of operations. If the university purchased 150 copiers, how many do you expect to require service during the first year of their operation? T Normal Distributions 39. Cholesterol levels The serum cholesterol levels of children aged 12 to 14 years follows a normal distribution with mean μ = 162 mg dl and standard deviation σ = 28 mg dl . In a population of 1000 of these children, how many would you expect to have serum cholesterol levels between 165 and 193? between 148 and 167? 40. Annual rainfall The annual rainfall in millimeters for San Francisco, California, is approximately a normal random variable with mean 510 mm and standard deviation 120 mm. What is the probability that next year’s rainfall will exceed 444 mm? 41. Manufacturing time The assembly time in minutes for a component at an electronic manufacturing plant is normally distributed with a mean of μ = 55 and standard deviation σ = 4. What is the probability that a component will be made in less than one hour? 42. Lifetime of a tire Assume the random variable L in Example 2f is normally distributed with mean μ = 35,000 kilometers and σ = 6,000 kilometers.

46. Brain weights In a population of 500 adult Swedish men, medical researchers find their brain weights to be approximately normally distributed with mean μ = 1400 gm and standard deviation σ = 100 gm. a. What percentage of brain weights are between 1325 and 1450 gm? b. How many men in the population would you expect to have a brain weight exceeding 1480 gm? 47. Blood pressure Diastolic blood pressure in adults is normally distributed with μ = 80 mm Hg and σ = 12 mm Hg. In a random sample of 300 adults, how many would be expected to have a diastolic blood pressure below 70 mm Hg? 48. Albumin levels Serum albumin in healthy 20-year-old men is normally distributed with μ = 4.4 and σ = 0.2. How likely is it for a healthy 20-year-old man to have a level in the range 4.3 to 4.45? 49. Quality control A manufacturer of generator shafts finds that it needs to add additional weight to its shafts in order to achieve proper static and dynamic balance. Based on experimental tests, the average weight it needs to add is μ = 35 gm with σ = 9 gm. Assuming a normal distribution, from 1000 randomly selected shafts, how many would be expected to need an added weight in excess of 40 gm? 50. Kilometers driven A taxicab company in New York City analyzed the daily number of kilometers driven by each of its drivers. It found the average distance was 300 km with a standard deviation of 50 km. Assuming a normal distribution, what prediction can we make about the percentage of drivers who will log in either more than 400 km or less than 250 km? 51. Germination of sunflower seeds The germination rate of a particular seed is the percentage of seeds in the batch which successfully emerge as plants. Assume that the germination rate for a batch of sunflower seeds is 80%, and that among a large population of n seeds the number of successful germinations is normally distributed with mean μ = 0.8n and σ = 0.4 n .

a. In a batch of 4000 tires, how many can be expected to last for at least 29,000 kilometers?

a. In a batch of n = 2500 seeds, what is the probability that at least 1960 will successfully germinate?

b. What is the minimum number of kilometers you would expect to find as the lifetime for 90% of the tires?

b. In a batch of n = 2500 seeds, what is the probability that at most 1980 will successfully germinate?

43. Height The average height of American women aged 18–24 is normally distributed with mean μ = 166 cm and σ = 6 cm. a. What percentage of women are taller than 172 cm? b. What is the probability a woman is between 155 cm and 163 cm tall? 44. Life expectancy At birth, a French citizen has an average life expectancy of 82 years with a standard deviation of 7 years. If 100 newly born French babies are selected at random, how many would you expect to live between 75 and 85 years? Assume life expectancy is normally distributed. 45. Length of pregnancy A team of medical practitioners determines that in a population of 1000 women with ages ranging from 20 to 35 years, the length of pregnancy from conception to birth is approximately normally distributed with a mean of 266 days and a standard deviation of 16 days. How many of these women would you expect to have a pregnancy lasting from 36 weeks to 40 weeks?

c. In a batch of n = 2500 seeds, what is the probability that between 1940 and 2020 will successfully germinate? 52. Suppose you toss a fair coin n times and record the number of heads that land. Assume that n is large and approximate the discrete random variable X with a continuous random variable that is normally distributed with μ = n 2 and σ = n 2. If n = 400, find the given probabilities. a. P (190 ≤ X < 210 ) c.

P( X

>

220 )

b. P ( X < 170 ) d. P ( X = 300 )

Discrete Random Variables

53. A fair coin is tossed four times and the random variable X assigns the number of tails that appear in each outcome. a. Determine the set of possible outcomes. b. Find the value of X for each outcome.

A.9  The Distributive Law for Vector Cross Products

c. Create a probability bar graph for X, as in Figure A.30. What is the probability that at least two heads appear in the four tosses of the coin?

AP-43

the product brand” (D), or “Undecided” (U). For each outcome, the random variable X assigns the number of L’s that appear. a. Find the set of possible outcomes and the range of X.

54. You roll a pair of six-sided dice, and the random variable X assigns to each outcome the sum of the number of dots showing on each face, as in Example 2e.

b. Create a probability bar graph for X. c. What is the probability that at least two people like the product brand?

a. Find the set of possible outcomes.

d. What is the probability that no more than one person dislikes the product brand?

b. Create a probability bar graph for X. c. What is the probability that X = 8?

56. Spacecraft components A component of a spacecraft has both a main system and a backup system operating throughout a flight. The probability that both systems fail sometime during the flight is 0.0148. Assuming that each system separately has the same failure rate, what is the probability that the main system fails during the flight?

d. What is the probability that X ≤ 5?   X > 9? 55. Three people are asked their opinion in a poll about a particular brand of a common product found in grocery stores. They can answer in one of three ways: “Like the product brand” (L), “Dislike

A.9 The Distributive Law for Vector Cross Products In this appendix we prove the Distributive Law u × ( v + w ) = u × v + u × w, which is Property 2 in Section 11.4. Proof To derive the Distributive Law, we construct u × v a new way. We draw u and v from the common point O and construct a plane M perpendicular to u at O (Figure A.37). We then project v orthogonally onto M, yielding a vector v 1 with length v sin θ. We rotate v 1 by 90 ° about u in the positive sense to produce a vector v 2 . Finally, we multiply v 2 by the length of u. The resulting vector u v 2 is equal to u × v since v 2 has the same direction as u × v by its construction (Figure A.37) and u v 2 = u v 1 = u v sin θ = u × v . M1

v u

u

u O

M v1

FIGURE A.37

v2

u×v

90°

As explained in the text, u × v = u v 2 .

Now each of these three operations, namely, 1. projection onto M 2. rotation about u through 90° 3. multiplication by the scalar u , when applied to a triangle whose plane is not parallel to u, will produce another triangle. If we start with the triangle whose sides are v, w, and s = v + w (Figure A.38) and apply these three steps, we successively obtain the following: 1. A triangle whose sides are v 1 , w 1 , and s 1 satisfying the vector equation v1 + w 1 = s1

AP-44

Appendix A

2. A triangle whose sides are v 2 , w 2 , and s 2 satisfying the vector equation v2 + w2 = s2

w

u

s

v

w1 M

v1 s1

The vectors v, w, and v + w and their projections onto a plane perpendicular to u.

FIGURE A.38

3. A triangle whose sides are u v 2 , u w 2 , and u s 2 satisfying the vector equation u v 2 + u w 2 = u s 2. Substituting u v 2 = u × v, u w 2 = u × w, and u s 2 = u × ( v + w ) from our discussion above into this last equation gives u × v + u × w = u × ( v + w ), which is the law we wanted to establish.

A.10 The Mixed Derivative Theorem and the Increment Theorem This appendix derives the Mixed Derivative Theorem (Theorem 2, Section 13.3) and the Increment Theorem for Functions of Two Variables (Theorem 3, Section 13.3). Euler first published the Mixed Derivative Theorem in 1734, in a series of papers he wrote on hydrodynamics.

THEOREM 2—The Mixed Derivative Theorem

If f ( x , y ) and its partial derivatives f x , f y , f xy , and f yx are defined throughout an open region containing a point ( a, b ) and are all continuous at ( a, b ) , then f xy ( a, b ) = f yx ( a, b ). Proof The equality of f xy ( a, b ) and f yx ( a, b ) can be established by four applications of the Mean Value Theorem (Theorem 4, Section 4.2). By hypothesis, the point ( a, b ) lies in the interior of a rectangle R in the xy-plane on which f , f x , f y , f xy , and f yx are all defined. We let h and k be the numbers such that the point ( a + h,  b + k ) also lies in R, and we consider the difference Δ = F ( a + h ) − F (a),

(1)

F ( x ) = f ( x , b + k ) − f ( x , b ).

(2)

where

We apply the Mean Value Theorem to F, which is continuous because it is differentiable. Then Equation (1) becomes Δ = hF ′(c1 ),

(3)

A.10  The Mixed Derivative Theorem and the Increment Theorem

AP-45

where c1 lies between a and a + h. From Equation (2), F ′( x ) = f x ( x , b + k ) − f x ( x , b ) , so Equation (3) becomes Δ = h[ f x ( c1 , b + k ) − f x ( c1 , b ) ].

(4)

Now we apply the Mean Value Theorem to the function g( y) = f x ( c1 , y ) and have g ( b + k ) − g(b) = kg′(d1 ), or f x ( c1 , b + k ) − f x ( c1 , b ) = k f xy ( c1 , d1 ) y

for some d1 between b and b + k. By substituting this into Equation (4), we get Δ = hkf xy ( c1 , d1 )

R k (a, b)

0

(5)

for some point ( c1 , d1 ) in the rectangle R′ whose vertices are the four points ( a, b ) , ( a + h, b ) , ( a + h, b + k ) , and ( a, b + k ). (See Figure A.39.) By substituting from Equation (2) into Equation (1), we may also write

R′ h

Δ = f ( a + h, b + k ) − f ( a + h, b ) − f ( a, b + k ) + f ( a, b )

x

= [ f ( a + h, b + k ) − f ( a, b + k ) ] − [ f ( a + h, b ) − f ( a, b ) ] = φ ( b + k ) − φ (b),

FIGURE A.39

The key to proving f xy ( a, b ) = f yx ( a, b ) is that no matter how small R′ is, f xy and f yx take on equal values somewhere inside R′ (though not necessarily at the same point).

(6)

where φ ( y) = f ( a + h, y ) − f ( a, y ).

(7)

The Mean Value Theorem applied to Equation (6) now gives Δ = kφ′(d 2 )

(8)

for some d 2 between b and b + k. By Equation (7), φ′( y) = f y ( a + h, y ) − f y ( a, y ).

(9)

Substituting from Equation (9) into Equation (8) gives Δ = k [ f y ( a + h, d 2 ) − f y ( a, d 2 ) ]. Finally, we apply the Mean Value Theorem to the expression in brackets and get Δ = khf yx ( c 2 , d 2 )

(10)

for some c 2 between a and a + h. Together, Equations (5) and (10) show that f xy ( c1 , d1 ) = f yx ( c 2 , d 2 ) ,

(11)

where ( c1 , d1 ) and ( c 2 , d 2 ) both lie in the rectangle R′ (Figure A.39). Equation (11) is not quite the result we want, since it says only that f xy has the same value at ( c1 , d1 ) that f yx has at ( c 2 , d 2 ). The numbers h and k in our discussion, however, may be made as small as we wish. The hypothesis that f xy and f yx are both continuous at ( a, b ) means that f xy ( c1 , d1 ) = f xy ( a, b ) + ε1 and f yx ( c 2 , d 2 ) = f yx ( a, b ) + ε 2 , where each of ε1 ,  ε 2 → 0 as both h,  k → 0. Hence, if we let h and k → 0, we have f xy ( a, b ) = f yx ( a, b ). The equality of f xy ( a, b ) and f yx ( a, b ) can be proved with hypotheses weaker than the ones we assumed. For example, it is enough for f ,  f x , and f y to exist in R and for f xy to be continuous at ( a, b ). Then f yx will exist at ( a, b ) and will equal f xy at that point.

AP-46

Appendix A

THEOREM 3—The Increment Theorem for Functions of Two Variables

Suppose that the first partial derivatives of f ( x , y ) are defined throughout an open region R containing the point ( x 0 , y 0 ) and that f x and f y are continuous at ( x 0 , y 0 ). Then the change Δz = f ( x 0 + Δx , y 0 + Δy ) − f ( x 0 , y 0 ) in the value of f that results from moving from ( x 0 , y 0 ) to another point ( x 0 + Δx , y 0 + Δy ) in R satisfies an equation of the form Δz = f x ( x 0 , y 0 ) Δx + f y ( x 0 , y 0 ) Δy + ε1Δx + ε 2Δy in which each of ε1 , ε 2 → 0 as both Δx , Δy → 0.

C(x0 + Δ x, y0 + Δ y)

Proof We work within a rectangle T centered at A( x 0 , y 0 ) and lying within R, and we assume that Δx and Δy are already so small that the line segment joining A to B ( x 0 + Δx , y 0 ) and the line segment joining B to C ( x 0 + Δx , y 0 + Δy ) lie in the interior of T (Figure A.40). We may think of Δz as the sum Δz = Δz 1 + Δz 2 of two increments, where

A(x0, y0 )

Δz 1 = f ( x 0 + Δx , y 0 ) − f ( x 0 , y 0 )

B(x0 + Δx, y0 )

is the change in the value of f from A to B and Δz 2 = f ( x 0 + Δx , y 0 + Δy ) − f ( x 0 + Δx , y 0 )

T

FIGURE A.40

The rectangular region T in the proof of the Increment Theorem. The figure is drawn for Δx and Δy positive, but either increment might be zero or negative.

is the change in the value of f from B to C (Figure A.41). On the closed interval of x-values joining x 0 to x 0 + Δx, the function F ( x ) = f ( x , y 0 ) is a differentiable (and hence continuous) function of x, with derivative F ′( x ) = f x ( x , y 0 ). By the Mean Value Theorem (Theorem 4, Section 4.2), there is an x-value c between x 0 and x 0 + Δx at which F ( x 0 + Δx ) − F ( x 0 ) = F ′(c) Δx or f ( x 0 + Δ x , y 0 ) − f ( x 0 , y 0 ) = f x ( c, y 0 ) Δ x or Δ z 1 = f x ( c, y 0 ) Δ x.

(12)

Similarly, G ( y) = f ( x 0 + Δx , y ) is a differentiable (and hence continuous) function of y on the closed y-interval joining y 0 and y 0 + Δy, with derivative G ′( y) = f y ( x 0 + Δx , y ). Hence, there is a y-value d between y 0 and y 0 + Δy at which G ( y 0 + Δy ) − G ( y 0 ) = G ′(d ) Δy or f ( x 0 + Δx , y 0 + Δy ) − f ( x 0 + Δx , y ) = f y ( x 0 + Δx , d ) Δy or Δz 2 = f y ( x 0 + Δx , d ) Δy.

(13)

A.10  The Mixed Derivative Theorem and the Increment Theorem

AP-47

z S Q P0

z = f (x, y)

Q′

Δz 2

P″

Δz P′

0

Δz 1

A(x0 , y 0 )

y0

y

B

x

y0 + Δ y C(x0 + Δ x, y 0 + Δy)

(x0 + Δ x, y 0 )

Part of the surface z = f ( x , y ) near P0 ( x 0 , y 0 , f ( x 0 , y 0 )). The points P0 , P ′, and P″ have the same height z 0 = f ( x 0 , y 0 ) above the xy-plane. The change in z is Δz = P ′ S. The change

FIGURE A.41

Δz 1 = f ( x 0 + Δx , y 0 ) − f ( x 0 , y 0 ) , shown as P ″Q = P ′Q′, is caused by changing x from x 0 to x 0 + Δx while holding y equal to y 0 . Then, with x held equal to x 0 + Δx, Δz 2 = f ( x 0 + Δx , y 0 + Δy ) − f ( x 0 + Δx , y 0 ) is the change in z caused by changing y 0 from y 0 + Δy, which is represented by Q′ S. The total change in z is the sum of Δz 1 and Δz 2 .

Now, as both Δx and Δy → 0, we know that c → x 0 and d → y 0 . Therefore, since f x and f y are continuous at ( x 0 , y 0 ) , the quantities ε1 = f x ( c , y 0 ) − f x ( x 0 , y 0 ) , ε 2 = f y ( x 0 + Δx , d ) − f y ( x 0 , y 0 )

(14)

both approach zero as both Δx and Δy → 0. Finally, Δz = Δz 1 + Δz 2 = f x ( c, y 0 ) Δ x + f y ( x 0 + Δ x , d ) Δ y

From Eqs. (12) and (13)

= [ f x ( x 0 , y 0 ) + ε1 ] Δx + [ f y ( x 0 ,  y 0 ) + ε 2 ] Δy

From Eq. (14)

= f x ( x 0 , y 0 ) Δ x + f y ( x 0 , y 0 ) Δ y + ε1 Δ x + ε 2 Δ y , where both ε1 and ε 2 → 0 as both Δx and Δy → 0, which is what we set out to prove.

AP-48

Appendix A

Analogous results hold for functions of any finite number of independent variables. Suppose that the first partial derivatives of w = f ( x , y, z ) are defined throughout an open region containing the point ( x 0 , y 0 , z 0 ) and that f x , f y , and f z are continuous at ( x 0 , y 0 , z 0 ). Then Δw = f ( x 0 + Δx , y 0 + Δy, z 0 + Δz ) − f ( x 0 , y 0 , z 0 ) = f x Δ x + f y Δ y + f z Δ z + ε1 Δ x + ε 2 Δ y + ε 3 Δ z ,

(15)

where ε1 , ε 2 , and ε 3 → 0 as Δx , Δy, and Δz → 0. The partial derivatives f x , f y , f z in Equation (15) are to be evaluated at the point ( x 0 , y 0 , z 0 ). Equation (15) can be proved by treating Δw as the sum of three increments, Δw1 = f ( x 0 + Δx , y 0 , z 0 ) − f ( x 0 , y 0 , z 0 )

(16)

Δ w 2 = f ( x 0 + Δ x , y 0 + Δ y, z 0 ) − f ( x 0 + Δ x , y 0 , z 0 )

(17)

Δ w 3 = f ( x 0 + Δ x , y 0 + Δ y, z 0 + Δ z ) − f ( x 0 + Δ x , y 0 + Δ y, z 0 ) ,

(18)

and applying the Mean Value Theorem to each of these separately. Two coordinates remain constant and only one varies in each of these partial increments Δw1 , Δw 2 , Δw 3 . In Equation (17), for example, only y varies, since x is held equal to x 0 + Δx and z is held equal to z 0 . Since f ( x 0 + Δx , y, z 0 ) is a continuous function of y with a derivative f y , it is subject to the Mean Value Theorem, and we have Δ w 2 = f y ( x 0 + Δ x , y1 , z 0 ) Δ y for some y1 between y 0 and y 0 + Δy.

Appendix B B.1 Determinants A matrix is a rectangular array of numbers. For example, 2 1 3  A =   1 0 −2  is a matrix with two rows and three columns. We call A a “2 by 3” matrix. More generally, an “m by n matrix” is one that has m rows and n columns. The element in the ith row and jth column of a matrix is represented by a ij . In the example above, we have a11 = 2,

a12 = 1,

a13 = 3,

a 21 = 1,

a 22 = 0,

a 23 = −2.

If A is an n by n matrix, then we associate A with a number called the determinant of A, written sometimes as det A and sometimes as A with vertical bars (which do not mean absolute value). For n = 1 and n = 2 we have these definitions: det [ a ] = a,

a b  = ad − bc.(1) det   c d 

For a 3 by 3 matrix, we define  a11 a12 a13  a 22 a 23 a 21 a 23 a 21 a 22   det  a 21 a 22 a 23  = a11 − a12 + a13 a 32 a 33 a 31 a 33 a 31 a 32   (2)  a 31 a 32 a 33  = a11 ( a 22 a 33 − a 23a 32 ) − a12 ( a 21a 33 − a 23a 31 ) + a13 ( a 21a 32 − a 22 a 31 ). In Equation (2) there are some determinants of 2 by 2 matrices – each of those matrices is obtained by deleting one row and one column of the original 3 by 3 matrix. For an n by n matrix A, we define the minor of the element a ij to be the determinant of the ( n − 1) by ( n − 1) matrix that remains when the row and the column that contain a ij are deleted from A. In particular, deleting the row and the column that contain the element a11 in the matrix  a11 a12 a13     a 21 a 22 a 23  , and then taking the determinant of the resulting matrix yields    a 31 a 32 a 33  a 22 a 23 , a 32 a 33

B-1

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B-2

Appendix B

There is a simple checkerboard pattern for the signs that correspond to (−1)i + j. This pattern can be applied to convert the minor of an element in a matrix to the cofactor of the element. For a determinant of a 3 by 3 matrix the sign pattern is + − + − + − + − + In the upper left corner, i = 1, j = 1, and (−1)1+1 = +1. In going from any element to an adjacent element in the same row or column, we change i by 1 or j by 1, but not both, so we change the exponent from even to odd or from odd to even, which changes the sign from + to − or from − to +.

a 21 a 23 a 21 a 22 which is the minor of a11. Similarly, and are the minors of a12 and a13, a 31 a 33 a 31 a 32 respectively. Given an n by n matrix A, the cofactor of a ij is defined as (−1)i + j times the minor of a ij . We denote the cofactor of a ij by Aij . We can express Equation (2) using cofactors:  a11 a12 a13    det   a 21 a 22 a 23  = a11 A11 + a12 A12 + a13 A13 .(3)    a 31 a 32 a 33 



EXAMPLE 1   Given the matrix  2 1 3   A =  3 −1 −2  , 2 3 0   (a) find the minors of a11, a12 , and a13 . (b) find the cofactors A11 ,  A12 , and A13 . (c) use Equation (3) to evaluate det A. Solution 

(a) To obtain the minor of a11 , we delete row 1 and column 1 from the matrix, and then compute the determinant by using Equation (1): minor of a11 =

−1 −2 3

0

= (−1)( 0 ) − ( −2 )( 3 ) = 6.

Likewise, we obtain minor of a12 =

3 −2 2

0

= ( 3 )( 0 ) − (−2 )( 2 ) = 4

and minor of a13 =

3 −1 2

3

= ( 3 )( 3 ) − (−1)( 2 ) = 11.

(b) The cofactor of a11 is (−1)1+1 = 1 times the minor of a11: A11 = ( −1)1+1 ( minor of a11 ) = (1)( 6 ) = 6. The cofactor of a12 is (−1)1+ 2 = −1 times the minor of a12 : A12 = ( −1)1+ 2 ( minor of a12 ) = ( −1)( 4 ) = −4 and the cofactor of a13 is (−1)1+ 3 = 1 times the minor of a13 : A13 = ( −1)1+ 3 ( minor of a13 ) = (1)(11) = 11.

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B.1  Determinants

B-3

(c) To find det A by using Equation (3), we multiply each element of the first row of A by its cofactor and add: det   A = a11 A11 + a12 A12 + a13 A13 = ( 2 )( 6 ) + (1)( −4 ) + ( 3 )(11) = 12 − 4 + 33 = 41.  For any integer n > 3, the determinant of an n by n matrix is defined to be the sum of the products of the elements in the first row by their cofactors:



 a11   a 21  det  a 31    a  n1

a12 a 22 a 32  a n2

a13  a1n   a 23  a 2 n   a 33  a 3n  = a11 A11 + a12 A12 + a13 A13 +  + a1n A1n .(4)      a n 3  a nn 

EXAMPLE 2   Use the cofactors of the elements in the first row of the matrix  1 −2   2 1 A =  2  −1  0 1 

1  2  1 −2  2 1  3

0

to evaluate det A. Solution  The cofactors are

1 0 1+1

A11 = (−1)

2 1 A13 = (−1)

2 0 1+ 2

2 1 −2 , 1 2

1+ 3

2 A12 = (−1)

1

1 −2 ,

0 2 2

1

2 1 0 1+ 4

−1 2 −2 , 0 1

−1

2

A14 = (−1)

−1 2 1 .

1

0

1 2

These are multiplied by the corresponding elements of A, and then the determinant is obtained by using Equation (4): det   A = a11 A11 + a12 A12 + a13 A13 + a14 A14 1 0

2

2 0

2

2 1

2

2 1 0

= 1 2 1 −2 − 2(−1) −1 1 −2 + 3 −1 2 −2 + 1(−1) −1 2 1 . 1 2

1

0 2

1

0 1

1

0 1 2

Each of the determinants on the right side can now be evaluated using Equation (2): det  A = 1 [ 1(1 + 4 ) − 0 + 2( 4 − 1) ] + 2 [ 2 (1 + 4 ) − 0 + 2 ( − 2 − 0 ) ] + 3 [ 2( 2 + 2 ) − 1( −1 − 0 ) + 2(−1 − 0 ) ] − 1 [ 2( 4 − 1) − 1(−2 − 0 ) + 0 ] 

Z02_HASS5901_15_GE_APPB.indd 3

= 11 + 12 + 21 − 8 = 36.

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B-4

Appendix B

It becomes very lengthy to compute determinants of large matrices by using the definition. In the mathematical field called linear algebra, additional methods are developed to speed up this process. We present a few of those results without proofs.

THEOREM 1—Cofactor Expansion  Suppose that A is an n by n matrix, where n > 2. For any i, the determinant of A is the sum of the products of the elements in the ith row (or column) with their cofactors.

EXAMPLE 3   Apply Theorem 1 to evaluate the determinant from Example 1 by a cofactor expansion along the third column. Solution 

2 1 3   det  3 −1 −2   0  2 3 Expand by cofactors of the third column (Theorem 1).

= a13 A13 + a 23 A23 + a 33 A33 = 3

3 −1 2

3

− 2 ( − 1)

2 1 2 3

+0

2

1

3 −1

= 3( 9 + 2 ) + 2 ( 6 − 2 ) + 0 ( − 2 − 3 )

Cofactor of a ij is (−1)i + j times the minor of a ij . Equation (1)

= 33 + 8 − 0 = 41 According to Theorem 1, it is no accident that the answers in Examples 1 and 3 are the same. If we were to multiply the elements of any row (or column) by the cofactors of the corresponding elements and add the products, we would get the same answer: the value of the determinant. In Example 3, we do not actually need to compute the cofactor A33 since it is multiplied by a 33 = 0. Generally, when calculating the determinant of a matrix containing some zero elements, we may want to choose to expand along a row or column with as many zeros as possible. In an n by n matrix  a11 a12   a 21 a22  A =  a 31 a 32      a a  n1 n2

a13 a 23 a 33

  

 a n3

 

a1n   a 2n   a 3n      a nn 

we refer to the elements a11 , a 22 , a 33 , … , a nn as the main diagonal of the matrix. If an n by n matrix has all elements below the main diagonal (or all elements above the main diagonal) equal to zero, then its determinant is easy to calculate as illustrated in our next example. EXAMPLE 4   Evaluate the determinant of the matrix 0 −1   7 −4    0 −3 16 8   . A =   0 0 − 2 12   0 0 2  0 

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B.1  Determinants

B-5

Solution 

0 −1   7 −4    0 −3 16 8  det  0 −2 12  0 0 0 2  0  = a11 A11 + a 21 A21 + a 31 A31 + a 41 A41 −3 = ( 7)

16

8

0 −2 12 + 0 + 0 + 0 0

0

2

Expand by cofactors of the first column (Theorem 1). We do not need the values of the cofactors A21 ,  A31 , and A41 since each of them is multiplied by 0.

−2 12   = ( 7 )−3 + 0 + 0  0 2  

Expand the 3 by 3 determinant by cofactors of the first column (Theorem 1).

= ( 7 )(−3 )(−2 )( 2 )

Equation (1)

= 84 The first part of the following theorem generalizes this result. The remaining parts give other methods that can simplify the computation of a determinant.

THEOREM 2—Properties of Determinants

Suppose that A is an n by n matrix, where n > 2. 1. If all elements of A below the main diagonal (or all above it) are zero, the determinant of A is the product of the elements on the main diagonal. 2. If each element of a row (or column) of A is multiplied by a constant c and the results are added to a different row (or column), the determinant is not changed. 3. If each element of a row (or column) of A is multiplied by a constant c, the determinant is multiplied by c. 4. If two rows (or columns) of A are interchanged, the determinant just changes its sign.

EXAMPLE 5   We can apply Theorems 1 and 2 to make the evaluation of the determinant from Example 2 easier. 1 −2 3

1

1 −2

2

2

0

−1 0

1 0

2 1 −2 1 2

=

1

3

1

5 −6

0

−1

2

1 −2

0

1

2

1 −2 =

1

3

1

0

5 −6

0

0

0

4 −1

0

1

2

5 −6 =1 0 1

0

Expand by cofactors of the first column (Theorem 1).

1

= 5( 4 + 2 ) − (−6 )( 0 + 1) + 0 = 36

Z02_HASS5901_15_GE_APPB.indd 5

We add 1 times row 1 to row 3. By part 2 of Theorem 2 the determinant does not change.

1

4 −1 + 0 + 0 + 0 2

We add −2 times row 1 to row 2. By part 2 of Theorem 2 the determinant does not change.

Equation (2)

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B-6

Appendix B

The determinant has several important applications, which are studied in a course in linear algebra. • The absolute value of the determinant gives the volume of the region spanned by the vectors that form the columns (or rows) of the matrix. We saw this in Section 11.4, where we encountered the box product. • A determinant can be used to indicate whether a system of n linear equations in n unknowns has a unique solution. The answer is yes when the determinant is not equal to zero, and in this case determinants can be used to solve the system. • We saw in Section 13.7 that the determinant of the Hessian matrix of a function of two variables can help us decide whether a critical point is a saddle point or a local extremum. In Appendix B.2, we will use determinants to extend this to functions of more than two variables.

EXERCISES 

B.1

Evaluating Determinants

0 −2 1 0

Evaluate the determinants in Exercises 1–14. 2 3

9.

2 −1 −2

1

1. 4 5 2 2. −1 2 1 1 2 3 5

3

2

0

3. 2 −1 3

5.

0 −3

4

1 4. 3 −2 2

3 −1 0

0

0 2

0

0

0 0 −2

11.

2 −1

1

5 0

0 0

0 −3

0 2 1 2 2 3

−1

0 0 2

−3 0

0

0

1 1

0

0

0 0 0 4

3

3

2 −2 1 −1 10. 0 1 2 0 2 −1 −1 0

−4 5

6 7

0

0 2

0 8

5 6

7

4 1 −5 0

0 12. 0 0 −2 5

6

0 2 −1 4

0

0 0

0 2

5

2 −1

0 0

0 0 −2

6

0 0 3 0 6. 0 0 2 0 0

0 −5

3 1

−1 −1

8

3

1 2 0 0

1

−2

0

0

1

2

0 2 1

0

1

2

1 0

1 0

2

1 0 0 0

13. 0 0 3 1

0 14. 0 −2

2 1

1 −1 2

3

0 2 0 2

2

1

0

1

2 1 2 6 7. 8. 0 0 2 1 1 0 2 3

1 0 0 0

1

1

0 −2 0

1 2 3 4 0 1 2 3

0 0 3 2

−2

1 −1 0

2 0 −5

B.2 Extreme Values and Saddle Points for Functions of More than Two Variables Many applications involve functions of multiple independent variables. For instance, to predict energy trends in the United States, an econometric model may use a function f ( p1 , p 2 , … , p 50 ) of fifty independent variables representing average prices of gasoline in the fifty states. It is often important to determine extreme values of such functions. In Section 13.1 we defined a function of n variables whose domain is a set of n-tuples of real numbers ( x 1 , x 2 , … , x n ). The domain is a region in n-dimensional space. We will use open balls to define some important concepts, such as interior and boundary points. An n-dimensional open ball (or open n-ball) of radius r centered at ( a1 , a 2 , … , a n ) is the set consisting of the points ( x 1 , x 2 , … , x n ) that satisfy ( x 1 − a1 ) 2 + ( x 2 − a 2 ) 2 +  + ( x n − a n ) 2 < r 2 .

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B.2  Extreme Values and Saddle Points for Functions of More than Two Variables

B-7

EXAMPLE 1   Determine whether the following points in 4-dimensional space are in the open 4-ball with radius 5 and center at (−2, 0, 3, 1). (a) (1, −2, 1, 0 ) (b) ( 0, 1, −1, 5 ) Solution 

(a) (1 − ( −2 )) 2 + ( −2 − 0 ) 2 + (1 − 3 ) 2 + ( 0 − 1) 2 = 9 + 4 + 4 + 1 = 18 < 25 = 5 2 ; therefore, (1, −2, 1, 0 ) is in the open 4-ball. (b) ( 0 − ( −2 )) 2 + (1 − 0 ) 2 + ( −1 − 3 ) 2 + ( 5 − 1) 2 = 4 + 1 + 16 + 16 = 37 ≥ 25 = 5 2 ; therefore, ( 0, 1, −1, 5 ) is not in the open 4-ball. Using open n-balls we can extend definitions from Section 13.1 to regions in n-dimensional spaces.

DEFINITIONS  A point ( x 1 , x 2 , … , x n ) in a region D in n-dimensional space is

an interior point of D if it is the center of an open n-ball of positive radius that lies entirely in D. A point ( x 1 , x 2 , … , x n ) is a boundary point of D if every open n-ball centered at ( x 1 , x 2 , … , x n ) contains points that lie outside D as well as points that lie inside D. The interior of D is the set of interior points of D. The boundary of D is the set of boundary points of D. A region is open if it consists entirely of interior points. A region is closed if it contains its entire boundary. A region is bounded if it lies inside an open n-ball of finite radius. A region is unbounded if it is not bounded.

Extreme Values and Saddle Points of Functions of More than Two Variables While Section 13.7 discussed extreme values and saddle points for functions of two variables, here we extend this treatment to functions of three or more variables.

DEFINITIONS Let f ( x 1 , x 2 , … , x n ) be defined on a set D of n-tuples of real numbers ( x 1 , x 2 , … , x n ) including a point ( a1 , a 2 , … , a n ). f ( a1 , a 2 , … , a n ) is a local maximum value of f if f ( a1 , a 2 , … , a n ) ≥ f ( x 1 , x 2 , … , x n ) for all domain points ( x 1 , x 2 , … , x n ) that lie in an open n-ball centered at ( a1 , a 2 , … , a n ). f ( a1 , a 2 , … , a n ) is a local minimum value of f if f ( a1 , a 2 , … , a n ) ≤ f ( x 1 , x 2 , … , x n ) for all domain points ( x 1 , x 2 , … , x n ) that lie in an open n-ball centered at ( a1 , a 2 , … , a n ).

EXAMPLE 2   The function f ( x ,  y,  z ) = 10 − x 2 − y 2 − z 2 is defined on all of three-dimensional space. Figure B.1 shows an assortment of level surfaces of this function. The level set f ( x ,  y,  z ) = 10 is a single point, the origin. At every point ( x ,  y,  z ) we have f ( x ,  y,  z ) = 10 − x 2 − y 2 − z 2 ≤ 10 = f ( 0, 0, 0 ); therefore, f has a local (and absolute) maximum value of 10 at ( 0, 0, 0 ). This function has no local minimum values.

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B-8

Appendix B

z

3

f (x, y, z) = 9 f(x , y, z) = 6 f (x, y, z) = 1

2 f (x , y, z) = 10 Local maximum value of 10 at (0, 0, 0)

1

1 2

1

2

3

y

3 x

FIGURE B.1  Level surfaces of the function

f ( x ,  y,  z ) = 10 − x 2 − y 2 − z 2 in Example 2.

Suppose f ( x 1 , x 2 , … , x n ) has a local maximum or minimum value at an interior point ( a1 , a 2 , … , a n ) of its domain and all of its first-order partial derivatives are defined there. The function g1 (t ) = f ( t , a 2 , … , a n ) must then have a local extremum at t = a1. Therefore, g1′ (a1 ) = 0 and consequently ∂f ( a , a , … , a n ) = 0. ∂ x1 1 2 Likewise, we can use the function g 2 (t ) = f ( a1 , t , a 3 , … , a n ) to show that ∂f ( a , a , … , a n ) = 0, ∂x2 1 2 and continue in the same manner to conclude that ∂f ( a , a , … , an ) = 0 ∂xi 1 2 for all i = 1, 2, … , n. The above discussion extends Theorem 10 of Section 13.7 to functions of n variables. We present this in a more succinct form by using the gradient notation ∇f =

∂f ∂f ∂f , , …, . ∂ x1 ∂ x 2 ∂xn

THEOREM 3—First Derivative Theorem for Local Extreme Values (General Version) If f ( x 1 , x 2 , … , x n ) has a local maximum or minimum value at an

interior point (a1 , a 2 , … , a n ) of its domain and if  ∇f   exists at that point, then ∇f ( a1 , a 2 , … , a n ) = 0, where 0 = 0, 0, … , 0 denotes the zero vector in n-dimensional space. The following definition of a critical point uses the gradient notation. DEFINITION  An interior point of the domain of f where either ∇f = 0 or ∇f

does not exist is called a critical point of f .

Note that a vector does not exist whenever one or more of its components does not exist.

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B.2  Extreme Values and Saddle Points for Functions of More than Two Variables

B-9

The notion of a saddle point generalizes to functions f of n variables. DEFINITION  A critical point ( a1 , a 2 , … , a n ) of a differentiable func-

tion tered f ( a1 , f ( a1 ,

f ( x 1 , x 2 , … , x n ) is a saddle point if every open n-ball cenat the point ( a1 , a 2 , … , a n ) contains points ( x 1 , x 2 , … , x n ) with a 2 , … , a n ) > f ( x 1 , x 2 , … , x n ) as well as points ( y1 , y 2 , … , y n ) with a 2 , … , a n ) < f ( y1 , y 2 , … , y n ).

The Hessian matrix of a function of two variables was introduced in Section 13.7. We now generalize this concept for functions of n variables. DEFINITIONS The Hessian matrix of f ( x 1 , x 2 , … , x n ) is the n × n matrix

(containing n rows and n columns)  fx x  11 f  x 2 x1 Hf =     f  x n x1

f x 1x 2



f x 2x 2







f xnx 2



f x 1x n   f x 2 x n  .    f x n x n  

The principal minor determinants of the Hessian matrix are ∆ 1 = f x1x1 , f x 1x 1

f x 1x 2

f x 2 x1

f x 2x 2

f x 1x 1

f x 1x 2

f x 1x 3

∆ 3 = f x 2 x1

f x 2x 2

f x 2x 3 ,

f x 3 x1

f x 3x 2

f x 3x 3

∆2 =

,

…

∆n = H f =

f x 1x 1

f x 1x 2



f x 2 x1

f x 2x 2

 f x 2x n





f x n x1



f x nx 2 

f x 1x n 

.

f x nx n

Theorem 11 in Section 13.7 gave us tools to decide whether a critical point of a function f ( x 1 ,  x 2 ) is a local maximum, local minimum, or a saddle point. The nature of a critical point of a function of more than two variables can be similarly determined using the principal minor determinants of the Hessian matrix. THEOREM 4—Second Derivative Test for Local Extreme Values (General Version)  Suppose that f ( x 1 , x 2 , … , x n ) and its first and second partial deriva-

tives are continuous throughout an open n-ball centered at ( a1 , a 2 , … , a n ) and suppose that ∇ f ( a1 , a 2 , … , a n ) = 0 . Then

i) f has a local maximum at ( a1 , a 2 , … , a n ) if (−1)i ∆ i > 0 at the point ( a1 , a 2 , … , a n ) for i = 1, 2, … , n. ii) f has a local minimum at ( a1 , a 2 , … , a n ) if ∆ i > 0 at ( a1 , a 2 , … , a n ) for i = 1, 2, … , n. iii) f has a saddle point at ( a1 , a 2 , … , a n ) if ∆ n ≠ 0 at ( a1 , a 2 , … , a n ) and neither of conditions (i) or (ii) hold. iv) the test is inconclusive if ∆ n = 0 at ( a1 , a 2 , … , a n ).

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B-10

Appendix B

For n = 2, Theorem 4 reduces to Theorem 11 of Section 13.7 (see Exercise 11). The principal minor determinants ∆ i keep track of directions along which f has a local maximum and directions along which f has a local minimum. If it has a local maximum (or local minimum) when traveling along each of the n possible axis directions, then f has a local maximum (or local minimum) in an open n-ball. Applying Theorem 4 to the function f ( x ,  y,  z ) = 10 − x 2 − y 2 − z 2 of Example 2,  −2 0 0   we get the Hessian matrix H f =  0 −2 0  . Its principal minor determinants are    0 0 −2   ∆ 1 = −2, ∆2 =

−2

0

0 −2 −2

∆3 =

= 4, and

0

0

0 −2 0

0 = −2

0 −2

−2

0

0 −2

−0

0

0

0 −2

+0

0 −2 0

0

= −8.

At the critical point ( 0, 0, 0 ) , we have ∆ 1 < 0,  ∆ 2 > 0, and ∆ 3 < 0, so Theorem 4 confirms the conclusion we reached in Example 2: f has a local maximum at ( 0, 0, 0 ). EXAMPLE 3   Find the local extreme values and saddle points of the function f ( x ,  y,  z ) = 2 x 3 y + 3 y 2 + 6 yz + 6 z 2 − 6 xy. Solution  The function f is differentiable, therefore its critical points are points at which ∇ f = 0. Setting each component of the gradient equal to zero yields the following equations:

f x = 6 x 2 y − 6 y = 6 y ( x 2 − 1) = 0 f y = 2 x 3 + 6 y + 6z − 6 x = 0 f z = 6 y + 12 z = 0. The equation f x = 0 can be satisfied in three possible ways: (i) if y = 0,  (ii) if x = 1, or (iii) if x = −1. Let’s consider these possibilities one at a time. (i) If y = 0, then the equation f z = 0 yields z = 0, and then the equation f y = 0 leads to 2 x 3 − 6 x = 2 x ( x 2 − 3 ) = 0. This equation has three solutions: x = 0, ± 3, so we obtain critical points 3, 0, 0 ) , and (− 3, 0, 0 ). (ii) If x = 1, then the equation f y = 0 becomes 6 y + 6 z − 4 = 0. Together with the equation f z = 6 y + 12 z = 0, this forms a system of two linear equations, which can be solved to obtain y = 4 3 and z = −2 3. The resulting critical point is (1, 4 3, −2 3 ). (iii) If x = −1, then similarly to case (ii) we solve the remaining two equations, which yields y = −4 3 and z = 2 3 and the final critical point is (−1, −4 3, 2 3 ). ( 0, 0, 0 ) , (

The Hessian matrix of f is  f xx  H f =  f yx f  zx

Z02_HASS5901_15_GE_APPB.indd 10

f xy f yy f zy

f xz   12 xy 6x 2 − 6 0     f yz  =  6 x 2 − 6 6 6,    12  0 6 f zz  

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B.2  Extreme Values and Saddle Points for Functions of More than Two Variables

B-11

and the principal minor determinants are ∆ 1 = 12 xy ∆2 =

∆3 =

12 xy

6x 2 − 6

6x 2 − 6

6

12 xy

6x 2 − 6

6x 2

−6

= 72 xy − ( 6 x 2 − 6 ) 2 0 6 = 12 xy

6

0

6

12

6

6

6 12

− ( 6x 2 − 6)

6x 2 − 6 6 0

12

+0

= 12 xy ⋅ 36 − ( 6 x 2 − 6 ) 2 (12 ). We list the values of these determinants at the critical points in the following table.

( 0, 0, 0 )

∆1 0

∆2 −36

∆3 −432

( 3, 0, 0 )

0

−144

−1728

(− 3, 0, 0 )

0

−144

−1728

(1,  43 , − 23 )

16

96

576

(−1, − 43 ,  23 )

16

96

576

Critical Point

According to Theorem 4, (i) f has a local maximum at a critical point if ∆ 1 < 0, ∆ 2 > 0, and ∆ 3 < 0. This pattern is not observed at any of our critical points; therefore, f has no local maximum values. (ii) f has a local minimum at a critical point if ∆ 1 > 0,  ∆ 2 > 0, and ∆ 3 > 0. The signs of the principal minor determinants listed in the table follow this pattern for the last two critical points, so f has local minimum values at (1, 4 3, −2 3) and at (−1, −4 3, 2 3 ). (iii) f has a saddle point at a critical point if ∆ 3 ≠ 0, but neither of the conditions (i) or (ii) hold. Based on our table, this is the case at each of the first three critical points. We conclude that f has saddle points at ( 0, 0, 0 ) , ( 3, 0, 0 ) , and (− 3, 0, 0 ).

EXERCISES 

B.2

Finding Local Extrema and Saddle Points

Find all the local maxima, local minima, and saddle points of the functions in Exercises 1–10. 1. f ( x ,  y,  z ) = x 3 y − 3 xy + 3 yz − 3 y + 6 z 2. f ( x ,  y,  z ) = xy + x 2 z 2 + x 2 + y 2 + z 2 3. f ( x ,  y,  z ) = x 3 − 3 x + y 2 + 2 y + 2 z 2 − 8 z + 6 x3 x − + xy − y 2 − z 2 8 2 12 z 5. f ( x ,  y,  z ) = 2 x + y2 + z 2 + 4 4. f ( x ,  y,  z ) =

7. f ( x ,  y,  z ,  w ) = x 2 + y 2 + z 2 + w 2 + xy + zw − y + 2 z 8. f ( x ,  y,  z ,  w ) = 3 x + yz − 4 z − w − x 2 − y 2 − z 2 − w 2 9. f ( x ,  y,  z ,  υ,  w ) = x 2 + 2 y 2 − z 2 + υ 2 + w 2 + x − 3υ + x υ 10. f ( x ,  y,  z ,  υ,  w ) = x + z + yυ − 4 w − 3 x 2 − 2 y 2 − z 2 − υ 2 − w 2 Theory

11. Show that for n = 2 Theorem 4 reduces to Theorem 11 of Section 13.7.

6. f ( x ,  y,  z ) = x 2 + 4 y 2 + xz − yz + 4

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B-12

Appendix B

B.3 The Method of Gradient Descent In Section 13.7 and in Appendix B.2 we discussed methods allowing us to analytically determine local maxima, local minima, and saddle points of a differentiable function f of n variables. While these methods apply for any integer n ≥ 2, they may not be practical when n is very large. Furthermore, there are important applications where we can evaluate (or approximate) f and its partial derivative values without having an explicit formula for f . These are among the reasons to numerically search for local extrema of a function of n variables. In this appendix, we will refer to values of a function of n variables f ( x 1 , x 2 , … , x n ) using the notation f ( x) where x = 〈 x 1 , x 2 , … , x n 〉 is the position vector of the point ( x 1 , x 2 , … , x n ). The directional derivative of f at that point in the direction of a unit vector u can be expressed as D u f ( x) = ∇f ( x) ⋅ u. In Section 13.5, where the notion of the directional derivative was introduced, we noticed that the function f increases most rapidly in the direction of ∇f while it decreases most rapidly in the direction of −∇f . This leads to the method of gradient descent (also known as the method of steepest descent) for numerically approximating a local minimum for a function f of n variables. The initial approximation x 0 can be chosen to be the best guess for where a minimum might be. (Alternatively a randomly chosen point can be used for x 0 .) We then take a step in the direction of −∇f ( x 0 ) to yield the next approximation x 1 = x 0 − h 0 ∇f ( x 0 ), where the positive quantity h 0 controls the size of the step. We obtain subsequent approximations x 2 ,  x 3 , … in the same manner. Method of Gradient Descent

1. Start with a first approximation to a local minimum of f ( x 0 ). 2. Use the first approximation to obtain a second, the second to obtain a third, and so on, using the formula x k +1 = x k − h k ∇f ( x k )(1)

with h k > 0.

In applications, including those in the field of machine learning, the method of gradient descent can be applied to approximate a local minimum of a function involving a large number of independent variables. To gain insight into this method, consider Figure B.2, where we show the method being applied to a function of two variables f ( x) = f ( x ,  y) (we take x = 〈 x ,  y〉 to be the position vector of the point ( x ,  y )). Notice that the vector −∇f ( x 0 ) is perpendicular to the level curve of f passing through the point x 0 , which consists of all points x in the plane that satisfy the equation f ( x) = f ( x 0 ). Adding −∇f ( x 0 ) to x 0 gives us our next approximation, x 1. Once again, −∇f ( x 1 ) is perpendicular to the level curve through that point. The process then continues in a similar manner. In addition to level curves, Figure B.2 shows the surface z = f ( x), which helps visualize the descent of the values f ( x 0 ),  f ( x 1 ),  f ( x 2 ), … . These values appear to be approaching a minimum value of f . In Figure B.2, we used a constant value h = 1 for all h k . Figure B.3 illustrates how different choices of h value could affect the method’s performance. When h is small (Figure B.3a), the method is forced to take more steps. It is often more efficient to take larger steps (Figure B.3b, same h as in Figure B.2). If the value of h is too large (Figure B.3c),

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B.3  The Method of Gradient Descent

B-13

z z = f (x)

x0

x

x1

−∇f (x 0 ) −∇f (x1) −∇f (x2)

x2 x3

y

f (x) = f(x2 ) f (x) = f (x1) f (x) = f (x 0 )

FIGURE B.2  The method of gradient descent

is applied to a function f of two variables using a constant h = 1. The level curves and the corresponding surface plot show how the value of f decreases at each step.

then the method may not converge. To understand why the method may fail when h is too large, recall that while at x k the function f decreases most rapidly in the direction of −∇f ( x k ), the function may start increasing if we continue in this direction and go too far from x k . Choosing a desirable value for the quantity h (which in machine learning is sometimes called the learning rate) that is neither too small nor too large could be challenging. We consider here only the case where the gradient descent method uses a constant, prechosen value h k = h. More advanced methods focus on identifying desirable h k values (which may not be constant) – one of those would involve performing a line search (see Exercise 5).

y

y

y

x3 − 2∇f (x2)

1

−−∇f (x2) 2 1 −−∇f (x1) 2 1 −−∇f (x 0) 2

− 2∇f (x1)

−∇f (x2)

x3

−∇f (x1) x3

f (x) = f (x2 ) f (x) = f (x1) f (x) = f (x 0 )

x2 x1 x0

−∇f (x 0)

x1

− 2∇f (x 0 )

x2

f (x) = f (x2 ) f(x) = f (x1) f (x) = f (x 0 )

x0

1

x2

(b) h = 1

f(x) = f(x2 )

f(x) = f(x1) f(x) = f(x 0 )

x0

x

x

x (a) h = − 2

x1

(c) h = 2

FIGURE B.3  Different choices of h affect the performance of the method of gradient descent.

Z02_HASS5901_15_GE_APPB.indd 13

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B-14

Appendix B

EXAMPLE 1  Apply the method of gradient descent to f ( x ,  y,  z ) = 2 x 3 y + 3 y 2 + 6 yz + 6 z 2 − 6 xy using the initial approximation x 0 = 〈2, 1, −1〉 and taking h k = h = 0.1. List the components of x k and f ( x k ) for k = 0, 1, 2, … , 10. Solution  At the initial approximation f ( 2, 1, −1) = 7. The gradient of the function f is

x 0 = 〈2, 1, −1〉,

we

have

f (x 0 ) =

∇f = 〈6 x 2 y − 6 y, 2 x 3 + 6 y + 6 z − 6 x , 6 y + 12 z〉. Evaluating the gradient at the initial approximation yields ∇f ( x 0 ) = 〈18, 4, −6〉. We now apply Equation (1) to obtain the second approximation x 1 = x 0 − h 0 ∇f ( x 0 ) = 〈2, 1, −1〉 − (0.1)〈18, 4, −6〉 = 〈0.2, 0.6, −0.4〉. At this point, f ( x 1 ) ≈ −0.11. Subsequent approximations x 2 ,  x 3 , … are then obtained using technology by repeatedly applying Equation (1). These approximations, along with the corresponding f values, are listed in the table below.

k

x component of x k

y component of x k

z component of x k

  0

2

1

  1

0.200000000000001

0.600000000000001

  2

0.545600000000001

0.598400000000001

  3

0.7977612025856

0.702237228236801

  4

0.950951612242085

0.839832907434822

  5

0.999170298397954

0.95095391811045

  6

1.00011671520043

1.03943288055905

  7

0.999971125564279

1.1063062093209

  8

1.0000094577719

1.15861170717517

  9

0.999996308247422

1.19849707338251

10

1.00000161770321

1.22948856581467

−1 −0.399999999999999 −0.280000000000001 −0.30304 −0.360734336942082 −0.431752877072477 −0.484221775451776 −0.526815373245074 −0.558420650943528 −0.583482894116396 −0.602401665206229

f (x k ) 7

−0.110400000000005 −1.22520965376901 −1.8946600083389 −2.26841318086511 −2.43587094205513 −2.5295411401069 −2.58519263199985 −2.61825804593691 −2.63790410815758 −2.64957702176556

The function f in Example 1 was analyzed in Example 3 of Appendix B.2 and was shown to have a local minimum f (1, 4 3, −2 3 ) = −8 3. The sequence of approximations generated in Example 1 appears to converge to this local minimum. The method of gradient descent is designed to approximate a local minimum of a function of many variables. If we are interested in approximating a local maximum instead, then we apply the method of gradient ascent x k +1 = x k + h k ∇f ( x k ), in which we follow the direction of the gradient ∇f, rather than −∇f .

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B.3  The Method of Gradient Descent

EXERCISES 

B-15

B.3

You may need to use appropriate technology (such as a calculator or a computer). Gradient Descent

1. Apply the method of gradient descent to f ( x ,  y,  z ) = 12 z using the initial approximation x 2 + y2 + z 2 + 4 x 0 = 〈1, 1, −1〉 and the given constant value h k = h = 1. List the components of x k and f ( x k ) for k = 0, 1, 2, … , 5. 2. Apply the method of gradient descent to f ( x ,  y,  z ) = x 3 − 3 x + y 2 + 2 y + 2 z 2 − 8 z + 6 using the initial approximation x 0 = 〈0, 0, 1〉 and the given constant value h k = h = 0.2. List the components of x k and f ( x k ) for k = 0, 1, 2, … , 5. 3. Apply the method of gradient descent to f ( x ,  y,  z ,  w ) = x 2 + y 2 + z 2 + w 2 + xy + zw − y + 2 z using the initial approximation x 0 = 〈−1, 1, −1, 1〉 and the given constant value h k = h = 0.5. List the components of x k and f ( x k ) for k = 0, 1, 2, … , 5.

4. Apply the method of gradient descent to f ( x ,  y,  z ,  υ,  w ) = −x − z − yυ + 4 w + 3 x 2 + 2 y 2 + z 2 + υ 2 + w 2 using the initial approximation x 0 = 〈1, 1, 1, 1, 1〉 and the given constant value h k = h = 0.2. List the components of x k and f ( x k ) for k = 0, 1, 2, … , 5. Line Search

5. In Example 1, we applied the method of gradient descent to the function f ( x ,  y,  z ) = 2 x 3 y + 3 y 2 + 6 yz + 6 z 2 − 6 xy with the initial approximation x 0 = 2, 1, −1 . Plot the function g(h 0 ) = f ( x 0 − h 0 ∇f ( x 0 ) ) to approximate a positive value of h 0 that minimizes g(h 0 ). (This process, known as the line search, can then be repeatedly used at subsequent steps of the gradient descent method to obtain the values h1 ,  h 2 , … . Using these values, rather than a constant h, can lead to a faster convergence to a minimum.)

ANSWERS TO ODD-NUMBERED EXERCISES

Appendix B

3. y z x w component component component component of x k of x k of x k of x k k

APPENDIX B.1, p. B-6

1. −5  3.  0  5.  100  7.  1  9.  22 11. −60  13.  −20 APPENDIX B.2, p. B-11

(

)

(

)

5 1 = 10, saddle point; f −1, −2,  = 2, 3 3 saddle point 3. f (1, −1, 2) = −5, local minimum; f (−1, −1, 2) = −1, saddle point 5. f ( 0, 0, 2 ) = 3, local maximum; f (0, 0, −2) = −3, local minimum 1 2 4 2 5 7. f − ,  , − ,  = − , local minimum 3 3 3 3 3 5 7 13 9. f − , 0, 0,  , 0 = − , saddle point 3 3 3 1. f 1, −2, 

( (

) )

f (x k )

0 −1

1

−1

1

−1

1 −0.5

1

−1.5

0.5

−1.5

2 −0.5

0.75

−1.25

0.75

−1.625

3 −0.375

0.75

−1.375

0.625

−1.65625

4 −0.375

0.6875

−1.3125

0.6875

−1.664062

0.6875

−1.34375

0.65625

−1.666016

5

−0.34375

5. g(h 0 ) 7 6 5 4

APPENDIX B.3, p. B-15

3

1.

2

y z x component component component of x k of x k of x k k

1 -1

f (x k )

0

1

1

−1

−1.714286

1

0.510204

0.510204

−2.22449

−2.81909

2

0.20641

0.20641

−2.167243 −2.961338

3

0.067207

0.067207

−2.072064 −2.994859

4

0.018721

0.018721

−2.022551 −2.999552

5

0.004841

0.004841

−2.006053 −2.999969

Z02_HASS5901_15_GE_APPB.indd 15

0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20

h0

h 0 L 0.058 minimizes g(h 0 )

07/03/2023 18:15

ANSWERS TO ODD-NUMBERED EXERCISES

Chapter 1

29. (a) f ( x ) =

SECTION 1.1, pp. 31–33

1. 3. 5. 7.

D : (−∞, ∞ ) , R : [ 1, ∞ ) D : [ −2, ∞ ) , R : [ 0, ∞ ) D : (−∞, 3 ) ∪ ( 3, ∞ ) , R : (−∞, 0 ) ∪ ( 0, ∞ ) (a) Not a function of x because some values of x have two values of y (b) A function of x because for every x there is only one possible y

3 2 x , p = 3x 4 d d3 11. x = , A = 2d 2 , V = 3 3 3 9. A =

20 x 2 − 20 x + 25 4 15. (−∞, ∞ )

-4

f(x) = 5 - 2x

2

g(x) = Í 0 x 0

y

y = -x

2

1 2 3 4 5

x

-2

2 1

3

-2

Dec. −∞ < x < ∞ 41. Symmetric about the y-axis

19. (−∞, 0 ) ∪ ( 0, ∞ )

y

y F(t) =

2

t 0t0

1

2

3

21. (−∞, −5 ) ∪ (−5, −3 ] ∪ [ 3, 5 ) ∪ ( 5, ∞ ) (b) For each value of x ≠ 0, 23. (a) For each positive value of there are two values of y. x, there are two values of y.

0 -1 4

x

6

-1

1

x

-2

-2

y f(x) =

x

2

-1

Inc. −∞ < x < ∞

1

2

3

x

y = - x 3>2

x,

-4

27.

0…x…1

-5

y

2 - x, 1 6 x … 2

y = x 2 + 2x

4

1

0

4

x 1

-3 -1

-4

25.

