N/A , Calculus 2012 9781478356882, 147835688X

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Calculus

The authol' advising the 1991 "SMALL" unde1·graduate 1·esearch Geometry Group, which incidentally edited the work which is described on page 67. At the head of the table, the author; to his right, Hugh Howards and Steve Root; to his left, Kathryn Kollett, Thomas Colthurst and Holly Lowy (not visible Chris Cox and Joel Foisy). The Williams College "SMALL" Project is a National Science Foundation site for Research Experiences for Undergi-aduates. (Photo by Che1-yl LeClaire.)

Calculus Frank Morgan Department of Mathematics and Statistics Williams Co11ege Williamstown, Massachusetts

Succeeding Calculus Lite. Copyright 2012, Frank Morgan. All rights reserved. ISBN- l 3: 978-1478356882 ISBN- 10: 147835688X

Cover picture shows one of infinitely many optimal convex pentagonal planar tilings. Figures I. I, 8.1, 8.2, 8.5, 8.9, 8. 10, 22.1 and 30.1 by James F. Bredt. Manuscript typed by Manuel Alfaro. Student editorial committee: Cordelia Aidkin, Joe Comeli, Chris Cox, Stephen Fiedler, Joel Foisy, Amy Huston, Kathryn Kollett, Holly Lowy, Jonathan Mattingly, Stephen Root, Eric Swanson.

This book is dedicated to all calculus teachers, especially my brother David. Here David a11d I are lea ving for school. (Photograph courtes.;r ofthe Morga,r family: taken by the anthor's fathe,·, Mr. Fra11k E. Mor11a11.)

Contents

Preface

ix

Introduction

xi

I

The Derivative 1 1 Instantaneous Velocity and the Deriva.t.ive . . . . . . . 3 2 Geometric Interpretation of the Derivative as the SlopP of the Graph . . . . . . . . . . . . . . . . . 11 3 Thf' Product and Quotient Rules . . . . . . 17 4 The Chain Rule and Implicit Differentiation 23 5 The Extended Power Huie . . . . . . . 29 6 Sines, Cosines, and Their Derivatives . 35 7 Maxima and Minima . . . . . . . . . . 43 8 Maxima-Minima Heal-World Problems 61 9 Expouentials and Logarithms . . . . . 77 10 Exponential Growth and Decay . . . . 85 11 The Second Derivative and Curve Sketching 93 12 Antidifferentiation . . . . . . . . 105 13 Differentiability and Continuit.y . 111 14 Review . . . . . . . . . . . . . . . 117 Vll

viii

II

III

Contents

The Integral 15 Area and the Riemann Integral . . . . . 16 The F\mdamental Theorem of Calculus. 17 Properties of the Definite Integral . 18 Recognition-Substitution (For Indefinite Integrals} .. 19 Integration by Substitution 20 Review of Integration 21 Trigonometric Functions and Their Inverses 22 Volume, Length, Avcrage . . . . . . . . . 23 Integration by Table . . . . . . . . . . . . 24 Partial Fractions aud Integration by Parts 25 N umerka.l Methodf; . 26 Review Problems

125 . 127 . 133 . 139

Infinite Series Infinite Series (Smns) . . . . . 28 Power Series and Taylor Series

201 . 203 . 211

27

IV

. 147 . 15~l . 159 . 161 . 169 . 175 . 181 . 189 . 193

Differential Equations 217 Differential Equations . . . . . . . . . . . . . . . . . . 219 30 Linear Second Order Homogeneous Const~nt-Coefficient Differential Equations . . . . . . . . . . . . . . . . . . 233 29

V

Multivariable Calculus 31 Partial Derivatives 32 Double Integrals .. 33 Critical Points 34 Maxima and l'vlinima An~wers to Odd-Numbered Exercises . GrePk Letters . . . . . . . . . . . . . .

. . . . . .

