Mathematics for Physicists Solutions to problems [1 ed.]

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SOLUTIONS TO PROBLEMS 1 1.1

(a) To two decimal places p = 124.68 (rounded up to the next even multiple); (b) To two significant figures q = 400 ; (c) To one significant figure r = 0.04 (the leading zeros are ignored).

1.2

1.3

(a) The prime factorisation in index form is 756 = 22337 ; (b) For a given integer m, the best choice of n is given by 0.57m rounded to the nearest integer. So we can construct the following table:

m=5

0.57m = 2.85

n=3

3 5 = 0.600

m=6

0.57m = 3.42

n=3

3 6 = 0.500

m=7

0.57m = 3.99

n=4

4 7 = 0.571

In scientific notation, (a) p = 1.246×103 to four significant figures; (b) (−1.2×10−2 )×(1.31×105 ) = (−1.2×1.31)×105−2 , which to two decimal places is

1.57 ×103 ; (c) 5×10−4 + 4 ×10−3 = 0.5×10−3 + 4 ×10−3 = 4.5×10−3. *1.4

We need to prove that this is not true, so that

m is irrational unless it is an integer. We start by assuming m is rational and we can write

m = p q , where p and q

have no common factors, and q ≠ 1 to rule out the case that

m is an integer.

Then, (a) p 2 and q 2 can have no common factors since p and q have no common factors, and (b) since p 2 = mq 2 , where m is an integer and q ≠ 1 , p 2 and q 2 have a common factor q. Statements (a) and (b) are contradictory, so the assumption that

m is a rational number when it is not an integer, is false. 1.5

(a)

123/49−1/361/2 = (2×2×3)3/4 (3×3)−1/3(3×2)1/2 = 23/4+3/4+1/233/4−1/3+1/2 = 25/2311/12, (b) 31/252 25

−1/5

32/3 = 31/2−2/352+2/5 = 3−1/6512/5 ,

(c) 493/27−1/2 141/2 7−1/3 = 7 37−1/2 71/221/27−1/3 = 77/32−1/2 .

1.6

(a)

(b)

3 − 51/2 3− 5 (3 − 5)(1 − 5 5) 8 5 −14 , = = = 1/2 31 1 + 125 1 + 5 5 (1 + 5 5)(1 − 5 5) 1 2 −( 3 − 5)

=

1 2 −( 3 − 5)

×

2 + ( 3 − 5) 2 + ( 3 − 5)

S1.1

=

−(2 + 3 − 5) 4 − 2 15

,

then rationalise a second time to give

−(2 + 3 − 5) (4 − 2 15) *1.7

=

−(2 + 3 − 5) (4 + 2 15) 4 − 3 3 + 5 + 2 15 × = . 22 (4 − 2 15) (4 + 2 15)

(a) We have,

10 n=3

n

2 R

= =

8 2

n =1

2n

=

2

R

=

0

so 10 = 10102 in base-2. Similarly,

0.31 n

n = −2 2 R

= 0.25 = 0.06

n = −4 2n R

≈ 0.06 ≈ 0

so 0.31 ≈ 01012 in base-2, and finally 10.31 = 1010.01012 … in base-2. (b) We have,

11012 = 23 + 22 + 20 = 13 in base-10 and

0.012 = 2−2 = 0.25 in base-10. So 1101.012 = 13.25 in decimals. *1.8

Firstly we find p and q in the decimal system. Thus,

p = 2014 = (2× 4 2 ) + (1× 4 0 ) = 33 and q = 1304 = (1× 4 2 ) + (3× 41 ) = 28 . (a) In decimals, p ×q = 924 and in base-4

p ×q =

201 130 20100 12030 32130

which in decimals is

S1.2

321304 = (3×4 4 ) + (2×4 3 ) + (1×4 2 ) + (3×4) = 768 + 128 + 16 + 12 = 924. (b) In decimals (p −q) = 5 and in base-4,

p −q = 201 130 11 and in decimals 114 = (1× 41 ) + (1× 4 0 ) = 5.

1.9

(a)

(b)

(c)

1.10

1.11

x x3 1 = x 1/2x 3/2x −5 = 3 , 5 x x 1+ 3 x 3+ x

=

(1 + 3 x )(3 − x ) (3 + x )(3 − x )

=

3 + 8 x − 3x , 9−x

3a(3a)−5/3a 3/2 = 31/23−5/332 a 1/2a 3/2a −5/3a −2 = 35/6 a −5/3 . 2 (a 3)

(

)(

)

(a) lhs =

1 1 2x + 1 + = 2 = rhs for all x ⇒ identity , x −2 x +3 x +x −6

(b) lhs =

1 1 2x + 5 + = 2 ≠ rhs for all x ⇒ equation . x + 2 x + 3 x + 5x + 6

(a) For x > 1.5 , the condition requires 2x − 3 < 5 , i.e. x < 4 . For x < 1.5 , it requires 3 − 2x < 5 , i.e. x > −1 . So −1 < x < 4 . (b) For x 2 > 10 , we must have x 2 −10 < 6 , i.e. x 2 < 16. For x 2 < 10 , we must have 10 − x 2 < 6 , i.e. x 2 > 4 . So −4 < x < −2 or 2 < x < 4. (c) We cannot multiply both sides by (x + 2) because this may be negative. But we can multiply by (x + 2)2 > 0 . Then

x(x + 2) < 2(x + 2)2 ⇒ x 2 + 6x + 8 = (x + 2)(x + 4) > 0 ⇒ x > −2 or x < −4. 1.12

The binomial expansion of (1 − x)5 is

(1 − x)5 = 1 − 5x +

5×4 2 5×4 ×3 3 x − x + − x 5 . 2! 3!

S1.3

Setting x = 0.002 gives

(1.996)5 = 25(1 − 0.002)5 ⎡ ⎤ 5×4 5×4 ×3 = 25 ⎢1 − 5(0.002) + (0.002)2 − (0.002)3 +⎥ ⎢ ⎥ 2! 3! ⎣ ⎦ = 32[1 − 0.01 + 0.00004 +] = 31.681 to 5 significant figures. 1.13

The binomial expansion is 14 ⎛ ⎞ (3a −b)14 = ∑ ⎜⎜⎜ 14 ⎟⎟⎟(3a)14−n (−b)n . ⎝ n ⎟⎠ n=0 ⎜

The term in a 3 therefore has 14 − n = 3 , i.e. n = 11 , and so

⎛ 14 ⎞⎟ 14! 3 3 11 3 11 3 11 ⎜⎜ ⎟ ⎜⎜⎝ 11 ⎟⎟⎠(3a) (−b) = − 11!3! 3 a b = −9828a b . 1.14

The binomial expansion is n ⎛ ⎞ (x + y)n = ∑ ⎜⎜⎜ n ⎟⎟⎟ x n−ky k . ⎝ k ⎟⎠ k=0 ⎜

Setting x = 1,y = 1 gives n ⎛ ⎞ 2n = ∑ ⎜⎜⎜ n ⎟⎟⎟ ⎝ k ⎟⎠ k=0 ⎜

as required, and setting x = 1,y = 2 gives

⎛ n ⎞⎟ k ⎟⎟ 2 = (1 + 2)n = 3n . k ⎟⎠ k=0 n

∑ ⎜⎜⎜⎜⎝ 1.15

50 + 7 = 12 , so 5n + 7 is divisible by 4 for n = 0 . To generalise to all n ≥ 0 , we just have to show that if 5n + 7 is divisible by 4 for any given n, then 5m + 7 is divisible by 4, where m = n + 1 . To see this, we write

5m + 7 = 5×5n + 7 = 4 ×5n + (5n + 7) which is divisible by 4 since both terms are divisible by 4. 1.16

One easily verifies that the equation is satisfied for n = 1 . This can be generalised to all n ≥ 1 by induction. i.e. by showing that if it is true for any given n ≥ 1 , it is true for m = n + 1 . To do this, we write

S1.4

m

n

k=1

k=1

∑ k 2 =∑ k 2 + (n + 1)2 =

n(n + 1)(2n + 1) (n + 1) ⎡⎢ 2n 2 + 7n + 6 ⎤⎥ + (n + 1)2 = ⎥ 6 6 ⎢⎢⎣ 6 ⎥⎦

=

(n + 1)(n + 2)(2n + 3) m(m + 1)(2m + 1) = , 6 6

as required. 1.17

(a) Squaring both sides and cross multiplying gives

y 2(x 2 − 2) = x 2 − 6 and hence x =

(b) y =

2(y − 3) , y 2 −1

(x −1)(x − 2) (x −1) 2y + 1 = , so after cross multiplying this gives x = . (x + 2)(x − 2) (x + 2) 1 −y 1/3

⎛(x + 2)(x − 2)⎞⎟ ⎟⎟ (c) y = ⎜⎜⎜ ⎟⎠ ⎜⎝ x +2 2

= (x − 2)1/3 , so cubing both sides and cross multiplying

gives x = y 6 + 2 . 1.18

We have

⎛ 4 − x ⎞⎟ ⎛ 4 + x ⎞⎟ ⎟⎟ and f (−x) = −⎜⎜ ⎟ f (x) = ⎜⎜⎜ ⎜⎜⎝ x + 3 ⎟⎟⎠ , ⎜⎝ x − 3 ⎟⎠ and

fS (x) = 12 [f (x) + f (−x)] and fA(x) = 12 [f (x)− f (−x)] . So,

fS (x) =

1 ⎡⎢ 4 − x 4 + x ⎤⎥ (x 2 −12) − =− 2 , 2 ⎢⎣ x − 3 x + 3 ⎥⎦ (x − 9)

and

fA(x) =

1 ⎡⎢ 4 − x 4 + x ⎤⎥ x . + = 2 ⎢ ⎥ 2 ⎣ x − 3 x + 3 ⎦ (x − 9)

To check,

(x 2 − x −12) (x 2 − 9) (x + 3)(x − 4) 4 − x =− = . (x − 3)(x + 3) x − 3

f (x) = fs (x) + fA(x) = −

1.19

(a) Setting y = f (x) = (x − 2) (x + 2) , gives x = 2(y + 1) (1 −y) , and hence the inverse function is

f −1(x) =

2(x + 1) , (1 − x)

S1.5

(x ≠ 1) .

(b) Setting y = f (x) = 13 (x + 4)1/3 , gives x = 27y 3 − 4 and hence f −1(x) = 27x 3 − 4 . 1.20

(a) If the straight line y = mx +c passes through the points (x,y) = (1, 2) and (4,−1) , this implies that 2 = m +c and −1 = 4m +c . Solving these gives m = −1, c = 3 and hence the equation is y = −x + 3 . (b) If the straight line y = mx +c passes through the point (2,–1) with slope 2 3 , this implies y = 2x 3 +c with −1 = 4 3 +c , i.e. c = −7 3 , so the equation is 3y = 2x − 7. (c) If the straight line y = mx +c is parallel to the line 2y + 5x = 7 , this implies

2y + 5x = c , and if it passes through the point (1, 4) , then 2y + 5x = 13 = c , so the equation is 2y + 5x = 13 . (d) Two perpendicular lines with slopes m1 and m2 have m1m2 = −1 , so the slope of the line is −2 5 , i.e. its equation is y = −2x 5 +c , or 5y + 2x = 5c . Since it passes through the point (1, 4) , 5c = 22 , i.e. 5y + 2x = 22. 1.21

The line y = 3x + 6 has slope 6 and so the slope of the perpendicular line to this has slope −1 3 . Since it passes through the point (2, 2) , its equation is 3y + x = 8 and this meets the line y = 3x + 6 at the point (−1,−3) . Finally, the distance between

(2, 2) and (−1,−3) is 1.22

(2 + 1)2 + (2 + 3)2 = 34 .

First we find the vertices of the triangle ABC. A is the intersection of 5y = 3x + 2 and y = −x + 2 , i.e. A = (1,1) . Likewise, B is the intersection of 5y = 3x + 2 and

y = 3x −14 , i.e. B = (6, 4) and C is the intersection of y = 3x −14 and y = −x + 2 , i.e. C = (4,−2) . The triangle is shown in the figure below.

S1.6

The area of the triangle ABC is

1 2

bh , where b is the base, taken to be the line BC

and h is the height shown on the figure. Then

b = BC = (6 − 4)2 + (4 + 2)2 = 40 To find h we need to find the length of the perpendicular from A to the line BC, i.e. y = 3x −14 . From the condition m1m2 = −1 , the perpendicular line must be of the form y = −x 3 +c and since it passes through A = (1,1) , it must be y = −x 3 + 4 3 . It meets the line BC when 3x −14 = −x 3 + 4 3 , i.e. when x = 23 5 and hence

y = −1 5 . So the length h is given by

h = (1 − 23 5)2 + (1 + 1 5)2 =

1 5

360 .

Finally,

area = 12 × 15 40 × 360 = 12 . 1.23

The equation of a circle with centre (x,y) = (a,b) and radius r is

(x −a)2 + (y −b)2 = r 2 . (a) The equation is (x −1)2 + (y − 3)2 = 4 (b) The radius is now the distance from (1, 3) to (−2,5) , i.e.

r 2 = (1 + 2)2 + (3 − 5)2 = 13 , so the equation is (x −1)2 + (y − 3)2 = 13 .

S1.7

1.24

Grouping the terms in x and y, we can write

(x 2 −ax) = (x −a 2)2 −a 2 4 and

(y 2 −bx) = (x −b 2)2 −b 2 4 , so that

x 2 + y 2 −ax −by +c 2 = (x −a 2)2 + (y −b 2)2 −a 2 4 −b 2 4 +c 2 = 0, that is,

(x −a 2)2 + (y −b 2)2 = (a 2 +b 2 − 4c 2 ) 4 , which is the equation of a circle centred at (a 2, b 2) and with radius

r=

a 2 +b 2 − 4c 2 , 4

provided a 2 +b 2 − 4c 2 > 0 . For x 2 + y 2 − 2x − 3y + 1 = 0 , we have a = 2, b = 3, c = 1 , so the circle is centred at (1, 3 2) , with radius r = 3 2 . 1.25

The functions y = f (x) = 2.5 − 0.5x and y = f (x) = x 2 − 5x + 6 figure below. The hatched area corresponds to the region where

(2.5 − 0.5x) ≥ y(x) ≥ x 2 − 5x + 6 .

S1.8

are shown in the

SOLUTIONS TO PROBLEMS 2 2.1

From (2.8),

⎛1 1⎞ ⎛ 1 ⎞ x 2 − ⎜⎜⎜ + ⎟⎟⎟ x + ⎜⎜⎜ ⎟⎟⎟ = 0 , i.e. αβx 2 −(α + β)x + 1 = 0 . ⎜⎝ α β ⎟⎠ ⎜⎝ αβ ⎟⎠ But α + β = 2 and αβ = −3 , so the required equation is 3x 2 + 2x −1 = 0 . 2.2

Rearranging the expression for p, we have the equation

5x 2 − 3px + (p + 10) = 0 . Then using the general formula for the solution of a quadratic equation, (2.6b), real roots are only possible if 9p 2 ≥ 20(p + 10) . 2.3

We need to solve the equation y = mx +c simultaneously with the equation for the circle. Substituting for y gives, after rearrangement,

(1 + m 2 )x 2 + 2mcx + (c 2 −r 2 ) = 0 , which is a quadratic of the form αx 2 + βx + γ = 0 . This will have a single real root if

β 2 = 4αγ , i.e. 1/2

2 2

2

2

2

4m c = 4(1 + m )(c −r ),

⎛c 2 −r 2 ⎞⎟ ⎟⎟ . or m = ± ⎜⎜⎜ ⎜⎝ r 2 ⎟⎠

Thus there are two tangents as shown in the figure below.

2.4

The equations of the two circles are

(x −1)2 + (y + 1)2 = 4,

i.e. x 2 + y 2 − 2x + 2y − 2 = 0 ,

S2.1

(1)

and

x2 + y2 = 4 ,

(2)

respectively. Subtracting (2) from (1) gives

−2x + 2y = −2,

i.e. y = x −1

(3)

and substituting in (2) gives 2x 2 − 2x − 3 = 0 , with solutions

x = (1 + 7) 2

and

x = (1 − 7) 2

respectively. From (3) the corresponding values of y are

y = (−1 + 7) 2

and

y = (−1 − 7) 2 .

Hence the co-ordinates of the points of intersection are

(x,y) =

(

1 2

+

7 2

The length of the chord is the

cosine

rule

(2.40a),

,− 12 +

7 2

)

and

(

1 2

−

7 2

,− 12 −

7 2

).

14 and the radius of the circle is 2 in each case. Hence with

a = 14, b = c = 2 ,

gives

cosA = −6 8

and



A = 2.42 rad = 139 . 2.5

(a) We have

(x 3 + x 2 − x − 4) = (x −1)(ax 2 +bx +c) + R(x) . Setting x = 1 gives R = −3 ; and multiplying out the bracket and equating powers of x gives a = 1, b = 2, c = 1 , so that

(x 3 + x 2 − x − 4) = (x −1)(x 2 + 2x + 1)− 3 . (b) By long division,

3x 2 + 5x + 5 (x 2 − 2x + 3) 3x 4 − x 3 + 4x 2 + 5x + 15 3x 4 −6x 3 + 9x 2 5x 3 − 5x 2 + 5x + 15 5x 3 −10x 2 + 15x 5x 2 − 10x + 15 5x 2 −10x + 15 0 so that

3x 4 − x 3 + 4x 2 + 5x + 15 = (x 2 − 2x + 3)(3x 2 + 5x + 5) ,

S2.2

with the remainder R(x) = 0 . Since both (x 2 − 2x + 3) and (3x 2 + 5x + 5) are of the form ax 2 +bx +c with b 2 < 4ac , both of them, and hence the quartic f (x) itself, have no real roots. 2.6

By inspection, one finds that x = 1 and x = 2 are roots. Hence, by the factor theorem,

x 4 − 2x 3 − 2x 2 + 5x − 2 = (x −1)(x − 2)(ax 2 +bx +c) . By comparing powers of x 4 on both sides one finds a = 1 , and by comparing the constant term, c = −1 . Hence

x 4 − 2x 3 − 2x 2 + 5x − 2 = (x −1)(x − 2)(x 2 +bx −1) . Also, comparing powers of x 3 gives b = 1 and hence

x 4 − 2x 3 − 2x 2 + 5x − 2 = (x −1)(x − 2)(x 2 + x −1) . The roots of (x 2 + x −1) are (−1 ± 5) 2 , and so finally the four roots are

x = 1, 2, (−1 + 5) 2, (−1 − 5) 2 . 2.7

Using the bisection method gives,

n

xn

f(xn )

1

x1 =

1.500000000 −0.12500000

2

x2 =

1.600000000

0.37600000

3

x 3 = 12 (x1 + x 2 ) =

1.550000000

0.11888000

4

x 4 = (x1 + x 3 ) =

1.525000000 –0.00467200

5

x 5 = 12 (x 3 + x 4 ) =

1.537500000

0.05669000

6

x 6 = 12 (x 4 + x 5 ) =

1.531250000

0.02591100

7

x 7 = 12 (x 4 + x 6 ) =

1.528125000

0.01059000

8

x 8 = 12 (x 4 + x 7 ) =

1.526562500

0.00295500

9

x 9 = (x 4 + x 8 ) =

1.525781250 −0.00086020

1 2

1 2

10 x10 = 12 (x 8 + x 9 ) =

1.526171875

0.00104700

11 x11 = (x 9 + x10 ) = 1.525976563

0.00009315

1 2

Thus the root correct to three decimal places is 1.526. 2.8

(a) Setting

2(x 2 − 9x + 11) A B C = + + , (x − 2)(x − 3)(x + 4) (x − 2) (x − 3) (x + 4)

S2.3

gives

2(x 2 − 9x + 11) = A(x − 3)(x + 4) + B(x − 2)(x + 4) +C(x − 2)(x − 3) and using x = 2 , x = 3 and x = −4 in turn, yields A = 1 , B = −2 and C = 3 , so that

2(x 2 − 9x + 11) 1 2 3 = − + . (x − 2)(x − 3)(x + 4) (x − 2) (x − 3) (x + 4) (b) Setting

7x 2 + 6x −13 Ax + B C , = 2 + 2 (2x + 1)(x + 2x − 4) x + 2x − 4 2x + 1 gives

7x 2 + 6x −13 = (Ax + B)(2x + 1) +C(x 2 + 2x − 4) and equating coefficients of powers of x yields A = 2, B = −1 and C = 3 , so that

7x 2 + 6x −13 2x −1 3 . = 2 + 2 (2x + 1)(x + 2x − 4) x + 2x − 4 2x + 1 (c) Setting

2(3x 2 + 4x + 2) A B C , = + + 2 x −1 2x + 1 (2x + 1)2 (x −1)(2x + 1) gives

2(3x 2 + 4x + 2) = A(2x + 1)2 + B(2x + 1)(x −1) +C(x −1) and equating coefficients of powers of x yields A = 2, B = −1 and C = −1 , so that

2(3x 2 + 4x + 2) 2 1 1 . = − − 2 x −1 2x + 1 (2x + 1)2 (x −1)(2x + 1) 2.9

(a) Performing a long division gives

x 3 − 2x 2 + 10 5x + 4 = (x − 3) + . (x −1)(x + 2) (x −1)(x + 2) Then, setting

5x + 4 A B = + (x −1)(x + 2) x −1 x + 2 gives

A(x + 2) + B(x −1) = 5x + 4 ⇒ A = 3, B = 2 and so

S2.4

x 3 − 2x 2 + 10 3 2 = (x − 3) + + . (x −1)(x + 2) (x −1) (x + 2) (b) Setting

3x 2 − 5x − 4 A Bx +C = + , 2 (x + 2)(3x + x −1) (x + 2) (3x 2 + x −1) gives

3x 2 − 5x − 4 = A(3x 2 + x −1) + (x + 2)(Bx +C ) , and choosing x = −2 , yields A = 2. Then

3x 2 − 5x − 4 = (6 + B)x 2 + (2B +C + 2)x + (2C − 2) , and equating coefficients of powers of x, gives B = −3 and C = −1 . So finally,

3x 2 − 5x − 4 2 3x + 1 = − . 2 (x + 2)(3x + x −1) (x + 2) (3x 2 + x −1) (c) Setting

3x 2 − x + 2 A B C D , = + + + 3 2 (x −1) (x − 3) (x − 3) (x −1)(x − 3) (x − 3)3 gives

3x 2 − x + 2 = A(x − 3)3 + B(x −1)(x − 3)2 +C(x −1)(x − 3) + D(x −1) . Letting x = 1 yields A = −1 2 , and letting x = 3 yields D = 13 . Then equating the coefficients of x 3 gives B = 1 2 and equating the constants on both sides gives C = 2 . So, finally

3x 2 − x + 2 1 1 2 13 . =− + + + 3 2 2(x −1) 2(x − 3) (x −1)(x − 3) (x − 3) (x − 3)3 2.10

(a) Using the double-angle formulas,

cos 4θ ≡ 2cos2 2θ −1 and

cos 2θ ≡ 2cos2 θ −1 ,

so that

cos 4θ ≡ 2(2cos2 θ −1)2 ≡ 8 cos4 θ − 8 cos2 θ + 1 . (b) Using the identities (2.36c) and (2.36d).

sin(nθ) + sin[(n + 4)θ] ≡ 2sin[(n + 2)θ]cos(2θ) and

cos(nθ) + cos[(n + 4)θ] ≡ 2cos[(n + 2)θ]cos(2θ) ,

S2.5

in the left-hand side of the identity gives

sin[(n + 2)θ][1 + 2cos(2θ)] ≡ tan[(n + 2)θ] . cos[(n + 2)θ][1 + 2cos(2θ)] (c) 2

2

⎛ sin 5θ ⎞⎟ ⎛ cos5θ ⎞⎟ ⎛ sin 5θ cos5θ ⎞⎟⎛ sin 5θ cos5θ ⎞⎟ ⎜⎜ ⎟⎟ − ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ − + ⎜⎜⎝ sin θ ⎟⎠ ⎜⎜⎝ cos θ ⎟⎠ ⎜⎜⎝ sin θ cos θ ⎟⎠⎜⎜⎝ sin θ cos θ ⎟⎠ ⎛ sin 5θ cos θ − cos5θ sin θ ⎞⎟⎛ sin 5θ cos θ + cos5θ sin θ ⎞⎟ ⎟⎟⎜⎜ ⎟⎟ = ⎜⎜⎜ ⎟⎠⎜⎜⎝ ⎟⎠ ⎜⎝ sin θ cos θ sin θ cos θ =

4 sin 4θ sin 6θ . sin 2θ sin 2θ

Then from (2.36c), sin 6θ may be written

sin 6θ = sin 4θ cos 2θ + cos 4θ sin 2θ , and using the double-angle formulas (2.37a) on the right-hand side,

sin 6θ = 3sin 2θ − 4 sin 3 2θ , so that

4 sin 4θ sin 6θ 8 sin 2θ cos 2θ(3sin 2θ − 4 sin 3 2θ) = , sin 2θ sin 2θ sin 2 2θ and finally 2

2

⎛ sin 5θ ⎞⎟ ⎛ cos5θ ⎞⎟ ⎜⎜ ⎟ − ⎜⎜ ⎟ =8 cos 2θ(3 − 4 sin 2 2θ) = 8 cos 2θ(4 cos2 2θ −1). ⎜⎜⎝ sin θ ⎟⎟⎠ ⎜⎜⎝ cos θ ⎟⎟⎠ 2.11

(a) Using the double-angle formula (2.37a), we have

2cos θ cos 2θ + 2sin θ cos θ = 2cos θ(3cos2 θ −1) , so the first solution is cos θ = 0 , i.e. θ = π2 ,

3π 2

. If cos θ ≠ 0 , we can divide by 2cos θ

to give

cos 2θ + sin θ = 3cos2 θ −1 , which again using the double-angle formula and writing everything in terms of sin θ , reduces to the quadratic sin 2 θ + sin θ −1 = 0 which leads directly to solutions θ = 0.666 and 2.475 radians. (b) Combining the first and third terms using (2.36) gives

2sin 2θ cos θ − 2sin θ cos θ = 0 ,

S2.6

so the first solution is

cos θ = 0 ⇒ θ = π2 ,

3π 2

.

If cos θ ≠ 0 , then dividing gives

sin θ(2cos θ −1) = 0 . The two possibilities are

sin θ = 0 ⇒ θ = π , or cos θ = 12 ⇒ θ = π3 ,

5π 3

.

So finally

θ = π3 , 2.12

π 2

, π,

3π 2

and

5π 3

.

We have

sin kθ − sin θ ≡ 2sin[ 12 (k −1)θ]cos[ 12 (k + 1)θ] . Thus one of the two brackets must be zero. The first bracket is zero if 1 2

(k −1)θ = ±nπ, i.e. kθ = ±nπ + θ ⇒ θ = ± 2nπ (k −1) ,

for all integer n and k ≠ 1 . The second bracket is zero if 1 2

(k + 1)θ = ± 12 (2n + 1)π, i.e. kθ = ±(2n + 1)π − θ ⇒ θ = ±(2n + 1)π (k + 1) ,

for all integer n and k ≠ −1 . 2.13

From the equation of the straight line x = (p −y cos θ) sin θ and substituting into the equation for the hyperbola gives

p 2 − 2py cos θ + y 2 cos2 θ y 2 − 2 =1, a 2 sin 2 θ b which is a quadratic in y of the form Ay 2 + By +C = 0 , where

⎡ cos2 θ ⎡ p2 ⎤ ⎡ 2p cos θ ⎤ 1⎤ A = ⎢⎢ 2 2 − 2 ⎥⎥ , B = − ⎢ 2 2 ⎥ and C = ⎢⎢ 2 2 −1⎥⎥ . ⎢ a sin θ ⎥ ⎢⎣ a sin θ b ⎥⎦ ⎢⎣ a sin θ ⎥⎦ ⎣ ⎦ If the line is to be a tangent, then there can be only one solution of this quadratic, the condition for which is that B 2 = 4AC , i.e. 2 2 2 ⎡ ⎤⎡ ⎤ ⎡ 2p cos θ ⎤ ⎥, ⎢ ⎥ = 4 ⎢ cos θ − 1 ⎥ ⎢ p −1 ⎢ a 2 sin 2 θ b 2 ⎥ ⎢ a 2 sin 2 θ ⎥ ⎢ a 2 sin 2 θ ⎥ ⎢ ⎥ ⎢ ⎥⎦ ⎣ ⎦ ⎣ ⎦⎣

which after simplifying gives a 2 sin 2 θ −b 2 cos2 θ = p 2 , as required.