2

1

2 2

0

- 1 - 1>8

y

y2 = x2

0y0 = x

-2

Dec. −∞ < x ≤ 0; Inc. 0 ≤ x < ∞ 45. No symmetry

y

y

0

-2

2 -4

x3 y = –– 8

1>8

y = Í0 x 0

t

-2

4

y

1

4

x

43. Symmetric about the origin

4

1 -4-3-2-1

1 2

Inc. −∞ < x < 0 and 0 < x < ∞

-2

-4

1 y=-x

- 2- 1 -1 -2

x

2

x

4

−1 ≤ x < 0 0 < x ≤1 1< x < 3

y

- 5 - 4 - 3 - 2 -1 -1 -2

-2

2

-1 1 y= x

0 -1

y= 1

2

1 x

-4

0

1 x-2

x

0

-2

2

1

3

4

-1 -2

51.

y

53.

y

4 3

2

1 y= x +2

1 -3 -2 -1

1 y = ––––––– (x - 1) 2

3

2

0

2

1

1 x

3

-1 0

1

2

x

3

y 5 4 3 2 1

y=

1

1 +1 x2

x

2

R : [ 2, 3 ]

(b) D : [ 0, 2 ], y

y = f (x) + 2

y = f(x) - 1 1

0

1

2

3

(c) D : [ 0, 2 ],

y = Íx + 4

R : [ −1, 0 ]

1

3

0

y 2

1 y-1= x-1

2 1 x-1 1

y=

1

1

37.

1 x −1 y

y-1=

x

2>3

y

2 -7

35. y − 1 =

y=1-x

x

(1, - 1)

-1

y

x

2

1

2 3

2

57. (a) D : [ 0, 2 ],

y = 2x

y = 2x - 7

4

x

1

3

0

7

1

1 0 -1

y = Íx - 1 - 1

x

33. y = 2 x

y = Íx

1

y = (x + 1) 2>3

0

x

5

y

49.

- 2- 1 0

y = Í x + 0.81

2

1

45.

y

55.

1

0

y

0

-2 -1 1

-2

x + 0.81

x

6

y

2 2 (x + 2) + (y + 3) = 49

y

4

43.

-1

31. y =

y = 1 + Íx - 1 (1, 1)

1

y + 1 = (x + 1) 3

1 (-2, -3) 0

2

2 -2

y 3

4

-3

(c) x 2

15.

y = 0x - 20

x−7 3x + 6

x 3x

41.

y

4

x

x

-1

R : [ 0, 2 ]

y

2

(d) D : [ 0, 2 ],

R : [ −1, 0 ]

y

x 1 2

y = 2 f(x)

x

y = - f(x)

1

0

0

1

2

3

x

-1

1

2

x

AN-3

Chapter 1: Answers to Odd-Numbered Exercises

(e) D : [ −2, 0 ],

R : [ 0, 1 ]

(f) D : [ 1, 3 ],

R : [ 0, 1 ]

y

y

SECTION 1.3, pp. 47–49

1. (a) 8π m

2

y = f(x - 1)

1

-2

5. θ

2

y = f(x + 2)

-1

1

x

0

(g) D : [ −2, 0 ],

R : [ 0, 1 ]

1

0

(h) D : [ −1, 1 ],

y

2

y 2

1

-1

1

x

0

59. y = 3 x 2 − 3

61. y =

x2 4− 4

65. y =

R : [ 0, 1 ]

y = - f (x + 1) + 1

y = f(- x)

-2

x

3

2

-1

1 1 + 2 2x 2

0

x

1

63. y =

(b)

−π

55π m 3. 8.4 cm (or 84 mm) 9 −2 π 3 0 π2 3 2 1 − 2

sin θ

0

cos θ

−1

tan θ

0

cot θ

UND

sec θ

−1

csc θ

UND

−

3 1 3 −2 2 − 3

4x + 1

0

0

UND

−1

UND

0

−1

1

UND

− 2

UND

1

y

y = cos px

1

y = sin 2x

5

2

2

15. Period 2

y 1 4

1 -2 -1

1

8 , tan x = − 8 3 1 2 11. sin x = − , cos x = − 5 5 13. Period π

71.

1

9. sin x = −

67. y = 1 − 27 x 3

y

1 2 1 − 2

0

7. cos x = − 4 5, tan x = − 3 4

y

69.

3π 4

3 1

2

3

4

x

2

-1

1

-2

p 2

y = (x - 1)3 + 2

-3-2-1

1

2

3

4

5

x

x

p

0

1

2

x

-1

-1

-3 y = - Í2x + 1

-4

19. Period 2π

17. Period 6 y

73.

75.

y

y

y = - sin px 3

y

p y = cos ax - b 2

1

1 4

4

3

y=

2

1 -1 2x

2

1

3

0 -1

3

x

p 2

0

p

x

2p

1

-4-3-2-1 -1

2

3

4

x

-4-3-2-1 -1 -2 -3

1

2

3 3

y = - Íx

4

x

-1

21. Period 2π

23. Period π 2, symmetric about the origin

y

-4

-4

s

p y = sin ax - b + 1 4

2

77.

6

y

2

s = cot 2t

1 1 3 2

-2

-1

1

p 0 4

p 4

3p 4

7p 4

x -p

-

p 2

0

y = 0x2 - 10

-1

x

-2

2

-1

79. (a) Odd (b) Odd (c) Odd (d) Even (e) Even (f) Even (g) Even (h) Even (i) Odd

p 2

p

t

AN-4

Chapter 1: Answers to Odd-Numbered Exercises

29. D : (−∞, ∞ ) , R : y = −1, 0, 1

25. Period 4, symmetric about the y-axis s

9.

y 3

y

2 s = sec p t 2 - 2p

1 -3 -2 -1

2

1

y = sin x

y = :sin x; -p

3

p

1

t

-4 -3 -2 -1 -1

2p

-2

x

2

3

x

4

y = 1 - ex

y = 1 - e -x - 3

-1

-1

1

-4 -5

39. − cos x

6 + 2 45. 4 π 2π 4 π 5π 51. , , , 3 3 3 3

41. − cos x

2+ 2 2− 3 49. 4 4 π π 5π 3π 53. , , , 59. c = 6 2 6 2 α sin (θ) 65. r = 1 − sin (θ)

47.

7 ≈ 2.65

y = 2sin (x + p) - 1

63. a ≈ 1.464

y

1

-1

p 2

3p 2

( )

1. One-to-one 3. Not one-to-one 7. Not one-to-one 9. One-to-one

2 69. A = − , B = 4, π 1 C = 0, D = π

y

p 2

6

SECTION 1.5, pp. 64–66

67. A = 2, B = 2π, C = −π , D = −1

-

2 + 4

43.

16 1 4 = 2 13. 4 1 2 = 2 15. 5 17. 14 3 19. 4 D : −∞ < x < ∞; R : 0 < y < 1 2 D : −∞ < t < ∞; R : 1 < y < ∞ x ≈ 2.3219 27. x ≈ −0.6309 29. After 19 years 1 t 14 (b) About 38 days later 31. (a) A(t ) = 6.6 2 33. ≈11.433 years, or when interest is paid 35. 2 48 ≈ 2.815 × 10 14 11. 21. 23. 25.

x

5p 2

3 p

11. D : ( 0, 1 ] R : [ 0, ∞ )

13. D : [ −1, 1 ] R : [ − π 2, π 2 ]

y

y=f

pt y = - 2 sin a b + 1 2 p p

-1

(x)

y

p 2 1

y=x

1

-3

-1

-

y = f(x)

1 p 1

3

15. D : [ 0, 6 ] R : [ 0, 3 ] y=x

3. y = (1>5)x y = 3-x

y = 4x

6

4 3

-4 -3 -2 -1 -1 -2 -3 -4 -5

2 1 -4 -3 -2 -1

0

5.

1

2

3

x

4

7.

y 1 y= x e

1

2

3

4

y = 2x - 1

3 3

2

2

-4 -3 -2 -1

-4 -3 -2 -1 0

1

2

3

4

x

-2

f −1 ( x ) f −1 ( x ) f −1 ( x ) f −1 ( x )

1

2

3

4

6

1

x

x

= = = =

33. f −1( x ) = 1 −

1

1

0

x −1 21. f −1 ( x ) = 3 x + 1 x −1 5 x ; D : −∞ < x < ∞; R : −∞ < y < ∞ 3 x − 1; D : −∞ < x < ∞; R : −∞ < y < ∞ 1 ; D : x > 0; R : y > 0 29. f −1 ( x ) = x 2x + 3 31. f −1( x ) = ; D : −∞ < x < ∞, x ≠ 1; x −1 R : −∞ < y < ∞, y ≠ 2 19. 23. 25. 27.

4

4

-1 p 2

y = Í1 - x 2 0…x…1

3

t

y = - 2t

5

y = ex

x

1

y = f -1(x)

y y = 2-x - 1

5

1 p 2

3

5 4 3 2 1

y = 2-t

y = 2x

5

y = f (x)

-

y = f(x)

y

y=x

y

6

y

y = f -1 (x)

17. Symmetric about the line y = x

y

SECTION 1.4, pp. 53–54

p -1 2

x

1

t

5

1 -p

1.

5. One-to-one

x

x + 1; D : −1 ≤ x < ∞; R : −∞ < y ≤ 1

2x + b ; x −1 D : −∞ < x < ∞, x ≠ 1,

35. f −1 ( x ) =

R : −∞ < y < ∞, y ≠ 2

1 37. (a) f −1 ( x ) = x m (b) The graph of f −1 is the line through the origin with slope 1 m .

AN-5

Chapter 1: Answers to Odd-Numbered Exercises

39. (a) f −1 ( x ) = x − 1 y

y=x+1

2 1 -2 -1

y=x

2

x

-1 -2

41.

43. 45. 49. 55. 57. 59. 67. 71. 75. 77. 79.

(c) 75 (d) 2 (e) 0.5 (f) −1 ln 3 2 (b) 3 (c) 2 (c) sin x 69. (a) (a) x (b) x ln 2 73. (a) π (b) π 2 (a) − π 6 (b) π 4 (c) − π 3 Yes, g( x ) is also one-to-one. Yes, f  g is also one-to-one. x x (b) f −1 ( x ) = log 1.1 (a) f −1 ( x ) = log 2 100 − x 50 − x 2x   x +1 (c) f −1 ( x ) = ln (d) f −1 ( x ) = e x +1 1− x (a) y = ln x − 3 (b) y = ln ( x − 1) (c) y = 3 + ln ( x + 1) (d) y = ln ( x − 2 ) − 4 (e) y = ln (−x ) (f) y = e x 1 t 12 ≈ −0.7667 85. (a) Amount = 8 (b) 36 hours 2 ≈44.081 years 2

(

(

81.

83. 87.

)

(

)

5. 13. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35.

-2 -1

1

(d)

C2 A = πr 2, C = 2πr ,  A = 3. x = tan θ,  y = tan 2 θ 4π Origin 7. Neither 9. Even 11. Even Odd 15. Neither (a) Even (b) Odd (c) Odd (d) Even (e) Even (a) Domain: all reals (b) Range: [ −2, ∞ ) (a) Domain: [ −4, 4 ] (b) Range: [ 0, 4 ] (a) Domain: all reals (b) Range: ( −3, ∞ ) (a) Domain: all reals (b) Range: [ −3, 1 ] (a) Domain: ( 3, ∞ ) (b) Range: all reals (a) Domain: ( −∞, −1 ]  and  [ 3, ∞ ) (b) Range: ( −∞, 5 ] (a) Domain: ( −∞, 0 ) and ( 0, ∞ ) (b) Range: [ −4, 4 ] (a) Increasing (b) Neither (c) Decreasing (d) Increasing (a) Domain: [ −4, 4 ] (b) Range: [ 0, 2 ]

⎪⎧1 − x , 37. f ( x ) = ⎨ ⎩⎪⎪ 2 − x ,

0 ≤ x 0 to make the new graph symmetric with respect to the y-axis. y y = 0x0 y=x

x

y=x

47. Reflects the portion for y < 0 across the x-axis 49. Reflects the portion for y < 0 across the x-axis 51. Adds the mirror image of the portion for x > 0 to make the new graph symmetric with respect to the y-axis 1 2 (b) y = g   x + −2 53. (a) y = g ( x − 3 ) + 2 3 (c) y = g(−x ) (d) y = −g( x ) (e) y = 5 g( x ) (f) y = g(5 x ) y y 55. 57.

(

)

x

2

-1

-2

-1

1

y = 12 + 1 2x

x

2

2 1

-1

-4-3-2-1 -1

-2

59. Period π 1

p 2

p

3p 2

2p

x

-1

p y = 2cos ax - b 3

1

p 6

-1 -2

1

-1

y

-

p 3

3

4

x

y = sin px

1

2

2

y

y = cos 2x

0

1

61. Period 2

y

63.

)

y = - Í1 + x 2

1

( )

PRACTICE EXERCISES, pp. 67–69

1.

-4

( )

(b)

(c) x ,  x ≠ 0

2

(b) f −1 ( x ) = x − b. The graph of f −1 is a line parallel to the graph of f . The graphs of f and f −1 lie on opposite sides of the line y = x and are equidistant from that line. (c) Their graphs will be parallel to one another and lie on opposite sides of the line y = x equidistant from that line. (a) ln 3 − 2 ln 2 (b) 2( ln 2 − ln 3 ) (c) −ln 2 2 1 1 (d) ln 3 (e) ln 3 + ln 2 (f) ( 3 ln 3 − ln 2 ) 3 2 2 2 t2 (a) ln 5 (b) ln ( x − 3 ) (c) ln b x 1 (a) 7.2 (b) 2 (c) 47. (a) 1 (b) 1 (c) −x 2 − y 2 y x e 2t + 4 51. e 5 t + b 53. y = 2 xe x + 1 (a) k = ln 2 (b) k = (1 10 ) ln 2 (c) k = 1000 ln a ln 2 ln .4 (a) t = −10 ln 3 (b) t = − (c) t = ln .2 k t = 4 ( ln x ) 2 61. t = ln 3 63. t = e 2 ( e 2 − 1)

65. (a) 7

2 5

41. (a) ( f  g)( x ) = −x , x ≥ −2, ( g  f )( x ) = 4 − x 2 (b) Domain ( f  g): [ −2, ∞ ), domain ( g  f ): [ −2, 2 ] (c) Range ( f  g): ( −∞, 2 ], range ( g  f ): [ 0, 2 ] y y 43.

y=x-1 1

1 = 2.5

39. (a) 1 (b)

5p 6

4p 3

11p 6

x

2

x

AN-6

Chapter 2: Answers to Odd-Numbered Exercises

2

-2

200 100 0 2017

1

19

20

x

21

(b) ≈$ 56,000 year (c) ≈$ 42,000 year 23. (a) 0.414213, 0.449489, ( 1 + h − 1) h (b) g( x ) = 1.1

1.01

1.001

1.0001

1.00001

x

1.000001

1.04880 1.004987 1.0004998 1.0000499 1.000005 1.0000005

( 1 + h − 1) h 0.4880 0.4987

x

2

18

Year

1+ h

y = x 1>3

-1

y

21. (a)

1+ h

y = x3

1

19. Your estimates may not completely agree with these. (a) PQ PQ PQ PQ 1 2 3 4 The appropriate units are m s. 43 46 49 50 (b) ≈50 m s or 180 km h Profit ($1000s)

65. (a) a = 1 b = 3 (b) a = 2 3 3 c = 4 3 3 b a (b) c = 67. (a) a = tan B sin A 69. ≈16.98 m 71. (b) 4π 73. (a) Domain: −∞ < x < ∞ (b) Domain: x > 0 75. (a) Domain: −3 ≤ x ≤ 3 (b) Domain: 0 ≤ x ≤ 4 77. ( f  g )( x ) = ln ( 4 − x 2 ) and domain: −2 < x < 2; ( g  f )( x ) = 4 − ( ln x ) 2 and domain: x > 0; ( f  f )( x ) = ln ( ln x ) and domain: x > 1; ( g  g )( x ) = −x 4 + 8 x 2 − 12 and domain: −∞ < x < ∞. −π π ⎤ , 83. (a) D: ( −∞, ∞ ) R: ⎡⎢ (b) D: [ −1, 1 ] R: [ −1, 1 ] ⎣ 2 2 ⎦⎥ 85. (a) No (b) Yes 3 87. (a) f ( g( x )) = ( 3 x ) = x ,   g( f ( x )) = 3 x 3 = x y (b)

0.4998

0.499

0.5

0.5

(c) 0.5 25. (a) 15 km/h, 3.3 km/h, 10 km/h (b) 10 km/h, 0 km/h, 4 km/h (c) 20 km/h when t = 3.5 hr

-1 -2

SECTION 2.2, pp. 87–90 ADDITIONAL AND ADVANCED EXERCISES, pp. 69–71

1. Yes. For instance: f ( x ) = 1 x and g( x ) = 1 x , or f ( x ) = 2 x and g( x ) = x 2, or f ( x ) = e x and g( x ) = ln x. 3. If f ( x ) is odd, then g( x ) = f ( x ) − 2 is not odd. Nor is g( x ) even, unless f ( x ) = 0 for all x. If ƒ is even, then g( x ) = f ( x ) − 2 is also even. y 5. 0x0 + 0y0 = 1 + x 1

x

1 -– 2 -1

19. (a) Domain: all reals. Range: If a > 0 , then ( d, ∞ ); if a < 0 , then ( −∞, d ). (b) Domain: ( c, ∞ ) , range: all reals 21. (a) y = 100,000 − 10,000 x , 0 ≤ x ≤ 10 (b) After 4.5 years ln ( 10 3 ) ≈ 15.6439 years. (If the bank only pays interest 23. After ln 1.08 at the end of the year, it will take 16 years.) 25. x = 2,  x = 1

1. (a) Does not exist. As x approaches 1 from the right, g( x ) approaches 0. As x approaches 1 from the left, g( x ) approaches 1. There is no single number L that all the values g( x ) get arbitrarily close to as x → 1. (b) 1 (c) 0 (d) 1 2 3. (a) True (b) True (c) False (d) False (e) False (f) True (g) True (h) False (i) True (j) True (k) False 5. As x approaches 0 from the left, x x approaches −1. As x approaches 0 from the right, x x approaches 1. There is no single number L that the function values all get arbitrarily close to as x → 0. 7. Nothing can be said. 9. No; no; no 11. −4 13. −8 15. 3 17. −25 2 19. 16 21. 3 2 23. 1 10 25. −7 27. 3 2 29. −1 2 31. −1 33. 4 3 35. 1 6 37. 4 39. 1 2 41. 3 2 43. −1 45. 1 47. 1 3 49. 4 − π 51. (a) Quotient Rule (b) Difference and Power Rules (c) Sum and Constant Multiple Rules 53. (a) −10 (b) −20 (c) −1 (d) 5 7 55. (a) 4 (b) −21 (c) −12 (d) −7 3 63. 5 57. 2 59. 3 61. 1 ( 2 7 ) 65. (a) The limit is 1. 67. (a) f ( x ) = ( x 2 − 9 ) ( x + 3 ) −3.1 −3.01 −3.001 −3.0001 −3.00001 −3.000001

x

f (x) −6.1 −6.01 −6.001 −6.0001 −6.00001 −6.000001

27. 1 2

−2.9 −2.99 −2.999 −2.9999 −2.99999 −2.999999

x

Chapter 2

f (x) −5.9 −5.99 −5.999 −5.9999 −5.99999 −5.999999 (c) lim f ( x ) = −6

SECTION 2.1, pp. 77–79

1. (a) 19 4 3. (a) − π 7. (a) 4 9. (a) 2 11. (a) 12 13. (a) −9 15. (a) −1 4 17. (a) 1 4

 x →−3

(b) 1 (b) − (b) (b) (b) (b) (b) (b)

y y y y y y

3 3 π = 4x − 9 = 2x − 7 = 12 x − 16 = −9 x − 2 = −x 4 − 1 = x 4+1

69. (a) G ( x ) = ( x + 6 ) ( x 2 + 4 x − 12 ) 5. 1

x

−5.9

G( x)

−.126582 −.1251564

−5.99999

−5.99

−5.999

−5.9999

−.1250156 −.1250015

−5.999999

−.1250001 −.1250000 x

−6.1

−6.01

−6.001

−6.0001

G( x)

−.123456 −.124843 −.124984 −.124998

AN-7

Chapter 2: Answers to Odd-Numbered Exercises

−6.00001

−6.000001

−.124999

−.124999

SECTION 2.4, pp. 104–106

1. (a) (d) (g) (j) 3. (a)

(c) lim G ( x ) = −1 8 = −0.125  x →−6

71. (a) f ( x ) = ( x 2 − 1 ) ( x − 1 ) −1.1 −1.01 −1.001 −1.0001 −1.00001 −1.000001

x f (x) x

2.1

2.01

2.001

2.0001

2.00001

2.000001

−.9

−.99

−.999

−.9999

−.99999

−.999999

1.9

1.99

1.999

1.9999

1.99999

1.999999

f (x)

True (b) True (c) False True (e) True (f) True False (h) False (i) False False (k) True (l) False 2, 1 (b) No, lim+ f ( x ) ≠ lim− f ( x )  x → 2

y

g(θ)

.998334 .999983 .999999

.999999 .999999 .999999

θ

−.1

−.0001 −.00001 −.000001

g(θ)

.998334 .999983 .999999 .999999 .999999

.01

.001

−.01

.0001

−.001

.00001

x 1 -1

.000001

.999999

x=1

0,

-1

.1

x 3, x Z 1

y=

1

(c) lim f ( x ) = 2  x →−1 73. (a) g(θ) = ( sin θ ) θ θ

 x → 2

(c) 3, 3 (d) Yes, 3 5. (a) No (b) Yes, 0 (c) No 7. (a)

(b) 1, 1 (c) Yes, 1 9. (a) D: 0 ≤ x ≤ 2, R: 0 < y ≤ 1 and y = 2 (b)  [ 0, 1 ) ∪ ( 1, 2 ] (c) x = 2 (d) x = 0 y

lim g(θ) = 1

 θ→ 0

75. (a) f ( x ) = x 1 (1− x ) x

.9

2

.99

.999

.9999

.99999

.999999

f (x) .348678

.366032 .367695 .367861

.367877 .367879

x

1.01

1.00001 1.000001

1.1

1.001

f (x) .385543 .369711

1.0001

.368063 .367897

.367881 .367878

lim f ( x ) ≈ 0.36788

  x →1

77. c = 0, 1, −1; the limit is 0 at c = 0, and 1 at c = 1, −1. 79. 7 81. (a) 5 (b) 5 83. (a) 0 (b) 0

3. δ = 1 2, or a smaller positive value 5. 7. 9. 11. 13. 15. 17. 19. 21.

(

1

( 5

-72 -3

δ = 1 18, or a smaller positive value 49 12 δ = 0.1, or a smaller positive value δ = 7 16, or a smaller positive value δ = 5 − 2 , or a smaller positive value δ = 0.36, or a smaller positive value ( 3.99, 4.01 ) , δ = 0.01, or a smaller positive value ( −0.19, 0.21 ) , δ = 0.19, or a smaller positive value ( 3, 15 ) , δ = 5, or a smaller positive value ( 10 3, 5 ) , δ = 2 3, or a smaller positive value

23. ( − 4.5, − 3.5 ) , δ = positive value

x

)

x=2

2,

1

x

11. 3 19. (a) 1 25. 3 4 37. 1 2

2

1

0

15. 2 5 17. (a) 1 (b) −1 21. (a) 1 (b) 2 3 23. 1 29. 1 2 31. 2 33. 0 35. 1 41. 3 8 43. 3 45. 0

13. 1 (b) −1 27. 2 39. 0

55. (a) 400

(b) 399

 x → 5

x−5 = 0

(c) The limit does not exist.

7 -12

x x

47

4.5 − 2 ≈ 0.12 , or a smaller

25. ( 15, 17 ) , δ = 17 − 4 ≈ 0.12, or a smaller positive value 0.03 0.03 0.03 , or a smaller positive value ,2 + , δ = 27. 2 − m m m 1 c c 1 c 29. − , + , δ = , or a smaller positive value 2 m m 2 m 31. L = −3,  δ = 0.01, or a smaller positive value 33. L = 4,  δ = 0.05, or a smaller positive value 35. L = 4,  δ = 0.75, or a smaller positive value 55.  [ 8.7332, 8.7476 ]. To be safe, the left endpoint was rounded up and the right endpoint was rounded down. 59. The limit does not exist as x approaches 3.

( (

Í1 - x 2 , 0 … x 6 1 1…x62 1,

51. δ = ε 2 , or a smaller positive value, lim+

SECTION 2.3, pp. 95–98

1. δ = 2, or a smaller positive value

y=

SECTION 2.5, pp. 117–120

1. (a) 0 (b) −2 (c) 2 (d) Does not exist (e) −1 (f) ∞ (g) Does not exist (h) 1 (i) 0 3. (a) −3 (b) −3 5. (a) 1 2 (b) 1 2 7. (a) −5 3 (b) −5 3 9. 0 11. −1 13. (a) 2 5 (b) 2 5 15. (a) 0 (b) 0 17. (a) 7 (b) 7 19. (a) 0 (b) 0 21. (a) ∞ (b) ∞ 23. 2 25. ∞ 27. 0 29. 1 31. ∞ 33. 1 35. 1 2 37. ∞ 39. −∞ 41. −∞ 43. ∞ 45. (a) ∞ (b) −∞ 47. ∞ 49. ∞ 51. −∞ 53. (a) ∞ (b) −∞ (c) −∞ (d) ∞ 55. (a) −∞ (b) ∞ (c) 0 (d) 3 2 57. (a) −∞ (b) 1 4 (c) 1 4 (d) 1 4 (e) It will be −∞. 59. (a) −∞ (b) ∞ 61. (a) ∞ (b) ∞ (c) ∞ (d) ∞ y

63. 10

)

y=

10

1 x-1

y= 5

5

-2

-1

-10

1 2x + 4

x = -2 1

-5

y

65.

2 x=1

3

4

x

-4

-3

-2

-1

0 -5 -10

1

2

x

AN-8

Chapter 2: Answers to Odd-Numbered Exercises

67.

109.

y x = -2

y=

y=x y = - 1x

-3

-2

x

0

y x = -2

y= x+3 x+2

1.5

y=1

111.

y

1

-1

1 x+2

y=

2 y= x -1 x

x

1

2

x Í4 -

-2

x2

1

-1

-1

1

x

2

-1 -2 x=2

69. Domain: ( −∞, ∞ ); horizontal asymptote: y = 7 71. Domain: ( −∞, ∞ ); horizontal asymptotes: y = −1, y = 4 73. Domain: ( −∞, 0 ) and ( 0, ∞ ); horizontal asymptotes: y = −1, y = 1; vertical asymptote: x = 0 75. Here is one possibility. 77. Here is one possibility.

113.

y

3 2

y

y

-4 -3 -2 -1

y = f (x)

(1, 2)

1

x

0

1. 3. 5. 7. 11. 15. 19. 23. 27. 33. 37. 41. 49. 77. 81.

x

0

3

-1

2 1 2

3

x

5

4

85. At most one 87. 0 89. −3 4 91. 5 2 99. (a) For every positive real number B there exists a corresponding number δ > 0 such that for all x, c − δ < x < c ⇒ f ( x ) > B. (b) For every negative real number −B there exists a corresponding number δ > 0 such that for all x, c < x < c + δ ⇒ f ( x ) < −B.

107.

y

y=x+1

At x = 1:

2 3 4 y= x -4 =x+1x-1 x-1

3

0

1 2

3

= x+1 + 4

5

2

3

x

x

-3

0

lim f ( x ) = 1, so

 x →−1+

 x →−1

lim f ( x ) = lim+ f ( x ) = 0, so lim f ( x ) = 0.

 x → 0 −

 x → 0

 x → 0

lim f ( x ) = −1 and lim+ f ( x ) = 1, so lim f ( x )

 x →1−

  x →1

  x →1

does not exist. The function is discontinuous at x = 1, and the discontinuity is not removable.

2 1

1 x-1

lim f ( x ) =

 x →−1−

However, f (0) ≠ 0, so f is discontinuous at x = 0. The discontinuity can be removed by redefining f (0) to be 0.

5

4

-2

At x = 0:

6 (2, 4)

x-1

1 3 y=- x - 2 x

2/3

lim f ( x ) = 1 = f ( −1 ); continuous at x = −1

y

6

y=

1

No; not defined at x = 2 Continuous (a) Yes (b) Yes (c) Yes (d) Yes (a) No (b) No 9. 0 1, nonremovable; 0, removable 13. All x except x = 2 All x except x = 3, x = 1 17. All x All x except x = 0 21. All x except n π 2, n any integer All x except nπ 2, n an odd integer 25. All x ≥ − 3 2 All x 29. All x 31. All x except x = 1 0; continuous at x = π 35. 1; continuous at y = 1 2 2; continuous at t = 0 39. 1; continuous at x = 0 −1 43. 0 45. g(3) = 6 47. f (1) = 3 2 a = 43 51. a = −2, 3 53. a = 5 2, b = −1 2 x ≈ 1.8794, −1.5321, −0.3473 79. x ≈ 1.7549 x ≈ 3.5156 83. x ≈ 0.7391

1. At x = −1:

c − δ < x < c ⇒ f ( x ) < −B.

2

-1

PRACTICE EXERCISES, pp. 133–134

(c) For every negative real number −B there exists a corresponding number δ > 0 such that for all x,

x2

-2

SECTION 2.6, pp. 130–132

x , xZ0 0x0 1

1 (x - 2)2

f (x) =

4

-3

1

y

5

y=x+1

2

81. Here is one possibility. h(x) =

5

3

3

-3

y

105.

2

-3

79. Here is one possibility.

1

1

-2

(-1, -2) -2 -3

0

4 x

-3 - 2 -1 -1

-1

x

1 2 3 4

5

y = x 2/3 + 1 x 1/3

1 3 2 (0, 0) 1

115. (a) y → ∞ (b) y → ∞ (c) cusps at ±1

y

1

3

4

5

y

x

y = f(x)

1

-2

-1

0 -1

1

x

AN-9

Chapter 3: Answers to Odd-Numbered Exercises

3. (a) −21 (b) 49 (c) 0 (d) 1 (e) 1 (f) 7 1 5. 4 (g) −7 (h) − 7 7. (a) ( −∞, ∞ ) (b)  [ 0, ∞ ) (c) ( −∞, 0 ) and ( 0, ∞ ) (d)  [ 0, ∞ ) 1 1 9. (a) Does not exist (b) 0 11. 13. 2x 15. − 4 2 19. 2 π 21. 1 23. 4 25. −∞ 17. 2 3 27. 0 29. 2 31. 0 35. No in both cases, because lim f ( x ) does not exist, and lim f ( x )   x →1  x →−1 does not exist. 37. Yes, f does have a continuous extension, to a = 1 with f (1) = 4 3. 39. No 41. 2 5 43. 0 45. −∞ 47. 0 49. 1 51. 1 53. − π 2 55. (a) x = 3 (b) x = 1 (c) x = −4 57. Domain: [ −4, 2 ) and ( 2, 4 ], Range: ( −∞, ∞ ) ADDITIONAL AND ADVANCED EXERCISES, pp. 134–137

3. 0; the left-hand limit was taken because the function is undefined for υ > c. 5. 17.5 < t < 22.5; within 2.5 °C 13. (a) B (b) A (c) A (d) A 21. (a) lim r+ (a) = 0.5, lim + r+ (a) = 1  a → 0

 a →−1

(b) lim r− (a) does not exist, lim + r− (a) = 1  a → 0

11. m = 4, y − 5 = 4 ( x − 2 ) 13. m = −2, y − 3 = −2( x − 3 ) 15. m = 12, y − 8 = 12( t − 2 ) 1 1 17. m = , y − 2 = ( x − 4 ) 4 4 19. m = −1 21. m = −1 4 23. (a) It is the rate of change of the number of cells when t = 5. The units are the number of cells per hour. (b) P ′(3) because the slope of the curve is greater there. (c) 51.72 ≈ 52 cells h 25. ( −2, −5 ) 27. y = −( x + 1 ) , y = −( x − 3 ) 31. 6π 35. Yes 37. No 29. 19.6 m s 39. (a) Nowhere 41. (a) At x = 0 43. (a) Nowhere 45. (a) At x = 1 47. (a) At x = 0 SECTION 3.2, pp. 147–151

1. −2 x , 6, 0, −2

 a →−1

0 27. 1 29. 4 31. −1 33. 1 2 35. 4 1 39. To make g continuous at the origin, define g(0) = 1. y = 2x 43. y = x , y = −x 47. −4 3 (a) Domain: { 0, 1, 1 2, 1 3, 1 4 , …} (b) The domain intersects any interval ( a, b ) containing 0. (c) 0 51. (a) Domain: ( −∞, −1 π ] ∪ [ −1 (2π ), −1 (3π ) ] ∪ [ −1 (4 π ), −1 (5π ) ] ∪  ∪ [ 1 (5π ), 1 (4 π ) ] ∪ [ 1 (3π), 1 (2π) ] ∪ [ 1 π , ∞ ) (b) The domain intersects any interval ( a, b ) containing 0. (c) 0

25. 37. 41. 49.

5. 11. 17. 21. 31.

Chapter 3 1. P1: m1 = 1, P2 : m 2 = 5 3. P1: m1 = 5 2, P2 : m 2 = −1 2 5. y = 2 x + 5 y

–8 –6 –4 –2 0

(−1, 3)

y

y = 2Íx

3

y = 4 − x2

1

0

1

x

2

y

(−2, −8)

0

1

2

3

4

x

0 −1.5

−2 −3

dT y = –– dt

2 4

6

8 10 12

t (h)

−3 −4.5

f ( 0 + h ) − f (0) =1 h f ( 0 + h ) − f (0) while lim− = 0, h→ 0 h f ( 0 + h ) − f (0) f ′(0) = lim does not exist and f ( x ) is h→ 0 h not differentiable at x = 0.

37. Since lim+ h→ 0

y = x3 x

−8

x

x

−6

9. y = 12 x + 16

−2

2 4 6 8

6 7 8 9 10 11

1.5 1

y = 12x + 16

0 −1

(ºC/h) 3

(1, 2)

2

−3

1

f′

4.5 2

−2 −1

29. (d)

35. (a)    i) 0.77 °C h ii) 1.43 °C h iii) 0 °C h iv) −1.88 °C h (b) 3.63 °C h at 12 P.M., −4 °C h at 6 P.M. Slope (c)

4

3

− 2t , 5

−5

7. y = x + 1 y=x+1

4

1 + 1) 2

−4

5 y = 2x + 5

1 2 2 , 2, − , − 4 3 3 t3

3 3 1 3 9. , , , 7. 6 x 2 ( 2t 2 3θ 2 3 2 2 2 3 12 9 q 13. 1 − 2 , 0 15. 3t 2 2 x −4 1 , y − 4 = − ( x − 6) 19. 6 ( x − 2) x − 2 2 −1 −1 18 23. 25. 27. (b) ( x + 2 )2 ( x − 1)2 y′ 33. (a) x = 0, 1, 4 y′ (b) 2 4 3 2 1

SECTION 3.1, pp. 140–142

3. −

AN-10

Chapter 3: Answers to Odd-Numbered Exercises

f ( 1 + h ) − f (1) = 2 while h f ( 1 + h ) − f (1) f ( 1 + h ) − f (1) 1 lim = , f ′(1) = lim h → 0− h→ 0 h 2 h does not exist and f ( x ) is not differentiable at x = 1. Since f ( x ) is not continuous at x = 0, f ( x ) is not differentiable at x = 0. f ( 0 + h ) − f (0) = 3 while Since lim+ h→ 0 h f ( 0 + h ) − f (0) lim = 0, f is not differentiable at x = 0. h → 0− h (a) −3 ≤ x ≤ 2 (b) None (c) None (a) −3 ≤ x < 0, 0 < x ≤ 3 (b) None (c) x = 0 (a) −1 ≤ x < 0, 0 < x ≤ 2 (b) x = 0 (c) None

39. Since lim+ h→ 0

41. 43.

45. 47. 49.

SECTION 3.3, pp. 159–161

1. 3. 5. 7. 9. 11. 13. 15. 17. 21.

dy d 2y = −2 x , 2 = −2 dx dx d 2s ds 2 = 15t − 15t 4 , 2 = 30 t − 60 t 3 dt dt dy d 2y = 4 x 2 − 1 + 2e x , 2 = 8 x + 2e x dx dx dw 6 1 d 2w 18 2 = − 3 + 2, 2 = 4 − 3 dz z z dz z z dy d 2y = 12 x − 10 + 10 x −3 , 2 = 12 − 30 x −4 dx dx dr −2 5 d 2r 2 5 = 3 + 2, 2 = 4 − 3 ds 3s 2s ds s s y ′ = −5 x 4 + 12 x 2 − 2 x − 3 1 y ′ = 3 x 2 + 10 x + 2 − 2 x −19 x2 + x + 4 19. g′( x ) = y′ = ( 3x − 2 )2 ( x + 0.5 ) 2 2 1 dv t − 2t − 1 = 23. f ′(s) = 2 2 dt (1 + t 2 ) s ( s + 1) −4 x 3 − 3 x 2 + 1 1 + 2 x −3 2 27. y ′ = 2 2 2 x2 ( x − 1) ( x 2 + x + 1) − x 3 x 2 x 3 x y ′ = −2e + 3e 31. y ′ = 3 x e + x e 9 54 ds 2 − x y′ = x − 2e 35. = 3t 1 2 4 dt dr se s − e s 2 y′ = − ex e−1 39. = 57 7x ds s2 y′ = 2 x 3 − 3 x − 1, y′′ = 6 x 2 − 3, y′′′ = 12 x , y (4) = 12, y ( n ) = 0 for n ≥ 5 y′ = 3 x 2 + 8 x + 1, y ″ = 6 x + 8, yt ′′′ = 6, y ( n ) = 0 for n ≥ 4 y′ = 2 x − 7 x −2 , y ″ = 2 + 14 x −3 dr d 2r = 3θ −4 , 2 = −12θ −5 dθ dθ dw d 2w = −z −2 − 1, 2 = 2 z −3 dz dz dw d 2w 2 z = 6 ze ( 1 + z ) , 2 = 6e 2 z ( 1 + 4 z + 2 z 2 ) dz dz (a) 13 (b) −7 (c) 7 25 (d) 20 x 5 (b) m = −4 at ( 0, 1 ) (a) y = − + 8 4 (c) y = 8 x − 15, y = 8 x + 17 y = 4 x, y = 2 59. a = 1, b = 1, c = 0 61. ( 2, 4 )

25. υ′ = − 29. 33. 37. 41. 43. 45. 47. 49. 51. 53. 55. 57.

( 0, 0 ) , ( 4, 2 ) 65. y = −16 x + 24 (a) y = 2 x + 2 (c) ( 2, 6 ) 50 71. a = −3 P ′( x ) = na n x n−1 + ( n − 1 ) a n−1 x n−2 +  + 2a 2 x + a1 The Product Rule is then the Constant Multiple Rule, so the latter is a special case of the Product Rule. d( uυ w ) = uυ w ′ + uυ ′ w + u ′ υ w 77. (a) dx d (b) ( u u u u ) = u1u 2u 3u 4 ′ + u1u 2u 3 ′u 4 + u1u 2 ′u 3u 4 + dx 1 2 3 4 u1 ′u 2u 3u 4 d( (c) u u n ) = u1u 2  u n−1u n ′ + u1u 2  u n−2u n−1 ′u n + dx 1  + u1 ′u 2  u n dP nRT 2an 2 79. = − 2 + ( V − nb ) dV V3

63. 67. 69. 73. 75.

SECTION 3.4, pp. 167–170

1. (a) −2 m, −1 m s (b) 3 m s, 1 m s; 2 m s 2, 2 m s 2 (c) Changes direction at t = 3 2 s 3. (a) −9 m, −3 m s (b) 3 m s, 12 m s; 6 m s 2 , −12 m s 2 (c) No change in direction 5. (a) −20 m, −5 m s (b) 45 m s, ( 1 5 ) m s; 140 m s 2, ( 4 25 ) m s 2 (c) No change in direction 7. (a) a (1) = −6 m s 2 , a (3) = 6 m s 2 (b) v (2) = 3 m s (c) 6 m 9. Mars: ≈ 7.5 s, Jupiter: ≈ 1.2 s 11. g s = 0.75 m s 2 13. (a) υ = −9.8t, υ = 9.8t m s, a = −9.8 m s 2 (b) t ≈ 3.4 s (c) υ ≈ −33.1 m s 15. (a) t = 2, t = 7 (b) 3 ≤ t ≤ 6 (c) (d) 0 y 0 (ms)

a 4 3 2 1

Speed

3

0

2

4

6

8

10

t (s)

0 −1 −2 −3 −4

dy a = –– dt

1 2 3 4 5 6 7 8 9 10

17. (a) 57 m s (b) 2 s (c) 8 s, 0 m s (d) 10.8 s, 27 m s (e) 2.8 s (f ) Greatest acceleration happens 2 s after launch (g) Constant acceleration between 2 and 10.8 s, −10 m s 2 19. C = position, A = velocity, B = acceleration 21. (a) $110 machine (b) $80 (c) $79.90 23. (a) b′(0) = 10 4 bacteria h (b) b′(5) = 0 bacteria h (c) b′(10) = −10 4 bacteria h dy t 25. (a) = −1 dt 12

t

Chapter 3: Answers to Odd-Numbered Exercises

(b) The largest value of

dy is 0 m h when t = 12 and the dt

⎛ 6 − 15 ⎟⎞ ⎛ 6 + 15 ⎤ , 2 ⎟⎟ ∪ ⎜⎜ , 4⎥ (d) The object speeds up on ⎜⎜ ⎝ ⎠ ⎝ ⎥⎦ 3 3 ⎡ 6 − 15 ⎟⎞ ⎛ 6 + 15 ⎞⎟   ⎟⎟.   ⎟⎟ ∪ ⎜⎜ 2, and slows down on ⎢ 0, ⎠ ⎠ ⎝ ⎢⎣ 3 3 (e) The object is moving fastest at t = 0 and t = 4 when it is 6 ± 15 s. moving 7 units second and slowest at t = 3 6 + 15 (f) When t = the object is at position s ≈ −6.303 3 units and farthest from the origin.

dy smallest value of is −1 m h when t = 0. dt y

(c) 6 5 4

y = 6 a1 − t b 2 12

3 2 1 −1

t

12 dy t = −1 dt 12

27. 0.846 m, 1.482 m, additional meters to stop car for 1 km/h speed increase 6250 m D = 29. t = 25 s, 9 s

31.

s = 60t − 4.9t2

180

t

12

  3. 2 x cos x − x 2 sin x

2 7 5. − csc x cot x − − x e x

  7. sin x sec 2 x + sin x

9. ( e −x sec x )( 1 − x + x tan x )

11.

13. 17.

31.

d2s = −9.8 dt2

−60

1. −10 − 3 sin x

25. 27.

ds = 60 − 9.8t dt

60

SECTION 3.5, pp. 174–176

19.

120

33.

(a) υ = 0 when t = 6.12 s. (b) υ > 0 when 0 ≤ t < 6.12 ⇒ the object moves up; υ < 0 when 6.12 < t ≤ 12.24 ⇒ the object moves down. (c) The object changes direction at t = 6.12 s. (d) The object speeds up on ( 6.12, 12.24 ] and slows down on [ 0, 6.12 ). (e) The object is moving fastest at the endpoints t = 0 and t = 12.24 when it is traveling 60 m s. It’s moving slowest at t = 6.12 when the speed is 0. (f) When t = 6.12 , the object is s = 183.6 m from the origin and farthest away.

− csc 2 x ( 1 + cot x ) 2 4 tan x sec x − csc 2 x 15. 0 3 x 2 sin x cos x   + x 3 cos 2 x − x 3 sin 2 x −2 csc t cot t sec 2 t + e −t 21. 23. −θ ( θ cos θ + 2 sin θ ) ( 1 − csc t ) 2 sec θ csc θ ( tan θ − cot θ ) = sec 2 θ − csc 2 θ sec 2 q 29. sec 2 q q 3 cos q − q 2 sin q − q cos q − sin q 2 ( q 2 − 1) (a) 2 csc 3 x − csc x (b) 2 sec 3 x − sec x y

35.

y=x

y = −x − p

10

y = sin x

1 −3p/2 −p −p/2 −1

p/2

p

ds = 3t 2 − 12t + 7 dt

(−p3, 2) 2 ap4, Í2b

y = −2Í3x −

1 Í2p 2Í3p +2 y = Í2 x − 4 + Í2 3

5 −p2 −p3

−10

(

s = t 3 − 6t 2 + 7t

(a) υ = 0 when t =

p4

0

p2

39. Yes, at x = π 41. No 43. Yes, at x = 0, π , and 2π π π 45. − , −1 ; ,1 4 4

−5

6±

15 3

)( ) y

y = tan x

s

6 − 15 6 + 15 ⇒ the < t < 3 3 6 − 15 object moves left; v > 0 when 0 ≤ t < or 3 6 + 15 < t ≤ 4 ⇒ the object moves right. 3 6 ± 15 (c) The object changes direction at t = s. 3

2p

y = sec x

Í2 d2s = 6t − 12 dt2

3p/2

y = −1 (3p2, −1)

y

37.

s

33.

AN-11

1

(b) υ < 0 when

(p4, 1) y = 2x − p + 1 2

−p2

−p4

p4

y = 2x + p − 1 2 −1 (−p4, −1)

p2

x

x

x

AN-12

Chapter 3: Answers to Odd-Numbered Exercises

47. (a) y = −x + π 2 + 2

(b) y = 4 −

67. 6 tan ( sin 3 t ) sec 2 ( sin 3 t ) sin 2 t cos t 1 2 6 1+ 69. 3( 2t 2 − 5 )3 ( 18t 2 − 5 ) 71. 3 1 + x x x 73. 2 csc 2 ( 3 x − 1 ) cot ( 3 x − 1 ) 75. 16( 2 x + 1 ) 2 ( 5 x + 1 )

3

(

49. − 2 m s, 2 m s, 2 m s 2 , 2 m s 3 51. c = 9 53. (a) sin x (b) 3 cos x − sin x (c) 73 sin x + x cos x 55. (a) i) 10 cm ii) 5 cm iii) −5 2 ≈ −7.1 cm (b) i) 0 cm/s

77. 79. 83. 89.

ii) −5 3 ≈ −8.7 cm s

iii) −5 2 ≈ −7.1 cm s SECTION 3.6, pp. 181–184

3. 3 cos( 3 x + 1 )

1. 12 x 3

5.

cos x 2 sin x

7. 2π x sec 2 ( π x 2 ) 9. With u = ( 2 x + 1 ) , y = u 5 : 10 ( 2 x + 1 ) 4

dy dy du = = 5u 4 ⋅ 2 = dx du dx

11. With u = ( 1 − ( x 7 ) ) , y = u −7 : −8

( 17 ) = (1 − 7x )

−7u −8 ⋅ − 13.

17.

19. 21. 23. 29. 31.

35. 39. 43. 47. 51. 55. 57. 59. 63. 65.

)

(

(

(

) (

( )) (

(

)(

)

(

)

)

)

( )

2( 2 x 2 + 1)e x 2 f ′( x ) = 0 for x = 1, 4 ; f ″( x ) = 0 for x = 2, 4 −π 4 85. 0 87. −5 (a) 2 3 (b) 2π + 5 (c) 15 − 8π (d) 37 6

)

81. 5 2

(e) −1 (f ) 2 24 (g) 5 32 (h) −5 ( 3 17 ) 91. 5 93. (a) 1 (b) 1 95. y = 1 − 4 x 97. (a) y = π x + 2 − π (b) π 2 99. It multiplies the velocity, acceleration, and jerk by 2, 4, and 8, respectively. 2 4 101. υ(6) = m s, a(6) = − m s2 5 125 SECTION 3.7, pp. 187–189

dy dy du With u = ( ( x 2 8 ) + x − ( 1 x ) ) , y = u 4 : = = dx du dx 1 3 x x2 1 1 x 4u 3 ⋅ +1+ 2 = 4 + x− +1+ 2 4 8 4 x x x dy dy du With u = tan x , y = sec u: =   = dx du dx ( sec u tan u )( sec 2 x ) = sec ( tan x ) tan ( tan x )sec 2 x dy dy du With u = tan x , y = u 3 : = = 3u 2 sec 2 x = dx du dx 3 tan 2 x ( sec 2 x ) dy y = e u , u = −5 x : = −5e −5 x dx dy y = e u , u = 5 − 7 x: = −7e ( 5−7 x ) dx csc θ 1 4 25. ( cos 3t − sin 5t ) − 27. π cot θ + csc θ 2 3−t 2 x sin 4 x + 4 x 2 sin 3 x cos x + cos −2 x + 2 x cos −3 x sin x ( 4 x + 3 )3 ( 4 x + 7 ) 1 ( 3x − 2 )5 − 33. 2 ( x + 1) 4 1 x3 4 − 2 2x 5 2 3 − x 2 x (1 − x ) e + 3x e 37. x − 3x + 3 e 5 x 2 2 x sec x tan x + sec x x sec 2 ( 2 x ) + tan ( 2 x ) 41. 2 7 + x sec x 2 sin θ 45. −2 sin (θ 2 )sin 2θ + 2θ cos( 2θ ) cos(θ 2 ) ( 1 + cos θ ) 2 t+2 t cos 49. 2θe −θ 2 sin ( e −θ 2 ) 2( t + 1 )3 2 t +1 8 sin (2t ) 2π sin ( πt − 2 ) cos( πt − 2 ) 53. ( 1 + cos 2t ) 5 10 t 10 tan 9 t sec 2 t + 10 t 9 tan 10 t dy 2 = −2π sin ( πt − 1 ) ⋅ cos( πt − 1 ) ⋅ e cos ( πt −1) dt −3t 6 ( t 2 + 4 ) 61. −2 cos( cos( 2t − 5 ) )( sin ( 2t − 5 ) ) 4 ( t 3 − 4t ) t 2 t t 1 + tan 4 tan 3 sec 2 12 12 12 t sin (t 2 ) − 1 + cos(t 2 )

(

15.

dy dy du =   = dx du dx

)(

( ))

−2 xy − y 2 x 2 + 2 xy −2 x 3 + 3 x 2 y − xy 2 + x 5. x 2y − x3 + y 1.

9. cos y cot y 13.

−y 2 ⎛ 1 ⎞⎟ ⎛1⎞ y sin ⎜⎜ ⎟⎟ − cos⎜⎜ ⎟⎟⎟ + xy ⎝ y⎠ ⎝ y⎠

1 − 2y 2x + 2y − 1 1   7. y( x + 1) 2 − cos 2 ( xy ) − y 11. x 2e 2 x − cos( x + 3 y ) 15. 3 cos( x + 3 y )   3.

−y 2 − x 2 −r x 21. y′ = − , y ″ = y y3 θ 2 x 2 2 2 x 2 xe + 1 d y ( 2 x y + y − 2 x )e 2 − x 2e 2 x 2 − 1 dy 23. = , = 2 dx y dx y3

17. −

r θ

25. y′ = 27. y′ =

35. 37. 39. 41. 43. 45. 47.

49. 55.

y 1 , y″ = 3 y +1 2( y + 1) 6 x ( 1 − cos y ) 2 − 9 x 4 sin y 3x 2 , y″ = 1 − cos y ( 1 − cos y )3

31. ( −2, 1 ): m = −1, ( −2, −1 ): m = 1 7 1 4 29 (a) y = x − (b) y = − x + 4 2 7 7 1 8 (a) y = 3 x + 6 (b) y = − x + 3 3 6 6 7 7 (a) y = x + (b) y = − x − 7 7 6 6 π 2 2 π (a) y = − x + π (b) y = x − + 2 π π 2 x 1 (a) y = 2π x − 2π (b) y = − + 2π 2π Points: ( − 7, 0 ) and ( 7, 0 ) , Slope: −2 ⎛ 3 3 ⎞⎟ ⎛ 3 1 ⎞⎟ , m = 3 at ⎜⎜ , m = −1 at ⎜⎜ , ⎝ 4 2 ⎠⎟⎟ ⎝ 4 2 ⎟⎟⎠ 27 27 27 ; ( −3, 2 ): m = − ; ( −3, −2 ): m = ; ( 3, 2 ): m = 8 8 8 27 ( 3, −2 ): m = − 8 ( 3, −1 ) dy y 3 + 2 xy x 2 + 3 xy 2 dx dx 1 = − 3 = , , = − 2 2 dx x + 3 xy dy y + 2 xy dy dy dx

29. −2 33.

19.

Chapter 3: Answers to Odd-Numbered Exercises

1 1 1 ⎤ 2 51. t ( t + 1 )( t + 2 ) ⎡⎢ + + ⎥ = 3t + 6t + 2 t +1 t + 2⎦ ⎣t θ+5⎡ 1 1 53. − + tan θ ⎤⎥ ⎢ θ cos θ ⎣ θ + 5 θ ⎦

SECTION 3.8, pp. 198–200

x 3 − 2 2

1. (a) f −1 ( x ) =

y

(b) 3

y = f(x) = 2x + 3 –1

y=f −32

55.

x 3 (x) = − 2 2

0

57.

x

3

−32

61. 67.

(c) 2,1 2 3. (a) f −1 ( x ) = −

73.

x 5 + 4 4

y

(b) 5

79. y = f(x) = −4x + 5

5 4 0

y=f

–1

5 4

85.

(x) = −

89.

x 5 + 4 4 x

5

93.

(c) −4, −1 4 2

⎤ x x2 + 1⎡ 1 x 2 − ⎢ + 2 ⎥ ( x + 1) 2 3 ⎣ x x + 1 3( x + 1 ) ⎦

(

)

1 3 x( x − 2) 1 1 2x   2   + − 2 3 x +1 x x−2 x +1 1−t 63. 1 ( 1 + e θ ) t y 2 − xy ln y ye y cos x dy 69. = 2 y dx x − xy ln x 1 − ye sin x ln 5 75. π x ( π −1) 5 s 2 s 2( ln r ) 3 81. x ln 4 r ( ln 2 )( ln 4 ) 1 sin ( log 7 θ ) + 87. cos( log 7 θ ) ln 7 1 1 ( log 2 3 )3 log 2 t 91. t t x ( x + 1) x 95. + ln ( x + 1 ) x +1

( )

(

)

59. −2 tan θ 65. e cos t ( 1 − t sin t ) 71. 2 x ln 2 77. 83.

1 θ ln 2 (x

−2 + 1 )( x − 1 )

1 ln 5

( t)

t

( ln2 t + 12 )

⎛ ln x 2 ⎞⎟ 99. ( x ln x )⎜⎜ ⎝ x ⎟⎟⎠ 1 − xy ln y 103. 2 x ( 1 + ln y )

97. ( sin x ) x ( ln sin x + x cot x )

y

5. (b)

AN-13

y = x3

1

101.

3 y − xy ln y x2 − x

y = x 13

SECTION 3.9, pp. 204–206 −2

−1

1

x

2

1. 3. 5. 7.

−1 −2

7. 13. 19. 25. 31.

(c) Slope of f at ( 1, 1 ): 3; slope of g at ( 1, 1 ): 1 3; slope of f at ( −1, −1 ): 3; slope of g at ( −1, −1 ): 1 3 (d) y = 0 is tangent to y = x 3 at x = 0; x = 0 is tangent to y = 3 x at x = 0. 19 9. 3 11. (a) 2 (b) 1 3 (c) 3 1 (a) 2 (b) 7 (c) 3 4 15. + 1 17. 2 t x 1 −1 x 21. 23. 3 x − eθ θ+1 1 − ln t 2( ln t ) + ( ln t ) 2 27. x 3 ln x 29. t2 1 1 33. 35. 2 cos( ln θ ) x ln x x ( 1 + ln x ) 2

37. −

3x + 2 2 x ( x + 1)

39.

2 t ( 1 − ln t ) 2

41.

tan ( ln θ ) θ

10 x 1 + x2 + 1 2( 1 − x ) 1 1 1 2x + 1 45. x ( x + 1) + = 2 x x +1 2 x ( x + 1) 43.

(

( )

47. 49.

( 12 )

(

)

)

t 1 1 1 − = t +1 t t +1 2 t ( t + 1)3 2 1 θ + 3 ( sin θ ) + cot θ 2( θ + 3 )

(

)

(a) (a) (a) (a)

9. 1

π4 −π 6 π3 3π 4 2

(b) (b) (b) (b)

−π 3 π4 3π 4 π6

11. −1

3

(c) (c) (c) (c)

π6 −π 3 π6 2π 3

13. π 2

15. π 2

17. π 2

−2 x 2 1    23.    25.  2s + 1 s 2 + s 1 − x4 1 − 2t 2 −2 x  −1  −1 27. 29. 31. 2 t (1 + t ) ( x 2 + 1) x 4 + 2 x 2 1 − t2 1 −e t −1 = 33. 35. − 1 2 ( tan x )( 1 + x ) e 2t − 1 e t ( e t )2 − 1 − 2s 2 37. 39. 0 41. sin −1 x 1 − s2 1 1 43.   45. − sin ( x − arccos x ) 1 + 1 − x2 2 arcsin x 1 − x 2 19. 0

   21.

(

47. 57. 59. 69.

71. 73.

8 2 −9 x 2 ( arccot ( x 3 ) ) 2 49. 0 51. x6 + 1 4 + 3π (a) Defined; there is an angle whose tangent is 2. (b) Not defined; there is no angle whose cosine is 2. (a) Not defined; no angle has secant 0. (b) Not defined; no angle has sine 2. (a) Domain: all real numbers except those having the form π + k π where k is an integer; range: −π 2 < y < π 2 2 (b) Domain: −∞ < x < ∞; range: −∞ < y < ∞ (a) Domain: −∞ < x < ∞; range: 0 ≤ y ≤ π (b) Domain: −1 ≤ x ≤ 1; range: −1 ≤ y ≤ 1 The graphs are identical.

)

AN-14

Chapter 3: Answers to Odd-Numbered Exercises

17. (a) 1.01

SECTION 3.10, pp. 211–214

dA dr 1. = 2 πr 7. −3 2 3. 10 5. −6 dt dt 11. (a) −180 m 2 min (b) −135 m 3 min

9. 31 13

dV dh dV dr = 2πhr = πr 2 (b) dt dt dt dt dV dh dr 2 = πr + 2πhr (c) dt dt dt 1 dR 1 dV V dI 15. (a) 1 volt s (b) − amp s (c) = − 3 dt I dt I dt (d) 3 2 ohms s, R is increasing. 13. (a)

(

17. (a) (b) (c) 19. (a) (b) (c) 21. (a)

ds = dt

)

x dx x 2 + y 2 dt x2

(c) −14 13 cm s, decreasing 23. (a) −3.6 m s (b) −5.355 m 2 s

(c) −1 rad s

25. 6 m s 27. (a) (b) 29. (a) (c) 31. 1 m

dh 1125 = ≈ 11.19 cm min dt 32π dr 375 = ≈ 14.92 cm min dt 8π −1 m min (b) r = 26 y − y 2 m 24 π dr 5 = − m min dt 288π min, 40 π m 2 min 33. 3.16 m s

35. Increasing at 466 1681 L min 2

(

)

( ) dx

y dy x dx + + y 2 dt x 2 + y 2 dt y dy dx = − dt x dt dA 1 dθ = ab cos θ dt 2 dt dA 1 dθ 1 da = ab cos θ + b sin θ dt 2 dt 2 dt dA 1 dθ 1 da 1 db = ab cos θ + b sin θ + a sin θ dt 2 dt 2 dt 2 dt 14 cm 2 s, increasing (b) 0 cm s, constant ds = dt

(b) 1.003 2 − 2x 2 3 − 21. 19. dx 2 dx 2 x (1 + x 2 ) 1− y 5 23. 25. cos ( 5 x ) dx dx 3 y + x 2 x 3x 2

37. −5 m s

39. −441 m s 5 10 41. cm min, cm 2 min 72π 3 43. (a) −10 13 ≈ −2.774 m s

x3 27. ( 4 x 2 ) sec 2 3 3 29. ( csc ( 1 − 2 x 1 ⋅ e x dx 31. 2  x −1 dx 37. e −2 x − 1 39. (a) 0.41 (b) 41. (a) 0.231 (b) 43. (a) −1 3 (b)

x ) cot ( 1 − 2 x ) ) dx 33.