241 243 247 251 253 257 293

Answers to Odd-Numbered Exercises

257

Greek Letters

293

Index

295

Preface

Many students anitaliLy during my year thrre and rectis perpcndicu•1j~-• .11f 1(7 I~ AB,-, KN, LO, J: KJ iZ:S oc~urrensm B, N,lJ, R, ~·. Qvo-_,;~fr nwn A•K dr 11t A I',,., cnr huJus momenruml r. ur 1llms mo-t:,.-,,. • ~ f.,'";,, aw., I • - fw....t ;J f•f" J ·MJ mrnrum 'l A f Q_, id dt ut A 1' in KC. Na1 velocif.iris inc re•~•~,. m(·nrum f Q, per morus Lcg.1,;,.P]i~Portionale cfc vi generantit:·..-~' KC. Compo1wltr,r:irio i1~11us f..: 1.., cu~tionc ipCius KN, & t• Al fi~·r rcctangulum Lx 4 N ~ A,l'x K'l:;x 'it. N; hoc cft, ob datlan rdbngulum K CxK N, ut A f. Atqui arci Hypcroolicl: I 1(/11 ,.. • ., •. . , "" i, i 61L. KN·

fc....,:.

~ . ~»:.--?J:!#. ,-, .~ ··""i·--'-r;. ~

-::i:~~:. ~-

....,...,!, - ~..i'""' 0)

4000

19. f(r) ~ 21l'r2 + - r X

(r > O)

2

20. f(:r;)

1 + ;z;2

21. J(x)

=l + x2

:,:'l

22.

a. J(x)

1 + x2 b. f(x) = 2 +:i:2 23.

a. J(x)

= 2x3 + 3:z:2 -

h. J(x)

2x 3 + 3.1:2

24. f (:1:)

6:.r; + 1 (x 2: 0)

+ 6x + 1

(x

:2:::

0)

= sinx

(You should just know this one without doing any work. } 25. f (:r) = cosx

27. f (x ) == sinx - cosx

57

7. Maxima and Minima

Compute the following limits.

28.

a.

. 5 lnn -

x-+8+ X -

8

5 b. lim - -

x ~8- X - 8

29.

a..

b. C.

30.

x -+ 4 -

(x + 3)(x

lim

x -+ 4+

+ 4)

(x - 3)(x - 4)

x 2 + 5x - 14 x -+3+ x 2 + X - 12

a.

b.

32.

(x - 3)(x - 4) (x + 3)(x + 4)

lim

a. lim

b.

31.

(x - 3)(x - 4) x➔-4+ (x + 3}(x + 4)

lim

l'

IIIl

x -+3 -

x 2 + 5x - 14 x 2 + x - 12

-

lim

x 2 + 3x + 2 X 2 + 9x + 18

lim

x 2 + 3x + 2 x 2 + 9x + 18

x -+-6+

x ➔ -6

. a. lun

x -+5+

9x2 x 2 - lOx

+ 25

9x2 b. lim _,,,...---x -+5- x 2 - 10x + 25 c.

x 2 + 7x + 10 :r.-+-3+ x 2 - lOx + 11

. l1m

I. The Derivative

58 33.

+ 2:1: + 5 , . x --+ l + J , - 3:1, + 2 . x 2 + 2x + 5 b. hm - 2 - - x --+ 1- X - 3:1; + 2 x 2 + 2.1.· + 5 c. Jim 2 :r.--+2+ x - 3:1; + 2 x 2 + 2:x; + 5 d. Jim 2 , x --+2- a; - 3x + 2 2 J' + 2x + 5 e. Jim 2 x --+3+ X - 3:1; + 2 . ;r 2 + 2x + 5 f • I1111 ·• x --+'.l- .i:'" - 3:t: + 2 a. Jim

:1;

2

,2

Find the maximum and mi11imum values and points. 34.

J (x ) =

1

- :c -:/: ± 1 1 - x2 X

35. f (:c) = l _ • 36.