S2.7

The y-co-ordinate of the point of intersection is y = −B 2A , i.e.

y=

2p cos θ . a 2b 2 sin 2 θ −b 2 cos θ = p 2a 2 sin 2 θ b 2 cos2 θ −a 2 sin 2 θ

and hence

x= 2.14

p −y cos θ a 2 sin θ = . sin θ p

Equation (2.37c) may be used on the left-hand side to give

1 + sin θ + cos θ 1 + (2sin θ 2cos θ 2) + (2cos2 θ 2 −1) ≡ 1 + sin θ − cos θ 1 + (2sin θ 2cos θ 2)−(1 − 2sin 2 θ 2) ≡ 2.15

cos θ 2 2cos2 θ 2 1 + cos θ ≡ ≡ . sin θ 2 2sin θ 2cos θ 2 sin θ

From the sine rule

a b a sin B . = ⇒ sin A = sin A sin B b So A = arcsin ⎡⎢5sin(0.5) 4 ⎤⎥ = 0.643 rad = 36.84 0 and hence C = 1.999 rad = 114.510 . ⎣ ⎦ Again using the sine rule, c = b sinC sin B = 7.59 cm . 2.16

The lengths of the sides are: a = BC = (5 − 7)2 + (6 − 2)2 = 20

and likewise

b = AC = 37 and c = AB = 25 . Then using the cosine rule

cosA =

giving

b 2 +c 2 −a 2 21 = , 2bc 5 37

A = 0.809 rad = 46.350 . In a similar way, B = 1.391 rad = 79.700

and

0

C = 0.942 rad = 53.97 . 2.17

For n = 0 the results are trivial. So we just need to show that if both are true for any n ≥ 0 they are true for m = n + 1 . Suppose both results are true for n. Then

sin[(2m + 1)θ] = sin[(2n + 1)θ + 2θ] = sin[(2n + 1)θ]cos 2θ + cos[(2n + 1)θ]sin 2θ ⎡ cos[(2n + 1)θ]⎤ ⎥ 2sin θ cos2 θ, = sin[(2n + 1)θ][1 − 2sin 2 θ]+ ⎢ ⎢ ⎥ cos θ ⎣ ⎦ which is a polynomial in sin θ since both sin[(2n + 1)θ] and cos[(2n + 1)θ] cos θ are polynomials and cos2 θ = 1 − sin 2 θ . Similarly,

S2.8

⎡ cos[(2m + 1)θ]⎤ ⎡ cos[(2n + 1)θ + 2θ]⎤ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥ cos θ cos θ ⎣ ⎦ ⎣ ⎦ ⎡ cos[(2n + 1)θ]⎤ ⎥ cos 2θ + sin[(2n + 1)θ]sin 2θ =⎢ ⎢ ⎥ cos θ cos θ ⎣ ⎦ ⎡ cos[(2n + 1)θ]⎤ ⎥ [1 − 2sin 2 θ]+ sin[(2n + 1)θ]2sin θ cos θ , =⎢ ⎢ ⎥ cos θ cos θ ⎣ ⎦

⎡ cos[(2n + 1)θ]⎤ ⎢ ⎥ ⎢ ⎥ cos θ ⎣ ⎦ polynomials. Hence both results are true for all n ≥ 0 . which is a polynomial in

2.18

sin θ

since

and

sin[(2n + 1)θ] are

(a)

log(xy) + 3 log(x y) + 2 log(y x) = log x + logy + 3 log x − 3 logy + 2 logy − 2 log x = 2 log x = log(x 2 ) (b) 6 log x 1/3 + 2 log(1 x) = 2 log x + 2 log1 − 2 log x = 0 2.19

(a) lhs = log(x 2 − 9) = 3 , so x 2 − 9 = 103 and x = 1009 = 31.765 . (b) ln(log x) = −3 , so log x = e −3 ≡ p and x = 10p . Using p = 0.049787068 , gives x = 1.122 .

2.20

(a) Using log x = ln x ln10 , we have

⎛ 1 ⎞⎟ 5 ln10 ⎟⎟ = 5, i.e. ln x = ln x ⎜⎜⎜1 + = 3.486 ⎜⎝ ln10 ⎟⎠ 1 + ln10 and hence x = exp(3.486) = 32.66. (b) We have

9 ln 2 − 9 ln x − ln x = −6, i.e. ln x =

9 10

(ln 2 + 23 ) = 1.224

and hence x = exp(1.224) = 3.401. 2.21 (a) Writing each of the hyperbolic functions in terms of exponentials gives:

sech x =

2 , x e +e −x

cosech x =

2 e x +e −x , coth x = . e x −e −x e x −e −x

Then cosech x > sech x if

S2.9

2 2 > x , or 4e −x > 0 , −x e −e e +e −x x

which is true for x > 0 . Likewise, coth x > cosech x if

e x +e −x 2 > x , or ex +e −x > 2 . x −x −x e −e e −e To prove the latter, consider the expression

ex +e −x − 2 = (e 2x + 1 − 2e x )e −x = (e x −1)2e −x > 0 , since both terms are greater than zero. So ex +e −x > 2 . (b) The right-hand side expressed in terms of exponentials is 3

4cosh 3x − 3cosh x = 4 ⎡⎢ 12 (e x +e −x )⎤⎥ − 3 ⎡⎢ 12 (e x +e −x )⎤⎥ ⎣ ⎦ ⎣ ⎦ 3x x −x −3x 1 = 2 (e + 3e + 3e +e )− 32 (e x +e −x ) = 12 (e 3x +e −3x ) = cosh 3x, where we have used the binomial expansion. 2.22

(a) In terms of exponentials, the equation is

3(e 2x +e −2x ) + 2(e 2x −e −2x ) = 6 , which gives

5e 2x − 6 +e −2x = (5e x −e −x )(e x −e −x ) = 0 . So, either (5e x −e −x ) = 0 ⇒ 5e x = e −x , which taking logs gives x = − 12 ln 5 , or

(e x −e −x ) = 0 , i.e. x = 0. (b) From (2.64),

arctanh x =

1 ⎛⎜1 + x ⎞⎟ ⎟⎟ , ln ⎜ 2 ⎜⎜⎝ 1 − x ⎟⎠

so we need to solve

⎛1 + x ⎞⎟ ⎟ = 2 ln 5 = ln 25 , ln ⎜⎜⎜ ⎜⎝ 1 − x ⎟⎟⎠ i.e. (1 + x) = 25(1 − x) , which gives x = 12 13 . (c) In terms of exponentials, the equation is

S2.10

⎡e ln x +e −ln x ⎤ − ⎡e ln(x/2) −e −ln(x/2) ⎤ − 2c ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎛ 1 ⎞ ⎛x 2 ⎞ = ⎜⎜⎜x + ⎟⎟⎟ − ⎜⎜⎜ − ⎟⎟⎟ − 2c = 0, ⎜⎝ x ⎟⎠ ⎜⎝ 2 x ⎟⎠ or x 2 − 4xc + 6 = 0 , which only has real roots if

16c 2 − 24 ≥ 0 ⇒ c ≥ 3 2 or c ≤ − 3 2 . 2.23

Using (2.66b) we have

cosh 4x = 2cosh 2(2x)−1 and hence

cosh 4x + 4 cosh 2x −125 = 0 may be written

cosh 2 2x + 2cosh 2x − 63 = 0 , i.e.

(cosh 2x + 9)(cosh 2x − 7) = 0 . But since cosh 2x > 0 for all x, the only solution is cosh 2x = 7 . Finally, using the definition (2.57),

x = ± 12 ln(7 + 2 12) = ±1.317 . 2.24

If the straight line is y = mx +c where m and c are constants, then at P1 ,

2at1 = mat12 +c and at P2 ,

2at2 = mat22 +c . Solving these equations gives

m=

2 t1 + t2

and

c = 2at1 −

and hence, after simplifying,

y=

2 (x + at1t2 ) . t1 + t2

But the line also passes through the focus (a, 0) , so

0=

2a (1 + t1t2 ) t1 + t2

and hence t1t2 = −1 .

S2.11

2at1 t1 + t2

2.25

The equation of the tangent is that of a straight line,

y = m1x +c1

(1)

which must pass through (x,y) = (1, 2) , so that

m1 +c1 = 2 .

(2)

In addition, it must be compatible with the equation of the parabola at the single point (1, 2) and no other. Substituting (1) into y 2 = 4x gives

m12x 2 + (2m1c1 − 4)x +c12 = 0, which is of the form ax 2 +bx +c = 0 , and has only a single real solution if b 2 = 4ac , i.e. if (2m1c1 − 4)2 = 4m12c12 , which reduces to

16(1 − m1c1 ) = 0, so that m1c1 = 1 . Together with (2), this requires m1 = c1 = 1 , so the equation of the tangent is y = x + 1. Since the normal y = m2x +c2 is at right angles to the tangent,

m1m2 = −1 , so that m2 = −1 ; and since it must pass through (x,y) = (1, 2) , then c2 = 3 . Hence the equation of the normal is

y = −x + 3.

S2.12

SOLUTIONS TO PROBLEMS 3 3.1

(a) (−1)2 = 1 , (b) 02 = 0 , (c) (2)2 = 4 .

3.2

(a)

(x + 5)2 − 25 x 2 + 10x = = x + 10 → 10 as x → 0 x x

(b) The bracket may be written

1 + cos(πx) cos2(πx) ⎡ = 1 + cos(πx)⎤⎥ ⎦ tan 2(πx) sin 2(πx) ⎢⎣ 2 cos2(πx) ⎡⎢1 + cos(πx)⎤⎥ ⎣ ⎦ = cos (πx) . = ⎡1 − cos2(πx)⎤ 1 − cos(πx) ⎢⎣ ⎥⎦ Then, since

lim cos(πx) = −1, x→1

it follows that

⎡ 1 + cos(πx) ⎤ 1 ⎥= . lim ⎢⎢ 2 ⎥ 2 x→1 tan (πx) ⎣ ⎦ (c) Let y = arcsin x , so that x = sin y . Then

⎛ arcsin x ⎞⎟ ⎛ y ⎞⎟ ⎟⎟ = lim ⎜⎜ ⎟⎟. lim ⎜⎜⎜ x→0 ⎜ ⎝ x ⎟⎠ y→0 ⎜⎜⎝ sin y ⎟⎠ But, from (3.15)

⎛ sin y ⎞⎟ ⎟⎟ = 1, lim ⎜⎜⎜ y→0 ⎜ y ⎟ ⎝ ⎠ 3.3

⎛ arcsin x ⎞⎟ ⎟⎟ = 1. so lim ⎜⎜⎜ x→0 ⎜ ⎝ x ⎟⎠

We have to show that given any ε > 0 , we can find a δ > 0 such that x 2 − 4 < ε when 0 < x − 2 < δ . Choose δ ≤ 1 , so that 0 < x − 2 < 1 . Then

x 2 − 4 = (x − 2)(x + 2) = (x − 2) (x + 2) = (x − 2) (x − 2) + 4 ≤ x − 2 ⎡⎢ (x − 2) + 4 ⎤⎥ < 5δ. ⎣ ⎦ Now take δ as the smaller of 1 or ε 5 . Then x 2 − 4 < ε whenever x − 2 < δ , which completes the proof. 3.4

(a) f (x) → −1, + 1 as x → 0−, 0+ , respectively. Hence this is a non-removable step discontinuity.

S3.1

(b) Since

x 3 − 3x 2 + 3x − 2 x 3 − 3x 2 + 3x − 2 = , (x + 5)(x − 2) x 2 + 3x −10 there are discontinuities at x = 2 and –5. At x = 2 , the numerator is zero by inspection and so

x 3 − 3x 2 + 3x − 2 = (x − 2)(ax 2 +bx +c) by the factor theorem and on equating coefficients,

x 3 − 3x 2 + 3x − 2 = (x − 2)(x 2 − x −1) . Hence

f (x) =

(x − 2)(x 2 − x −1) , (x + 5)(x − 2)

and the discontinuity at x = 2 , where f (2) is ill-defined, can be removed by setting

f (x = 2) =

x 2 − x −1 1 = . x + 5 x=2 7

At x = −5 , the discontinuity is non-removable because x 3 − 3x 2 + 3x − 2 ≠ 0 as x → −5 , so that the limits as x → 5−, 5+ diverge. 3.5

(a) There is a discontinuity at x = 0 where f (0) = 0 0 is ill-defined. Since f (x) → 0 as x → 0−, 0+ , it can be removed by defining f (0) ≡ 0 . (b) There are discontinuities at x = ±3 , where the denominator vanishes. For x ≠ ±3 we can write

f (x) =

x +3

4 + x2 + 7

4 − x2 + 7 4 + x2 + 7

=

4 + x2 + 7 . 3−x

For x → +3 , f (x) diverges, so the discontinuity is non-removable. For f (x) → −3 ,

f (x) → −(4 + 2) 6 and the discontinuity can be removed by defining f (−3) ≡ −(4 + 2) 6 . 3.6

(a) Continuity requires that

lim f (x) = lim− f (x)

x→0+

x→0

and hence A = 1 . (b) Differentiability requires continuity, i.e. A = 1 , plus

S3.2

lim f ′(x) = lim− f ′(x) ,

x→0+

x→0

i.e. nBx n−1 = 0 at x = 0 . This is satisfied for B = 0 (all n), or for n ≥ 2 (all B). 3.7

(a) f (x) = 2x 3 + 4x + 3 , so that

⎡ 2(x + δx)3 + 4(x + δx) + 3 − 2x 3 − 4x − 3 ⎤ df ⎥ = lim ⎢⎢ ⎥ dx δx→0 ⎢⎣ δx ⎥⎦ 2 2 ⎡ 6x δx + 4δx +O(δx ) ⎤ ⎥ = 6x 2 + 4. = lim ⎢⎢ ⎥ δx→0 δx ⎢⎣ ⎥⎦ (b) f (x) = 1 x 2 , so that

δf (x) =

1 1 −2xδx − δx 2 − = , (x + δx)2 x 2 x 2(x + δx)2

and

df (x) δf 2 = lim =− 3. δx→0 dx δx x (c) From (3.10), and using (2.36e), we have

δf = 5 ⎡⎢cos(3x + 3δx)− cos 3x ⎤⎥ = −10 ⎡⎢sin(3x + 3δx 2)sin(3δx 2)⎤⎥ . ⎣ ⎦ ⎣ ⎦ Hence, using (3.15),

⎡ sin(3δx 2) ⎤ df δf ⎥ = lim = −10 sin 3x lim ⎢ ⎥ δx→0 ⎢ dx δx→0 δx δx ⎣ ⎦ ⎡ sin(3δx 2) ⎤ ⎥ = −15sin 3x. = −15sin 3x lim ⎢ ⎥ δx→0 ⎢ 3δx 2 ⎣ ⎦ 3.8

(a) By the product rule,

(b) By the quotient rule,

d 3 x (x e ) = x 3e x + 3x 2e x . dx

d ⎛⎜ sinh x ⎞⎟ x cosh x − sinh x ⎟⎟ = . ⎜ dx ⎜⎜⎝ x ⎟⎠ x2

(c) y = arccosx , therefore x = cosy and dx dy = −sin y , so that

dy 1 1 1 . =− =− =− dx sin y 1 − cos2 y 1−x2 (d) y = arcsinh x , therefore x = sinh y , dx dy = cosh y , and using (2.58e),

S3.3

dy 1 1 1 . = = = 2 dx cosh y 1 + sinh y 1+ x2 (e) y = e z where z = 2x 3 , dz dx = 6x 2 . Then, using the chain rule, 3 dy dy dz = = 6x 2e 2x . dx dz dx

(f) y = 3 ln z where z = 1 + x 2 , dz dx = 2x . Then, by the chain rule,

dy dy dz 3 6x = = 2x = . dx dz dx z 1+ x2 3.9

(a) Let f (z) = sin z and z = ln x . Then by the chain rule,

df df dz cosz cos(ln x) = = = . dx dz dx x x (b) By the quotient rule,

df (1 − x 2 )[−x sin x + cosx ]− x cosx[−2x ] = dx (1 − x 2 )2 =

(1 + x 2 )cosx − x(1 − x 2 )sin x . (1 − x 2 )2

(c) Let f (z) = arctan z and z(x) = 1 + x 2 . Then by the chain rule,

df df dz 1 dz = = , dx dz dx dz df dx where dz dx = 2x . To find dz df , we use z = tan f . So

dz = sec2 f = 1 + tan 2 f = 1 + z 2 df and

df 2x . = dx 1 + (1 + x 2 )2

(d) Let Let f (z) = ln z and z = ln x . Then by the chain rule,

df df dz 11 1 = = = . dx dz dx z x x ln x

S3.4

3.10

(a) y = x x ⇒ ln y = x ln x . So,

1 dy = ln x + 1 y dx

dy = x x (ln x + 1). dx

and

(b) y = x cosx ⇒ ln y = cosx ln x . So,

1 dy ⎛⎜ 1 ⎞⎟ = ⎜ ⎟⎟ cosx − sin x ln x , y dx ⎜⎜⎝ x ⎟⎠ and

⎛ cosx ⎞ dy = x cosx ⎜⎜⎜ − sin x ln x ⎟⎟⎟ . ⎟⎠ ⎜⎝ x dx (c) y = ln ⎡⎢sin(1 x 2 )⎤⎥ , so ⎣ ⎦

⎛ 1 ⎞⎟ ⎛ 2 ⎞⎟ ⎛ 1 ⎞⎟ dy 1 d 1 ⎜⎜ ⎟ = ⎜ ⎟ ⎜ ⎟ = sin ⎜⎜⎝ x 2 ⎟⎟⎠ sin(1 x 2 ) ⎜⎜⎜⎝− x 3 ⎟⎟⎠ cos ⎜⎜⎜⎝ x 2 ⎟⎟⎠ dx sin(1 x 2 ) dx =−

2cot(1 x 2 ) . x3

(d) y = ln ⎡⎢sec(x)⎤⎥ = ln(1 cosx) . So, ⎣ ⎦

dx = cos(−1 cos2 x)(−sin x) = tan x . dy 3.11

(a) y = a x ⇒ ln y = x lna and therefore

1 dy = lna y dx

and

dy = y lna = (lna)a x . dx

For a = 2 , this gives

dy = (ln 2)2x = (0.693…)2x , dx in agreement with Example 3.6. (b) y = loga x , so x = a y and dx dy = a y lna from the previous result. Hence

dy 1 1 . = y = dx a lna x lna 3.12

Differentiating x and y give

S3.5

dx dy 4t , = (1 + t 2 ) + 2t 2 = 1 + 3t 2; =− dt dt (1 + t 2 )2 and so

dy dy ⎛⎜ 1 ⎞⎟ 4t ⎟⎟ = − . = ⎜⎜ 2 2 ⎟ ⎜ dx dt ⎝ dx dt ⎠ (1 + t ) (1 + 3t 2 ) But t = xy 2 , therefore

dy = dx ⎡1 + (x 2y 2 ) ⎢⎣ 3.13

−2xy 4 ⎤⎥ ⎦

2

⎡1 + (3x 2y 2 ) ⎢⎣

4 ⎤⎥ ⎦

=

−128xy . (4 + x y ) (4 + 3x 2y 2 ) 2 2 2

The angle of flight is the angle θ between the tangent to the curve and the horizontal at time t, as shown in the figure below.

Hence

tan θ =

dy dy dt w − gt = = , dx dx dt u

i.e.

⎛ w − gt ⎞⎟ ⎟. θ = arctan ⎜⎜⎜ ⎜⎝ u ⎟⎟⎠ The kinetic energy 2

2

1 ⎛ dx ⎞ 1 ⎛ dy ⎞ T = m ⎜⎜⎜ ⎟⎟⎟ + m ⎜⎜⎜ ⎟⎟⎟ = 12 mu 2 + 12 m(w − gt)2 . 2 ⎜⎝ dt ⎟⎠ 2 ⎜⎝ dt ⎟⎠ Hence the rate of loss of kinetic energy is

dT 1 = m 2(w − gt)(−g) = −mg(w − gt) , dt 2

S3.6

where we have used the chain rule to differentiate (w − gt)2 . 3.14

Differentiating with respect to x gives

6x −12

dy dy + 3x + 3y − 8 = 0 , dx dx

i.e. the gradient is

dy 6x + 3y − 8 = dx 3(4 − x) and at (x,y) = (1,1) , dy dx = 1 9 . So the equation of the tangent at this point is

y −1 = 19 (x −1) ⇒ 9y − x − 8 = 0. The tangent therefore intercepts the x axis at −8 and the y axis at 8 9 . 3.15

Differentiating with respect to x gives

4x + 3y 2

dy dy dy − 2xy −y 2 − =0 dx dx dx

so that

dy y 2 − 4x = 2 , dx 3y − 2xy −1 which at (x,y) = (2,−1) is dy dx = −7 6 . The normal to the curve at (x,y) = (2,−1) therefore has a gradient 6 7 and an equation 7y = 6x +c , where c is a constant. This can be found from the fact that the normal passes through the point (x,y) = (2,−1) , which gives c = −19 and hence 7y = 6x −19 . 3.16

Using the notation of (3.40b), we have as x → 0+ ,

f (0)(x) = x 2 → 0; f (1)(x) = 2x → 0; f (2)(x) = 2 → 2 , and

f (n) = 0 → 0, (n ≥ 3), whereas for x < 0 , we have as x → 0− ,

f (0)(x) = x sin x → 0; f (1)(x) = x cosx + sin x → 0; f (2)(x) = 2cosx − x sin x → 2 , and

f (3)(x) = −3sin x − x cosx → 0; f (4)(x) = −4 cosx + x sin x → −4 . Only the derivatives f (1), f (2) and f (3) exist as x → 0 , because

S3.7

⎡ d4 f ⎤ ⎡ d4 f ⎤ lim− ⎢⎢ 4 ⎥⎥ ≠ lim+ ⎢⎢ 4 ⎥⎥ . x→0 dx ⎢⎣ ⎥⎦ x→0 ⎢⎣ dx ⎥⎦ 3.17

(a)

f (0)(x) = sinh 2x, f (1)(x) = 2cosh 2x, f (2)(x) = 22 sinh 2x, f (3)(x) = 23 cosh 2x … from which we deduce that

2n−1e 2x + (−1)n+1 2n−1e −2x . (b)

f (1)(x) = x −1, f (2)(x) = −x −2, f (3)(x) = 2x −3, f (4)(x) = −(3×2)x −4 , f (5)(x) = (4 ×3×2)x −5 … from which we deduce that

f (n)(x) =

3.18

Differentiating

(−1)n+1(n −1)! for all n ≥ 1. xn

dx 1 using the chain rule gives, = dy (dy dx )

d2x d ⎛⎜ 1 ⎞⎟ dx ⎛⎜ d ⎞⎟⎛⎜ 1 ⎞⎟ ⎟⎟ = ⎟⎟ , = ⎜ ⎜ ⎟⎟⎜ dy 2 dy ⎜⎜⎝ dy dx ⎟⎠ dy ⎜⎜⎝ dx ⎟⎠⎜⎜⎝ dy dx ⎟⎠ which using the quotient rule is

⎛ d2y ⎞⎟ d2x 1 ⎜⎜ ⎟. = − 3⎜ 2 ⎟ ⎜⎝ d x ⎟⎠ d2y dy dx ( ) Similarly, differentiating again, using the product and quotient rules, gives

⎛ dx ⎞⎟⎛⎜ d d2y ⎞⎟ d3x d2y dx d 1 1 ⎜⎜ ⎟⎜ ⎟⎟ = − − ⎟ 3 2 3 3⎜ dy dx dy dx (dy dx ) (dy dx ) ⎜⎝ dy ⎟⎠⎜⎜⎝ dx dx 2 ⎟⎠ 2

⎛ d2y ⎞⎟ 1 d3y ⎜⎜ ⎟⎟ − = . 5⎜ 4 3 ⎜⎝ dx 2 ⎟⎠ dx dy dx dy dx ( ) ( ) 3

3.19

(a) The result is trivially satisfied for n = 1 , using (1.27a). The proof then follows by induction. Suppose it is valid for a given n. Then

F (n) ≡

n ⎛ ⎞ dn (fg) ⎜⎜ n ⎟⎟ f n−r (x)g r (x) , = ∑ ⎜⎝ r ⎟⎟⎠ dx n r=0 ⎜

S3.8

and differentiating with respect to x, we have n ⎛ n ⎛ ⎞ F (n+1) = ∑ ⎜⎜⎜ n ⎟⎟⎟ f n−r+1(x)g r (x) + ∑ ⎜⎜⎜ n ⎝ r ⎟⎠ ⎝ r r=0 ⎜ r=0 ⎜ N −1 ⎛ N −1 ⎛ ⎞ = ∑ ⎜⎜⎜ N −1 ⎟⎟⎟ f N −r (x)g r (x) + ∑ ⎜⎜⎜ ⎟⎠ ⎝ r ⎝ r=0 ⎜ r ′=1 ⎜

⎞⎟ n−r ⎟⎟ f (x)g r+1(x) ⎟⎠ N −1 r ′ −1

⎞⎟ N −r ′ ′ ⎟⎟ f (x)g r (x), ⎟⎠

where we have set N = n + 1 and r ′ = r + 1 . On relabeling, the dummy index r ′ → r in the second term, this becomes N −1 ⎡⎛ ⎞ ⎛ F (N ) = f N −r g r + ∑ ⎢⎢⎜⎜⎜ N −1 ⎟⎟⎟ + ⎜⎜⎜ N −1 ⎟⎠ ⎜⎝ r −1 ⎝ r r=1 ⎢⎣⎜ N ⎛ ⎞ = ∑ ⎜⎜⎜ N ⎟⎟⎟ f N −r (x)g r (x), ⎝ r ⎟⎠ r=0 ⎜

⎞⎟⎤ N −r ⎟⎟⎥ f (x)g r (x) ⎟⎠⎥⎥ ⎦

where we have used (1.27a,c). Hence if the Leibniz formula holds for any n it holds for N = n + 1 ; and since it holds for n = 1 , it therefore holds for all n. (b) Setting f = ln x and g = x 2 , we have

f (0) = ln x, f (1) = 1 x, f (2) = −1 x 2 , f (3) = 2 x 3 , f (4) = −6 x 4 and

g (0) = x 2, g (1) = 2x, g (2) = 2, g (n) = 0 (n ≥ 3) , so that

d 4 (fg) ⎛⎜ 4 ⎞⎟ (4) (0) ⎛⎜ 4 ⎞⎟ (3) (1) ⎛⎜ 4 ⎞⎟ (2) (2) ⎟ f g +⎜ ⎟ ⎟ = ⎜⎜ ⎜⎜⎝ 1 ⎟⎟⎠ f g + ⎜⎜⎜⎝ 2 ⎟⎟⎠ f g ⎜⎝ 0 ⎟⎟⎠ dx 4 ⎛ 1⎞ 6 2 2 = − 4 x 2 + 4 3 2x + 6 ⎜⎜⎜− 2 ⎟⎟⎟ 2 = − 2 . ⎟ ⎜ x x x ⎝ x ⎠ 3.20

2

(a) If f (x) = x 2e −x , then 2

df (x) dx = 2x(1 − x 2 )e −x = 0 at x = 0, ± 1 and

⎪⎧⎪ 2 > 0 d2 f (x) at x = 0 2 4 −x 2 = (2 −10x + 4x )e = ⎨ 2 ⎪ dx ⎪⎩ −4 e < 0 at x = ±1 Hence there is a minimum at x = 0 and maxima at x = ±1 . (b) The shape of the curve can be found from the positions of the maxima and the minimum, plus the fact that f (x) > 0 for x ≠ 0 and f (x) → 0 as x → ±∞ . It is shown in the figure below.