2x dx 1 + x2

47. dS = 12 x 0 dx

45. dV = 4 πr0 2 dr 49. dV = 2πr0 h dr

51. (a) 0.08π m 2 (b) 2% 53. dV ≈ 565.5 cm 3 55. (a) 2% (b) 4% 1 59. 3% 57. % 3 61. The ratio equals 37.52, so a change in the acceleration of gravity on the moon has about 38 times the effect that a change of the same magnitude has on Earth. 63. Increase V ≈ 40% f ′′(a) 65. (a) i) b 0 = f (a) ii) b1 = f ′(a) iii) b 2 = 2 (b) Q( x ) = 1 + x + x 2 (d) Q( x ) = 1 − ( x − 1 ) + ( x − 1 ) 2 x x2 (e) Q( x ) = 1 + − 2 8 (f) The linearization of any differentiable function u( x ) at x = a is L ( x ) = u(a) + u′(a)( x − a ) = b 0 + b1 ( x − a ), where b 0 and b1 are the coefficients of the constant and linear terms of the quadratic approximation. Thus, the linearization for f ( x ) at x = 0 is 1 + x; the linearization for g( x ) at x = 1 is 1 − ( x − 1 ) or 2 − x; and the linearization for x h( x ) at x = 0 is 1 + . 2 67. (a) L ( x ) = x ln 2 + 1 ≈ 0.69 x + 1 y y (b) y = 2x 3

(c) d θ1 dt = 1 6 rad s, d θ 2 dt = −1 6 rad s

2

45. −5.5 deg min

1.4

−3 −2 −1

3. L ( x ) = 2 5. L ( x ) = x − π 1 4 9. −x − 5 11. x+ 13. 1 − x 7. 2 x 12 3 k −1 15. f (0) = 1. Also, f ′( x ) = k ( 1 + x ) , so f ′(0) = k. This means the linearization at x = 0 is L ( x ) = 1 + kx.

y = 2x

y = (ln 2)x + 1

1

0

1 0.8

y = (ln 2)x + 1

0.4

47. 12π km min 1. L ( x ) = 10 x − 13

2 xe x 2 dx 1 + e 2x 2

0.4 (c) 0.01 0.2 (c) 0.031 −2 5 (c) 1 15

(b) d θ1 dt = 5/39 rad s, d θ 2 dt = −5 39 rad s

SECTION 3.11, pp. 224–226

35.

1

2

3

x

−1

−0.5

0

PRACTICE EXERCISES, pp. 227–231

1. 5 x 4 − 0.25 x + 0.25 5. 2( x + 1 )( 2 x 2 + 4 x + 1 )

  3. 3 x ( x − 2 )

7. 3( θ 2 + sec θ + 1 ) 2 ( 2θ + sec θ tan θ ) 1 9. 11. 2 sec 2 x tan x 2 2 t (1 + t )

0.5

1

x

Chapter 3: Answers to Odd-Numbered Exercises

15. 5( sec t )( sec t + tan t ) 5

13. 8 cos 3 ( 1 − 2t ) sin ( 1 − 2t )

θ cos θ + sin θ cos 2θ 19. 2θ sin θ 2θ 2 2 2 21. x csc + csc cot x x x 1 12 2 23. x sec(2 x ) [ 16 tan (2 x ) 2 − x −2 ] 25. −10 x csc 2 ( x 2 ) 2 −( t + 1 ) 27. 8 x 3 sin (2 x 2 ) cos(2 x 2 ) + 2 x sin 2 (2 x 2 ) 29. 8t 3 − 2 sin θ 1− x   −1 35. 31. 33. 12 ( x + 1)3 1 ( cos θ − 1 ) 2 2x 2 1 + x ⎡ 5 x + cos 2 x ⎤ ⎥ 37. 3 2 x + 1 39. −9 ⎢ 41. −2e −x 5 ⎢⎣ ( 5 x 2 + sin 2 x ) 5 2 ⎥⎦

1 y = −–x +p + 1 2 8

)

43. xe 4 x 2 sin θ cos θ 45. = 2 cot θ sin 2 θ

47.

  −1 1 − x 2 arccos x 1− z 61. + arcsec z z2 − 1 −3 x 2 − 4 y + 2 67. 4 x − 4 y1 3

59. tan −1 (t ) +

57.

63. −1 69. −

73. −1 2

y x

81.

d 2y −2 xy 3 − 2 x 4 = 2 dx y5

1 1 − u2

t 1 − 1 + t2 2t y+2 65. − x+3 71.

1 2 y( x + 1) 2

77. −

75. y x

dp 6q − 4 p = 3 p 2 + 4q dq

83. (a)

49. −8 −t ( ln 8 )

   53. ( x + 2 ) x + 2 [ ln ( x + 2 ) + 1 ]    55. −

51. 18 x 2.6

79.

2 ( ln 2 ) x

2e −arctan x 1 + x2

dr = ( 2r − 1 )( tan 2s ) ds (b)

d 2y −2 xy 2 − 1 = 2 dx x4y3

85. (a) 7 (b) −2 (c) 5 12 (d) 1 4 (e) 12 (f ) 9 12 (g) 3 4 87. 0

89.

32

3 2e 4

cos ( e

32

)

y

95. (a)

1 −2 93. ( 2t + 1 ) 2 2 (b) Yes (c) Yes 91. −

1

−1

0

1

−p2

−p4

p4

−1

(−p4, −1)

p2

−1

p 1 y=−–x− −1 2 8

1 9 111. Tangent: y = − x + , normal: y = 4 x − 2 4 4 1 7 113. Tangent: y = 2 x − 4, normal: y = − x + 2 2 4 11 5 115. Tangent: y = − x + 6, normal: y = x − 5 5 4 1 117. ( 1, 1 ): m = − ; ( 1, −1 ): m not defined 2 119. B = graph of  f , A = graph of  f ′ 121.

y

(−1, 2)

2

(6, 1)

−1

123.

y = f (x)

(4, 3)

3

1

4

6

x

2( x 2 + 1 ) ⎡ 2 x  ⎢ + tan 2 x ⎤⎥ ⎦ cos 2 x ⎣   x 2 + 1

1 1 1 ⎤ ⎡ ( t + 1 )( t − 1 ) ⎤ 5 ⎡ 1 125. 5 ⎢ + − − ⎥  ⎢ ⎥ t −1 t − 2 t + 3⎦ ⎣ ( t − 2 )( t + 3 ) ⎦ ⎣ t + 1 ln sin θ 1 ( sin θ ) θ   + θ cot θ 127. 2 θ dS dr dS dh = 2πr 129. (a) = ( 4 πr + 2 π h ) (b) dt dt dt dt

(

(c)

)

dS dr dh dr r dh = − = ( 4 πr + 2πh ) + 2πr   (d) dt dt dt dt 2r + h dt

131. −40 m 2 s

133. 0.02 ohm s

135. 2 m s

y f(x) = y

97. (a)

x 2, −1 ≤ x < 0 −x 2, 0 ≤ x < 1

(b) Yes

x, 0≤x≤1 2 − x, 1 < x ≤ 2

y=

(c) No

1

0

99.

2

( 52 , 49 ) and ( 32 , − 14 )

y = 2x + (p − 2)2 p4

x

101. ( −1, 27 ) and ( 2, 0 )

103. (a) ( −2, 16 ) , ( 3, 11 ) (b) ( 0, 20 ) , ( 1, 7 )

y = tan x

1

−p4 1

109. 4

x

2 5 137. (a) r = h (b) − m min 5 16π 18 3 139. (a) km s or 600 m s (b) rpm 5 π π−2 141. (a) L ( x ) = 2 x + 2

x

1 4

(p4, 1)

1

( ) ( )

(

107. y = tan x

17.

( )

y

105.

AN-15

(−p4, −1)

−1

x

AN-16

Chapter 4: Answers to Odd-Numbered Exercises

17. Absolute maximum at x = 2; no absolute minimum

15. Absolute minimum at x = 0; no absolute maximum

2( 4 − π ) 4

(b) L ( x ) = − 2 x + y

y

y

y = g(x)

f(x) = 0 x 0

1

2

1

1

x

2

Í2 y = sec x

−p4, Í2

−p2

−p4

p2

0

−1

147. (a) 4%

(b) 8%

145. dS =

πrh 0 dh r 2 + h 0 2

y 3

(c) 12%

1. (a) sin 2θ = 2 sin θ cos θ; 2 cos 2θ = 2 sin θ ( − sin θ ) + cos θ ( 2 cos θ ); 2 cos 2θ = −2 sin 2 θ + 2 cos 2 θ; cos 2θ = cos 2 θ − sin 2 θ (b) cos 2θ = cos 2 θ − sin 2 θ; −2 sin 2θ = 2 cos θ ( − sin θ ) − 2 sin θ ( cos θ ); sin 2θ = cos θ sin θ + sin θ cos θ; sin 2θ = 2 sin θ cos θ 1 3. (a) a = 1, b = 0, c = − (b) b = cos a, c = sin a 2 9 5 5 5. h = −4, k = , a = 2 2 7. (a) 0.09y (b) Increasing at 1% per year 9. Answers will vary. Here is one possibility.

p/2

p

3p/2

21. Absolute maximum: −3; absolute minimum: −19 3 y

−2 −1 0 −1

1

2

11. (a) 2 s, 19.6 m s (b) 12.25 s, 120 m x y−1 y 2 − ye x ( x y + 1 ) ln y 15. y′ = e x ( x y + 1 ) − x y y ln x b 17. (a) m = − (b) m = −1, b = π π 3 9 19. (a) a = , b = 21. f  odd ⇒ f ′ is even 4 4 25. h′ is defined but not continuous at x = 0; k ′ is defined and continuous at x = 0. −7 rad s 27. 75 31. (a) 0.248 m (b) 0.02015 s (c) It will lose about 28.3 min day.

−1 −2 −3 −4

Absolute minimum at x = c 2 ; absolute maximum at x = b Absolute maximum at x = c; no absolute minimum Absolute minimum at x = a; absolute maximum at x = c No absolute minimum; no absolute maximum Absolute maximum at ( 0, 5 ) 11. (c) 13. (d)

2 y=x −1 −1 ≤ x ≤ 2

1

y = 23 x − 5

−6

−1

−2 ≤ x ≤ 3

1

x

2

(0, −1) Abs min

−7

y

1

(2, 3) Abs max

3

Abs max

−5

27. Absolute maximum: 2; absolute minimum: −1 y

x

2

(2, −0.25) Abs max

3

y = Îx −1 ≤ x ≤ 8

1 −1 −1 (−1, −1) Abs min

y = − 1 , 0.5 ≤ x ≤ 2 x2

1

2

3

4

5

6

7

8

(8, 2) Abs max x

(0.5, −4) Abs min

29. Absolute maximum: 2; absolute minimum: 0

31. Absolute maximum: 1; absolute minimum: −1

y

Chapter 4 SECTION 4.1, pp. 241–243

x

3

25. Absolute maximum: −0.25; absolute minimum: −4

0

y

(3, −3)

−3

(−2, −19/3) Abs min

23. Absolute maximum: 3; absolute minimum: −1

2

−2

−4

t

x

2p

−3

y

1. 3. 5. 7. 9.

−1

19. Absolute maximum at x = π 2; absolute minimum at x = 3π 2

ADDITIONAL AND ADVANCED EXERCISES, pp. 231–234

0

x

2

x

y = −Í 2x + Í2 Q4 − pR 4

143. L ( x ) = 1.5 x + 0.5

1

y = Î4 − x2 −2 ≤ x ≤ 1

y

(p2, 1) Abs max

(0, 2) Abs max 1 1

(−2, 0) −1 Abs min

0 −1

−p/2 1

x

−1 (−p2, −1) Abs min

p/2

5p/6

y = sin u, −p2 ≤ u ≤ 5p6

u

Chapter 4: Answers to Odd-Numbered Exercises

33. Absolute maximum: 2 absolute minimum: 1 y

1.2 1.0 0.8 0.6 0.4 0.2

Abs max Qp3, 2Î3R

3;

p3

p2

(0, 2) Abs max

−1 x

2p3

max Q1, e1R

1 −3 −2 −1 −1 y = xe−x−2

1

2

y=2−0t0 −1 ≤ t ≤ 3

1

37. (a) Absolute maximum is 1 e  at  x = 1; absolute minimum is −e at x = −1. y (b) Absolute 2

0

1

2

i)

x

(−1, −e) −3 Absolute −4 min

t

3

ii)

Abs min (3, −1)

−1

2

3

7.

1 (1 + 3

3. 1

5. ± 1 −

1 7 ) ≈ 1.22, ( 1 − 3

33. 35.

4

37.

5

71. Maximum value is 11 at x = 5; minimum value is 5 on the interval [ −3, 2 ]; local maximum at ( −5, 9 ). 73. Maximum value is 5 on the interval [ 3, ∞ ); minimum value is −5 on the interval ( −∞, −2 ].

1. 1 2

29. Yes

b

4

41. Increasing on ( 0, 8 ), decreasing on ( −1, 0 ); absolute maximum: 16 at x = 8; absolute minimum: 0 at x = 0 43. Increasing on ( −32, 1 ); absolute maximum: 1 at θ = 1; absolute minimum: −8 at θ = −32 45. x = 3 47. x = 1, x = 4 49. x = 1 51. x = 0, x = 4 53. x = 0, x = 1 55. x = 2, x = −4 57. (a) No (b) The derivative is defined and nonzero for x ≠ 2. Also, f (2) = 0 and f ( x ) > 0 for all x ≠ 2. (c) No, because ( −∞, ∞ ) is not a closed interval. (d) The answers are the same as parts (a) and (b), with 2 replaced by a. 59. y is increasing on ( −∞, ∞ ) so has no extrema. 61. y is decreasing on ( −∞, ∞ ) so has no extrema. 63. Yes 65. g assumes a local maximum at −c. 67. (a) Maximum value is 144 at x = 2. (b) The largest volume of the box is 144 cubic units, and it occurs when x = 2. v0 2 69. + s0 2g

SECTION 4.2, pp. 248–249

4 ≈ ±0.771 π2 7 ) ≈ −0.549

−5

−4

0

x

39. 43. 47. 51.

53.

57.

x

−1

71.

0

x

2

x 0

31. (a) 4

4

(b) 3

9

18

24

(c) 3

x2 x3 x4 + C (b) + C (c) +C (a) 2 4 3 1 1 1 (a) + C (b) x + + C (c) 5 x − + C x x x t 1 (a) − cos 2t + C (b) 2 sin + C 2 2 1 t (c) − cos 2t + 2 sin + C 2 2 e 2x f (x) = x 2 − x 41. f ( x ) = 1 + 2 1 − cos( π t) s = 4.9t 2 + 5t + 10 45. s = π s = e t + 19t + 4 49. s = sin (2t ) − 3 If T (t ) is the temperature of the thermometer at time t, then T (0) = −19 °C and T (14) = 100 °C. From the Mean Value Theorem, there exists a 0 < t 0 < 14 such that T (14) − T (0) = 8.5 °C s = T ′(t 0 ), the rate at which the 14 − 0 temperature was changing at t = t 0 as measured by the rising mercury on the thermometer. Because its average speed was approximately 7.667 knots, and by the Mean Value Theorem, it must have been going that speed at least once during the trip. The conclusion of the Mean Value Theorem yields 1 1 − b a = − 1 ⇒ c 2 a − b = a − b ⇒ c = ab . b−a c2 ab f ( x ) must be zero at least once between a and b by the Intermediate Value Theorem. Now suppose that f ( x ) is zero twice between a and b. Then, by the Mean Value Theorem, f ′( x ) would have to be zero at least once between the two zeros of f ( x ), but this can’t be true since we are given that f ′( x ) ≠ 0 on this interval. Therefore, f ( x ) is zero once and only once between a and b. 1.09999 ≤ f (0.1) ≤ 1.1

(

61.

x

2

−3

iv)

1.5 1.25 y = x1 + ln x 1 Abs min at (1, 1) 0.75 0.5 0.25 1

−2

iii)

39. (a) Absolute maximum value is ( 1 4 ) + ln 4 at  x = 4; absolute minimum value is 1 at x = 1; local maximum at ( 1 2, 2 − ln 2 ). (b) y 1 Abs max at 4, + ln 4 a

3

9. Does not; f is not differentiable at the interior domain point x = 0. 11. Does 13. Does not; f is not differentiable at x = −1. 17. (a)

y

Abs max Q2p3, 2Î3R

y = csc x (p2, 1) p3 ≤ x ≤ 2p3 Abs min

0

35. Absolute maximum: 2; absolute minimum: −1

AN-17

)

SECTION 4.3, pp. 253–255

1. (a) (b) (c) 3. (a) (b) (c) 5. (a) (b) (c)

0, 1 Increasing on ( −∞, 0 ) and ( 1, ∞ ); decreasing on ( 0, 1 ) Local maximum at x = 0; local minimum at x = 1 −2, 1 Increasing on ( −2, 1) and ( 1, ∞ ); decreasing on ( −∞, −2 ) No local maximum; local minimum at x = −2 Critical point at x = 1 Decreasing on ( −∞, 1 ), increasing on ( 1, ∞ ) Local (and absolute) minimum at x = 1

AN-18

Chapter 4: Answers to Odd-Numbered Exercises

7. (a) 0, 1 (b) Increasing on ( −∞, −2 ) and ( 1, ∞ ); decreasing on ( −2, 0 ) and ( 0, 1 ) (c) Local minimum at x = 1 9. (a) −2, 2 (b) Increasing on ( −∞, −2 ) and ( 2, ∞ ); decreasing on ( −2, 0 ) and ( 0, 2 ) (c) Local maximum at x = −2; local minimum at x = 2 11. (a) −2, 0 (b) Increasing on ( −∞, −2 ) and ( 0, ∞ ); decreasing on ( −2, 0 ) (c) Local maximum at x = −2; local minimum at x = 0 π 2π 4 π , 13. (a) , 2 3 3 2π 4 π π π 2π ; decreasing on 0, , , , , (b) Increasing on 3 3 2 3 2 4π , 2π and 3 4π (c) Local maximum at x = 0 and x = ; local minimum at 3 2π and x = 2π x = 3 15. (a) Increasing on ( −2, 0 ) and ( 2, 4 ); decreasing on ( −4, −2 ) and ( 0, 2 ) (b) Absolute maximum at ( −4, 2 ); local maximum at ( 0, 1 ) and ( 4, −1 ); absolute minimum at ( 2, −3 ); local minimum at ( −2, 0 ) 17. (a) Increasing on ( −4, −1 ) , ( 1 2, 2 ) , and ( 2, 4 ); decreasing on ( −1, 1 2 ) (b) Absolute maximum at ( 4, 3 ); local maximum at ( −1, 2 ) and ( 2, 1); no absolute minimum; local minimum at ( −4, −1 ) and ( 1 2, −1 ) 19. (a) Increasing on ( −∞, −1.5 ); decreasing on ( −1.5, ∞ ) (b) Local maximum: 5.25 at t = −1.5 21. (a) Decreasing on ( −∞, 0 ); increasing on ( 0, 4 3 ); decreasing on ( 4 3, ∞ ) (b) Local minimum at x = 0 ( 0, 0 ); local maximum at x = 4 3( 4 3, 32 27 ) 23. (a) Decreasing on ( −∞, 0 ); increasing on ( 0, 1 2 ); decreasing on ( 1 2, ∞ ) (b) Local minimum at θ = 0 ( 0, 0 ); local maximum at θ = 1 2( 1 2, 1 4 ) 25. (a) Increasing on ( −∞, ∞ ); never decreasing (b) No local extrema 27. (a) Increasing on ( −2, 0 ) and ( 2, ∞ ); decreasing on ( −∞, −2 ) and ( 0, 2 ) (b) Local maximum: 16 at x = 0; local minimum: 0 at x = ±2 29. (a) Increasing on ( −∞, −1 ); decreasing on ( −1, 0 ); increasing on ( 0, 1 ); decreasing on ( 1, ∞ ) (b) Local maximum: 0.5 at x = ±1; local minimum: 0 at x = 0 31. (a) Increasing on ( 10, ∞ ); decreasing on ( 1, 10 ) (b) Local maximum: 1 at x = 1; local minimum: −8 at x = 10 33. (a) Decreasing on ( −2 2, −2 ); increasing on ( −2, 2 ); decreasing on ( 2, 2 2 ) (b) Local minima: g(−2) = −4, g ( 2 2 ) = 0; local maxima: g ( −2 2 ) = 0, g(2) = 4 35. (a) Increasing on ( −∞, 1 ); decreasing when 1 < x < 2; decreasing when 2 < x < 3; discontinuous at x = 2; increasing on ( 3, ∞ ) (b) Local minimum at x = 3( 3, 6 ); local maximum at x = 1( 1, 2 )

(

)

(

)

(

)(

)

37. (a) Increasing on ( −2, 0 ) and ( 0, ∞ ); decreasing on ( −∞, −2 ) (b) Local minimum: −6 3 2 at x = −2 39. (a) Increasing on ( −∞, −2 7 ) and ( 2 7 , ∞ ); decreasing on ( −2 7 , 0 ) and ( 0, 2 7 ) (b) Local maximum: 24 3 2 7

76

≈ 3.12 at x = − 2

7 ; local

76

minimum: −24 2 7 ≈ −3.12 at x = 2 7 41. (a) Increasing on ( ( 1 3 ) ln ( 1 2 ) , ∞ ) , decreasing on ( −∞, ( 1 3 ) ln ( 1 2 ) ) 3 (b) Local minimum is 2 3 at x = ( 1 3 ) ln ( 1 2 ); no local 2 maximum 43. (a) Increasing on ( e −1 , ∞ ); decreasing on ( 0, e −1 ) (b) A local minimum is −e −1 at x = e −1; no local maximum 45. (a) Increasing on ( 0, e −2 ) and ( 1, ∞ ); decreasing on ( e −2 , 1 ) (b) Local maximum is 4 e 2 at x = e −2 47. (a) Local maximum: 1 at x = 1; local minimum: 0 at x = 2 (b) Absolute maximum: 1 at x = 1; no absolute minimum 49. (a) Local maximum: 1 at x = 1; local minimum: 0 at x = 2 (b) No absolute maximum; absolute minimum: 0 at x = 2 51. (a) Local maxima: −9 at t = −3 and 16 at t = 2; local minimum: −16 at t = −2 (b) Absolute maximum: 16 at t = 2; no absolute minimum 53. (a) Local minimum: 0 at x = 0 (b) No absolute maximum; absolute minimum: 0 at x = 0 55. (a) Local maximum: 5 at x = 0; local minimum: 0 at x = −5 and x = 5 (b) Absolute maximum: 5 at x = 0; absolute minimum: 0 at x = −5 and x = 5 57. (a) Local maximum: 2 at x = 0; 3 local minimum: at x = 2 − 3 4 3−6 (b) No absolute maximum; absolute minimum at x = 2 − 3 59. (a) Local maximum: 1 at x = π 4; local maximum: 0 at x = π; local minimum: 0 at x = 0; local minimum: −1 at x = 3π 4 61. (a) Local maximum: 2 at x = π 6; local maximum: 3 at x = 2π ; local minimum: −2 at x = 7π 6; local minimum: 3 at x = 0 63. (a) Local minimum: ( π 3 ) − 3 at x = 2π 3; local maximum: 0 at x = 0; local maximum: π at x = 2π 65. (a) Local minimum: 0 at x = π 4 67. Local minimum at x = 1; no local maximum 69. (a) Increasing on ( −1, 2 ); decreasing on ( −2, −1 ) (b) The sign of f cannot be determined solely from the graph of f ′. 71. Local maximum: 3 at θ = 0; local minimum: −3 at θ = 2π 73. 3

y

y

y

y

y = f (x)

y = f(x)

1

0

1

1

(a)

x

0

1

y= f (x) 1

(b)

x

0

1

y = f (x) 1

(c)

x

0

1

(d)

x

AN-19

Chapter 4: Answers to Odd-Numbered Exercises

75. (a)

(b)

y

13.

y

15.

y

y

(2, 5) Loc max 3

y = g(x)

2

y = g(x)

2

0

2 1

1

x

2

0

−3

x

2

79. a = −2, b = 4 81. (a) Absolute minimum occurs at x = π 3 with f ( π 3 ) = − ln 2, and the absolute maximum occurs at x = 0 with f (0) = 0. (b) Absolute minimum occurs at x = 1 2 and x = 2 with f ( 1 2 ) = f (2) = cos ( ln 2 ) , and the absolute maximum occurs at x = 1 with f (1) = 1. 83. Minimum of 2 − 2 ln 2 ≈ 0.613706 at x = ln 2; maximum of 1 at x = 0 85. Absolute maximum value of 1 2e assumed at x = 1 e df −1 1 = x −2 3 89. Increasing; 9 dx df −1 1 91. Decreasing; = − x −2 3 3 dx SECTION 4.4, pp. 264–268

1. Local maximum: 3 2 at x = −1; local minimum: −3 at x = 2; point of inflection at ( 1 2, −3 4 ); rising on ( −∞, −1 ) and ( 2, ∞ ); falling on ( −1, 2 ); concave up on ( 1 2, ∞ ); concave down on ( −∞, 1 2 ) 3. Local maximum: 3 4 at x = 0; local minimum: 0 at x = ±1; ⎛ ⎛ 3 3 4 ⎞⎟ 3 3 4 ⎞⎟ and ⎜⎜ 3, points of inflection at ⎜⎜ − 3, ⎟; ⎟ ⎟ ⎝ ⎝ 4 ⎠ 4 ⎟⎠ rising on ( −1, 0 ) and ( 1, ∞ ); falling on ( −∞, −1 ) and ( 0, 1 ); concave up on ( −∞, − 3 ) and ( 3, ∞ ); concave down on

Infl (2, 1)

2

Infl (1, 1) −1

1

x

2

−2

−1

1

0

2

x

4

−1 (0, −3) Loc min

y = −2x3 + 6x 2 − 3

17.

−2

19.

y

y = (x − 2) 3 + 1

y

y = x 4 − 2x2

(3, 27) Abs max

27 21 (2, 16) 1

Loc max (0, 0) −2

−1

1

21.

x

2

Abs min (−1, −1)

1

23.

Loc max (0, 0)

3

x

4

y

y = x 5 − 5x 4

0

1

2

5

x

5

4

3

y=

4

−100 −200

2

Q1Î3 , −59R Infl

y

−2

Infl

9 Infl (0, 0) 3

Abs min (1, −1)

Q−1Î3 , −59R Infl

y = 4x 3 − x 4

15

2x 2 + x − 1 x2 − 1

3

y=2

2

(3, −162) Infl

1 (4, −256) Loc min

−300

−1 0 −1

( − 3, −1), ( −1, 1), and ( 1, 3 )

1

−2

− 2π 3 π 3 + at x = − 2π 3, + at 3 2 3 2 π 2π 3 3 at x = − π 3, − x = π 3; local minima: − − 3 2 3 2 at x = 2 π 3; points of inflection at ( − π 2, − π 2 ) , ( 0, 0 ), and ( π 2, π 2 ); rising on ( − π 3, π 3 ); falling on ( −2 π 3, −π 3 ) and ( π 3, 2 π 3 ); concave up on ( − π 2, 0 ) and ( π 2, 2 π 3 ); concave down on ( −2 π 3, −π 2 ) and ( 0, π 2 ) 7. Local maxima: 1 at x = − π 2 and x = π 2, 0 at x = −2π and x = 2π; local minima: −1 at x = −3 π 2 and x = 3 π 2, 0 at x = 0; points of inflection at ( −π , 0 ) and ( π , 0 ); rising on ( −3 π 2, − π 2 ) , ( 0, π 2 ) , and ( 3 π 2, 2π ); falling on ( −2π, −3 π 2 ) , ( − π 2, 0 ), and ( π 2, 3 π 2 ); concave up on ( −2π , −π ) and ( π , 2π ); concave down on ( −π , 0 ) and ( 0, π ) 11. 9.

3

2

x

3

x=1

5. Local maxima:

y

y

Loc max (−1, 5)

4 3 y = x 2 − 4x + 3

25.

27.

y

y= 4 y= x +1 x2 (1, 2) Abs min

2 Abs min (−1, 2) 1 −1

−1

1 −1

x= −1

29.

y = x 3 − 3x + 3

y x = −1

2 y = −x − 2 x2 − 1

x (0, −1) Loc max x= 1

31.

y

y=

1

2

3

4

−Î2

Infl

x

2

(2, −1) Abs min

−1

y = −1

1 −1

x2 x+1

y=x−1

4

1

−2

1

x

1

1 x2 − 1

5

2

−4 −3 −2 −1 0 −1

y

(1, 1) Loc min 1

x

x = −1

1 −1 −2

Î2

x

(0, −2) Loc max x=1

−4 −3 −2

1

2

3

(0, 0) Loc min

Loc max (−2, −4) −4

x

AN-20

Chapter 4: Answers to Odd-Numbered Exercises

33.

35.

y

x = −2

x−1

37.

−6

x=1

−4

−2

x2 + x − 2

0 −4

1

1

1 0

x

1

−1

y = 8(x2 + 4)

0

59.

6

y= x −3 x−2

−8 −6 −4

x=1

8x y= 2 x +4

(3, 6) Loc min

(2, 2) Abs max Q2Î3, Î3R Infl

2 1

4

x

(1, 2) Loc max

2 x

1

y

8

(0, 2) Abs max

x

(4, 0) Abs min

y = xÎ8 − x2

y

y 2

(−4, 0) Abs min

−4

(−2, −4) Abs min

57.

39.

x

2 Q2 2 , 0R Î Loc min

−3

2

−1

(0, 0) Infl

−2 −1

−16

y= x x2 − 1

Infl (0, 0)

1

Q−2Î2 , 0R

x

4 y=x−4

2

Loc max

−12

(−5, −12) Loc max

y

x = −1

(0, 4) Abs max y = Î16 − x2

−8

(0, −1) Loc max

y

Abs max (2, 4)

3

(x − 1)3

x

2

−1 y=x

y=

9 2 1

55.

y 4

3

x2 − x + 1

y=

53.

y

Loc min (2, 3)

2 4

6

8

1 2 (0, 0) Infl

−1 −2 Q−2Î3, −Î3R (−2, −2) Infl Abs min

x

−2 −4 −6 −8

41.

43.

y

2p

8 6

(p, p)

4

Infl

p

0

Infl

Qp2, Î3p2R

x

2p

p2

0 (0, −2) Abs min

p

3p2

(1, 0) Abs min

65.

y

Abs max (p4, 12)

1

p4

(0, 0) Loc min

2 Infl (p2, 0)

p2

Vert tan at x = 0

Loc max (p, 0)

3p4

x

p

−3

−1

2

x 15

67. y=

3

1 2 (0, 0) Infl

3

x 9 − x2

−2

−2

−1 −2

2 (0, 0) Infl

x

4

2

−3 −2 −1 −1

Î3 1

2

3

−2 −3

−3

49.

51.

y

y



y = 2 x − 3x 2 3 Cusp, Loc max (0, 0) −1 −1

−5

1 (1, −1) Loc min

y=x 4

5

x

3 2

Infl

23

5 Q–2

y

− xR

Loc Loc 3 max max (−4p, 0) (−2p, 0) 2 1 −4p −2p

(1, 32) Loc max

3

Q−12, 3 Î4R

−2

−1

71. y′′ = 1 − 2 x

69.

4

(0, 0) 1 Cusp Loc min

2

3

Loc max (0, 0) Loc max (2p, 0) 2p

Infl Loc max (4p, 0) 4p x

x y = ln (cos x)

4

Loc max (0, ln 3)

1

−Î 3

1 −4

3

y

y = ln (3 − x 2)

2 x

(0, 0) 1 2 Cusp Abs min

x

y

1

−2 −1

(3p4, −12) Abs min

−1

y=

−4 −3 −2 −1

(−1, 0) Abs min

y = sin x cos x

y = Î0x0

1

Loc max (0, 1)

y = Î3x − 2 cos x

47.

y

y = ∣ x 2 − 1∣

2

−2

45.

y 2

3

x

2p

63.

y

Q5p3, 5Î3p3 − 1R Infl Loc min Q3p2, 3Î3p2R

2

Abs min

61. Q2p, 2Î3p − 2R Abs max

Q4p3, 4Î3p3 + 1R

10

y = x + sin x

p

y Loc max

Abs max (2p, 2p)

Loc min x = −1

Loc max x=2

x = 12

x

x

Chapter 4: Answers to Odd-Numbered Exercises

73. y′′ = 3( x − 3 )( x − 1 )

75. y′′ = 3( x − 2 )( x + 2 )

y

95.

Loc max Infl Infl Loc min

Abs min

x=1

x=0

P

x=0 Infl x = −2

x=3

y

Infl x = 2 Abs min

Infl

Infl y′

x = 2Î 3

x = −2Î3

AN-21

Loc min

77. y′′ = 4 ( 4 − x )( 5 x 2 − 16 x + 8 )

x

y″

Loc max x = 85 Infl

Infl x = 8 − 2 Î6 5

79. y′′ =

97.

x = 8 + 2Î6 5

Loc min x=0

y′

y′′

P

−

+

Q

+

0

R

+

−

S

0

−

1

T

−

−

0

Infl x=4

2 sec 2

θ 1 81. y′′ = − csc 2 , 2 2 0 < θ < 2π

x tan x

u=p

Infl

y

99.

Point

(6, 7) 7 (4, 4)

4 (2, 1) 2

4

6

x

Abs max x=0

101.

103.

y

(0, 3)

f 0: (1, 2)

2

π π 83. y′′ = 2 tan θ sec 2 θ, − < θ < 2 2

−3

u = − p Loc max 4

105. 1 1 1 1 f 9:

u=p 4 Loc min

f 0:

85. y′′ = − sin t , 0 ≤ t ≤ 2π t=p 2

Loc max t = 2p

Abs max t=0 Loc min

t=p Infl

87. y′′ = −

2( x + 1 )−5 3 3

t = 3p 2 Abs min

89. y′′ =

1 −2 3 2 + x −5 3 x 3 3

x = −2 Infl

x = −1 Infl Vert tan

Infl vert tan x=0

x=1 Abs min

⎪⎧ −2, x < 0 91. y′′ = ⎨ ⎪⎪⎩ 2, x > 0

93.

y

Loc max

y″

y′

(2, 0)

22

0

−1

22

22

0

−3

3

0

−1

22

0

1111

2

x

0

1

2 0 0

11

0

22

x

0

There are local maxima at x = −1 and x = 4. There is a local minimum at x = 2. There are points of inflection at x = 0 and x = 3.

x

4

222

x

3

107. Concave up on ( −2, 0 ) and ( 1, 2 ); concave down on ( 0, 1 ) 109. (a) Toward origin: 0 ≤ t < 2 and 6 ≤ t ≤ 10; away from origin: 2 ≤ t ≤ 6 and 10 ≤ t ≤ 15 (b) t = 2, t = 6, t = 10 (c) t = 5, t = 7, t = 13 (d) Positive: 5 ≤ t ≤ 7, 13 ≤ t ≤ 15; negative: 0 ≤ t ≤ 5, 7 ≤ t ≤ 13 111. ≈ 60 thousand units 113. Local minimum at x = 2; inflection points at x = 1 and x = 53 117. b = −3 121. x = −1, x = 2 123. a = 1, b = 3, c = 9 125. The zeros of y′ = 0 and y ″ = 0 are extrema and points of inflection, respectively. Inflection at x = 3; local maximum at x = 0; local minimum at x = 4 y

y

Infl P

x=0

1

−1

1

(−2, −1) −2

u=0 Infl

Infl

−2

0

−3

There are points of inflection at x = −3, x = −1, and x = 2.

1

(−1, 0)

22

200

y″ = 20x 2 (x − 3)

Loc min 0 x

−200 −400

3

4

y′ = 5x 3 (x − 4) y = x5 − 5x4 − 240

5

x

AN-22

Chapter 4: Answers to Odd-Numbered Exercises

127. The zeros of y′ = 0 and y ″ = 0 are extrema and points of inflection, respectively. Inflection at x = − 3 2; local maximum at x = −2; local minimum at x = 0. y

y′ = 4x(x 3 + 8)

17. (a) V ( x ) = 2 x ( 60 − 2 x )( 45 − 2 x ) (b) Domain: ( 0, 22.5 ) V 20000

Maximum x = 8.4861217, V = 20,468.041645

15000

100 10000 50

−50 y″ = 16(x 3 + 2)

5000

x 2 3 y = 45 x5 + 16x2 − 25

−3

5

−100

3. 5 7

5. 1 2

11. 5 7

13. 0

15. −16

21. 1 4

23. 2

25. 3

33. ln 2

35. 1

37. 1 2

43. −1 2

45. −1

55. 1

57. 1 e

67. 1

69. 3

x

20

17. −2

19.

27. −1

71. 1

49. 0

51. 2 63. e 3

61. 1 73. 0

79. (d) is correct.

31.

41. −∞

39. ln 2

47. 1

32

29. ln 3

1 ln 2

53. 1 e 65. 0

Abs max 10000 8000 6000

75. ∞ 81. c =

35 − 5 13 , which confirms the result in part (c). 2 (e) x = 12.5 cm or x = 5 cm 19. Radius = 10 2 3 cm, height = 20 3 cm, volume = 4000 π ( 3 3 ) cm 3 21. (a) h = 24, w = 18 (24, 10368) (b) V x =

9. −23 7

7. 1 4

59. e 1 2

77. (b) is correct.

27 10

−1 3 85. −1 89. (a) y = 1 (b) y = 0, y = 2 2 91. (a) We should assign the value 1 to f ( x ) = ( sin x ) x to make it continuous at x = 0.

83. (b)

y

V = 54h2 − 3 h3 2

4000 2000 0

5

10

15 20

30

h

35

( )

27. Radius =

0.8

25

23. If r is the radius of the hemisphere, h the height of the cylinder, 3V 1 3 3V 1 3 and V the volume, then r = . and h = π 8π 25. (b) x = 16.2

1

y = (sin x)

2 m, height = 1 m, volume =

0.2 0.5

1

1.5

2

2.5

3

x

(c) The maximum value of f ( x ) is close to 1 near the point x ≈ 1.55 (see the graph in part (a)). SECTION 4.6, pp. 282–289

1. 16 cm, 4 cm by 4 cm 3. (a) ( x , 1 − x ) (b) A( x ) = 2 x (1 − x ) 1 1 square units, 1 by 2 2

5. 14 × 35 × 5 cm, 2450 cm 3 7. 80,000 m 2 ; 400 m by 200 m 9. (a) The optimum dimensions of the tank are 2 m on the base edges and 1 m deep. (b) Minimizing the surface area of the tank minimizes its weight for a given wall thickness. The thickness of the steel walls would likely be determined by other considerations such as structural requirements. π 11. 45 × 22.5 cm 13. 15. h:r = 8:π 2

2π 3 m 3

9b b 3π m , triangle; m , circle 9 + 3π 9 + 3π 3 37. (a) 16 (b) −1 35. × 2 33. 5 + 2 3 m 2 29. 1

0.4

( )

(c) L ≈ 28 cm

x

0.6

(b)

15

(c) Maximum volume ≈ 20, 468 cm 3 when x ≈ 8.49 cm (d) V ′( x ) = 24 x 2 − 840 x + 5400, so the critical point is at

SECTION 4.5, pp. 275–276

1. −1 4

10

31.

4 3 1 ( 4 + ln 2 ) 43. Area 8 when a = 2 41. (a) 4 + ln 2 (b) 2 45. (a) v(0) = 29.4 m s (b) 78.4 m at t = 3 s (c) Velocity when s = 0 is v(7) = −39.2 m s. 47. ≈ 9.58m 39. r =

2 2 3

h =

( )

49. (a) 4 m 51. (a) 10 3 × 10 6 cm 53. (a) 10 π ≈ 31.42 cm s; when t = 0.5 s, 1.5 s, 2.5 s, 3.5 s; s = 0, acceleration is 0. (b) 10 cm from rest position; speed is 0. 12

55. (a) s = ((12 − 12t ) 2 + 64 t 2 ) (b) −12 knots; 8 knots (c) No (e) 4 13; this limit is the square root of the sum of the squares of the individual speeds. c ka 2 a 57. x = , v = 59. + 50 2 4 2

Chapter 4: Answers to Odd-Numbered Exercises

2 km h C 67. M = 2

2 km h

(b)

61. (a)

65. 2 × 2 ×

3 m, $720 2

73. (a) y = −1

5 . 2 (b) The minimum distance is from the point ( 3 2, 0 ) to the point (1, 1) on the graph of y = x , and this occurs at the value x = 1, where D( x ), the distance squared, has its minimum value.

75. (a) The minimum distance is

y, D(x)

9. (a) x 2 3 (b) x 1 3 (c) x −1 3 11. (a) ln x (b) 7 ln x (c) x − 5 ln x 1 13. (a) cos(πx ) (b) −3 cos x (c) − cos(π x ) + cos(3 x ) π x 2 3x (c) − tan 15. (a) tan x (b) 2 tan 3 3 2 πx 1 17. (a) − csc x (b) csc (5 x ) (c) 2 csc 5 2 1 3x x 2 − x 19. (a) e (b) −e (c) 2e 3 1 x −1 −x 1 5 x 3 2 21. (a) (b) (c) ln 2 ln 3 ln ( 5 3 ) 3

( )

2.5

1 1 arctan x (c) arctan 2 x 2 2 2 5x 2 x2 x4 t + x + C  27. t 3 + + C  29. − + 7x + C 2 2 2 4 x3 x 1 3 − − − +C 33. x 2 3 + C x 3 3 2 2 32 3 43 8 x + x +C 37. 4 y 2 − y 3 4 + C 3 4 3 2 2 2 x + +C 41. 2 t − +C x t θ −2 sin t + C 45. −21 cos + C 3 1 3 cot x + C 49. − csc θ + C 2 1 3x 4x − x e − 5e + C 53. −e −x + +C ln 4 3 1 4 sec x − 2 tan x + C 57. − cos 2 x + cot x + C 2 sin 4 t t + +C 61. ln x − 5 arctan x + C 2 8

23. (a) 2 arcsin x 2

25.

y = Îx

31.

1 Dmin= Î5 2

0.5

0.5

1

1.5

35.

2

2.5

x

39.

SECTION 4.7, pp. 292–294

5 13 51 5763 2387 1. x 2 = − , 3. x 2 = − , 5. x 2 = 3 21 31 4945 2000 9. x 2 is approximately 0.68394 7. x 2 is approximately 2.20794 11. x 1, and all later approximations will equal x 0 . y 13.

43. 47. 51. 55.

−h y=

x

h Îx , x ≥ 0 Î−x, x < 0

(d) For x 0 = − 21 7 or x 0 = 21 7, Newton’s method does not converge. The values of x i alternate between − 21 7 and 21 7 as i increases. 33. Answers will vary with machine speed. SECTION 4.8, pp. 300–304

3. (a) x −3 5. (a) − 7. (a)

1 x x3

x3 − x2 + x 3 1 1 (b) − x −3 (c) − x −3 + x 2 + 3 x 3 3 5 5 (c) 2 x + (b) − x x

(b)

x3 3

(b)

(c)

x

59.

(b)

3 x ( 3 +1 ) 65. tan θ + C 67. − cot x − x + C +C 3+1 69. − cos θ + θ + C 2x x2 d x2 83. (a) Wrong: sin x + C = sin x + cos x = 2 2 dx 2 2 x x sin x + cos x 2 d (b) Wrong: ( −x cos x + C ) = − cos x + x sin x dx d (c) Right: ( −x cos x + sin x + C ) = − cos x + x sin x + dx cos x = x sin x 63.

15. The points of intersection of y = x 3 and y = 3 x + 1 or of y = x 3 − 3 x and y = 1 have the same x-values as the roots of part (i) or the solutions of part (iv). 17. x ≈ 1.165561185, y ≈ 2.33112237 19. (a) Two (b) 0.35003501505249 and −1.0261731615301 21. ±1.3065629648764, ± 0.5411961001462 23. x ≈ 0.45 25. 0.8192 27. 0, 0.53485 29. The root is 1.17951. 31. (a) For x 0 = −2 or x 0 = −0.8, x i → −1 as i gets large. (b) For x 0 = −0.5 or x 0 = 0.25, x i → 0 as i gets large. (c) For x 0 = 0.8 or x 0 = 2, x i → 1 as i gets large.

1. (a) x 2

( ) ( )

( )

9 D(x) = x2 − 2x + – 4

1.5

AN-23

2 x3 (c) +2 x 3

(

(

)

)

d ( 2 x + 1)3 3( 2 x + 1 ) 2 ( 2 ) = +C = dx 3 3 2 2( 2 x + 1)

85. (a) Wrong:

d ( ( 2 x + 1 ) 3 + C ) = 3( 2 x + 1 ) 2 ( 2 ) = dx 6( 2 x + 1) 2

(b) Wrong:

(c) Right: 87. Right 93. y = −

d ( ( 2 x + 1)3 + C ) = 6( 2 x + 1) 2 dx 89. (b) 91. y = x 2 − 7 x + 10

1 x2 1 + − x 2 2

95. y = 9 x 1 3 + 4

AN-24

Chapter 4: Answers to Odd-Numbered Exercises

97. s = t + sin t + 4 1 1 101. v = sec t + 2 2

99. r = cos(πθ) − 1

41. (a) t = 0, 6, 12 (b) t = 3, 9 (d) 0 < t < 6, 12 < t < 14

103. v = 3 arcsec t − π

105. y = x 2 − x 3 + 4 x + 1 109. y = x 3 − 4 x 2 + 5

1 + 2t − 2 t 111. y = − sin t + cos t + t 3 − 1 107. r =

y

43.

(c) 6 < t < 12 y

45.

15 3

y = −x 3 + 6x2 − 9x + 3 3

113. y = 2 x 3 2 − 50 115. y

8 3

3

y = x2 − x 6

1

y = f (x)

−2 −1 0

1

2

4

1 1

x

6

2

3

x

4

−1

1 1

4

6

8

−2

x

500

1

400 2

4

6

8

x

−2

200

−3 −4

100

1 2 y = − sin x − cos x − 2 (a) (i) 33.2 units, (ii) 33.2 units, (iii) 33.2 units (b) True t = 88 k , k = 16 (a) v = 10 t 3 2 − 6t 1 2 (b) s = 4 t 5 2 − 4 t 3 2 (a) − x + C (b) x + C (c) x + C (d) −x + C (e) x − x + C (f ) −x − x + C

1. Minimum value is 1 at x = 2.

5. 7. 9. 11. 13. 15. 17. 21. 23. 25. 33. 39.

(6, 432) y = x 3(8 − x)

−3 −4

9

27

4

6

x

8

y

11 9 Q1 − Î2, Q6 + 4Î2R e1−Î2R

2

x

−1

2

1 −1

(1, 4e) y = (x − 3)2e x

5 3 1

1

−4−3−2−1 0 1 2 3 4

3

x

Q1 + Î2, Q6 − 4Î2R e1+Î2R

y = x Î3 − x

−2

)

4 41 ,− Local maximum at (−2, 17 ); local minimum at 3 27 Minimum value is 0 at x = −1 and x = 1. There is a local minimum at ( 0, 1 ). 1 1 Maximum value is at x = 1; minimum value is − at 2 2 x = −1. The minimum value is 2 at x = 0. 1 1 The minimum value is − at x = . e e π The maximum value is at x = 0; an absolute minimum value 2 is 0 at x = 1 and x = −1. No 19. No minimum; absolute maximum: f (1) = 16; critical points: x = 1 and 11 13 Absolute minimum: g(0) = 1; no absolute maximum; critical point: x = 0 Absolute minimum: 2 − 2 ln 2 at x = 2; absolute maximum: 1 at x = 1 Yes, except at x = 0 27. No 31. (b) one (b) 0.8555 99677 2 35. NA 1 Global minimum value of at x = 2 2

ln 3

−5 −3

x

(8, −4)

53.

y

55.

(

2

y

51.

18

(4, 256)

−2 −1 0 −100

PRACTICE EXERCISES, pp. 305–308

3.

y

300

119. y = x − x 4 3 +

125. 127. 131.

49.

y = x − 3x 23

2

−1

121. 123.

y

47.

y

117.

−1 −2 −3 −4

y

57.

5 4 3 2 1

p 2

y = ln( x 2 − 4 x + 3)

−3 −2 −1 2

5

7

x

2

y = arcsin(1x)

1 1

−1 −2

−

2

3

x

p 2

59. (a) Local maximum at x = 4, local minimum at x = −4, inflection point at x = 0 x=4 (b) Loc max x=0 Infl Loc min x = −4

61. (a) Local maximum at x = 0, local minima at x = −1 and x = 2, inflection points at x = (1 ± 7 ) 3 Loc max (b) Infl

Loc min x = −1

x=0 x = 1 − Î7 3

x = 1 + Î7 3 Loc min Infl x=2

Chapter 4: Answers to Odd-Numbered Exercises

63. (a) Local maximum at x = − 2, local minimum at x = inflection points at x = ±1 and 0 Loc max (b)

2,

x=0 x = Î2 Infl x=1 Loc min y

y

71.

y = x+1 = 1 + 4 x−3 x−3 5

y=

x2 + 1

5

x = x + 1x

4 (1, 2)

3 2

2 1

y=x

1

−1

2 3 4

x

6

−4 −3 −2 −1 –1

1

2

3

x

4

−2

−3

−3

(−1, −2)

y

(c) Absolute maximum of 1 at x = 0, inflection point ( 1, 2 e ) , concave up on ( 1, ∞ ) , concave down on ( −∞, 1 )

−4

x2 − 4 y= 2 x −3

4 3

2 y= x 2

3 2

y=1 x

1 0

y

75.

4

1 −1

2

3

x = −Î3

−4 −3

x

x = Î3 y=1

2

−1 0 −1

ADDITIONAL AND ADVANCED EXERCISES, pp. 308–311

1

2 3

4

x

−2

3 2 y= x +2= x + 1 2 2x x

−3

77. 5 79. 0 81. 1 83. 3 7 85. 0 87. 1 89. ln 10 91. ln 2 93. 5 95. −∞ 97. 1 99. e bk 101. −∞ 103. (a) 0, 36 (b) 18, 18 105. 54 square units 2

109. x = 5 − 5 hundred ≈ 276 tires, y = 2( 5 − 5 ) hundred ≈ 553 tires 111. Dimensions: base is 15 cm by 30 cm, height = 5 cm;  maximum volume = 2250 cm 3 113. x 5 = 2.1958 23345 117. 2t 3 2 −

4 +C t

121. ( θ 2 + 1)3 2 + C 125. 129. 133. 137.

s 10 tan +C 10 1 x x − sin + C 2 2 1 t e + e −t + C 2 3 arcsec x + C 2

1. 3. 5. 9.

The function is constant on the interval. The extreme points will not be at the end of an open interval. NA No 11. a = 1, b = 0, c = 1 13. Yes

15. Drill the hole at y = h 2. 17. r =

−3

107. height = 2,  radius =

115.

5 x4 + x 2 − 7x + C 4 2

1 +C r +5 1 123. (1 + x 4 )3 4 + C 3 1 127. − csc 2 θ + C 2 x2 131. 3 ln x − +C 2 θ 2− π +C 135. 2−π 1 139. y = x − − 1 x 119. −

2e ≈ 0.43 units 2

147. Absolute maximum = 0 at x = e 2, absolute minimum = −0.5 at x = 0.5 149. x = ±1 are the critical points; y = 1 is a horizontal asymptote in both directions; absolute minimum value of the function is e − 2 2 at x = −1, and absolute maximum value is e 2 2 at x = 1. 151. (a) Absolute maximum of 2 e at x = e 2, inflection point ( e 8 3, ( 8 3 ) e −4 3 ) , concave up on ( e 8 3, ∞ ) , concave down on ( 0, e 8 3 ) (b) Absolute maximum of 1 at x = 0, inflection points (±1 2 , 1 e ), concave up on ( −∞, −1 2 ) ∪ ( 1 2 , ∞ ), concave down on ( −1 2 , 1 2 )

−5

73.

e units high,

2 units long by 1

A =1

Infl

69.

141. r = 4 t 5 2 + 4 t 3 2 − 8t 143. Yes, arcsin x and − arccos x differ by the constant π 2. 145. 1

Infl x = −1

x = −Î2

AN-25

RH  for  H > 2 R ,   r = R  if  H ≤ 2 R 2( H − R )

12 and 5 5 10 5 1 1 21. (a) (b) (c) (d) 0 (e) − (f) 1 3 3 2 2 1 (h) 3 (g) 2 c−b c+b b 2 − 2bc + c 2 + 4 ae 23. (a) (b) (c) 2e 2 4e c+b+t (d) 2 1 1 25. m 0 = 1 − , m1 = q q 19.

27. s = ce kt

s s = soe kt

1000 800 600

s = 16t 2

400 200 0

29. (a) k = 15

1

2

3

4

5

6

7

t

(b) 7.5 m

31. Yes, y = x + C

33. v 0 =

2 2 34 b 3

39. 3

AN-26

Chapter 5: Answers to Odd-Numbered Exercises

39. (a)

Chapter 5

(b)

y

y

SECTION 5.1, pp. 320–322 f (x) = sin x, −p ≤ x ≤ p Left-hand 1

1. 3. 5. 9. 13.

(a) 0.125 (b) 0.21875 (c) 0.625 (d) 0.46875 (a) 1.066667 (b) 1.283333 (c) 2.666667 (d) 2.083333 0.3125, 0.328125 7. 1.5, 1.574603 (a) 87 cm (b) 87 cm 11. (a) 1180 m (b) 1300 m (a) 22.862 m s (b) 13.867 m s (c) 44.893 m 31 17. 1 15. 16 19. (a) Upper = 758 L,  lower = 543 L (b) Upper = 2363 L,  lower = 1693 L (c) ≈31.4 h, ≈32.4 h π ≈ 3.061 21. (a) 2 (b) 2 2 ≈ 2.828 (c) 8 sin 8 (d) Each area is less than the area inside the circle, π. As n increases, the polygon area approaches π.

c1 = −p

−p

49.

(2, 3)

(2, 3)

3

f (x) = x 2 − 1, 0≤x≤2 Left-hand

2

2

x

y

2

−1

c2

c3

x

2

1.

∫0

7.

∫−π 4 sec x dx

9. 11. 13. 17. 21. 27. 35. 43. 51.

f (x) = x 2 − 1, 0≤x≤2 Right-hand

x 2 dx

3.

5

∫−7 ( x 2 − 3x ) dx

x

3

∫2

1 dx 1− x

(a) 0 (b) −8 (c) −12 (d) 10 (e) −2 (f) 16 (a) 5 (b) 5 3 (c) −5 (d) −5 (a) 4 (b) −4 15. Area = 21 square units Area = 9π 2  square units 19. Area = 2.5 square units Area = 3 square units 23. b 2 4 25. b 2 − a 2 (a) 2π (b) π 29. 1 2 31. 3π 2 2 33. 7 3 1 24 37. 3a 2 2 39. b 3 41. −14 −2 45. −7 4 47. 7 49. 0 Using n subintervals of length Δx = b n and right-endpoint values: Area =

b

∫0 3x 2 dx

= b3

53. Using n subintervals of length Δx = b n and right-endpoint values: Area = c1 c2 = 1 c3 c4 = 2

x

55. 61. 63. 69. 73.

77.

x

5.

0

b

∫0

2 x dx = b 2

av( f ) = 0 57. av( f ) = −2 59. av( f ) = 1 (a) av( g) = −1 2 (b) av( g) = 1 (c) av( g) = 1 4 c( b − a ) 65. b 3 3 − a 3 3 67. 9 b4 4 − a4 4 71. a = 0 and b = 1 maximize the integral. Upper bound = 1,  lower bound = 1 2 0

c4

c4 = p

1 1 1 1 , + + 2 n 2n 2 2

1

f (x) = x 2 − 1, 0≤x≤2 Midpoint

c1

c2 = 0 c3

41. 1.2 2 2 1 1 , − 43. − 3 2n 6n 2 3 27n + 9 , 12 45. 12 + 2n 2 5 6n + 1 5 , 47. + 6 6n 2 6

75. For example, ∫ sin ( x 2 ) dx ≤

(2, 3)

1

0

0 −1

−1

3

c3 p2 c4 p

1

1

(c)

c1

SECTION 5.3, pp. 338–341

(a) −15 (b) 1 (c) 1 (d) −11 (e) 16 (a) 55 (b) 385 (c) 3025 21. −56 23. −73 240 27. 3376 29. (a) 21 (b) 3500 (c) 2620 (a) 4n (b) cn (c) ( n 2 − n ) 2 33. 2600 35. −2 3 y y (a) (b)

c1 = 0 c2 c3 = 1 c4

−p

−1

3. cos(1) π + cos( 2 ) π + cos( 3 ) π + cos( 4 ) π = 0 π π 3−2 5. sin π − sin + sin = 2 3 2 7. All of them 9. b 6 4 5 1 1 13. ∑ k 15. ∑ (−1) k +1 11. ∑ k 2 k k =1 k =1 k =1

2

x

−1

f (x) = sin x, −p ≤ x ≤ p Midpoint 1 c1 −p2 c2

6(1) 6( 2 ) + = 7 1+1 2+1

3

p

y

(c)

SECTION 5.2, pp. 328–329

17. 19. 25. 31. 37.

c3 = 0 c4 –1

( )

1.

c2

f (x) = sin x, −p ≤ x ≤ p Right-hand 1

b

∫a

f ( x ) dx ≥

b

∫a

0 dx = 0

1

∫0 dx

=1

79. Upper bound = 1 2

89. Since f is continuous, the Extreme Value Theorem implies that the continuous function f has a maximum value M and a minimum value m (Chapter 4, Theorem 1). The values of f and g are equal except at a single point c where g (c) differs from f (c). The magnitude of the difference is some number between the two numbers g (c) − m and g (c) − M . Call the larger of these two numbers D. In the formula for a Riemann sum given in Equation (1), for a given n, the difference between the Riemann sum for f and the

Chapter 5: Answers to Odd-Numbered Exercises

Riemann sum for g is at most D ( b − a ) n. Taking the limit as n → ∞, we see that the difference between the two Riemann sums approaches 0, so that the two limits are equal. A similar bound holds for a general partition, where D P → 0 . SECTION 5.4, pp. 351–353

1. −10 3 11. 0

37. 43. 51.