:i;Z

X

"'f' ± 1

f ('.1.·) -- 1cos :1;2 +x

• 37 . J('r )

:i: + 2x + 5 =--:t2 - 2a: + 2

38 . f (:i;)

2 :i: + 2:1; + 5 = X 2 - 3:i; + 2

40. f (x)

= 8:1:;i - xa

2

.·

2

4

59

7. Maxima and Minima

41. The temperature T in degrees centigrade on the orbiting space shuttle from noon until 8:00 p.m. is given by the formula

T(t) = t 3 - 9t 2 + 15t, where t is the time in hours since noon. What arP the highest and lowest temperatures during that period?

8 Maxima-Minima Real-World Problems

This chapter will consider the more interesting kind of maximaminima problems that occur in real life. Reasoning through a realworld minimization problem requires the following steps: 8.1. Procedure for Solving Maxima-Minima Real-World Problems 1. Identify the quantity y you are trying to maximiie or minimi~e (such as cost), the variable x you can adjust to minimize y (such

as the radius of some cylinder), and any prescribed quantity (such as volume) . 2. Get a formula. for y as a function of :c: y putting everything in terms of x.

= f(x).

This involves

3. As in the previous section, look at. extreme cases and critical cases (or where y or y1 is undefined). In most real-world problems, the maxima and minima are extremf' cases, often 61

62

I. The Derivative

obviously; in others, the maxima and minima are obviously not extreme cases.

8.2. The Garden Problem You have an opportunity to work in your neighbor's garden for up to four hours Saturday at $6 per hour. How long should you work to earn the most money? Solution. Here the answer is obviously the extreme case of working all four hours. For practice we will now see how the ahovP three-step procedure leads to this answer. The quantity y to be maximized is the earnings; the variable x is the number of hours to work. The formula for y in terms of ;,: is y

=

f(x)

= 6x.

In the ext reme cases, when ;c = 0, y == 0, and when ;r: 4, y = 24. Since y 1 ·- 6 is never 0, there arc no critical cases. The maximum vah1P of $24 occnrs at x = 4. The minimum value of $0 occurs at J' 0.

8.3. The Rectangular Pen Problem Find the shape of the biggest rectangular pen for your pct that you can make with 36 feet of fencing. Sec Figlll'P 8.1. Solution. The quantity to be maximized is the area A. For the variable, we could use the length f or the width w; then the unused variable is determined by the prescribed fence perimeter 2f + 2w 36. Let's use fas the variable. Theu w - (36 - 21',)/2 = 18 - eand

=

A = tw = f(l8 - f)

18£ -

e2 .

Fur this problem the ext reme cases of long, narrow pens (see Figure 8.2) arc obviously no good, so we turn to the critical cases:

o = A' == 1s - 2e Hence

e= 9,

w

1s -

2(9 - e).

e= 9.

Th(~ maximum area is given by the!) x 9 squa.r 1,

loga x is

positive if X if .i: 0 { nc>gative if X

>1 1

(since a0 = 1)

< 1;

on the other hand if a < 1, negative if x > 1 if X = 1 0 positive if .i: < 1.

9.1. Logarithm Rules Logarithms obey the following rulE>s: Product Rulf':

Q uotient Rules: Power Rule: nasc Cl1angc Rulc:

log" .i:y logn x + log0 y log ay£ loga X - loga Y, loga .rP .:. p • loga x logb x - ~ log,;l,

Not 0, 1-he graph is rnucave up.

95

ll. The Second Derivative and Curve Sketching

To distinguish them from relative maxima and minima, genuine maxima and minima are often called absolute maxima and minima. Inflection point. Au inflection point is a point where J" changes sign, i.e., concavity changes from up to down or from down to up. (Often occurs if f"(x) 0.) Asymptote. An asymptote is a line approached by the graph.