S3.9

3.21

On differentiating,

f ′(x) = (cosx − sin x)e −x

and

f ′′(x) = −2cosx e −x .

Using (2.29) and (2.31), f ′(x) = 0 gives

sin x = cosx = cos(π 2 − x) , implying x = π 2 − x + 2πn , i.e. x = π 4 + nπ as the positions of the extrema. For

x = π 4 + 2πn , cosx is positive and, and hence f ′′(x) , are negative and so the extrema are maxima. For x = 5π 4 + 2πn , cosx is negative, and hence f ′′(x) is positive, so they are minima. 3.22

This is shown in the figure below. By writing the function as

y(x) =

x3 (x − 2)(x + 1)

it can be seen that there are singular points at x = 2 and x = −1 . Also as x → ∞ , y(x) → x and so y(x) → ±∞ as x → ±∞ . Finally, differentiating y(x) and setting the result to zero, yields three turning points: x = 1 + 7, 1 − 7, 0 . By forming second derivatives these can be shown to be minimum, a maximum and a point of inflexion, respectively.

S3.10

3.23

Differentiating using the quotient rule gives

dy (x 2 + x − 2)(4x − 5)−(2x 2 − 5x − 25)(2x + 1) = dx (x 2 + x − 2)2 7x 2 + 42x + 35 7(x + 1)(x + 5) = = (x 2 + x − 2)2 (x 2 + x − 2)2

,

so there stationary points at x = −1 and x = −5 . To investigate their nature, we find the second derivative. Using the quotient rule again this is

d2y 14(x 2 + x − 2)2(x + 3)−14(x + 1)(x + 5)(2x + 1)(x 2 + x − 2) = dx 2 (x 2 + x − 2)4 14(x + 3) = 2 at x = −1 and x = −5. (x + x − 2)2 At x = −1 , d2x dx 2 > 0 and so this is a minimum, and at x = −5 , d2x dx 2 < 0 and so this is a maximum. The function is discontinuous at points where the denominator vanishes, i.e. at x = 1 and x = −2 . Finally, to find the limits as x → ±∞ only the highest powers of x need be retained, which gives

lim f (x) = 2.

x→∞

The function is sketched in the figure below.

S3.11

3.24

The figure below shows a graph of the function y(x) = x 3 − 4x 2 + x + 4 for the range −1 ≤ x ≤ 4 . The solutions of the equation are where the curve cuts the x-axis, i.e. at the approximate values x = −0.8, 1.5 and 3.4 .

3.25

For all x, f (x) ≤ exp(−x 2 ) , with

f (x) = exp(−x 2 ) at 10πx = 2πn , i.e., x = ±0.2n, (n = 0,1, 2,…) , and

f (x) = −exp(−x 2 ) at 10πx = π(2n + 1) , i.e., x = ±0.1(2n + 1), (n = 0,1, 2,…) . Thus f (x) oscillates between ± exp(−x 2 ) , with zeros at

10πx = (2n + 1)π 2 , i.e. x = ±0.05(2n + 1),(n = 0,1, 2,…) . The function is plotted below in the figure below as a solid line. Because cos(10πx) oscillates rapidly compared to exp(−x 2 ) , the maxima and minima are close to, but not exactly at, the points where f (x) = ± exp(−x 2 ) , the latter being shown as dashed lines.

S3.12

3.26

Differentiating the standard equation of an ellipse

x2 y2 + =1 a 2 b2 gives

2x 2y dy + = 0, a 2 b 2 dx so that at P,

b 2x dy =− 2 1 . dx a y1 The gradient of the normal at P is therefore (a 2y1 ) (b 2x1 ) and hence the equation of the normal is

y −y1 =

a 2y1 b 2x1

(x − x1 ).

Since A lies on the major axis, y = 0 and so at A ,

x=

(a 2 −b 2 ) x1 = e 2x1 , a2

where we have used (a 2 −b 2 ) = a 2e 2 , and since F is the point (ae, 0) ,

AF = ae −e 2x1 . By definition,

PF = ePN = e (a e − x1 ) = a −ex1

so

S3.13

ePF = ae −e 2x1 and hence AF = ePF . 3.27

The configuration is shown in the figure below.

(

)

Let P have co-ordinates (x1,y1 ) = x1, 2(ax1 )1/2 . The tangent at P is

y = m1x +c1 where 1/2

dy 2a ⎛⎜ a ⎞⎟⎟ m1 = = = ⎜⎜ ⎟ . ⎜⎝ x1 ⎟⎟⎠ dx y Substituting y = 2(ax1 )1/2 , x = x1 gives c1 = (ax1 )1/2 , so that the tangent is 1/2

⎛ a ⎞⎟ y = x ⎜⎜⎜ ⎟⎟ ⎝⎜ x1 ⎟⎟⎠

+ (ax1 )1/2 .

This meets the directrix x = −a at 1/2

⎛ a ⎞⎟ y = −a ⎜⎜⎜ ⎟⎟ ⎝⎜ x1 ⎟⎟⎠

+ (ax1 )1/2

(1)

which is the y-co-ordinate of R. The normal to P is

y = m2x +c , where m2 = −1 m1 = −(x1 a ) , since it is perpendicular to the tangent. Substituting 1/2

(

)

the co-ordinates x1, 2(ax1 )1/2 at P gives 1/2

1/2

c = 2(ax1 )

⎛ x ⎞⎟ + ⎜⎜⎜ 1 ⎟⎟ x1 , ⎜⎝ a ⎟⎟⎠

so that the normal is

S3.14

1/2

⎛ x ⎞⎟ y = −x ⎜⎜⎜ 1 ⎟⎟ ⎝⎜ a ⎟⎠⎟

1/2

⎛ x ⎞⎟ + x1 ⎜⎜⎜ 1 ⎟⎟ ⎝⎜ a ⎟⎠⎟

+ 2(ax1 )1/2 .

At R, the y co-ordinate is given by (1), and substituting (1) into (2) gives

x=

a2 + x1 + a , x1

y2 =

a3 + ax1 − 2a 2 x1

whereas squaring (1) gives

so that

y 2 = a(x − 3a) , which is a parabola whose vertex is at x = 3a, y = 0 .

S3.15

(2)

SOLUTIONS TO PROBLEMS 4 4.1

The two curves are shown in the figure below and the required area is shown shaded. Note that it consists of two positive and two negative regions. Its value is 3.0

∫ ⎡⎢⎣(1 + 4x + 2x 1.5

2

)

− x 3 −(6 − 5x − 2x 2 + x 3 )⎤⎥ dx ⎦ 3.0

= ⎡⎢−5x + 92 x 2 + 43 x 3 − 12 x 4 ⎤⎥ = 16.41. ⎣ ⎦1.5

−10

x 4.2

3.0

(a) Applying the first mean value theorem for integration (4.16) to f ′(x) gives b

∫ f ′(x)dx = (b −a) f ′(ζ) , a

while (4.11) applied to f ′(x) gives b

∫ f ′(x)dx = f (b)− f (a) . a

Equating the two results gives the first mean valued theorem, as required. (b) The mean value of a function f (x) over the range a ≤ x ≤ b is

1 b −a

b

∫ f (x)dx , a

which for the given function is 1

1

0

0

1 1 ⌠ x2 ⌠ 1 dx = ⎡⎢x ⎤⎥ − ⎮ dx = ⎡⎢x − arctan x ⎤⎥ = 1 − π 4. ⎮ 2 2 ⎣ ⎦ ⎣ ⎦ 0 0 ⌡ 1+x ⌡ 1+x

The first mean value theorem states that there exists a value of x, say ζ , in the interval a ≤ x ≤ b such that f (ζ) is equal to the mean value. Setting

S4.1

ζ2 π = 1− 2 4 1+ζ

f (ζ) =

gives ζ = 0.523 , which is in the range 0 ≤ x ≤ 1 and so verifies the theorem. 4.3

(a) Substituting z = 2 + x 2 with dz dx = 2x , gives

∫ x(2 + x

2 3/2

)

dx =

1 2

∫z

3/2

dz

= 15 z 5/2 +c = 12 (2 + x 2 )5/2 +c . (b) Substituting z = 3 + x with dz dx = 1 , gives 2 ⌠⎛ 1 ⌠ x2 6 9⎞ ⌠ z − 6z + 9 dx = dz = ⎮ ⎜⎜⎜ 2 − 3 + 4 ⎟⎟⎟ dz ⎮ ⎮ 4 4 ⌡ z z z ⎟⎠ ⌡ (3 + x) ⌡ ⎜⎝ z 1 3 3 1 3 3 = − + 2 − 3 +c = − + − +c . 2 z z 3 + x (3 + x) (3 + x)3 z

(c) Substituting z = x −1 , x = 1 + z 2 , with dx dz = 2z , gives

⌠ ⌠ dx z z ⌠ = 2⎮ dz = 2⎮ dz. ⎮ 2 ⌡ 1 + z + 2z ⌡ x + 2 x −1 ⌡ (1 + z)2 Now set u = 1 + z , z = u −1 , with dz du = 1 , gives

⌠ dx du ⌠ u −1 ⌠ du = 2⎮ 2 du = 2⌠ − 2⎮ 2 ⎮ ⎮ ⌡ ⌡ ⌡u u u ⌡ x + 2 x −1 2 = 2 ln u + 2 u = 2 ln 1 + z + +c 1+z 2 = 2 ln 1 + x −1 + +c. 1 + x −1

(

4.4

(a) Set 3x = sin θ , so that

)

1 − 9x 2 = cos θ and dx dθ = 13 cos θ . Then, substituting

gives

1 cos2θ dθ 3∫ θ 1 (1 + cos 2θ) dθ = + sin 2θ +c, 6 12

I = ∫ 1 − 9x 2 dx = =

1 6∫

where c is a constant. Finally, using sin 2θ = 2sin θ cos θ = 2sin θ 1 − sin 2 θ , gives

S4.2

I=

1 x arcsin(3x) + 1 − 9x 2 +c , 6 2

(b) Put 3 + x − x 2 = t 2 , so that dx dt = 2t (1 − 2x) . Then,

⌠ 1 − 2x t 2 1/2 I =⎮ dx = 2⌠ ⎮ dt = 2t +c = 2(3 + x − x ) +c . 2 1/2 ⌡ t ⌡ (3 + x − x ) where c is a constant. (c) Substituting z = x − 2 , with dz = dx , gives

⌠ ⌠ x x I =⎮ 2 dx = ⎮ dx ⌡ (x − 2)2 ⌡ x − 4x + 4 ⌠ ⎛⎜ 1 2 ⎞⎟ 2 ⌠ (z + 2) =⎮ dz = ⎮ ⎜⎜ + 2 ⎟⎟⎟ dz = ln z − +c 2 ⌡ z z ⌡ ⎝⎜ z z ⎠ = ln(x − 2)−

4.5

2 +c. (x − 2)

(a) Substitute z = sin x and dz dx = cosx , so that cos2 x = 1 − sin 2 x = 1 − z 2 . Then,

∫ cos x dx = ∫ (1 − z 5

=z−

) dz = ∫ (1 − 2z 2 + z 4 )dz

2 2

2z 3 z 5 2sin 3 x sin 5 x + +c = sin x − + +c . 3 5 3 5

(b) Use cos 2x = 1 − 2sin 2 x = 2cos2 x −1 , i.e. 2sin 2 x = 1 − cos 2x . Then,

∫ 4 sin x dx = ∫ (1 − cos 2x) dx = ∫ (1 − 2cos 2x + cos 2x)dx 4

2

2

⌠ ⎛3 cos 4x ⎞⎟ 3 sin 4x ⎟⎟ dx = x − sin 2x + = ⎮ ⎜⎜⎜ − 2cos 2x + +c , ⎟ ⎜ 2 2 2 8 ⎠ ⌡⎝

where we have used

cos 4x = 2cos2 2x −1 , i.e. cos2 2x = 12 (1 + cos 4x) . 4.6

(a) Using z = sinh x , cosh 2 x = 1 + sinh 2 x , with dz dx = cosh x , gives

∫ sinh

6

x cosh 3 x dx = ∫ z 6 (1 + z 2 )dz =

z7 z9 sinh 7 x sinh 9 x + +c = + +c . 7 9 7 9

(b) Substituting z = 3 + sin x , sin x = z − 3 , with dz dx = cosx , gives

S4.3

cos2 x = 1 − sin 2 x = 1 −(z − 3)2 = 6z − z 2 − 8 and so

∫ (3 + sin x) cos 5

=

3

x dx = ∫ z 5(6z − z 2 − 8)dz

6z 7 z 8 4z 6 6(3 + sin x)7 (3 + sin x)8 4(3 + sin x)6 − − +c = − − +c , 7 8 3 7 8 3

where c is a constant. 4.7

(a) This may be written in partial fractions, so that

5x − 4 ⌠ ⌠ 5x − 4 I =⎮ 2 dx = ⎮ dx ⌡ x −x −2 ⌡ (x − 2)(x + 1) ⌠⎛ 2 3 ⎞⎟ ⎟⎟ dx = 2 ln(x − 2) + 3 ln(x + 1) +c , = ⎮ ⎜⎜⎜ + ⌡ ⎜⎝ x − 2 x + 1 ⎟⎠ where c is a constant. (b) This may be written in partial fractions as follows.

1 A B C = + + , (x −1)(x + 2)(x − 3) x −1 x + 2 x − 3 i.e.

A(x + 2)(x − 3) + B(x −1)(x − 3) +C(x −1)(x + 2) = 1. then setting x = 1, − 2, 3 in turn gives A = −1 6, B = 1 15, C = 1 10 and so

1⌠ 1 1 ⌠ 1 1 1 I =− ⎮ dx + ⎮ dx − ⌠ dx ⎮ ⌡ 6 ⌡ x −1 15 ⌡ x + 2 10 x − 3 = − 16 ln(x −1) + 151 ln(x + 2)− 101 ln(x − 3) +c, where c is a constant. 4.8

(a) This may be written in partial fractions as follows.

2−x2 A B C , = + + 2 x + 3 x − 4 (x − 4)2 (x + 3)(x − 4) so that

2 − x 2 = A(x − 4)2 + B(x + 3)(x − 4) +C(x + 3) . Then setting x = −3 gives A = −1 7 , setting x = 4 gives C = −2 , and setting

x = 0 and using the values of A and C, gives B = −6 7 . So, the integral is

S4.4

⌠⎡ 1 6 2 ⎤⎥ 1 6 2 −⎮ ⎢⎢ + + dx = − ln(x + 3)− ln(x − 4) + +c. 2⎥ 7 7 (x − 4) ⌡ ⎣ 7(x + 3) 7(x − 4) (x − 4) ⎦ where c is an integration constant. (b) In terms of partial fractions,

1 A Bx +C , = + 2 2 (x + 1)(x + 1) x + 1 x +1 i.e.

1 = A(x 2 + 1) + (Bx +C )(x + 1) . Then setting x = −1 gives A = 1 2 , setting x = 0 gives A +C = 1 , so that C = 1 2 , and likewise setting x = 1 gives B = −1 2 . Hence 1

1 ⌠⎛ 1 x 1 ⎞⎟ ⎟⎟ dx I = ⎮ ⎜⎜⎜ − 2 + 2 2 ⌡ ⎜⎝ x + 1 x + 1 x + 1 ⎟⎠ 0 1

⎤ 1⎡ 1 1 π = ⎢ ln(x + 1)− ln(x 2 + 1) + arctan x ⎥ = ln 2 + . ⎥ 2 ⎢⎣ 2 4 8 ⎦0 4.9

(a) Substituting x = tan(θ 2) and using Equations (4.28) and (4.29), gives

⌠ dx ⌠ dt = 2⎮ ⎮ ⌡ (1 + t)2 ⌡ 1 + sin x 2 2 =− +c = − +c . (1 + t) 1 + tan(x 2) (b) Substituting x = tan θ and using Equations (4.30) and (4.31), gives

dx ⌠ dt ⌠ = ⎮ ⌡ 3 − 2sin 2 x ⎮ ⌡ 3 + t2 =

4.10

⎛ t ⎞ ⎛ tan x ⎞⎟ 1 ⎟ +c . tan−1 ⎜⎜⎜ ⎟⎟⎟ +c = tan−1 ⎜⎜⎜ ⎜⎝ 3 ⎟⎠ ⎜⎝ 3 ⎟⎟⎠ 3 3

1

(a) Setting z = tan x with dz dx = sec2 x , gives

∫ sec x ln(tan x)dx = ∫ ln z dz = z ln z − z +c 2

= tan x ln(tan x)− tan x +c , where we have integrated ln z by parts using u = ln z and υ = z (cf. section 4.4.) (b) Substituting t = tan(x 2) and using Equations (8.28) and (4.29), gives

S4.5

⌠ 1 −(1 −t 2 ) (1 + t 2 ) 2 ⌠ 1 − cosx dx = ⎮ dt ⎮ ⌡ 1 + cosx ⌡ 1 + (1 −t 2 ) (1 + t 2 ) (1 + t 2 ) ⌠ 2t 2 ⌠ 1+ t2 ⌠ 1 =⎮ dt = 2 dt − 2⎮ dt ⎮ 2 2 ⌡ 1+ t2 ⌡ 1+t ⌡ 1+t = 2t − 2 tan−1 t +c = 2 tan(x 2)− x +c . 4.11

⌠ cosh(4x) (a) ∫ coth(4x)dx = ⎮ dx ⌡ sinh(4x) 1⌠ 1 d sinh(4x) 1 = ⎮ = ln [sinh(4x)] +c. 4 ⌡ sinh(4x) dx 4 where c is a constant. (b) Substituting t = tan(x 2) so that dx dt = 2 (1 + t 2 ) , and using the half-angle formulas from Section 4.3.6 gives

1 ⌠ ⎛⎜ 1 1 ⎞⎟⎟ ⌠ dt = 2 = + ⎜ ⎮ ⎮ ⎟⎟ dt ⌡ 5 −t 2 ⎠ (1 + t 2 ) ⎡⎢2 + 3(1 −t 2 ) (1 + t 2 )⎤⎥ ⌡ ⎜⎜⎝ 5 −t 5 5 + t ⎣ ⎦ ⎛ t + 5 ⎞⎟ ⎡ ⎤ 1 ⎜ ⎟⎟ = 1 ln ⎢ tan(x 2) + 5 ⎥ +c, = ln ⎜⎜ ⎢ ⎥ ⎟ 5 ⎜⎜⎝ t − 5 ⎟⎠ 5 ⎢⎣ tan(x 2)− 5 ⎥⎦

I =∫

2dt

where c is a constant. 4.12

(a) Set x = a sin θ so that dx dθ = a cos θ . Then,

⌠ dx 1 ⌠ dθ 1 I =⎮ 2 = 2⎮ = 2 ∫ sec2θ dθ 2 3/2 2 a ⌡ cos θ a ⌡ (a − x ) 1 x = 2 tan θ +c = 2 2 +c , a a (a − x 2 )1/2 where c is a constant. (b) The condition b 2 < 4ac ensures that the integrand has a constant sign. We start by completing the square in the quadratic as 2 ⎡⎛ ⎤ b ⎞⎟ 4ac −b 2 ⎥ ⎢⎜ ax +bx +c = a ⎢⎜⎜x + ⎟⎟ + ⎥. 2a ⎟⎠ ⎢⎜⎝ 4a 2 ⎥ ⎣ ⎦ 2

⎛ b ⎞ Then setting ⎜⎜⎜x + ⎟⎟⎟ = t so that dx = dt , ⎜⎝ 2a ⎟⎠

S4.6

⎡ 4ac −b 2 ⎤⎥ ax 2 +bx +c = a ⎢⎢t 2 + 4a 2 ⎥⎥⎦ ⎢⎣ and using Table 4.2

⎛t ⎞ 1 ⌠ dt 1 ⌠ dt ⎮ = = arctan ⎜⎜⎜ ⎟⎟⎟ +c ⎮ 2 2 2 ⎮ 2 ⎜⎝ h ⎟⎠ ⌡ t +h h h ⌡ 1 + (t h ) where 1/2

⎛ 4ac −b 2 ⎞⎟ ⎟⎟ h = ⎜⎜⎜ 2 ⎝⎜ 4a ⎟⎠ and c is an integration constant. So finally,

I =2

4.13

⎡ (x +b 2a ) 2a ⎤ ⎥ +c = arctan ⎢⎢ ⎥ 2 2 ⎢ 4ac −b ⎥ 4ac −b ⎣ ⎦ 1

⎛ 2ax +b ⎞⎟ ⎟⎟ +c . arctan ⎜⎜⎜ ⎜⎝ 4ac −b 2 ⎟⎟⎠ 4ac −b 2 2

(a) Set (x −1) = u 2 , so that (10 − x) = 9 −u 2 and dx du = 2u . Then, 1

1

⌠ ⎛⎜ 1 ⌠ 2u 1 ⎞⎟ ⎟⎟ du I =⎮ du = 2 + ⎜ ⎮ ⌡ (9 −u 2 )u ⌡ ⎜⎜⎝ 3 −u 3 + u ⎟⎠ 0

=

0

1 ⎡⎢ ⎛⎜ 3 + u ⎞⎟ ⎟⎟ ln ⎜ 3 ⎢⎢ ⎜⎜⎝ 3 −u ⎟⎠ ⎣

1

⎤ ⎥ = 1 ln 2. ⎥ ⎥⎦ 0 3

(b) Set u = arcsin x and dυ dx = 1 , i.e. υ = x . Then, 1

d arcsin x I = ⎡⎢x arcsin x ⎤⎥ − ⌠ x dx, ⎣ ⎦0 ⎮ ⌡ dx 1

0

which using Equation (3.27) becomes 1

1 π ⌠ x π π I = −⎮ dx = + ⎡⎢(1 − x 2 )1/2 ⎤⎥ = −1. ⎦0 2 ⌡ 1−x2 2 ⎣ 2 0

(c) π/2

π/2

⌠ ⌠ 1 1 I=⎮ dx = ⎮ dx 2 2 ⌡ (sin x − cosx) ⌡ cos x(tan x −1)2 0 π/2

0 ∞

⌠ sec2 x ⌠ 1 =⎮ dx = ⎮ d(tanx). 2 ⌡ (tan x −1)2 ⌡ (tan x −1) 0

0

S4.7

Setting tan x = u , gives ∞

∞

⌠ 1 1 I =⎮ du = − = −1. 2 (u −1) 0 ⌡ (u −1) 0

4.14

We start by using the usual formula for integration by parts with

u(x) = sinn x

υ(x) = e ax a .

and

Then, π/2

π/2

ax ⎡e ax ⎤ ⌠e I n = ⎢⎢ sinn x ⎥⎥ − ⎮ n sinn−1 x cosx dx ⌡ a ⎢⎣ a ⎥⎦ 0 0

e aπ/2 n = − a a

π/2

∫e

ax

sinn−1 x cosx dx .

0

To evaluate the new integral, set u(x) = sinn−1 x cosx and υ(x) = e ax a . Then,

In =

π/2 e aπ/2 n ⎡ ax n−1 − ⎢e sin x cosx ⎤⎥ ⎦0 a a⎣

n + 2 a e

π/2

∫e 0

ax

⎡−sinn−1 x sin x + (n −1)sinn−2 x cos2 x ⎤ dx ⎣⎢ ⎦⎥

aπ/2

n n(n −1) = − 2 In + a a a

2 π/2

∫e

ax

sinn−2 x cos2 x dx .

0

The latter integral can be evaluated by setting cos2 x = 1 − sin 2 x and gives

In =

e aπ/2 n n(n −1) n(n −1) − 2 In + I n−2 − In . 2 a a a a2

Finally, collecting terms,

In = 4.15

ae aπ/2 n(n −1) + 2 I . 2 2 n +a n + a 2 n−2

Using integration by parts, gives

⌠ 2nx 2 ⌠ dx x In = ⎮ = + dx ⎮ ⌡ (1 + x 2 )n (1 + x 2 )n ⌡ (1 + x 2 )n+1 ⌠ x2 +1 x = + 2n dx − 2nI n+1 ⎮ (1 + x 2 )n ⌡ (1 + x 2 )n+1 =

x + 2nI n − 2nI n+1 , (1 + x 2 )n

and so

S4.8

2nI n+1 = (2n −1)I n +

x , (1 + x 2 )n

as required. To find I 3 , we start from

⌠ dx I1 = ⎮ = arctan x . ⌡ (1 + x 2 ) Then,

2I 2 = I 1 +

x x , = arctan x + 2 (1 + x ) (1 + x 2 )

and

I3 = 4.16

3 x x 3 x 3 I2 + = + + arctan x . 2 2 2 2 2 4 8 (1 + x ) 8 4(1 + x ) 4(1 + x )

Use the general formula

⌠ u(x) dυ(x) dx = u(x)υ(x)− ⌠ υ(x) du(x) dx . ⎮ ⎮ ⌡ ⌡ dx dx (a) Set u = e −x and υ = −cosx . Then

I = −e −x cosx − ∫ cosx e −x dx. To evaluate the new integral, set u = e −x and υ = sin x . This gives

I = −e −x cosx −e −x sin x − ∫ e −x sin x dx = −e −x cosx −e −x sin x − I +c , and so

I = − 12 e −x (sin x + cosx) +c . where c is a constant. (b) Set u = ln x and υ = (1 − x)−1 . Then

I= =

ln x ⌠ ⎛1 1 1 ⎞⎟ ⌠ ⎟⎟ dx −⎮ dx = − ⎮ ⎜⎜⎜ + (1 − x) ⌡ x(1 − x) (1 − x) ⌡ ⎜⎝ x 1 − x ⎟⎠ ln x

ln x (1 − x)

− ln x + ln (x −1) +c =

where c is a constant. (c) Set x = e t so that dx dt = e t and

S4.9

x ln x (1 − x)

+ ln (x −1) +c ,

I = ∫ e t sin t dt . Then integrate by parts using u = sin t and dυ dt = e t . This gives

I = e t sin t − ∫ e t cost dt and integrating by parts again, this time with u = cost and dυ dt = e t , gives

I = e t sin t − ⎡⎢e t cost + ∫ e t sin t dt ⎤⎥ , ⎣ ⎦ i.e.