13. −π 4

8 +1 25. 1 2 27. 16 1( π 1 π 4 −2 ) 35. ( e − 1) 73 31. 2π 3 33. 2 π 1 5 26 − 5 39. ( cos x ) 41. 4 t 2 x 1 3 x 2e −x 3 45. 1 + x 2 47. − x −1 2 sin x 49. 0 2 2 1 53. 2 xe (1 2 )x 55. 1 57. 28 3 59. 1 2 23.

2−

4

(

2π 2 1 65. d, since y′ = and y ( π ) = x

61. π

)

π

1 dt − 3 = −3 t

67. b, since y′ = sec x and y ( 0 ) =

∫0 sec t dt + 4

73. 77. 79. 83.

85.

5.

13. 15. 17. 21. 25. 29.

63.

∫π

69. y =

1.

3. 124 3

21. −3 4 29.

SECTION 5.5, pp. 360–361

9.

5. 753 16 7. 1 9. 2 3 π 2− 2 17. 19. −8 3 15. 1 − 4 4

x

∫2 sec t dt + 3

0

33. = 4

71. (1 + ln t ) 2

2 bh 75. $9.00 3 (a) T ( 0 ) = 20 °C, T (16 ) = 24 °C , T ( 25 ) = 30 °C (b) av T = 23.33 °C 2x − 2 81. −3 x + 5 (a) True. Since f is continuous, g is differentiable by Part 1 of the Fundamental Theorem of Calculus. (b) True: g is continuous because it is differentiable. (c) True, since g′(1) = f (1) = 0. (d) False, since g ″(1) = f ′(1) > 0. (e) True, since g′(1) = 0 and g ″(1) = f ′(1) > 0. (f) False: g ″( x ) = f ′( x ) > 0, so g ″ never changes sign. (g) True, since g′(1) = f (1) = 0 and g′( x ) = f ( x ) is an increasing function of x (because f ′( x ) > 0). ds d t (a) υ = = ∫ f ( x ) dx = f (t ) ⇒ υ ( 5 ) = f (5) = 2 m s dt dt 0 (b) a = df dt is negative, since the slope of the tangent line at t = 5 is negative. 3 1 9 (c) s = ∫ f ( x ) dx = ( 3 )( 3 ) = m , since the integral is 0 2 2 the area of the triangle formed by y = f ( x ), the x-axis, and x = 3. (d) t = 6, since after t = 6 to t = 9, the region lies below the x-axis. (e) At t = 4 and t = 7, since there are horizontal tangents there. (f) Toward the origin between t = 6 and t = 9, since the velocity is negative on this interval. Away from the origin between t = 0 and t = 6, since the velocity is positive there. (g) Right or positive side, because the integral of f from 0 to 9 is positive, there being more area above the x-axis than below.

AN-27

37. 41. 45. 47.

1( 1 2 x + 4 )6 + C 3. − ( x 2 + 5 )−3 + C 6 3 1 1( 2 5 7. − cos 3 x + C 3x + 4 x ) + C 3 10 1 ( sec 2t + C 11. −6 1 − r 3 )1 2 + C 2 1 32 1 (x − 1) − sin ( 2 x 3 2 − 2 ) + C 3 6 1 1 2 (a) − ( cot 2θ ) + C (b) − ( csc 2 2θ ) + C 4 4 2 1( 32 − 3 − 2s ) + C 19. − (1 − θ 2 ) 5 4 + C 5 3 1 ( 23. tan 3 x + 2 ) + C ( − 2 (1 + x )) + C 3 6 1 6 x r3 sin +C 27. −1 + C 2 3 18 2 1 32 − cos ( x + 1) + C 31. +C 3 2 cos( 2t + 1) sin 2 (1 θ ) 1 +C − sin − 1 + C 35. − 2 t 2( 2 1 32 1 + x )3 2 − 2(1 + x )1 2 + C 39. 2− +C 3 3 x 32 2 3 1( 1 1− 3 +C 43. x − 1)12 + ( x − 1)11 + C 27 12 11 x 4 2 1 − (1 − x ) 8 + (1 − x ) 7 − (1 − x ) 6 + C 7 3 8 −1 1( 2 1 52 3 2 x + 1) − ( x 2 + 1 ) + C 49. +C 5 3 4( x 2 − 4 )2

(

( )

(

)

)

(

(

)

)

53. 2 tan ( e

51. e sin x + C 57. z − ln (1 +

ez

+ 1) + C 55. ln ln x + C 5 2r − 1 59. tan +C 6 3

)+C

x

( )

1 63. ( arcsin x )3 + C 3

61. e arcsin x + C

65. ln arctan y + C 6 6 + C (b) − +C 67. (a) − 2 + tan 3 x 2 + tan 3 x 6 +C (c) − 2 + tan 3 x 1 1 69. sin 3( 2r − 1) 2 + 6 + C 73. s = ( 3t 2 − 1) 4 − 5 2 6 π 75. s = 4 t − 2 sin 2t + +9 6 π 77. s = sin 2t − + 100 t + 1 79. 6 m 2

(

(

)

)

SECTION 5.6, pp. 367–371

1. (a) 14 3 (b) 2 3 3. (a) 1 2 (b) −1 2 5. (a) 15 16 (b) 0 9. (a) 4 (b) 0 7. (a) 0 (b) 1 8 506 375 86, 744 375 (b) 13. (a) 0 (b) 0 11. (a) 17. 3 4

15. 2 3 25. e

27. ln 3

35. ln ( 2 + 43.

3 −1

19. 3 5 2 − 1 29. ( ln 2 )

2

21. 3 1 31. ln 4

23. π 3 33. ln 2

3 39. π 12 41. 2π 3 37. π 2 47. π 2 32 49. 16 3 45. −π 12

3) −

AN-28

Chapter 6: Answers to Odd-Numbered Exercises

53. π 2

51. 2 5 2

55. 128 15

63. 49 6

61. 38 3 71. 8

59. 5 6

57. 4 3

65. 32 3

67. 48 5

69. 8 3 75. 18

73. 5 3 (There are three intersection points.)

77. 243 8

79. 8 3 81. 2 83. 104 15 85. 56 15 4 4 91. π 2 87. 4 89. − 93. 2 95. 1 2 3 π 97. 1 99. ln 16 101. 2 103. 2 ln 5 23 105. (a) ( ± c , c ) (b) c = 4 (c) c = 4 2 3 107. 11 3

109. 3 4

111. Neither

113. The dark blue region has area 2, which is larger than the light blue region, which has an area ( 0.6 ) π ≈ 1.88. 115. The dark blue region has area e − 2, or approximately 0.72, which is larger than 1 2, which is the area of the light blue region. 117. F (6) − F (2) 119. (a) −3 (b) 3 121. I = a 2 PRACTICE EXERCISES, pp. 372–375

1. (a) About 680 m

4

21. 29. 37. 41.

5

∫1 ( 2 x − 1)

−1 2

t (s)

(c) 13

(d) 0

dx = 2

7.

13. 62

15. 1

−8

0

x

∫−π cos 2 dx

( )

(b)

3 arcsin 2( r − 1) + C 2 1 2x − 1 +C 69. sec −1 4 2 65.

81. 1

15. 13 3

75. 83. 8

91. 6 3 − 2π

y

67.

y=t

2 2−1 2

−2

− ln 7 3 1 63. (3 x 2 ) + C 2 ln 3

x

95. 2

y = sin pt

2

t

17. 1 2 y=2

−1

1

27.

35. (a) 41. 1 6

1 ln 2

2

19. π 2

x

sin 4 y sin y − y 2 y

29. 2 x ln x − x ln

x 2

33. x = 1 1 2 ln 2

(b)

43.

1

∫0

(c) 2:1

f ( x ) dx

37. 2 17

45. (b) πr 2

)

Chapter 6

+C

85. 27 3 160

1

21. ln 2 23. (a) 0 (b) −1 (c) −π (d) x = 1 (e) y = 2 x + 2 − π (f) x = −1, x = 2 (g) [ −2π , 0 ]

y = 1 − x2

25. 2 x

57.

1 +C 4 ( sin 2θ + cos 2θ ) 2 93. −1

0

y

y=1

2 x −1 tan −1 +C 2 2

71. e arcsin

x

3 y = −4

31. ( sin x ) x

(

1 2 − 2 π

1

2

47. θ 2 + θ + sin ( 2θ + 1) + C

1 61. − ( ln x )−2 + C 2

12

−1

43. f ( x ) = e x

55. e tan x + C

73. 2 arctan y + C

−4

y = arcsin x

53. tan ( e x − 7 ) + C 59. ln ( 9 25 )

0

3

9. y = x 3 + 2 x − 4

4

−4

8 2−7 25. 4 27. 6 31. 6 5 33. 1

1 51. − cos ( 2t 3 2 ) + C 3

3

5. (a) 1 4

13.

= 2

19. 18

t3 4 + +C 3 t

89.

(b) No x x2 + 1

(e) 8 5

17. 1 6

π2 2 + −1 98 23. 32 2 Min: −4; max: 0; area: 27 4 x sin t y = ∫ dt − 3 39. 5 t 2π ,x >1 y = sec −1 x + 3

79. 2

133. Yes 135. − 1 + x 2 137. Cost ≈ $13,897.50 using a lower sum estimate

2

(c) −2 (d) −2π

(b) 2

45. −4 ( cos x )1 2 + C 49.

d 1 ( x ln x − x + C ) = x ⋅ + ln x − 1 + 0 = ln x dx x 1 (b) e −1 −6 123. −4 °C 125. 2 + cos 3 x 127. 3 + x4 1 dy −2 cos( 2 ln x ) 131. dy = 129. e = 2 dx dx x 1 − x   1 − 2( arcsin x ) 2 121. (a)

y = x 23

8

6

113. π 6

3

y 2

105. 9 14

(b) b

11. 36 5

3. (a) −1 2 (b) 31

11. 8 3

111. π

117. (a) b

7. f ( x ) =

0

9. (a) 4

109. π

1. (a) Yes

h (meters)

5.

9 ln 2 4 115. π 12 107.

103. 1 6

ADDITIONAL AND ADVANCED EXERCISES, pp. 375–379

(b) h (meters)

700 600 500 400 300 200 100

101. e − 1

99. 15 16 + ln 2

87. π 2 97. 1

SECTION 6.1, pp. 387–391

77. 16

1. 16

3. 16 3

9. 8π 2π 17. 3

11. 10

5. (a) 2 3

(b) 8

7. (a) 60

13. (a) s 2 h (b) s 2 h 15. 8 3 32π 21. 19. 4 − π 23. 36π 25. π 5

(b) 36

AN-29

Chapter 6: Answers to Odd-Numbered Exercises

27.

(

1 π 1− 2 2 e

33. 2π

)

55. 59. 63.

π ln 4 2

31. π

37. 4 π ln 4

35. 2π

117π 43. 5 53.

29.

( π2 + 2

39.

π2

2−

11 3

− 2π

4π − 47. 45. 49. 8π 3 32π 8π 224 π (a) 8π (b) (c) (d) 5 3 15 16π 56π 64 π (a) (b) (c) 57. V = 2a 2 bπ 2 15 15 15 πh 2 ( 3a − h ) 1 ms (a) V = (b) 120 π 3 V = 3308 cm 3 65. ( 4 − b + a ) 2 π( π

2)

)

2π 41. 3 7π 51. 6

(

1. (a) 2π ∫

11.

53 6

17. (a)

0.6 0.4

2

π

19. (a)

∫0

21. (a)

∫−1

23. (a)

123 5. 32

3

π6

∫0

1 + 4 x 2 dx 1 + cos 2 y dy

99 7. 8

0.2

0 2

3. (a) 2π ∫

1

0.4

0.6

x

0.8

1 1 + y −4 dy y

(c) S ≈ 5.02

y

(b) 2 1.8 1.6

xy = 1

1.4 1.2 1 0.5

0.6

0.7

0.8

4

π 6 π 6

5. (a) 2π ∫ ( 3 − x 1 2 )

0.9

x

1

2

2

1 + (1 − 3 x −1 2 ) dx

1

y

(b)

(c) S ≈ 63.37

4

3 x12 + y12 = 3 2

1

1

2

7. (a) 2π ∫ y

(b)

π3 0

(∫

3 y 0

4

x

)

(c) s ≈ 2.08

tan t dt sec y dy

1 0.8

0.4

3 9. ln 2 + 8

(c) ≈3.82 (c) ≈9.29

(c) ≈0.55

25. (a) y = x from (1, 1) to (4, 2) (b) Only one. We know the derivative of the function and the value of the function at one value of x. x 2 (10 3 2 − 1) 27. 1 29. 2 5 37.   ∫ 1 + 9t dt , 0 27

x=

y

L0

tan t dt

0.2 0

15. 2

(c) ≈6.13

1 + ( y + 1) 2 dy sec x dx

y = tan x

0.2

0.6

13. 2 3 2 − ( 5 4 )3 2

∫−1

(c) S ≈ 3.84

1

SECTION 6.3, pp. 403–405

1. 12

( tan x ) 1 + sec 4 x dx

y

(b)

)

53 3. 6

π4 0

0.8

SECTION 6.2, pp. 396–399

1. 6π 3. 2π 5. 14 π 3 7. 8π 9. 5π 6 7π 16π 15. 11. 13. (b) 4π (3 2 + 5) 15 15 8π 4π 16π 17. 19. 21. 3 3 3 23. (a) 16π (b) 32π (c) 28π (d) 24π (e) 60π (f) 48π 27π 27π 72π 108π 25. (a) (b) (c) (d) 2 2 5 5 6π 4π 27. (a) (b) (c) 2π (d) 2π 5 5 2π 29. (a) About the x-axis: V = ; about the y-axis: V = 15 2π (b) About the x-axis: V = ; about the y-axis: V = 15 5π 4π 2π 31. (a) (b) (c) 2π (d) 3 3 3 4π 7π 24 π 48π 33. (a) (b) 35. (a) (b) 15 30 5 5 9π 9π 37. (a) (b) 16 16 39. Disk: 2 integrals; washer: 2 integrals; shell: 1 integral 256π 244 π (b) 41. (a) 3 3 1 47. π 1 − 49. 2 e

SECTION 6.4, pp. 408–410

9. 4 π 5

0.1 0.2 0.3 0.4 0.5 0.6 0.7

11. 3π 5

x

13. 98π 81

15. 2π 15 19. 35π 5 3 17. π ( 8 − 1) 9 21. π + ln 2 16 27. Order 226.2 liters of each color. 23. 253π 20

(

SECTION 6.5, pp. 415–419

1. 7. 9. 15.

116 J 3. 400 N m 5. 4 cm, 0.08 J (a) 12,000 N/cm (b) 6000 N-cm, 18,000 N-cm 780 J 11. 108,000 J 13. 234 J (a) 235,200,000 J (b) 17 h, 46 min (c) 266 min (d) At 9780 N m 3 : a) 234,720,000 J b) 17 h, 44 min At 9820 N m 3 : a) 235,680,000 J b) 17 h, 48 min 17. 5,772,676.5 J 19. 8,977,666 J 21. 385,369 J 23. 15,073,099.75 J 27. 120 J 29. 151.3 J

)

AN-30

Chapter 7: Answers to Odd-Numbered Exercises

2.175 J 33. 5.144 × 10 10 J 35. 9146.7 N (a) 26,989.2 N (b) 25,401.6 N (a) 182,933 N (b) 187,600 N 41. 5808 N (a) 19,600 N (b) 14,700 N (c) 16,135 N wb 45. (a) 14,933 N (b) 2.94 m 47. 2 49. No. The tank will overflow because the movable end will have moved only 2.18 m by the time the tank is full.

31. 37. 39. 43.

SECTION 6.6, pp. 429–431

1. 5. 9. 13. 17. 21. 23. 25.

M = 14 3, x = 93 35 3. M = ln 4, x = ( 3 − ln 4 ) ( ln 4 ) M = 13, x = 41 26 7. x = 0, y = 12 5 x = 1, y = − 3 5 11. x = 16 105, y = 8 15 x = 0, y = π 8 15. x ≈ 1.44, y ≈ 0.36 ln 4 15 15 ,y = x = ,y = 0 19. x = 4 ln 2 128 ln 2 π x = 5 7, y = 10 33. ( x ) 4 < y , so the center of mass is outside the region. x = 3 2, y = 1 2 224 π (a) (b) x = 2, y = 0 3 y (c) 4 y=

4 Îx

1

5.

2x − a π

π 30 2

4

7. 28 3

Chapter 7 SECTION 7.1, pp. 445–447

1. ln

( 23 )

3. ln y 2 − 25 + C

7. ln ( 1 +

x) + C

9π 3. π 2 280 7. (a) 2π (b) π

72π 35 (c) 12π 5

13. 2

17. −e +C 19. −e 1 x + C 15. 2e r + C 1 21. e sec πt + C 23. 1 25. ln ( 1 + e r ) + C π 1 6 29. 31. 33. 32760 35. 3 2 +1 ln 2 ln 7 2 3 ln 2 1 ⎛⎜ ( ln x ) ⎞⎟⎟ +C 39. 2( ln 2 )2 37. 41. ⎜⎜ ln 10 ⎝ 2 ⎟⎟⎠ 2 −t 2

27.

43. ln 10

51. y = x + ln x + 2 57. (b) 0.00469

9.

2 32 y − x1 2 = C 3

(d) 26π 5

28π 3 m 3 10 1 ( π 3 )( a 2 + ab + b 2 ) h 19. 21. 3 + ln 2 8 3 2 25. 28π 2 3 27. 4π 29. 4640 J w( 33. 65,680,230 J 2ar − a 2 ) 2 3,375,000 π J, 257 s 37. (a) 11.31 J (b) 29.18 J x = 0, y = 8 5 41. x = 3 2,   y = 12 5 15.

45. 52,267 J

11. e y − e x = C

13. −x + 2 tan y = C 17. y = sin ( x 2 + C )

15. e − y + 2e 19.

x

= C

1 ln y 3 − 2 = x 3 + C 3

21. 4 ln ( y + 2 ) = e x 2 + C 23. (a) −0.00001 (b) 10,536 years 27. 19.9 m

(c) 82%

29. 2.8147498 × 10 14

(b) 32.02 years

33. Yes, y ( 20 ) < 1

35. 15.28 years 37. 56,562 years 41. (a) 17.5 min (b) 13.26 min 43. −3 °C 45. About 6693 years 47. 54.62% 49. ≈15,683 years

(c) 512π 15

13. π ( e − 1 )

43. x = 9 5, y = 11 10

1 2 ln 2

47. y = 1 − cos ( e t − 2 )

55. 6 + ln 2

31. (a) 8 years

5.

(b) 1088π 15

11. π ( 3 3 − π ) 3

35. 39.

11. 2( ln 2 ) 4

9. 1

SECTION 7.2, pp. 454–456

1.

31.

5. ln 6 + 3 tan t + C

77. (a) 1.89279 (b) −0.35621 (c) 0.94575 (d) −2.80735 (e) 5.29595 (f) 0.97041 (g) −1.03972 (h) −1.61181

PRACTICE EXERCISES, pp. 432–434

23.

4 h 3mh 3

9.

n ,   ( 0, 1 2 ) 2n + 1 15. (a) x = y = 4 ( a 2 + ab + b 2 ) ( 3π ( a + b )) (b) ( 2a π , 2a π ) 17. ≈365.867 N

25. 54.88 g

17.

C 2 − 1 x + a, where  C ≥ 1

11. x = 0, y =

53. π ln 16

29. x = y = 1 3 31. x = a 3, y = b 3 33. 13δ 6 aπ 35. x = 0, y = 37. x = 1 2, y = 4 4 39. x = 6 5, y = 8 7 43. V = 32π ,   S = 32 2π 2a 4b 45. 4 π 2 47. x = 0, y = 49. x = 0, y = π 3π a b 51. 2πa 3 ( 4 + 3π ) 6 53. x = , y = 3 3

9. (a) 8π

3. f ( x ) =

49. y = 2( e −x + x ) − 1

x

y=− 4 Îx

−4

1. f ( x ) =

45. ( ln 10 ) ln ln x + C

(2, 0) 0

ADDITIONAL AND ADVANCED EXERCISES, pp. 434–435

47. 344,960 N

SECTION 7.3, pp. 462–465

1. cosh x sech x 3. sinh x sech x 1 5. x + x

= 5 4 , tanh x = − 3 5, coth x = −5 3, = 4 5, csch x = −4 3 = 8 15, tanh x = 8 17, coth x = 17 8, = 15 17, csch x = 15 8 x 7. e 5 x 9. e 4 x 13. 2 cosh 3

tanh t 17. coth z t 19. ( ln sech θ )( sech θ tanh θ ) 21. tanh 3 υ

15. sech 2 t +

23. 2

Chapter 8: Answers to Odd-Numbered Exercises

25.

1 2 x (1 + x )

29.

1 − coth −1 t 2 t

35. sec x

27.

41.

25. It could take one million steps for a sequential search; at most 20 steps for a binary search.

1 − tanh −1 θ 1+θ 31. − sech −1 x

( 2x − ln 3 ) + C 45. 7 ln e + e 1 47. tanh ( x − ) + C 49. −2 sech t + C 2 53.

x 7

3 + ln 2 32

61. ln ( 2 3 )

ln 2

33.

( )

1 1+ 2

2θ

cosh 2 x +C 2

43. 12 sinh

55. e − e −1 63.

− ln 3 2

57. 3 4

−x 7

+C

51. ln

59.

5 2

3 + ln 2 8

65. ln 3

71. (a)

− sech −1

( )

73. (a) 0 77. (b)

3. ln 8 5. 2 ln 2 1. − cos e x + C 1 ( 2 7. ( ln x − 5 ) ) + C 9. 3 ln 7 11. 2( 2 − 1 ) 2 ln 2 1 15. y = ln x − ln 3 17. y = 13. y = 1 − ex ln ( 3 2 ) 19. (a) (e) 21. (a) (e) 23. 1 3

Same rate (b) Same rate (c) Faster (d) Faster Same rate (f ) Same rate True (b) False (c) False (d) True True (f ) True 25. 1 e m s 27. ln 5 x − ln 3 x = ln ( 5 3 )

(

31. y = tan −1

( x +2 C ))

2

33. y 2 = sin −1( 2 tan x + C ) 35. y = −2 + ln ( 2 − e −x )

( 12 ) ln ( 13 )

39. 19,035 years

( )

12 4 + sech −1 13 5

⎛ 1 + 1 − ( 12 13 ) 2 (b) − ln ⎜⎜⎜ ( 12 13 ) ⎜⎝ = − ln

(b)

PRACTICE EXERCISES, pp. 471–472

29. 1 2

67. (a) sinh −1 ( 3 ) (b) ln ( 3 + 2 ) 69. (a) coth −1 (2) − coth −1 ( 5 4 )

37. y = 4 x − 4 x + 1

41. ln ( 16 9 )

ADDITIONAL AND ADVANCED EXERCISES, p. 472

2 ⎞⎟ ⎛ ⎟⎟ + ln ⎜⎜ 1 + 1 − ( 4 5 ) ⎜⎜ ⎟⎟⎠ (4 5) ⎝

⎞⎟ ⎟⎟ ⎟⎟⎠

( 32 ) + ln (2) = ln( 4 3 )

(b) π 2 (c) π 1 π 3. x + tan −1 is a constant and the constant is for 2 x π x > 0; it is − for x < 0. 2 1. (a) 1

( )

tan −1

y

(b) 0 mg k

y=p 2

2

(c) 70 30 47 ≈ 55.93 m s

1 −4

6 81. 5

79. 2π

SECTION 7.4, pp. 469–470

1. (a) Slower (b) Slower (c) Slower (d) Faster (e) Slower (f ) Slower (g) Same (h) Slower 3. (a) Same (b) Faster (c) Same (d) Same (e) Slower (f ) Faster (g) Slower (h) Same 5. (a) Same (b) Same (c) Same (d) Faster (e) Faster (f ) Same (g) Slower (h) Faster 7. d, a, c, b 9. (a) False (b) False (c) True (d) True (e) True (f ) True (g) False (h) True 13. When the degree of f is less than or equal to the degree of g. 15. 1, 1 6 21. (b) ln (e 17000000 ) = 17,000,000  1 1 85. 2π 87. ln 2 (a) 1 (b) π 3 (c) Diverges (a) π 2 (b) π 93. (b) ≈0.88621 (a)

55. π 3

57. 2π 3 + π 2  ln ( 2 + 61. 7.62

63. 69. 75. 81. 83. 89. 91. 95.

9. ln 3 11. ln 4 π 17. π 19. ln 1 + 0 15. 3 2 −1 23. 1 25. −1 4 27. π 2 29. π 3 6 33. ln 2 35. Diverges 37. Diverges Diverges 41. Diverges 43. Converges Converges 47. Diverges 49. Converges Converges 53. Diverges 55. Converges Converges 59. Diverges 61. Converges

(

)

-2

-1

0

1

2

3

x

(b) ≈0.683, ≈0.954, ≈0.997 103. ≈0.16462 PRACTICE EXERCISES, pp. 532–534

1. ( x + 1 )( ln ( x + 1 ) ) − ( x + 1 ) + C 1 3. x arctan ( 3 x ) − ln ( 1 + 9 x 2 ) + C 6 5. ( x + 1 ) 2 e x − 2( x + 1 )e x + 2e x + C 7. 9. 11. 13. 15. 17.

2e x sin 2 x e x cos 2 x + +C 5 5 2 ln x − 2 − ln x − 1 + C 1 ln x − ln x + 1 + +C x +1 cos θ − 1 1 − ln +C 3 cos θ + 2 1 4 ln x − ln ( x 2 + 1 ) + 4 arctan x + C 2 ( υ − 2 )5 ( υ + 2 ) 1 ln +C 16 υ6

1 3 t arctan t − tan −1 +C 2 6 3 x2 4 2 21. + ln x + 2 + ln x − 1 + C 2 3 3 x2 9 3 23. − ln x + 3 + ln x + 1 + C 2 2 2 1 x +1−1 25. ln +C 27. ln 1 − e −s + C 3 x +1+1 1 29. − 16 − y 2 + C 31. − ln 4 − x 2 + C 2 1 1 x+3 33. ln 35. ln +C +C 6 x−3 9 − x2 19.

20

25

t

Chapter 9: Answers to Odd-Numbered Exercises

cos 5 x cos 7 x tan 5 x + +C 39. +C 5 7 5 cos θ cos 11θ − +C 43. 4 1 − cos( t 2 ) + C 2 22 49. −4  °C At least 16 47. All three rules yield π (a) ≈2.42 L (c) ≈24.83 km L π2 55. 6 57. ln 3 59. 2 61. π 6 Diverges 65. Diverges 67. Converges 1 2x 1 xe − e 2 x + C 71. 2 tan x − x + C 2 4 1 x tan x − ln sec x + C 75. − ( cos x )3 + C 3 1 2 1 4x + 1 79. 2 ln 1 − + +C 1 + ln 1 + e2 2x 2 2 x

37. − 41. 45. 51. 53. 63. 69. 73. 77. 81. 87. 89. 91. 93. 95. 97. 99.

(

)

e 2x − 1 83. 9 4 85. 256 15 +C ex 1 − csc 3 x + C 3 2x 3 2 − x + 2 x − 2 ln ( x + 1 ) + C 3 1 −1 ( 1 sin x − 1) + ( x − 1) 2 x − x 2 + C 2 2 −2 cot x − ln csc x + cot x + csc x + C 1 3+υ 1 υ ln + tan −1 + C 12 3−υ 6 3 θ sin ( 2θ + 1 ) cos ( 2θ + 1 ) + +C 2 4 ⎛ ( 2 − x )3 ⎟⎞ 1 sec 2 θ + C 101. 2⎜⎜⎜ − 2 2 − x ⎟⎟⎟ + C 4 3 ⎜⎝ ⎟⎠

( )

1 1 1 1 1 z ⎤ +C 105. ln z − −   ⎡⎢ ln ( z 2 + 4 ) + arctan 4 4z 4 ⎣2 2 2 ⎥⎦ 1 et + 1 107. − 9 − 4t 2 + C 109. ln t +C 4 e +2 2 1 111. 1 4 113. x 3 2 + C 115. − arctan ( cos 5t ) + C 3 5

(

)

117. 2 r − 2 ln ( 1 + r ) + C 119. 1 2 1 x − ln ( x 2 + 1 ) + C 2 2 2 1 1 2x − 1 121. ln x + 1 + ln x 2 − x + 1 + arctan +C 3 6 3 3 72 52 32 4 8 4 123. ( 1 + x ) − ( 1 + x ) + ( 1 + x ) + C 7 5 3

(

x +

1+ x +C

127. ln x − ln 1 + ln x + C

1 ln x 1 1 − 1 − x4 x +C 131. ln +C 2 2 x2 1 π 133. x − arctan ( 2 tan x ) + C 135. 4 2 129.

ADDITIONAL AND ADVANCED EXERCISES, pp. 534–537

1. x ( arcsin x ) 2 + 2( arcsin x ) 1 − x 2 − 2 x + C 3. 5.

x2

arcsin x x 1− + 2

x2

)

− arcsin x +C 4

1 ( ln ( t − 1 − t 2 ) − arcsin t ) + C 2

7. 0

15. 2π

21.

(e

2

+1 e−2 ,  4 2

)

⎛ 1 + e2 1⎞ 1 + e 2 − ln ⎜⎜⎜ + ⎟⎟⎟ − 2 + ln ( 1 + 2 ) ⎝ e e⎠   ln 2 1 1 12π 25. 29. < p ≤1 27. a = , − 2 2 5 4 3π 2 33. 35. 1 37. +C 9 1 − tan ( x 2 ) 23.

39.

tan ( t 2 ) + 1 − 1 ln tan ( t 2 ) + 1 + 2

41. ln

2 +C 2

1 + tan ( θ 2 ) +C 1 − tan ( θ 2 )

Chapter 9 SECTION 9.1, pp. 547–551

1. a1 = 0, a 2 = −1 4 , a 3 = − 2 9, a 4 = −3 16 3. a1 = 1, a 2 = −1 3, a 3 = 1 5, a 4 = −1 7 5. a1 = 1 2, a 2 = 1 2, a 3 = 1 2, a 4 = 1 2 3 7 15 31 63 127 255 511 1023 7. 1, , , , , , , , , 2 4 8 16 32 64 128 256 512 1 1 1 1 1 1 1 1 , 9. 2, 1, − , − , , , − , − , 2 4 8 16 32 64 128 256 11. 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 13. a n = ( −1 ) n +1 , n ≥ 1

103. arctan ( y − 1 ) + C

125. 2 ln

9. ln (4) − 1 11. 1 13. 32π 35 17. (a) π (b) π ( 2e − 5 ) ⎛ 8( ln 2 ) 2 16( ln 2 ) 16 ⎞⎟⎟ 19. (b) π ⎜⎜ − + ⎜⎝ 3 9 27 ⎟⎟⎠

AN-35

15. a n = ( −1 ) n +1 (n) 2 , n ≥ 1 19. a n = n 2 − 1, n ≥ 1

2 n−1 ,n ≥1 3( n + 2 ) 21. a n = 4 n − 3, n ≥ 1 17. a n =

3n + 2 1 + ( −1 ) n +1 ,n ≥1 ,n ≥1 25. a n = 2 n! ⎛ 1 n + 1 ⎞⎟ 27. a n = 29. a n = sin ⎜⎜ ⎝ 1 + ( n + 1 ) 2 ⎟⎟⎠ ( n + 1 )( n + 2 ) 31. Converges, 2 33. Converges, −1 35. Converges, −5 37. Diverges 39. Diverges 41. Converges, 1 2 43. Converges, 0 45. Converges, 2 47. Converges, 1 49. Converges, 0 51. Converges, 0 53. Converges, 0 55. Converges, 1 57. Converges, e 7 59. Converges, 1 61. Converges, 1 63. Diverges 65. Converges, 4 67. Converges, 0 69. Diverges 71. Converges, e −1 73. Diverges 75. Converges, 0 77. Diverges 79. Converges, e 2 3 81. Converges, x ( x > 0 ) 83. Converges, 0 85. Converges, 1 87. Converges, 1 2 89. Converges, 1 91. Converges, π 2 93. Converges, 0 95. Converges, 0 97. Converges, 1 2 99. Converges, 0 101. 8   103. 4   105. 5   107. 1 + 2   109. x n = 2 n−2 23. a n =

111. (a) f ( x ) = x 2 − 2, 1.414213562 ≈

2

(b) f ( x ) = tan ( x ) − 1, 0.7853981635 ≈ π 4 (c) f ( x ) = e x , diverges 113. (b) 1

AN-36 121. 123. 125. 127. 129. 131. 133. 145.

Chapter 9: Answers to Odd-Numbered Exercises

Monotonic, bounded Not monotonic, bounded Monotonic, bounded, converges Monotonic, bounded, converges Not monotonic, bounded, diverges Monotonic, bounded, converges Monotonic, bounded, converges (b) 3

SECTION 9.3, pp. 565–567

1. Converges 9. Converges

2( 1 − ( 1 3 ) n ) ,3 1 − (1 3 )

3. s n =

5. Converges

13. Converges; geometric series, r =

1 − ( −1 2 ) n , 2 3 1 − ( −1 2 )

1 4 1 1 1 1 1 , − 7. 1 − + − + , 5 2 n+2 2 4 16 64 3 9 57 249 − + + + + , diverges. 4 16 64 256 5 1 5 1 5 1 23 ( 5 + 1) + + + + + + + , 2 3 4 9 8 27 2 1 1 1 1 1 1 17 (1 + 1) + − + + + − + , 2 5 4 25 8 125 6 Converges, 5 3 17. Converges, 1 7 e 21. Diverges 23. 23 99 Converges, e+2 79 27. 1 15 29. 41333 33300 31. Diverges Inconclusive 35. Diverges 37. Diverges 1 ; converges, 1 sn = 1 − n+1

n =1≠ 0 n+1 Diverges; p-series, p < 1 1 Converges; geometric series, r = < 1 8 Diverges; Integral Test Converges; geometric series, r = 2 3 < 1 Diverges; Integral Test 2n ≠ 0 Diverges; lim n →∞ n + 1

17. 19. 21. 23. 25. 27.

9.

29. Diverges; lim ( n ln n ) ≠ 0

13. 15. 19. 25. 33. 39.

) ( ) (

( (

) ( ) (

)

)

n →∞

31. Diverges; geometric series, r = 33. 35. 37. 39. 41. 43. 45. 49.

49. 55. 59. 61. 67. 73.

(

1

)

79. a = 3, r = ( x − 1 ) 2; converges to 6 ( 3 − x ) for x in ( −1, 3 ) 1 1 1 81. x < , 83. −2 < x < 0, 2+ x 2 1 − 2x π 1 85. x ≠ ( 2 k + 1 ) , k an integer; 1 − sin x 2 ∞

∑ ( n + 4 )(1 n + 5 )

n =−2 ∞

(c)

∞

(b)

∑ ( n + 2 )(1 n + 3 ) n=0

1

∑ ( n − 3 )( n − 2 ) n=5

1 + 2r 97. (a) r = 3 5 (b) r = −3 10 99. r < 1, 1 − r2 101. (a) 16.84 mg, 17.79 mg (b) 17.84 mg 1 2 1 2 7 8 1 2 7 8 , , , , , , , , , ,1 103. (a) 0, 27 27 9 9 27 27 3 3 9 9 ∞

(b)

12

π π 1 ; converges, − s n = − arccos    45. 1    47. 5 6 3 n+2 1 53. Converges, 2 + 2 1 51. − ln 2 Converges, 1 57. Diverges e2 Converges, 2 e −1 Converges, 2 9 63. Converges, 3 2 65. Diverges π Converges, 4 69. Diverges 71. Converges, π−e Converges, −5 6 75. Diverges

87. (a)

( )

∑ 12 23 n =1

n −1

=1

y=

1

77. a = 1, r = −x; converges to 1 ( 1 + x ) for x < 1

105. ( 4 3 ) π

1 >1 ln 2

Converges; Integral Test Diverges; nth-Term Test Converges; Integral Test Diverges; nth-Term Test Converges; by taking limit of partial sums Converges; Integral Test Converges; Integral Test 47. a = 1 y (a)

41. s n = ln n + 1; diverges 43.

1 1 (d) Boundary: x 2 + y 2 = 1 (e) Open (f) Unbounded (f), (h) 33. (a), (i) 35. (d), (j) (a) (b) z

43. (a)

4 4

z = 16 z=4

2

2

x

1 2

z=0 4

2

1 y

0 x

45. (a)

z

(b)

y

z = 1 – 0y0

z = –1

(0, 0, 1) 2

z=0

1

z=1

0 y

1

x

z=0

–1

x

z = –1

–2

47. (a)

(b)

z

y

z = Î x2 + y2 + 4

4

z = Î 20

3 z = Î13

z=4 z=1 x z=1 z=4

z=0

y

z = 4x2 + y2

16

y

z = y2

(b)

z

2 z = Î8 1 z = Î5

2

z=2

y

–4 –3 –2 –1

y

1

2

3

4

x

–1

x

–2

x

–3

39. (a)

z

(b)

–4

y

z = x2 + y2

z=4

49. x 2 + y 2 = 10

z=1 z=0 –2

–1

1

51. x + y 2 = 4

y

2

x

y

Î 10

2

–Î 10

Î 10

x

4

y x

41. (a)

–2

(b)

z

z = –3

z=x –y

y

z = –1 z=0

y

3

z=1

2

z=2

1

z=3

0 –1 x

–Î 10

z = –2

2

–2 –3

53.

55.

z

z

1

1

x 1

1

x

y

1 x

f(x, y, z) = x 2 + y2 + z2 = 1

f(x, y, z) = x + z = 1

y

x

Chapter 13: Answers to Odd-Numbered Exercises

57.

59.

z f(x, y, z) = x2 + y2 = 1

5

2 2

1

1

1 y

1 f(x, y, z) = z – x 2 – y 2 = 1 or z = x 2 + y2 + 1

1

x

y

x

y 1

–1

1

x

31. f x = −2 xe −( x –1

level curve: π sin −1 y − sin −1 x = 2 SECTION 13.2, pp. 781–784

3. 2 6 5. 1 7. 1 2 9. 1 13. 0 15. −1 17. 2 19. 1 4 23. 3 25. 19 12 27. 2 29. 3 All ( x , y ) (b) All ( x , y ) except ( 0, 0 ) All ( x , y ) except where x = 0 or y = 0 (b) All ( x , y ) All ( x , y, z ) All ( x , y, z ) except the interior of the cylinder x 2 + y 2 = 1 All ( x , y, z ) with z ≠ 0 (b) All ( x , y, z ) with x 2 + z 2 ≠ 1 All points ( x , y, z ) satisfying z > x 2 + y 2 + 1

(b) All points ( x , y, z ) satisfying z ≠ + Consider paths along y = x , x > 0, and along y = x , x < 0. Consider the paths y = kx 2 , k a constant. Consider the paths y = mx, m a constant, m ≠ −1. Consider the paths y = kx 2 , k a constant, k ≠ 0. Consider the paths x = 1 and y = x. Along y = 1 the limit is 0; along y = e x the limit is 1 2. Along y = 0 the limit is 1; along y = − sin x the limit is 0. (a) 1 (b) 0 (c) Does not exist The limit is 1. 61. The limit is 0. (a) f ( x , y ) y = mx = sin 2θ , where tan θ = m 65. 0 x2

41. 43. 45. 47. 49. 51. 53. 55. 59. 63.

69. π 2 75. δ = 0.005

67. Does not exist 73. δ = 0.1 79. δ = 0.015

y2

71. f ( 0, 0 ) = ln 3 77. δ = 0.04

81. δ = 0.005

SECTION 13.3, pp. 793–796

∂f ∂f ∂f ∂f = 2 x ( y + 2 ), = 4 x, = −3 3. = x2 − 1 ∂x ∂y ∂y ∂x ∂f ∂f = 2 y ( xy − 1 ), 5. = 2 x ( xy − 1 ) ∂x ∂y 1.

yz xy xz ,  f y = ,  f z = 1 − x 2 y 2z 2 1 − x 2 y 2z 2 1 − x 2 y 2z 2 2 3 1 29. f x = ,  f = ,  f = x + 2 y + 3z y x + 2 y + 3z z x + 2 y + 3z 27. f x =

level curve: y = 2 x

52 14 1 (a) (a) (a) (b) 37. (a) 39. (a)

∂f = 2 sin ( x − 3 y ) cos( x − 3 y ), ∂x ∂f = −6 sin ( x − 3 y ) cos( x − 3 y ) ∂y ∂f ∂f ∂f ∂f 21. = yx y−1 , = x y ln x = −g( x ), = g ( y) 19. ∂y ∂x ∂y ∂x 23. f x = y 2 ,   f y = 2 xy,   f z = −4 z 25. f x = 1,   f y = − y ( y 2 + z 2 )−1 2 ,   f z = −z ( y 2 + z 2 )−1 2

x

1. 11. 21. 31. 33. 35.

∂f y x , = x 2 + y2 ∂y x 2 + y2 ∂f −1 −1 = , = ( x + y )2 ∂ y ( x + y )2 −y 2 − 1 ∂ f −x 2 − 1 = = , ( xy − 1 ) 2 ∂ y ( xy − 1 ) 2 ∂f ∂f ∂f 1 1 = e x + y +1 , = e x + y +1 15. , = = ∂x x + y ∂y x+y ∂y =

17.

61. x − y − ln z = 2 63. x 2 + y 2 + z 2 = 4 65. Domain: all points ( x , y ) 67. Domain: all points ( x , y ) satisfying x < y satisfying −1 ≤ x ≤ 1 and y y=x −1 ≤ y ≤ 1

y = –x

∂f ∂x ∂f 9. ∂x ∂f 11. ∂x ∂f 13. ∂x 7.

z

AN-55

2 + y2 +z 2 )

,   f y = −2 ye −( x

2 + y2 +z 2 )

,   f z = −2 ze −( x

2 + y2 +z 2 )

33. f x = sech 2 ( x + 2 y + 3z ),   f y = 2 sech 2 ( x + 2 y + 3z ), f z = 3 sech 2 ( x + 2 y + 3z ) ∂f ∂f = −2π sin ( 2πt − α ), = sin ( 2πt − α ) ∂t ∂α ∂h ∂h ∂h 37. = sin ø cos θ, = ρ cos ø cos θ, = −ρ sin ø sin θ ∂ρ ∂ø ∂θ δυ 2 , 39. W P ( P , V ,  δ ,  υ,  g ) = V , WV ( P , V ,  δ ,  υ,  g ) = P + 2g Vυ2 V δυ , Wυ ( P , V ,  δ ,  υ,  g ) = , Wδ ( P , V ,  δ ,  υ,  g ) = 2g g 2 V δυ W g ( P , V ,  δ ,  υ,  g ) = − 2g 2 35.

41. 43.

45.

47.

∂f ∂f ∂2 f ∂2 f ∂2 f ∂2 f = 1 + y, = 1 + x, = 0, = 0, = =1 ∂x ∂y ∂x 2 ∂y2 ∂y ∂x ∂x ∂y ∂g ∂g = 2 xy + y cos x , = x 2 − sin y + sin x , ∂x ∂y ∂2g ∂2g = 2 y − y sin x , = − cos y, 2 ∂x ∂y2 ∂2g ∂2g = = 2 x + cos x ∂y ∂x ∂x ∂y 1 ∂r ∂r , = = ∂x x + y ∂y x ∂ 2r ∂ 2r = = ∂y ∂x ∂x ∂y ( x

1 ∂ 2r −1 ∂ 2r −1 , , , = = ( x + y )2 ∂ y 2 ( x + y )2 + y ∂x 2 −1 + y )2

∂w ∂w = x 2 y sec 2 ( xy ) + 2 x tan ( xy ), = x 3 sec 2 ( xy ), ∂x ∂y ∂2w ∂2w = 2 x 3 y sec 2 ( xy ) tan ( xy ) + 3 x 2 sec 2 ( xy ) = ∂y ∂x ∂x ∂y ∂2w = 4 xy sec 2 ( xy ) + 2 x 2 y 2 sec 2 ( xy ) tan ( xy ) + 2 tan ( xy ) ∂x 2 ∂2w = 2 x 4 sec 2 ( xy ) tan ( xy ) ∂y2

AN-56

49.

51.

53.

55. 57.

Chapter 13: Answers to Odd-Numbered Exercises

∂w ∂w = sin ( x 2 y ) + 2 x 2 y cos( x 2 y ), = x 3 cos( x 2 y ), ∂x ∂y ∂2w ∂2w = = 3 x 2 cos( x 2 y ) − 2 x 4 y sin ( x 2 y ) ∂y ∂x ∂x ∂y ∂2w = 6 xy cos( x 2 y ) − 4 x 3 y 2 sin ( x 2 y ) ∂x 2 ∂2w = −x 5 sin ( x 2 y ) ∂y2 ∂f ∂f = 3x 2 y 2 + 5 y 4 , = 2 xy 3 − 4 x 3 , ∂x ∂y ∂2 f ∂2 f = 6 x 2 y + 20 y 3 , = 2 y 3 − 12 x 2 , ∂x 2 ∂y2 ∂2 f ∂2 f = = 6 xy 2 ∂y ∂x ∂x ∂y ∂z = 2 x cos ( 2 x − y 2 ) + sin ( 2 x − y 2 ), ∂x ∂z = −2 xy cos ( 2 x − y 2 ), ∂y ∂2z = 4 cos ( 2 x − y 2 ) − 4 x sin ( 2 x − y 2 ), ∂x 2 ∂2z = −4 xy 2 sin ( 2 x − y 2 ) − 2 x cos ( 2 x − y 2 ), ∂y2 ∂2z ∂2z = = 4 xy sin ( 2 x − y 2 ) − 2 y cos ( 2 x − y 2 ) ∂x ∂y ∂y ∂x ∂w ∂w 3 ∂2w ∂2w −6 2 = , = , = = ∂ x 2 x + 3 y ∂ y 2 x + 3 y ∂ y ∂ x ∂ x ∂ y ( 2 x + 3 y )2 ∂w ∂w = y 2 + 2 xy 3 + 3 x 2 y 4 , = 2 xy + 3 x 2 y 2 + 4 x 3 y 3 , ∂x ∂y ∂2w ∂2w = = 2 y + 6 xy 2 + 12 x 2 y 3 ∂y ∂x ∂x ∂y

81. f x ( x , y ) = 0 for all points ( x , y ), ⎧⎪ 3 y 2 , y ≥ 0 , f y ( x , y ) = ⎪⎨ ⎪⎪ −2 y, y < 0 ⎪⎩ f xy ( x , y ) = f yx ( x , y ) = 0 for all points ( x , y ) 99. Yes SECTION 13.4, pp. 803–805

dw dt dw 3. (a) dt dw 5. (a) dt ∂ 7. (a) z ∂u ∂z ∂υ

77.

∂2 f 2x − 2 y ∂2 f 2 y − 2x 79. = ≠ = so impossible ( x + y )3 ( x + y )3 ∂y ∂x ∂x ∂y

= −4u sin υ ln ( u sin υ ) +

∂z = −2 2 ( ln 2 − 2 ) ∂υ

2 ( ln 2 + 2 ),

=

4u cos 2 υ sin υ

∂w = −2υ + 2u 2 ∂υ ∂w 3 = 3, = − ∂υ 2 −y ∂u z ∂u = 0, = , = ( z − y )2 ∂ z ( z − y )2 ∂y = 2u + 4uυ,

∂u ∂u ∂u = 0, = 1, = −2 ∂x ∂y ∂z dz ∂ z dx ∂ z dy = + 13. dt ∂ x dt ∂ y dt (b)

z

'z 'x

'z 'y

x

y dy dt

dx dt t

15.

∂w ∂w ∂ x ∂w ∂ y ∂w ∂z , = + + ∂u ∂ x ∂u ∂ y ∂u ∂ z ∂u ∂w ∂w ∂x ∂w ∂ y ∂w ∂z = + + ∂υ ∂ x ∂υ ∂ y ∂υ ∂ z ∂υ w

x

(b) 2

∂f = 3x 2 y 2 − 2 x   ⇒ ∂x f ( x ,  y ) = x 3 y 2 − x 2 + g ( y )   ⇒ ∂f = 2 x 3 y + g′( y) = 2 x 3 y + 64  ⇒ ∂y g′( y) = 6 y   ⇒ g( y) = 3 y 2  works  ⇒ f ( x ,  y ) = x 3 y 2 − x 2 + 3 y 2  works

dw ( ) 1 = π+1 dt

= 4 cos υ ln ( u sin υ ) + 4 cos υ,

∂z ∂u ∂w 9. (a) ∂u ∂w (b) ∂u ∂u 11. (a) ∂x

67. 12 69. −2 c cos A − b ∂A a ∂A 71. = , = ∂a bc sin A ∂ b bc sin A ln υ 73. υ x = ( ln u )( ln υ ) − 1 75. (a) 3

(b)

= 4 t tan −1 t + 1, (b)

(b)

2x ∂w ∂w −3 x 2 = 3, = 59. y ∂y y4 ∂x 2 2 ∂ w −6 x ∂ w −6 x , = = y4 ∂x ∂y y4 ∂y ∂x 61. (a) x first (b) y first (c) x first (d) x first (e) y first (f) y first 63. f x ( 1, 2 ) = −13,   f y ( 1, 2 ) = −2 65. f x ( −2, 3 ) = 1 2,   f y ( −2, 3 ) = 3 4

dw (π ) = 0 dt dw ( ) = 1, (b) 3 =1 dt = 0,

1. (a)

'w 'x 'w 'y y

'x 'u

w

'w 'z 'y '  u 'z 'u

z

x

'w 'x 'w '  y y

'x 'y

'y 'y 'z 'y

z

y

u

17.

'w 'z

∂w ∂w ∂ x ∂w ∂ y ∂w ∂w ∂ x ∂w ∂ y , . = + = + ∂u ∂ x ∂u ∂ y ∂u ∂ υ ∂ x ∂υ ∂ y ∂υ w

'w 'x

w

'w 'y

x

'w 'x y

'x 'u

'y 'u u

'w 'y

x

y

'x 'y

'y 'y y

Chapter 13: Answers to Odd-Numbered Exercises

19.

∂z ∂z ∂ x ∂z ∂ y ∂z ∂z ∂ x ∂z ∂ y = + = + , ∂t ∂ x ∂t ∂ y ∂t ∂s ∂ x ∂s ∂ y ∂s z

'z 'y

x

'z 'x y

'x 't

∇f = 1 i + 3 j 2 4

'z 'y

x

'y 't

y

'x 's

(–1, 2)

4 = 2x + 3y x 2

s

7. ∇f = 3i + 2 j − 4 k

∂w dw ∂u ∂ w dw ∂u = , = ∂s du ∂ s ∂ t du ∂ t

u

1 5 1 i− j− k, ( D u   f ) P0 = 3 3; 3 3 3 3 3 3 1 5 1 −u = − i+ j+ k, ( D−u   f ) P0 = −3 3 3 3 3 3 3 3 23. u = 1 ( i + j + k ), ( D u   f ) P = 2 3; 0 3 1 ( −u = − i + j + k ), ( D−u   f ) P0 = −2 3 3 25. 27. 21. u =

'u 't

'u 's s

t

∂y ∂w ∂w ∂x ∂w ∂ y ∂w ∂ x  since  = + = = 0, 23. ∂r ∂ x ∂r ∂ y ∂r ∂ x ∂r ∂r ∂w ∂w ∂ x ∂w ∂ y ∂w ∂ y ∂x  since  = + = = 0 ∂s ∂ x ∂s ∂ y ∂s ∂ y ∂s ∂s w

'w 'x

w

'w 'y

x

'w 'x y

'x 'r

33. 39. 43. 53. 55.

57. 61.

y

29. 20

2

31.

x2 + y2 = 4

∂z 1 ∂z 3 = , = − ∂x 4 ∂y 4

∂z ∂z = −1, 35. 12 37. −7 = −1 ∂x ∂y ∂z ∂z ∂w ∂w = 2, =1 41. = 2t e s 3 + t 2 , = 3s 2 e s 3 + t 2 ∂u ∂υ ∂t ∂s 23 45. −16, 2 47. −0.00005  amp s ( cos 1, sin 1, 1 )  and  ( cos(−2), sin (−2), −2 ) ⎛ ⎛ 2 2 2 ⎞⎟ 2 ⎟⎞ (a) Maximum at ⎜⎜ − , , − ⎟; minimum ⎟ and ⎜⎜⎝ ⎝ 2 2 ⎟⎠ 2 2 ⎟⎠ ⎛ ⎛ 2 2 ⎞⎟ 2 ⎞⎟ 2 , − , at ⎜⎜ ⎟ ⎟ and ⎜⎜⎝ − ⎝ 2 2 ⎟⎠ 2 ⎟⎠ 2 (b) Max = 6; min = 2 x2 3x 2 5  °C s 59. 2 x x 8 + x 3 + ∫ dt 0 2 t4 + x3 0.19 m min 63. −8 61 rad min 3.

y

x

∇f = –2i + 2j –2

31. 33. 41. 43. 45. 49.

y

y – x = –1

1 0 –1

x= 2 y2

(2, 1) 1

2

x

x

(2, –1)

∇f = i – 4j

2

x

(2, –2)

3 4 i − j,   D u   f ( 1, −1 ) = 5 5 5 3 4 (b) u = − i + j,   D u   f ( 1, −1 ) = −5 5 5 4 3 4 3 (c) u = i + j,   u = − i − j 5 5 5 5 24 7 (d) u = − j,   u = i− j 25 25 7 24 (e) u = −i,   u = i+ j 25 25 7 2 7 2 u = i− j, −u = − i+ j 53 53 53 53 No, the maximum rate of change is 185 < 14. 35. −7 5 r (t ) = ( −3 − 6t ) i + ( 4 + 8t ) j, −∞ < t < ∞ r (t ) = ( 3 + 6t ) i + ( −2 − 4 t ) j + ( 1 + 2t ) k, −∞ < t < ∞ 1 14 1 4 ,  , −12, − 47. 30, 1, 90, , 9 3 6 9 e π 3 1 − 51. + 2 20e 2 2 12

SECTION 13.6, pp. 822–825 ∇f = – i + j 2

y=x–4

y = –x + 2Î2

29. (a) u =

SECTION 13.5, pp. 814–815

1.

xy = –4 (Î2, Î2)

s

27. −4 5

y

∇f = 2Î2i + 2Î2j

2

'y 's

'x = 0 's

r

y

'w 'y

x

'y =0 'r

25. 4 3

26 23 23 i+ j− k 27 54 54

13. 21 13 15. 3 17. 2 1 1 ( 1 1 i+ j, D u   f ) P0 = 2; −u = i− j, 19. u = − 2 2 2 2   ( D−u   f ) P0 = − 2

dw du

dw du

9. ∇f = −

11. −4

w

u

4 3

'y 's

t

w

y

z

'z 'x

21.

5.

AN-57

1. (a) (b) 3. (a) (b) 5. (a) (b) 7. (a) (b)

x+ y+z = 3 x = 1 + 2t , y = 1 + 2t , z = 1 + 2t 2x − z − 2 = 0 x = 2 − 4 t , y = 0, z = 2 + 2t 2x + 2 y + z − 4 = 0 x = 2t , y = 1 + 2t , z = 2 + t x + y + z −1 = 0 x = t, y = 1 + t, z = t

AN-58

Chapter 13: Answers to Odd-Numbered Exercises

9. (a) −x + 3 y + z e = 2 (b) x = 2 − t , y = 1 + 3t , z = e + ( 1 e ) t 11. 2 x − z − 2 = 0 13. x − y + 2 z − 1 = 0 15. x = 1, y = 1 + 2t , z = 1 − 2t 1 17. x = 1 − 2t , y = 1, z = + 2t 2 19. x = 1 + 90 t , y = 1 − 90 t , z = 3 9 ≈ 0.0008 23. dg = 0 21. df = 11,830 3 1 sin 3 − cos 3 ≈ 0.935  °C m 25. (a) 2 2 (b) 3 sin 3 − cos 3 ≈ 1.87  °C s 27. (a) L ( x , y ) = 1 (b) L ( x , y ) = 2 x + 2 y − 1 29. (a) L ( x , y ) = 3 x − 4 y + 5 (b) L ( x , y ) = 3 x − 4 y + 5 31. (a) L ( x , y ) = 1 + x π (b) L ( x , y ) = − y + 2 33. (a) W ( 30, −5 ) = − 13 °C, W ( 50, − 25 ) = −42.2 °C, W ( 30, −10 ) = −19.5 °C

35. 37. 41. 43.

45.

47. 49. 51. 53. 55. 61.

(b) W ( 15, −40 ) ≈ −53.7  °C, W ( 80, −40 ) ≈ −66.6  °C, W ( 90, 0 ) ≈ −10.2  °C (c) L ( υ, T ) ≈ 1.336 T − 0.11 υ − 2.952 (d) i) L ( 39, 9 ) ≈ −19.3  °C ii) L ( 42, −12 ) ≈ −23.6  °C iii) L ( 10, − 25 ) ≈ −37.5  °C L ( x , y ) = 7 + x − 6 y; 0.06 L ( x , y ) = x + y + 1; 0.08 39. L ( x , y ) = 1 + x; 0.0222 (a) L ( x , y, z ) = 2 x + 2 y + 2 z − 3 (b) L ( x , y, z ) = y + z (c) L ( x , y, z ) = 0 (a) L ( x , y, z ) = x 1 1 x+ y (b) L ( x , y, z ) = 2 2 1 2 2 (c) L ( x , y, z ) = x + y + z 3 3 3 (a) L ( x , y, z ) = 2 + x π (b) L ( x , y, z ) = x − y − z + + 1 2 π (c) L ( x , y, z ) = x − y − z + + 1 2 L ( x , y, z ) = 2 x − 6 y − 2 z + 6, 0.0024 L ( x , y, z ) = x + y − z − 1, 0.00135 Maximum error (estimate) ≤ 0.31 in magnitude Pay more attention to the smaller of the two dimensions. It will generate the larger partial derivative. f is most sensitive to a change in d. (a) 1.75% (b) 1.75%

SECTION 13.7, pp. 832–834

1. f ( −3, 3 ) = −5, local minimum 3. f ( −2, 1 ), saddle point 3 17 = , local maximum 5. f 3, 2 2 9. f ( 1, 2 ), saddle point 7. f ( 2, − 1) = − 6, local minimum 16 16 , 0 = − , local maximum 11. f 7 7 2 2 170 13. f ( 0, 0 ), saddle point; f − , = , local maximum 3 3 27 15. f ( 0, 0 ) = 0, local minimum; f ( 1, −1 ), saddle point 17. f ( 0, ± 5 ), saddle points; f ( −2, −1 ) = 30, local maximum; f ( 2, 1 ) = −30, local minimum 19. f ( 0, 0 ), saddle point; f ( 1, 1 ) = 2,   f ( −1, −1 ) = 2, local maxima

( )

(

)

(

)

21. 23. 25. 27. 29. 31. 33. 35. 37. 39. 41.

f ( 0, 0 ) = −1, local maximum f ( nπ , 0 ), saddle points, for every integer n f ( 2, 0 ) = e −4 , local minimum f ( 0, 0 ) = 0, local minimum; f ( 0, 2 ), saddle point 1 1 f , 1 = ln − 3, local maximum 2 4 Absolute maximum: 1 at ( 0, 0 ); absolute minimum: −5 at ( 1, 2 ) Absolute maximum: 4 at ( 0, 2 ); absolute minimum: 0 at ( 0, 0 ) Absolute maximum: 11 at ( 0, −3 ); absolute minimum: −10 at ( 4, − 2 ) 3 2 Absolute maximum: 4 at ( 2, 0 ); absolute minimum: at 2 π π π π 3, − ,  3,  ,  1, − , and 1,  4 4 4 4 a = −3,  b = 2 ⎛ 1 1° ⎛ 1 3 ⎞⎟ 3 ⎞⎟ Hottest is 2 at ⎜⎜ − ,  and ⎜⎜ − , − ⎟; coldest is ⎝ 2 2 ⎟⎟⎠ ⎝ 2 4 2 ⎟⎠ ° 1 1 − at , 0 . 4 2 (a) f ( 0, 0 ), saddle point (b) f ( 1, 2 ), local minimum (c) f ( 1, −2 ), local minimum; f ( −1, −2 ), saddle point 1 1 355 9 6 3 ,  ,  51. ,  ,  53. 3, 3, 3 55. 12 6 3 36 7 7 7 4 4 4 × × 59. 2 m × 2 m × 1 m 3 3 3 Points ( 0, 2, 0 ) and ( 0, −2, 0 ) have distance 2 from the origin. (a) On the semicircle, max f = 2 2 at t = π 4 , min f = −2 at t = π. On the quarter circle, max f = 2 2 at t = π 4 , min   f = 2 at t = 0,  π 2. (b) On the semicircle, max g = 2 at t = π 4 , min g = −2 at t = 3π 4. On the quarter circle, max g = 2 at t = π 4, min g = 0 at t = 0,  π 2. (c) On the semicircle, max h = 8 at t = 0,  π; min h = 4 at t = π 2. On the quarter circle, max h = 8 at t = 0, min h = 4 at t = π 2.