Figures 11.3(a- f) illustrate important features of graphs. Figure 11.3{a) shows y = 2x + 1, which you immediately recognize from the formula as a line y = rnx + b with slope m ::::: 2 and y-intercept b I (i.e., it rrosses the y-axis at y = l). Figure ll.3(b) shows y = - x 2 + 2x + 1, which you immediately recognize from the formula as a. parabola, since only x 2 's, :,;'s, and constants occur. Since the x 2 term grows fastest, the negative sign on the x 2 means that as :c ➔ ± oo, y ➔ -oo. Hence it must be a downward para.bola. Another wa.y to sec this is that the second derivative y" = - 2, so that the graph is concave downward. The most important feature is the maximum, which occurs where O = y' = - 2x + 2, so that x = I, and hence by the formula y = 2. Figure 11.3(c) shows the graph of a cubic (with an x 3 term). Since the :i:a term grows fastest, as x ➔ +oo, y ➔ +oo, and as x ➔ - oo, y ➔ - oo. The relative extrema can be found by setting y' 0. Th 0 on this graph y

= f (x) .

1 - - · - -·- - -

Sketch the graphs of y and inflection points. 18. y

x- 2

= -:c -- 5

as x

➔

= f(x).

(Note th hori7,ontal asymptote y == 1, approa.c-hcd

± oo.)

19. y

= x:l -

8x2

+ 5x ..J.. 1

20. y

= x4 -

2x 2

+1

21. y

=x

- 4x x2

22. y = - J; 4 23.

a. y

= ✓:i

b. y

~

c. y

= x!

24. y - xex

25. y

Label relative maxima and minima

1

{Note the asymptot formula for (x 4 )' from the definition of the derivative. 32. Simplify or say "cannot be simplified:" 272/3, 27- 4/3, 8 3,(if;i:)12, (x- 3y-6)- t/3, (x6 + y12)1/3 _

33. Find the equation of the line tangent to the graph y x -= 4.

34. Find y'(l) if

Find the absolute extrema: 35. y

1 3 = -x 3

36. y

2x3

-

1 - x2

2

-

2x ➔ 3

15x2 + 6x - 1

for x ~ O; for O ::S x ::; 5;

=N

at

124

I. T he Derivative

37. y == x 3

+ x2 + x + 1

for x ~ O;

.l:3

38. y

= 27 + x4.

Find the derivatives (from an old fi11al ):

w w+ l 39. J(w) = - - + - - ; w- 1 3'I.I) 40. y

=

41. g(x )

j cos Ji;; = xvlux;

Part II

The Integral

15 Area and the Riemann Integral

The ancient Greeks knew the area of a right triangle is ½bh and used a limit argument to deduce that the area of a disc {the area inside a circle) is ½C1· = 7rr 2 >where C is the cirrumference. See Figure 15.1. Using a more complicated argument, they decided that the volume of a ball is ~ 7rr3 . They never knew that the volume of a 4-dimensional ball is ½71"2 r 4 . Integral calculus provides a. systematic method for computing areas and volumes. We begin by asking for the area under the graph of a function> betwec>n x = a and x = b, such as the area under y = x 2 for x between O and 1, as in Figme 15.2. The method is to cut the x-interval from a to b into lit tle subintervals and estimate the area over each subinterval by say a circumscribed rectaugle. The length L\x of the subinterval becomes the width of the rectangle. The height of each rectangle is f (x), where x is the place in the subinterval where f is biggest. See Figure 15.3. The area of each rectangle is f (x) L1'J:. The estimate given by the sum of the areas of the rectangles is the sum of all the f (x) L1x, abhreviated I:, f (x )L1x. The capita.I Greek letter Sigma (I:,) is used to 127

128

[[. The Integral

Figure 15. 1. A d isc is composed of infinitely many infinitesimal triangles of height ,. a n,I lot al lmsc C for a total an•a of ½Cr - 1rr 2 .

dxactly 1/3. Notice that this process involves some formula, here the formula for the sum of the first n squares.