2I = e t (sin t − cost), and so

I = 12 x ⎡⎢sin(ln x)− cos(ln x)⎤⎥ +c , ⎣ ⎦ where c is a constant. 4.17

(a) From the definition, ∞

∞

Γ(x + 1) = ∫ t xe −t dt = −∫ t x 0

0

∞

d −t (e )dt dt

∞

= ⎡⎢−t xe −t ⎤⎥ + ∫ x t x−1e −t dt = xΓ(x). ⎣ ⎦0 0

For x = 1 , ∞

∞

Γ(1) = ∫ e −t dt = ⎡⎢e −t ⎤⎥ = 1. ⎣ ⎦0 0

Hence

Γ(2) = 1, Γ(3) = 2Γ(2) = 2, Γ(4) = 3Γ(3) = (3×2)Γ(2) = 6 , etc. i.e. Γ(n + 1) = n ! for integer n ≥ 1 as required. (b) We have

Γ(x) =

1 Γ(x + 1) , x

so that the gamma function diverges for x = 0 and all negative integers. To find Γ(−5 2) , we have

⎛ 5⎞ 2 ⎛ 3⎞ 2 ⎛ 2⎞ ⎛ 1 ⎞ Γ ⎜⎜⎜− ⎟⎟⎟ = − Γ ⎜⎜⎜− ⎟⎟⎟ = − ⎜⎜⎜− ⎟⎟⎟ Γ ⎜⎜⎜− ⎟⎟⎟ ⎜⎝ 2 ⎟⎠ 5 ⎜⎝ 2 ⎟⎠ 5 ⎜⎝ 3 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎛1⎞ 2 ⎛ 2⎞ 8 π = − ⎜⎜⎜− ⎟⎟⎟(−2)Γ ⎜⎜⎜ ⎟⎟⎟ = − . ⎜⎝ 2 ⎟⎠ 5 ⎜⎝ 3 ⎟⎠ 15 4.18

The velocity of the rocket is given by

S4.10

υ=

dx dt = 2g ⌠ − gt = −20g ln(1 − 0.1t)− gt, ⎮ ⌡ dt 1 − 0.1t

t ≤5

where we have used υ = 0 at t = 0 to fix the integration constant as zero. Integrating again gives

x = −20g ∫ ln(1 − 0.1t)dt − gt 2 2

= 200g ⎡⎢(1 − 0.1t)ln(1 − 0.1t)− ln(1 − 0.1t)⎤⎥ − gt 2 2 +c ⎣ ⎦ 2 ⎡ ⎤ = 200g ⎢(1 − 0.1t)ln(1 − 0.1t) + 0.1t)⎥ − gt 2, ⎣ ⎦

t ≤5

where we have substituted z = 1 − 0.1t and used the integral of ln z given in Section 4.4, and set x = 0 at t = 0 . Setting g = 9.81ms−2 and t = 5 gives

υ(t = 5) = 86.95 ms−1,

x(t = 5) = 178.4 m.

For t ′ = t − 5 ≥ 0 , the rocket moves with constant acceleration –g, so that

υ(t ′) = 86.95 − gt ′ and using x(t ′ = 0) = 178.27 m to fix the constant of integration,

x(t ′) = 86.95t ′ − 12 g(t ′)2 + 178.4. The maximum height is reached when υ(t ′) = 0 , i.e. at t ′ = 86.95 g = 8.86 s, when

x max ≡ x(8.86 s) = 564 m . 4.19

(a) Setting the upper limit to b and substituting z = x 2 gives b

∫ 0

xe

−x 2

b2

dx =

1 2

∫ 0

b2

e −z dz = 12 ⎡⎢−e −z ⎤⎥ ⎣ ⎦0

2 = ⎡⎢1 −e −b ⎤⎥ → 12 , ⎣ ⎦ b→∞

1 2

so the integral is convergent and has a value of 1 2 . (b) Setting the lower limit to ε gives

S4.11

1

∫ ε

1

⎞ ⌠ d ⎛⎜ x 2 (x − 2)ln x dx = ⎮ ln x ⎜⎜ − 2x ⎟⎟⎟ dx ⎟⎠ dx ⎜⎝ 2 ⌡ ε 1

1

1

⎡⎛ x 2 ⎤ ⎡⎛ x 2 ⎤ ⎞ ⎞ ⎞ ⌠ ⎛x x2 = ⎢⎢⎜⎜⎜ − 2x ⎟⎟⎟ ln x ⎥⎥ − ⎮ ⎜⎜⎜ − 2⎟⎟⎟ dx = ⎢⎢⎜⎜⎜ − 2x ⎟⎟⎟ ln x − + 2x ⎥⎥ . ⎟⎠ ⎟⎠ ⎟⎠ 4 ⎢⎣⎜⎝ 2 ⎥⎦ ε ⌡ ⎜⎝ 2 ⎢⎣⎜⎝ 2 ⎥⎦ ε ε Since x ln x → 0 as x → 0 [cf. Equation (3.6b)], this expression is well behaved as ε → 0 , giving 1

∫ (x − 2)ln x dx = 7

4 .

0

(c) The integral ∞

⌠ dx ⎮ ⌡ x 1

is divergent, so since ln x > 1 for large x, the given integral is also divergent. (d) Writing

1 1 = , x + 2x − 3 (x + 3)(x −1) 2

it follows that the given integral is convergent only if all three integrals in the decomposition ∞

−3

1

∞

−∞

−∞

−3

1

⌠ dx dx dx dx ⌠ ⌠ ⌠ ≡⎮ +⎮ +⎮ ⎮ 2 ⌡ (x + 2x − 3) ⌡ (x + 3)(x −1) ⌡ (x + 3)(x −1) ⌡ (x + 3)(x −1) are convergent. Consider the integral ∞

∞

1+ε

ε

dx ⌠ ⌠ dz =⎮ , ⎮ ⌡ (x + 3)(x −1) ⌡ (z + 4)z where z = x −1 . For small ε , (z + 4)z → 4z and the integral diverges, so the given integral also diverges. 4.20

(a) Using sin x x → 1 as x → 0 , we have

1 − cosx 2sin 2(x 2) 1 = → 1/2 as x → 0. 5/2 5/2 x x 4x

S4.12

This is the same limiting behaviour as (4.49) for α = 1 2 , and the integral is therefore convergent. (b) Using sin x ≤ 1 , we have a

a

a

1

1

1

⌠ ⎛⎜ 1 ⎞⎟ ⌠ sin x ⌠ dx ⎮ ⎜⎜− 2 ⎟⎟⎟ dx < ⎮ 2 dx < ⎮ 2 . ⎜ ⌡ ⌡x x ⌡⎝ x ⎠ Since the two outer integrals converge as a → ∞ , so must the given integral. (c) We need to consider both limits. As x → 0 ,

tan x sin x 1 = → 1, x x cosx so there are no convergence problems. For x = π 2 , change variables to x = π 2 − ε . The integral then becomes π/2

π/2

⌠ tan x dx = ⌠ tan(π / 2 − ε) dε ⎮ ⎮ ⌡ x ⌡ π / 2−ε 0

0 π/2

π/2

dε dε ⌠ sin(π / 2 − ε) ⌠ cos(ε) =⎮ =⎮ . ⌡ cos(π / 2 − ε) (π / 2 − ε) ⌡ sin(ε) (π / 2 − ε) 0

0

As ε → 0 ,

cos ε dε 2dε → , sin ε (π / 2 − ε) επ so the integral is divergent. 4.21

(a) If α ≥ 0 , then x α → ∞ or 1 as x → ∞ , and the integral does not converge. If

α < 0 , then x α → 0 and sin x α → x α as x → ∞ , and the integral converges for all α < −1 . (b) Set z = ln x , so that x = e z and dx dz = e z = x . The integral then becomes ∞

∫e

(α+1)z β

z dz ,

0

which converges provided β > −1, α < −1 [cf. Exercises 4.13b and Equation (4.49).] 4.22

The work done is in three parts and is given by

S4.13

a

2a

W = ∫ F(x)dx + ∫ a/2

a

a

3a

2a

3a

kx kx 2 ⌠ k F(x)dx + ∫ F(x)dx = ⎮ 2 dx + ∫ 3 dx + ∫ dx 4 ⌡ x a 2a 2a a 2a 1

2a 3a k k 17k = k ⎡⎢−1 x ⎤⎥ + 3 ⎡⎢x 2 ⎤⎥ + 4 ⎡⎢x 3 ⎤⎥ = . ⎣ ⎦a/2 2a ⎣ ⎦a ⎣ ⎦ 2a 3a 6a a

4.23

The length is given by (4.52): 1/2

1

⌠ ⎡ ⎛ dy ⎞ ⎢ ⎜ ⎟ L=⎮ ⎮ ⎢⎢1 + ⎜⎜⎜⎝ dx ⎟⎟⎟⎠ ⌡⎣

2

0

⎤ ⎥ ⎥ dx , ⎥ ⎦

where

dy ⎛⎜ 1 ⎞⎟ (−2x) x = ⎜⎜ ⎟⎟ =− . 2 1/2 ⎟ ⎜ dx ⎝ 2 ⎠ (1 − x ) (1 − x 2 )1/2 So, 1

1

1/2

1/2 ⌠ ⎛⎜ ⌠ ⎛⎜ 1 ⎞⎟ x 2 ⎞⎟⎟ ⎟ dx, L = ⎮ ⎜⎜1 + ⎟ dx = ⎮ ⎮ ⎜⎜⎜⎝1 − x 2 ⎟⎟⎠ ⎮ ⎜⎝ 1 − x 2 ⎟⎠ ⌡ ⌡ 0

0

which using Table 4.2 is 1

L = ⎡⎢arcsin x ⎤⎥ = π 2. ⎣ ⎦0 4.24

The hyperbola has two regions for any positive value of x, and since these are symmetric about the x axis, using (4.52) we have 2.5

1/2

⌠ ⎡ ⎛ ⎞2 ⎤ ⎢ ⎜ dy ⎟ ⎥ L = 2⎮ ⎮ ⎢1 + ⎜⎜⎜ ⎟⎟⎟ ⎥ dx , ⎮ ⎢ ⎝ dx ⎠ ⎥ ⌡⎣ ⎦ 1.5

where

dy x = 2 . dx (x −1)1/2 So 2.5

1/2

⌠ ⎛ 2x 2 −1 ⎞⎟ ⎟ dx . L = 2⎮ ⎜⎜⎜ 2 ⎮ ⎜⎝ x −1 ⎟⎟⎠ ⌡ 1.5

To evaluate L using a 4-interval rule, we need the following values of the integrand fn :

S4.14

n xn

0 1.50

1 1.75

2 2.00

3 2.25

4 2.50

fn

1.6733

1.5763

1.5275

1.4987

1.4800

Then, using the trapezium rule (4.34) give L = 3.0896 and using Simpson’s rule (4.37) gives L = 3.0847 . (An evaluation precise to four decimal places is 3.0844.) 4.25

From Table 4.2, 0.5

0.5

I ≡ ∫ (1 − x 2 )−1/2 dx = ⎡⎢sin−1 x ⎤⎥ = π 6 , ⎣ ⎦0 0

so the value of π may be estimated from 6I . Using n = 2 and n = 4 , we find the following values of the integrand fn :

n=2

fn

n=4

fn

x 0 = 0.00

6.00000

x 0 = 0.000

6.00000

x1 = 0.25

6.19677

x1 = 0.125

6.04743

x 2 = 0.50

6.92820

x 2 = 0.250

6.19677

x 3 = 0.375

6.47232

x 4 = 0.500

6.92820

Using Simpson’s rule (4.37) then gives the estimates π = 3.14294 for n = 2 and 3.14170 for n = 4 . So four intervals are needed to satisfy the requirement. *4.26

Using the result of Problem 4.23, we have

⎡ ⎛ ⎞2 ⎤ ⎢ ⎜ dy ⎟⎟ ⎥ ⎢1 + ⎜⎜ ⎟⎟ ⎥ ⎢ ⎜⎝ dx ⎠ ⎥ ⎣ ⎦

1/2

1/2

⎛ 1 ⎞⎟ ⎟ = ⎜⎜⎜ ⎜⎝1 − x 2 ⎟⎟⎠

and

⎡ ⎛ ⎞2 ⎤ dy ⎥ ⎢ y ⎢1 + ⎜⎜⎜ ⎟⎟⎟ ⎥ ⎜ ⎢ ⎝ dx ⎟⎠ ⎥ ⎣ ⎦

1/2

= 1.

So from (4.55), 1

A= ∫ 0

⎡ ⎛ ⎞2 ⎤ dy ⎥ ⎢ 2πy ⎢1 + ⎜⎜⎜ ⎟⎟⎟ ⎥ ⎢ ⎜⎝ dx ⎟⎠ ⎥ ⎣ ⎦

1/2

dx = 2π .

Also, from (4.54), 1

1

V = ∫ πy 2dx = π ∫ (1 − x 2 )dx 0

0

1

= π ⎡⎢x − 13 x 3 ⎤⎥ = 2π 3. ⎣ ⎦0

S4.15

Finally, the total surface area is obtained by adding to A the area of the circular end of radius 1, so that S = π + 2π = 3π . *4.27

From (4.54), we have a

⌠⎛ x2 ⎞ V = ∫ πy 2 dx = πb 2 ⎮ ⎜⎜⎜1 − 2 ⎟⎟⎟ dx ⌡ ⎜⎝ a ⎟⎠ −a a

−a

a

⎡ x3 ⎤ 4 = πb 2 ⎢⎢x − 2 ⎥⎥ = πab 2 . 3a ⎥⎦ −a 3 ⎢⎣ From (4.55) we have a

1/2

⎡ ⎛ ⎞2 ⎤ ⌠ dy ⎥ ⎢ ⎮ A = ⎮ 2πy ⎢1 + ⎜⎜⎜ ⎟⎟⎟ ⎥ dx . ⎜ ⎢ ⎝ dx ⎟⎠ ⎥ ⎮ ⌡ ⎣ ⎦ −a

Substituting 1/2

⎛ x2 ⎞ y = b ⎜⎜⎜1 − 2 ⎟⎟⎟ ⎝⎜ a ⎟⎠

with

dy bx 1 =− 2 2 dx a (1 − x a 2 )1/2

gives, after some algebra, a

1/2

⌠ ⎛ e 2x 2 ⎞ A = 2πb ⎮ ⎜⎜⎜1 − 2 ⎟⎟⎟ dx . ⎮ ⎜⎝ a ⎟⎠ ⌡ −a

Substituting

x=

a dx a sin θ with = cos θ e dθ e

gives θ(x = ±a) = ± sin−1 e , and using 1 − sin 2 θ = cos2 θ gives sin−1 e

sin−1 e

2πba πba 2 A= ∫−1 cos θ dθ = e ∫−1 (1 + cos 2θ) dθ e −sin e −sin e sin−1 e

πba ⎡⎢ sin 2θ ⎤⎥ = θ+ . e ⎢⎣ 2 ⎥⎦ −sin−1 e But sin 2θ = 2sin θ cos θ = 2sin θ(1 − sin 2 θ)1/2 , therefore sin−1 e πba ⎡ 2 1/2 ⎤ θ + sin θ(1 − sin θ) ⎦⎥ −sin−1 e e ⎣⎢ 2πba −1 2πba = sin e + e(1 −e 2 )1/2 . e e

A=

S4.16

.

Since (1 −e 2 )1/2 = b a , this gives

A = 2πb 2 +

2πab −1 sin e e

as required. *4.28

(a) The figure below shows the plate oriented with its y axis as the axis of rotation. Then, if σ = m a 2 is the mass density per unit area, the contribution from the strip shown is

dI = σa dx x 2 , so that a/2

I = σa

∫

x 2 dx =

−a/2

σa 4 1 = ma 2 . 12 12

(b) The figure below shows the plate oriented so that the y axis is the axis of rotation. Then the contribution from the strip shown is

dI = σ2y dx x 2 , where y = a

2 − x , so that a/ 2

a/ 2

⎛a ⎞ ⎞ ⌠ ⌠ ⎛a I = ⎮ 2σx 2 ⎜⎜⎜ − x ⎟⎟⎟ dx = 4σ ⎮ x 2 ⎜⎜⎜ − x ⎟⎟⎟ dx ⎟⎠ ⎟⎠ ⎜⎝ 2 ⌡ ⌡ ⎜⎝ 2 0

−a/ 2 a/ 2

⎡ a x3 x4 ⎤ = 4σ ⎢⎢ − ⎥⎥ 4 ⎥⎦ 0 ⎢⎣ 2 3

=

σa 4 1 = ma 2 . 12 12

S4.17

*4.29

The disk is shown in the figure below. The mass of the annulus is σ2πx dx , where

σ = M πr 2 is the mass density per unit area.

The contribution to the moment of inertia from the annulus is

dI = 2πσx 3 dx and so summing over such contributions gives the total moment of inertia of the disk as r

I = 2πσ ∫ x 3 dx = 0

*4.30

πσr 4 Mr 2 . = 2 2

To find the moment of inertia of a cone about its axis, we consider the thin disk shown in the figure below.

S4.18

The radius of the disk is r = xR h and its mass is ρπr 2 dx , where ρ is the mass density per unit volume. The moment of inertia of the disk is obtained using the result from Problem 4.24 and is

dI =

1 ⎛⎜ x 2R 2 ⎞⎟⎟ ⎛⎜ x 2R 2 ⎞⎟⎟ ρπ ⎜ ⎟ dx ⎜ ⎟. 2 ⎜⎜⎝ h 2 ⎟⎠ ⎜⎜⎝ h 2 ⎟⎠

Summing over the contributions from all the disks gives

I=

ρπR 4 2h 4

h

∫

x 4 dx =

o

ρπR 4h . 10

Finally, using (4.56a), the mass of the cone is

M=

ρπR 2h , 3

so that the moment of inertia of the cone about its axis of symmetry is I = 3MR 2 10 .

S4.19

SOLUTIONS TO PROBLEMS 5 5.1

The series is N

S = 23 + 25 + 27 ++ 771 = ∑ (23 + 2n) , n=0

where

23 + 2N = 771, i.e. N = 374.

This is an arithmetic series, and in the notation of (5.7), a = 23, x = 2 . The sum S is therefore, from (5.8),

S = 12 (N + 1)(2a + Nx) = 148875 . 5.2

Writing out the terms, we have N ⎞ ⎛ 2r 2 ⎞⎟ ⎛ 3⎞ ⎛ ⎛r ⎞ ⎟⎟ + ln ⎜⎜ 3r ⎟⎟⎟ ++ ln ⎜⎜ Nr ⎟⎟⎟ SN = ln ⎜⎜⎜ ⎟⎟⎟ + ln ⎜⎜⎜ ⎜⎜ 4 ⎟ ⎜⎜ N + 1 ⎟ ⎜⎝ 2 ⎟⎠ ⎜⎝ 3 ⎟⎠ ⎝ ⎠ ⎝ ⎠ Z ⎛ r ⎞⎟ ⎟⎟, = ln ⎜⎜⎜ ⎜⎝ N + 1 ⎟⎠

where

Z = 1 + 2 + 3 + + N = 12 N(N + 1) is an arithmetic series. Hence

⎛ r N (N +1)/2 ⎞⎟ ⎟⎟ . SN = ln ⎜⎜⎜ ⎜⎝ N + 1 ⎟⎠ If r > 1 , the term in brackets diverges and hence SN diverges as N → ∞ . If r ≤ 1 , the term in brackets tends to zero as N → ∞ , so that again SN diverges. Hence there are no values of r for which the infinite series converges. 5.3

The series can be rewritten in the form N

SN = ∑ e −x/2(e −x )n , n=0

i.e. as a geometric series of the form (5.9). The sum is given from (5.10) as

e −x/2 ⎡⎢1 −(e −x )N +1 ⎤⎥ ⎣ ⎦ SN = 1 −e −x and as N → ∞ ,

SN →

e −x/2 , 1 −e −x

S5.1

and so is convergent. 5.4

Explicitly,

SN = a + (a + x)y + (a + 2x)y 2 ++ (a + Nx)y N and

ySN = ay + (a + x)y 2 + (a + (N −1)x)y N +1 ++ (a + Nx)y N +1 , so that

(1 −y)SN = a + xy + xy 2 ++ xy N −(a + Nx)y N +1 = a −(a + Nx)y N +1 + xy(1 + y ++ y N −1 ) = a −(a + Nx)y N +1 +

xy(1 −y N ) (1 −y)

using (5.10). Hence

SN =

a −(a + Nx)y N +1 xy(1 −y N ) + , (1 −y) (1 −y)2

and if y < 1 ,

lim SN =

N →∞

a xy , + 1 −y (1 −y)2

as required. 5.5

Writing the series as

∞

∑u n=0

n

and rn =

un+1 un

in each case, we have the following.

(a)

rn =

1 [2 + 3(n + 1)2 ] 1 → < 1 as n → ∞, 2 3 3 2 + 3n

so by d’Alembert’s test the series converges. (b) 2

⎛ n ⎞⎟ ⎟ → 2 > 1 as n → ∞ , rn = 2 ⎜⎜⎜ ⎜⎝ n + 1 ⎟⎟⎠ so by d’Alembert’s test the series diverges. (c)

in this case

rn → 1 , so d’Alembert’s test cannot be used. However,

un → 1 4 ≠ 0 as n → ∞ , so the series diverges by (5.15). (d) r

1 ⎛⎜ n + 1 ⎞⎟ ⎟⎟ → 0 as n → ∞ , rn = ⎜ (n + 1) ⎜⎜⎝ n ⎟⎠ so by d’Alembert’s test the series converges.

S5.2

5.6

(a)

un+1 un

=

(n + 1)2 x → x as n → ∞ , n2

and so the series converges for x < 1 . (b)

un+1 un

⎛ n + 3 ⎞⎟ ⎟ → 2(x −1) as n → ∞ , = 2(x −1)⎜⎜⎜ ⎜⎝ n + 4 ⎟⎟⎠

and so the series converges for 2(x −1) < 1 , i.e. for 1 2 < x < 3 2 . (c) un+1 un = 2x . So the series converges if 2x < 1 , i.e. x ln 2 < 0 or all x < 0 . 5.7

In each case, the functions are of the form f g , with f → 0 and g → 0 in the limit. So we use l’Hôpital’s rule

⎡ f (x) ⎤ ⎡ ′ ⎤ ⎥ = lim ⎢ f (x) ⎥ . lim ⎢ ⎢ ⎥ x→1 g(x) x→1 ⎢ g ′(x) ⎥ ⎣ ⎦ ⎣ ⎦ (a)

⎡ 4 ⎤ ⎡ f ′(x) ⎤ ⎥ = lim ⎢ 5x + 6x ⎥ = 11 . lim ⎢ ⎥ x→1 ⎢ g ′(x) ⎥ x→1 ⎢ 2 ⎢⎣ 2x ⎥⎦ ⎣ ⎦

(b)

⎡ f ′(x) ⎤ ⎡ ⎤ ⎥ = lim ⎢ π cos πx ⎥ = −π . lim ⎢ ⎥ x→1 ⎢ g ′(x) ⎥ x→1 ⎢ ⎣ ⎦ ⎣ 1x ⎦

(c) We need to differentiate g = arcsin(x −1) , i.e. x −1 = sin g . This gives

dx dg = cosg = (1 − sin 2 g = 2x − x 2 and so

g′ = 1

(2x − x 2 ) → 1 as x → 1 .

Therefore,

⎡ 2 ⎤ ⎡ arcsin x ⎤ ⎥ = lim ⎢⎢ 1 2x − x ⎥⎥ = −1 . lim ⎢ x→1 ⎢ sinh(1 − x) ⎥ x→1 −cosh(1 − x) ⎢⎣ ⎥⎦ ⎣ ⎦ (d)

5.8

⎡ f ′(x) ⎤ ⎡ ⎤ ⎥ = lim ⎢ −π sin πx ⎥ = 1 lim ⎢ x→1 ⎢ g ′(x) ⎥ x→1 ⎢ 2π sin πx cos πx ⎥ ⎣ ⎦ ⎣ ⎦ 2

(a) Writing the functions in the form f g , we have

S5.3

f ′ = 12 (x + 5)−1/2 → 1 (2 5)

f = x + 5 − 5 → 0, g = x → 0, g ′ = 1 → 1. Therefore, by l’Hôpital’s rule,

⎡ x +5 − 5⎤ ⎥ = lim ⎡⎢ f ′ ⎤⎥ = 1 . lim ⎢⎢ ⎥ x→0 ⎢ g ′ ⎥ x→0 x ⎢⎣ ⎥⎦ ⎣ ⎦ 2 5 (b) Similarly,

2 ln(1 + x) → 0, 1+x x g ′ = arcsin x + → 0, 1−x2 f′=

f = ln 2(1 + x) → 0, g = x arcsin x,

using equation (3.27). So we must use (5.29a), where

2 2 ln(1 + x) − → 2 as x → 0, 2 (1 + x) (1 + x)2 2 1 x g ′′ = − → 2 as x → 0. 2 1/2 2 (1 − x 2 )3/2 (1 − x ) f ′′ =

Therefore

⎡ ln 2(1 + x) ⎤ ⎥ = 1. lim ⎢⎢ x→0 x arcsin x ⎥ ⎢⎣ ⎥⎦ 5.9

If f (x) = cosx , then for all n ≥ 0 ,

f (2n)(x) = (−1)n cosx ,

f (2n+1)(x) = (−1)n+1 sin x

by (3.42). Hence on substituting in (5.37b), we have f (N +1)(x) ≤ 1 for all N and

RN ≤

(x − π 4)N +1 →0 (N + 1)!

as N → ∞ for all x by (5.20). Hence the Taylor series is valid for all x. Substituting for the derivatives in (5.37a) using cos(π 4) = sin(π 4) = 1

2 gives

2n 2n+1 ⎤ ⎡ (−1)n (x − π 4) (−1)n+1(x − π 4) ⎥ ⎢ cos(x − π 4) = ∑ ⎢ + ⎥ (2n)! ⎥ n=0 ⎢ 2 2(2n + 1)! ⎣ ⎦ 1 (x − π 4) (x − π 4)2 (x − π 4)3 = − − + + 2 2 2 2 6 2 ∞

5.10

Taylor’s theorem (5.21) gives

S5.4

(−1)n x 2n+1 + RN n=0 (2n + 1)! N

sin x = ∑ for x 0 = 0, x = h , where

RN =

x 2N +2 (−1)N +1 sin(θx) , (2N + 2)!

0 < θ 1 , so that

(cosh α)n → ∞ as n → ∞ n and the series diverges by (5.15). Thus (c) is conditionally convergent for α = 0 . 5.23

(a)

1 → 0 as n → ∞ , ln(αn)

so that the series converges by (5.61). However,

1 1 > ln(αn) n for large n, and, as shown in Example 5.12, ∞

1

∑n n=1

diverges. Hence ∞

1

∑ ln αn n=1

diverges by the comparison test and therefore the series is only conditionally convergent. (b)

(n + α) 1 → ≠ 0 as n → ∞ , (2αn + 3) 2α

so the series does not converge. n

(c)

(ln αn)n ⎛⎜ ln α + ln n ⎞⎟ ⎟⎟ → 0 as n → ∞ , = ⎜⎜ ⎟⎠ ⎜⎝ n n/2 n

so that the series converges. However, for large n, n

⎛ ln α + ln n ⎞⎟ 1 ⎜⎜ ⎟⎟ > , ⎟ ⎜⎜⎝ n ⎠ n

S5.12

so that the series is conditionally convergent. (d) By d’Alembert’s ratio test,

rn =

un+1 un

=

(n + 2)α → 0 as n → ∞ , (n + 1)2

so that the series is absolutely convergent.

S5.13

SOLUTIONS TO PROBLEMS 6 6.1

(a) z1 + z 2 − z 3 = −(1 + 2i) , (b) (z1 − z 2 )* + z 3 = −5(1 + 2i) , (c) (z1 + z 3 )(z 2 + z 3 ) = 6(5 + 7i) , (d) (z 2 + z1 )(z 3 − z 2 )* = 4(3 − 4i) , (e)

z2 * 1 3

zz

=

2 − 3i (2 − 3i)(16 −13i) 7 74 = = + i, −16 −13i −(16 + 13i)(16 −13i) 425 425

(f)

z1 + z 2

=

z1 + z 3

3+i (3 + i)(5 − 7i) 11 8 = = − i. 5 + 7i (5 + 7i)(5 − 7i) 37 37

(g) z1 = 12 + 4 2 = 17 and using z −1 = z ∗ z 6.2

2

gives z1−1 = 1 17 −(4 17)i .

Let z1 = x1 + y1 and z 2 = x 2 + y 2 . (a)

(z1z 2 ) = (x1x 2 − y1y 2 ) + i(x1y 2 − x 2y1 ) and (z1z 2 )* = (x1x 2 − y1y 2 )− i(x1y 2 − x 2y1 ) .

But

z1*z 2* = (x1 − iy1 )(x 2 − iy 2 ) = (x1x 2 − y1y 2 )− i(x1y 2 − x 2y1 ). (b) 2

z1z 2 = (x1x 2 − y1y 2 )2 + (x1y 2 − x 2y1 )2 = x12x 22 + y12y 22 + x12y 22 + x 22y12 . But,

(z

1

z2

)

2

= (x12 + y12 )(x 22 + y 22 ) = x12x 22 + y12y 22 + x12y 22 + x 22y12 .