(

(

)

( )

)(

)(

(

43. 49. 57. 61. 63.

)

(

)

)

(

)

(

)

65.    i) min f = −1 2 at t = −1 2; no max   ii) max f = 0 at t = −1, 0; min f = −1 2 at t = −1 2 iii) max f = 4 at t = 1; min   f = 0 at t = 0 20 9 71 69. y = − x + ,  y = − 13 13 13 x=4 SECTION 13.8, pp. 841–844

1. 7. 9. 13. 15. 17. 23. 25. 29. 33. 39. 43.

( ± 12 ,  12 ), ( ± 12 , − 12 )

3. 39

5. ( 3, ±3 2 )

(a) 8 (b) 64 r = 2 cm,  h = 4 cm 11. Length = 4 2, width = 3 2 f ( 0, 0 ) = 0 is minimum; f ( 2, 4 ) = 20 is maximum. Lowest = 0 °, highest = 125 ° 3 5 , 2,  19. 1 21. ( 0, 0, 2 ), ( 0, 0, −2 ) 2 2 f ( 1, −2, 5 ) = 30 is maximum; f ( −1, 2, −5 ) = −30 is minimum. 2 2 2 3, 3, 3 27.  by   by   units 3 3 3 ( ±4 3, −4 3, −4 3 ) 31. (b) ≈24,322 units 4 ( ) U 8, 14 = $128 37. f ( 2 3, 4 3, −4 3 ) = 3 ( 2, 4, 4 ) 41. Maximum is 1 + 6 3 at ( ± 6,  3, 1 ); minimum is 1 − 6 3 at ( ± 6, − 3, 1 ). Maximum is 4 at ( 0, 0, ±2 ); minimum is 2 at ( ± 2, ± 2, 0 ).

(

)

Chapter 13: Answers to Odd-Numbered Exercises

SECTION 13.9, p. 848

1. Quadratic: x + xy; cubic: x + xy +

1 2 xy 2

3. Quadratic: xy; cubic: xy 1 5. Quadratic: y + ( 2 xy − y 2 ); 2 1( 1 cubic: y + 2 xy − y 2 ) + ( 3 x 2 y − 3 xy 2 + 2 y 3 ) 2 6 1( 2 2 7. Quadratic: 2 x + 2 y ) = x 2 + y 2  ; cubic: x 2 + y 2 2 9. Quadratic: 1 + ( x + y ) + ( x + y ) ; cubic: 1 + ( x + y ) + ( x + y ) 2 + ( x + y )3 1 1 11. Quadratic: 1 − x 2 − y 2 ;  E ( x , y ) ≤ 0.00134 2 2

AN-59

7. Domain: all ( x , y, z ) such that ( x , y, z ) ≠ ( 0, 0, 0 ); range: positive real numbers. Level surfaces are spheres with center (0, 0, 0) and radius r > 0. h(x, y, z) = 2 12 =1 x + y + z2 or x2 + y2 + z2 = 1

z 1

1

1

x

y

2

SECTION 13.10, p. 852

1. (a) 0 (b) 1 + 2 z (c) 1 + 2 z ∂U ∂U V ∂U nR ∂U 3. (a) + (b) + ∂P ∂T nR ∂P V ∂T 5. (a) 5 (b) 5 ∂x ∂r x 7. = cos θ = ∂r θ ∂x y x 2 + y2

( )

( )

( )

( )

PRACTICE EXERCISES, pp. 853–856

1. Domain: all points in the xy-plane; range: z ≥ 0. Level curves are ellipses with major axis along the y-axis and minor axis along the x-axis. y 3

9. −2 17. No;

1

∂g ∂g = cos θ + sin θ,  = −r sin θ + r cos θ ∂θ ∂r ∂f ∂f ∂f 1 1 1 21. = − 2 ,  = − 2 ,  = − 2 ∂ R1 R1 ∂ R 2 R 2 ∂ R3 R3 ∂P RT ∂ P nT ∂ P nR ∂ P nRT = ,  = ,  = ,  = − 2 ∂n V ∂R V ∂T V ∂V V ∂2g ∂2g ∂2g 2x ∂ 2 g 1 = 0,  2 = 3 ,  = = − 2 25. ∂x 2 ∂y y ∂y ∂x ∂x ∂y y ∂2 f ∂2 f ∂2 f 2 − 2x 2 ∂ 2 f = −30 x + 2 = 0,  = =1 27. 2 ,  ∂x 2 ∂y ∂x ∂x ∂y ( x + 1) ∂ y 2 23.

dw dt ∂w 31. ∂r 29.

33.

x

–3

35. 3. Domain: all ( x ,  y ) such that x ≠ 0 and y ≠ 0; range: z ≠ 0. Level curves are hyperbolas with the x- and y-axes as asymptotes. y

z=1 x

5. Domain: all points in xyz-space; range: all real numbers. Level surfaces are paraboloids of revolution with the z-axis as axis. z

( x ,  y )→( 0,  0 )

19.

z=9 –1

11. 1 2 13. 1 15. Let y = kx 2 ,  k ≠ 1 lim f ( x , y ) does not exist.

t=0

= −1

( r ,  s )=( π ,  0 )

df dt

t =1

dy dx

( x ,  y )=( 0,

= 2, 

∂w ∂s

( r ,  s )=( π ,  0 )

= −( sin 1 + cos 2 )( sin 1 ) + ( cos 1 + cos 2 )( cos 1 ) −2( sin 1 + cos 1 )( sin 2 ) 1)

= −1

2 2 i− j; 2 2 2 2 i+ j; decreases most rapidly in the direction −u = 2 2 v 2 2 7 Du   f = ;  D−u   f = − ; D u1   f = − where u 1 = v 2 2 10 2 3 6 39. Increases most rapidly in the direction u = i + j + k; 7 7 7 2 3 6 decreases most rapidly in the direction −u = − i − j − k; 7 7 7 v D u   f = 7;  D−u   f = −7;  D u1 f = 7 where u 1 = v 37. Increases most rapidly in the direction u = −

41. π 45.

2

43. (a) f x ( 1, 2 ) = f y ( 1, 2 ) = 2 z

x2 + y + z2 = 0

∇f 0 (0, –1, 1) = j + 2k f(x, y, z) = x 2 + y2 – z = –1 or z = x2 + y2 + 1

1 ∇f 0 (0, 0, 0) = j y

1

–1 y

x

= 2−π

x ∇f 0 (0, –1, –1) = j – 2k

(b) 14 5

AN-60

Chapter 14: Answers to Odd-Numbered Exercises

47. Tangent: 4 x − y − 5 z = 4; normal line: x = 2 + 4 t ,  y = −1 − t ,  z = 1 − 5t 49. 2 y − z − 2 = 0 51. Tangent: x + y = π + 1; normal line: y = x − π + 1 y

SECTION 14.1, pp. 863–864

y=x–p+1

1

z

y = 1 + sin x 1

2

x

p

z = 9 − x 2 − y2 (paraboloid)

9 8

53. x = 1 − 2t ,  y = 1,  z = 1 2 + 2t 55. Answers will depend on the upper bound used for f xx , f xy , f yy . With M = 2 2,  E ≤ 0.0142. With M = 1,  E ≤ 0.02. 57. L ( x , y, z ) = y − 3z ,  L ( x , y, z ) = x + y − z − 1 59. Be more careful with the diameter. 61. dI = 0.038, % change in I = 15.83%, more sensitive to voltage change 63. (a) 5% 65. Local minimum of −8 at ( −2, −2 ) 67. Saddle point at ( 0, 0 ),   f ( 0, 0 ) = 0; local maximum of 1 4 at ( −1 2, −1 2 ) 69. Saddle point at ( 0, 0 ),   f ( 0, 0 ) = 0; local minimum of −4 at ( 0, 2 ); local maximum of 4 at ( −2, 0 ); saddle point at ( −2, 2 ),   f ( −2, 2 ) = 0 71. Absolute maximum: 28 at ( 0, 4 ); absolute minimum: −9 4 at ( 3 2, 0 ) 73. Absolute maximum: 18 at ( 2, − 2 ); absolute minimum: −17 4 at ( −2, 1 2 ) 75. Absolute maximum: 8 at ( −2, 0 ); absolute minimum: −1 at ( 1, 0 ) 77. Absolute maximum: 4 at ( 1, 0 ); absolute minimum: −4 at ( 0, −1 ) 79. Absolute maximum: 1 at ( 0, ±1 ) and ( 1, 0 ); absolute minimum: −1 at ( −1, 0 ) 81. Maximum: 5 at ( 0, 1 ); minimum: −1 3 at ( 0, −1 3 ) 1 1 1 83. Maximum: 3 at , − ,  ; minimum: − 3 at 3 3 3 1 1 1 − ,  , − 3 3 3 13 2 c V b 2V 1 3 a 2V 1 3 , depth = , height = 85. Width = ab ac bc 1 1 3 1 1 87. Maximum: at , − , − 2 ; ,  ,  2 and − 2 2 2 2 2 1 1 1 1 1 minimum: at − ,  , − 2 and , − ,  2 2 2 2 2 2 sin θ ∂ w ∂ w cos θ ∂ w ∂w ∂w ∂w = cos θ − ,  = sin θ + 89. ∂r r ∂θ ∂r r ∂θ ∂ y ∂x 95. ( t , −t ± 4, t ),  t a real number ⎛ z ⎞⎟ 101. (a) ( 2 y + x 2 z )e yz (b) x 2e yz ⎜⎜ y − ⎟ (c) ( 1 + x 2 y )e yz ⎝ 2 y ⎟⎠

(

(

3. 1 5. 16 7. 2 ln 2 − 1 9. ( 3 2 )( 5 − e ) 13. ln 2 15. 3 2, −2 17. 14 19. 0 23. 2 ln 2 25. ( ln 2 ) 2

1. 24 11. 3 2 21. 1 2 27.

y = –x + p + 1

2

0

Chapter 14

)

)

( ) ( (

( ) ) ( ) (

(

5 4

2

1

y

x

(1, 2)

29. 8 3 31. 1 33. 2 3 39. (a) 1 3 37. ln 3 − 1 2 1.

3.

y

y = 2x

3abc r2 1 = ( x 2 + y2 + z 2 ) 13. V = 2 2 2 y x 9 17. f ( x , y ) = + 4,  g ( x , y ) = + 2 2 2 19. y = 2 ln sin x + ln 2 1 ( −1 ( 21. (a) 2 i + 7 j ) (b) 98 i − 127 j + 58k ) 53 29,097 23. w = e −c 2π 2t sin π x

y x = y2

6

2

4

x

x

3

−2

5.

7. y

y

y = ex

e

1

y=e x = arcsin y x

1

2

8

3

3x

9. (a)

∫0 ∫ x

11. (a)

∫0 ∫ x

13. (a)

∫0 ∫0

15. (a)

∫ 0 ∫e

17. (a)

∫0 ∫ x

(b)

∫0 ∫0

ADDITIONAL AND ADVANCED EXERCISES, pp. 856–858

7. (c)

(b) 2 3

SECTION 14.2, pp. 870–873

) ) )

1. f xy ( 0, 0 ) = −1,   f yx ( 0, 0 ) = 1

35. 2 27

19. 42

dy dx 3

2

9

ln 3

x

dy dx

1 −x

1

3− 2 x

1

y

8

∫0 ∫0

(b)

dy dx

dy dx

p 2 y1 3

dx dy

y

9

(b)

∫0 ∫ y 3 dx dy

(b)

∫0 ∫ y

(b)

3

9 2

dx dy

1

ln 3

∫1 3 ∫−ln y dx dy

dy dx

dx dy +

21. 21 40

3

( 3− y )

∫1 ∫0

23. π 2

2

dx dy 25. 49 5

x

AN-61

Chapter 14: Answers to Odd-Numbered Exercises

π2 +2 2

27.

29. 8 ln 8 − 16 + e y

y

ln 8

y

(ln ln 8, ln 8)

(p, p)

p

e3

3

∫1 ∫ln x ( x + y ) dy dx

53.

x=1

x=

3

55. 2 y

ey

p

(p, p)

x = ln y

1

y=x x

p

0

0

x

ln ln 8

1

31. e − 2 y

x

e3

e−2 2

57.

(1, 1)

1

59. 2 y

y x = y2

3 33. ln 2 2 37. 8

0

35. −1 10 39. 2π

y

−2

p

2

(−p3, 2)

y = −p

y=p

x

1

−2

61. 1 ( 80 π )

u 2

63. −2 3 y

1 16

1

(2, −2)

1

1 1 Q 2 , 16R

41.

( 4− y )

2

∫2 ∫0

dx dy

43.

1

∫0 ∫ x

x

y = 4 − 2x 2

−1

e

65. 4 3 75.

2

y=x x

0

1

∫1 ∫ln y dx dy

47.

67. 625 12

69. 16

y

( 9− y ) 2

∫0 ∫0

z

2

1

e

(1, e)

2

9

3

y = ex

y = 9 − 4x

x 3 x

x

1

0

1− x 2

1

∫−1 ∫0

0

3 y dy dx

x

3 2

51.

y

1

∫ 0 ∫e

e y

79. π 2

77. 1 85.

xy dx dy

1

2− x

∫0 ∫ x

81. −

3 32

( x 2 + y 2 ) dy dx =

83.

20 3 9

4 3

y

y

y=

2

2

y = ln x x 2 + y2 = 1

− x

1

y

2

(1, 1)

1

z = 1 − 1x − 1y 3 2

16 x dx dy

y

y

73. 2( 1 + ln 2 )

71. 20

x

1

9

x−y=1 −1

(1, 1) y=x

1

x

1

−x − y = 1

(1, 2)

0

x+y=1

dy dx

2

1

4

49.

x

1 2

y

y

45.

−x + y = 1

t

p 3

−p 3

x

y

u = sec t (p3, 2)

0 4

(Îln 3, 2Îln 3)

Îln 3

0

y = x4 (−2, −2)

y = 2x

x=y

x

1

2Îln 3

(1, 1)

1

0

x

p

0

y=

x

1

−1

0

1

1

x 1

e

x

x

87. R is the set of points ( x , y ) such that x 2 + 2 y 2 ≤ 4. 89. No, by Fubini’s Theorem, the two orders of integration must give the same result. 93. 0.603 95. 0.233

AN-62

Chapter 14: Answers to Odd-Numbered Exercises

SECTION 14.3, pp. 875–876 2

1.   ∫

0

2− x

∫0

2− y

2

∫0 ∫0

17.

dy dx = 2 or

3.

−y 2

1

∫−2 ∫ y−2 dx dy

9 2

=

y (−1, 2)

y

dx dy = 2

(−1, 1)

1

−4 y=x+2

2

−2

0

ex

ln 2

x = −y

∫0 ∫0

7.

2 y− y 2

1

∫0 ∫ y

y

2

dx dy =

SECTION 14.4, pp. 881–883

1 3

(1, 1)

1

x = y2

1 x = 2y − y2

9.

x

ln 2

0

2

3y

∫0 ∫ y 2

x

1

0

1 dx dy = 4 or

x

6

19. ( 2 ln 2 − 1 )( π 2 )

2

∫0 ∫ x 3 1 dy dx + ∫2 ∫ x 3 1 dy dx

y = Î1 − x2 or x = Î1 − y2

y=2

2

1

x

6

2x

2

3− x

0

∫ x 2 1 dy dx + ∫1 ∫ x 2

x

1

3 or 2 1 2y 2 3− y 3 ∫0 ∫ y 2 1 dx dy + ∫1 ∫ y 2 1 dx dy = 2 1 dy dx =

1− x 2

1

∫0 ∫0 y

25.

1

∫0 ∫0

xy dy dx or

xy dx dy

y=x

y = 2x or x = 1 y 2

2 2

y = 1 x or x = 2y 2 2

x

∫0 ∫0

1

1 2 3 y = 3 − x or x = 3 − y

13. 12

2 −1

y y = cos x y 2 = 3x

(12, 6)

y= x 2 12 NOT TO SCALE

Î2

2

(p4, Î2/2) y = sin x

x 0

p 4

x

y 2 ( x 2 + y 2 ) dy dx or

2

2

∫0 ∫ y

y 2 ( x 2 + y 2 ) dx dy

2a 27. 2( π − 2 ) 29. 12π 31. ( 3π 8 ) + 1 33. 3 2a 4 5π 39. + 35. 37. 2π ( 2 − e ) 3 8 3 1 π 41. (a) (b) 1 43. π ln 4, no 45. ( a 2 + 2h 2 ) 2 2 8( 47. 3π − 4 ) 9

x

15.

y

0

1− y 2

x=2

2

y

6

2)

y = 1x 3

y=x

2

3

2( 1 + 3

21.

y

23.

= 4

1

y

11.   ∫

π π 3π ≤ θ ≤ 2π ,  0 ≤ r ≤ 9 3. ≤ θ ≤ ,  0 ≤ r ≤ csc θ 2 4 4 π 5. 0 ≤ θ ≤ ,  1 ≤ r ≤ 2 3 sec θ; 6 π π ≤ θ ≤ ,  1 ≤ r ≤ 2 csc θ 6 2 π π π 9. 7. − ≤ θ ≤ ,  0 ≤ r ≤ 2 cos θ 2 2 2 13. 36 15. 2 − 3 17. ( 1 − ln 2 ) π 11. 2π 1.

y

y=e

x 2

19. (a) 0 (b) 4 π 2 21. 8 3 23. π − 2 25. 40,000 ( 1 − e −2 ) ln ( 7 2 ) ≈ 43,329

(ln 2, 2)

x

x

2 (2, −1)

−2

x

dy dx = 1

(0, 0)

x

2

(−4, −2)

5.

y=1−x

y=−

y=2−x

2

2

y = −2x

y

0

3 2

x

Chapter 14: Answers to Odd-Numbered Exercises

27. IL = 1386

SECTION 14.5, pp. 891–893

1. 1 6 3.   ∫

0

2− 2 x

∫0

3− 3 x − 3 y 2

∫0

2− 2 x − 2 z 3

3− 3 x

1

∫0 ∫0

∫0

3− 3 y 2

2

∫0 ∫0

dz dy dx ,  ∫

dy dz dx ,  ∫

1− y 2− z 3

∫0

2 0

3 0

dx dz dy,  ∫

1− y 2

∫0

1− z 3

3

∫0

2− 2 x − 2 z 3

∫0

0

3− 3 x − 3 y 2

∫0

2− 2 z 3

∫0

dz dx dy,

dy dx dz ,

1− y 2− z 3

∫0

dx dy dz.

The value of all six integrals is 1. 5.

(b) x = 4 5  cm,  y = z = 2 5  cm

29. (a) 4 3  gm 1

4− x 2

2

∫−2 ∫−

4− x 2

8− y 2

2

∫−2 ∫4

∫−

8− z

8

∫4 ∫−

8− z 8− x 2

2

∫−2 ∫4 8

∫4 ∫−

∫− ∫−

8− z 8− z

8− x 2 − y 2

∫x

∫−

8− z − y 2 8− z − y

1 dx dz dy + 2

8− z − y 2 8− z − y

1 dx dy dz + 2

8− z − x 2 8− z − x

2

1 dz dx dy,  ∫

−2

2 + y2

1 dy dz dx + 2

8− z − x 2 8− z − x 2

∫−

4− y 2

2

8− x 2 − y 2

4− y 2

∫x

4

z−y2

∫−2 ∫ y ∫−

z−y

2

∫0 ∫− z ∫− 4

2

∫−2 ∫ x ∫−

1 dy dx dz +

z−y

z

4

∫0 ∫− z ∫−

abc ( a 2 + b 2 ) 12 2 ( abc a + 7b 2 ) (b) IL = 3 39. μ x = μ y = 7 12 37. (a) I c.m. =

41. μ x = 3 4,  μ y = 2 3

SECTION 14.7, pp. 911–915

9. 6

15. 7 6 21. (a) (c) (e)

11.

17. 0 1

1− x 2

∫−1 ∫0 1

1− z

∫x

1− z

∫0 ∫0 ∫− 1

y

2

y y

1− y

∫0 ∫− y ∫0

23. 2 3

5( 2 − 3 ) 4 1 π 19. − 2 8 dy dz dx

z−x 2 z−x 2

−2

dx dy dz

(d)

1

1− z

1− z

1

1− y

∫0 ∫0 ∫−

y y

2

z = −1: z plane parallel to the xy-plane

−1

dy dx dz

dx dz dy

29. 16 3

31. 8π −

37. 31 3 39. 1 41. 2 sin 4 33. 2 35. 4π 43. 4 45. a = 3 or a = 13 3 47. The domain is the set of all points ( x ,   y,   z ) such that 4 x 2 + 4 y 2 + z 2 ≤ 4.

y

5.

32 3

9. x = −1,   y = 1 4 13. x = 3 8,   y = 17 16

11. Ix = 64 105 15. x = 11 3,   y = 14 27,   Iy = 432

17. x = 0,   y = 13 31,   Iy = 7 5 19. x = 0,   y = 7 10;   Ix = 9 10 kg m 2,   Iy = 3 10 kg m 2,

z

r = u: spiral in xy-plane

2p

x

− 3p 2 x

− 3p 2

7.

−2 2

y

−2

π 9. r ≤ z ≤ 9 − r 2 : cone with vertex angle below a sphere of 2 radius 3 centered at (0, 0, 0), and its interior z

I0 = 6 5 kg m 2

z = "9 − r 2

M( 2 M( 2 M( 2 a + b2 ) b + c 2 ) ,   Iy = a + c 2 ) ,   Iz = 3 3 3 23. x = y = 0,   z = 12 5,   Ix = 7904 105 ≈ 75.28,

3

21. Ix =

Iy = 4832 63 ≈ 76.70,   Iz = 256 45 ≈ 5.69 25. (a) x = y = 0,   z = 8 3

y

r 2 + z2 = 4 sphere of radius 2 centered at (0, 0, 0)

2

2

p 2

2p

5p 2

z

x

r = u: spiral cylinder

p 2

−p

SECTION 14.6, pp. 900–903

1. x = 5 14 ,   y = 38 35 3. x = 64 35,   y = 5 7 5. x = y = 4 a (3π ) 7. Ix = Iy = 4 π gm cm 2,   I0 = 8π gm cm 2

y

x

5p 2

27. 1

y

x

dz dx dy

25. 20 3

2

2

1 dy dx dz.

3.

∫0 ∫− 1−z ∫ x

r = 2: cylinder of radius 2 centered on the z-axis

1 dy dz dx ,

13. 18

(b)

z

1.

1 dx dy dz , 2

z−x 2

(c) Ix = Iy = Iz = 11 6

43. f ( x , y ) = 1 6,  P ( X < Y ) = 2 3

1 dx dz dy, 2

z−x 2

2

1 dz dx dy,

z−y2

z

4

2 + y2

31. (a) 5 2 (b) x = y = z = 8 15 33. 3 kg

The value of all six integrals is 16π. 7. 1

AN-63

(b) c = 2 2

−

9

Ä2

z=r

p 2 9

Ä2

x

y

AN-64

Chapter 14: Answers to Odd-Numbered Exercises

π , 0 ≤ z ≤ 5: half-cylinder of 2 height 5, radius 2, and tangent to the z-axis, and its interior

11. 0 ≤ r ≤ 4 cos θ, 0 ≤ θ ≤

z

5

4π ( 2 − 1) 17π 25. 3 5 3π 29. 31. π 3 10 33. (a) (b)

2

(c)

y

4 x

13. ρ = 3: sphere of radius 3 centered at ( 0, 0, 0 ) z

3

3 x

15. θ =

y

2π

2π

2 π: closed half-plane along the z-axis 3 z

3

1

∫0 ∫0 ∫0 4 −r 2

1

∫0 ∫0

π2

cos θ

37.

∫0 ∫0

39.

∫−π 2 ∫1

41.

∫0 ∫0 ∫0

π

2 sin θ

r dz dr d θ

r dr dz d θ + 2π

∫0

3r 2

∫−π 2 ∫0 ∫0

π4

sec θ

4

∫0

2

2−r sin θ

47. 5π

2π

π6

2

2π

π2

csc φ

2π

2

sin −1( 1 ρ )

2π

2

π6

2π

1

π2

∫0 ∫0 ∫0

r cos f = 4

x

19. 0 ≤ ρ ≤ 3 csc φ ⇒ 0 ≤ ρ sin φ ≤ 3: cylinder of radius 3 centered on the z-axis, and its interior z

31π 6 π 2π 1− cos φ 8 π 57. ∫ ∫ ∫ ρ 2 sin φ dρ d φ d θ = 0 0 0 3 π2 2π 2 cos φ π 59. ∫ ∫ ∫ ρ 2 sin φ dρ d φ d θ = π4 0 0 3 2π

π2

61. (a) 8  ∫

(c) 8  ∫

−3 y

3

63. (a)

x

21. 0 ≤ ρ cos θ sin φ ≤ 2,  0 ≤ ρ sin θ sin φ ≤ 3, 0 ≤ ρ cos φ ≤ 4: rectangular box 2 × 3 × 4 , and its interior z

73.

2

3 y

π2 0 π2 0

π2

2

2

4 −r 2

=

∫0 ∫0 ρ 2 sin φ dρ dφ d θ ∫0 ∫0 4− x 2

2

0 ∫0

2π

π3

2π

3

(b)

∫0 ∫0 ∫1

(c)

∫− 3 ∫−

3

∫0

r dz dr d θ

4− x 2 − y 2

dz dy dx

2

4 −r 2

3− x 2 3− x 2

67. 9 4

( 4 2 − 5 )π 3

∫1

r dz dr d θ 4− x 2 − y 2

69. 75. π 2

dz dy dx

3π − 4 18

(d) 5π 3

2 πa 3 3 4 ( 2 2 − 1) π 77. 3 71.

4π ( 8 − 3 3 ) 85. 2 3 87. 3 4 3 89. x = y = 0,   z = 3 8 91. ( x , y , z ) = ( 0, 0, 3 8 ) 81. 5π 2

x

2

∫0 ∫0 ∫sec φ ρ 2 sin φ dρ dφ d θ

65. 8π 3

4

ρ 2 sin φ d φ dρ d θ +

∫0 ∫0 ∫cos φ ρ 2 sin φ dρ dφ d θ

(b) 8  ∫

3

ρ 2 sin φ d φ dρ d θ +

∫0 ∫0 ∫π 6 ρ 2 sin φ dφ dρ d θ 55.

y

⎛ 8 − 5 2 ⎟⎞ 51. ⎜⎜ ⎟⎟⎠ π ⎝ 2

49. 2π

ρ 2 sin φ dρ d φ d θ

∫0 ∫1 ∫π 6

(b)

17. ρ cos φ = 4: plane with z-intercept 4 and parallel to the xy-plane

4

f ( r , θ, z ) r dz dr d θ

∫0 ∫0 ∫0 ρ 2 sin φ dρ dφ d θ +

x

z

r dr dz d θ

f ( r , θ, z ) r dz dr d θ

45. π 3

53. (a)

y

∫0 ∫ 3 ∫0

4−z 2

f ( r , θ, z ) r dz dr d θ

∫0 ∫π 6 ∫0 u = 3p

2

f ( r , θ, z ) r dz dr d θ

4 −r sin θ

1+ cos θ

2π

r d θ dz dr

∫0

π2

43. π 2

−3

4 −r 2

1

∫0 ∫0 ∫0

35.

3

−3

27. π ( 6 2 − 8 )

23.

83.

79. 16π

Chapter 14: Answers to Odd-Numbered Exercises

a 4 hπ 93. x = y = 0,   z = 5 6 95. Ix = π 4 97. 10 π 4 99. (a) ( x , y , z ) = 0, 0, ,   Iz = 5 12 π 5 (b) ( x , y , z ) = 0, 0, ,   Iz = 6 14 3M 101. πR 3 103. The surface’s equation r = f ( z ) tells us that the point ( r , θ, z ) = ( f ( z ), θ, z ) will lie on the surface for all θ. In particular, ( f ( z ), θ + π, z ) lies on the surface whenever ( f ( z ), θ, z ) lies on the surface, so the surface is symmetric with respect to the z-axis.

( (

5.

) )

z

( f(z), u, z) f(z)

z

( f(z), u + p, z) f(z)

u+υ υ − 2u 1 ,  y = ;  3 3 3 (b) Triangular region with boundaries u = 0,   υ = 0, and u + υ = 3 1 1( 1 (a) x = ( 2u − υ ),   y = 3υ − u );   5 10 10 (b) Triangular region with boundaries 3υ = u,   υ = 2u, and 3u + υ = 10 2 3 2u 52 64 5 9. ∫ ∫ ( u + υ ) du d υ = 8 + ln 2 1 1 3 υ πab ( a 2 + b 2 ) 1 3 13. 1 + 2 ≈ 0.4687 4 3 e 225 a 2 b 2c 2 17. 12 19. 16 6

3.

7. 11. 15.

(

21. (a)

−u sin υ

sin υ

u cos υ

sin υ

(b) 27.

cos υ

ln 17 4 π−2 19. π 21. 4 2( 31 − 3 5 2 ) 29. 3

u cos υ

cos υ −u sin υ

= u

υ+u

sin 2

11.

2

∫− 2 ∫− 2π

π4

2π

π4

1

3− x 2

2

sec φ

∫0 ∫ 1− x ∫1

∫

3

23. 0

ρ 2 sin φ dρ d φ d θ =

3− x 2

∫1

0

3

27. π 2

3 dz dx dy

x 2 + y2

4− x 2 − y 2

x 2 + 4y2 = 9

17. 1 4

25. 8 35

4− x 2 − y 2

3 2

15. 4 3

13. 4 3

(c) 2π ( 8 − 4 2 )

3 ρ 2 sin φ dρ d φ d θ

2

∫1 ∫0

y dy dx =

π 3

z 2 xy dz dy dx

4− x 2 − y 2

z 2 xy dz dy dx

8π ( 4 2 − 5 ) 8π ( 4 2 − 5 ) (b) 3 3 8πδ ( b 5 − a 5 ) 1 39. Iz = 41. x = y = 15 2 − ln 4

37. (a)

43. I0 = 104

45. Ix = 2δ

47. M = 4,   Mx = 0,   My = 0 51. (a) x =

49. x =

15π + 32 ,  y = 0 6π + 48

3 3 ,  y = 0 π

y

(b)

r = 1 + cos u

υ = u

1

r=1

c.m.

= −u sin 2 υ − u cos 2 υ = −u

1 ≈ 1.18

2

x

−1

ADDITIONAL AND ADVANCED EXERCISES, pp. 927–929

PRACTICE EXERCISES, pp. 925–927

1. 9e − 9

3. 9 2

y

1. (a)

t

10 (110, 10)

3 2

y = 1x

0

2− y 2

∫0 ∫0 ∫0

3 ln 2 2

1

2− y 2

∫0 ∫0 ∫0

 +

)

cos 2

−3

x

9. sin 4

35.

1. (a) x =

y = 2x + 4

−2

33.

SECTION 14.8, pp. 923–924

( 1 2 ) 9− x 2

3

∫−3 ∫0

7.

−3

s2 + 4t 2 = 9

(1, 1) 1

3

x NOT TO SCALE

s

2

6− x 2

∫−3 ∫ x

9 2

y

4

(b) y

4 3

x = −Î 4 − y

u u+p

=

y

31. (a)

x

4− x 2

0

∫−2 ∫2 x + 4 dy dx

AN-65

x 2 dy dx

(b)

2

6− x 2

∫−3 ∫ x

x2

∫0

dz dy dx

(c) 125 4 3. 2π 5. 3π 2 (b) 4 3π 7. (a) Hole radius = 1,  sphere radius = 2 b 4 11. ln 15. 1 3 9. π 4 a 17. Mass = a 2 cos −1  b − b a 2 − b 2 , a a4 b b3 b3 1  − I0 = − a 2 − b 2 − ( a 2 − b 2 )3 2 cos a 2 2 6

( )

( ) ( )

x

AN-66

19.

Chapter 15: Answers to Odd-Numbered Exercises

1 a 2b 2 − 1) (e ab

21. (b) 1

25. h = 180  cm,   h = ⎡1 1 2⎤ 27. 2π ⎢ −   ⎥ ⎢⎣ 3 3 2 ⎥⎦

SECTION 15.3, pp. 959–961

(c) 0

1. Conservative

3. Not conservative 3y 2 + 2z 2 + C 7. f ( x , y, z ) = x 2 + 2 9. f ( x , y, z ) = xe y + 2 z + C

540  cm

( )

15. −16 17. 1 ⎛ x 2 − 1 ⎞⎟ F = ∇⎜⎜ 29. (a) 1 (b) 1 (c) 1 ⎝ y ⎟⎟⎠ (a) 2 (b) 2 33. (a) c = b = 2a (b) c = b = 2 It does not matter what path you use. The work will be the same on any path because the field is conservative. The force F is conservative because all partial derivatives of M, N, and P are zero. f ( x , y, z ) = ax + by + cz + C;   A = ( xa, ya, za ) and B = ( xb, yb, zb ). Therefore, ∫ F ⋅ d r = ( ) ( ) ( ) f ( B) −  f ( A) = a xb − xa + b yb − ya + c zb − za = F ⋅ AB.

13. 49

SECTION 15.1, pp. 935–937

27.

1. Graph (c)

17. 21. 25. 31. 35.

3. Graph (g) 5. Graph (d) 7. Graph (f) 1 13 2 11. 15. ( 5 5 + 9 ) 13. 3 14 6 2 1 b 32 ( 17 − 1 ) 3 ln 19. (a) 4 5 (b) a 12 15 ( 16 1 ( 40 3 2 − 13 3 2 ) e − e 64 ) 23. 32 27 1 32 10 5 − 2 27. 29. 8 ( 5 + 7 2 − 1) 6 3 1 ( 17 3 2 − 1 ) 33. 2 2 − 1 6 (a) 4 2 − 2 (b) 2 + ln ( 1 + 2 ) 37. Iz = 2πδ a 3

( )

39. (a) Iz = 2π 2δ

(b) Iz = 4 π 2δ

41. I x = 2π − 2

SECTION 15.2, pp. 947–950

1. ∇f = −( xi + yj + zk )( x 2 + y 2 + z 2 )−3 2 ⎛ 2 x ⎞⎟ ⎛ 2 y ⎞⎟ 3. ∇g = −⎜⎜ 2 i − ⎜⎜ 2 j + e zk ⎝ x + y 2 ⎟⎟⎠ ⎝ x + y 2 ⎟⎟⎠ ky kx 5. F = − 2 32 i − 3 2 j, any k > 0 ( x + y2 ) ( x 2 + y2 )

7. 9. 11. 13. 19. 29. 31. 35. 39. 43. 47.

(a) 9 2 (b) 13 3 (c) 9 2 (a) 1 3 (b) −1 5 (c) 0 (a) 2 (b) 3 2 (c) 1 2 −15 2 15. 36 17. (a) −5 6 (b) 0 (c) −7 12 12 21. −π 23. 69 4 25. −39 2 27. 25 6 (a) Circ 1 = 0,  circ 2 = 2π ,  flux 1 = 2π ,  flux 2 = 0 (b) Circ 1 = 0,  circ 2 = 8π ,  flux 1 = 8π,  flux 2 = 0 Circ = 0,  flux = a 2 π 33. Circ = a 2 π ,  flux = 0 π (a) − (b) 0 (c) 1 37. ( .0001 ) π kg s 2 (a) 32 (b) 32 (c) 32 41. 115.2 J 5 3 − ( 3 2 ) ln 2 m 2 s 45. 5 3 g s

x2 + y2 = 4

49. (a) G = − yi + xj xi + yj 51. F = − x 2 + y2

x

(b) G = 55. 48

x 2 + y2F 57. π

59. 0

61.

37.

SECTION 15.4, pp. 971–973

2y − 1 3. ye x − xe y 5. sin y − sin x Flux = 0,  circ = 2πa 2 9. Flux = −πa 2 ,  circ = 0 Flux = 2,  circ = 0 13. Flux = −9,  circ = 9 Flux = −11 60 ,  circ = −7 60 Flux = 64 9,  circ = 0 19. Flux = 1 2,  circ = 1 2 Flux = 1 5,  circ = − 1 12 23. 0 25. 2 33 0 29. −16π 31. πa 2 33. 3π 8 (a) 0 if C is traversed counterclockwise (b) ( h − k )( area of the region ) 37. 0 45. (a) 4 π (b) 4 π if ( 0, 0 ) lies inside K, 0 otherwise 1. 7. 11. 15. 17. 21. 27. 35.

1. r ( r , θ ) = ( r cos θ ) i + ( r sin θ ) j + r 2 k,  0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π 3. r ( r , θ ) = ( r cos θ ) i + ( r sin θ ) j + ( r 2 ) k,  0 ≤ r ≤ 6, 0 ≤ θ ≤ π 2 5. r ( r , θ ) = ( r cos θ ) i + ( r sin θ ) j + 9 − r 2 k, 0 ≤ r ≤ 3 2 2,  0 ≤ θ ≤ 2π; Also: r ( φ, θ ) = ( 3 sin φ cos θ ) i + ( 3 sin φ sin θ ) j + ( 3 cos φ ) k,  0 ≤ φ ≤ π 4 ,  0 ≤ θ ≤ 2π 7. r ( φ, θ ) = ( 3 sin φ cos θ ) i + ( 3 sin φ sin θ ) j + ( 3 cos φ ) k,   π 3 ≤ φ ≤ 2π 3,  0 ≤ θ ≤ 2π 9. r ( x ,   y ) = xi + yj + ( 4 − y 2 ) k,  0 ≤ x ≤ 2, −2 ≤ y ≤ 2 11. r ( u, υ ) = u i + ( 3 cos υ ) j + ( 3 sin υ ) k,  0 ≤ u ≤ 3, 0 ≤ υ ≤ 2π 13. (a) r ( r , θ ) = ( r cos θ ) i + ( r sin θ ) j + ( 1 − r cos θ − r sin θ ) k,  0 ≤ r ≤ 3,  0 ≤ θ ≤ 2π (b) r ( u, υ ) = ( 1 − u cos υ − u sin υ ) i + ( u cos υ ) j + ( u sin υ ) k, 0 ≤ u ≤ 3,  0 ≤ υ ≤ 2π 15. r ( u, υ ) = ( 4 cos 2 υ ) i + u j + ( 4 cos υ sin υ ) k,  0 ≤ u ≤ 3, −( π 2 ) ≤ υ ≤ ( π 2 ); Another way: r ( u, υ ) = ( 2 + 2 cos υ ) i + u j + ( 2 sin υ ) k,  0 ≤ u ≤ 3,  0 ≤ υ ≤ 2π

2

2

31. 35.

SECTION 15.5, pp. 981–983

y

0

1 ( 2 ln y + z 2 ) + C 2 19. 9 ln 2 21. 0 23. −3

11. f ( x , y, z ) = x ln x − x + tan ( x + y ) +

Chapter 15

9.

5. Not conservative

1 2

2π

1

2π

3

17.

∫0 ∫0

19.

∫0 ∫1

5 π 5 r dr d θ = 2 2 r 5 dr d θ = 8π 5

Chapter 16: Answers to Odd-Numbered Exercises

2π

4

2π

1

21.

∫0 ∫1

23.

∫0 ∫0

25.

2π

17. −π 25. −8π

1 du d υ = 6π

( 5 5 − 1)

u 4u 2 + 1 du d υ =

π

∫0 ∫π 4

27.

6

π

29. 2

z = Îx + y

1. 9. 17. 25. 27.

z

2

x 2 + ( y – 3)2 = 9 Á3x + y = 9 x + y − Î2z = 0

(Î2, Î2, 2)

x

x

y

y

 

2π

π

∫0 ∫0

a 2c 2

[ a 2 b 2 sin 2 φ cos 2 φ + b 2c 2 sin 4 φ cos 2 θ +

sin 4 φ sin 2 θ ]1 2 dφ dθ

35. x 0 x + y 0 y = 25

37. 13π 3

39. 4

43. π +1 41. 6 6 − 2 2 π 45. ( 17 17 − 5 5 ) 47. 3 + 2 ln 2 6 2 π 49.   ( 13 13 − 1 ) 51. 5π 2 53. ( 5 5 − 1 ) 3 6 c2

1.

∫∫

x dσ =

∫∫

x 2 dσ =

∫∫

z dσ =

S

3.

∫0 ∫0 2π

S

π

1

17 17 − 1 4

sin 3 φ cos 2 θ d φ d θ =

1

∫0 ∫0 ( 4 − u − υ )

∫∫

x 2 5 − 4z dσ =

S

u 4u 2 + 1 d υ du = 9. 9a 3

11.

2π

1

∫0 ∫0

45. 49.

4π 3

55.

3 d υ du = 3 3

1

2π

∫0 ∫0

u 2 cos 2 υ ⋅

abc ( ab + ac + bc ) 4

13. 2

1 πa 3 17. 6 30 19. −32 21. ( 2 + 6 6) 30 6 4 23. 13a 6 25. 2π 3 27. −73π 6 29. 18 15.

πa 3 37. −32 2 a a a 41. 3a 4 43. ,  ,  39. −4 2 2 2 14 15π 2 45. ( x ,   y ,   z ) = 0,  0,   ,  I z = δ 9 2 8π 4 47. 49. 70 3 mg a δ 3 31.

πa 3 6

33.

πa 2 4

(

35.

(

)

)

37. (a) 1 − e −2 π

41. (a) 4 2 − 2 (b) 2 + ln ( 1 + 2 ) 16 2 232 64 56 ( x ,   y ,   z ) = 1, , ;   Ix = ,  I = ,  I = 15 3 45 y 15 z 9 3 7 3 z = ,   Iz = 47. ( x , y , z ) = ( 0, 0, 49 12 ), Iz = 640 π 2 3 Flux: 3 2;  circ: −1 2 53. 3 2π 57. 0 59. π (7 − 8 2) 3

(

)

11π 12

1. 6π 3. 2 3 5. (a) F ( x ,   y,   z ) = zi + xj + yk (b) F ( x ,   y,   z ) = zi + yk (c) F ( x ,   y,   z ) = zi 16πR 3 7. 9. a = 2,   b = 1. The minimum flux is −4. 3 16 16 11. (b) g (c) Work = ∫ gxy ds y = g ∫ xy 2 ds = g C C 3 3 4 13. (c) πw 19. False if F = yi + xj 3

(

Chapter 16 SECTION 16.1, pp. 1030–1032

1. (d) 5.

3. (a) y

x

SECTION 15.7, pp. 1004–1006

1. −i − 4 j + k 3. ( 1 − y ) i + ( 1 − z ) j + ( 1 − x ) k 5. x ( z 2 − y 2 ) i + y ( x 2 − z 2 ) j + z ( y 2 − x 2 ) k 7. 4 π 9. −5 6 11. 0 13. −6π 15. 2πa 2

(b) 1 − e −2 π

ADDITIONAL AND ADVANCED EXERCISES, pp. 1021–1023

4u 2 + 1 ⋅

u 3 ( 4u 2 + 1 ) cos 2 υ d υ du =

)

39. 0

(for x = u,   y = υ ) 7.

1. Path 1: 2 3; path 2: 1 + 3 2 3. 4 a 2 5. 0 7. 8π sin(1) 9. 0 11. π 3 1 abc 1 1 1 15. + 2 + 2 17. 50 13. 2π 1 − 2 a2 b c 2 19. r ( φ, θ ) = ( 6 sin φ cos θ ) i + ( 6 sin φ sin θ ) j + ( 6 cos φ ) k, π 2π ≤φ ≤ ,  0 ≤ θ ≤ 2π 6 3 21. r ( r , θ ) = ( r cos θ ) i + ( r sin θ ) j + ( 1 + r ) k,  0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π 23. r ( u, υ ) = ( u cos υ ) i + 2u 2 j + ( u sin υ ) k,  0 ≤ u ≤ 1, 0 ≤ υ ≤ π 25. 6 27. π [ 2 + ln ( 1 + 2 ) ] 29. Conservative 31. Not conservative 33. f ( x ,   y,   z ) = y 2 + yz + 2 x + z

43.

u 4u 2 + 1 du d υ =

∫0 ∫0

S

5.

2

PRACTICE EXERCISES, pp. 1019–1021

35. Path 1: 2; path 2: 8 3

SECTION 15.6, pp. 991–993 3

0 3. ( y 2 z + xz 2 + x 2 y )e xyz 5. 0 7. 0 −16 11. −8π 13. 3π 15. −40 3 12π 19. 12π ( 4 2 − 1 ) 23. No The integral’s value never exceeds the surface area of S. 184 35

(

6 3Á3 , 9/2, 0 2

33. (b) A =

23. −15π

SECTION 15.8, pp. 1016–1018

2 sin φ d φ d θ = ( 4 + 2 2 ) π z

19. 12π 21. −π 4 33. 16 I y + 16 Ix

AN-67

7. y′ = x − y; y(1) = −1 9. y′ = −( 1 + y ) sin x; y(0) = 2 11. y′ = 1 + x e y ; y ( −2 ) = 2

)

AN-68

Chapter 16: Answers to Odd-Numbered Exercises

13.

f ( y) =

y −3

15. 17. 19. 21. 23. 27.

SECTION 16.3, pp. 1043–1044

dy dx

1. (a) 168.5 m (b) 41.13 s 3. s(t ) = 4.91( 1 − e −( 22.36 39.92 )t )

y 3

2

5. x 2 + y 2 = C

2

−2

   1.5

−1

0

0

−1

1

−0.75

2

0

3

1

1 −3 −2 −1

7. ln y −

1 2 1 y = x2 + C 2 2 y

y 1

2

x

3

−1 −2 x

−3

x

y ( exact ) = −x 2 , y1 = − 2, y 2 = − 3.3333, y 3 = −5 y ( exact ) = 3e x( x + 2 ) , y1 = 4.2, y 2 = 6.216, y 3 = 9.697 y ( exact ) = e x 2 + 1, y1 = 2.0, y 2 = 2.0202, y 3 = 2.0618 y ≈ 2.48832; exact value is e. y ≈ −0.2272; exact value is 1 ( 1 − 2 5 ) ≈ −0.2880.

kx 2 + y 2 = 1

9. y = ± 2 x + C y

x

29.

31.

y

y

4

4

3

3

2

2

1 −4 −3 −2 −1 −1

1 1

2

3

4

x

−4 −3 −2 −1 −1

−2

−2

−3

−3

−4

−4

1

2

3

4

x

39. Euler’s method gives y ≈ 3.45835; the exact solution is y = 1 + e ≈ 3.71828. 41. y ≈ 1.5000; exact value is 1.5275. SECTION 16.2, pp. 1036–1038

ex + C , x > 0 x 1 1 C = − + 2, x > 0 2 x x = x ( ln x ) 2 + C x t3 t = − + ( ( t − 1) 4 3 t − 1) 4 = ( csc θ )( ln sec θ + C ) , 3 1 = − e −2 t 2 2 2 ex 2 = 6e x − x +1

1. y = 5. y 9. y 11. s 13. r 15. y 19. y

V V − e −3 R R 1 29. y = 1 + Ce −x

27. (a) i =

C − cos x , x > 0 x3 1 7. y = xe x 2 + Ce x 2 2

3. y =

C ( t − 1) 4 0 < θ < π2 1 π 17. y = − cos θ + θ 2θ 21. y = y 0 e kt

L ln 2 sec R V V = ( 1 − e −3 ) ≈ 0.95 amp (b) 86% R R

23. (b) is correct, but (a) is not.

(b) ( 400 + 4t ) L

13. (a) 4 kg min

( 100y+ t ) kg min

4y dy = 4− , y (0) = 20, 100 + t dt 60 y = 0.8( 100 + t ) − t 4 1+ 100 y (25) (e) Concentration = amt. brine in tank

(d)

(

)

75.424 ≈ 0.151 kg L 500 15. y (27.8) ≈ 5.93 kg, t ≈ 27.8 min =

SECTION 16.4, pp. 1052–1053

1. y′ = ( y + 2 )( y − 3 ) (a) y = −2 is a stable equilibrium value, and y = 3 is an unstable equilibrium value. 1 (b) y ″ = 2( y + 2 ) y − ( y − 3 ) 2

(

)

y′ > 0 −4

y′ < 0 −2

y″ < 0

0

y′ > 0 2 y″ < 0

y″ > 0

4 y″ > 0

y

0.5 y

(c) 4

25. t =

31. y 3 = 1 + Cx −3

(c) 4

y = 12

y′ > 0, y″ > 0

2

−0.5

y′ < 0, y″ < 0 0.5

−2

1

1.5

x

y′ < 0, y″ > 0 y′ > 0, y″ < 0

Chapter 16: Answers to Odd-Numbered Exercises

3. y′ = y 3 − y = ( y + 1 ) y ( y − 1 ) (a) y = −1 and y = 1 are unstable equilibria, and y = 0 is a stable equilibrium. (b) y ″ = ( 3 y 2 − 1 ) y′ = 3( y + 1 )( y + 1

3 ) y( y − 1

AN-69

dP 1 = 1 − 2 P has a stable equilibrium at P = ; dt 2 d 2P dP = −2 = −2( 1 − 2 P ). dt 2 dt

9.

P

3 )( y − 1 ) 1.5

y′ < 0 −1.5

y′ > 0 −1

y″ < 0

y′ < 0

y′ > 0

−0.5 0 0.5 y″ > 0 y″0 y″ < 0

1.5 y″ > 0

y

1 P′ < 0, P″ > 0 0.5

−

1

1

Ë3

Ë3 0.25

y

(c)

y′ > 0, y″ > 0

0.75

1

1.25

1.5

1.75

t

dP = 2 P ( P − 3 ) has a stable equilibrium at P = 0 and an dt dP d 2P = unstable equilibrium at P = 3; 2 = 2( 2 P − 3 ) dt dt 4 P ( 2 P − 3 )( P − 3 ).

11.

y′ < 0, y″ < 0 0.5 −0.5 −0.5

0.5

1

1.5

2

y′ < 0, y″ > 0 x

2.5

y′ > 0, y″ < 0

P′ > 0

y′ > 0, y″ > 0

−1 −0.5 P″ < 0

y′ < 0, y″ < 0

−1.5

1

2

y

3

y

4

4

y′ < 0

6

8

P′ < 0, P″ < 0

1

P′ < 0, P″ > 0 0.1

0.2

0.3

0.4

0.5

0.6

P

t 0.7 P′ > 0, P″ < 0

Before Catastrophe

x

y′ > 0 y″ > 0

y″ < 0

y′ < 0 2

P

M0 Pc M1

y″ > 0

3 y″ < 0

y

mg dυ k = g − υ2 = 0 ⇒ υ = dt m k d 2υ k dυ k k υ g − υ2 υ Concavity: 2 = −2 = −2 dt m dt m m (a) (b)

y″ > 0

Equilibrium:

( )

6 + Ë3 < 2.58 3

4 3.5 3 2.5 2 1.5

y′ > 0, y″ > 0 y′ < 0, y″ < 0 y′ < 0, y″ > 0 y′ > 0, y″ < 0 y′ > 0, y″ > 0 y′ < 0, y″ < 0

1 0.5 2

3

x

t

tcatastrophe

dυ k = g − υ 2, g, k , m > 0 and υ(t ) ≥ 0 dt m

15.

y′ > 0 4

t

tcatastrophe

y

1

After Catastrophe

P

6 − Ë3 < 1.42 3

(c)

3.5 4 P″ > 0

2

⎛ ⎛ 6 − 3 ⎞⎟( 6 + 3 ⎞⎟( 3( y − 1 )⎜⎜ y − ⎟ y − 2 )⎜⎜⎝ y − ⎟⎟⎠ y − 3 ) ⎝ 3 ⎟⎠ 3

1

3

−2

7. y′ = ( y − 1 )( y − 2 )( y − 3 ) (a) y = 1 and y = 3 are unstable equilibria, and y = 2 is a stable equilibrium. (b) y′′ = ( 3 y 2 − 12 y + 11 )( y − 1 )( y − 2 )( y − 3 ) =

y″ < 0

2 2.5 P″ < 0

13. Before the catastrophe, the population exhibits logistic growth and P (t ) increases toward M 0 , the stable equilibrium. After the catastrophe, the population declines logistically and P (t ) decreases toward M 1 , the new stable equilibrium.

2

0

1.5

P′ > 0, P″ > 0

−1

y′ > 0 y″ > 0

−2

0.5 1 P″ > 0

3

y″ > 0

17.5 15 12.5 10 7.5 5 2.5

0

P′ > 0

4

y′ > 0 0

P′ < 0

p

5. y′ = y , y > 0 (a) There are no equilibrium values. 1 (b) y′′ = 2

−1

0.5

−0.5

1.5

(c)

P′ > 0, P″ < 0

0

dy >0 dt 2 d y 0 dt d 2i k 2 and m n > k1 , we ensure that an equilibrium point exists inside the first quadrant. PRACTICE EXERCISES, pp. 1059–1060

(

2( 4 x − 2 ) 5 2 − ( x − 2 )3 2 5 3 3. tan y = −x sin x − cos x + C 5. ( y + 1 ) e − y = − ln x + C 1. y = − ln C −

31. y(3) ≈ 0.9131 33. (a)

[−0.2, 4.5] by [−2.5, 0.5]

(b) Note that we choose a small interval of x values because the y values decrease very rapidly and our calculator cannot handle the calculations for x ≤ −1. (This occurs because the analytic solution is y = −2 + ln ( 2 − e −x ) , which has an asymptote at x = − ln 2 ≈ −0.69. Obviously, the Euler approximations are misleading for x ≤ −0.7.)

[−1, 0.2] by [−10, 2]

35. y ( exact ) =

1 1 2 3 x − ; y(2) ≈ 0.4; exact value is . 2 2 2

37. y ( exact ) = −e ( x −1) 2 ; y(2) ≈ − 3.4192; exact value is −e 3 2 ≈ −4.4817. 39. (a) y = −1 is stable and y = 1 is unstable. d 2y dy (b) = 2y = 2 y( y 2 − 1) dx 2 dx 2

y = −1

)

y=1

dy >0 dx

dy 3

2 1 1 2 3 4 5

x

-2-1 -1

1

2

4

5

x

6

-4 -5

13.  [ −2, 8 ]  by  [ −5, 10 ]

ADDITIONAL AND ADVANCED EXERCISES, p. 1061 1. (a) y = c + ( y 0 − c ) e −k( A V )t

15.  [ −3, 3 ]  by  [ 0, 10 ]

y

(b) Steady-state solution: y ∞ = c 5. x 2( x 2 + 2 y 2 ) = C

y

10 9 8 7 6 5 4 3 2

10 8

= C

x

4 3

-2-1

43. (a) 1 min: 608 L; 10 min: 680 L; 60 min: 1080 L (b) S (1) ≈ 11.76 kg, S (10) ≈ 57.89 kg, S (60) ≈ 225.56 kg

7. ln x + e − y

f(x) = x Í9 - x 2

9. ln x − ln sec ( y x − 1) + tan ( y x − 1) = C

y = 5x 2>5 - 2x

2 -2

Appendices

-2

2

4

6

8

x

-4

y = 0x2 - 10

-3 -2 -1

APPENDIX A.1, p. AP-6

1

2

x

3

1. 0.1, 0.2, 0.3, 0.8, 0.9 or 1 5. x ≤ −

3. x < −2 x

−2

7. 3, −3 11.  [ −2, 4 ]

−

17.  [ −10, 10 ]  by  [ −10, 10 ]

x

1 3

t

0

15. ( −∞, −2 ] ∪ [ 2, ∞ ) −2

17. ( −∞, −3 ] ∪ [ 1, ∞ ) s

2

−3

19. ( −3, −2 ) ∪ ( 2, 3 ) -2

-1

0

z

10

- 10 - 8- 6 - 4

r

1

21. ( 0, 1 )

1

2

3

x

0

- 10 - 8- 6 - 4

x

APPENDIX A.2, p. AP-10

-

y

+ 15

-2

30 20

1.0

f(x) = x - 5x + 10

1 - 50 - 100

10

- 10

2 4 6 8 10

2

- 150 1

3

4

x

- 200 - 250

2

3

4

5

6

x

4

23.  [ −6, 10 ]  by  [ −6, 6 ] y

x-1 f (x) = x2 - x - 6

-2 -4 -6

4

3

0.5

x

f (x) =

6 4 -5

5

-2 -4 -6

6x 2 - 15x + 6 4x 2 - 10x x

10

27. [ −100 π , 100 π ]  by   [ −1.25, 1.25 ] y

x y = cos a50b

y = sin 250x

0.5

50

40

-2 -1

5

1

1.0

1.0

7.  [ −3, 6 ]  by  [ −250, 50 ]

x2 + 2 x2 + 1

x

y

1. d 3. d 5.  [ −3, 5 ]  by  [ −15, 40 ] f(x) =

2 4 6 8 10

-2 -4 -6 -8

π π ⎤ 25. ⎡⎢ − ,  by   ⎣ 125 125 ⎦⎥ [ −1.25, 1.25 ]

−1

f (x) =

2.0

6 4 2

0x0 + 0y0 ≤ 1

4x 3

2.5

y

1

x4

3.0

-4-3-2-1

y

y

x+3 x+2

21.  [ −10, 10 ]  by  [ −6, 6 ]

27. The graph of x + y ≤ 1 is the interior and boundary of the “diamond-shaped” region.

1

y

x

1

23. ( −∞, 1 ]

−1

y=

8 6 4

13.  [ 0, 10 ] 4

19.  [ −4, 4 ]  by  [ 0, 3 ]

y

9. 7 6, 25 6

−2

-3

1 3

x

- 0.02

0.02

x

- 300

300 - 0.5 - 1.0

x

Appendix A: Answers to Odd-Numbered Exercises

y

y=x +

1 10

25. x 2 + ( y − 3 2 ) 2 = 25 4

31. y

sin 30x

0.2

2

4 3

1

2

- 0.2

- 0.1

0.1

-2 -1

x

2

(2, 0) 3

x

4

1

2

35.