Exercises 15 1. Following the example in the text, compute the area under

y = x for O ~ x ~ l, and check that you get the expected answer 1/ 2. For the Riemann sum you should get

t (~) (!) =-;.

k= l

n

n

n

n (n + 1) _ 2

You will need to use the formula that the sum of the first n · · n(n+l ) ( t l 1e numl)er nof m · t egers tunes · · average mtegers 1s t l1e1r 2

!!±1) 2 .

2. Compute

J24 xd:r. For the Riemann sum you should get

E 2 + -;;-2k) . (;;2) n

(

=

4

4

n (n

+ 1)

- •n +-2 · - - - . n n 2

Check your answer against the formula for the area of a trape• zoid: base times average height. 3. Memorize: We approximate an area by a Riemann sum of areas of n skinny rectangles of height f (x ) and width Llx b~a. The exact area is given by the limit as n ➔ oo, called the Riemann integral.

16 The Fundamental Theorem of Calculus

Areas or definite integrals are hard to compute directly as limits of Riemann sums. The Fundamental Theorem of Calculus tells us that definite integrals rau be easily computed by antidifferentiatin_g. Our work on dm·ivatives comes from out of the blue to save us!

16.1. The Fundamental Theorem of Calculus

t

To compute J(x) dx for any continuou.~ function f on {a,b], take any antidcrivative F off. Then

1b

J(x) dx

= F(b) -

F (t f xx and f xy ( pronounced "fxx," "fxy"). Or you can s tart with fy and take it.s partial derivatives fy~· and fw- For om example:

D (y,2( ' 'r1•) .,

f.u J.,·y

-. V:r

O ( f :r ) = = -Dy

.11:iexy, .

O (lJ 2 e·"Y) = 2yPTY -;Dy

,

+ y 2 .i:eXl/

fyx

fyy

f)

{)

= {)y (fy) = -;((l + :i:y )e ry ) uy

xery + (1 + .i:y).ce"11

= (2:r; + x 2y)c?Y. =

f ry? It is not. It is g 0 for O < x < 3/2 or 3 < x < 4,

f"(x) < 0 for 3/2 < x < 3

11.

=

4:t + 2y 6 i.e., y = -2:t + 3

-2

267

268

Answers to Odd-Numherecl Exercises

(Chavter 11, continued)

13.

15.

I

I

-- - --

Relative maximum

I

- - --r

-

-

. Inflectiim point

i

No absolute maximum or minimum

269

Answers to Odd-Numbered Exercises

17

--------

1-1/>0-J (excluding the endpoints, where 1/ == 0, and the vertical asymptotes, where r/ is undefined)

19. Relative maximum

x=}

1

\

-12 ,

·l

-

Inflection point at x

11

=f

= x 3 - 8x 2 + 5x + 1

Relative minimum z 5

=

270

Answers to Odd-Numbered Exercises

( Chapter 11, continurd} 21.

/

-a

-4

-6

/

/

/

/

-2

/

/ /

-4

/ /

/ /

/

/

/ -8

No maxima or minima 23. ................. _.......

... -- .... -4

-8

--- - - --

,, , •2

'

Minimum for y = x 2l 3 and 11 inflection point for y = 4'x

=../z,

271

Answers to Odd-Numbered Exercises

25.

J 1f,3!')

W!,ct;on poin> at

Maxim~ ( z/= .....-..--......__-'

{iif11

i::s

1.12

Minimum (0, 0)

27.

I

I

Maximum (1,5/4)

I

+ I

_l!o_!i~~nJ~ !8}'.~~t~t~

4

_

~ Inflection point at x

: 2+3s+l

11=~ -1

-2

=2

272

Answcn; to Odd-Numbered Exercises

Chapter 12

l. ix6+ C

½x

2

3. jx4 5.

+3x+ C

1x + C 2

7. x+ C

11.