(c) 2

z1 + z 2 = (x1 + x 2 )2 + (y1 + y 2 )2 = x12 + x 22 + 2x1x 2 + y12 + y 22 + 2y1y 2 .

(A)

But,

(z

1

+ z2

) = ⎡⎢⎣(x 2

2

+ y12 )1/2 + (x 22 + y 22 )1/2 ⎤⎥ ⎦ = x12 + x 22 + y12 + y 22 + 2(x12 + y12 )1/2(x 22 + y 22 )1/2 . 2 1

So the inequality is true if (A)−(B) ≤ 0 , i.e.

x12x 22 + y12y 22 −(x12 + y12 )1/2(x 22 + y 22 )1/2 ≤ 0, or, on squaring,

(x1x 2 + y1y 2 )2 ≤ (x12 + y12 )(x 22 + y 22 ),

S6.1

(B)

i.e.

2x1x 2y1y 2 ≤ x12y 22 + x 22y12 ⇒ (x1y 2 − x 2y1 )2 ≥ 0, which is always true. (d) z1 = z1 − z 2 + z 2 , which using (c) implies z1 ≤ z1 − z 2 + z 2 , so that

z1 − z 2 ≥ z1 − z 2 . 6.3

6.4

(a)

(1 + 2i)(3 − 2i) 5 + 4i 5 + 4i (5 + 4i)(18 + i) 86 77 = = = = + i, 2 (18 −i)(18 + i) 325 325 (2 + i) (2 − 3i) (3 + 4i)(2 − 3i) 18 −i

(b)

(2 + i) (2 + i) (2 + i)(2 + 4i) 3 = = = i, (1 −i)(3 −i) (2 − 4i) (2 − 4i)(2 + 4i) 10

(c)

(3 + i)2 6 + 6i 6(1 + i)(7 + i) 18 24 = = = + i. (2 −i)(3 + i) 7 −i (7 −i)(7 + i) 25 25

The general polynomial equation is

anz n + an−1z n−1 + an−2z n−2 ++ a 0 = 0 , and taking its complex conjugation gives

(anz n + an−1z n−1 + an−2z n−2 ++ a 0 )* = 0 . Then using the result (z1z 2 )* = z1*z 2* and the fact that the coefficients are real, gives

an (z * )n + an−1(z * )n−1 + an−2(z * )n−2 ++ a 0 = 0 , which shows if z is a root then so is its complex conjugate z * . 6.5

i = x + iy ⇒ i = x 2 + 2ixy − y 2 and by equating real and imaginary

(a) Setting

parts, x 2 −y 2 = 0 and 2xy = 1 . Thus, x = y = ±1 modulus

is

therefore

r = 12 + 12

2 and

2 =1

and

arg z = arctan(y x) = arctan(1) , i.e. arg z = π 4 . (b) Setting z = (1 + i) ( 3 − i) , gives

r = z = (1 + i) ( 3 −i) = 2 2 = 1 and

S6.2

2

i = ±(1 + i) its

2 . The

argument

is

arg z = arg(1 + i)− arg( 3 − i) = π 4 + π 6 = 5π 12 . (c) Setting z = (1 + i)( 3 − i) , gives

r = z = (1 + i) ( 3 −i) = 2 2 and

arg z = arg(1 + i) + arg( 3 − i) = π 4 − π 6 = π 12 . 6.6

(a)

z −1 = (z −1)(z * −1) = (x + iy −1)(x − iy −1) = (x −1)2 + y 2 = 2, which is the equation of a circle centre (1,0) and radius 2. (b) z + 1 = (x + 1)2 + y 2 and z −i = x 2 + (y −1)2 . Squaring both sides gives

x 2 + y 2 + 2x + 1 = x 2 + y 2 − 2y + 1 , with solution x = −y , which is a straight line passing through (0, 0) with gradient

−1 . 6.7

(a) The ratio of successive terms is

Rn+1 Rn

=

(n + 1) (z − 3i) , 2(2n + 1)

and as n → ∞ , this gives a radius of convergence z − 3i < 4 , i.e. a circle with radius 4 centred on (0, 3) . (b) The ratio of successive terms is

Rn+1 Rn

=

(n + 4)2 (2iz)n , (n + 3)2

and as n → ∞ , this gives a radius of convergence z < 1 2 , i.e. a circle with radius

1 2 centred on (0, 0) . (c) The ratio of successive terms is

Rn+1 Rn

=

−z , (n + 1)

S6.3

and as the modulus of this ratio is always less than 1 as n → ∞ , the series converges for all values of z. 6.8

(a) Using Euler’s formula,

e iπ/2 + 2e iπ/4 = i + 2[cos(π 4) + i sin(π 4)] = 1 + 2i. Therefore z = 5 and arg(z) = tan−1 2 = 1.11 rad . (b)

(1 + i)e iπ/6 = (1 + i)[cos(π 6) + i sin(π 6)] = (1 + i)( 3 + i) 2 =( 3 −1) 2 + i ( 3 + i) 2. Hence z = 2 and arg(z) = tan−1[( 3 + 1) ( 3 −1)] = 1.31 rad . (c)

(2 + i) iπ 3 (1 + i) iπ 3 e =− e (i − 3) 2 ⎛ π ⎞⎤ (1 + i) ⎡⎢ ⎛⎜ π ⎞⎟ (1 + i) ⎟⎟ + i sin ⎜⎜ ⎟⎟⎟⎥ = − =− cos (1 + i 3) ⎜ ⎢ ⎥ ⎜ ⎜ ⎜⎝ 3 ⎟⎠⎥ 2 ⎢⎣ ⎜⎝ 3 ⎟⎠ 4 ⎦ 1 = ⎡⎢(1 − 3) + i(1 + 3)⎤⎥ . ⎦ 4⎣

z=

Therefore, z = 1 4 and arg(z) = tan−1 (1 − 3)2 + (1 + 3)2 = −0.939 rad . 6.9

(a) Write

⎛ 3 ⎞⎟ 3(1 − 2i) ⎜⎜ ⎟⎟ = = re iθ , ⎜⎜ ⎟ ⎟ 1 + 2i (1 + 2i)(1 − 2i) ⎜⎝ ⎠ from which, r = 3 5 and θ = arctan(−2) = −1.107 rad . So 4

⎛ 3 ⎞⎟ ⎜⎜ ⎟⎟ = 9 e 4iθ = 9 e −4.428i = −0.101 − 0.346i . ⎜⎜ ⎟⎟ 1 + 2i 25 25 ⎜⎝ ⎠ (b) First write each factor in exponential form. This gives

S6.4

(2 + 3i) : r1 = 13, θ1 = arctan(3 2) = 0.983; (1 − 4i) : r2 = 17, θ2 = arctan(−4) = −1.3260; (1 − 2i) : r3 = 5, θ3 = arctan(−2) = −1.107; (3 + 2i) : r4 = 13, θ4 = arctan(2 3) = 0.588; (4 + i) : r5 = 17, θ5 = arctan(1 4) = 0.245. Hence,

rr (2 + 3i)(1 − 4i)* = 1 2 exp[i(θ1 − θ2 + θ3 − θ4 − θ5 )] * (1 − 2i) (3 + 2i)(4 + i) r3r4r5 = 0.447e 0.369i = 0.417 + 0.161i . (c) Write (1 − 2i) = re iθ . Then, r = 5 and θ = arctan(−2) = −1.107 rad , and

⎡ i(θ + 2πk) ⎤ ⎥ , k = 0,1 . (1 − 2i)1/2 = r 1/2 exp ⎢ ⎢ ⎥ 2 ⎣ ⎦ So the two roots are

k = 0 : 1.272 − 0.786i, 6.10

k = 1 : −1.272 + 0.786i .

(a) Write

⎛ i ⎞⎟ i(1 −i) i iθ ⎜⎜ ⎟ ⎜⎜⎝1 + i ⎟⎟⎠ = (1 + i)(1 −i) = 2 (1 −i) = re , from which r = 1

2 and θ = arctan(1) = π 4 , so that 10

10 ⎛ 1 ⎞⎟ 5iπ/2 ⎛ i ⎞⎟ ⎜ ⎟ ⎜⎜ ⎟ = 0.0313i. ⎜⎜⎝1 + i ⎟⎟⎠ = ⎜⎜⎜⎝ ⎟⎟⎠ e 2

(b) Using the same method as in 6.9(c), we write (6 + 3i) = re iθ . Then,

r = 45 = 6.708 and θ = arctan(1 2) = 0.4636 rad , and

⎡ i(θ + 2πk) ⎤ ⎥ , k = 0,1,…, 2 . (6 + 3i)1/3 = r 1/3 exp ⎢ ⎢ ⎥ 3 ⎣ ⎦ So the three roots are

k = 0 : 1.864 + 0.290i , k = 1 : −1.183 + 1.468i ,

(c)

k = 2 : − 0.680 −1.758i .

2 2(i + 1) (i + 1) π = =− = re iθ , where r = 1 and θ = arctan(1) = . So, i −1 (i −1)(i + 1) 4 2

S6.5

6

⎛ 2 ⎞⎟ ⎜⎜ ⎟⎟ = e iπ/4 ⎜⎜ ⎜⎝ i −1 ⎟⎟⎠

(

6.11

)

6

= e 1.5iπ = −i .

(a) Using the same method as in 6.9(c), we write i = re iθ . Then,

r = 1 and θ = arctan(1) = π 2 rad , and

⎡ i(π 2 + 2πk) ⎤ ⎥ , k = 0,1,…, 4 . i 1/5 = exp ⎢ ⎢ ⎥ 5 ⎣ ⎦ So the five roots are

k = 0 : 0.951 + 0.309i , k = 1 : i , k = 2 : − 0.951 + 0.309i , k = 3 : − 0.588 − 0.809i , k = 4 : 0.588 − 0.809i . (b) First write each factor in exponential form. This gives

(2 + i) : r1 = 5, θ1 = arctan(1 2) = 0.464; (1 −i) : r2 = 2, θ2 = arctan(−1) = −0.785; (1 + i) : r3 = 2, θ3 = arctan(1) = 0.785; (3 − 4i) : r4 = 5, θ4 = arctan(−4 3) = −0.927. Hence,

rr (2 + i)(1 −i)* = 1 2 exp[i(θ1 − θ2 + θ3 − θ4 )] * (1 + i) (3 − 4i) r3r4 1

=

(c)

3i 1+ 2

=

5

e 1.391i = 0.080 + 0.440i .

3i( 2 − i) ( 2 + i)( 2 − i)

=

i( 2 − i) 3

= re iθ ,

where r = 1 and θ = arctan( 2) = 0.955 rad . So 7

⎛ 3i ⎞⎟ ⎜⎜ ⎟⎟ = e 0.955i ⎜⎜ ⎜⎝ i + 2 ⎟⎟⎠

(

6.12

)

7

= 0.920 + 0.391i .

(a) Using equation (2.61b), arcsinh(z) = ln(z + z 2 + 1) , gives

S6.6

⎡ ⎛ 1 ⎞⎤ i 3) = ln ⎢⎢ + ⎜⎜⎜− + 1⎟⎟⎟ ⎥⎥ ⎟⎠ ⎥ ⎜⎝ 3 ⎢ 3 ⎣ ⎦ ⎡ 1 ⎤ = ln ⎢⎢ 2 + i ⎥⎥ = −ln 3 + ln( 2 + i). ⎣ 3 ⎦

arcsinh(i

(

)

Now in polar form ( 2 + i) = 5e 0.615i , so

(

)

arcsinh i

3 = −ln 3 + ln

( 5) cos(0.615) + i ln ( 5) sin(0.615)

= 0.1080 + 0.4643i. (b) Using Euler’s formula,

sin(π − i ln 2) =

1 ⎡ (π−i ln 2) i e −e −(π−i ln 2) ⎤⎥ = ⎡⎢e ln 2 −e −ln 2 ⎤⎥ . ⎢ ⎣ ⎦ ⎦ 2i 2⎣

(c) In polar form,

1 + i : r = 2, θ = arctan(1) = 0.7853 − i and

3 − i : r = 10, θ = arctan(−1 3) = −0.322. So, (1 + i) (3 − i) = (1 6.13

5)e 1.007i and ln[(1 + i) (3 −i)] = −0.805 + 1.007i .

(a) (2i)1+i = exp[(1 + i)ln(2i)] , where from (6.23), ln i = ln1 + i(π 2 ± 2nπ) . If we use the principal value with n = 0 , then

(2i)1+i = exp{(1 + i)[ln(2) + iπ 2]} = exp[(ln 2 − π 2) + i(ln 2 + π 2)] = exp[−0.878 + 2.264i] = e −0.878[cos 2.264 + i sin 2.264] = −0.266 + 0.320i.

(b) Firstly,

1+i 2

1 2 2 =− + i = re iθ , where 3 3 1 −i 2 r = 1 and θ = arctan(−2 2) = −1.231 rad .

Secondly,

sin i =

1 −1 e −e = 1.175i. 2i

(

)

So, sini

⎛1 + i 2 ⎞⎟ ⎜⎜ ⎟⎟ ⎜⎜ ⎜⎝ 1 −i 2 ⎟⎟⎠

(

= e −1.231

)

1.175i

= 4.248.

(c) Set z = ln[(1 −i) ( 2 + i)] and use cos(iz) = 12 (e −z +e z ) to give

S6.7

⎡ ⎛ 2 + i ⎞⎤ ⎡ ⎤⎫ ⎪ 1⎧ ⎟⎟⎥ + exp ⎢ ln ⎛⎜⎜ 1 −i ⎞⎟⎟⎥ ⎪ ⎪ cos(iz) = ⎪ ⎨exp ⎢⎢ ln ⎜⎜⎜ ⎟⎟⎥ ⎢ ⎜⎜⎝ 2 + i ⎟⎟⎠⎥ ⎬ ⎜ 2⎪ 1 −i ⎝ ⎠ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎪ ⎪ ⎪ ⎩ ⎭ ⎛ ⎞ ⎛ ⎞ 1 ⎜ 2 + i 1 −i ⎟ 1 ⎜ 3 + 2i ⎟ ⎟⎟ = ⎜ ⎟⎟. = ⎜⎜ + 2 ⎝⎜ 1 −i 2 + i ⎠⎟ 2 ⎜⎜⎝ 3 −i ⎟⎠ Finally, rationalising gives

⎡ ⎛ 1 −i ⎞⎤ ⎟⎟⎥ = 7 + 9 i. cos ⎢⎢i ln ⎜⎜⎜ ⎟⎥ ⎜ ⎢⎣ ⎝ 2 + i ⎟⎠⎥⎦ 16 16 6.14

(a) Using de Moivre’s theorem,

(cos 2θ + i sin 2θ)3(cos 3θ + i sin 3θ)2 = (e 2iθ )3(e 3iθ )2 = e 12iθ = cos(12θ) + i sin(12θ). (b) From (6.26), 7

⎛ 1⎞ (2i sin θ) = ⎜⎜⎜z − ⎟⎟⎟ ⎜⎝ z ⎟⎠ 7

35 21 7 1 − 3+ 5− 7 z z z z ⎛ 7 1 ⎞⎟ ⎛ 5 1 ⎞⎟ ⎛ 3 1 ⎞⎟ ⎛ 1⎞ = ⎜⎜⎜z − 7 ⎟⎟ − 7 ⎜⎜⎜z − 5 ⎟⎟ + 21⎜⎜⎜z − 3 ⎟⎟ − 35 ⎜⎜⎜z − ⎟⎟⎟. ⎜⎝ ⎜⎝ ⎜⎝ ⎜⎝ z ⎟⎠ z ⎟⎠ z ⎟⎠ z ⎠⎟ = z 7 − 7z 5 + 21z 3 − 35z +

Then again using (6.26),

(2i sin θ)7 = 2i sin 7θ − 7(2i sin 5θ) + (2i sin 3θ)− 35(2i sin θ) and finally

sin 7 θ = 6.15

1 64

(35sin θ − 21sin 3θ + 7 sin 5θ − sin 7θ) .

(a) Using de Moivre’s theorem,

cos 2θ −i sin 2θ = cos(−2θ) + i sin(−2θ) = e −2iθ and

cos5θ + i sin 5θ = e 5iθ , so that

cos 2θ − i sin 2θ e −2iθ = 5iθ = e −7iθ = cos(7θ)− i sin(7θ). cos5θ + i sin 5θ e (b) From de Moivre’s theorem,

S6.8

cos 4θ + i sin 4θ = (c + is)4 = c 4 + i4c 3s + 6i 2c 2s 2 + 4i 3cs 3 + i 4s 4 = (c 4 − 6c 2s 2 + s 4 ) + i(4c 3s − 4i 3cs 3 ), where c = cos θ and s = sin θ . Now cos 4θ = Re[(c + is)4 ] = c 4 − 6c 2s 2 + s 4 , and

sin 4θ = Im[(c + is)4 ] = 4c 3s − 4cs 3 , So, finally,

tan 4θ = *6.16

4c 3s − 4cs 3 4t(1 −t 3) = . c 4 − 6c 2s 2 + s 4 1 − 6t 2 + t 4

By Euler’s formula, ∞

I = Re ∫ xe −(1+2i )x dx , 0

which may be integrated by parts as follows. ∞

∞

∫ xe

−(1+2i )x

0

∞ −(1+2i )x ⎡ ⎤ x e −(1+2i )x ⎥ ⎢ dx = − e +∫ dx ⎢ (1 + 2i) ⎥ 1 + 2i ⎣ ⎦0 0 2 ∞ 1 ⎡e −(1+2i )x ⎤ = (1 − 2i) . =− ⎥⎦ 0 25 (1 + 2i)2 ⎢⎣

So I = −3 25. *6.17

In summation notation,

⎡ ∞ 2n ⎤ 2n sin(nx) = Im ⎢⎢ ∑ e inx ⎥⎥ . n=0 n ! ⎢⎣ n=0 n ! ⎥⎦ ∞

S(x) = ∑ Using 22 = e ln 2 gives,

⎡ ∞ (e y )n ⎤ ⎥, S(x) = Im ⎢⎢ ∑ ⎥ n ! ⎢⎣ n=0 ⎥⎦ where y = ln 2 + ix . Now using the exponential expansion

yn n=0 n ! ∞

ey = ∑ gives

{

}

S(x) = Im exp ⎡⎢e ln 2e ix ⎤⎥ = Im {exp[2(cosx + i sin x)]} ⎣ ⎦ 2cosx =e sin(2sin x). *6.18

Using the binomial theorem and Euler’s formula,

S6.9

.

n ⎛ n ⎛ n ⎛ ⎞ ⎞ ⎞ (1 +e ix )n = ∑ ⎜⎜⎜ n ⎟⎟⎟e ikx = ∑ ⎜⎜⎜ n ⎟⎟⎟ coskx + i ∑ ⎜⎜⎜ n ⎟⎟⎟ sin kx. ⎝ k ⎟⎠ ⎝ k ⎟⎠ ⎝ k ⎟⎠ k=0 ⎜ k=0 ⎜ k=0 ⎜

Similarly, the left-hand side is given by

(1 +e ix )n = (1 + cosx + i sin nx)n = 2n [cos(x 2)]n [cos(x 2) + i sin(x 2)]n = 2n [cos(x 2)]n e inx/2 = 2n [cos(x 2)]n [cos(nx 2) + i sin(nx 2)], where we have used trigonometric identities together with de Moivre’s theorem. Equating the imaginary parts of both expressions, we finally obtain

⎛ n ⎞⎟ ⎜⎜ ⎟⎟ sin kx = 2n cosn (x 2)sin(nx 2). ∑ ⎜ ⎟⎠ ⎜ k ⎝ k= n

S6.10

SOLUTIONS TO PROBLEMS 7 7.1

(a)

∂f ∂f = 2xy + 2x . Hence = 3x 2 + y 2 + 2y + 6x and ∂y ∂x ∂ ⎛⎜ ∂f ⎞⎟ ∂ ⎛⎜ ∂f ⎞⎟ ⎜⎜ ⎟⎟ = 2y + 2 = ⎜ ⎟⎟ . ∂x ⎜⎝ ∂y ⎟⎠ ∂y ⎜⎜⎝ ∂x ⎟⎠

(b) Using the quotient rule (3.21),

∂f xy(2x)−(x 2 + y 2 )y 1 y = = − 2 ∂x y x x 2y 2

and

∂f 1 x = − 2 ∂y x y

Hence,

∂ ⎛⎜ ∂f ⎞⎟ 1 1 ∂ ⎛⎜ ∂f ⎞⎟ ⎜⎜ ⎟⎟ = − 2 − 2 = ⎜ ⎟⎟ ∂x ⎜⎝ ∂y ⎟⎠ ∂y ⎜⎜⎝ ∂x ⎟⎠ x y (c) Using the product rule (3.20) and the chain rule (3.29),

⎛y ⎞ ⎛ x ⎞⎛ y ⎞ ⎛ y ⎞ (x + y) ∂f = ln ⎜⎜⎜ ⎟⎟⎟ + (x + y)⎜⎜⎜ ⎟⎟⎟⎜⎜⎜− 2 ⎟⎟⎟ = ln ⎜⎜⎜ ⎟⎟⎟ − ⎜⎝ x ⎟⎠ ⎜⎝ y ⎟⎠⎜⎝ x ⎟⎠ ⎜⎝ x ⎟⎠ ∂x x and

⎛ y ⎞ (x + y) ∂f . = ln ⎜⎜⎜ ⎟⎟⎟ + ⎜⎝ x ⎟⎠ ∂y y Hence

∂ ⎛⎜ ∂f ⎞⎟ 1 1 ∂ ⎛⎜ ∂f ⎞⎟ ⎜⎜ ⎟⎟ = − = ⎜ ⎟⎟ . ∂x ⎜⎝ ∂y ⎟⎠ y x ∂y ⎜⎜⎝ ∂x ⎟⎠ (d) Putting z = sin−1 y , we have y = sin z , and

dy = cosz = 1 − sin 2 z = 1 −y 2 dz so that

d sin−1 y 1 . = dy 1 −y 2 Then

∂f ∂f exp(x 2 ) = 2x exp(x 2 )sin−1 y and = ∂x ∂y 1 −y 2 and so

∂ ⎛⎜ ∂f ⎞⎟ 2x exp(x 2 ) ∂ ⎛⎜ ∂f ⎞⎟ = ⎜⎜ ⎟⎟ = ⎜ ⎟⎟ . ∂x ⎜⎝ ∂y ⎟⎠ ∂y ⎜⎜⎝ ∂x ⎟⎠ 1−y2 7.2

Set z = y x , and using the prime notation to denote differentiation with respect to z, we have

7.1

∂g 1 = g ′, ∂y x

∂g y = − 2 g ′, ∂x x

∂g ′ 1 ∂g ′ y = g ′′ , = − 2 g ′′ . ∂y x ∂x x

Therefore

∂f 1 y = − 2 g − 3 g ′, ∂x x x

∂f 1 = 2 g ′, ∂y x

∂2 f 2 y = − 3 g ′ − 4 g ′′ , ∂x ∂y x x

and

∂2 f 2 4y y2 ′ = g + g + g ′′, ∂x 2 x 3 x4 x5

∂2 f 1 = 3 g ′′. 2 ∂y x

Hence by substitution

x2

2 ∂2 f ∂2 f ∂f ∂f 2 ∂ f + 2xy + y +x +y = f, 2 2 ∂x ∂y ∂x ∂y ∂x ∂y

as required. 7.3

At the point of contact, the plane and the sphere must have the same slopes ∂z ∂x and ∂z ∂y . For the plane,

∂z ∂z = α, = β, ∂x ∂y and for the sphere,

2z

∂z ∂z = −2x , 2z = −2y . ∂x ∂y

Equating at the point of contact (1, 2, 3) , then gives α = −1 3, β = −2 3 and from

z = αx + βy + γ , γ = 14 3 , so that the equation of the plane is x + 2y + 3z = 14 7.4

(a) The first partial derivatives are:

∂F = k(a 2 +b 2 )1/2 sin(ax)sin(by)cos ⎡⎢kz(a 2 +b 2 )1/2 ⎤⎥ , ⎣ ⎦ ∂z ∂F = b sin(ax)cos(by)cos ⎡⎢kz(a 2 +b 2 )1/2 ⎤⎥ , ⎣ ⎦ ∂y ∂F = a cos(ax)sin(by)cos ⎡⎢kz(a 2 +b 2 )1/2 ⎤⎥ , ⎣ ⎦ ∂x and hence

7.2

∂2 F ∂2 F ∂2 F 2 2 2 2 = −k (a +b )F , = −b F, = −a 2F . ∂z 2 ∂y 2 ∂x 2 Thus, by substitution

⎛ 2 ∂F 2 ∂ 2 F ⎞⎟⎟ 2 ⎜∂ F = k + ⎜ ⎟, ⎜⎜ ∂x 2 ∂z 2 ∂y 2 ⎟⎠ ⎝ as required. (b) The first partial derivatives are:

∂F ∂F ∂F = ae −kz cosax, = −be −kz sinby, = −ke −kz (sinax + cosby), ∂x ∂y ∂z and hence

∂2 F ∂2 F 2 −kz = −a e sinax, = −b 2e −kz cosby. 2 2 ∂x ∂y Thus, by substitution

∂ 2 F ∂ 2 F ⎛⎜a 2 +b 2 ⎞⎟⎟ ∂F , + =⎜ ⎟ ∂x 2 ∂y 2 ⎜⎜⎝ k ⎟⎠ ∂z as required. 7.5

Taking differentials of the defining relations gives

du + dw = 2xdx + 2ydy

and udw + wdu = 2a 2xdx + 2b 2ydy .

Now ∂u ∂x is the change in u with respect to x with y constant, i.e. dy = 0 . So setting dy = 0 in these equations and eliminating dw gives

(u −w)du = 2x(u −a 2 )dx , and

⎛a 2 − u ⎞⎟ ∂u ⎟⎟. = −2x ⎜⎜⎜ ⎜⎝ u − w ⎟⎠ ∂x Similarly, setting dx = 0 and eliminating du , gives

⎛b 2 −w ⎞⎟ ∂w ⎟⎟ . = 2y ⎜⎜⎜ ⎜⎝ u −w ⎟⎠ ∂y To find ∂x ∂u , we need to find the change in x with respect to u with w fixed. dw = 0 . Returning to the original relations for differentials and setting dw = 0 and eliminating dy , gives

(b 2 −w)du = 2x(b 2 −a 2 )dx,

7.3

and

∂x 1 ⎛ b 2 −w ⎞⎟⎟ = − ⎜⎜⎜ 2 ⎟ . ∂u 2x ⎜⎝a −b 2 ⎟⎠ Similarly, setting du = 0 and eliminating dx , gives

∂y 1 ⎛ a 2 −u ⎞⎟⎟ = ⎜⎜⎜ 2 ⎟. ∂w 2y ⎜⎝a −b 2 ⎟⎠ 7.6

(a) From (1) we have

⎛ ∂E ⎞⎟ ⎜⎜ ⎟ ⎜⎜⎝ ∂S ⎟⎟⎠ = T,

⎛ ∂E ⎞⎟ ⎜⎜ ⎟ ⎜⎜⎝ ∂V ⎟⎟⎠ = −P ,

V

S

so that

⎛ ∂T ⎞⎟ ∂2 E ⎟⎟ and = ⎜⎜⎜ ∂V ∂S ⎜⎝ ∂V ⎟⎠S

⎛ ∂P ⎞⎟ ∂2 E ⎟ . = −⎜⎜⎜ ⎜⎝ ∂S ⎟⎟⎠ ∂S ∂V V

Hence, using (7.5)

⎛ ∂T ⎞⎟ ⎛ ∂P ⎞⎟ ⎜⎜ ⎜ ⎟ ⎟ ⎜⎜⎝ ∂V ⎟⎟⎠ = −⎜⎜⎜⎝ ∂S ⎟⎟⎠ , S V as required. (b) From the definition of G, we have

dG = dE −TdS −SdT + PdV +VdP = −SdT +VdP, on substituting for dE from (1). We then have

⎛ ∂G ⎞⎟ ⎜⎜ ⎟ ⎜⎜⎝ ∂T ⎟⎟⎠ = −S, P

⎛ ∂G ⎞⎟ ⎜⎜ ⎟ ⎜⎜⎝ ∂P ⎟⎟⎠ =V, T

and hence

⎛ ∂S ⎞⎟ ∂ 2G ⎟ and = −⎜⎜⎜ ⎜⎝ ∂P ⎟⎟⎠ ∂P ∂T T

⎛ ∂V ⎞⎟ ∂ 2G ⎟⎟ = ⎜⎜⎜ ∂T ∂P ⎜⎝ ∂T ⎟⎠P

and the desired result follows directly from (7.5).