(0, −3)

2.0

3

1.5

2

1.0

1

2

3

-6

4

f(x) = sin 2x + cos 3x (−5, 0) −6

-2

2

4

6

x

−3 (−1, 0) 0

0.5 x

V(−3, 4)

y

f(x) = - tan 2x

-1

x

(0, −5)

(−6, −5)

-2 -3 -4

31. Exterior points of a circle of radius 7, centered at the origin 33. The washer between the circles x 2 + y 2 = 1 and x 2 + y 2 = 4 (points with distance from the origin between 1 and 2) 35. ( x + 2 ) 2 + ( y − 1 ) 2 < 6 1 2 1 2 ,  , − , − 37. 5 5 5 5 1 1 1 1 , − , , − 39. − 3 3 3 3 mm (b) ≈ −0.31 °C mm −0.1 °C 41. (a) ≈ (c) ≈ −0.15 ° C mm

- 2.0

APPENDIX A.4, pp. AP-21–AP-22

1. 2, −4; 2 5 1 5. m ⊥ = − 3

( (

3. Unit circle

y A(−1, 2) y = 3x + 5

2 1

Slope = 3 −1

B(−2, −1)

) ( ) (

)

)

42. 604.9 kPa 45. Yes: C = F = −40°

x

0 −1

C

C=F

7. (a) x = −1 (b) y = 4 3 9. y = −x 5 x 11. y = − x + 6 13. y = 4 x + 4 15. y = − + 12 4 2 17. x -intercept = 3,   y-intercept = − 2

−40

32 C=

F

5 (F − 32) 9

y −40 (−40, −40) 1

−1

x

2

51. k = −8, k = 1 2

Î2 x − Î3 y = Î6

APPENDIX A.8, pp. AP-40–AP-43

−2

1. No

19. ( 3, −3 ) 21. x 2 + ( y − 2 ) 2 = 4 y

23. ( x +

2

3 ) + ( y + 2 )2 = 4 y

Q−Î3, 0R

(0 , 4)

(0, −1)

C(0 , 2) C Q−Î 3, –2R

(0, 0) 1

2

x

(0, −3) −4

3. Yes

13. ≈0.688

(

5. Yes

15. ≈0.0502

)

7. Yes 17.

21

11. ≈0.537 1 19. ln 2 2

1 1 π ≈ 0.10242 , tan −1 2 − π π 4 8 25. mean = ≈ 2.67, median = 8 ≈ 2.83 3 27. mean = 2,  median = 2 ≈ 1.41 21.

x

−4

−2 −1

x

y y = −x2 − 6x − 5

1

0

(3, 0)

V(1, −4)

4

−2

C(0, 32)

1 2 −1 (0, −1) −2 2 2 x + (y – 32) = 254

29. y

-2 -1

0

−2 −1 0

- 0.2

33.

y = x2 − 2x − 3

(−1, 0)

1

(−2, 0)

x

0.2

y (0, 4)

(x + 1)2 + ( y - 2)2 = 9

0.1 -4

27.

y

Axis: x = 1

π π 29.  ⎡⎢ − , ⎤⎥   by  [ −0.25, 0.25 ] ⎣ 15 15 ⎦

Axis: x = −3

AN-72

29. P ( X < 12 ) ≈ 0.3935 31. (a) ≈0.57, so about 57 in every 100 bulbs will fail. (b) ≈832 hr

Appendix A: Answers to Odd-Numbered Exercises

33. ≈60 hydra 35. (a) ≈0.393 (b) ≈0.135 (c) 0 (d) The probability that any customer waits longer than 3 minutes is 1 − ( 0.997521) 200 ≈ 0.391 < 1 2. So the most likely outcome is that all 200 would be served within 3 minutes. 37. $10,256 39. ≈323, ≈262 41. ≈0.89435 43. (a) ≈16% (b) ≈0.23832 45. ≈618 women 47. ≈61 adults 49. ≈289 shafts

AN-73

51. (a) ≈0.977 (b) ≈0.159 (c) ≈0.838 55. (a) {LLL, LLD, LDL, DLL, LLU, LUL, ULL, LDD, LDU, LUD, LUU, DLD, DLU, ULD, ULU, DDL, DUL, UDL, UUL, DDD, DDU, DUD, UDD, DUU, UDU, UUD, UUU} (c) 7 27 ≈ 0.26

(d) 20 27 ≈ 0.74

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Applications Index Note: • Page numbers with the prefix “17” are in the online Chapter 17 Second-Order Differential Equations • Page numbers with the prefix “18” are in the online Chapter 18 Complex Functions • Page numbers with the prefix “19” are in the online Chapter 19 Fourier Series and Wavelets • Page numbers with the prefix “AP” are in Appendix A • Page numbers with the prefix “B” are in the online Appendix B Business and Economics Cobb-Douglas production function, 842 Cost, 287–288 Continuous price discounting, 456 Depreciation of 38-wheel truck, 71 Gross national product growth, 148 Industrial costs, 33 Inventory management, 161, 288 Inventory replenishment, 288 Investment growth, 50, 53, 54, 66, 71, 472 Marginal costs, 166–167, 169, 267, 280–282, 352 Marginal profit, 280–282 Marginal rates and taxes, 167 Marginal revenue, 169, 232, 280–282, 352 Maximizing utility, 843 Maximum profit for backpack sales, 287 for manufacturing tires, 307 for tour service rates, 287 Maximum revenue, 288 Order size, 824 Production growth, 232 Production levels, 288 Profit rate of change, 78 Revenue as function of price increase, 33 Rule of 86, 71, 472 Single-family home prices, 149 Engineering and Physical Sciences Acceleration, 168, 183, 232, 286, 303, 310–311, 464 Aircraft flight distance, 212 Airplane flight off course, 63 Airplane ground speed and direction, 686 Airplane landing path, 287 Airplane takeoff, 170 Angle for pipes, 311 Arc length, 464, 491, 509

Area, 211, 320, 373, 464, 491, 496, 882 change with respect to diameter of circle, 161–162 change with respect to radius of circle, 225, 230 changing, of a disk, 141 of circular washer, 876 between curves, 370 of surfaces of revolution, 433 of a triangle, 211 Atmospheric pressure and altitude, 455 Average temperature, 183, 687, 876 Average values, 882 Balloon/bicycle distance increase, 212 Beam motion, 214 stiffness, 286 strength, 286 under a load, 898 Boring a cylinder, 212 Bouncing ball, 554 Boyle’s Law, 33 Building’s shadow, 213 Buoyant force, 1053 Carbon dating, 456 Cardiac output, 213 Center of gravity, 491 Center of mass, 894–895 fluid forces and, 427 of slender metal arch, 933–934 of thin, flat plate, 423–425, 472 for thin plate bounded by curves, 426 of thin plate with constant density, 429–430, 433 of thin plate with varying density, 430 of thin shell, 990–991 of thin wire, 421, 426–427, 429 Centroid, 509, 895, 906–907 of a semicircular region of radius, 428 of thin plates, 431, 433 of a triangle, 427, 430

Chemical mixture problem, 1041–1043, 1060 Chemical reaction, 455, 504 Circulation around bounding circle, 996–997 density, 963–964 of paddle wheel, 1000 Circumference of a circle, 404 Coffee draining from conical filter, 212–213 Conformal maps of a square, 18-33–18-34 Conical sand pile, 212 Constructing a cone, 285–286 Constructing a cylinder, 284 Cooling, 453–454, 456, 471 Cost to build parking lot, 533 for isosceles right triangle shaped fence, 33 Cylinder pressure, 161 Damped motion, 17-17–17-18 Deceleration to stop moving vehicle, 303, 310 Density of thin wires, 430 Designing a box with lid, 283 a can, 283 a gondola for hot air balloon, 232 a plumb bob, 390 a suitcase, 283 a tank, 283 a wok, 390 Diameter of a tree, 225 Dimensions of fence, 282 of silo, 284 Distance traveled, 78, 320–321, 518 between two ships, 287 Draining hemispherical reservoir, 212 a swamp, 517 a tank, 135, 169, 212, 230 Drilling boreholes, 858 Edges of a cube, 230 Electrical resistance, 824

Electrical resistors, manufacturing, 97 Electric circuit, 17-18–17-19, 211, 230, 855, 1036, 1037, 1053 Electric force, 418 Electricity, 211 Electromagnetic theory, 1013, 1021 Elevation angle of hot air balloon, 67, 207–208 Erecting radio telescope, 843 Error compounding, 231 controlling, 231 Escape velocity, 1060 Falling object, 140, 141, 168, 183, 303, 304 resistance, 1052 skydiving, 1052 Fermat’s principle in optics, 287 Filling a conical tank with water, 206–207 Fluid force center of mass and, 427 on a curved surface, 1021 on end plates of a trough, 419 out of hole in water tank, 309 on parabolic gate, 419 on rectangular plate, 418, 419, 435 of seawater, 419 on semicircular plate, 418, 419 on square plate in different positions, 419 streamlines, 973 in a swimming pool, 415 on triangular plate, 418, 419, 435 on viewing portion of fish tank, 419 Flux, 946, 969, 988, 989, 1008, 1012 Flying a kite, 212 Force of attraction, 416 below Earth’s surface, 433 due to gravity pulling down, 17-15–17-16 of wind passing over boat’s sail, 699 Free-falling object, 78, 164, 167, 168, 321

AI-1

AI-2

Applications Index

Free-falling object (continued ) in the fourteenth century, 310 on the moon, 249 near the surface of a planet, 304 Glider position, 735–736, 743 Graphing, 287 Gravitational fields, 952 Grinding engine cylinders, 97 Haar coefficients, 19-27–19-29 Hanging cables, 464 Harmonic motion, 17-16–17-17, 172–173 Hauling in a dinghy, 212 Heat capacity of a gas, 532–533 Heat diffusion equation, 1016 Heat equation, 796, 858 Heating a plate, 211 Heat transfer (cooling), 453–454, 456 Height of a baseball, 739, 741 of a building, 225 of a bullet shot straight up, 168 of a lamppost, 231 of a palm tree, 352 of a paper clip shot into the air, 232 of a pole, 69 of a rocket, 1061 of a weather balloon, 69 Highway patrol plane, 214 Impedance in a series circuit, 230 Industrial costs, 33 Inertia, moments of, 895–897 Inflating balloon, 170, 206 Inner product, 19-15–19-16 Intensity of illumination, 286 Kinetic energy and potential energy, 33 and work, 417–418, 434 Length of airplane gasoline tank, 518–519 of an asteroid, 404 of an object, 135 of a beam, 843 of curves, 400–401, 403–405, 433 of a fish, 472 of a graph, 401 of a ladder carried around a corner, 307 of a line segment, 404 of a rocket, 135 Lighthouse beam, 214 Linearizations, 226, 231 Locating a planet, 293 Mass, 893–895 Maximum volume of a box, 277, 307 of a can, 278, 821 Measuring cup stripes, 135 Melting ice layer, 214 Minimizing perimeter, 282 Motion of an object along a line, 267 of a beam, 214 of a body along a line, 168, 169 coasting battleship, 1043 coasting bicycle, 1043 with constant acceleration, 304 damped, 17-17–17-18

of a helicopter, 707–708 of a model rocket, 168 of a particle, 168, 169, 213, 230, 232, 286, 303, 307 in the plane, 213 of a projectile, 167, 286, 321, 740–742 of a vehicle, 213 Moving shadow, 213 Norm of a function, 19-13–19-14 Ohm’s law for electrical circuits, 97 Oil consumption rate, 519 Oil slick spread, 214 Orthogonal trajectory of curves, 1040–1041 Osculating circle, 232 Parametrization of a cone, 974 of a cylinder, 974 of a sphere, 974 Particle motion, 167, 183, 209, 213, 303, 307 Path of a ball, 738–739, 741 of a baseball player, 214 of a football, 309 of a glider, 735–736, 743 of a heat-seeking particle, 857 of a javelin, 763 of a projectile, 737–738, 740–741 of a shot-put throw, 741 of a water skier, 496 Periodic functions, 19-1–19-2 Planetary orbits, 669 Pressure of fluid at bottom of dam, 414 Pressure waves produced by tuning fork, 24 Projectile motion, 167, 321, 698, 740–742, 763 Pumping oil to the rim of the tank, 413 Quadratic approximations, 225–226 Quickest route, 286 Radioactive decay rate, 53, 54, 66, 452–453, 456 Radius increase of raindrop, 212 of inflating balloon, 212 Radon-238 decay, 66 Reservoir water, 305 Resistors connected in parallel, 230, 789 Resultant force, 686–687 Rolle’s Theorem, 249 Rotating spool speed, 230 Satellite orbit, 418 Searchlight beam motion, 231 Shortest beam, 286 Skaters stopping distances, 1039, 1043–1044, 1060 Skylab 24 view, 764 Sliding ladder, 211–212 Sonobuoy submarine location, 294 Specific heat of a gas, 375 Speed of aircraft approaching island, 209–210

of a frictionless cart, 286 of marathoner, 249 of object dropped from tower, 141 during parachute jump, 232 of particle, 230, 232 of a rocket, 141 of rotating spool, 230 of ships moving apart, 214 underwater transmission cable, 308 of a vehicle, 78, 208, 249 velocity and, 162–163 of warship, 249 Speed of light, 279–280 Spring’s force constant, 416 Standard unit vectors, 19-7, 19-11 Stopping distance, 303 Stream flow, 810 Stretching a rubber band, 106, 416 Submarine intercept, 722 Surface area, 509 of an asteroid, 410 of an oil spill, 214 of a cone, 230, 231, 409, 976 cut from bottom of paraboloid, 979–980 of an infinite paint can or Gabriel’s horn, 530 of a melting ice cube, 234 of a right circular cylinder, 230 of sliced bread, 410 of a sphere, 211, 977 of a torus, 429, 982 of Weather Service dome, 409 of a wok, 409 Suspension bridge cables, 662 Temperature air, 375 below Earth’s surface, 772 in Fairbanks, Alaska, 183 greatest, 939 mean, 532 and period of a pendulum, 183 rate of change, 149, 249 of a room, 352 Thermal expansion in laboratory equipment, 135 Tin pest, 287 Tolerance, 225 Torque, 702–704, 722 Underwater transmission cable speed, 308 Vectors, 19-7, 19-11 Vehicular stopping distance, 169–170 Velocity after ricochet, 858 average and instantaneous, 233 displacement from an antiderivative of, 303 of falling meteorite, 183 fields, fluid, 944–945, 966 of moving object, 178 of a particle, 233 of particle moving along a horizontal line, 162–163, 168 in a resisting medium, 1038–1039

of a rocket, 168 of a skydiver, 464 of two particles, 232 Velocity and acceleration, 164–165, 361 of a piston, 182 of weight on spring, 172–173, 175 Vertical motion, 286 Vertical spring with weight, 172–173 Vibrating springs, 17-15–17-16 Videotaping a moving car, 213 Volcanic lava fountains, 170 Voltage changes, 211, 455, 804–805 Volume, 191, 464 of an ice cream cone, 909–910 of a ball, 141 of a balloon, 39 of a bowl, 390 of a box, 33 of a Bundt cake, 398 of a football, 432 of gasoline tanks, 509–510 of a hemisphere, 390 of material in a cylindrical shell, 225 of noncircular right cylinder, 882 of a pipeline, 855 of a pyramid, 388 of a right circular cone, 211 of right prism, 866 of a solid, 380–400 of a solid region, 880 of a solid sphere hole, 432 of storage tank for molasses, 821 of a swamp, 517 of a tetrahedron, 888 of a torus (doughnut), 390, 428 of a trough, 284 using cylindrical shells, 391–396 of water in swimming pool, 518 of a wedge, 434 of a wedgelike solid beneath surface, 869–870 Water flow, 135 Water main, 697–698 Wave equations, 795 Weight on a pulley, 210 Wind chill factor, 823 Work along different paths, 960 to carry water up mountain in a leaky tank, 433 to compress a bathroom scale, 416 done by a force field, 943–944 done by a jack, 311 done by a locomotive, 699 done by the conservative field, 954 done in lifting a bucket and rope, 412–413 to fill a water tower, 418 to haul up a rope, 416, 433 and kinetic energy, 417–418, 434 lifting a bag of sand, 416 lifting an elevator cable, 416 lifting water in a leaking bucket, 416 to move electrons, 418 to pump a reservoir, 433

Applications Index

to pump a tank, 433 to pump liquids from containers, 416–417 to pump oil to the rim of a tank, 413 to put a satellite in orbit, 418 required to compress a spring, 412, 416 to stretch a spring, 416, 433 to suck milkshake through a straw, 417–418 General Box dimensions, 33, 805, 831 Clock’s moving hands, 214 Computer graphics hidden lines, 714 perspective, 713 Designing a poster, 283 Endangered species, 455 Helicopter rescue, 722

Leibniz’s Rule, 233–234, 377 Linearizations, 226, 231 Mixtures, 1041–1044, 1060 Oil depletion, 455 Paper folding, 284 Period of a clock pendulum, 225, 234 Random number generator, 899–900 Rolle’s Theorem, 249 Sugar, inversion of, 455 Swing of a clock pendulum, 628 Urban gardening, 472 U.S. Postal Service box size, 284 Well depth, 222 Window proportions, 284 Working underwater, 455 Life Sciences Antihistamine effects, 520 Blood flow, 225, 311

Blood tests, 310 Blood types, 843 Body surface area, 169 Cardiac output, 213, 855 Diseases, eliminating, 54, 455 Diseases dying out, 451–452 Drug absorption, 71 Drug assimilation, 520 Drug concentrations, 225, 455 Drug dosages, 623 Drug effectiveness, 141 Endangered species, 455 Flow velocity of a cough, 288 Heart, effect of flight maneuvers on, 225 Hybridization, 167 Medicine dosages, body’s reaction to, 160 Population density of bacteria, 876 regional, 876

AI-3

Population growth of bacteria, 54, 169, 455 of fruit flies, 148–149 of midwestern city, 54 of rabbits and foxes, 1058–1059 of town in California, 66 of trout vs bass, 1054–1055, 1057 in U.S., 455 of world, 1039–1040 of yeast cells, 141 of yeast culture, 450–451 Sensitivity to medicine, 288 Tooth size reduction, 454–455 Transport through cell membrane, 1061 Social and Behavioral Sciences Information diffusion, 504, 1052 Population, 1052 World population, 1039–1040

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Subject Index Note: • Page numbers with the prefix “17” are in the online Chapter 17 Second-Order Differential Equations • Page numbers with the prefix “18” are in the online Chapter 18 Complex Functions • Page numbers with the prefix “19” are in the online Chapter 19 Fourier Series and Wavelets • Page numbers with the prefix “AP” are in Appendix A • Page numbers with the prefix “B” are in the online Appendix B Absolute change, 222 Absolute convergence, 572–577 Absolute Convergence Test, 573 definition of, 573 Ratio Test, 574–575 Root Test, 576–577 Absolute Convergence Test, 573 Absolute extrema, of continuous functions, 829–832 Absolute (global) maxima or minima extrema boundary points of, 830–831 on closed bounded regions, 829–832 definitions of, 235, 825 Extreme Value Theorem and, 236–237 finding, 238–240 interior points of, 830–831 Absolute maximum, 825–826, 829–832 Absolute minimum, 825–826, 829–832 Absolute value of complex numbers, 18-5 definition of, AP-4–AP-5, 18-5 properties, AP-4 Abstract algebra, AP-28 Acceleration definition of, 163 formula for calculating the normal component of, 754 in free fall, 164 jerk associated with, 163–164, 173 normal component of, 753–754 in polar coordinates, 759–762 in simple harmonic motion, 172–173 tangential and normal components of, 753–754 torsion and, 754–757 vector, 729, 753 velocity in, 163–165 Addition of complex numbers, 18-3–18-4 formulas in trigonometric functions, 45 of functions, 34–35

parallelogram law of, 682–683, 692 of terms to series, 557 of vectors, 682–683 Addition property of real numbers, AP-27 Additivity, line integrals and, 932 Additivity Rule for definite integrals, 334 for double and triple integrals, 869 Agnesi, Maria Gaetana, 159 Agnesi, Mary, 632 Albert of Saxony, 568 Algebra abstract, AP-28 Fundamental Theorem of Algebra, 18-9 linear, B-4 operations, vectors in, 682–683, 687 rules for finite sums, 324 for gradient vectors, 811 Algebraically representing vectors, 681 Algebraic functions, 29–30 Algebraic numbers, defined, 445 Algebraic properties of exponential functions, 442–443 of ln x, 438 of the natural logarithm, 441–442 of natural logarithmic functions, 58–59 of real numbers, AP-1, AP-27 Alternating harmonic series, 579–583 Alternating series, 579–583 Alternating Series Estimation Theorem, 581 Alternating Series Test, 579–580, 583 Amperes, 17-19 Amplitude, 19-4 Angle convention, 42 Angle of elevation, 737 Angle of inclination, AP-15–AP-16 Angles Angle Between Two Vectors, 691–693 of elevation, 737

firing, 737 of inclination, AP-15–AP-16 measured in degrees and radians, 41 phase, 17-17 between planes, 711–712 in standard position, 42 in trigonometric functions, 41–42 between two curves, computing, 18-32 between vectors, 711–712 Angular momentum, 766 Angular velocity, 1000 Antiderivatives, 294–300 Constant Multiple Rules for, 296–297, 299 definition of, 294 Difference Rule for, 296–297, 299 finding, 294–297 indefinite integrals of, 299–300 initial value problems and differential equations, 297 integration formulas, 300 linearity rules of, 296–297 motion and, 297–298 Sum Rule for, 296–297, 299 of vector functions, 734–735 Antidifferentiation, 294, 297, 300 Aphelion, 669 Applied optimization, 277–282 examples from mathematics and physics, 279–282 problems, solving, 277–278 Approximations convergence of, 291–292 in converting mass to energy, 223 differential, 220–221 error in, 220–221 finite for area, 315–320, 437 theory of limits of (Riemann sums), 326–328, 331 of functions, 19-16–19-19 in higher dimension vectors, 19-10–19-11 linear, 215–217 error formula for, 845 standard, 819–820

lower sum, 313–314 Midpoint Rule for, 332–333, 510 polynomial, of two-variable functions, 847 quadratic, 601 at resolution level zero, 19-29–19-31 for roots and powers, 217 sensitivity to change in, 222–223 of a series, error estimation in, 564–565 by Simpson’s Rule, 512–514 tangent line in, 215 tangent planes, 792, 819–820 trapezoidal, 511–512 by Trapezoidal Rule, 511–512 by trigonometric polynomials, 19-3–19-5 upper sum, 313–314 using the Haar system, 19-27, 19-29–19-31 Arbitrary constant, 295 Arccosecant function derivatives of, 200–201 graph of, 200 identities involving, 204 Arccosine derivatives of, 200, 201 identities involving, 204 Arccotangent function derivatives of, 200–201 graph of, 200 identities involving, 204 Archimedes’ principle, 1053, 1021 Archimedes spiral, 672 Arc length, 399–403 arc length function, 402–403 of a curve y " f (x), 399–401 differential formula for, 402–403 discontinuities in dy/dx, 402 Arc length differential, 638–639 Arc length in space, 743–745 arc length formula, 743 arc length parameter, 744 definition of, 743 of a function, 744 speed on a smooth curve, 744–745 unit tangent vector, 745

I-1

I-2

Subject Index

Arcsecant function derivatives of, 200–201, 203 graph of, 200 graphs/graphing of, 492 identities involving, 204 Arcsine and arccosine functions, 61–64 arc in, 62 definition of, 61–62 derivatives of, 189, 201, 202 graphs/graphing of, 62–63, 492 identities involving, 63–64, 204 values for, 63 Arctangent functions derivatives of, 200–203 graphs/graphing of, 200, 492 identities involving, 204 in Taylor series, 612–613 values of, 201 Area between curves, 364–366 definition of, 873 differential, change substituting, 916 by double integrals, 873–875 average value of, 874–875 bounded regions in a plane, 873–874 Riemann sums in, 873 finite approximations for, 315–320 finite sums for estimating, 312–320 Green’s Theorem formula for, 972 infinite, 522 lower sum approximation of, 313–314 Midpoint Rule for approximating, 314, 332–333 Pappus’s Theorem for Surface Areas, 429 in a plane, 650–652 in polar coordinates, 650–653, 878–879 of rectangles (See Rectangles, area of) surface (See Surface area) of surfaces of revolution, 405–408, 639–640 tangent, 633–634 total, 349–350 upper sum approximation of, 313–314 Argand Diagrams, 18-4 Argument, 18-5 Arrow diagrams, 851 Associative laws of real numbers, AP-28 ASTC rule, 43 Asymptotes horizontal, 108–111 of hyperbolas, 659 oblique, 111–112 slant line, 111 vertical, 115–116 Autonomous differential equations, 1044–1051 definition of, 1045 equilibrium values or rest points in, 1044–1046

falling body encountering resistance, 1047–1049 graphical solutions of, 1044–1051 Logistic Equation in Neural Networks and Machine Learning, 1050–1051 logistic model for population growth, 1049–1050 Newton’s Law of Cooling, 1047, 1048 phase lines, 1044–1046 stable and unstable equilibria, 1046–1047 Auxiliary equation, 17-3 Average daily cost, 281–282 Average speed, 72–74 Average value for approximating area of rectangles, 319–320 of continuous functions, 337–338 definition of, 338 of a function in space, 890 of nonnegative continuous functions, 318–320 Average velocity, 162 Axis coordinate, AP-14 of ellipses, AP-21 focal axis, 657 major axis, 658 minor axis, 658 semimajor axis, 658 semiminor axis, 658, 666–667 focal, of hyperbolas, 658 imaginary, 18-4 major, AP-21 minor, AP-21 moments of inertia about, 934, 990 of parabolas, AP-19–AP-20 real, 18-4 spin around, 961–964 Base a logarithm, 444 Basis defined, 19-8 orthonormal, for three-dimensional space, 19-8–19-9 Behavior, exponential, 49–51 Bendixson’s criterion, 973 Bernoulli, Daniel, 236 Bernoulli, Jakob, 1038 Bernoulli, John (Johann), 179, 268 Bernoulli differential equation, 1038 Binary search, 468–469 Binomial series, 609–611 Binormal vectors, 753, 757 Biology, derivatives in, 167 Birkhoff, George David, 357 Bolzano, Bernard, 163 Boundary definition of, B-7 of intervals, AP-3 in a plane, 769 in space, 771–772 in substitution, 917–919, 922 Boundary points of absolute (global) maxima or minima extrema, 830–831

closed bounded, absolute maxima and minima on, 829–832 definition of, B-11 of intervals, AP-3 in a plane, 769 in space, 771 Boundary values, 17-6 Bounded from above/below (sequences), 546 Bounded function, 81, 136 Bounded region absolute maxima and minima on closed, 829–832 definition of, B-7 nonrectangular, 864–865 in a plane, 769, 873–874 in space, 771 Bounded sequences, 546–547 Bounds for the remainder in the Integral Test, 565 Bowditch curve, 642 Box function, 19-25 Box product. See Triple scalar product Boyle’s Law, 33 Brachistochrones, 628–629 Branches of hyperbolas, 655 Brief Table of Integrals, 504 Calculators to estimate limits, 84–85 Calculus of variations, 629 Calculus with parametric curves, 633–640 arc length differential, 638–639 areas of surfaces of revolution, 639–640 Chain Rule in, 633 complete elliptic integral of the second kind in, 637 continuously differentiable functions, 635 differentiable parametrized curves, 633–634 length of parametrically defined curves, 635–638 parametric formulas, 633–634 Riemann sums in, 635 for smooth curves, 635–636 tangent lines and areas, 633–634 twice-differentiable functions, 633–634 Cantor, Georg, 560 Cantor set, 560 Capacitance, 17-19 Carbon-14 decay, 452 Cardioid curve, 648 Carrying capacity, 1049 Cartesian coordinates equations relating spherical and cylindrical coordinates to, 907–909 in a plane, AP-13–AP-14 polar coordinates related to, 643–645 system of, AP-14 in three-dimensional coordinate system, 675

Cartesian equations for hyperbolas, 665 parametric, 627 Cartesian integrals, changing into polar integrals, 879–881 Cartographers, 18-32 CAS. See Computer algebra systems (CAS) Catenary, 31, 464 Cauchy, Augustin-Louis, 273 Cauchy condensation test, 567 Cauchy-Riemann Equations, 18-20–18-22 Cauchy’s Mean Value Theorem, 273–274 Cavalieri, Bonaventura, 382 Center of an ellipse, AP-21 of curvature for plane curves, 749 of ellipses, 657 of hyperbolas, 658 of linear approximation, 215, 217 Centered difference quotient, 175–176 Center of mass, 894–895 along a line, 420–421 centroids, 426–427 coordinates of, 934, 990 definition of, 421, 422 distributed over a plane region, 422 moments and, 420–429 plates and, 422–426 of plates bounded by two curves, 425–426 system torque, 420–421 theorems of Pappus, 427–429 of thin, flat plates, 422–425 thin wires and, 421 torque, 420–421 Center-to-focus distance of ellipses, 658 Centroids, 426–427, 894–895 Chain curve, 464 Chain Rule, 354–359, 796–803 in calculus with parametric curves, 633 dependency diagrams for, 800, 801 derivatives and, 176–184 in composition of functions, 176–178 in differentials, 219, 221–222 of exponential functions, 438, 441–443 in implicitly defined functions, 185–186 of inverses of differentiable functions, 191 of the natural logarithm function, 192–193 outside-inside rule and, 178–179 with powers of a function, 179–181 in proving Derivative Power Rule, 196 in related rates equations, 206, 208 repeated use of Chain Rule in, 179

Subject Index

in trigonometric functions, 202–203 ways to write, 178 of ax and logax, 194–195 diagram for remembering, 797 differentiable parametrized curves and, 633–634 for finding complex derivatives, 18-19 functions and curvature, 747, 748 defined on surfaces, 799–801 Implicit Function Theorem, 802–803 of many variables, 803 of three variables, 798–799 of two variables, 796–799 vector functions, 730, 731 implicit differentiation and, 801–803, 979 motion in polar coordinates and, 759 “outside-inside” rule and, 814 for paths, 813–814 Power Chain Rule and, 179–181 proof of, 178, 221–222, 731 theorems, 798, 799, 801–803 Chain Rule for Functions of One Independent Variable and Three Intermediate Variables, 798 Chain Rule for Functions of One Independent Variable and Two Intermediate Variables, 796 Chain Rule for Two Independent Variables and Three Intermediate Variables, 799 Formula for Implicit Differentiation, 801 Implicit Function Theorem, 802–803 Change. See also Rates of change exponential, 447–448 sensitivity to, 222–223 in approximations, 222–223 genetic data and, 167 Change-of-base formula, 59 Characteristic equation, 17-3 Charge, 17-18–17-19 Circles of convergence, 18-26 of curvature for plane curves, 749–750 exterior regions of, AP-19 interior regions of, AP-19 osculating, 232, 749–750 in a plane, AP-17–AP-19 standard equation of, AP-18 unit, AP-18 Circular cylinder, 714 Circular functions, 458. See also Trigonometric functions Circulation density, 963–964 flux differentiated from, 945 in Green’s Theorem, 966–968 for velocity fields, 944–945

Cissoid of Diocles, 188 Clairaut, Alexis, 791 Clairaut’s Theorem, 790–791 Closed curves, 945 Closed intervals, AP-3, 122, 127 Closed region bounded, absolute maxima and minima on, 829–832 definition of, B-7 in a plane, 769 in space, 771–772 Coefficients drag, 742 Fourier, 19-18–19-19 Haar, 19-27–19-29 linear second-order constantcoefficient models, 17-20 method of undetermined, 17-8–27-31 of polynomials, 29 in power-series solutions, 17-24–17-29 rational functions in, determining by differentiating, 502 undetermined, 497 Cofactor, B-2–B-5 Cofactor Expansion, B-4 Common logarithmic functions, 58 Commutativity laws of real numbers, AP-28 Comparison tests, 18-24, 567–571 Direct Comparison Test, 568–569, 583 Limit Comparison Test, 569–571, 583 Competitive-hunter model, 1054–1055 Complementary solution, 17-8 Complete elliptic integral of the second kind, 637 Completeness property, 51, 547 Complete ordered field, AP-28 Completing the square, AP-18–AP-19 Complex analysis, 18-1 Complex conjugate, 18-4 Complex derivatives definition of, 18-17–18-19 differentiable, 18-18–18-19 formula for, 18-20–18-21 rules for finding, 18-19 Complex differentiable function, 18-18–18-19 Complex functions, 18-1–18-36 Cauchy-Riemann Equations, 18-20–18-22 complex numbers, 18-1–18-9 complex power series, 18-22– 18-24, 18-25 conformal maps, 18-31–18-34 derivatives of, 18-17–18-19 Euler’s Formula, 18-27–18-28 logarithmic, 18-28–18-29 variables of, 18-10–18-17 Complex numbers, 18-1–18-9 absolute value of, 18-5 adding, 18-3–18-4 Argand Diagrams, 18-4 complex conjugate, 18-4

complex plane and, 18-4–18-5 computations with, 18-4 computing reciprocals, 18-6 definition of, 18-1 De Moivre’s Theorem, 18-7 dividing, 18-4 equality in, 18-3 Euler’s Formula, 18-5 Fundamental Theorem of Algebra, 18-9 geometric representations of, 18-4–18-5 hierarchy of, 18-1–18-3 imaginary parts of, 18-3–18-4 multiplying, 18-3–18-4 powers of, 18-7 products of, 18-5–18-6 quotients in, 18-6–18-7 real parts of, 18-3–18-4 roots in, 18-7–18-9 Complex plane, 18-4–18-5 Complex power series, 18-22–18-24 complex sequences, 18-23 complex series, 18-23–18-24 Convergence Theorem for Complex Power Series, 18-25 Complex variables, 18-10–18-17 complex domains, 18-10–18-11 complex limits, 18-11–18-15 continuous, 18-15–18-17 functions of, 18-12–18-15 Component equation for a plane, 709 Component form of vectors, 680–682 Component functions, 725, 937 Component test for conservative fields, 955–957 for exact differential form, 958 Composite functions Chain Rule in, 176–178, 814 continuity of, 125–126 continuous, 125–126 derivative of, 176–178 evaluating, 35–36 Composition Rule, 777 Computer algebra systems (CAS) improper integrals and, 525–526 improper integrals in integration evaluated with, 525–526 integrate command in, 506–507 integration with, 506–507 Computer graphing, 772 Computing reciprocals, 18-6 Concavity curve sketching and, 255–268 definition of, 256 graphing y " f(x), 261–263 points of inflection and, 256–259 Second Derivative Test for, 256, 259–261 Conditionally convergent series, 581–582 Cones elliptical, 714–717 parametrization of, 974 surface area of, 976 Conformal maps, 18-31–18-34 Conic sections, 655–660

I-3

ellipses, 657–658 hyperbolas, 658–660 illustrated, 655 parabolas, 656–657 Conics in polar coordinates, 663–668 directrices for, 663–665 eccentricity of, 663–667 polar equations for, 665–667 Connected domains and regions, 951 Connectedness, 127 Conservation of angular momentum, 766 Conservative fields component test for, 955–957 line integrals in, 952–955 potentials for, finding, 955–957 properties of, 951 Stokes’ Theorem and, 1003–1004 Constant-depth surface, fluid force on, 414–415 Constant force, work done by, 410–411 Constant Function Rule, 730 Constant functions, 27, 81 norm of, 19-13–19-14 zero derivatives in, 246 Constant length, vector functions of, 731 Constant Multiple Rule, 82–83, 556–557 for antiderivatives, 296–297, 299 to calculate limits of sequences, 542 for definite integrals, 334 for derivatives (See Derivative Constant Multiple Rule) for double and triple integrals, 869 for finding complex derivatives, 18-19 for finite sums, 324 for gradients, 811 for limits, AP-23 for limits for functions of two variables, 777 for series, 556–557 Constant Value Rule, 324 Constrained variables arrow diagrams for, 851 constrained by another equation, determining, 849–850 dependent/independent, determining, 848–849 notation for, 850–851 partial derivatives with, 848–851 Containers, lifting objects and pumping liquids from, 412–413 Continuity, 120–132. See also Continuous functions on an interval, 124 of compositions of functions, 125–126 continuous derivatives, 273 continuous extension, 128–129, 783, 853 continuously differentiable functions, 635 continuous partial derivatives, 1003

I-4

Subject Index

Continuity (continued ) continuous vector fields, 937 discontinuity and, 122–123 at interior point, 121–122 inverse functions and, 124–125 at left endpoint, 121 limits and, 106–132 for multivariable functions, 779–781 one-sided, 122 partial derivatives and, 779–781, 789–790 at a point, 121–123 test, 122 two-sided, 122 of vector functions, 726–727 Continuous functions, 123–129, 779–781 absolute extrema of, 829–832 of compositions, 781 compositions of, 125–126 continuous at the point, 779–780 continuous extension of, 783, 853 definition of, 123, 779 differentiable, 147, 407, 408, 429, 635, 638, 639 extension to a point, 128–129 graphing, 127–128 Intermediate Value Theorem for, 126–128 left (or continuous from the left), 121–122 limits of, 126 over a closed interval, 122, 127 partial derivatives and, 789–790 properties of, 124 right (or continuous from the right), 121–122 Contour curves of functions of two variables, 770 Convergence absolute, 572–577 of approximations, 291–292 circle of, 18-26 conditional, 581–582 of improper integrals, 523 tests for convergence, 526–528 interval of, 591 in Newton’s method, 291–292 of power series, 586–591 Convergence Theorem for Power Series, 589–591 operations on, 591–594 radius of, 589–591 testing for convergence, 591 radius of, 589–591 of Riemann sums, 331 testing for alternating series, 579–583 Alternating Series Estimation Theorem, 581 Alternating Series Test, 579–580 Direct Comparison Test, 568–569, 583 geometric series, 553–555, 583 improper integrals, 526–528

Limit Comparison Test, 569–571, 583 power series, 591 p-Series Test, 563, 583 Ratio Test, 574–575, 583 Rearrangement Theorem for Absolutely Convergent Series, 582, 583 Root Test, 576–577, 583 series with some negative terms, 573, 581–583 summary of, 583 Convergence Theorem for Complex Power Series, 18-25 Convergence Theorem for Power Series, 589–591 Convex, 256 Coordinate axes, AP-14 Coordinate pair, AP-14 Coordinate planes in threedimensional coordinate system, 675–676 Coordinates of center of mass, 934, 990 cylindrical (See Cylindrical coordinates) polar (See Polar coordinates) rectangular (See Rectangular coordinates) Coordinate systems Cartesian, AP-14 rectangular, AP-14 three-dimensional, coordinate planes in, 675–678 Coplanar vectors, 683 Corner, 146 Cosecant function, 42–43 derivatives of, 173–174 graphs of, 44, 62 integrals of, 359 inverse of, 200–201 odd, 44 periodicity of, 44 Cosine function, 19-2, 42–43 arccosine function and, 61–62 derivatives of, 171–172 to eliminate a square root, 488 even, 44 graphs of, 30, 44, 62 inequalities and, 46 periodicity of, 44 product of, 490 products of powers of, 486–487 Cosines direction, 697 law of, 45–46 summing, 19-3–19-5, 19-3–19-5 values of, 43 Costs fixed, 166 marginal, 165–166, 280–281 of production, 165–166 variable, 166 Cotangent function, 42–43 derivatives of, 173–174 graphs of, 44, 62 integrals of, 359 inverse of, 200–201

odd, 44 periodicity of, 44 Coulombs, 17-19 Courant, Richard, 151 Cramer, Gabriel, 188 Cramer’s Rule, 17-12 Critical damping, 17-17–17-18 Critical points definition of, B-8 extreme values and, 239–240 to find increasing or decreasing functions, 250–253 on graphs, 259–262 in local (relative) maxima or minima extrema, 826–829 in Max-Min Tests, 831 in optimization problems, 80, 277, 279, 282 Cross Product Rule, 730–731 Cross products, 699–704 area of a parallelogram in, 700–701 calculating, as a determinant, 701–702 Cross Product Rule, proof of, 730–731 definition of, 699–700 determinant formula for, 701 distributive law for, AP-30–AP-31 pairwise, 700 parallel vectors in, 700 properties of, 700 torque vector in, 702–703 triple scalar product (box product) in, 703–704 of two vectors in space, 699–700 Cross-sections definition of, 380 for finding limits of integration, 868 method of slicing, 380–382 solids of revolution, 382–387 disk method, 382–385 washer method, 385–387 volumes using, 380–387 Cube root function, 28 Cubic function, 29 Curl vector divergence and, 1009–1010 field, 993–995 in Green’s Theorem, 966–968 k-component of, 961–964 Current, 17-18–17-19 Curvature calculating, formula for, 747–749 center of, 749 circle of, for plane curves, 749–750 computing, formulas for, 756–757 definition of, 747 normal vectors for space curves and, 750–751 normal vectors of a curve and, 747–751 of a plane curve, 747–750 radius of, 749 vector formula for, 756

Curve arc length definition of, 399–401 differential formula for, 402–403 discontinuities in, 402 discontinuities in dy/dx, 402 formula for, 399–401 function, 403 plates bounded by two, 425–426 smooth, 399 areas between, 364–366 assumptions on, 951 Bowditch, 642 cardioid, 648 chain, 464 closed, 945 contour, of functions of two variables, 770 generating, 714 infinite area under, 522 level, 810 level, of functions of two variables, 770 normal vectors of, 747–751 parametric, 1054 parametric, arc length of, 743–745 parametric, calculus with (See Calculus with parametric curves) parametrization of (See Parametrizations of plane curves) piecewise smooth, 728 plane circle of curvature for, 749–750 curvature of, 747–750 plane, flux across, 945–946 plates bounded by two, 425–426 polar, length of a, 652–653 principal unit normal vector for smooth, 748 sigmoid shape of, 1050 sinusoid, 642 sketching, 255–264 slope of, 138–139 smooth, 635–636, 728–729 snowflake, 560 space arc length and, 743–745 binormal vectors for, 753 computation formulas for, 757 curvature and normal vectors for, 750–751 definition of, 750 normal vectors for, 750–751 parametric equations for, 725, 736–738 tangents of, 725–731 work done by a force over a curve in, 942–944 synchronous, 764 tangent line to a level curve, equation for, 810 tangents to, 745 torsion along, 755

Subject Index

Cusp, 146 Cycloid brachistochrones, 628–629 calculus of variations in, 629 definition of, 628 Huygens’ pendulum clock and, 628, 629 parametric equations of, 628–629 tautochrones, 628–629 Cylinders definition of, 714 hyperbolic, 835–836 parametrization of, 974 slicing with, 391–393 Cylindrical coordinates, 903–907 definition of, 903 equations relating rectangular coordinates to, 903–905 equations relating spherical and Cartesian coordinates to, 907–909 integration in, 903–907 limits of integration in, finding, 905–907 motion in, 759 parametrization by, 974 Cylindrical shells, volumes using, 391–396 shell method, 393–396 slicing with cylinders, 391–393 Damped motion, 17-17–17-18 Damped vibrations, 17-18 Daubechies, Ingrid, 19-33 Daubechies scaling function, 19-33–19-34 Daubechies wavelet, 19-33–19-34 Decay exponential, 52–53, 448 radioactive, 452–453 rate, 53 Decreasing functions, 25–26 Dedekind, Richard, 327, 557 Definite integrals, 329–338 applications of, 380–435 arc length, 399–403 areas of surfaces of revolution, 405–408 moments and centers of mass, 420–429 volumes using cross-sections, 380–387 volumes using cylindrical shells, 391–396 work and fluid forces, 410–415 area under the graph of a nonnegative function, 336–337 average value of a continuous function, 337–338 definition of, 312, 329–331 existence of, 330–331 in integrable and nonintegrable functions, 331–332 integration by parts for evaluating, 482–483

as a limit of Riemann sums with equal-width subintervals, 331 Mean Value Theorem for, 342–343, 345–347 Midpoint Rule for approximating, 314, 332–333, 510 notation for, 330 properties of, 333–335 rules satisfied by, 334 shift property for, 371 substitution for, 361–367 of symmetric functions, 363–364 Trapezoidal Rule for the value of, 511–512 of vector functions, 735 Degrees angles measured in, 41 of polynomials, 29 ), 91, 93–95 De Moivre’s Theorem, 18-7, 19-5 Density circulation, 963–964 flux, 964–966, 1006–1007 mass, of surface integrals, 983, 985 of plates, 422 Dependent variables constrained, 848–849 of functions, 21 in functions of several variables, 767 Derivative Constant Multiple Rule, 152–153, 155 Derivative Power Rule applying, 152 generalizing, 160, 196 general version, 152, 195–197 irrational exponents and, 195–197 for negative integers, 160 positive integer, 151 Power Chain Rule and, 179–180 in solving a differentiation problem, 158 Derivative Product Rule definition of, 155–156 picturing, 156 proof of, 156 for trigonometric functions, 171, 172, 174 Derivative Quotient Rule, 157–158, 173, 180 Derivatives, 138–234 of absolute value function, 180 antiderivatives, 294–304 applications of, 235–311 applied optimization, 277–289 arctangent, 202–203 in biology, 167 calculating (See Differentiation) Chain Rule and, 176–184 complex (See Complex derivatives) of composite functions, 176–178, 814 concavity and curve sketching, 255–268 of a constant function, 151

continuous, 273 Cross Product Rule, 730–731 definition of, 142 calculating derivatives from, 143–144 (See also Differentiation) difference quotient and, 139–140 difference rules for (See under Differentiation) differentials and, 218–223 directional (See Directional derivatives) in direction of vectors, 728–729 Dot Product Rule, 730 in economics, 165–167 First Derivative Test for Local Extreme Values (General Version), B-8 of functions, 139, 142–151 cosecant, 173–174 cosine, 171–172 cotangent, 173–174 definition of, 142–143 differentiable (See Differentiable functions) exponential, 154–155 extreme values of, on closed intervals, 235–243 f at a point x0, 139 graphical behavior of, 263–264 hyperbolic, 458–462 inverse, 189–191 monotonic, and the first derivative test, 250–255 natural exponential, 438, 440–441 natural logarithm, 192 negative, 153 with no derivative at a point, 146 reciprocal, 143 recovering, from its derivative, 148 secant, 173–174 sine, 170–171 trigonometric, 170–176, 200–206 u and v for denoting, 153 vector, 727–729 General Power Rule for, 152, 195–197 graphing, 144–145 higher-order, 158 history of, 142 implicit differentiation and, 184–189 as instantaneous rate of change, 140, 161–162 intermediate forms and l’Hôpital’s Rule, 268–276 involving logax, 444–445 in irrational exponents and the Power Rule, 195–197 left-hand, 145–146 Leibniz’s Rule, 233–234, 377 linearization and, 214–217 in logarithmic differentiation, 195 Mean Value Theorem in, 245–249

I-5

Mixed Derivative Theorem, AP-31–AP-32, 790–791 motion and, 727–729 Newton’s dot notation for, 756 Newton’s method in, 289–294 notations for, 144 nth, 158 number e expressed as a limit and, 197 one-sided, 145–146 partial (See Partial derivatives) at a point, 138–142 rates of change and, 139–140 of a positive integer power, 151 of power series, 592 as a rate of change, 161–170 of related rates, 206–214 right-hand, 145–146 Rolle’s Theorem in, 243–244 second, 158 Second Derivative Test for Local Extreme Values (General Version), B-9 of the sine function, 170–171 of the square root function, 144, 146, 191 symbols for, 158 of the tangent function, 173–174 tangent lines and, 138–142 definition of, 138 finding, to graph a function, 138–139 slope of, 138–139, 144 of tangent vector, 748 third, 158 of trigonometric functions, 170–176 as velocity, 162, 729 of ax and logax, 193–194 Derivative Sum Rule, AP-12–AP-13, 153–154, 171 Descartes, René, AP-14, 184, 188 Determinants, B-1–B-6 absolute value of, 704 calculating a cross product as, 701–702 calculating the triple scalar product as, 703–704 cofactors in, B-2–B-5 definition of, B-1 evaluating, 701 formula, 701 Jacobian, 916, 920 matrix in, B-1–B-6 method of, 17-12 properties of, B-5 vector as, 703–704 Difference. See Subtraction Difference quotient, 139–140 definition of, 139 forms for, 143 limit of, 143 Difference Rule, 82–83, 153–154, 171 for antiderivatives, 296–297, 299 to calculate limits of sequences, 542 definition of, 153

I-6

Subject Index

Difference Rule (continued ) for double and triple integrals, 869 to find derivatives, 153–154, 171 for finite sums, 324 for gradients, 811 for limits, AP-23 for limits for functions of two variables, 777 for series, 556–557 for vector functions, 730 Differentiability on an interval, 145–146 definition of, 792–793 for functions of several variables, 792–793 partial derivatives and, 792–793 Differentiable functions absolute maxima and minima, 239–240 Cauchy’s Mean Value Theorem, 273–274 Chain Rule formula for functions of many variables, 803 functions of three variables, 798–799 functions of two variables, 796–798 implicit differentiation, 802 complex, 18-18–18-19 continuous, 147, 407, 408, 429, 635, 638, 639 with continuous first partial derivatives, 793 continuously, 635 definition of, 143, 792–793 domain of, 810 Extreme Value Theorem, 240 First Derivative Test for Local Extrema, 251–253 graphical behavior, 263–264 increasing and decreasing functions, 250–251 l’Hôpital’s Rule, 268–274 Mean Value Theorem, 245–246 of more than two variables, 821–822 negative, 153 with no derivative at a point, 146 on one-sided derivatives, 145–146 optimization problems, 278, 280 parametrized curves, 633–634 plane tangent to the surface of, 815–817 points of, 827 Rolle’s Theorem, 243 rules for (See under Differentiation) Second Derivative Test for concavity, 256 local extrema, 259–263 twice-, 633–634 vector functions, 728 Differentiable Implies Continuous Theorem, 793 Differential equations definition of, 297

first-order, 1024–1061 applications of, 1038–1043 autonomous differential equations, 1044–1051, 1053 autonomous systems of, 1053 Bernoulli differential equation, 1038 carrying capacity, 1049 competitive-hunter model, 1054–1055 curve, sigmoid shape of, 1050 definition of, 1024–1025 equilibrium values, 1044–1046 Euler’s method, 1027–1030 Euler solution, 1029–1030 exponential population growth model, inaccuracy of, 1039–1040 falling body encountering resistance, 1047–1049 first-order initial value problem, 1025–1026 first-order linear equations, 1032–1036 graphical solutions of autonomous, 1044–1051 homogeneous, 1061 initial value problems, 1025–1026 integrating factor, 1033–1035 Law of Exponential Change, 1039–1040 limit cycle, 1056 limiting population, 1049 logistic equation in Neural Networks and Machine Learning, 1050–1051 logistic growth, 1049 logistic model for population growth, 1049–1050 mixture problems, 1041–1043 motion with resistance proportional to velocity, 1038–1039 Newton’s Law of Cooling, 1047 Newton’s second law of motion, 1038, 1048 numerical method, 1027–1030 numerical solution of the problem, 1027–1030 orthogonal trajectories, 1038, 1040–1041 particular solution, 1025–1026 phase-plane analysis method, limitations of, 1056 phase planes, 1053–1054 predator-prey, 1058–1059 resistance proportional to velocity, 1038–1039 rest points, 1044–1046 RL circuits, 1036 slope fields, 1026–1027 solution curves, 1026–1027 solutions, 1024–1026 standard form of linear equations, 1032–1033 steady-state value, 1036

systems of, 1053–1056 terminal velocity, 355, 1049 initial value problems and, 297 second-order, 17-1–17-29 applications, 17-15–17-20 auxiliary equation, 17-3 boundary value problems, 17-6 boundary values, 17-6 constant-coefficient homogeneous equations, 17-2–17-5 damped motion, 17-17–17-18 damped vibrations, 17-18 definition of, 17-1 electric circuits, 17-18–17-19 Euler equations, 17-22–17-23 general solution, 17-2 homogeneous linear equations, 17-1 initial conditions, 17-5 initial value and boundary value problems, 17-5–17-6 linear combination, 17-1 linearly independent solutions, 17-2 linear second-order constantcoefficient models, 17-20 nonhomogeneous linear equations, 17-1, 17-7– 17-14 overdamping, 17-18 power-series solutions, 17-24–17-29 simple harmonic motion, 17-16–17-17 superposition principle, 17-1 trivial solution, 17-2 underdamping, 17-18 variation of parameters, 17-7, 17-12–17-14 vibrations, 17-15–17-16 separable, 448–450 solution of, 449 solving, 297 for vector functions, 735–736, 740 Differential formula for arc length, 402–403 Differentials arc length, 638–639 area change substituting, 916 Chain Rule in, 219, 221–222 definition of, 218, 820 error in approximation of, 220–221 estimating with, 219–220 forms of, 943, 958–959 geometric meaning of, 218 of integration by parts formula, 479–482 linearization and, 218–223 for a parametrized surface, 976 in partial derivatives, 820–821 surface area, for a parametrized surface, 976 tangent plane, 820–821 total, 820–822

Differentiation Chain Rule for implicit, 979 coefficients determined by, 502 definition of, 143 implicit, 184–189 integration and, relationship between, 349 logarithmic, 195 in related rates equations, 209 rules, 151–160, 729–731 Chain Rule, 176–184, 731 choice of, in solving a differentiation problem, 157–158 for constant function derivatives, 151 Cross Product Rule, proof of, 730–731 Derivative Constant Multiple Rule, 152–153, 155 Derivative Power Rule (See Derivative Power Rule) Derivative Product Rule, 155–156, 171, 172, 174 Derivative Quotient Rule, 157–158, 173, 180 Derivative Sum Rule, 153–154, 171 Difference Rule, 153–154, 171 Dot Product Rule, proof of, 730 for exponential function derivatives, 154–155 for second-and higher-order derivatives, 158 for vector functions, 729–731 Diocles, 188 Direct Comparison Test, 526–528, 568–569, 583 Directed line segment to represent vectors, 680 Direction along a path, 931 cosines, 697 downhill, 806 estimating change in a specific, 818 field, 1026–1027 forward, 939 rate of change in, 806 scalar component in, 694–695 of vectors, 684–685 Directional derivatives, 806–814 calculations of, 808–810 definition of, 806 as a dot product, 808–809 estimating change in a specific direction, 818 gradient vectors in, 808–814 interpretation of, 807–808 in a plane, 806–807 properties of, 809–810 zero, 814 Directrices for ellipses, 664–665 for hyperbolas, 664–665 of parabolas, 656–657 Dirichlet, Lejeune, 522 Dirichlet ruler function, 136

Subject Index

Discontinuity, 146, 402 Discontinuous functions, 122–123 Discriminant of local (relative) maxima or minima extrema, 827–829 Disk method, 382–385 Displacement, 162 definition of, 317 versus distance traveled, 317–318 work by force through, 696 Distance formula for points in a plane, AP-18 in a plane, AP-17–AP-19 from a point to a line, 708 from a point to a plane, 711 spheres in space and, 677–678 squares of, 897 traveled, 315–318 displacement versus, 317–318 total distance traveled, 317–318, 348 velocity function, 315–318 between two functions, 19-13 between two points in the xyplane, formula for, 677–679 Distributive law for cross products, AP-30–AP-31 Divergence curl and, 1009–1010 Divergence Theorem, 1006–1014 form of Green’s Theorem, 967–968, 1006 partial sums in, 555–556 of series, 552, 555–556 tests to determine for alternating series, 579–583 Alternating Series Estimation Theorem, 581 Alternating Series Test, 579–580 Direct Comparison Test, 568–569, 583 geometric series, 553–555, 583 Limit Comparison Test, 569–571, 583 nth-Term Test, 555–556, 583 p-Series Test, 563, 583 Ratio Test, 574–575, 583 Root Test, 576–577, 583 for series with some negative terms, 573, 581–583 summary of, 583 in three dimensions, 1006–1007 of a vector field, 964–966, 1006–1007 Divergence Theorem, 1006–1014 corollary, 1007 Green’s Theorem compared to, 1006–1007, 1014 for other regions, 1011–1012 proof of, for special regions, 1010–1011 for special regions, 1010–1011 statement of, 1007 Divergent improper integrals, 523, 526–528

Division of complex numbers, 18-4 of functions, 34–35 Domain assumptions on, 951 connected, 951 of differentiable functions, 810 of functions, 21–22, 34 of functions of several variables, 767, 768 natural, 22 parameter, 973 restrictions, applied to trigonometric functions, 60–61 simply connected, 951 Dominant terms, 116–117 Domination Rule for definite integrals, 334 for double and triple integrals, 869 Dot Product Rule, 730 Dot products, 691–697 Angle Between Two Vectors, 691–693 angles and, 692–693 definition of, 691 directional derivatives as, 808–809 Dot Product Rule, 730 of gradient vectors, 808–809 inner product of functions, 19-12, 19-14–19-15, 19-24 orthogonal vectors and, 693 properties of, 693–696 of three-dimensional vectors, 19-6 of two n-dimensional vectors, 696–697 work by force through displacement, 696 Double-angle formulas, 45 Double integrals area by, 873–875 definitions of, 860, 865 Fubini’s Theorem for calculating first form, 861–863 stronger form, 865–868 iterated or repeated, 861 limits of integration, finding (See Limits of integration, finding) over general regions, 864–870 over rectangles, 859–863 partitions formed by, 859–860 in polar form, 876–881 properties of, 869–870 Riemann sums for, 859–861, 869, 873 substitutions in, 915–919 volumes over general regions, 865–868 over rectangles, 860–861 of a solid region in space, 883–884 Downhill direction, 806 Drag coefficient, 742 Drag force, 741–742 D-tuple of numbers, 19-9 Dummy variable, 331