13. 5111 !.ri+ C 15. 1111k11ow11 17. 1111kuow11 1!). , 2

t

+ l(j

21. i;iu t +t

2:J. J(t) - - 16t2

25. a. h -

+ 15

16t2 + 40t + 200 (anticliffcrentiate a -

h. h' - 0 => t

c. h

= 5/4 =>

32 twice)

h - 225

= 0 ==> t == 5

Chapter 13 1. differentiable• except at multiples of 1/2; continuous cvrywhcre

3. differentiable except at 0, 1, 2, 3; continuous everywhere 5. differentiable except at oO

f(x)

= lim4.r-->0 (x+4lr - x·'

x·• + 4x 34 x+6x 2(4 x) 2+ 4x(4 x)3+(4 x) 4 -x4

2x

3

- lim4,r➔o (4x

+ 6x

2

Llx + 4x(L1x) 2

+ (L1x)3 ) = 4x3

276

Answers to Odd-Nurnbered Exercises

"''3. y I -

'-

3

2X:!

Line 11 =- 20x + b

y'(4) - ~ · 8 - 20.

= 4, y = 4 ! - 32; 32 20 -4 = b => b = 32 80 = - 48 Line y = 20x 48 Extreme cases: x = 0 => y = 3 When x

35.

➔

As x

...c

+oo, y ➔ +oo

Critical cases: 0 = y' = y

=~-

2- 4+3

:1: ➔

- oo, y

:i:2 -

=! - 3 =

x - 2 = (x - 2)(:i: + I ) ⇒ ;r = 2 (- 1 :j: U) 8 9 ;

= - ½ MIN

= O =} y = I

37. Extreme cases: x

As

NO MAX

>-

oo

MAX

NO MIN

Critical cases: 0 =- y' - 3.r2 + 2x + 1, no solution I

I

(w- t):! -

41. 4:$.

~

v1lix + 1/ 2 ~ S( cos x sin x + :i; 2 - !• 2

3. 6 5. 109

7. -::l8 1 9. 396.8 11. 2(/3

/2)

Answers

Lo

Odd-Numbered Exercises

13. 2(e~

ei)

15 . ./2/ 4 17. - c:osx+ C

½cos(2J· -

19.

f )+

C

Chapter 17 34 1· 3

3.. 567 (exactly)

7. 2 - ln4

9. 2./2 - 2

J01 (cosx -

(-

sin :z:) dx + J} (siux - c:osx) dx )

•

11. (4./2 - 2)/'J (= J/'!(c:os'Jx

+ J5t (cos3x 12 13.

½e5 -

sin 3x) dx

+ f n~ (sin 3x - cos 3x) d:c

sin'Jx) dx)

5e! - fe 25

+ 5e

15. 3/2 17. -4 Chapter 18

l.

f (5x + 3)99 dx = ¼f (5x + 3)99(5dx) = ½(5xio3J100 + C = 5tio(5x + 3)100 +

C

3. !,j e'1x- t+ C 5.

- ½cos{3x + 7r)+

C

7. i~o(5x2 + 3)100+ C 9. /0 siu( 4x4

-

3)+ C

278

Answers to Odd- Numbered Exercises

13. ~ sin3 x+ C

17.

I

3

In ( x·'I

25. ,~(.r 2

Wti = lu 3' + I )] x=0

+ 3)6 + C

C hapter 19

1. 1r/G Since you see J 4 - :i:2 , use x when .c = 1, sin0 = 1/ 2, 0 = 1r/ G.

= 2sin0. When x = 0, 0 - 0:

J 4 - x 2 becomes j 4 -(2sin0)2 - / 4

= ✓4 /1 -

fxI= 0

siu 2 0

= 2cos0, d:c becomes 2cos0d0, and

dx _ r"/ 6 2cos 0d0 _ r:r/ 61. d(J V•l - x :! - J0= 0 2cos0 - J0= 0

/6 = ()10=0 = 1