7.7

From (7.13),

⎛ ∂V ⎞⎟ ⎜⎜ ⎟ ⎜⎜⎝ ∂T ⎟⎟⎠

P

⎛ ∂T ⎞⎟ ⎜⎜ ⎟ ⎜⎜⎝ ∂P ⎟⎟⎠

V

⎛ ∂P ⎞⎟ ⎜⎜ ⎟ = −1 , ⎜⎜⎝ ∂V ⎟⎟⎠ T

so that by (7.12)

1 ⎛ ∂V ⎞⎟ 1 ⎛ ∂P ⎞⎟ ⎟⎟ = − ⎜⎜ ⎟⎟ α = ⎜⎜⎜ T ⎜⎝ ∂T ⎟⎠P V ⎜⎜⎝ ∂T ⎟⎠V

7.4

⎡⎛ ∂P ⎞ ⎤ ⎟ ⎥ ⎢⎜⎜ ⎢⎜⎜⎝ ∂V ⎟⎟⎟⎠ ⎥ ⎢⎣ T ⎥⎦

−1

.

(3)

From Dieterici’s equation

P=

⎛ −a ⎞⎟ RT ⎟ exp ⎜⎜⎜ ⎜⎝ RTV ⎟⎟⎠ V −b

so that

⎛ ∂P ⎞⎟ ⎡ R ⎛ ⎞ RT a ⎤⎥ ⎜⎜ ⎜⎜− a ⎟⎟ ⎟⎟ = ⎢ + exp ⎜⎜⎝ ∂T ⎟⎠ ⎜⎜⎝ RTV ⎟⎟⎠ ⎢V −b (V −b) RT 2V ⎥ ⎣ ⎦ V ⎛ R ⎡⎢ a ⎤⎥ a ⎞⎟ ⎟, = 1+ exp ⎜⎜⎜− ⎜⎝ RTV ⎟⎟⎠ V −b ⎢⎣ RTV ⎥⎦ and

⎡ ⎛ ∂P ⎞⎟ ⎛ ⎞ RT RT a ⎤⎥ ⎜⎜ ⎜⎜− a ⎟⎟ ⎟⎟ = ⎢− + exp ⎜⎜⎝ ∂V ⎟⎠ ⎜⎜⎝ RTV ⎟⎟⎠ ⎢ (V −b)2 (V −b) RT 2V ⎥ ⎣ ⎦ T ⎤ ⎛ ⎞ RT ⎡⎢ a ⎥ exp ⎜⎜− a ⎟⎟⎟, =− 1− (V −b) ⎜⎜⎝ RTV ⎟⎠ ⎥ V −b ⎢⎣ RTV 3 ⎦ and substituting in (3) gives (2) as required. (b) From the definition of A and Dieterici’s equation,

⎛ a ⎞⎟ ⎟ , A = P(V −b) = RT exp ⎜⎜⎜− ⎜⎝ RTV ⎟⎟⎠ so that

⎡ ⎤ ⎛ a a a ⎞⎟ ⎟ . dA = PdV + (V −b)dP = ⎢RdT + dT + 3 dV ⎥ exp ⎜⎜⎜− ⎢ ⎥ ⎜⎝ RTV ⎟⎟⎠ TV V ⎣ ⎦ Setting dP = 0 then gives

⎡ ⎤ ⎡ ⎤ ⎢P − a e −a/RTV ⎥ dV = ⎢R + a ⎥ e −a/RTV dT ⎢ ⎥ ⎢ TV ⎥⎦ V2 ⎣ ⎦ ⎣ and hence

⎛ ∂V ⎞⎟ ⎡ a ⎤⎥ −a/RTV ⎜⎜ ⎢ ⎟ ⎜⎜⎝ ∂T ⎟⎟⎠ = R ⎢1 + RTV ⎥ e ⎣ ⎦ P

⎡ ⎤ ⎢P − a e −a/RTV ⎥ 2 ⎢ ⎥ V ⎣ ⎦

−1

Multiplying top and bottom by e a/TRV and using

Pe a/TRV =

RT V −b

from Dieterici’s equation, gives

1 ⎛⎜ ∂V ⎞⎟ (V −b) ⎡⎢ a ⎤⎥ ⎡⎢ a(V −b) ⎤⎥ ⎟⎟ = 1+ 1− ⎜⎜ V ⎜⎝ ∂T ⎟⎠P VT ⎢⎣ RTV ⎥⎦ ⎢⎣ RTV 2 ⎥⎦ as required.

7.5

−1

.

7.8

For (a), (b) and (c), (7.19a) may be used to test whether the differentials are exact. We find, in the notation of (7.19a), (a)

(b)

∂A(x,y) ∂B = 3x 2 − 3xy 2 = ⇒df is exact , ∂y ∂x

∂A(x,y) ∂B = x − 6x 2y ≠ = x + 6x 2y ⇒df is inexact , ∂y ∂x (c)

∂A(x,y) ∂B = sin x cosy = ⇒df is exact . ∂y ∂x

For (d), (7.19b) may be used, and we find

∂C y ⎫⎪⎪ ⎪⎪ = 2y = ∂y ∂x ⎪⎪ ⎪ ∂C x ∂C z ⎪⎪ ⎬ ⇒ df is exact . = 2x = ∂z ∂x ⎪⎪ ⎪ ∂C y ∂C z ⎪⎪⎪ =y = ⎪ ∂z ∂y ⎪⎭⎪⎪

∂C x (d)

7.9

We will use the condition (7.19a) to establish whether a differential is exact, using the notation of that equation. (a) We have,

A=

y2 x2 and , B = (x + y)2 (x + y)2

so that

∂A 2y 2y 2 2xy ∂B , = − = = 2 3 3 ∂y ∂x (x + y) (x + y) (x + y) which implies that the differential is exact, with

∂f y2 ∂f x2 =A= and = B = . ∂x ∂y (x + y)2 (x + y)2 Integrating ∂f ∂x gives

f (x,y) = −

y2 +c(y), x +y

where c(y) is an arbitrary function of y. Hence differentiating with respect to y gives

7.6

∂f 2y y2 x2 ′ . =− + + c (y) = B = ∂y (x + y) (x + y)2 (x + y)2 Hence

c ′(y) =

2y (x 2 −y 2 ) + = 1, (x + y) (x + y)2

so that c(y) = y + k , where k is an integration constant. Finally,

f (x,y) =

−y 2 xy +y +k = + k. (x + y) (x + y)

(b) We have,

A = 2x ln(xy) + x,

B = x2 y ,

so that

∂A 2x ∂B = = ∂y y ∂x and df is an exact differential with

∂f = 2x ln(xy) + x, ∂x

∂f x2 = . ∂y y

Integrating the second of these relations gives

f = x 2 ln y +c(x) , so that

∂f = 2x ln y + c ′(x) = 2x ln(xy) + x ∂x on comparing with B. Hence

c ′(x) = 2x ln x + x and

c(x) = x 2 ln x + k , giving finally

f (x,y) = x 2 ln x + x 2 ln y + k = x 2 ln(xy) + k, where k is an arbitrary constant. 7.10

(a) We have,

∂z ∂z dx dy = 4x + 3y 3, = 9xy 2 + 16y 3, = cost, = −sin t . ∂x ∂y dt dt

7.7

So, using (7.20), with x1 = x and x 2 = y , 2 dz ∂f dx i =∑ = y(4x + 3y 3 )− xy 2(9x + 16y). dt i=1 ∂x i dt

(b) In the same way as (a), we have

∂z −2 ∂z 2x 2y dx dy = , = , = et , = −e −t . 2 2 2 2 ∂x x(1 + x y ) ∂y (1 + x y ) dt dt Hence from (7.20),

dz 2e t 2x 2ye −t 2e −t (e 2t + x 3y) =− − = − = −2, dt x(1 + x 2y 2 ) (1 + x 2y 2 ) x(1 + x 2y 2 ) on substituting x = e t and y = e −t . (c) Again, we have

∂z 1 ∂z x dx 1 dy 1 = y + + y ln (x y ), = −x − 2 + x ln (x y ), = , =− . ∂x y ∂y dt t dt t y Hence from (7.20),

dz = e −x dt 7.11

⎡ ⎛ ⎞ ⎛ ⎞⎤ ⎢(y − x)ln ⎜⎜ x ⎟⎟ + ⎜⎜y + 1 + x + x ⎟⎟⎥ . ⎟ ⎢ ⎜⎜⎝ y ⎟⎟⎠ ⎜⎜⎝ y y 2 ⎟⎠⎥⎥⎦ ⎢⎣

The equation is satisfied if f is homogeneous of degree k, i.e. if

f (λx,λy,λz) = λk f (x,y,z) . We then apply this to the given forms. (a) f is homogeneous of order k = 3 , and hence the equation is satisfied. (b) f is not homogeneous, and hence the equation is not satisfied. (c) f may be written

⎛ xy 2 ⎞ f = ln x + 2 ln y − 3 ln z + 4 = ln ⎜⎜⎜ 3 ⎟⎟⎟ + 4 , ⎜⎝ z ⎟⎠ and so f is homogeneous of order zero, and the equation is satisfied.

7.8

(d) The proof of Euler’s theorem in the text does not rely on k being an integer. In this case the equation is satisfied with k = 1 2 , as may be verified by direct substitution. 7.12

If f (x1,x 2,…,x n ) is a homogeneous function of order k, then n

∑x i=1

i

∂f ∂f ∂f ∂f = x1 + x2 ++ x n = kf ∂x i ∂x1 ∂x 2 ∂x n

by Euler’s theorem. Hence

⎛ n n ∂ ⎞⎟⎟⎛⎜ n ∂f ⎞⎟⎟ ∂f ⎜⎜ x = k 2f . ⎟⎟⎜⎜∑ x i ⎟⎟ = ∑ kx j ⎜⎜∑ j ⎟ ⎟ ∂x j ⎝ j =1 ∂x j ⎠⎜⎝ i=1 ∂x i ⎠ j =1

(1)

But,

xj

∂ ⎛⎜ ∂f ⎞⎟⎟ ∂2 f ∂f ⎜⎜x i , + δij x i ⎟⎟ = x i x j ∂x j ⎜⎝ ∂x i ⎟⎠ ∂x i ∂x j ∂x i

where δij is the Kronecker delta symbol (9.24b). Substituting into (1) gives

∑ ∑ xix j i

j

n n ∂2 f ∂f ∂2 f + ∑ xi = ∑ xix j + k f = k 2f , ∂x i ∂x j ∂x ∂x ∂x i=1 i,j =1 i i j

where we have again used Euler’s theorem. Hence,

∑∑x x i

j

i

j

∂2 f = k(k −1)f , ∂x i ∂x j

as required. On setting n = 2, x1 = x, x 2 = y, this becomes

x2

2 ∂2 f ∂2 f 2 ∂ f + 2xy + y = k(k −1)f , ∂x ∂y ∂x 2 ∂y 2

assuming

∂2 f ∂2 f = ∂x1 ∂x 2 ∂x 2 ∂x1 as usual. 7.13

By (7.24),

∂z ∂z ∂x ∂z ∂y 1 u ⎛⎜ ∂z ∂z ⎞⎟ = + = e ⎜ + ⎟⎟ ∂u ∂x ∂u ∂y ∂u 2 ⎜⎜⎝ ∂x ∂y ⎟⎠ and

∂z ∂z ∂x ∂z ∂y 1 ⎛ ∂z ∂z ⎞ = + = e w ⎜⎜⎜ − ⎟⎟⎟ . ∂w ∂x ∂w ∂y ∂w 2 ⎜⎝ ∂x ∂y ⎟⎠ So adding gives

7.9

∂z ∂z ∂z ∂z + =x +y , ∂u ∂w ∂x ∂y while eliminating ∂z ∂y gives directly

e u+w 7.14

∂z ∂z ∂z . = ew +e u ∂x ∂u ∂w

Setting u1 = x −ct and u 2 = x +ct , and then taking derivatives with respect to x, gives

∂f ∂f ∂u1 ∂f ∂u 2 = + = φ1′(u1 ) + φ2′(u 2 ) ∂x ∂u1 ∂x ∂u 2 ∂x and

∂2 f ∂ ⎛⎜ ∂f ⎞⎟ = ⎜ ⎟⎟ ∂x ⎜⎜⎝ ∂x ⎟⎠ ∂x 2 ⎡ ∂ ⎛ ∂f ⎞⎤ ∂u ⎡ ⎛ ⎞⎤ ⎜⎜ ⎟⎟⎥ 1 + ⎢ ∂ ⎜⎜ ∂f ⎟⎟⎥ ∂u 2 = φ′′(u ) + φ′′(u ). = ⎢⎢ ⎟ 1 1 2 2 ⎥ ⎢ ∂u ⎜⎜⎝ ∂x ⎟⎟⎠⎥ ∂x ⎜ ⎢⎣ ∂u1 ⎜⎝ ∂x ⎟⎠⎥⎦ ∂x ⎢⎣ 2 ⎥⎦ In a similar way,

∂f ∂f ∂u1 ∂f ∂u 2 = + = −c φ1′(u1 ) +c φ2′(u 2 ) ∂t ∂u1 ∂t ∂u 2 ∂t and

∂2 f ∂ ⎛⎜ ∂f ⎞⎟ = ⎜ ⎟⎟ ∂t ⎜⎜⎝ ∂t ⎟⎠ ∂t 2 ⎡ ∂ ⎛ ∂f ⎞⎤ ∂u ⎡ ⎛ ⎞⎤ ⎜⎜ ⎟⎟⎥ 1 + ⎢ ∂ ⎜⎜ ∂f ⎟⎟⎥ ∂u 2 = c 2φ′′(u ) +c 2φ′′(u ). = ⎢⎢ ⎟ 1 1 2 2 ⎥ ⎢ ∂u ⎜⎜⎝ ∂t ⎟⎟⎠⎥ ∂t ⎜ ⎢⎣ ∂u1 ⎜⎝ ∂t ⎟⎠⎥⎦ ∂t ⎢⎣ 2 ⎥⎦ Hence,

∂2 f 1 ∂2 f − =0 ∂x 2 c 2 ∂t 2 as required. 7.15

From the relations for x and y, we can form the first derivatives

∂x ∂x ∂y ∂y = u 2 − w 2, = −2uw = 2uw, = u2 −w2 , ∂u ∂w ∂u ∂w and so, using g when differentiating with respect to u or w, and f when differentiating with respect to x or y, gives

∂g ∂f ∂x ∂f ∂y ∂f ∂f = + = (u 2 −w 2 ) + 2uw ∂u ∂x ∂u ∂y ∂u ∂x ∂y

7.10

and

∂g ∂f ∂x ∂f ∂y ∂f ∂f = + = −2uw + (u 2 −w 2 ) . ∂w ∂x ∂w ∂y ∂w ∂x ∂y The second derivatives may then be found using (7.24), i.e.

⎛ ⎞ ∂2 g ∂ ⎛⎜ ∂g ⎞⎟ ∂f ∂f ∂ ⎛⎜ ∂f ⎞⎟ 2 2 ∂ ⎜ ∂f ⎟ ⎟ ⎟ = = 2u + (u −w ) + 2w + 2uw ⎜ ⎜ ⎜ ⎟⎟, ⎟ ⎟ ∂x ∂u ⎜⎜⎝ ∂x ⎟⎠ ∂y ∂u ⎜⎜⎝ ∂y ⎟⎠ ∂u 2 ∂u ⎜⎜⎝ ∂u ⎟⎠ and

∂2 g ∂ ⎛⎜ ∂g ⎞⎟ ∂f ∂ ⎛⎜ ∂f ⎞⎟ ∂f ∂ ⎛⎜ ∂f ⎞⎟ = − 2uw + (u 2 −w 2 ) ⎜⎜ ⎟⎟ = −2u ⎜⎜ ⎟⎟ − 2w ⎜ ⎟⎟. 2 ∂w ⎜⎝ ∂w ⎟⎠ ∂x ∂w ⎜⎝ ∂x ⎟⎠ ∂y ∂w ⎜⎜⎝ ∂y ⎟⎠ ∂w To find the second derivatives on the right-hand side, we use (7.24) to give 2 ∂ ⎛⎜ ∂f ⎞⎟ ∂ 2 f ∂x ∂ 2 f ∂y ∂2 f 2 2 ∂ f . + = (u −w ) + 2uw ⎜⎜ ⎟⎟ = ∂u ⎜⎝ ∂x ⎟⎠ ∂x 2 ∂u ∂x ∂y ∂u ∂x ∂y ∂x 2

Similarly,

∂ ⎛⎜ ∂f ⎞⎟ ∂2 f ∂2 f 2 2 + 2uw 2 , ⎜⎜ ⎟⎟ = (u −w ) ∂u ⎜⎝ ∂y ⎟⎠ ∂x ∂y ∂y ∂ ⎛⎜ ∂f ⎞⎟ ∂2 f ∂2 f 2 2 ⎟ , = −2uw + (u −w ) ⎜ ⎟ ∂w ⎜⎜⎝ ∂x ⎟⎠ ∂x ∂y ∂x 2 and

∂ ⎛⎜ ∂f ⎞⎟ ∂2 f ∂2 f + (u 2 −w 2 ) 2 . ⎜⎜ ⎟⎟ = −2uw ∂w ⎜⎝ ∂y ⎟⎠ ∂x ∂y ∂y Substitution and adding, then gives 2 ⎞ ⎛ ∂ 2 g ∂ 2 g ⎞⎟ ⎛ 2 ⎜⎜ ⎟⎟ = (u 2 + w 2 )2 ⎜⎜ ∂ f + ∂ f ⎟⎟⎟ , + ⎜⎜ ∂u 2 ∂w 2 ⎟ ⎜⎜ ∂x 2 ∂y 2 ⎟ ⎝ ⎠ ⎝ ⎠

as required. 7.16

The derivatives that are need at x 0 = 2, y 0 = 1 are

7.11

∂f 1 ∂2 f 1 = e x/y → e 2, = 2 e x/y → e 2, 2 ∂x y ∂x y ⎡⎛ ⎞2 ⎛ ⎞⎤ ∂f x ∂2 f ⎢⎜ −x ⎟⎟ ⎜ 2x ⎟⎟⎥ x/y 2 = − 2 e x/y → −2e 2, = ⎢⎜⎜ 2 ⎟⎟ + ⎜⎜ 3 ⎟⎟⎥ e → 8e , 2 ⎜ ⎜ ∂y y ∂y ⎢⎝ y ⎠ ⎝ y ⎠⎥ ⎣ ⎦ ⎛ x ∂2 f 1⎞ = ⎜⎜⎜− 3 − 2 ⎟⎟⎟e x/y → −3e 2 . ∂y ∂x ⎜⎝ y y ⎟⎠ These are then used in the Taylor expansion n

1 ⎛⎜ ∂ ∂⎞ + k ⎟⎟⎟ f (x 0 ,y 0 ) , ⎜⎜h ∂y ⎟⎠ ⎝ ∂x n=0 n ! ⎜ ∞

f (x,y) = ∑

where h = x − x 0 = x − 2 and k = y − y 0 = y −1 . Keeping terms up to n = 2 gives

e x/y = e 2 ⎡⎢1 + (x − 2)− 2(y −1) + 12 (x − 2)2 + 4(y −1)2 − 3(x − 2)(y −1)⎤⎥ . ⎣ ⎦ 7.17

The Taylor series is about the point x = 0, y = 0 , so the derivatives up to third order that are needed are

∂f 1 ∂f −ln(1 + x) = → 1, = → 0, ∂x (1 + x)(1 + y) ∂y (1 + y)2 ∂2 f −1 ∂2 f 2 ln(1 + x) = → −1, = → 0, 2 2 2 ∂x (1 + x) (1 + y) ∂y (1 + y)3 ∂2 f −1 = → −1, ∂x ∂y (1 + x)(1 + y)2 ∂3 f 2 ∂3 f −6 ln(1 + x) = → 2, = → 0, 3 3 3 ∂x (1 + x) (1 + y) ∂y (1 + y)4 ∂3 f 2 ∂3 f 1 = → 2, = → 1. 2 3 2 2 ∂x ∂y (1 + x)(1 + y) ∂y ∂x (1 + x) (1 + y)2 These are then used in the Maclaurin expansion n

1 ⎛⎜ ∂ ∂⎞ + k ⎟⎟⎟ f (x 0 ,y 0 ) , ⎜⎜h ∂y ⎟⎠ ⎝ ∂x n=0 n ! ⎜ ∞

f (x,y) = ∑

where h = x and k = y . Keeping terms up to n = 3 gives

7.12

f (x,y) = x + 12 (−x 2 − 2xy) + 16 (2x 3 + 3x 2y + 6xy 2 )+  = x − 12 x 2 − xy + 13 x 3 + 12 x 2y + xy 2 + . 7.18

At stationary points

∂f = cosx sin y sin(x + y) + sin x sin y cos(x + y) ∂x = sin y cos(2x + y) = 0, and similarly

∂f = sin x cos(x + 2y) = 0. ∂y Since sin x ≠ 0 and sin y ≠ 0 within the square, we have

cos(2x + y) = 0 = cos(x + 2y) and hence the possibilities

2x + y = π, 2π and x + 2y = π, 2π. If 2x + y = π, x + 2y = 2π , then x = 0,y = π , while if 2x + y = 2π, x + 2y = π we have x = π,y = 0 . Neither lie inside the square. If 2x + y = π = x + 2y , then

x = y = π 3 , and f = sin 2(π 3)sin(2π 3) = 3 3 8. If 2x + y = 2π = x + 2y , then x = y = 2π 3 and

f = sin 2(2 π 3)sin(4π 3) = −3 3 8. Finally, since f = 0 along the boundaries x = 0,y = 0,x = π,y = π , these are the maximum and minimum values possible. 7.19

The first derivatives of f are

∂f = e x−y ⎡⎢(2x + y) + (x 2 + xy + y 2 )⎤⎥ , ⎣ ⎦ ∂x and

∂f = e x−y ⎡⎢(x + 2y)−(x 2 + xy + y 2 )⎤⎥ . ⎣ ⎦ ∂y The condition for a stationary point is ∂f ∂x = ∂f ∂y = 0 , so adding the two equations above gives

7.13

3e x−y (x + y) = 0 , i.e. x = −y . Substituting this back into the equation ∂f ∂x = 0 , gives

(2x − x) + (x 2 − x 2 + x 2 ) = 0, i.e. x = 0 or x = −1 . Thus there are stationary points at (x,y) = (0, 0) and (−1,1) . To classify these, we first calculate the second derivatives,

∂2 f = e x−y ⎡⎢(2 + 2x + y) + (2x + y + x 2 + xy + y 2 )⎤⎥ , ⎣ ⎦ ∂x 2 ∂2 f = e x−y ⎡⎢(2 − x − 2y)−(x + 2y − x 2 − xy −y 2 )⎤⎥ , ⎣ ⎦ ∂y 2 ∂2 f = e x−y ⎡⎢(1 − 2x −y) + (x + 2y − x 2 − xy −y 2 )⎤⎥ . ⎣ ⎦ ∂x ∂y At (x,y) = (0, 0) ,

∂ 2 f ∂x 2 = 2 , ∂ 2 f ∂y 2 = 2 and ∂ 2 f ∂x ∂y = 1 , so from (7.40a,b), this is a maximum. At (x,y) = (−1,1) ,

∂ 2 f ∂x 2 = e −2 , ∂ 2 f ∂y 2 = e −2 and ∂ 2 f ∂x ∂y = 2e −2 , so from (7.41), this is a saddle point. *7.20

There is only one constraint, so from (7.45) we define

F(x,y) = f + λg , where g(x,y) = x 2 − 2y − 2 . Then the stationary points are solutions of (7.46), i.e.

∂F ∂f ∂g ∂F ∂f ∂g = +λ = 2x + 2xλ = 0, and = +λ = −2y − 2λ = 0. ∂x ∂x ∂x ∂y ∂y ∂y The first of these equations gives either x = 0 or λ = −1 . While the second gives y = −λ . For x = 0 , the constraint equation gives y = −1 , so the stationary point is

(x,y) = (0,−1) . For λ = −1 , we have y = −λ and the constraint equation gives

x 2 = 4 , so x = ±2 and the stationary points are (x,y) = (±2,1) . *7.21

The ellipsoid is centred on the origin, and the largest box will have its vertices (±x,± y,± z) on the surface of the ellipsoid, so we have to maximise

7.14

V = 8xyz subject to the constraint

g(x,y,z) =

x2 y2 z2 + + −1 = 0 . a 2 b2 c 2

Forming the function

F =V + λg , we have

∂F 2λx = 8yz + 2 = 0 ∂x a or

V + 2λ x 2 a 2 = 0. Similarly, from ∂F ∂y = 0, ∂F ∂z = 0 , one finds

V + 2λ

y2 z2 = 0 and V + 2λ = 0. b2 c2

Hence

x2 y2 z2 V 1 = 2 = 2 =− = , 2 2λ 3 a b c on applying the constraint. So as x = a

V = 8xyz = *7.22

3, y = b

8abc 3 3

3, and z = c

.