Eccentricity, 663–665 definition of, 664 of ellipses, 663–667 of hyperbolas, 663–667 of parabolas, 663–667 Economics, derivatives in, 165–167 Electric circuits, 17-18–17-19 Electric field, 1013 Electromagnetic theory, 1013 Elements of sets, AP-2 Elevation, angle of, 737 Ellipse, AP-20–AP-21, 657–658 axis of, AP-21 center of, AP-21, 657 center-to-focus distance of, 658 definition of, 657 directrices for, 664–665 eccentricity of, 663–667 equations for, 658 focal axis of, 657 law (Kepler’s first law), 760 major axis of, 658 minor axis of, 658 polar equations for, 665–667 semimajor axis of, 658 semiminor axis of, 658, 666–667 standard equation of, AP-21 tangent line to, 810–811, 817 vertices of, 657 Ellipsoid of revolution, 715 Ellipsoids, 714–718 Elliptical cones, 714–717 Elliptical paraboloid, 714–717 Elliptic integrals, 519 Empty set, AP-2 Endpoints of an interval, limits at, 100–101 continuity at, 120 extreme values and, 239 Energy, approximations in converting mass to, 223 Engineering, vectors in, 686 E, 91, 93–95 Equal area law (Kepler’s second law), 760–761 Equality in complex numbers, 18-3 of vectors, 680, 681 Equations Bernoulli differential, 1038 differential (See Differential equations) for ellipses, 658 homogeneous, 1061 for hyperbolas, 659–660 ideal projectile motion, 737 inverse (See Inverse equations) for parabolas, 656–657 parametric (See Parametric equations) parametric, for lines, 706–707 for a plane, 708–709 predator-prey, 1058–1059 separable, 448–450 solution of, 449 for the sphere of radius and center, 678–679

I-7

for the tangent line to a level curve, 810 vector for lines, 706 for planes, 709 Equilibria, stable and unstable, 1046–1047 Equilibrium value in graphical solutions of autonomous equations, 1044–1046 in systems of differential equations, 1054–1055 Error analysis, in numerical integration, 514–517 estimation, in approximating the sum of a series, 564–565 formula for linear approximations, 845 function, 530 in standard linear approximation, 820 term (or remainder of order n), 603–604 Euler, Leonhard, 1029 Euler equations, 17-22–17-23 Euler’s constant, 567 Euler’s Formula, 18-5, 18-27– 18-28, 19-5, 19-22 Euler’s gamma function, 536 Euler’s identity, 614–615 Euler’s method, 1027–1030 Euler’s solution, 1029–1030 Evaluation Theorem, 346–347 Net Change Theorem, 347–348 proof of, 346–348 Even functions, symmetry properties of, 26–27 Exact differential forms, 958–959 Expected values in multiple integrals, 900 Exponential change, 447–454 definition of, 448 heat transfer and, 453–454 radioactive decay and, 452–453 separable differential equations and, 448–450 unlimited population growth and, 450–452 Exponential decay, 52–53, 448 Exponential functions, 30, 49–54 algebraic properties of, 442–443 with base a, 30, 50 behavior of, 49–51 Chain Rule and, 438, 441–443 definition of, 49–50 derivatives of, 154–155, 194, 440–441 exponential decay, 52–53 exponential growth, 52–53 general, 443–444 graphs of, 30 integral of, 440–441 natural (See Natural exponential functions) Exponential growth, 52–53, 448

I-8

Subject Index

Exponential population growth model, inaccuracy of, 1039–1040 Exponents growth in, 448 irrational, Derivative Power Rule and, 195–197 laws of, 441–443 rules for, 51 Exterior regions of circle, AP-19 Extrema, 237–240 absolute (global) maxima or minima, 235–240 critical points and, 239 finding, 238–240 First Derivative Test for Local Extrema Values, 238–239 local (relative) maxima or minima, 237–240 Extreme values. See also Extrema absolute maxima and minima on closed bounded regions, 829–832 of continuous functions on closed, bounded sets, 781 critical points and, 239 definition of, 235 endpoints and, 239 Extreme Value Theorem, 236–237 finding extrema, 238–240 of functions of two variables, B-6–B-11 of functions on closed intervals, 235–243 inflection points and, 239 Lagrange multipliers for finding, 834–841 local, for functions of two variables, 825–829 local (relative) extreme values, 237–238 local extreme values for functions of two variables, 825–829 relative, 826 saddle points, 825–832 Extreme Value Theorem, 236–237 Falling body encountering resistance, 1047–1049 Faraday’s law, 1021 Fermat, Pierre de, 75 Fermat’s principle in optics, 279–280 Fibonacci numbers in sequences, 545 Fields in algebraic properties, AP-28 conservative component test for, 955–957 line integrals in, 952–955 potentials for, finding, 955–957 properties of, 951 Stokes’ Theorem and, 1003–1004 curl vector, 993–995 electric, 1013 gradient, 938–939 gravitational, 950 ordered, AP-28

Finite approximations for area, 315–320, 437 theory of limits of (Riemann sums), 326–328, 331 Finite intervals, AP-3 Finite limits as x q ±∞, 107–108 Finite sums, 322–328 algebra rules for, 324 for estimating area, 312–320 estimating area with, 312–320 limits of, 325–326 Riemann sums and, 326–328 sigma notation and, 322–325 Firing angle, 737 First Derivative Test for Local Extreme Values, B-8, 238–239, 251–253, 826, 836 First moments about coordinate planes, 934, 990 about the coordinate planes, 894 definition of, 893–894 formulas for, 894–895 of a line integral, 934, 990 masses and, 893–895 for three-dimensional solid, 894 for two-dimensional plate, 894 First octant, 675 First-order differential equations. See under Differential equations First-order initial value problem, 1025–1026 First-order linear equations, 1032–1036 definition of, 1032–1033 integrating factor for, 1033–1035 RL circuits, 1036 solving, 1033–1035 Fixed costs, 166 Fixed point, 132 Flight time, for ideal projectile motion, 738 Flow integrals, 944–945 Fluid flow rates, 965 force centroids and, 427 on a constant-depth surface, 414–415 integral for, against a vertical flat plate, 415 pressure and, 414–415 pressures, force and, 414–415 weight-density, 414 Flux across a plane curve, 945–946 across a rectangle boundary, 965 calculation of, 945, 987 circulation differentiated from, 945 definition of, 945, 987 density, 964–966, 1006–1007 in Divergence Theorem, 1006–1010 evaluating the integral for, 945–946 in Green’s Theorem, 967–968 magnetic, 1021 surface integrals for, 987–988

Focal axis of ellipses, 657 of hyperbolas, 658 Focal length of parabolas, 656 Focus of parabolas, 656–657 Folium, 184, 188 Force constant, work done by, 410–411 drag, 741–742 fluid centroids and, 427 on a constant-depth surface, 414–415 pressure and, 414–415 work and, 410–415 done by constant force, 410–411 done by variable force along a line, 411 Hooke’s Law, 411–412 lifting objects and pumping liquids from containers, 412–413 work done by, through displacement, 696 Force constant, 412 Formula for Implicit Differentiation, 801–803 Forward direction, 939 Fourier, Jean-Baptiste Joseph, 19-1 Fourier coefficients, 19-18–19-19 Fourier series, 19-1 coefficients in, 19-18–19-19 convergence of, 19-19–19-21 Parseval Equality for, 19-21–19-22 Fractions method of partial, 497–502 partial, definition of, 497 Free fall, 72 Frenet, Jean-Frédéric, 753 Frenet frame, 753 Frequency in periodic functions, 19-2 Frequency zero, 19-2 Fubini, Guido, 862 Fubini’s Theorem for calculating double integrals over general regions (stronger form), 865–868 for calculating double integrals over rectangles (first form), 861–863 Functions, 21–71 absolute value, 25 adding, 34–35 approximation of, 19-16–19-19 arc length, 402–403, 744 arcsine and arccosine, 61–64 area under the graph of nonnegative, 336–337 arrow diagram of, 22 average value of for approximating area of rectangles, 319–320 continuous, 337–338 nonnegative continuous, 318–320 in space, 890

bounded, 81, 136 box, 19-25 circular, 458 on closed intervals, extreme values of, 235–243 combining, 34–41 common, 27–31 (See also individual functions) algebraic, 29–30 exponential, 30, 49–54 inverse, 31, 55–57, 60–61 linear, 27 logarithmic, 31, 57–60 polynomials, 29 power, 27–28 rational, 29 transcendental, 31 trigonometric, 30, 41–49 complex differentiable, 18-18–18-19 of complex variables, 18-12– 18-15 component, 725, 937 compositions of, 35–36, 125–126 constant, 27, 81 continuity and, 779–781 of compositions, 781 continuous, definition of, 779 continuous at the point, 779–780 continuous extension, 128–129, 783, 853 partial derivatives and, 789–790 continuously differentiable, 635 critical points of (See Critical points) cube root, 28 cubic, 29 Daubechies scaling, 19-33–19-34 decreasing, 25–26, 250–251 defined by formulas, 34 defined on surfaces, 799–801 definition of, 21 dependent variable of, 21 derivative of, 139, 142–151 differentiable, 786, 792–793 discontinuous, 122–123 distance between two, 19-13 dividing, 34–35 domain of, 21–22, 34, 768 dominant terms in, 116–117 error, 530 even, symmetry properties of, 26–27 exponential (See Exponential functions) extreme values on closed intervals, 235–243 of continuous functions on closed, bounded sets, 781 saddle points and, 825–832 finitely many, sums of, AP-12–AP-13 Fourier series convergence of, 19-19–19-21 Parseval Equality for, 19-21–19-22 general sine, 47

Subject Index

gradient of, 811–813 graphing with software, 772 graphs of, 23 of arcsine and arccosine functions, 62–63 of catenary, 31 exponential functions, 30 logarithmic functions, 31 reflecting, 36–38 scaling, 36–38 shifting, 36 symmetric, 26–27 transformations of trigonometric, 46–47 trigonometric functions, 30, 44, 46–47, 62–63 greatest integer, 25 growth rates of, 465–469 Hessian of, 827–829 hyperbolic (See Hyperbolic functions) identity, 27, 57, 59, 80–81, 1003 Implicit Function Theorem, 802–803 implicitly defined, 184–186 increasing, 25–26, 250–251 Increment Theorem of, AP-33–AP-35 independent variable of, 21 infinitely many orthonormal, 19-20 inner product of, 19-12, 19-14–19-15, 19-24 input variable of, 21, 36 integer ceiling, 25 integer floor, 25 integrable, 19-12–19-13, 331–332 joint probability density, 899 least integer, 25 limits of, 72–137 linearization of, 819–820 definition of, 819 function of two variables, 818–820 functions of more than two variables, 821–822 standard linear approximation and, 819–820 logarithmic, 18-28–18-29 as a machine, diagram of, 22 of many variables, 803 marginal cost, 165–166, 280–281 maximum and minimum values of, 825–837 monotonic, 250–255 of more than two variables differential, 821–822 linearization of, 821–822 in tangent planes, 821–822 multiplying, 34–35 natural domain of, 22 natural exponential, 52 natural logarithm, 58 nonintegrable, 331–332 nonnegative area under graph of, 337–338 norm of, 19-12–19-14 numerical representation of, 24

odd, symmetry properties of, 26–27 one-to-one, 54–55 orthogonal, 19-15, 19-24 orthogonality of the trigonometric system, 19-15–19-16 orthonormal, 19-12, 19-15–19-16, 19-20, 19-24, 19-27 output variable of, 21, 36 of partial derivatives, 784–793 periodic, 19-1–19-2 piecewise continuous, 331, 376 piecewise-defined, 25 piecewise-smooth, 728 position, 25 positive, area under graph of, 319 potential, 951 probability density (See Probability density functions) proportional relationship of, 27 Pythagorean Theorem for, 19-15 quadratic, 29 range of, 21–22, 768 real-valued, 22, 725–726, 767–768 reciprocal, derivative of, 143 representation as power series, 586–594 root of, 289 scalar, 725–726, 930–935, 943, 984 scaling of, 36–38 scatterplot of, 24 of several variables, 767–772 definitions of, 767 dependent/independent variables, 767 differentiability for, 792–793 domain of, 767, 768 input/output variables, 767 partial derivatives of, 767–772 range of, 767, 768 real-valued function, 767 shift formulas for, 36 sine-integral, 530 smooth, 399 square root, 28 subtracting, 34–35 symmetric, 26–27, 363–364 of three variables, 770–772 boundary of, 771 boundary points of, 771 bounded/unbounded regions of, 771 Chain Rule for, 798–799 definitions of, 770–772 gradient vectors in, 811–812 interior points of, 771 interior regions of, 771 level surface of, 770 open/closed regions of, 771–772 torsion, of a smooth curve, 755 total area under graph of, 349–350 total cost, 166 transcendental, 31, 445

trigonometric (See Trigonometric functions) twice-differentiable, 633–634 of two variables boundary of, 769 boundary point of, 769 Chain Rule for, 796–798 Composition Rule for, 777 Constant Multiple Rule for, 777 contours of, 770 definitions of, 769 extreme values of, B-6–B-11 graphs, level curves, and contours of, 770 graphs/graphing, 770 Increment Theorem for Functions of Two Variables, AP-33–AP-35, 793 interior of the region of, 769 interior point of, 769 level curves of, 770 limits for, 775–779 linearization of, 818–820 more than, 781, 788–789 open/closed regions of, 769 partial derivatives of, 784–786 properties of limits of, 777 saddle points of, B-6–B-11 Sandwich Theorem for, 783 Taylor’s formula for, 846–847 values of local extreme for, 825–829 unit, 19-13, 19-24 unit step, 81–82 values, 22 limits of, 79–80 zero, 127 vector (See Vector functions) velocity, 315–318, 738–739 vertical line test for, 24 Fundamental period, 19-1 Fundamental Theorem of Algebra, 18-9 Fundamental Theorem of Calculus, 342–350 for arc length differential, 638–639 for continuous vector functions, 735–736 to define the natural logarithm function ln x as an integral, 436–438 differential formula for arc length, 402–403 Divergence Theorem and, 1014 double integrals evaluated with, 862 Fundamental Theorem of Line Integrals, 952–954 Green’s Theorem and, 967 integration and differentiation in, relationship between, 349 for line integrals, 930, 931, 952–954 Part 1 (Mean Value Theorem), 342–345, 438 proof of, 345

I-9

Part 2 (Evaluation Theorem), 346–347 Net Change Theorem, 347–348 proof of, 346–348 path independence and, 951 total area and, 349–350 Fundamental Theorem of Line Integrals, 952–954 Galileo Galilei, 72 free fall formula, 72, 164, 168 law of, 72 Gauss, Carl Friedrich, 324, 681 Gauss’s Law, 1013 General exponential function, 443–444 General linear equation, AP-17 General quadric surfaces, 716, 718 General sine function, 47 General solution, 449 General solution of differential equations, 297 Generating curve, 714 Genetic data and sensitivity to change, 167 Geometric series, 553–555, 583 Geometry of space, 675–724 angles between planes, 711–712 cross products, 699–704 cylinders, 714 distance and spheres in space, 677–678 distance from a point to a line, 708 distance from a point to a plane, 711 dot products, 691–697 equation for a plane, 708–709 line of intersection, 709–710 lines and line segments, 706–708 quadric surfaces, 714–716 general, 716, 718 graphs of, 717 three-dimensional coordinate systems, 675–678 vectors, 680–688 Gibbs, Josiah Willard, 744 Gibbs phenomenon, 19-21 Global (absolute) maxima or minima extrema. See Absolute (global) maxima or minima extrema Gradient ascent, method of, B-14 Gradient descent, method of, B-12–B-14 Gradient fields, 938–939 Gradient vectors algebra rules for, 811 in Chain Rule for paths, 813–814 curl of, 1003 definition of, 808 in directional derivatives, 808–814 dot product of, 808–809 in functions of more than three variables, 812–813 in functions of three variables, 811–812 to level curves, 810–811 nonzero, 810–811 Orthogonal Gradient Theorem, 837–838

I-10

Subject Index

Graphs/graphing of an equation or inequality, AP-14 asymptotes of horizontal, 109–111 oblique, 111–112 slant line, 111 vertical, 115–116 of autonomous differential equations, 1044–1051 equilibrium values or rest points in, 1044–1046 falling body encountering resistance, 1047–1049 Logistic Equation in Neural Networks and Machine Learning, 1050–1051 logistic model for population growth, 1049–1050 Newton’s Law of Cooling, 1047, 1048 phase lines, 1044–1046 stable and unstable equilibria, 1046–1047 terminal velocity, 355, 1049 connectedness and, 127 curve sketching, 255–264 of functions, 23 arcsine and arccosine, 62–63 catenary, 31 continuous, 127–128 exponential, 30 logarithmic, 31 reflecting, 36–38 scaling, 36–38 shifting, 36 symmetric, 26–27 trigonometric f, 30, 44, 46–47, 49, 62–63 to verify limit statements for, 91–93 of functions of two variables, 770 of polar equations, 643–644, 646–649 of quadric surfaces, 717 secant lines on, 74–77 with software, 772 surface area of, formula for, 981 tangent lines on, 74–77 Grassmann, Hermann, 684 Gravitational fields, 950 definition of, 950 vectors in, 938 Greatest integer, 25 Greatest integer function, 122, 127 Greatest lower bound sequence, 546 Greek letters to describe prescribed errors, 91 Green’s first formula, 1016 Green’s second formula, 1016 Green’s Theorem area formula, 972 divergence form of, 967–968, 1006 Divergence Theorem compared to, 1006–1007, 1014 to evaluate line integrals, 968–969

generalizations in three dimensions, 1014 generalization to three dimensions, 1014 normal form of, 1013, 1014 in a plane, 961–970 for special regions, 969–970 Stokes’ Theorem compared to, 995–996, 1002–1003, 1014 tangential form of, 966–967, 1014 Growth, exponential, 52–53 Growth models, exponential population, 1039–1040 Growth rates of functions, 465–469 definition of, 465–467 order and oh-notation, 467–468 sequential vs. binary search, 468–469 Haar coefficients, 19-27–19-29 Haar scaling function, 19-24– 19-29 Haar system, 19-27, 19-29–19-31 adding details at resolution level zero, 19-30–19-31 approximations using, 19-27, 19-29–19-31 box function in, 19-25 coefficients in, 19-27–19-29 definition of, 19-27 multiresolution recursion, 19-31–19-32 scaling function in, 19-24–19-29 theorems for, 19-27, 19-29, 19-32 wavelets in, 19-24–19-26, 19-31, 19-33–19-34 Haar wavelet, 19-24–19-26, 19-31, 19-33–19-34 Half-angle formulas, 45 Half-life, 60, 452 Half-open intervals, AP-3 Harmonic analysis, 19-1 Harmonic series, 563, 579–583 alternating, 579–583 definition of, 579–583 p-Series Test and, 563 Heat equation, 796 Heat transfer, 453–454 Heaviside, Oliver, 501 Height, for ideal projectile motion, 738 Helix, 726 Hessian matrix, B-9–B-11, 828 Hessian of a function, 827–829 Higher dimension vectors, 19-9–19-11 definition of, 19-9–19-10 orthonormal bases and approximation in, 19-10–19-11 theorems, 19-10–19-11 Homogeneous equations, 1061 Homogeneous linear equations, 17-1 Hooke’s Law, 17-15–17-16, 172, 411–412 Horizontal asymptotes, 108–111

Horizontal cross-sections, for finding limits of integration, 868 Horizontal line test, for one-to-one functions, 55 Horizontal scaling and reflecting formulas, 37 Horizontal shifts, 36 Huygens, Christiaan, 628 Huygens’ pendulum clock, 628, 629 Hyperbolas, 658–660 asymptotes of, 659 branches of, 655 Cartesian equation for, 665 center of, 658 definition of, 658 directrices for, 664–665 eccentricity of, 663–667 equations for, 659–660 focal axis of, 658 polar equations for, 665–667 vertices of, 658 Hyperbolic cylinder, 835–836 Hyperbolic functions, 457–462 definition of, 457–458 derivatives of, 458–459 inverse hyperbolic functions, 460–462 identities of, 457–458, 460 integrals of, 458–459 inverse, 459–460 derivatives of, 460–462 Hyperbolic paraboloid, 715–717 Hyperboloids, 714–717 Hypocycloids, 632 i-component of vectors, 684–685 Ideal projectile motion. See under Projectile motion Identity Euler’s Formula and, 18-5 function, 27, 57, 59, 80–81, 1003 of hyperbolic functions, 457–458, 460 in inverse trigonometric functions, 204 involving arcsine and arccosine functions, 63–64 trigonometric, 44–46 Image, 915 Imaginary axis, 18-4 Imaginary parts of complex numbers, 18-3–18-4 Implicit differentiation, 184–189, 801–803 Formula for Implicit Differentiation, 801–803 Implicit differentiation, Chain Rule for, 979 Implicit Function Theorem, 802–803, 978 Implicitly defined functions, 185–186 Implicit surfaces, 978–981 Improper integrals with CAS, 525–526 convergent and divergent, 523 tests for, 526–528 infinite limits in, 521–523

integrands with vertical asymptotes, 524–525 in integration, 520–528 of Type I, 521–524 of Type II, 524–525 Increasing functions, 25–26 Increments, AP-14–AP-17 Increment Theorem for Functions of Two Variables, AP-33–AP-35 Indefinite integrals, 299–300, 354–359 definition of, 299–300, 354 of exponential function, 441 of natural exponential function, 441 substitution for, 354–359 of vector functions, 734 Independence, path, 951 Independent constrained variables, 848–849 Independent variables definition of, 21 for functions of one independent variable and three intermediate variables, 798 of one independent variable and two intermediate variables, 796 of several variables, 767 for two independent variables and three intermediate variables, 799 Indeterminate forms ∞/∞,∞ . 0, ∞–∞, 270–272 0/0, 268–270 Indeterminate forms, in Taylor series, 613–614 Indeterminate powers, 272–273 Index of sequences, 539–540 of summation, 323–324 Inductance, 17-19 Induction, mathematical, AP-10– AP-13 Inequality graphs/graphing of, AP-14 rules for, AP-1 solving, AP-3–AP-4 in trigonometric functions, 46 Inertia, moments of, 895–898, 934, 990 Infinite area, 521–523 Infinite discontinuity, 123 Infinite intervals, AP-3 Infinite limits, 521–523 Infinitely many orthonormal functions, 19-20 Infinite right-hand limit, 120 Infinite sequences, 539–540 Infinite series, 538, 551–558 Infinity divergence of sequences to, 542 limits involved in, 107–132 finite limits as, 107–108 finite sums and, 325–326

Subject Index

horizontal asymptotes, 108–111 infinite limits, 111–114 limits at infinity of rational functions, 108 oblique asymptotes, 111 vertical asymptotes, 115–117 negative, 107, 114 Inflection points, 239, 256–259 Initial conditions, 17-5 Initial point of vectors, 680 Initial ray, in polar coordinates, 642 Initial speed in ideal projectile motion, 737 Initial value problems boundary value and, 17-5–17-6 definition of, 297 first-order, 1025–1026 Inner products. See Dot products Input variable, 21, 36, 767 Instantaneous rates of change, 76–77, 79, 102, 140, 161–162. See also Rates of change Instantaneous speed, 73 Integer ceiling, 25 Integer floor, 25 Integers, 18-1–18-2 function, greatest, 122, 127 greatest, 25, 122, 127 least, 25 negative, Derivative Power Rule for, 160 starting, AP-12 Integrable functions, 331–332 Integrals area, 312–315 Brief Table of Integrals, 504 complete elliptic, of the second kind, 637 definite (See Definite integrals) distance traveled, 315–318 elliptic, 519 exponential change and, 447–448 of exponential function, 440–441 finite sums, 322–328 flow, 944–945 for fluid force against a vertical plate, 415 Fundamental Theorem of Calculus, 342–350 of hyperbolic functions, 458–459 indefinite (See Indefinite integrals) inside integral (functions), 866, 885 iterated (See Iterated integrals) logarithm defined as, 436–445 of lower sums, 313–314 Midpoint Rule, 314, 332–333, 510 multiple (See Multiple integrals) nonelementary, 508, 611–612 nonnegative continuous function, average value of, 318–320 outside (constants), 866, 885 in polar coordinates, 876–878 Product Rule in integral form, 442–482 of a rate, 347–348

repeated (See Iterated integrals) sine-integral function, 530 substitution in, 354–367 for definite integrals, 361–367 for indefinite integrals, 354–359 surface, 983–991, 1003 trigonometric (See Trigonometric integrals) of upper sums, 313–314 of vector functions, 734–739 work, 942–944 Integral sign, 299 Integral tables, 504–505 Integral Test, 18-24, 561–565 Integrands, 299, 524–525 Integrate command, in CAS, 506–507 Integrating factor, for first-order linear equations, 1033–1035 Integration basic formulas in, 473–477 with CAS, 506–507 in cylindrical coordinates, 903–907 definition of, 312 differential version, 479–482 differentiation and, relationship between, 349 formulas, 300 improper integrals in, 520–528 with CAS, 525–526 convergent and divergent, 523 tests for, 526–528 infinite limits in integration, 521–523 integrands with vertical asymptotes, 524–525 integral tables, 504–505 limits of, finding (See Limits of integration, finding) limits of, for multiple integrals (See Multiple integrals) nonelementary integrals in, 508 numerical, 510–517 error analysis in, 514–517 Midpoint Rule for approximating integrals, 510 parabolas used for approximations, Simpson’s Rule and, 512–514 trapezoidal approximations, 511–512 by parts, 478–483 evaluating definite integrals, 482–483 formula, 479–482 Product Rule in integral form, 442–482 by parts formula, 479 of rational functions by partial fractions, 497–502 coefficients, determining by differentiating, 502 method used in, general description of, 497–501 reduction formulas, 505–506 with respect to y, area between curves in spherical coordinates, 907–911

with substitution (See Substitution, in integrals) techniques of, 473–537 term-by-term for power series, 592–594 by trigonometric substitutions, 492–495 variable of, 299, 330–331 in vector fields, 930–1021 of vector function, 734–739 Interest, compounding continuously, 52–53 Interior definition of, B-7 intervals, AP-3 region of a circle, AP-19 Interior points of absolute (global) maxima or minima extrema, 830–831 closed bounded, absolute maxima and minima on, 829–832 definition of, B-7 of intervals, AP-3 in a plane, 769 in space, 771 Intermediate Value Property, 126 Intermediate Value Theorem, 126–128 Intermediate Value Theorem for Continuous Functions, 342 Intermediate variables for functions of one independent variable and three intermediate variables, 798 of one independent variable and two intermediate variables, 796 for two independent variables and three intermediate variables, 799 Intersection of sets, AP-2 Intervals closed, continuous functions over, 122, 127 closed, extreme values of functions on, 235–243 continuity on, 124 of convergence, 591 definition of, AP-3 differentiability on, 145–146 limits at endpoints of, 100–101 Inverse finding, 56–57 of hyperbolic functions, 460–462 of ln x and the number e, 439–440 in logarithmic functions, 18-29–18-29 properties of logarithms, 59 Inverse equations, 440 definition of, 440 for ln x, 440 for natural exponential functions, 440 for ax and logax, 444 Inverse functions, 31, 55–57 Chain Rule in differentiable functions, 191

I-11

continuity and, 124–125 definition of, 55 trigonometric, 60–61 arccosecant, 200–201 arccosine, 201 arccotangent, 200–201 arcsecant, 200–201, 203 arcsine, 189, 201, 202 arctangent, 200–203 derivatives of, 200–206 identities involving, 204 Inversely proportional, 27 Irrational exponents, 195–197 Irrational numbers, AP-2 Irreducible quadratic polynomial, 498 Isolated singularity, 18-25 Iterated integrals, 884–890 definition of, 861 finding, 884–890 volume of a solid region in space, 883–884 Jacobi, Carl Gustav Jacob, 915 Jacobian determinant, 916, 920 Jacobian of the transformation, 917–919, 922 j-component of vectors, 684–685 Jerk, in acceleration, 163–164, 173 Joint probability density function, 899 Joule (J), 410–411 Joule, James Prescott, 411 Jump discontinuity, 121, 123 k-component of, 961–964 k-component of vectors, 684–685 Kepler, Johannes, 761 Kepler’s first law (ellipse law), 760 Kepler’s second law (equal area law), 760–761 Kepler’s third law (time-distance law), 761–762 Kinetic energy, 33 converting mass to, approximations in, 223 Kirchhoff, Gustav R., 17-19 Kirchhoff’s law, 17-19 Koch, Helge von, 560 Korteweg-de Vries equation, 796 Kovalevsky, Sonya, 461, 825 Kth subinterval, 327 Lagrange, Joseph-Louis, 245 Lagrange, Joseph Louis, 834 Lagrange multipliers, 834–841 constrained maxima and minima, 834–837 definition of, 837 geometry of the solution, 839–840 method of, 837–840 Orthogonal Gradient Theorem and, 837–838 partial derivatives and, 834–841 with two constraints, 840–841 Laplace, Pierre-Simon, 790 Laplace equations, 795, 805

I-12

Subject Index

Laws of cosines, 45–46 of exponents, 441–443 Law of Exponential Change, 1039–1040 Law of Refraction, 280 Limit Laws, 82–83 Learning rate, B-13 Least integer, 25 Least squares line, 834 Least squares solution, 19-10–19-11 Least upper bound for a sequence, 546 Least upper bound for a set of numbers, AP-28 Left-continuous function (or continuous from the left), 121–122 Left-hand derivatives, 145–146 Left-hand limit (or limit from the left), AP-25, 99–104, 121, 122 Legendre, Adrien-Marie, 1034 Leibniz, Gottfried, 330, 342 Leibniz’s formula, 613 Leibniz’s notation, 144, 330 in Chain Rule, 177 for definite integral, 330 in differentials, 214, 218 in outside-inside rule, 178–179 for writing the formula for arc length, 635 Leibniz’s Rule, 233–234, 377 Leibniz’s substitution method, 354–359 Length arc, in space, 743–745 arc length differential, 638–639 constant, vector functions of, 731 focal, of parabolas, 656 of parametrically defined curves, 635–638 in polar coordinates, 650–653 of a polar curve, 652–653 of a smooth curve, 743 of three-dimensional vectors, 19of vectors, 680, 681 Level curves of functions of two variables, 770 Level surface of functions of three variables, 770–771 Level surfaces, 770 L’Hôpital, Guillaume François Antoine de, 268 L’Hôpital’s Rule, 268–274 Cauchy’s Mean Value Theorem, 273–274 indeterminate forms, 268–274 indeterminate powers, 272–273 proof of, 273–274 for sequences, 543–544 Limaçons, graphing, 650 Limit Comparison Test, 569–571, 583 Limit Product Rule, AP-23–AP-24 Limit Quotient Rule, AP-24–AP-25 Limits, 72–137

commonly occurring, AP-26– AP-27, 545 complex, 18-2–18-3 Constant Multiple Rule for, 82–83 continuity and, 120–132 of continuous functions, 126 for cylindrical coordinates, 905–907 for definite integrals, 331 definition of general, 137 informal, 79–82 precise, 90–95 to prove theorems, 95 testing, 91–93 deltas for given epsilons, finding algebraically, 93–95 of difference quotient, 139–140 Difference Rule for, 82–83 estimating, by using calculators and computers, 84–85 finding by eliminating common factors from zero denominators, 83–84 for multiple integrals, 868 process of, 80–82 for functions of two variables, 775–779 definition of, 775–776 properties of, 777–779 two-path test for nonexistence of a limit, 780 of function values, 79 graphs to verify limit statements for functions, 91–93 history of, 79 indeterminate forms of, 268–275 infinite, in improper integrals, 521–523 infinity involved in, 106–132 finite limits as x q ±h, 106–108 finite sums and, 325–326 horizontal asymptotes, 108–111 infinite limits, 112–113 infinite right-hand limit, 120 limits at infinity of rational functions, 108 oblique asymptotes, 111–112 vertical asymptotes, 115–117 laws, AP-23 left-hand (or limit from the left), 99–104, 121, 122 Limit Comparison Test, 527–528 Limit Laws, 82–83 number e as, 197 one-sided, 99–104 approaching, 99–100 at endpoints of an interval, 100–101 involving sine, 102–104 precise definitions of, 101–102 of polynomials, evaluating, 83 Power Rule for, 82–83 precise definitions of, 90–95 Product Rule for, 82–83 Quotient Rule for, 82–83

of rational functions, evaluating, 83 for rectangular coordinates, 884–890 of Riemann sums, 326–328, 331, 394 right-hand (or limit from right), 99–104, 121, 122 r-limits of integration, finding of, 878, 904, 906 Root Rule for, AP-23, 82–83, 777 Sandwich Theorem for calculating, 85–87 of sequences, 540 calculating, 542–543 commonly occurring, 545 definition of, 540 of sine, 102–104 for spherical coordinates, 909–910 Sum Rule for, 82–83, 95 theorems, proofs of, AP-23–AP25 Left-Hand Limits, AP-25 Limit Product Rule, AP-23–AP-24 Limit Quotient Rule, AP-24–AP-25 Right-Hand Limits, AP-25 Sandwich Theorem, AP-25 Two-Sided Limits, AP-25 theory of (Riemann sums), 326–328, 331 two-sided, 99, 103, 116, 122, 126 of vector-valued functions, 726–727 Limits of integration, finding for double integrals, 859, 861, 864, 866, 868 in polar form, 878–879 in Fubini’s Theorem, 861, 867 in iterated integrals, 884–890 for iterated triple integrals, 884–890 lemniscate for, 879 for multiple integrals (See Multiple integrals) F-limits, 909, 910 in polar coordinates, 878–879 in rectangular coordinates, 878–879 r-limits, 909, 910 r-limits, 878, 904, 906 by sketching, 868, 878, 884, 905, 909 u-limits, 878, 905, 906, 909, 910 for triple integrals in cylindrical coordinates, 905–907 in rectangular coordinates, 884–890 in spherical coordinates, 909–911 using horizontal cross-sections, 868 using vertical cross-sections, 868 x-limits, 868, 885, 886, 889

y-limits, 868, 885–889 z-limits, 885–889, 904–906 Linear algebra, B-4 Linear approximation, 215–217 center of, 215, 217 error formula for, 845 for roots and powers, 217 standard, 819–820 Linear combination, 17-1, 19-7 Linear drag, 742 Linear equations, AP-17, 1032–1033 Linear functions, 27 Linearization, 214–217 definition of, 215, 819 differentials and, 218–223 of a function of two variables, 818–820 of functions of more than two variables, 821–822 linear approximation and, 215–217 standard linear approximation and, 819–820 Linearly independent solutions, 17-2 Linear regression, 1051 Linear second-order constantcoefficient models, 17-20 Linear transformations, 916 Line integrals additivity and, 932 in conservative fields, 952–955 definitions of, 931, 940 evaluating, 931, 940 formulas for, 941 Fundamental Theorem of, 952–954 mass calculations and, 933–934 moment calculations, 933–934 in a plane, 934–935 with respect to dx, dy, or dz, 940–941 of scalar functions, 930–935 of vector fields, 939–941 work done by a force over a curve in space, 942–944 Lines center of mass along, 420–421 of intersection, 709–710 least squares, 834 line segments and, 706–708 normal, tangent planes and, 815–817 parallel, AP-17 parametric equation for, 706–707 perpendicular, AP-17 regression, 834 slope of a nonvertical, AP-15 straight, increments, AP-14–AP-17 tangent, 633–634 trend, 834 vector equation for, 706 work done by variable force along, 411 Line search, B-13 Line segments, 706–708 directed, to represent vectors, 680 midpoint of, 685

Subject Index

Line through u1, 19-11 Lissajous figures, 642 ln x algebraic properties of, 438 derivative of, 438 graph and range of, 438–439 integral of, 439 inverse equation for, 440 inverse function of, 439–440 number e and, 439–440 2-place values of, 437 Local (relative) maxima or minima extrema critical points in, 826–829 definitions of, 237, 825 discriminant or Hessian of, 827–829 finding, 238–240 finding the absolute extrema of a continuous function f on a finite closed interval, 239–240 First Derivative Test for Local Extrema Values, 238–239, 251–253, 826 Second Derivative Test for Local Extreme Values, 259–261, 827, 829, 831, 835, 844–845 values for functions of two variables, 825–829 Local maximum, B-7, 825–829, 831 Local minimum, B-7, 825–829, 831 Logarithmic differentiation, 195 Logarithmic functions, 31, 57–60 algebraic properties of, 58–59 applications, 59–60 with base a, 57 common, 58 of complex numbers, 18-28–18-29 definition of, 57 graphs of, 31 inverses in, 18-29–18-29 natural, 58–59 principal branch of, 18-29 Logarithms base a, 444 defined as integrals, 436–445 laws of exponents and, 441–443 properties of, 58–59 algebraic, 58–59 change-of-base formula, 59 inverse, 59 Logax derivatives and integrals in, 444–445 inverse equation for, 444 Logistic Equation in Neural Networks and Machine Learning, 1050–1051 Logistic growth, 1049 Logistic model for population growth, 1049–1050 Loops, 945 Lotka-Volterra equations, 1058 Lower bound sequence, 546 Lower sums, 313–314

Machine Learning, Logistic Equation in, 1050–1051 Maclaurin, Colin, 598 Maclaurin series, 598–600, 604 Magnetic flux, 1021 Magnitude, 680, 681 Main diagonal, B-4 Major axis, AP-21 Major axis of ellipses, 658 Marginal costs, 165–166, 280–281 Marginal profit, 280–281 Marginal revenue, 280–281 Marginals, 165–167 Mass along a line, 420–421 calculations, line integrals and, 933–934 center of, 894–895 centers of, moments and, 420–429 centroids, 426–427 definition of, 426 fluid forces and, 427 converting to energy, approximations in, 223 definition of, 893 density, of surface integrals, 983, 985 distributed over a plane region, 422 first moments and, 893–895 formulas, 894–895 in multiple integrals, 893–895 plates bounded by two curves, 425–426 thin, flat, 422–425 of surface integrals, 983, 985 theorems of Pappus Pappus’s Theorem for Surface Areas, 429 Pappus’s Theorem for Volumes, 427–428 of thin shells, 989–991 thin wires, 421 Mathematical induction, AP-10–AP-13 Mathematics, applied optimization in, 279–282 Matrix, B-1–B-6 cofactor in, B-2–B-5 definition of, B-1 Hessian, B-9–B-11 main diagonal of, B-4 minor defined in, B-1–B-2 Maximum absolute, 825–826, 829–932 constrained, 834–837 local, 825–829, 831 Max-Min Inequality Rule, 334–335 Max-Min Tests, 831 Mean. See also Average value definition of, 338 in multiple integrals, 900 Mean Value Theorem, 245–249, 342–345, 438 arbitrary constants, 295 Cauchy’s, 273–274 in defining surface area, 407

definition of, 245 for derivatives, 245–249 to find velocity and position from acceleration, 247 mathematical consequences of, 246–247 for parametrically defined curves, 635 physical interpretation of, 246 proof of, 245–246 Rolle’s Theorem, 243–246 Taylor’s Theorem and, 602 Mendel, Gregor Johann, 167 Mercator Projection, 18-32 Mesh size, 511 Meter (m), 410 Method of determinants, 17-12 Method of gradient ascent, B-14 Method of gradient descent, B-12–B-14 Method of Lagrange multipliers, 837–840 Method of partial fractions, 497–502 Method of slicing with cylinders, 391–393 definition of, 380–381 by parallel planes, 381–382 volume and, 381–382 Midpoint, 685 Midpoint Rule, 314, 332–333 for approximating area, 314, 332–333 for approximating definite integrals, 510 for approximating integrals, 510 error analysis and, 514–517 Minimum absolute, 825–826, 829–932 constrained, 834–837 local, 825–829, 831 Minor axis, AP-21, 658 Mixed Derivative Theorem, AP-31– AP-32, 790–791 Mixture problems, 1041–1043 Moments calculations of, 933–934 center of mass and, 420–429 definition of, 420 first, 893–895 of inertia, 895–898, 934, 990 polar, 897 of the system about the origin, 420–421 of thin shells, 989–991 Monotonic functions, 250–255 Monotonic sequences, 546–547 Monotonic Sequence Theorem, 547, 561 Motion along a line, 162–165 acceleration, 163–164 average velocity, 162 displacement, 162 simple harmonic motion, 172–173 speed, 162–163 velocity, 162–165 antiderivatives and, 297–298

I-13

in cylindrical coordinates, 759 damped, 17-17–17-18 direction of, 728–729 Kepler’s first law (ellipse law), 760 Kepler’s second law (equal area law), 760–761 Kepler’s third law (time-distance law), 761–762 Newton’s second law of, 736, 1038, 1048 of planets in planes, 760 in polar coordinates, 759 projectile (See Projectile motion) with resistance proportional to velocity, 1038–1039 simple harmonic, 17-16–17-17, 172–173 vibrations, 17-15–17-16 Multiple integrals, 859–929 applications, 893–900 masses and first moments, 893–895 means and expected values, 900 moments of inertia, 895–898 probability, 898–900 double integrals area by, 873–875 over general regions, 864–870 over rectangles, 859–863 in polar form, 876–881 substitutions in, 915–922 triple integrals in cylindrical coordinates, 903–907 iterated integrals, 884–890 in rectangular coordinates, 883–890 in spherical coordinates, 907–911 Multiplication of complex numbers, 18-3–18-4 of functions, 34–35 positive definite, 698 scalar, of vectors, 682–683, 694–695 Multiplication property of real numbers, AP-27 Napier, John, 58 Natural domain of functions, 22 Natural exponential functions, 52, 59, 155 algebraic properties of, 442–443 definition of, 440 derivative of, 155, 440–441 indefinite integral of, 441 integral of, 440–441 inverse equation for, 440 laws of, 441–443 Natural logarithm function algebraic properties of, 58–59, 441–442 Chain Rule in, 192–193 definition of, 436–437 derivative and integral of ex in, 440–441 derivative of, 192–193

I-14

Subject Index

Natural logarithm function (continued ) derivative of y " ln x in, 438 derivatives and integrals involving logax in, 444–445 general exponential function ax in, 443 graph and range of ln x in, 438–439 integral ∫ 1/x dx in, 439 inverse of ln x and the number e in, 439–440 laws of exponents for ex in, 442–443 logarithms with base a, 444 number e in, 437 Natural logarithms algebraic properties of, 441–442 definition of, 436–437 Natural parametrization, 626 Navigation, vectors in, 686 N dimensional vectors, 687–688, 696–697 Negative infinity, 107, 114, 542 Negative of differentiable functions, 153 Negative reciprocal of a slope, AP-17 Net Change Theorem, 347–348 Neural Networks, Logistic Equation in, 1050–1051 Newton (N), 410 Newton, Sir Isaac, 342 Newton-meter (N · m), 410–411 Newton’s Law of Cooling, 453–454, 1047, 1048 Newton’s method, 289–292 applying, 290–291 convergence of the approximations, 291–292 procedure for, 289–290 sequences generated by, 538, 546, 549, 550 Newton’s second law of motion, 223, 736, 1038, 1048 Newton’s second law, 223 Newton’s serpentine, 159, 206 Nicole Oresme’s Theorem, 540, 622 Nondecreasing partial sums, 561 Nondecreasing sequences, 546–547 Nonelementary integrals, 508 Nonhomogeneous linear equations, 17-1, 17-7–17-14 form of the general solution, 17-7–17-8 method of undetermined coefficients, 17-8–27-31 method of variation of parameters, 17-12–27-34 Nonincreasing sequences, 546–547 Nonintegrable functions, 331–332 Nonlinear transformations, 918–919 Nonzero gradient vectors, 810–811 Nonzero vectors, 691–694, 699–700, 706, 709–710 Norm of functions, 19-12–19-14, 19-24 of partitions, 328 Normal components of acceleration, 753–754

Normal form of Green’s Theorem, 967–968, 1013, 1014 Normalized trigonometric system, 19-16 Normal line, 186, 815–817 Normal probability density function, 530 Norm of partitions, 860 Notation average value of a continuous function, 337 for constrained variables, 850–851 definite integrals, 330 Evaluation Theorem, 346 indefinite integrals, 354 for intervals, AP-3 Leibniz’s, 330 for partial derivatives, 785 sigma, 19-5, 322–325 for vectors, 694 Nth derivative, 158 Nth partial sums, 551–553 Nth term of a series, 552 Nth term of sequences, 539, 544, 551–553 Nth-Term Test, 555–556, 583 Numbers algebraic, 445 completeness property of real, 547 complex (See Complex numbers) counting (natural), 18-1 d-tuple of, 19-9 e definition of, 437 inverse of ln x and, 439–440 as a limit, 197 Fibonacci in sequences, 545 hierarchy of, 18-1–18-3 integers in, AP-2, 18-1–18-2 irrational, AP-2 rational, AP-2, 18-2 real (See Real numbers) systems of, successive stages of invented, 18-1–18-2 transcendental, 445 Numerical integration, 510–517 error analysis in, 514–517 Midpoint Rule for approximating integrals, 510 parabolas used for approximations, Simpson’s Rule and, 512–514 trapezoidal approximations, 511–512 Numerical method, 1027–1030 Numerical solution of the problem, 1027–1030 Oblique asymptotes, 112 Octants, 675–676 Odd functions, symmetry properties of, 26–27 Oh-notation, 467–468 One-dimensional heat equation, 796, 858 One-dimensional wave equation, 795 One-sided continuity, 122 One-sided derivatives, 145–146

One-sided limits, 99–104 approaching, 99–100 at endpoints of an interval, 100–101 involving sine, 102–104 precise definitions of, 101–102 One-to-one functions, 54–55 horizontal line test for, 55 trigonometric, 60–61 Open intervals, AP-3 Open region definition of, B-7 in a plane, 769 in space, 771–772 Optimization problem, 19-9 Or abscissa (x-coordinate), AP-14 Order, oh-notation and, 467–468 Ordered field, AP-28 Order of Integration Rule, 334 Order properties of real numbers, AP-1, AP-28 Oresme, Nicole, 540, 622 Oriented surface, 987 Origin, AP-14 graph of an odd function as symmetric about, 26–27 of planes, 675–676 Taylor’s formula at, 847 Origin (pole), in polar coordinates, 642 Or ordinate (y-coordinate), AP-14 Orthogonal functions, 19-15, 19-24 Orthogonal Gradient Theorem, 837–838 Orthogonality of the trigonometric system, 19-15–19-16 Orthogonal projections, 19-11 Orthogonal trajectories, 1038, 1040–1041 Orthogonal vectors, 693 Orthonormal basis for threedimensional space, 19-8–19-9 Orthonormal functions, 19-12, 19-15–19-16, 19-20, 19-24, 19-27 Oscillating discontinuity, 123 Osculating circle, 232, 749–750 Output variables, 767 Outside-inside rule, 178–179, 814 Overdamping, 17-18 Paddle wheel interpretation, 999–1002 Pairs, polar coordinate, 642–643 Pairwise cross products, 700 Pappus, 427 Pappus’s Theorem for Surface Areas, 429 Pappus’s Theorem for Volumes, 427–428 Parabolas, 656–657 areas between curves, 365, 367 axis of, AP-19–AP-20 definition of, 656–657 directrix of, 656–657 eccentricity of, 663–667 equations for, 656–657

focal length of, 656 focus of, 656–657 as graphs of equations, AP-19–AP-20 normal lines to, 188 polar equations for, 665–667 reflective property of, 663 semicubical, 189 used for approximations, Simpson’s Rule and, 512–514 vertex of, 656 y-axis of, 656 Paraboloids definition of, 714 elliptical, 714–717 hyperbolic, 715–717 region enclosed by, 889 Parallel Axis Theorem, 902 Parallel lines, AP-17 Parallelograms area of, cross product in, 700–701 law of addition, 682 Parallel vectors, 699, 700 Parameter definition of, 973 domain, 973 interval in parametric equations, 624 method of variation of, 17-7, 17-12–27-34, 17-12–17-14 Parametric curves, 743–745, 1054 Parametric equations, 624–628 Cartesian, 627 of cycloids, 628–629 definition of, 624 examples of, 624–628 for ideal projectile motion, 736–738 for lines, 706–707 for natural parametrization, 626 parameter interval in, 624 parametrization of the curve, 624–628 terminal point in, 624 values of, 625, 627 Parametric formulas, 633–634 Parametrization arc length, 744 of surfaces, 973–981 of cones, 974 of cylinders, 974 by cylindrical coordinates, 974 of implicit surfaces, 978–981 of a sphere, 974 of surface area, 974–978 Parametrization, natural, 626 Parametrizations of plane curves, 624–629 area in a plane, 650–652 areas and lengths in polar coordinates, 650–653 calculus with (See Calculus with parametric curves) conic sections, 655–660 conics in polar coordinates, 663–668

Subject Index

equations for (See Parametric equations) length of a polar curve, 652–653 Parseval Equality, 19-21–19-22 Partial derivatives, 784–793 calculations of, 787–788 Chain Rule for, 796–803 with constrained variables, 848–851 continuity and, 779–781, 789–790 continuous, identity function with, 1003 definitions of, 785–786 differentiability and, 792–793 Differentiable Implies Continuous Theorem, 793 differentials in, 820–821 directional derivatives and, 806–814 extreme values and, 781, 825–832 of a function of two variables, 784–786 of functions of more than two variables, 788–789, 821–822 of functions of several variables, 767–772 gradient vectors and, 806–814 Increment Theorem for Functions of Two Variables, 793 Lagrange multipliers, 834–841 limits for functions of two variables, 775–779 linearization of a function of two variables, 818–820 Mixed Derivative Theorem, 790–791 normal lines and, 815–817 notation for, 785 second-order, 790 of still higher order, 791 tangent planes and, 815–822 Partial fractions definition of, 497 integration of rational functions by, 497–502 method of, 497–502 Partial sums in divergent series, 555–556 nondecreasing, 561 nth, 551–553 of sequences, 551–552 sequences of, 551–552 Particles path of, 725 position vector of, 725 Particular solution, 17-8, 297, 1025–1026 Partitions formed by double integrals over rectangles, 859–860 kth subinterval of, 327 norm of, 328, 860 for Riemann sums, 326–328 Parts, integration by, 478–483 evaluating definite integrals, 482–483 formula, 479–482

Product Rule in integral form, 442–482 Pascal, Blaise, 552 Paths Chain Rule for, 813–814 derivative along, 814 direction along, 931 independence, 951 of particles, 725 Percentage change, 222 Perihelion, 669 Period definition of, 19-1 fundamental, 19-1 Periodic functions, 19-1–19-2, 44 Periodic trigonometric functions, 44 Periods, of trigonometric functions, 44 Perpendicular lines, AP-17, 186 Perpendicular vectors, 693 Phase angle, 17-17 Phase lines, 1044–1046 Phase-plane analysis method, limitations of, 1056 Phase planes, 1053–1054 F, 907 F-limits of integration, 909, 910 Physics applied optimization in, 279–282 vectors in, 686 Piecewise continuous functions, 331, 376 Piecewise-defined functions, 25 Piecewise smooth curves, 728 Pinching Theorem. See Sandwich Theorem Plane angles between, 711–712 area in, 650–652 Cartesian coordinates in, AP-13–AP-14 circle of curvature for plane curves, 749–750 circles in, AP-17–AP-19 coordinate, 675–676 curvature of, 747–749 curve, flux across, 945–946 determined by coordinate axes, 675 determined by or spanned by r and s, 19-9 distance from a point to, 711 distance in, AP-17–AP-19 equation for, 708–709 flux across, 946 line integrals in, 934–935 origin of, 675–676 points in, AP-18 tangent (See Tangent planes) vector equation for, 709 Plane region boundary of, 769 boundary point of, 769 bounded, 769 closed, 769 definitions of, 769 interior of, 769

interior point of, 769 mass distributed over, 422 open, 769 unbounded, 769 Planetary motion Kepler’s first law (ellipse law), 760 Kepler’s second law (equal area law), 760–761 Kepler’s third law (time-distance law), 761–762 Plates bounded by two curves, 425–426 density of, 422 integral for fluid force against a vertical, 415 thin, flat, 422–425 two-dimensional, formulas for, 894 Points boundary, of a region closed bounded, absolute maxima and minima on, 829–832 in Max-Min Tests, 831 in a plane, 769 in space, 771 continuity at, 120–123 continuous extension to, 128–129 continuous functions at, 779–780 critical (See Critical points) of differentiable functions, 827 distance between two, in the xyplane, formula for, 677–679 distance from a line to, 708 distance from a plane to, 711 fixed, 106 functions that are continuous at, 779 of inflection, 239, 256–259 initial, of vectors, 680 interior, of a region closed bounded, absolute maxima and minima on, 829–832 in a plane, 769 in space, 771 in a plane, distance formula for, AP-18 saddle, 716, 825–832 standard position for, 681 Taylor’s formula at, 846 terminal, of vectors, 680 Point-slope equation, AP-16–AP-17 Poisson, Siméon-Denis, 825 Polar coordinates acceleration in, 759–762 area in, 878–879 area of polar region, 650–652 areas and lengths in, 650–653 Cartesian coordinates related to, 643–645 conic sections, 655–660 conics in, 663–668 definition of, 642–643 equations (See Polar equations) initial ray in, 642 integrals in, 876–878

I-15

length of polar curve, 652–653 of lines, 667 motion in, 759 pairs in, 642–643 polar equations of, 643 pole (origin) in, 642 slope of polar curve, 647–648 symmetry tests for, 647 velocity in, 759–762 Polar curve, length of, 652–653 Polar equations for circles, 667–668 for conics, 665–667 for ellipses, 665–667 graphs/graphing, 643–644, 646–649 Archimedes spiral in, 672 converting a graph from the rq-plane to the xy-plane, 647–648 lemniscate curves, 649 limaçons, 650 slope, 647–648 symmetry, 647 technology for, 647–648 for hyperbolas, 665–667 for lines, 667 for parabolas, 665–667 relating polar and cartesian coordinates, 644–645 Polar form, double integrals in, 876–881 Polar moment, 897 Polar rectangles, 876 Pole (origin), in polar coordinates, 642 Polyhedral surfaces, 1002–1003 Polynomials, 29 coefficients of, 29 degrees of, 29 irreducible quadratic, 498 limits of, evaluating, 83 Taylor, 19-1, 599–601 trigonometric, 19-3–19-5 of degree n, 19-18–19-19 Population growth logistic model for, 1049–1050 unlimited, 450–452 Position, acceleration for finding, 247 Position vector, 725 Positive definite multiplication, 698 Potential function, 951 Potentials for conservative fields, finding, 955–957 Power Chain Rule, 179–181 Power functions, 27–28, 179–181 Power Rule, 58, 82–83 for base a logarithm, 444 for derivatives (See Derivative Power Rule) for finding complex derivatives, 18-19 for limits, AP-23 for limits for functions of two variables, 777 for natural logarithms, 441–442

I-16

Subject Index

Powers binomial series for, 609–611 of complex numbers, 18-7 linear approximation for, 217 of sines and cosines, products of, 486–487 of tan x and sec x, 488–489 Power series, 586–594 convergence of, 586–591 Convergence Theorem for Power Series, 589–591 interval of, 591 radius of, 589–591 testing for, 591 definition of, 586 method, 17-24–17-29 operations on, 591–594 convergence of power series, 592 Series Multiplication for Power Series, 591 Term-by-Term Differentiation, 592–593 Term-by-Term Integration, 593–594 solutions, 17-24–17-29 Predator-prey model, 1058–1059 Preimage, 915 Prescribed errors, constants to describe, 91 Pressure-depth equation, 414 Pressure of fluid, force and, 414–415 Principal branch of logarithmic functions, 18-29 Principal minor determinants, B-9–B-11 Principal nth root, 18-8 Principal unit normal vector, 748, 757 Principal value, 18-25, 18-29 Probability density functions (See Probability density functions) joint probability density function, 899 in multiple integrals, 898–900 Probability density functions normal, 530 Production, costs of, 165–166 Product Rule, 58, 82–83 for base a logarithm, 444 to calculate limits of sequences, 542 for derivatives (See Derivative Product Rule) for finding complex derivatives, 18-19 for gradients, 811 in integral form, 442–482 for limits, AP-23 for limits for functions of two variables, 777 for natural logarithms, 441–442 Products of complex numbers, 18-5–18-6 derivatives of, 155–156 of trigonometric integrals, 490 powers of, 486–487

triple scalar, 703–704 vectors of, 700 Projectile motion, 734–739 ideal equation for, 737 height, flight time, and range for, 738 vector and parametric equations for, 736–738 integrals of vector functions, 734–736 with wind gusts, 738–739 Projections of vectors, 693–696 Proof by mathematical induction, AP-10–AP-13 Proportional relationship, of functions, 27 P-series, 563–565 P-Series Test, 563, 583 Pythagorean Theorem, 19-6–19-7, 19-15, 44, 46, 63 Pythagorean triples, 549 Quadrants, AP-14 Quadratic approximation, 601 Quadratic functions, 29 Quadratic polynomial, irreducible, 498 Quadric surfaces, 714–717 definition of, 714 ellipsoids, 714–718 elliptical cones, 714–717 general, 716, 718 graphs of, 717 hyperboloids, 714–717 saddle point of, 716 Quotient Rule, 58, 82–83 for base a logarithm, 444 to calculate limits of sequences, 542 for derivatives (See Derivative Quotient Rule) for finding complex derivatives, 18-19 for gradients, 811 for limits, AP-23 for limits for functions of two variables, 777 for natural logarithms, 441–442 Quotients in complex numbers, 18-6– 18-7 derivatives of, 157–158 Radians, 41–43 Radioactive decay, 452–453 Radioactive elements, half-life of, 60, 452 Radioactivity, 452–453 Radius of convergence, 589–591 of curvature, 749 Range of functions, 21–22 of functions of several variables, 767, 768 for ideal projectile motion, 738 Rate constant, 448

Rates of decay, 53 fluid flow, 965 growth, 465–469 integrals of, 347–348 Rates of change, 72–77, 161–170 average, 74, 76–77 instantaneous, 76–77, 79, 102, 161–162 marginals, 165–167 secant lines and, 74 sensitivity to, 167 slope of a curve and, defining, 74–76 in speed, 72–74 tangent lines and, 76–77 in velocity, 162–165 Ratio, in geometric series, 553 Rational functions, 29 coefficients in, determining by differentiating, 502 integration of, by partial fractions, 497–502 method used in, general description of, 497–501 limits at infinity of, 108 limits of, evaluating, 83 Rational numbers, AP-2, 18-2 Ratio Test, 18-24, 574–575, 583 Real Analysis, 19-13 Real axis, 18-4 Real line, AP-1 Real numbers addition property of, AP-27 algebraic properties of, AP-1, AP-27 associative laws of, AP-28 commutativity laws of, AP-28 completeness property of, AP-1– AP-2, AP-27, AP-28, 547 construction of reals and, AP-29– AP-30 definition of, AP-1, AP-2 irrational, AP-2 multiplication property of, AP-27 order properties of, AP-1, AP-28 real line and, AP-1 sets of, AP-2 subsets of, AP-2 theory of, AP-27–AP-30 Real parts of complex numbers, 18-3–18-4 Reals, construction of, AP-29–AP-30 Real-valued function, 22, 725–726, 767–768 Rearrangement Theorem for Absolutely Convergent Series, 582, 583 Reciprocal computing, 18-6 derivative of, 143 of sine function, 61 Reciprocal Rule, 58, 160 for base a logarithm, 444 for natural logarithms, 441–442 Rectangles, area of average value of the function for approximating, 319–320

between curves, 364–367 finite approximations, 315–320, 437 finite sums for estimating, 313–315, 325–326 Fundamental Theorem of Calculus, 343 under the graph of a nonnegative function, 336 integration with respect to y, 366–367 Mean Value Theorem, 342 Riemann sums for approximating, 327–328, 349 Riemann sums in defining, 328 total, 349–350 Rectangles, polar, 876 Rectangular coordinates, 675, 883–890 average value of a function in space, 890 definition of, 883 equations relating cylindrical coordinates to, 903–905 iterated integrals, 884–890 properties of, 890 volume of a solid region in space, 883–884 Rectangular coordinates, triple integrals in, 883–890 average value of a function in space, 890 in cylindrical coordinates, 903–907 definition of, 883 iterated integrals, 884–890 properties of, 890 Riemann sums for, 904, 908 in spherical coordinates, 907–911 volume of a solid region in space, 883–884 Rectangular coordinate system, AP-10 Recursion formula, in sequences, 545 Recursive definitions, of sequences, 545–546 Reduction formula, 482, 505–506 Reflection of a graph, 36–38 Reflective property of parabolas, 663 Region connected, 951 Divergence Theorem for other regions, 1011–1012 for special regions, 1010–1011 exterior, of a circle, AP-19 Green’s Theorem for special, 969–970 interior, of a circle, AP-19 open, 1003–1004 simply connected, 951 Regression lines, 834 Reindexing series, 557–558 Related rates equations, 206–210. See also Related rates problem Chain Rule in, 206, 208 differentiation in, 209