The function to me maximised is n

P = ∏ x i = x1x 2 …x n i=1

and the constraint is n

g = ∑ xi − N . i=1

So we define the function

F = P + λg

and maximise this imposing the constraint. Imposing

∂F = x 2x 3 …x n + λ = 0 , ∂x1 gives

P + λx1 = 0 , and similarly

P + λx i = 0 ,

7.15

3,

for all x i . Hence all the x i are equal, and imposing the constraint gives x i = N n , and hence n

⎛N ⎞ P = ∏ x i = ⎜⎜⎜ ⎟⎟⎟ . . ⎜⎝ n ⎟⎠ n

i=1

*7.23

(a) Using (7.56) gives ∞

∞

∞ dI ∂ −xy −xy −xy −x =⌠ ⎮ xe dy = ∫ e (1 − xy)dy = ⎡⎢⎣ye ⎤⎦⎥1 = −e dx ⌡ ∂x 1 1

because x > 0 . So

I(x) = e −x +c = e −x −e −1 , since I(1) = 0 . (b) Using (7.55), we obtain t2

t2

t

t

d ⌠ dx 2t 1 ∂ I(t) = ⎮ = − +⌠ ⎮ 2 dt ⌡ ln(x + t) ln(t + t) ln(2t) ⌡ ∂t

⎡ ⎤ 1 ⎢ ⎥ dx. ⎢ ln(x + t) ⎥ ⎣ ⎦

Putting z = ln(x + t) and using

∂ ∂z ∂f (z) , f (z) = ∂t ∂t ∂z one obtains t2

2t 1 dx I(t) = − −∫ 2 ln(t + t) ln(2t) t (x + t)ln 2(x + t) t2

2

z(t ) ⎡ ⎤ 2t 1 dz 2t 1 1 ⎥ , = − − = − +⎢ ∫ 2 2 2 ln(t + t) ln(2t) z(t ) z ln(t + t) ln(2t) ⎢⎣ ln(x + t) ⎥⎦ t

and hence t2

d ⌠ dx 2t + 1 2 = − . ⎮ 2 dt ⌡ ln(x + t) ln(t + t ) ln(2t) t

*7.24 (a) Using (7.55) gives

7.16

ey

e y sin(ye y ) sin y 2 I(y) = − + ∫ cos(xy) dx y ey y ey

sin y 2 ⎡⎢ sin(xy) ⎤⎥ = sin(ye )− + ⎢ y ⎥ y ⎣ ⎦y 2 ⎛ ⎞ 1 2sin y = sin(ye y )⎜⎜⎜1 + ⎟⎟⎟ − . ⎟ ⎜⎝ y⎠ y y

(b) Differentiating gives ∞

∞

dI(α) ⌠ sin x ∂e −αx =⎮ dx = −∫ e −αx sin x dx. ⌡ x dα ∂α 0 0

Integrating by parts twice then gives ∞

∞

∞ dI(α) d cosx = ∫ e −αx dx = ⎡⎢cosx e −αx ⎤⎥ + α ∫ cosx e −αx dx ⎣ ⎦0 dα dx 0 0 ∞

∞

= −1 + α ⎡⎢sin x e −αx ⎤⎥ + α2 ∫ sin x e −αx dx ⎣ ⎦0 0

dI(α) = −1 − α2 . dα Hence

dI(α) 1 =− dα 1 + α2 and integrating, using the standard integral of Table 4.2, gives

I(α) = c − tan−1 α , where c is an integration constant, Using the fact I(α) → 0 as α → ∞ , gives c = π 2 , and hence I(α) = π 2 − tan−1 α . Then, from (2.31), we have

sin θ = cos(π 2 − θ) and cos θ = sin(π 2 − θ) and hence from (2.32a),

α ≡ tan θ = cot(π 2 − θ). thus tan−1 α = θ, cot−1 α = π 2 − θ = π 2 − tan−1 α , so that finally

7.17

I(α) = cot−1 α . *7.25

Differentiating I k (a) with respect to a gives

dI k (a) da Writing I 0 (a) = (a π)

−1/2

∞

2

= − ∫ x 2k+2e −ax dx = −I k+1. −∞

, gives −3/2

I 1(a) = −

d 1 1 ⎛⎜ a ⎞⎟ I 0 (a) = ⎜ ⎟⎟ da π 2 ⎜⎜⎝ π ⎟⎠

, −5/2

d 1 1 3 ⎛⎜ a ⎞⎟ I 2(a) = − I 1(a) = 2 ⎜ ⎟⎟ da π 2 2 ⎜⎜⎝ π ⎟⎠

,

and −(2k+1/2)

d 1 1 3 (2k −1) ⎛⎜ a ⎞⎟ I k (a) = − I k−1(a) = k  ⎜ ⎟⎟ da 2 ⎜⎜⎝ π ⎟⎠ π 22

,

or equivalently, −(2k+1/2)

1 ⋅ 3 ⋅5 ⋅…⋅(2k −1) ⎛⎜ a ⎞⎟ I k (a) = ⎜⎜ ⎟⎟ (2π)k ⎝⎜ π ⎟⎠

7.18

.

SOLUTIONS TO PROBLEMS 8 8.1

Referring to the figure below, the unit vectors i, j and k are chosen to lie along OA, OB and OC. Then vectors r1, r2 and r3 are drawn as shown. From the diagram,

a

r1 = ai cos(π 4) + aj sin(π 4) =

2

r2 = 2ai cos(π 4) + 2ak sin(π 4) = r3 = 3aj cos(π 4) + 3ak sin(π 4) =

(i + j), 2a 2 3a 2

(i + k), (j + k),

and

r = r1 + r2 + r3 =

3a 2

i+

4a 2

j+

5a 2

k.

Finally, the magnitude of r is

(

r = r = r12 + r22 + r32

8.2

)

1/2

= 5a .

Referring to the figure below, BD = b − a and so BP = x(b − a) , where x is a real number in the range 0 < x < 1 . Also, AC = a + b and AP = y(a + b) , where y is also a real number in the range 0 < y < 1 . But,

AB = AP + PB = AP − BP , that is,

a = y(a + b)− x(b − a) = (x + y)a + (y − x)b . But since a and b are not collinear, x + y = 1 and y − x = 0 , i.e. x = y = 1 2 , so the diagonals bisect one another.

S8.1

8.3

The position vector of a point on the line corresponding to a 0 is given by

rA = (1 + λ)i + (2 − 2λ)j + (3 + 3λ)k and similarly the position vector of a point on the line corresponding to b0 is

rB = (3 − 3µ)i + (2 + 2µ)j + (1 − µ)k . If the lines intersect, rA = rB , i.e., the equations

1 + λ = 3 − 3µ,

2 − 2λ = 2 + 2µ,

3 + 3λ = 1 − µ,

must be satisfied simultaneously. Solving the first two equations give µ = 1 and λ = −1 and by substitution this is consistent with the third equation. Thus the point of intersection is (0, 4, 0). 8.4

(a) From (8.5c), unit vectors aˆ1 and aˆ2 , corresponding to the vectors a1 and a 2 , may be written in the form

aˆ1 = a i +b j +c k and aˆ2 = α i + β j + γ k . Then

aˆ1 ⋅ aˆ2 = cos θ = aα +bβ +cγ . (b) The two vectors are

r1 = 3i − 4 j with r1 = 5, and r2 = −2i + j with r2 = 5. Then

cos θ =

r1 ⋅ r2 r1 r2

=−

2 5

, i.e. θ = 2.68 rad = 153.4 0.

If the required vector is r = ai +bj +ck , then r ⋅ r1 = r ⋅ r2 = 0 , which gives a = b = 0 . But a 2 +b 2 +c 2 = 1 , so c = 1 and the direction cosines are therefore (0, 0,1). 8.5

Referring to the figure below,

AP = AO + OP = −a + r

and

S8.2

BP = BO + OP = (a + r),

so that

(AP)2 = AP ⋅ AP = (a − r)⋅(a − r) = a 2 + r 2 − 2a ⋅ r and

(BP)2 = BP ⋅ BP = (a + r)⋅(a + r) = a 2 + r 2 + 2a ⋅ r . Then (AP)2 + (BP)2 = 2(a 2 + r 2 ) = 4a 2 , because a = r , and so (AP)2 + (BP)2 = (AB)2 which is Pythogoras’ theorem and hence the angle subtended at P is a right angle.

8.6

(a) Three vectors form a triangle if (a) one is the sum of the other two, or (b) the sum of the vectors is zero. By inspection, a = b + c , so the vectors do form a triangle. The condition that two non-zero vectors are at right angles is that their scalar product vanishes, i.e. θ = π 2 . We find, a ⋅ b = 14, a ⋅ c = 0 and b ⋅ c = −21 . Hence a is perpendicular to c and the triangle is right-angled. (b) From Figure 8.2, s = a + b , i.e. a = s − b . Squaring gives,

a ⋅ a = (s − b)⋅(s − b) = s ⋅ s + b ⋅ b − 2s ⋅ b , i.e.

a 2 = s 2 +b 2 − 2sb cos θ , where θ is the angle between s and b. The other relations may be found in a similar way. 8.7

Using (8.16a), we have

a ×b = −i − 4 j − 5k and a × b = 1 + 16 + 25 = 42 . So

ˆ=± n

a×b a×b

is a unit vector perpendicular to both a and b. Thus

ˆ= n

1 42

i+

4 42

j+

5 42

k or −

S8.3

1 42

i−

4 42

j−

5 42

k

The area of the triangle is given by the two vectors. This gives an area 8.8

1 2 21 2

ab sin θ = 12 a ×b , where θ is the angle between .

Consider the tetrahedron shown in the figure below. Using the fact that the area of a triangle with sides a and b is

1 2

a ×b , we have (note the directions are important)

v1 = 12 (a × b), v2 = 12 (b×c), v3 = 12 (c×a) . But,

v4 = 12 (c − a)×(b − a) = 12 c×b − 12 c×a − 12 a ×b = −v2 − v3 − v1, and so

v1 + v2 + v3 + v4 = 0 .

8.9

From the co-ordinates of the vertices we find the following vectors:

DA = −3i − 3j − 3k, DB = −4 j − k, DC = −4i − j + 3k . Therefore, using (8.16)

DA×DB = −9i − 3j + 12k

and

(DA×DB)⋅ DC = (−9i − 3j+ 12k)⋅(−4i − j + 3k) = 75 . The volume is 1 6 of this, i.e. 75 6 . 8.10

(a) using (8.27) we have

a ×(b×c) = (a ⋅ c)b −(a ⋅ b)c = −b − 2c = −6i − 3j. (b) From (1),

c×(r×a) = c×(b×a) which, on using (8.27) and r ⋅ c = 0 , gives

(a ⋅ c)r = (a ⋅ c)b −(b ⋅ c)a . Hence,

S8.4

r = b− 8.11

(b ⋅ c) a . (a ⋅ c)

(a) From (8.32), the angular momentum is

L = mr×(ω ×r) , and expanding this using (8.31) gives

L = m(r ⋅ r)ω −m(r ⋅ ω)r . But if r and ω are perpendicular,

(r ⋅ ω) = 0 , and v = ω ∧ r ⇒ υ = ωr , so that L = mr 2ω and hence L = mrυ . (b) Expanding the expression for a using (8.31) gives

a = ω ×(ω ×r) = (ω ⋅ r)ω − (ω ⋅ ω)r = −ω 2r again because (r ⋅ ω) = 0 . Thus the direction of a is opposite to that of r, i.e. towards the centre of the circle. Its magnitude is a = ω 2r . But ω = υ r and hence a = υ 2 r . 8.12

(a) r = 3i + 2j + 3k , so that τ = r×F = −9i + 6j + 5k . (b) τ = (r − a)×F , where a = i − j + 2k , giving τ = 3i − 2j . (c) τ z = τ ⋅ k = 5 , where τ is given in (a) above.

ˆ= (d) A unit vector in the direction specified is n

1 2

(j + k) . So the moment about the

given line is

ˆ⋅τ = τl = n

1 2

(6 + 5) =

11 2

,

where τ is again given in (a) above. *8.13

If a, b, and c are an orthogonal set of vectors, then

a ⋅b = a ⋅c = b⋅c = 0 and we need to show that

a ′ ⋅ b′ = a ′ ⋅ c ′ = b′ ⋅ c ′ = 0 .

S8.5

(1)

Using (8.36) and (8.33a), we have

a ′ ⋅ b′ =

(b×c)⋅(c×a) (b ⋅ c)(c ⋅ a)−(b ⋅ a)c 2 = =0 [abc]2 [abc]2

by (1). The other results, a ′ ⋅ c′ = b′ ⋅ c′ = 0 , follow in a similar way. *8.14

Using (8.36) and (8.20), we have

[ a ′b′c′] =

(b×c)⋅(c×a)×(a ×b) . [abc]3

(1)

Then using (8.33c),

[ a ′b′c′] =

1 , [abc]

and hence

a ′′ ≡

b′ × c′ (c×a)×(a ×b) = =a . [ a ′b′c′] [abc]

The corresponding relations b′′ = b and c′′ = c follow in a similar way. *8.15

(a) First form the triple scalar product

a ⋅(b×c) = (3j − k)⋅[(j + k)×(i + 2j)] = −5. Then using the relations given in (8.36) gives

a ′ = − 15 (j + k)×(i + 2j) = − 15 (2i + j − k), b′ = − 15 (i + 2j)×(3i − k) = − 15 (−2i + j− 6k), c′ = − 15 (3i − k)×(j + k) = − 15 (i − 3j + 3k). (b) Using (8.18) and (8.21), gives a ⋅ b×c = a 3 4 , while (8.36) and (8.16) give

1 1 a ′ = (j + k)×(i + k) = (i + j − k), a a 1 1 b′ = (i + k)×(i + j) = (−i + j + k), a a 1 1 c′ = (i + j)×(j + k) = (i − j + k). a a 8.16

From (8.41b), with appropriate changes in notation, we see that the minimum distance between them vanishes, i.e. they intersect if

a ⋅(c×d) = b ⋅(c×d) .

S8.6

Using the cyclic property of scalar triple products (8.20), then gives the desired result

d ⋅(a ×c) = d ⋅(b×c) . At the point of intersection, a + λc = b + µd , and hence

a ×d + λc×d = b×d , since d×d = 0 , and λa ⋅(c×d) = a ⋅(b×d) because a ⋅(a ×d) = 0 . Therefore,

λ=

a ⋅(b×d) a ⋅(c×d)

µ=

b ⋅(a ×c) . b ⋅(d×c)

and a similar argument gives

8.17

The corresponding position vectors are

p = 3i + 2j + 4k,

a = i − k,

b = −i + 2j + k .

(1)

The equation of the line through AB is

r = a + s(b − a) = (i − k) + s(−2i + 2j + 2k) and the shortest distance is given by (cf. (8.41a))

d=

(p − a)×(b − a) b−a

.

On substituting, we have

(p − a) = 2i + 2j + 3k , so that

(p − a)×(b − a) = −2i −10j + 8k and (2) gives d = 168

12 = 14 .

S8.7

(2)

From the figure above, we see that the distance

AD = p − a cos θ =

(p − a)⋅(b − a) b−a

,

and hence

s=

(p − a)⋅(b − a) AD = . (b − a)2 b−a

On substituting for (p − a) and (b − a) , we obtain s = 1 2 , so that from (1)

d = a + 12 (b − a) = j and D is the point (0,1, 0) on the y axis. 8.18

(a) The distance from a to r is a constant, and so the surface is spherical, centre A, radius λ . (b) (r − a) is perpendicular to a. Hence the surface is a plane perpendicular to a and passing through the point (ax ,ay ,az ) .

ˆ)ˆ ˆ and r −(r ⋅ n ˆ) is the vector projection of r in the direction of n n is the (c) (r ⋅ n ˆ . Therefore the surface is that of a circular cylinder component of r perpendicular to n ˆ. of radius λ with its axis parallel to n 8.19

Referring to the figure below, r = x i + y j + z k is the position vector of the point

P(x,y,z) in the plane and Q is the end point of the vector a that is also in the plane. Then PQ = a − r is perpendicular to n and hence (a − r)⋅ n = 0, i.e. r ⋅ n = n ⋅ a ,

S8.8

which is the vector form of the plane. Evaluating this in Cartesian co-ordinates using the given forms for n and a, gives

x − 2y − 3z = 7 .

The distance from the origin (0, 0, 0) to the plane is the projection of a onto n. A unit vector in the direction of n is

n 1 = (i − 2j − 3k) , n 14

ˆ= n and so

ˆ= a ⋅n 8.20

1 14

(i + 3j − 4k)⋅(i − 2j − 3k) =

7 2

.

The three position vectors are:

r1 = i − 2j + 2k, r2 = 3i + 2j − k, r3 = −i + 3j − k , and the general point in the plane is P(x,y,z) . Any of the lines P1P2,PP2,P1P3 etc lie in the plane. In particular,

PP1 = r − r1; P2P1 = r2 − r1; P3P1 = r3 − r1 all lie in the plane. Then by (8.22a)

(r − r1 )⋅ ⎡⎢(r2 − r1 )×(r3 − r1 )⎤⎥ = 0 ⎣ ⎦ and evaluating this gives

x + 4y + 6z = 5 , which is the equation for the plane in Cartesian co-ordinates. 8.21

On comparing to (8.45a,b) we see that the normal to the planes are

S8.9

n1 = i + 2j + 3k,

n 2 = 2i + 3j + 4k .

Since the angle θ between the planes is the same as the angle between the normals,

cos θ =

n1 ⋅ n 2 n1 n 2

=

20 , 14 ×29

giving θ = 1.52 rad = 87.20 . Along the line of intersection,

x + 2y + 3z + 4 = 0 and 2x + 3y + 4z + 5 = 0. Solving these gives x = z + 2 and y = −2z − 3 . So, setting z = s , we obtain

r = xi + yj + zk = (s + 2)i −(2s + 3)j + sk = (2i − 3j) + s(i − 2j + k), as the equation of the line in vector form. 8.22

The curve r is given by

r = (t 2 + 1)i + (4t − 3)j + (2t 2 − 6t)k and the tangent is therefore

dr = 2t i + 4 j + 2(2t − 3)k . dt

T= At the point t = 2 ,

T= and the unit tangent is

8.23

dr = 4i + 4j + 2k dt

ˆ = 2 i + 2 j + 1 k. T 3 3 3

We can write

d ⎛⎜ da ⎞ d2a da da ⎜⎜a × ⎟⎟⎟ = a × 2 + × . dt ⎜⎝ dt ⎟⎠ dt dt dt But the second term is zero, and hence

⌠ ⎛ da ⎞ da I = ⎮ d ⎜⎜⎜a × ⎟⎟⎟ = a × + c ⎟ dt ⎠ dt ⌡ ⎜⎝ where c is a constant vector.

S8.10

SOLUTIONS TO PROBLEMS 9 9.1

The cross product is

i

j

k

a×b = 1

2

3 = 7i + 10j − 9k .

4 −1 2 The triple product is

4 −1 2 b ⋅ a ×c = 1 2 3 = 4(6 + 18) + 1(3 −15) + 2(−6 −10) = 52. 5 −6 3 9.2

(a) (i) The Laplace expansion about the third column gives

det A = 3

1 i −4 2 2 −i − 2i 2 2 −i 1+i 1 1+i 1 1 i

= 3(2 −i) + 4(1 + i) + 2i(2 − 3i) = 16 + 5i. (ii) About the first row gives,

4 +3 1 i 4 −(2 −i) 1 det A = 2 i 1 + i −2i 1+i 1 1 −2i = 2(2 − 4)−(2 −i)(4 + 6i) + 3(2 −i) = 16 + 5i. (b) Labelling the rows and columns by r and c, we have, for example,

14 7 r1 → r1 −r3 gives det A =

8 3 −1 , 13 7

c3 → c3 +c1 gives det A =

2 3

14 7 16 8 3 7 , 13 7 16

c1 → c1 −c2

7 7 16 gives det A = 5 3 7 , 6 7 16

7 7 9 c3 → c3 −c2 gives det A = 5 3 4 , 6 7 9 and finally

S9.1

1 0 0 r1 → r1 −r3 gives det A = 5 3 4 6 7 9 = 3×9 − 4 ×7 = −1. 9.3

This may be obtained in several ways using the rules for manipulating determinants. For example, adding multiples of row 1 to rows 2, 3, and 4, reduces column 3 to 1, 0, 0, 0:

3

7

1

2

9

18

0

3

2

4

0 −1

−14 −35

0 −6

Δ=

9

18

3

2

4

−1

−14 −35

−6

=

,

where we have expanded about column 3. Replacing c2 → c2 − 2c1 , then gives

Δ=

9

0

3

2

0 −1 = 7

−14 −7 −6 9.4

9

3

2 −1

= −105.

(a) By expanding along row 1, for example, one obtains

Δ1 = −x(x −1)(x + 2) = 0, so that x = 0,1 or − 2 . (b) If α = β , β = γ or α = γ , the determinant vanishes, since it would then have identical columns. So we can write

Δ2 = k(α − β)(β − γ)(γ − α) . Since all the terms in Δ2 are of the fourth power in α, β and γ , and since Δ2 is antisymmetric, k must be symmetric and of the first power in α, β and γ . Hence

Δ2 = c(α + β + γ)(α − β)(β − γ)(γ − α). Comparing the coefficient of, for example, βγ 3 gives c = 1 . 9.5

Expanding along the first row gives

S9.2

1

1

0

0 

0

0 −2

1

0 

0

1 −2

1 

0 = −2Δn−1 − Δn−2 .













0

0

0

0

1

−2

Δn = −2Δn−1 − 0

Hence Sn = Δn + Δn−1 = −(Δn−1 + Δn−2 ) = −Sn−1 for n > 2 , so that

Sn = −Sn−1 = (−1)2 Sn−2 = (−1)n−2 S 2 = (−1)n on evaluating S 2 explicitly. Hence

Δn = (−1)n − Δn−1 = (−1)n −[(−1)n−1 − Δn−2 ] n

= 2(−1)n + Δn−2 = 3(−1) − Δn−3 etc. = m(−1)m + (−1)m Δn−m . Putting m = n − 2 , and evaluating Δ2 = 3 explicitly, gives

Δn = (n − 2)(−1)n−2 + 3(−1)n−2 = (n + 1)(−1)n . 9.6

(a) Using c2 → c2 +c3 gives

2 −5

5

2

0

5

2 5 = −1 1 −2 = 4 −1 −2 = − 1 3 1 0 3 1 −3 3 4

As this is non-zero, the set (a) has no non-trivial solutions. (b) Using r1 → r1 − r3 gives

0 −2 2 3 5 4 1 1 2 , = 1 1 2 3 7 2 3 7 2 and then using c2 → c2 +c3 ,

0 −2 2 0 0 2 1 1 2 = 1 3 2 =2 1 3 =0. 3 9 3 7 2 3 9 2

S9.3

As this is zero, the set (b) does have non-trivial solutions. By (9.13a), they are

x :y :z = 1 2 :− 1 2 : 1 1 7 2 3 2 3 7 = −12 : 4 : 4 = −3 : 1 : 1. 9.7

This is equivalent to finding a solution of the homogeneous equations

αx + 3y − 2z = 0, −3x + αy + (α + 4)z = 0, − x + 3y + 4z = 0,

(1)

with z = 1 . Such a solution exists if

α 3 −2 α 3 −5 −3 α α + 4 = −3 α 4 −1 3 1 −1 3 4 = α2 −17α + 42 = (α − 3)(α −14) = 0, i.e. if α = 3 or 14. For the latter, (1) becomes

14x + 3y − 2z = 0 −3x + 14y + 18z = 0 − x + 3y + 4z = 0 with solutions

x :y :z =

14 18 −3 18 −3 14 :− : 3 4 −1 4 −1 3

= 2 : −6 : 5 by (9.13a). Since z = 1 , the values of x and y are x = 2 5, y = −6 5 . 9.8

(a) 2

a + b = ([a + b ],[a + b ]) = (a, a) + (a, b ) + (b, a) + (b, b ), which using

(b, a) = (a, b )* is 2

2

2

2

2

2

a + b = a + b + 2Re(a, b). Likewise,

a − b = a + b − 2Re(a, b), and hence 2

2

2

2

a + b + a − b = 2[ a + b ] .

S9.4

(b) Using, from (a), 2

2

2

a + b = a + b + 2Re(a, b) and the Schwarz inequality

(a, b) ≤ a b , gives 2

2

2

a+b ≤ a + b

2

+ 2 a b = ⎡⎢ a + b ⎤⎥ . ⎣ ⎦

So

a+b ≤ a + b .

9.9

(a)

⎛ 1 −2 0 ⎞⎟ ⎛ 9 3 −6 ⎞⎟ ⎛ −8 −5 6 ⎞⎟⎟ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ ⎜ ⎟ ⎜ ⎟ ⎜ A − 3B = ⎜⎜ 3 2 5 ⎟⎟ − ⎜⎜ 3 0 6 ⎟⎟ = ⎜⎜ 0 2 −1 ⎟⎟ , ⎟ ⎟ ⎟⎟ ⎜⎜ ⎟ ⎜ ⎟ ⎜⎜ ⎜⎜⎝ −1 3 1 ⎟⎟⎠ ⎜⎜⎜⎝ −6 12 9 ⎟⎟⎠ ⎜⎜⎝ 5 −9 −8 ⎟⎟⎠ ⎛ 1 1 −6 ⎞⎟⎟ ⎜⎜ ⎟ ⎜ AB = ⎜⎜ 1 23 13 ⎟⎟⎟, ⎟⎟ ⎜⎜ ⎜⎜⎝ −2 3 11 ⎟⎟⎠

⎛ 8 −10 3 ⎞⎟⎟ ⎜⎜ ⎟⎟ ⎜ BA = ⎜⎜ −1 4 2 ⎟⎟. ⎟⎟ ⎜⎜ ⎜⎜⎝ 7 21 23 ⎟⎟⎠

(b) XY is defined only if the number of columns in X is the same as the number of columns in Y. So AC and DA are not defined, but

⎛ 1 −7 −2 ⎞⎟ ⎜ ⎟⎟, CA = ⎜⎜ ⎜⎜ 21 4 28 ⎟⎟⎟ ⎝ ⎠ ⎛ 25 25 ⎜ CD = ⎜⎜ ⎜⎜ 17 −16 ⎝

⎞⎟ ⎟⎟, ⎟⎟ ⎟⎠

⎛ 3 6 ⎜⎜ AD = ⎜⎜⎜ 2 17 ⎜⎜ ⎜⎝ −5 −5 ⎛ 37 7 ⎜⎜ ⎜⎜ DC = ⎜ 5 −13 ⎜⎜ ⎜⎜⎝ −18 21

⎞⎟ ⎟⎟ ⎟⎟, ⎟⎟ ⎟⎟ ⎠ 11 ⎞⎟⎟ ⎟⎟ 7 ⎟⎟, ⎟⎟ −15 ⎟⎟⎠

are defined. 9.10

(a) The scalar product of σ and a is

⎛ 0 ⎜ (σ ⋅ a) = ⎜⎜ ⎜⎜ a ⎝ x

⎛ ax ⎞⎟⎟ ⎜⎜ 0 ⎟⎟ + ⎜ 0 ⎟⎟⎠ ⎜⎜⎜⎝ iay

−iay ⎞⎟⎟ ⎛⎜ az ⎟⎟ + ⎜⎜ 0 ⎟⎟⎟⎠ ⎜⎜⎝ 0

Thus

S9.5

⎞⎟ ⎛⎜ az ax −iay ⎟⎟ = ⎜⎜ ⎟⎟ ⎜ −az ⎟⎠ ⎜⎜⎝ ax + iay −az 0

⎞⎟ ⎟⎟ ⎟⎟ . ⎟⎟ ⎠

⎛ az ax −iay ⎞⎟⎟ ⎛⎜⎜ az ax −iay ⎜ ⎟⎟ ⎜ (σ ⋅ a)(σ ⋅ a) = ⎜⎜⎜ ⎜⎜ ax + iay −az ⎟⎟⎟⎠ ⎜⎜⎜⎝ ax + iay −az ⎝ ⎛ ⎞ ⎛ ⎞ = ⎜⎜⎜ a ⋅ a 0 ⎟⎟⎟ = a 2 ⎜⎜⎜ 1 0 ⎟⎟⎟ = a 2 I. ⎜⎝ 0 a ⋅ a ⎟⎠ ⎜⎝ 0 1 ⎟⎠

⎞⎟ ⎟⎟ ⎟⎟ ⎟⎟ ⎠

(b) We have

⎛ 0 1i 0 ⎜ 1 ⎜⎜ M± = ⎜⎜ 1 ± i 0 1i 2 ⎜⎜ 1±i 0 ⎝ 0

⎞⎟ ⎟⎟ ⎟⎟ , ⎟⎟ ⎟⎟ ⎠

so that

⎛ 2 0 0 ⎜⎜ ⎡ M , M ⎤ = (M M ) − (M M ) = ⎜⎜ 0 0 0 − ⎦⎥ + − − + ⎜⎜ ⎣⎢ + ⎜⎜ ⎝ 0 0 −2 9.11

⎞⎟ ⎟⎟ ⎟⎟ = 2M . z ⎟⎟ ⎟⎟ ⎠

The rotation does not change z, while x and y transform as in (9.48) and Figure 9.1, so that