Subject Index

Related rates problem consistent units in solving, 210 definition of, 206 strategy, 207 Relative change, 222 Relative extrema, 826 Relative (local) maxima or minima extrema. See Local (relative) maxima or minima extrema Remainder in Integral Test, 564–565 of order n (or error term), 603–604 Remainder Estimation Theorem, 604 Remainder Formula, 564 Removable discontinuity, 120, 123 Representations, series, 597–598 Resistance definition of, 17-19 falling body encountering, 1047–1049 proportional to velocity, 1038–1039 Resolution level zero adding details at, 19-30–19-31 approximation at, 19-29 Rest points, 1044–1046 in graphical solutions of autonomous equations, 1044–1046 in systems of differential equations, 1054–1055 unstable, 1055 Resultant vectors, 682 Revolution solids of, 382–387 disk method, 382–385 washer method, 385–387 surface area of, 405–408 Revolution, areas of surfaces of, 639–640 Riemann, Georg Friedrich Bernhard, 326 Riemann sums, 326–328 for approximating area of rectangles, 327–328, 349 for arc length, 400 in calculus with parametric curves, 635 convergence of, 331 in defining area of rectangles, 328 in defining the definite integral, 329–330 for double integrals, 859–861 over area, 873 over general regions, 869 over rectangles, 860–861 for integrals, 380, 422 limits of, 326–328, 331, 394 for mass of thin, flat plates, 422 for mass of thin wires, 421 notion of, 326–328 for parametrically defined curves, 635 partitions for, 326–328 in shell method, 394 slicing by parallel lines, 381

slicing with cylinders, 392–393 for spherical coordinates, 904, 908 total area of rectangles, 349–350 for triple integrals, 904, 908 for work done by a variable force, 411, 413, 415 Right-continuous function (or continuous from the right), 120–122 Right-hand derivatives, 145–146 Right-hand limit (or limit from right), AP-25, 99–104, 120, 122 Right-hand rule for vectors, 699 Rise, AP-15 RL circuits, 1036 R-limits of integration, 878, 904, 906 Rolle, Michel, 236, 244 Rolle’s Theorem, 243–244 Root finding, 127 Root Rule, AP-23, 82–83, 777 Roots binomial series for, 609–611 in complex numbers, 18-7–18-9 finding, 127 Intermediate Value Theorem and, 127–128 linear approximation for, 217 nth, 18-7–18-8 principal nth, 18-8 Root Test, 576–577, 583 Rule of 70, 71 Run, AP-15 Saddle point, 825–832 definition of, B-9 of functions of two variables, B-6–B-11 St. Vincent, Gregory, 655 Sandwich Theorem, AP-25, 543 Scalar, definition of, 682 Scalar components of acceleration, 753–754 Scalar differential form, 943 Scalar functions, 725–726 line integrals of, 930–935 surface integrals of, formulas for, 984 Scalar Multiple Rules, 730 Scalar multiplication of vectors, 682–683, 694–695 Scalar products. See Dot products Scaling of a graph, 36–38 Scatterplot, of functions, 24 Secant function, 42–43 derivatives of, 173–174 even, 44 graphs of, 44, 62 integrals of, 359 integrals of powers of, 488–489 inverse of, 200–201 periodicity of, 44 Secant lines, 74–77 Second derivative, 158 Second Derivative Test for concavity, 256 for differentiable functions, 256, 259–263

for local (relative) maxima or minima extrema, 259–261, 827, 829, 831, 835, 844–845 Taylor’s formula to derive, 844–845 Second Derivative Test for Local Extreme Values (General Version), B-9 Second-order differential equations. See under Differential equations Second-order partial derivatives, 790 Semicubical parabola, 189 Semimajor axis of ellipses, 658 Semiminor axis of ellipses, 658, 666–667 Separable differential equations, 448–450 Separable equation, 448–450 Sequences, 538–547 bounded, 546–547 Continuous Function Theorem for, 543 convergence of, 540–73 converging to a complex number, 18-23 divergence of, 540–542 diverging to a complex number, 18-23 diverging to infinity, 542 diverging to negative infinity, 542 essay on, 538 Fibonacci numbers in, 545 generated by Newton’s method, 538, 546, 549, 550 history of, 538 index of, 539–540 infinite, 539–540 L’Hôpital’s Rule for, 543–544 limits of, 542–543 calculating, 542–543 commonly occurring, 545 definition of, 540 monotonic, 546–547 Monotonic Sequence Theorem, 547 nondecreasing, 546–547 nonincreasing, 546–547 nth term of, 539, 544, 551–553 of partial sums, 551–552 recursion formula in, 545 recursive definitions of, 545–546 representing, 539–540 Sandwich Theorem for, 543 terms of, 539 zipper theorem for, 550 Sequential search, 468–469 Series absolute convergence, 572–577 adding or deleting terms to/from, 557 alternating, 579–583 binomial, 609–611 combining, 556–557 comparison tests for convergence of, 567–571 complex, 18-23–18-24 conditionally convergent, 581–582

I-17

Constant Multiple Rule for, 556–557 convergent, 552 converging absolutely, 18-23– 18-24 Difference Rule for, 556–557 divergent, 552, 555–556 error estimation, 564–565 essay on, 538 Fourier, 565 geometric, 553–555, 583 harmonic, 563, 579–583 history of, 538 infinite, 538, 551–558 Integral Test, 561–565 Maclaurin, 598–600, 604 Monotonic Sequence Theorem, 547, 561 nth partial sums in, 551–552 nth term of, 552 nth-Term Test for divergent, 555–556, 583 p-, 563–565 of partial sums, 551–553, 555–556, 561 power, 586–594 p-Series Test, 563, 583 rearranging, 582–583 reindexing, 557–558 representations, 597–598 sigma notation to write, 552 sum of, 552 Sum Rule for, 556–557 Taylor, 598–615 test to verify convergence, 18-24 Series Multiplication for Power Series, 591 Sets of real numbers, AP-2, AP-28 Shell method, 393–396 Shift formulas, for functions, 36 Shifting of a graph, 36 Shift property for definite integrals, 371 Sigma notation, 322–325 in trigonometric polynomials, 19-5 to write series, 552 ¨, 323 Sigmoid shape of curves, 1050 Simple harmonic motion, 17-16–17-17, 172–173 derivatives of, 172–173 Simply connected domains and regions, 951 Simpson, Thomas, 501 Simpson’s Rule, 512–517 approximations by, 512–514 error analysis and, 514–517 Sine limits involving, 102–104 summing, 19-3–19-5 values of, 43 Sine function, 19-2, 42–43 arcsine function and, 61–62 derivatives of, 170–171 general, 47 graphs of, 30, 44, 62 inequalities and, 46

I-18

Subject Index

Sine function (continued ) odd, 44 one-to-one, 55, 60–61 periodicity of, 44 product of, 490 products of powers of, 486–487 reciprocal of, 61 Sine-integral function, 530 Sinusoid, 642 Sinusoid formula, 47 SI units (Système International or International System), 410–411 Slant line asymptote, 111 Slicing. See Method of slicing Slope fields, 1026–1027 Slope-intercept equation, AP-17 Slopes of the curve, definition of, 138 negative reciprocal of, AP-17 of a nonvertical line, AP-15 of polar curve, 647–648 on sine curves, 182 of tangent lines, 138–139, 144 Smooth curves, 728–729 curvature of, 747–749 length of, 743 principal unit normal vector for, 748 speed on, 729 torsion of a, 755 Smooth function, 399 Smooth surface, 975–976, 984, 987 Snell’s Law, 280 Snell van Royen, Willebrord, 279 Snowflake curve, 560 Solids, formulas for threedimensional, 894 Solids of revolution, 382–387 Cavalieri’s principle, 382 definition of, 382 disk method, 382–385 washer method, 385–387 Solution curves, 1026–1027 Solutions of differential equations, 449 of first-order differential equations, 1024–1026 of first-order linear equations, 1033–1035 general, 449 for inequalities, AP-3–AP-4 particular, 1025–1026 steady-state, 1053 Space arc length in, 743–745 average value of a function in, 890 curves, 725–731, 750–751 motion in, 734–739 Space regions boundary of, 771–772 boundary point of, 771 bounded, 771 closed, 771–772 interior point of a, 771 open, 771–772 unbounded, 771 Speed, 162–163 along smooth curves, 729

average, 72–74 initial, in ideal projectile motion, 737 instantaneous, 73 Spheres parametrization of, 974 of radius and center, equations for, 678–679 in space, 677–678 surface area of, 977 Spherical coordinates, 907–911 algebra for finding, 908 definition of, 907 equations relating Cartesian and cylindrical coordinates to, 907–909 geometry for finding, 908 integration in, 907–911 limits of integration in, finding, 909–911 Riemann sums for, 904, 908 Spin around axis, 961–964 Spring constant, 412 Square completing, AP-18–AP-19 conformal maps of, 18-33–18-34 of the distance, 897 Square root functions definition of, 28 derivatives of, 144, 146, 191 linearization and, 216 Square roots, eliminating, 488 Square wavelet, 19-24–19-26, 19-31, 19-33–19-34 Squeeze Theorem. See Sandwich Theorem Stable equilibria, 1046–1047 Standard equation of an ellipse, AP-21 of a circle, AP-18 Standard form of linear equations, 1032–1033 Standard linear approximations definition of, 215, 819 error in, 820 line, 215, 819 linearization and, 215–217 tangent-plane/tangent-line approximation, 819–820 Standard position angles in, 42 for a point, 681 Standard unit vectors, 19-7, 684 Starting integers, AP-12 Steady-state solution, 1053 Steady-state values, 1036 Steepest descent, method of, B-12–B-14 Step size, 511 Still higher order, partial derivatives of, 791 Stirling, James, 536–537 Stirling’s formula, 536–537 Stokes’ Theorem, 993–1004 conservative fields and, 1003–1004 definition of, 995 Green’s Theorem compared to, 995–996, 1002–1003, 1014

identity arising in mathematics and physical sciences, 1003 paddle wheel interpretation, 999–1002 for polyhedral surfaces, proof outline of, 1002–1003 for surfaces with holes, 1003 Subsets of real numbers, AP-2 Subspace, 19-11 Substitution boundaries in, 917–919, 922 differential area change, 916 in double integrals, 915–919 trigonometric, 492–495 trigonometric, integration by, 492–495 in triple integrals, 920–922 Substitution, in integrals for definite integrals, 361–367 areas between curves, 364–366 integration with respect to y, 366–367 Substitution Formula, 361–363 for indefinite integrals, 354–359 Chain Rule, 354–359 Substitution Rule, 334, 355–356, 358 of tangent, cotangent, secant, and cosecant functions, 359 trying different substitutions, 359 Substitution Formula, 361–363 Substitution Rule for definite integrals, 334 definition of, 916 for indefinite integrals, 355–356, 358 use of the variable u in, 356 Subtraction derivatives of, 153–154 of functions, 34–35 of terms from series, 557 of vectors, 683 Sum Rule for antiderivatives, 296–297, 299 to calculate limits of sequences, 542 for definite integrals, 334 derivative, AP-12–AP-13 for derivatives, 153–154, 171 for double and triple integrals, 869 for finding complex derivatives, 18-19 for finite sums, 324 for gradients, 811 for limits, AP-23, 82–83, 95 for limits for functions of two variables, 777 for series, 556–557 for vector functions, 730 Sums derivatives of, 153 of series, 552 Superposition principle, 17-1 Surface area, 405–408, 973–981 of an implicit surface, formula for, 979 definition of, 405–407

differential for a parametrized surface, 976 element, 976 of a graph, formula for, 981 parametrization of, 974–978 of revolution, 405–408 for revolution about the x-axis, 407 for revolution about the y-axis, 408 of a smooth surface, 975–976 of a sphere, 977 Surface integrals, 983–991 definitions of, 984 for a level surface, computing, 989 mass density of, 983, 985 orientation of a surface, 987 for a parametrized surface, computing, 988–989 of a scalar function, formulas for, 984 in Stokes’ Theorem, 1003 thin shells, moments and masses of, 989–991 of vector fields, 987 Surfaces area (See Surface area) functions defined on, 799–801 with holes, in Stokes’ Theorem, 1003 integrals of (See Surface integrals) level, 770 orientation of, 987 parametrization of, 973–981 of cones, 974 of cylinders, 974 by cylindrical coordinates, 974 of implicit surfaces, 978–981 of a sphere, 974 of surface area, 974–978 plane tangent to, 815–817 polyhedral, 1002–1003 saddle points of, 825–832 smooth, 975–976, 984, 987 of two-variable functions, 770 Symmetric functions, definite integrals of, 363–364 Symmetry tests for polar coordinates, 647 Synchronous curves, 764 System of first-order differential equations, 1053–1056 System torque, 420–421 Tangent function, 42–43 derivatives of, 173–174 graphs of, 44, 62 integrals of, 359 integrals of powers of, 488–489 inverse of, 200–201 odd, 44 periodicity of, 44 Tangential components of acceleration, 753–754 Tangential form of Green’s Theorem, 966–967, 1014 Tangent lines to an ellipse, 810–811, 817

Subject Index

approximating functions based on, 214 (See also Linearization) approximation, 819–820 areas and, 633–634 to a curve, 74–75 to a level curve, 810 to a parabola, 75–76 rates of change and, 76–77 vertical, 141–142, 146 Tangent planes, 815–822 approximation, 792, 819–820 definition of, 815, 816 differentials, 820–821 estimating change in a specific direction, 818 functions of more than two variables, 821–822 linearization of a function of two variables, 818–820 normal lines and, 815–817 to a surface, 815–817 Tangents area, 633–634 to curves, 745 of curves in space, 725–731 definition of, 42–43 unit vectors, 745, 757 values of, 43 Tautochrones, 628–629 Taylor, Brook, 598 Taylor polynomial, 19-1, 599–601 Taylor series, 598–615 applications of, 609–615 arctangents, 612–613 binomial series for powers and roots, 609–611 Euler’s identity, 614–615 evaluating indeterminate forms, 613–614 evaluating nonelementary integrals, 611–612 Leibniz’s formula, 613 convergence of, 18-18, 602–607 Remainder Estimation Theorem, 604 remainder of order n (or error term), 603–604 Taylor’s Formula, 602–603 Taylor’s Theorem, 602, 607 using Taylor series in, 605–606 definition of, 598–599 frequently used, 615 Taylor polynomials in, 599–601 Taylor’s Formula, 844–847 definition of, 602, 603 to derive the Second Derivative Test, 844–845 error formula for linear approximations, 845 for functions of two variables, 846–847 at the origin, 847 at the point, 846 Taylor’s Theorem, 514–515 definition of, 602 Mean Value Theorem and, 602 proof of, 607 Remainder Estimation Theorem and, 604

Term-by-Term Differentiation, 592–593 Term-by-Term Integration, 593–594 Terminal point in parametric equations, 624 Terminal point of vectors, 680 Terminal velocity, 355, 1049 Terms, of sequences, 539 Theorems absolute convergence of series, 18-24 Absolute Convergence Test, 573 Algebraic Properties of the Natural Logarithm, 58, 441–442 Alternating Series Estimation Theorem, 581 Alternating Series Test, 579–580, 583 Angle Between Two Vectors, 691–693 approximations of functions, 19-17, 19-18 approximations using the Haar system, 19-27, 19-29– 19-30 Area of Surface of Revolution for Parametrized Curves, 640 for calculating limits of sequences, 542 Cauchy’s Mean Value Theorem, 273–274 Chain Rule, 177 Chain Rule for Two Independent Variables and Three Intermediate Variables, 799 Formula for Implicit Differentiation, 801 Functions of One Independent Variable and Three Intermediate Variables, 798 Functions of One Independent Variable and Two Intermediate Variables, 796 Implicit Function Theorem, 802–803 Clairaut’s Theorem, 790–791 Cofactor Expansion, B-4 of commonly occurring limits, 545 of complex functions, 18-25 composition of continuous functions, 125 conformal maps, 18-32–18-33 constant-coefficient homogeneous equations, 17-2–17-5 Constant Multiple Rule, 556– 557 Convergence Theorem for Complex Power Series, 18-25 Convergence Theorem for Power Series, AP-23–AP-25, 589–592 corollary to, 590 for convergent series, 556–557 curl F " 0 related to the closedloop property, 1003

De Moivre’s Theorem, 18-7 derivative rule for inverses, 190 Differentiable Implies Continuous Theorem, 147, 793 Direct Comparison Test, 526–528, 568–569, 583 Directional Derivative Is a Dot Product, 808 divergence and the curl, 1009– 1010 Divergence Theorem, 1006–1014 Extreme Value Theorem, 236–237 for finding antiderivatives, 295 First Derivative Test for Local Extreme Values, B-8, 238–239, 251–253, 826, 836 Formula for Implicit Differentiation, 801–803 Fourier coefficients, 19-18 Fourier series, convergence of, 19-19–19-21 Fubini’s Theorem for calculating double integrals over general regions (stronger form), 865–868 for calculating double integrals over rectangles (first form), 861–863 Fundamental Theorem of Algebra, 18-9 Fundamental Theorem of Calculus, 299 Fundamental Theorem of Line Integrals, 952–954 Green’s Theorem, 961–970, 972 Haar system, 19-27, 19-29, 19-32 Implicit Function Theorem, 802–803 Increment Theorem for Functions of Two Variables, AP-33– AP-35 initial value and boundary value problems, 17-5 Integral Test, 562 interior points of a function’s domain, 100 Intermediate Value Theorem for continuous functions, 127 Laws of Exponents for ex, 442–443 l’Hôpital’s Rule, 268–274, 543–544 for limits of continuous functions, 126 Limit Comparison Test, 527–528, 569–571, 583 Limit Laws, 82, 108 of polynomials, 83 to prove definition of, 95 of rational functions, 83 of the ratio sin uu as u q0, 102 linearly independent solutions, 17-2 Mean Value Theorem, 245–249, 602 Mixed Derivative Theorem, AP-31–AP-32, 790–791

I-19

Monotonic Sequence Theorem, 547, 561 Nicole Oresme’s, 540, 622 nonhomogeneous linear equations, 17-8 nth-Term Test for Divergence, 555–556 number e as a limit, 197 one-sided limits, 100 Orthogonal Gradient Theorem, 837–838 orthonormal bases and approximation in higher dimensions, 19-10–19-11 orthonormal basis for threedimensional vectors, 19-9 Pappus’s Theorem for Surface Areas, 429 Theorem for Volumes, 427–428 Parallel Axis Theorem, 902 Parseval Equality, 19-21–19-22 properties of continuous functions, 124 Properties of Determinants, B-5 Properties of Limits of Functions of Two Variables, 777 Pythagorean Theorem, 19-6–19-7, 19-15, 44, 46, 63 Ratio Test, 574–575, 583 Rearrangement Theorem for Absolutely Convergent Series, 582, 583 Remainder Estimation Theorem, 604 Rolle’s Theorem, 243–244 Root Test, 576–577, 583 Sandwich Theorem, 85, 543 Second Derivative Test for concavity, 256 for local extreme values, B-9, 259–261, 827, 829, 831, 835, 844–845 Substitution for Double Integrals, 916 Taylor’s Formula, 602, 603 Taylor’s Theorem, 514–515, 602, 604, 607 Unifying Fundamental Theorem of Vector Integral Calculus, 1014 zipper theorem, 550 Theory of real numbers, AP-27–AP-30 6-limits of integration, 878, 905, 906, 909, 910 Thickness variable, 394–395 Thin shells, moments and masses of, 989–991 Third derivative, 158 Three-dimensional coordinate systems, 675–678 Cartesian coordinates in, 675 coordinate planes in, 675 planes determined by coordinate axes in, 675 rectangular coordinates in, 675 Three-dimensional Laplace equation, 795

I-20

Subject Index

Three-dimensional solid, formulas for, 894 Three-dimensional vectors, 19-5–19-9, 681 definitions of, 19-6, 19-7 distance between, 19-6 dot product of, 19-6 length of, 19-6 linear combination of, 19-7 orthogonal (or orthonormal) set of, 19-6, 19-7–19-8 orthonormal basis for, 19-8–19-9 Pythagorean Theorem for, 19-6–19-7 standard unit vectors in, 19-7 theorem, 19-9 unit vectors, 19-6 Time constant, 1037 Time-distance law (Kepler’s third law), 761–762 Time measurement, fixed error in, 222 TNB frame, 753 Torque, 420–421 Torque vectors, 702–703 Torsion along the curve, 755 definition of, 755 formulas for computing, 756–757 function of a smooth curve, 755 Total area, 349–350 Total differentials, 820–822 Total distance traveled, 317–318, 348 Training set, 1051 Trajectories orthogonal, 1038, 1040–1041 systems of differential equations, 1053–1054 Transcendental functions, 31, 445 catenary, 31 Transcendental numbers, 445 Transformations Jacobian of, 917–919, 922 linear, 916 nonlinear, 918–919 rules applied to sine function, 47 of trigonometric graphs, 46–47 Trapezoidal approximations, 511–512 Trapezoidal Rule, 511–512 approximations by, 511–512 error analysis and, 514–517 Trend lines, 834 Trigonometric functions, 30, 41–49 addition formulas in, 45 angles in, 41–42 arcsine and arccosine, 61–64 ASTC rule for remembering, 43 Chain Rule in, 202–203 cosecant, 42–43 cosine, 42–43 cotangent, 42–43 derivatives of, 170–176 cosecant function, 173–174 cosine function, 171–172 cotangent function, 173–174 secant function, 173–174 simple harmonic motion, 172–173 sine function, 170–171

tangent function, 173–174 double-angle formulas in, 45 half-angle formulas in, 45 inequalities in, 46 integrals of, 359 inverse, 60–61 law of cosines, 45–46 one-to-one, 60–61 periodic, 44 periods of, 44 secant, 42–43 sine, 42–43 tangent, 42–43 Trigonometric graphs of functions, 44 of transformations, 46–47 Trigonometric identities, 44–45 Trigonometric integrals, 486–490 graphed as functions, 492 of powers of tan x and sec x, 488–489 products of powers of sines and cosines, 486–487 products of sines and cosines, 490 square roots, eliminating, 488 Trigonometric polynomials, 19-3–19-5 amplitude in, 19-4 definition of, 19-3 of degree n, 19-18–19-19 sigma notation in, 19-5 Trigonometric substitutions, 492–495 Trigonometric system, 19-2, 19-15–19-16 Triple integrals in cylindrical coordinates, 903–907 iterated integrals, 884–890 properties of, 869, 891 in rectangular coordinates, 883–890 in spherical coordinates, 907–911 substitutions in, 920–922 Triple scalar product, 703–704 Trivial solution, 17-2 Trochoids, 632 Twice-differentiable functions, 633–634 Two-dimensional Laplace equation, 795 Two-dimensional plate, formulas for, 894 Two-dimensional vectors, 681, 696–697 Two-path test for nonexistence of a limit, 780 Two-sided continuity, 122 Two-sided limits, AP-25, 99, 103, 116, 122, 126 Type I, improper integrals of, 521–524 Type II, improper integrals of, 524–525 Unbounded region definition of, B-7 in a plane, 769 in space, 771 Unbounded sequence, 546 Underdamping, 17-18

Undetermined coefficients, 497 Undetermined coefficients, method of, 17-8–27-31 Unified theory of vector field integrals, 1013–1014 Union of sets, AP-2 Unit circle, AP-18 Unit functions, 19-13, 19-24 Unit normal vectors, 699 Units of measurement for fluid’s weight-density, 414 in work, 410–411 Unit step function, 81–82 Unit tangent vector, 745, 757 Unit vectors, 19-6, 684–685 definition of, 684 tangent, 745, 757 writing vectors in terms of, 684–685 Universal gravitational constant, 760 Unstable equilibria, 1046–1047 Upper bound for a sequence, 546 for a set of numbers, AP-28 Upper sums, 313–314 Values absolute, 18-5 absolute maximum, 825–826, 829–932 absolute minimum, 825–826, 829–932 boundary, 17-6 equilibrium, 1044–1046 extreme (See Extreme values) of functions, 22, 127 average, in space, 890 limits of, 79–80 of two variables, B-6–B-11 zero, 127 initial value and boundary value problems, 17-5–17-6 Intermediate Value Theorem and, 127–128 of a line integral, 933 local maximum, 825–829, 831 local minimum, 825–829, 831 of logarithmic functions, 18-29 needed by the Second Derivative Test, 829 of parametric equations, 625, 627 principal, 18-25, 18-29 steady-state, 1036 of trigonometric ratios, 43 Variable costs, 166 Variables complex, complex variables dependent, 21 dummy, 331 force along a line, work done by, 411 functions of many, 803 functions of more than two differential, 821–822 linearization of, 821–822 in tangent planes, 821–822 functions of several, 767–772 definitions of, 767

dependent/independent variables, 767 differentiability for, 792–793 domain of, 767, 768 input/output variables, 767 partial derivatives of, 767–772 range of, 767, 768 real-valued function, 767 functions of three, 770–772 boundary of, 771 boundary points of, 771 bounded/unbounded regions of, 771 Chain Rule for, 798–799 definitions of, 770–772 gradient vectors in, 811–812 interior points of, 771 interior regions of, 771 level surface of, 770 open/closed regions of, 771–772 functions of two boundary of, 769 boundary point of, 769 Chain Rule for, 796–798 Composition Rule for, 777 Constant Multiple Rule for, 777 contours of, 770 definitions of, 769 graphs/graphing, 770 Increment Theorem for Functions of Two Variables, 793 interior of the region of, 769 interior point of, 769 level curves of, 770 limits for, 775–779 linearization of, 818–820 more than, 781, 788–789 open/closed regions of, 769 partial derivatives of, 784–786 properties of limits of, 777 Sandwich Theorem for, 783 Taylor’s formula for, 846–847 values of local extreme for, 825–829 independent, 21 input, 21, 36 of integration, 299, 330–331 output, 21, 36 proportional relationship of, 27 in related rates problem, 207 thickness, 394–395 u, in Substitution Rule, 356 Vector fields assumptions on, 951 circulation for, 944–945 continuous, 937 curl, 993–995 definition of, 937–938 differentiable, 937 divergence of, 964–966, 1006–1007 flux across a simple closed plane curve, 945–946 gradient fields, 938–939 gravitational, 950 integrals flow, 944–945

Subject Index

line, 930–935, 939–941 surface, 987 unified theory of, 1013–1014 integration in, 930–1021 potential function for, 951 work done by a force over a curve in space, 942–944 Vector functions, 725–766 antiderivatives of, 734–735 arc length in space, 743–745 circle of curvature for plane curves, 749–750 of constant length, 731 continuity of, 726–727 curvature and normal vectors for space curves, 750–751 curvature and normal vectors of a curve, 747–751 curvature of a plane curve, 747–749 curves in space and their tangents, 725–731 definite integrals of, 735 derivatives and motion, 727–729 differentiable, 728 differential equations for, 735–736, 740 differentiation rules for, 729–731 equations for differential, 735–736, 740 of the path for ideal projectile motion, 737 formula for calculating curvature, 747–749 formulas for computing curvature and torsion, 756–757 Fundamental Theorem of Calculus for continuous, 735–736 indefinite integrals of, 734 initial value problems for, 736–737, 740 integrals of, 734–739 limits of, 726–727 motion in space and, 734–739 tangential and normal components of acceleration, 753–754 TNB frame, 753 torsion, 754–757 vector functions of constant length, 731 velocity and acceleration in polar coordinates, 759–762 Vectors, 680–688 acceleration, 729, 753 addition of, 682–683 in algebra operations, 682–683, 687 Angle Between Two Vectors, 691–693 angles between, 711–712 applications of, 686–687 binormal vector, 753, 757 component form of, 680–682 coplanar, 683 cross product as area of parallelogram, 700–701 cross product of two, 699–700 curl, 961–964, 966–968, 1009–1010

of a curve binormal, 753, 757 normal, 747–751 tangent, 725–731 definition of, 680 derivative in direction of, 728–729 as determinant, 703–704 difference (subtraction) of, 683 differential form, 943 directed line segment to represent, 680 direction of, 684–685 in engineering, 686 equality of, 680, 681 fields, 725 formula for curvature, 756 geometry in space and, 675–724 in gravitational field, 938 in higher dimensions (See Higher dimension vectors) i-component of, 684–685 initial point of, 680 j-component of, 684–685 k-component of, 684–685 length (magnitude) of, 680, 681 midpoint of line segments, 685 in navigation, 686 n dimensional, 687–688 nonzero, 691–694, 699–700, 706, 709–710 notation for, 694 orthogonal (perpendicular), 693 parallel, 699, 700 parallelogram law of addition, 682 in physics, 686 position, 725 principal unit normal, 748, 757 product, 700 projections of, 693–696 resultant, 682 right-hand rule for, 699 scalar component in direction of, 694–695 scalar multiplication of, 682–683, 694–695 standard position, for a point, 681 standard unit, 19-7, 684 terminal point of, 680 three-dimensional, 681 in three dimensions (See Threedimensional vectors) torque, 702–703 triple scalar product of, 703–704 two-dimensional, 681, 696–697 unit, 19-6, 684–685 unit normal, 699, 753 unit tangent, 745, 757 velocity, 680 writing, 681 zero vector, 681 Velocity in acceleration, 163–165 acceleration for finding, 247 angular, 1000 average, 162 definition of, 162 displacement and, 162 distance traveled, 315–318, 348

fields, circulation for, 944–945 motion with resistance proportional to, 1038–1039 in polar coordinates, 759–762 rates of change in, 162–165 resistance proportional to, 1038–1039 in speed, 162–163 terminal, 355, 1049 vector, 680 Vertical asymptotes, 115–116, 524–525 Vertical cross-sections, for finding limits of integration, 868 Vertical line test, for functions, 24 Vertical scaling and reflecting formulas, 37 Vertical shifts, 36 Vertical tangent lines, 141–142, 146 Vertices of ellipses, 657 of hyperbolas, 658 of parabolas, AP-19–AP-20, 656 Vibrations, 17-15–17-16 Volts (V), 17-19 Volumes of double integrals over general regions, 865–868 over rectangles, 860–861 of a solid region in space, 883–884 of solids definition of, 381 Pappus’s Theorem for Volumes, 427–428 of revolution, 382–387 of solids, calculating Cavalieri’s principle, 382 by disks for rotation about the x-axis, 383–384 by disks for rotation about the y-axis, 384–385 shell formula for revolution about a vertical line, 394–395 shell method, 393–396 using cross-sections, 380–387 using cylindrical shells, 391–396 using disk method, 382–385 using washer method, 385–387 by washers for rotation about the x-axis, 386–387 by washers for rotation about the y-axis, 387 Washer method, 385–387 Wave equation, 795 Wavelets, 19-24 Daubechies, 19-33–19-34 definition of, 19-26 Haar, 19-24–19-26, 19-31, 19-33–19-34 Weierstrass, Karl, 526 Weierstrass function, 150 Weight-density, of fluid, 414 Wild oscillation, 146 Wilson lot size formula, 288, 794, 824

I-21

Wires, thin, 421 Witch of Agnesi, 159, 632 Work definition of, 410 done by a force over a curve in space, 942–944 force and, 410–415 constant force, 410–411 fluid force on a constant-depth surface, 414–415 fluid pressures, 414–415 force constant or spring constant, 412 Hooke’s Law, 411–412 integral for fluid force against a vertical plate, 415 lifting objects and pumping liquids from containers, 412–413 pressure-depth equation, 414 variable force along a line, 411 by force through displacement, 696 integrals, 942–944 units of measurement in, 410–411 Writing vectors, 681 x-axis curve as symmetric about, 33 by disks for rotation about, 383–384 moment about, 422–425 revolution about, 407, 640 x-coordinate (or abscissa), AP-14 x-intercept, AP-17 x-limits of integration, 868, 885, 886, 889 xy-plane, 675, 677 xz-plane, 675 y, integration with respect to, 366–367 y-axis by disks for rotation about, 384–385 graph of an even function as symmetric about, 26–27 moment about, 422–425 of the parabola, 656 revolution about, 408, 640 y-coordinate (or ordinate), AP-14 y-intercept, AP-17 y-limits of integration, 868, 885–889 yz-plane, 675 z-axis, 675–676, 714, 716 Zero, frequency, 19-2 Zero denominators, eliminating common factors from, 83–84 Zero derivatives in constant functions, 246 Zero directional derivatives, 814 Zero value of a function, 127 Zero vectors, 681 Zero Width Interval Rule, 334 Zipper theorem, 550 z-limits of integration, 885–889, 904–906

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C-1

A Brief Table of Integrals Basic Forms

x n +1 + C , n ≠ −1 n+1

1.

∫

k dx = kx + C,

3.

∫

dx = ln x + C x

5.

∫

a x dx =

7.

∫

cos x dx = sin x + C

9.

∫

csc 2 x dx = − cot x + C

10.

∫ sec x tan x dx

11.

∫

csc x cot x dx = − csc x + C

12.

∫

13.

∫

cot x dx = ln sin x + C

14.

∫ sinh x dx

15.

∫

cosh x dx = sinh x + C

16.

∫

17.

∫

1 dx x = arctan + C a2 + x2 a a

18.

∫

19.

∫

20.

∫

k any number

ax +C ln a

(a

> 0,  a ≠ 1)

dx x = sinh −1 + C 2 a + x

a2

(a

> 0)

2.

∫

4.

∫ e x dx

6.

∫ sin x dx

8.

∫ sec 2 x dx

x n dx =

= ex + C = − cos x + C = tan x + C = sec x + C

tan x dx = ln sec x + C = cosh x + C

dx x = arcsin + C 2 a −x

a2

dx 1 x = arcsec +C 2 2 a a x x −a dx x = cosh −1 + C 2 a −a

x2

(x

> a > 0)

Forms Involving ax + b

21.

∫ ( ax + b )

22.

∫

23.

∫ ( ax + b )

25.

∫

27.

T-1

n

dx =

x ( ax + b ) n dx = −1

dx =

+ b ) n +1 + C , n ≠ −1 a ( n + 1)

( ax

( ax

1 ln ax + b + C a

x ( ax + b )−2 dx =

∫(

ax + b )

n

+ b ) n +1 ⎡ ax + b b ⎤ − ⎢ ⎥ + C , n ≠ −1, −2 a2 n + 1⎦ ⎣ n+2

1 a2

b ⎤ ⎡ ⎢ ln ax + b + ⎥+C ax + b ⎦ ⎣

2 ( ax + b ) dx = a n+2

∫

x ( ax + b )−1 dx =

26.

∫

dx 1 x = ln +C x ( ax + b ) b ax + b

28.

∫

n+2

+ C , n ≠ −2

x b − 2 ln ax + b + C a a

24.

ax + b dx dx = 2 ax + b + b ∫ x x ax + b

A Brief Table of Integrals

29. (a) 30.

ax + b − ax + b +

dx 1 = ln x ax + b b

∫

a ax + b ax + b dx = − + 2 x x 2

∫

b +C b

dx +C x ax + b

∫

(b)

∫

ax − b +C b

2 dx arctan = x ax − b b

dx ax + b a = − − bx 2b ax + b

31.

∫

x2

33.

∫

(a2 + x 2 )

T-2

∫

dx +C x ax + b

Forms Involving a 2 + x 2

dx 1 x = arctan + C a2 + x2 a a

32.

∫

34.

∫

35.

∫

a 2 + x 2 dx =

36.

∫

x 2 a 2 + x 2 dx =

37.

∫

a2 + x2 dx = x

38.

∫

a2 + x2 dx = ln ( x + x2

39.

∫

x2 a2 dx = − ln ( x + 2 a2 + x2

40.

∫

41.

∫

dx x = sinh −1 + C = ln ( x + 2 a + x

a2

1 a+ dx = − ln a + x2

x

=

1 x x + arctan + C 2a 2 ( a 2 + x 2 ) 2a 3 a

=

x 1 x+a ln + +C 2a 2 ( a 2 − x 2 ) 4a 3 x−a

a2 + x 2 ) + C

x 2 a4 ( a + 2x 2 ) a 2 + x 2 − ln ( x + 8 8

a2

2

a2 + x 2 ) + C

x a2 a2 + x2 + ln ( x + 2 2

a+

a2 + x 2 +C x

a 2 + x 2 )  −

a2 + x 2 +C x

a 2 + x 2 − a ln

dx

a2 + x2 ) +

a2 + x 2 ) + C

x a2 + x2 +C 2

a2 + x2 +C x

a2 + x2 dx = − +C a 2x a2 + x2

x2

Forms Involving a 2 − x 2

dx 1 x+a = ln +C a2 − x2 2a x−a

42.

∫

44.

∫

46.

∫

x 2 a 2 − x 2 dx =

47.

∫

a2 − x2 dx = x

49.

∫

a2

51.

∫

dx x = arcsin + C 2 2 a a −x

43.

∫

45.

∫

a 2 − x 2 − a ln

a+

a2 − x2 + C 48. x 50.

a2 − x2 dx = − +C a 2x a2 − x2

Forms Involving x 2 − a 2

52.

∫

53.

∫

dx = ln x + − a2

x2

x 2 − a 2 dx =

2

a 2 − x 2 dx =

x a2 x a2 − x2 + arcsin + C a 2 2

a4 x 1 arcsin − x a 2 − x 2 ( a 2 − 2 x 2 ) + C 8 a 8

a2 x 1 x2 dx = arcsin − x a 2 − x 2 + C 2 a 2 2 −x

x2

dx (a2 − x 2 )

x2 − a2 + C

x a2 x2 − a2 − ln x + 2 2

x2 − a2 + C

a2 − x2 x dx = − arcsin − 2 x a

∫ ∫

x

1 a+ dx = − ln a − x2

a2

a2 − x2 +C x

a2 − x 2 +C x

T-3

A Brief Table of Integrals

x( x 2 − a 2 ) na 2 dx = − n+1 n+1 n

54.

55.

∫(

x2

−

a2

)

n

dx

∫(

x2

−

a2

)

x( x 2 − a 2 ) ( 2 − n )a 2

=

n

(

x2 − a2 ) n+2

56.

∫

x ( x 2 − a 2 ) dx =

57.

∫

x 2 x 2 − a 2 dx =

58.

∫

x2 − a2 dx = x

59.

∫

x2 − a2 dx = ln x + x2

60.

∫

a2 x2 dx = ln x + 2 x2 − a2

61.

∫

62.

∫

n

2− n

−

(n

∫(

n−3 − 2)a 2

x2 − a2 )

n−2

dx , n ≠ −1

dx

∫(

x2

− a2 )

n−2 ,

n ≠ 2

n+2

+ C , n ≠ −2

x a4 ( 2x 2 − a 2 ) x 2 − a 2 − ln x + 8 8 x 2 − a 2 − a arcsec

x2 − a2 + C

x +C a x2 − a2 +C x

x2 − a2 −

x2 − a2 +

x x2 − a2 + C 2

1 1 a dx x = arcsec + C = arccos +C a x a a x x2 − a2 x2

x2 − a2 +C a 2x

dx = x2 − a2

Trigonometric Forms

1 = − cos ax + C a

63.

∫ sin ax dx

65.

∫ sin 2 ax dx

=

67.

∫ sin n ax dx

= −

68.

∫

cos n ax dx =

sin 2ax x − +C 2 4a sin n −1 ax cos ax n −1 + na n

cos n −1 ax sin ax n −1 + na n

69. (a)

∫ sin ax cos bx dx

= −

(b)

∫ sin ax sin bx dx

=

(c)

∫

cos ax cos bx dx =

∫

cos ax dx =

66.

∫

cos 2 ax dx =

1 sin ax + C a sin 2ax x + +C 2 4a

cos n − 2 ax dx

cos ( a + b ) x cos ( a − b ) x + C, a 2 ≠ b 2 − 2( a + b ) 2( a − b )

sin ( a − b ) x sin ( a + b ) x + C, a 2 ≠ b 2 − ( ) 2 a−b 2( a + b ) sin ( a − b ) x sin ( a + b ) x + C, a 2 ≠ b 2 + 2( a − b ) 2( a + b )

cos 2ax +C 4a

∫ sin ax cos ax dx

72.

∫

cos ax 1 dx = ln sin ax + C sin ax a

74.

∫

75.

∫

sin ax 1 dx = − ln cos ax + C cos ax a sin n −1 ax cos m +1 ax n −1 sin n ax cos m ax dx = − + ( ) a m+n m+n

76.

∫ sin n ax cos m ax dx

=

∫

∫ sin n−2 ax dx

70.

= −

64.

sin n +1 ax cos m −1 ax m −1 + a( m + n ) m+n

71.

∫ sin n ax cos ax dx

73.

∫

=

sin n +1 ax + C , n ≠ −1 ( n + 1) a

cos n ax sin ax dx = −

∫ sin n−2 ax cos m ax dx ,

∫ sin n ax cos m −2 ax dx ,

cos n +1 ax + C , n ≠ −1 + 1) a

(n

n ≠ −m ( reduces sin n ax ) m ≠ −n ( reduces cos m ax )

A Brief Table of Integrals

77.

∫

⎡ b−c π −2 dx ax ⎤ = arctan ⎢ tan − ⎥ + C, b 2 > c 2 2 2 4 2 ⎥⎦ b + c sin ax a b −c ⎣⎢ b + c

78.

∫

c + b sin ax + c 2 − b 2 cos ax dx −1 ln = + C, b 2 < c 2 b + c sin ax b + c sin ax a c2 − b2

79.

∫

1 ax π dx = − tan − +C a 1 + sin ax 4 2

81.

∫

⎡ b−c 2 dx ax ⎤ arctan ⎢ tan ⎥ + C , b 2 > c 2 = 2 2 2 ⎥⎦ b + c cos ax b c + ⎢ a b −c ⎣

82.

∫

c + b cos ax + c 2 − b 2 sin ax dx 1 ln = + C, b 2 < c 2 b + c cos ax b + c cos ax a c2 − b2

83.

∫

1 ax dx = tan +C a 1 + cos ax 2

85.

∫

x sin ax dx =

87.

∫

x n sin ax dx = −

89.

∫

tan ax dx =

91.

∫

tan 2 ax dx =

93.

∫

tan n ax dx =

95.

∫ sec ax dx

97.

∫ sec 2 ax dx

99.

∫

(

(

)

)

80.

sec n ax dx =

)

dx 1 ax = − cot +C 1 − cos ax a 2

86.

∫

x cos ax dx =

88.

∫

x n cos ax dx =

90.

∫

cot ax dx =

1 tan ax − x + C a

92.

∫

1 cot 2 ax dx = − cot ax − x + C a

tan n −1 ax − a ( n − 1)

94.

∫

cot n ax dx = −

96.

∫

1 csc ax dx = − ln csc ax + cot ax + C a

98.

∫

1 csc 2 ax dx = − cot ax + C a

1 x sin ax − cos ax + C 2 a a

∫

x n −1 cos ax dx

∫

tan n − 2 ax dx , n ≠ 1

1 ln sec ax + tan ax + C a

=

(

dx 1 ax π = tan + +C a 1 − sin ax 4 2

∫

1 ln sec ax + C a

=

∫

84.

xn n cos ax + a a

1 tan ax + C a sec n − 2 ax tan ax n−2 + ( ) a n −1 n −1

∫ csc n ax dx

101.

∫ sec n ax tan ax dx

= −

∫ sec n−2 ax dx ,

csc n − 2 ax cot ax n−2 + a ( n − 1) n −1

100.

=

∫

1 x cos ax + sin ax + C 2 a a xn n sin ax − a a

∫

x n −1 sin ax dx

1 ln sin ax + C a

cot n −1 ax − a ( n − 1)

∫

cot n − 2 ax dx , n ≠ 1

n ≠1

csc n − 2 ax dx , n ≠ 1

sec n ax + C, n ≠ 0 na

102.

∫ csc n ax cot ax dx

104.

∫ arccos ax dx

= −

csc n ax + C, n ≠ 0 na

Inverse Trigonometric Forms

103.

∫ arcsin ax dx

= x arcsin ax +

1 1 − a 2x 2 + C a

105.

∫ arctan ax dx

= x arctan ax −

1 ln (1 + a 2 x 2 ) + C 2a

106.

∫

x n arcsin ax dx =

x n +1 a arcsin ax − n+1 n+1

∫

T-4

x n +1 dx 1 − a 2x 2

, n ≠ −1

= x arccos ax −

1 1 − a 2x 2 + C a

T-5

107. 108.

A Brief Table of Integrals

∫

x n arccos ax dx =

x n +1 a arccos ax + n+1 n+1

∫

∫

x n arctan ax dx =

x n +1 a arctan ax − n+1 n+1

∫

x n +1 dx 1 − a 2x 2

, n ≠ −1

x n +1 dx , n ≠ −1 1 + a 2x 2

Exponential and Logarithmic Forms

1 ax e +C a

109.

∫ e ax dx

111.

∫

xe ax dx =

113.

∫

x n b ax dx =

114.

∫ e ax sin bx dx

=

115.

∫ e ax cos bx dx

=

117.

∫

x n ( ln ax ) m dx =

118.

∫

x −1 ( ln ax ) m dx =

=

Forms Involving

e ax ( ax − 1) + C a2 x n b ax n − a ln b a ln b a2

∫

b ax dx =

112.

∫

x n e ax dx =

∫

ln ax dx = x ln ax − x + C

1 n ax n x e − a a

∫

x n −1e ax dx

e ax ( a sin bx − b cos bx ) + C + b2

e ax ( a cos bx + b sin bx ) + C a2 + b2 x n +1 ( ln ax ) m m − n+1 n+1

116.

x n ( ln ax ) m −1 dx , n ≠ −1

∫

( ln ax ) m +1 + C , m ≠ −1 m+1

∫

119.

dx = ln ln ax + C x ln ax

2 ax − x 2 ,  a > 0

(

)

∫

dx x−a = arcsin +C a 2ax − x 2

121.

∫

2ax − x 2 dx =

∫(

∫

x n −1b ax dx , b > 0,  b ≠ 1

120.

122.

1 b ax + C , b > 0,  b ≠ 1 a ln b

110.

(

2ax − x 2 ) dx = n

dx

(x

(x

123.

∫(

124.

∫

x 2ax − x 2 dx =

125.

∫

2ax − x 2 dx = x

126.

∫

2ax − x 2 dx = −2 x2

127.

∫

2ax − x 2

2ax −

x2

x dx

)

n

=

)

x−a a2 x−a 2ax − x 2 + arcsin +C a 2 2

(x

− a )( 2ax − x 2 ) na 2 + n+1 n+1 n

− a )( 2ax − x 2 ) ( n − 2)a 2

2− n

+

(n

∫(

n−3 − 2)a 2

2ax − x 2 )

∫

n−2

dx

dx

(

2ax − x 2 )

(

n−2

)

+ a )( 2 x − 3a ) 2ax − x 2 a3 x−a + arcsin +C 6 2 a

2ax − x 2 + a arcsin

= a arcsin

( x −a a ) + C

(

)

2a − x x−a − arcsin +C x a

( x −a a ) −

2ax − x 2 + C

128.

∫

dx 1 2a − x +C = − a x x 2ax − x 2

130.

∫

cosh ax dx =

132.

∫

cosh 2 ax dx =

Hyperbolic Forms

129.

∫ sinh ax dx

131.

∫ sinh 2 ax dx

= =

1 cosh ax + C a sinh 2ax x − +C 4a 2

1 sinh ax + C a sinh 2ax x + +C 4a 2

A Brief Table of Integrals

∫

sinh n ax dx =

sinh n −1 ax cosh ax n −1 − na n

134.

∫

cosh n ax dx =

cosh n −1 ax sinh ax n −1 cosh n − 2 ax dx , n ≠ 0 + na n ∫

135.

∫

x sinh ax dx =

x 1 cosh ax − 2 sinh ax + C a a

137.

∫

x n sinh ax dx =

139.

∫

tanh ax dx =

141.

∫

tanh 2 ax dx = x −

143.

∫

tanh n ax dx = −

tanh n −1 ax + ( n − 1) a

∫

tanh n − 2 ax dx , n ≠ 1

144.

∫

coth n ax dx = −

coth n −1 ax + ( n − 1) a

∫

coth n − 2 ax dx , n ≠ 1

145.

∫ sech ax dx

147.

∫ sech 2 ax dx

=

1 tanh ax + C a

149.

∫ sech n ax dx

=

sech n − 2 ax tanh ax n−2 + ( n − 1) a n −1

150.

∫

csch n ax dx = −

151.

∫

sech n ax tanh ax dx = −

153.

∫ e ax sinh bx dx

=

e ax ⎡ e bx e −bx ⎤ − ⎢ ⎥ + C, a 2 ≠ b 2 a − b⎦ 2 ⎣a + b

154.

∫ e ax cosh bx dx

=

e ax ⎡ e bx e −bx ⎤ + ⎢ ⎥ + C, a 2 ≠ b 2 a − b⎦ 2 ⎣a + b

133.

=

xn n cosh ax − a a

∫

∫ sinh n−2 ax dx ,

n ≠ 0

∫

x cosh ax dx =

138.

∫

x n cosh ax dx =

140.

∫

coth ax dx =

142.

∫

coth 2 ax dx = x −

146.

∫

csch ax dx =

148.

∫

1 csch 2 ax dx = − coth ax + C a

1 ln ( cosh ax ) + C a 1 tanh ax + C a

1 arcsin ( tanh ax ) + C a

∫ sech n−2 ax dx ,

csch n − 2 ax coth ax n−2 − ( n − 1) a n −1

∫

x 1 sinh ax − 2 cosh ax + C a a

136.

x n −1 cosh ax dx

xn n sinh ax − a a

157.

∞

∫0

x n −1e − x dx = Γ(n) = ( n − 1)!, n > 0

π 2

∫0

sin n x dx =

π 2

∫0

x n −1 sinh ax dx

1 ln sinh ax + C a 1 coth ax + C a

1 ax ln tanh +C a 2

n ≠1

152.

∫

156.

∫0

csch n ax coth ax dx = −

csch n ax + C, n ≠ 0 na

Some Definite Integrals

155.

∫

csch n − 2 ax dx , n ≠ 1

sech n ax + C, n ≠ 0 na

T-6

⎧⎪ 1 ⋅ 3 ⋅ 5 ⋅  ⋅ ( n − 1) π ⎪⎪ ⋅ , 2 ⋅ 4 ⋅ 6 ⋅⋅ n 2 cos n x dx = ⎪⎨ ⎪⎪ 2 ⋅ 4 ⋅ 6 ⋅  ⋅ ( n − 1) , ⎪⎪ 3 ⋅ 5 ⋅ 7 ⋅⋅ n ⎪⎩

∞

e −ax 2 dx =

1 π , a > 0 2 a

if  n  is an even integer ≥2 if  n  is an odd integer ≥3

Basic Algebra Formulas Arithmetic Operations

a c ac ⋅ = b d bd

a ( b + c ) = ab + ac, a c ad + bc + = , b d bd

ab a d = ⋅ cd b c

Laws of Signs

−a a a = − = b b −b

−( −a ) = a, Zero Division by zero is not defined.

0 = 0, a 0 = 1, 0 a = 0 a For any number a: a ⋅ 0 = 0 ⋅ a = 0

If a ≠ 0:

Laws of Exponents

a ma n = a m+n ,

( ab ) m

= a mb m ,

(a m ) n = a mn ,

am n =

n

am = ( n a)

m

If a ≠ 0, then am = a m−n , an The Binomial Theorem (a

a 0 = 1,

a −m =

1 . am

For any positive integer n,

n( n − 1) n−2 2 a b 1⋅2 n ( n − 1 )( n − 2 ) n −3 3 + a b +  + nab n −1 + b n . 1⋅2⋅3

+ b ) n = a n + na n −1b +

For instance, (a

+ b ) 2 = a 2 + 2ab + b 2 ,

(a

− b ) 2 = a 2 − 2ab + b 2

(a

+ b )3 = a 3 + 3a 2 b + 3ab 2 + b 3 ,

(a

− b )3 = a 3 − 3a 2 b + 3ab 2 − b 3 .

Factoring the Difference of Like Integer Powers, n > 1

a n − b n = ( a − b )( a n −1 + a n −2 b + a n −3b 2 +  + ab n −2 + b n −1 ) For instance, a 2 − b 2 = ( a − b )( a + b ), a 3 − b 3 = ( a − b )( a 2 + ab + b 2 ), a 4 − b 4 = ( a − b )( a 3 + a 2 b + ab 2 + b 3 ). Completing the Square

If a ≠ 0, then

ax 2 + bx + c = au 2 + C

( u = x + ( b 2a ), C = c − 4ba ). 2

The Quadratic Formula

If a ≠ 0 and ax 2 + bx + c = 0, then x =

−b ±

b 2 − 4 ac . 2a

Geometry Formulas A = area, B = area of base, C = circumference, S = surface area, V = volume Triangle

Similar Triangles c′

c h

a′

Pythagorean Theorem c

a

b

b′

b

a

b a′ = b′ = c′ b c a

A = 1 bh 2

Parallelogram

a2 + b2 = c2

Trapezoid

Circle a

h

h

b

A = pr 2, C = 2pr

r

b

A = bh

A = 1 (a + b)h 2

Any Cylinder or Prism with Parallel Bases

Right Circular Cylinder r

h

h

h

V = Bh

B

B

V = pr2h S = 2prh = Area of side

Any Cone or Pyramid

Right Circular Cone

h

h

B

Sphere

V=

1 Bh 3

B

V = 1 pr2h 3 S = prs = Area of side

V = 43 pr3, S = 4pr2

tan A + tan B 1 − tan A tan B tan A − tan B tan ( A − B ) = 1 + tan A tan B π π sin A − = − cos A, cos A − = sin A 2 2 π π sin A + = cos A, cos A + = − sin A 2 2 1 1 sin A sin B = cos ( A − B ) − cos ( A + B ) 2 2 1 1 cos A cos B = cos ( A − B ) + cos ( A + B ) 2 2 1 1 sin A cos B = sin ( A − B ) + sin ( A + B ) 2 2 1( 1 sin A + sin B = 2 sin A + B ) cos ( A − B ) 2 2 1 1 sin A − sin B = 2 cos ( A + B ) sin ( A − B ) 2 2 1 1 cos A + cos B = 2 cos ( A + B ) cos ( A − B ) 2 2 1 1 cos A − cos B = −2 sin ( A + B ) sin ( A − B ) 2 2

Trigonometry Formulas

tan ( A + B ) =

y

Definitions and Fundamental Identities y 1 = Sine: sin θ = r csc θ

x 1 = cos θ = r sec θ

Cosine:

tan θ =

Tangent:

P(x, y)

r u x

0

( (

y x

y 1 = x cot θ

Identities

sin (−θ) = − sin θ, cos(−θ) = cos θ sin 2 θ + cos 2 θ = 1, sec 2 θ = 1 + tan 2 θ, csc 2 θ = 1 + cot 2 θ sin 2θ = 2 sin θ cos θ, cos 2θ = cos 2 θ − sin 2 θ 1 + cos 2θ 1 − cos 2θ , sin 2 θ = 2 2 sin ( A + B ) = sin A cos B + cos A sin B sin ( A − B ) = sin A cos B − cos A sin B cos 2 θ =

cos ( A + B ) = cos A cos B − sin A sin B cos ( A − B ) = cos A cos B + sin A sin B

) )

(

)

(

)

Trigonometric Functions y Degrees

Radian Measure s 1

u

C ir

it circ

1

p 4

1

l

cle of radiu

p 4

"2

90

e

Un

1

45

r

y = sin x

Radians

45 "2

p 2

1

–p – p 2

p 2

0

y

"3

60

90 1

3p 2p 2

x

–p – p 2

p 2

0

2

"3

p 3

p 2

– 3p –p – p 2 2

p

3p 2p 2

x

Domain: (−∞, ∞) Range: [−1, 1]

y

y = tan x

p 6

30 2

p

y = cos x

Domain: (−∞, ∞) Range: [−1, 1]

sr

s s θ = = θ or θ = , r 1 r 180 ° = π radians.

y

0 p p 3p 2 2

x

y = sec x

1 – 3p –p – p 0 2 2

p p 3p 2 2

x

1

The angles of two common triangles, in degrees and radians.

Domain: All real numbers except odd integer multiples of p2 Range: (−∞, ∞) y

y

y = csc x

1 –p – p 0 2

Domain: All real numbers except odd integer multiples of p2 Range: (−∞, −1] ´ [1, ∞)

p 2

p 3p 2p 2

Domain: x ≠ 0, ±p, ±2p, . . . Range: (−∞, −1] ´ [1, ∞)

x

y = cot x

1 –p – p 0 2

p 2

p 3p 2p 2

Domain: x ≠ 0, ±p, ±2p, . . . Range: (−∞, ∞)

x

Series Tests for Convergence of Infinite Series

5. Series with some negative terms: Does ∑ a n converge? If yes, so does ∑ a n because absolute convergence implies convergence. 6. Alternating series: ∑ a n converges if the series satisfies the conditions of the Alternating Series Test.

1. The nth-Term Test: Unless a n → 0, the series diverges. 2. Geometric series: ∑ ar n converges if r < 1; otherwise it diverges. 3. p-series: ∑ 1 n p converges if p > 1; otherwise it diverges. 4. Series with nonnegative terms: Try the Integral Test, Ratio Test, or Root Test. Try comparing to a known series with the Comparison Test or the Limit Comparison Test. Taylor Series

1 = 1 + x + x2 +  + xn +  = 1− x

∞

∑ x n,

1 = 1 − x + x 2 −  + (−x ) n +  = 1+ x ex = 1 + x + sin x = x − cos x = 1 −

x2 xn ++ += 2! n!

∞

∑

n=0

∞

∑ (−1)n x n,

x DQGFLVDQ\DQWLGHULYDWLYHRIf RQ < a b >WKHQ b

∫a

f  x dx = F b − F a 

6XEVWLWXWLRQLQ'H¿QLWH,QWHJUDOV b

∫a

f  g x ⋅ g′ x dx 

,QWHJUDWLRQE\3DUWV gb

∫ga

f u du

b

∫a

b

⎤ u x υn x dx  u x υ x ⎥ − ⎦a

b

∫a

υ x u′ x dx