⎛ cos θ −sin θ 0 ⎞⎟ ⎜⎜ ⎟ R(θ) = ⎜⎜ sin θ cos θ 0 ⎟⎟⎟. ⎜⎜ ⎟⎟ ⎜⎝ 0 0 1 ⎟⎠ Then

⎛ cos θ −sin θ 0 ⎞⎟⎛ cos θ −sin θ 0 ⎞⎟ ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ 1 1 2 2 ⎜⎜ ⎟⎟⎜⎜ ⎟ R(θ1 )R(θ2 ) = ⎜ sin θ1 cos θ1 0 ⎟⎟⎜⎜ sin θ2 cos θ2 0 ⎟⎟⎟ . ⎜⎜ ⎟⎟⎜ ⎟⎟ ⎜⎜ 0 0 1 ⎟⎟⎠⎜⎜⎝ 0 0 1 ⎟⎟⎠ ⎝ Multiplying out, and using the trigonometric identities for cos(θ1 + θ2 ) and

sin(θ1 + θ2 ) , gives R(θ1 )R(θ2 ) = R(θ1 + θ2 ) = R(θ2 + θ1 ) = R(θ2 )R(θ1 ) . Similarly, evaluating the inverse gives

⎛ cos θ sin θ 0 ⎜⎜ R−1(θ) = ⎜⎜⎜ −sin θ cos θ 0 ⎜⎜ 0 0 1 ⎜⎝

⎞⎟ ⎛⎜ cos(−θ) −sin(−θ) 0 ⎞⎟ ⎟⎟ ⎟⎟ ⎜⎜ ⎟ ⎟⎟ = ⎜ sin(−θ) cos(θ) 0 ⎟⎟ = R(−θ) ⎟⎟ ⎜⎜ ⎟⎟ ⎟⎟ ⎜⎜ 0 0 1 ⎟⎟⎠ ⎠ ⎜⎝

and also R−1(θ) = RT (θ) as required. 9.12

(a) Since the rotation about the x axis acts first, we have

S9.6

R(θ1 , θ2 ) = Ry (θ2 )R x (θ1 ) ⎛ cos θ ⎜⎜ 2 ⎜ = ⎜⎜ 0 ⎜⎜ ⎜⎝ −sin θ2 ⎛ cos θ ⎜⎜ 2 ⎜ = ⎜⎜ 0 ⎜⎜ ⎜⎜ −sin θ 2 ⎝

0 sin θ2 ⎞⎟⎟⎛⎜⎜ 1 0 0 ⎟⎟⎜ 0 cos θ −sin θ1 ⎜ 1 0 ⎟⎟⎟⎜⎜ 1 ⎟⎟⎜ 0 cos θ2 ⎟⎠⎜⎝ 0 sin θ1 cos θ1 sin θ1 sin θ2 sin θ2 cos θ1 ⎞⎟⎟ ⎟⎟ cos θ1 −sin θ1 ⎟⎟⎟, ⎟⎟ sin θ1 cos θ2 cos θ1 cos θ2 ⎟⎟⎠

⎞⎟ ⎟⎟ ⎟⎟ ⎟⎟ ⎟⎟ ⎟⎠

as required. If instead, one considers

⎛ 1 0 0 ⎜⎜ ⎜ R x (θ1 )Ry (θ2 )= ⎜⎜ 0 cos θ1 −sin θ1 ⎜⎜ cos θ1 ⎜⎝ 0 sin θ1

⎞⎟⎛ cos θ 0 sin θ ⎟⎟⎜⎜ 2 2 ⎟⎟⎜⎜ 0 1 0 ⎟⎟⎜⎜ ⎟⎟⎜ ⎟⎠⎜⎝ −sin θ2 0 cos θ2

⎞⎟ ⎟⎟ ⎟⎟, ⎟⎟ ⎟⎟ ⎟⎠

then, for example, the 12 element of R x (θ1 )Ry (θ2 ) is zero, whereas for Ry (θ2 )R x (θ1 ) it is sin θ1 sin θ2 . So in general the two rotation matrices do not commute (unless θ1 or

θ2 is zero). (b) Since R(θ1 , θ2 ) = Ry (θ2 )R x (θ1 ) , then

R−1(θ1, θ2 ) = R−1 (θ1 )R−1 (θ2 ) = R x (−θ1 )Ry (−θ2 ) . x y But from the explicit expressions for R x and Ry , one sees that R x (−θ1 ) = RTx (θ1 ) and

Ry (−θ2 ) = RTy (θ2 ) , so that R−1(θ1, θ2 ) = RTx (θ1 )RTy (θ2 ) = [Ry (θ2 )R x (θ1 )]T = RT (θ1 , θ2 ), as required. Finally,

det R(θ1 , θ2 ) = det[Ry (θ2 )R x (θ1 )] det Ry (θ2 )det R x (θ1 ) = 1, since det R x (θ1 ) = det Ry (θ2 ) = 1 by explicit evaluation. 9.13

(a)

(A + B)3 = (A + B)(A + B)(A + B) = (A + B)(A2 + AB + BA + B2 ) = (A3 + A2B + ABA + AB2 + BA2 + BAB + B2A + B3 )

S9.7

(b)

e Ae B = (I + A + 12 A2 +)(I + B + 12 B2 +) = I + A + B + 12 (A2 + 2AB + B2 ) + e (A+B) = I + (A + B) + 12 (A + B)2  = I + (A + B) + 12 (A2 + AB + BA + B2 ) + so that

e Ae B = e (A+B)

(1)

requires that A and B commute, i.e. AB = BA . If this is satisfied, then A and B behave algebraically like ordinary numbers and (1) is satisfied to all orders in A and B. 9.14

The three matrices are:

⎛ i 2i 2 ⎜⎜ AT = ⎜⎜ 2 1 1 + i ⎜⎜ ⎜⎝ −3 + i 3 2

⎞⎟ ⎟⎟ ⎟⎟, ⎟⎟ ⎟⎠

⎛ −i 2 −3 −i ⎜⎜ A* = ⎜⎜ −2i 1 3 ⎜⎜ ⎜⎝ 2 1 −i 2

⎞⎟ ⎟⎟ ⎟⎟, ⎟⎟ ⎟⎠

and

⎛ −i −2i 2 ⎜⎜ A = ⎜⎜ 2 1 1−i ⎜⎜ ⎜⎝ −3 − i 3 2 †

9.15

⎞⎟ ⎟⎟ ⎟⎟. ⎟⎟ ⎟⎠

(a) A matrix A is unitary if it satisfies the relation AA† = A†A = I . Now A† is

⎛ ⎜⎜ 0 (−1 −i) 6 2 6 ⎜ † A = ⎜⎜ 1 0 0 ⎜⎜ ⎜⎜ 0 (1 + i) 3 1 3 ⎝

⎞⎟ ⎟⎟ ⎟⎟ ⎟⎟ , ⎟⎟ ⎟⎟⎠

so

⎛ ⎞⎟ ⎛ ⎞ 0 1 0 ⎜⎜ ⎟⎟ ⎜⎜ 0 (−1 −i) 6 2 6 ⎟⎟⎟ ⎜ ⎟⎜ ⎟ AA† = ⎜⎜ (−1 + i) 6 0 (1 −i) 3 ⎟⎟ ⎜⎜ 1 0 0 ⎟⎟⎟ ⎟⎟ ⎜⎜ ⎜⎜ ⎟⎟ ⎟⎟ ⎜ ⎜⎜ 2 6 0 1 3 ⎟⎠ ⎜⎝ 0 (1 + i) 3 1 3 ⎟⎟⎠ ⎝ ⎛ 1 0 1 ⎜⎜ ⎜ 0 (1 −i)(1 + i) 6 + (1 −i)(1 + i) 3 (−1 + i) 3 + (1 −i) 3 = ⎜⎜ ⎜⎜ −(1 + i) 3 + (1 + i) 3 2 3+1 3 ⎜⎝ 0 Likewise, A†A = I . (b) By construction,

⎛ A = ⎜⎜⎜ 1 2 ⎜⎝ 3 2 T

T

⎞⎟ ⎛ ⎞ ⎟⎟ = ⎜⎜ 1 3 ⎟⎟⎟, ⎟⎠ ⎜⎜⎝ 2 2 ⎟⎠

S9.8

⎞⎟ ⎛ ⎟⎟ ⎜⎜ 1 0 0 ⎟⎟ = ⎜ ⎟⎟ ⎜⎜ 0 1 0 ⎟⎟ ⎜⎜⎝ 0 0 1 ⎠

⎞⎟ ⎟⎟ ⎟⎟. ⎟⎟ ⎟⎠

and AS = 12 (A + AT ) is a symmetric and AAS = 12 (A − AT ) is an anti-symmetric matrix. So,

⎞ 1⎛ AS = ⎜⎜⎜ 2 5 ⎟⎟⎟, 2 ⎜⎝ 5 4 ⎟⎠

⎞ 1⎛ AAS = ⎜⎜⎜ 0 −1 ⎟⎟⎟, 2 ⎜⎝ 1 0 ⎟⎠

and

and

1⎛ A = ⎜⎜⎜ 2 5 2 ⎜⎝ 5 4 9.16

⎞⎟ 1 ⎛ 0 −1 ⎞⎟ ⎟⎟ + ⎜⎜ ⎟. ⎟⎠ 2 ⎜⎜⎝ 1 0 ⎟⎟⎠

Using Equations (9.51), (9.53a) and the definitions given in Section 9.4.1, it follows that A is symmetric, B is hermitian, D is unitary and E is orthogonal. Matrix C has none of these properties. Using C, an anti-symmetric matrix can be formed from

Mantisymmetric = C − C T ⎛ 1 1 −i 2i ⎞⎟⎟ ⎛⎜ 1 ⎜⎜ −3i 1 + 3i ⎟ ⎜ = ⎜⎜⎜ −3i −2i 1 ⎟⎟⎟ − ⎜⎜ 1 −i −2i 2 ⎟⎟ ⎜⎜ ⎜⎜ 2 3 ⎟⎠ ⎜⎝ 2i 1 3 ⎜⎝ 1 + 3i

⎞⎟ ⎛⎜ 0 1 + 2i −1 −i ⎟⎟ ⎜ ⎟⎟ = ⎜⎜ −1 − 2i 0 −1 ⎟⎟ ⎜⎜ ⎟⎠ ⎝⎜ 1 + i 1 0

⎞⎟ ⎟⎟ ⎟⎟, ⎟⎟ ⎟⎠

and anti-hermitian matrix from

Manti-hermitian = C − C† ⎛ 1 1 −i 2i ⎜⎜ = ⎜⎜ −3i −2i 1 ⎜⎜ ⎜⎝ 1 + 3i 2 3 9.17

⎞⎟ ⎛ 1 3i 1 − 3i ⎟⎟ ⎜⎜ ⎟⎟ − ⎜⎜ 1 + i 2i 2 ⎟⎟ ⎜⎜ ⎟⎠ ⎜⎝ −2i 1 3

⎞⎟ ⎛⎜ 0 1 − 4i −1 + 5i ⎟⎟ ⎜ ⎟⎟ = ⎜⎜ −1 − 4i −4i −1 ⎟⎟ ⎜⎜ ⎟⎠ ⎜⎝ 1 + 5i 1 0

⎞⎟ ⎟⎟ ⎟⎟. ⎟⎟ ⎟⎠

(a) From the definition, S = ST and A = −AT , so that

Tr(SA) = ∑ (SA)ii = ∑ ∑ Sik Aki = ∑ ∑ Ski Aik i

i

T

k

k

i

T

= Tr(S A ) = −Tr(SA), which implies that Tr(SA) = 0 . (b) Let the two diagonal matrices be Dij = dii δij and Fij = fii δij . Then,

(DF)ij = ∑ Dik Fkj = ∑ dii δik f jj δkj = dii f jj ∑ δik δkj k

k

k

= dii f jj δij = dii fii . Similarly, by interchanging F and D we can show that (FD)ij = fii dii so that

FD = DF and F and D commute. 9.18

The cofactors of A are

S9.9

A11 = −1, A12 = 10, A13 = −7, A21 = 6, A22 = 14, A23 = 5, A31 = 10, A32 = 11, A33 = −4, and so the transposed matrix of cofactors is

⎛ −1 ⎜⎜ ⎜ adjA = ⎜⎜ 10 ⎜⎜ ⎜⎜⎝ − 7

10 ⎞⎟⎟ ⎟⎟ 11 ⎟⎟ . ⎟⎟ − 4 ⎟⎟⎠

6 14 5

The determinant of A is

det A = −3 − 20 −14 = −37.

So

A−1

⎛ −1 ⎜ adjA −1 ⎜⎜⎜ = = ⎜ 10 detA 37 ⎜⎜ ⎜⎜⎝ − 7

6 14 5

10 ⎞⎟⎟ ⎟⎟ 11 ⎟⎟ . ⎟⎟ − 4 ⎟⎟⎠

To check, we form

AA−1

9.19

⎛ 3 ⎜ −1 ⎜⎜⎜ = ⎜ 1 37 ⎜⎜ ⎜⎜⎝ −4

2 ⎞⎟⎟⎛⎜⎜ −1 ⎟⎟⎜ − 2 −3 ⎟⎟⎜⎜ 10 ⎟⎟⎜⎜ 1 2 ⎟⎟⎠⎜⎜⎝ − 7

−2

6 14 5

10 ⎞⎟⎟ ⎛⎜ 1 0 0 ⎞⎟ ⎟⎟ ⎜ ⎟ 11 ⎟⎟ = ⎜⎜ 0 1 0 ⎟⎟⎟ = I. ⎟⎟ ⎜⎜ ⎟⎟ − 4 ⎟⎟⎠ ⎜⎝ 0 0 1 ⎟⎠

The determinant of A is

−2 1

2

−2 1

A = −1 1

1

= −1 1

0 2 −1

0

−2 1 = 1, 0 =− −1 1 0 2 −1

so A is non-singular. The co-factors are:

A11 = −3, A12 = −1, A13 = −2, A21 = 5,

A22 = 2,

A31 = −1, A32 = 0,

A23 = 4, A33 = 1.

So, by (9.65),

A−1

⎛ −3 5 −1 ⎞⎟ ⎜⎜ ⎟⎟ ⎟⎟ = ⎜⎜ −1 2 0 ⎜⎜ ⎟ ⎜⎝ −2 4 −1 ⎟⎟⎟⎠

and

S9.10

⎛ 6 −2 ⎞⎟ ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎜⎜ ⎟⎟ ⎜⎜ −3 5 −1 ⎟⎟⎟⎜⎜ 6 −2 ⎟⎟⎟ ⎜⎜ 1 3 ⎟⎟ ⎜ ⎜ 0 ⎟⎟⎟ = ⎜⎜ −1 2 0 ⎟⎟⎟ = ⎜⎜ 2 2 ⎟⎟⎟ . X = A ⎜⎜ 4 0 ⎟⎟⎟⎜⎜ 4 ⎟⎟ ⎟⎟ ⎜⎜ ⎟ ⎜ ⎜⎜ ⎜ ⎟ 3 ⎟⎠ ⎜⎝ −2 4 −1 ⎟⎟⎠⎜⎜⎝ 1 3 ⎟⎟⎠ ⎜⎜⎝ 3 1 ⎟⎠ ⎜⎝ 1 −1

9.20

In matrix notation Ax = b , where

⎛ 3 1 −1 ⎞⎟⎟ ⎜⎜ ⎟⎟ ⎜ A = ⎜⎜ 1 −1 1 ⎟⎟ , ⎟⎟ ⎜⎜ ⎜⎜⎝ −2 2 2 ⎟⎟⎠

⎛ x ⎞⎟ ⎛ 1 ⎞⎟ ⎜⎜ ⎜⎜ ⎟⎟ ⎟ x = ⎜⎜ y ⎟⎟ and b = ⎜⎜ 2 ⎟⎟⎟. ⎜⎜ ⎟ ⎜⎜ ⎟ ⎜⎝ 3 ⎟⎟⎠ ⎜⎝ z ⎟⎟⎠

The solution is therefore x = A−1b and we find

A−1

⎛ 1 1 ⎜ 1 ⎜⎜⎜ = ⎜ 1 −1 4 ⎜⎜ ⎜⎜⎝ 0 2

0 ⎞⎟⎟ ⎟⎟ 1 ⎟⎟ , ⎟⎟ 1 ⎟⎟⎠

so that,

⎛ 1 ⎛ x ⎞⎟ 1 ⎜ ⎜⎜ ⎟⎟ 1 ⎜⎜ ⎜⎜ y ⎟⎟ = ⎜ 1 −1 ⎜⎜ ⎟ 4 ⎜⎜ ⎜⎜ ⎜⎝ z ⎟⎟⎠ 2 ⎜⎝ 0

0 ⎞⎟⎟ ⎛⎜ 1 ⎞⎟ ⎟⎟ ⎜ ⎟ 1 ⎟⎟ ⎜⎜ 2 ⎟⎟⎟ = ⎟⎟ ⎜⎜ ⎟⎟ 1 ⎟⎟⎠ ⎜⎝ 3 ⎟⎠

⎛ ⎞ 1 ⎜⎜⎜ 3 ⎟⎟⎟ ⎜ 2 ⎟⎟⎟, 4 ⎜⎜ ⎜⎝ 7 ⎟⎟⎠

and hence x = 3 4, y = 1 2 and z = 7 4. 9.21

Using the notations of equation (9.72), we have

⎛ 2 3 −1 ⎞⎟⎟ ⎜⎜ ⎟⎟ ⎜ A = ⎜⎜ 1 −1 1 ⎟⎟ with det A ≡ Δ = −15. ⎟⎟ ⎜⎜ ⎜⎜⎝ −1 1 2 ⎟⎟⎠ Also

⎛ 0 3 −1 ⎞⎟⎟ ⎜⎜ ⎟⎟ ⎜ Δx = det ⎜⎜ 1 −1 1 ⎟⎟ = −3, ⎟⎟ ⎜⎜ ⎜⎜⎝ 2 1 2 ⎟⎟⎠

⎛ 2 0 −1 ⎞⎟ ⎜⎜ ⎟⎟ ⎜ ⎟ Δy = det ⎜⎜ 1 −1 1 ⎟⎟ = −3 ⎟⎟ ⎜⎜ ⎜⎜⎝ −1 1 2 ⎟⎟⎠

and

⎛ 2 3 0 ⎞⎟⎟ ⎜⎜ ⎟ ⎜ Δz = det ⎜⎜ 1 −1 1 ⎟⎟⎟ = −15. ⎟⎟ ⎜⎜ ⎜⎜⎝ −1 1 2 ⎟⎟⎠ so that

x= 9.22

Δ Δ 1 1 = , y = y = , z = x = 1. Δ 5 Δ 5 Δ

Δx

Since the mass of a sample is proportional to the number of atoms it contains, from the definition of the half-life, its mass at time t is given by

S9.11

m(t) = m0 2−t

τ

,

where m0 = m(t = 0) is the initial mass. Thus at t = 6 hr,

m(6) = ma 2−4 + mb 2−2 = 90, i.e.

ma + 4mb = 90×24 . Similarly at t = 12 hr,

ma + 8mb = 30×26 . To use Cramer’s rule, we need the determinant of coefficients

Δ = 1 4 = 4. 1 8 Then, using Equations (9.72) gives

ma =

1 90×24 4 30×26

mb =

1 1 90×24 4 1 30×26

4 = 10×24 9 1 = 960 gm, 8 12 2

and

9.23

=

30×24 4

1 3 1

4

= 120 gm.

(a) the condition for a unique solution is

4 2 α det A = 7 3 4 ≠ 0 1 1 2 i.e. 4a −12 ≠ 0 . Thus, unique solutions exist for α ≠ 3 , independent of the value of β. (b) For α = 2 det A = −4 and

A−1

⎛ 2 ⎜ 1 ⎜⎜⎜ = − ⎜ −10 4 ⎜⎜ ⎜⎜⎝ 4

−2 6 −2

2 ⎞⎟⎟ ⎟⎟ − 2 ⎟⎟ , ⎟⎟ − 2 ⎟⎟⎠

where the matrix is the transposed matrix of the co-factors of A. Thus,

S9.12

⎛ x ⎞⎟ ⎛ 3 ⎞⎟ ⎛⎜ 12 ⎜⎜ ⎜ ⎜ ⎟⎟ ⎜⎜ y ⎟⎟ = A−1 ⎜⎜ 8 ⎟⎟⎟ = ⎜⎜ −5 2 ⎜⎜ ⎟ ⎜ ⎜⎜ ⎟ ⎜⎜⎝ 4 ⎟⎟⎟⎠ ⎜⎜⎜ ⎜⎝ z ⎟⎟⎠ 3 ⎜⎝

⎞⎟ ⎟⎟ ⎟⎟ . ⎟⎟ ⎟⎟ ⎟⎠

(c) For α = 3 , multiplying the first equation by 2 and comparing to the sum of the second and third equations, gives

8x + 4y + 6z = 2β

and

8x + 4y + 6z = 12 ,

respectively. So in case (i) β = 6 , a solution exists, but is not unique, while in case (ii) β = 2 , no solutions exist.

S9.13

SOLUTIONS TO PROBLEMS 10 10.1

The eigenvalues are the solutions of

1 −λ 2 −1 = −(λ 3 − 5λ 2 + 5λ + 3) = 0. 1 2 −λ 0 −1 1 2 −λ Since λ = 3 is a solution, we find

λ 3 − 5λ 2 + 5λ + 3 = (λ − 3)(λ 2 − 2λ −1) = 0 , with solutions λ1 = 3, λ2 = 1 − 2 , λ3 = 1 + 2 . The associated eigenvectors u(1, 2, 3) are found by solving the equations Au(1, 2, 3) = λ1, 2,3u(1, 2, 3) and then normalising the results. This gives

⎛ ⎞ 1 ⎜⎜⎜ 1 ⎟⎟⎟ (2) (1) u = ⎜ 1 ⎟⎟⎟ , u = ⎟ 2 ⎜⎜⎜ ⎝ 0 ⎟⎠

⎛ ⎜⎜ −(1 + 2) 1 ⎜⎜ ⎜⎜ 1 6 + 2 2 ⎜⎜⎜ − 2 ⎝

⎞⎟ ⎟⎟ ⎟⎟ , u(3) = ⎟⎟ ⎟⎟ ⎟⎠

⎛ ⎞ ⎜⎜ −(1 − 2) ⎟⎟ ⎟⎟ 1 ⎜⎜ ⎟⎟. ⎜⎜ 1 ⎟⎟ ⎜ 6 − 2 2 ⎜⎜ ⎟⎟ 2 ⎝ ⎠

Using (9.22) it is straightforward to show, for example, that (u(1), u(2) ) ≠ 0 and hence the eigenvectors are not orthogonal. 10.2

The eigenvalues are obtained from

1 −λ 3 2 = 0. 1 2 −λ 3 3 2 1 −λ Expanding the determinant gives

(1 −λ)(λ 2 − 3λ − 4) + (9λ + 16) = λ 3 − 4λ 2 −10λ −12 , = (λ − 6)(λ + 1 −i)(λ + 1 + i) = 0 with solutions λ1 = 6, λ2 = −1 + i , λ3 = −1 − i and hence 3

∑ λi = 4 = Tr A

and

i=1

10.3

3

∏λ i=1

The eigenvalues are obtained from

S10.1

i

= 12 = det A .

2 −λ 1 1 =0, −1 −λ 1 1 2 2 −λ i.e.,

(2 −λ)(λ 2 − 2λ − 2)−(λ − 3) + (λ − 2) = λ 3 − 4λ 2 + 2λ + 3 = (λ − 3)(λ 2 −λ −1) = 0 . The eigenvalues are thus λ1 = 3, λ2 = 1.618, λ3 = −0.618 . The inverse of A is

A−1

⎛ 2 0 −1 ⎞⎟⎟ ⎜ ⎟⎟ 1 ⎜⎜⎜ = ⎜ −3 −3 3 ⎟⎟ ⎟⎟ 3 ⎜⎜ ⎜⎜⎝ 2 3 −1 ⎟⎠⎟

from which we can find the eigenvalues in the same way as for A. They are

µ1 = 0.33 = 1 λ1 , µ2 = 0.618 = 1 λ, µ3 = −1.618 = 1 λ3 , as required. 10.4

Since taking the transpose leaves the corresponding determinant unchanged, the characteristic equation

A −λi I = 0

(1)

implies

(A −λi I)T = AT −λi I = 0 ;

so that λi is an eigenvalue of AT if it is an eigenvalue of A (and vice versa). If A−1 exists, A ≠ 0 and (1) implies

⎛ 1 −λi A ⎜⎜⎜A−1 − ⎜⎝ λi

⎞⎟ 1 I⎟⎟ = (−λi )n A A−1 − I = 0 , ⎟⎟⎠ λi

and hence

A−1 −λi−1I = 0 , so that if λi is an eigenvalue of A, λi−1 is an eigenvalue of A−1 (and vice versa). 10.5

(a) Suppose x is a non-zero eigenvector x of the matrix A with an eigenvalue λ . Then the eigenvector and its hermitian conjugate satisfy

Ax = λx,

x † A† = λ * x † ,

so that

S10.2

x†A†Ax = λ*λx†x . But A†A = I for a unitary matrix, so that 2

ˆ †x = λ*λx†x , i.e. λ = 1, x as required. (b) Suppose we have a non-zero eigenvector x such that

Ux = λx .

(1)

Then x† U† = λ*x† , which together with (1) implies 2

x†U†Ux = λ x†x . 2

2

But U† = −U* , so that λ = −1 , which is impossible since λ > 0 . Therefore no solution (1) exists. 10.6

Consider an eigenvector a such that

Aa = λa a .

(1)

Taking the hermitian conjugate gives

(Aa)† = a†A† = −a†A = λa*a† .

(2)

Multiplying (1) by a† on the left and (2) by a on the right, gives

a†Aa = λa a†a and

a†Aa = −λa*a†a, and since a†a ≠ 0 , we must have λa = −λa* , i.e. pure imaginary. Now assume there is a second eigenvector b satisfying

Ab = λb b,

λb ≠ λa .

Multiplying (3) by a† on the left and (2) by b on the right, gives

a†Ab = λb a†b = λb (a, b), and

a†Ab = −λa*a†b = λa (a, b) ,

S10.3

(3)

where in the second of these equations we have used the previous result λa = −λa* . Since λa ≠ λb , these results are only compatible if (a, b) = 0 , as required. 10.7

The characteristic equation is

i −λ 0 −i = (1 −λ)λ 2 = 0, 0 1 −λ 0 i 0 −i −λ so that the eigenvalues are λ = 0, 1 . For λ = 0 , the eigenvalue equation (A −λI)x = 0 then gives

ix −iz = 0, y = 0, ix −iz = 0, so that x = z ,y = 0 . The only linearly independent eigenvector for λ = 0 is therefore

⎛ ⎞ 1 ⎜⎜⎜ 1 ⎟⎟⎟ ⎜ 0 ⎟⎟⎟ . ⎟ 2 ⎜⎜⎜ ⎝ 1 ⎟⎠ For λ = 1 , the eigenvalue equation gives

(i −1)x − iz = 0, ix −(i + 1)z = 0, with no condition on y. Since

i −1 −i = 1 ≠ 0, i −(i + 1) (1) has no non-trivial solutions, so x = z = 0 and the eigenvector is

⎛ 0 ⎞⎟ ⎜⎜ ⎟ ⎜⎜ 1 ⎟⎟⎟ . ⎜⎜ ⎟ ⎜⎝ 0 ⎟⎟⎠ The matrix is defective because there are only two independent eigenvectors. 10.8

The eigenvalues are the solutions of

2 −λ 1 2 = −(λ 3 − 5λ 2 −λ + 5) = 0. 1 2 −λ 2 2 2 1 −λ

S10.4

(1)

with solutions λ1 = 5, λ2 = 1, λ3 = −1 , which are all real. The associated eigenvectors

u(1, 2, 3) are found by solving the equations Au(1, 2, 3) = λ1, 2,3u(1, 2, 3) and then normalising the results. This gives

u

(1)

⎛ 1 ⎞⎟ ⎛ −1 ⎞⎟ ⎛ −1 ⎞⎟ ⎜⎜ ⎟ ⎜⎜ ⎜⎜ ⎟⎟ ⎟ (2) (3) ⎟ ⎜ ⎜ = ⎜ 1 ⎟⎟ , u = ⎜ 1 ⎟⎟ , u = ⎜⎜ −1 ⎟⎟⎟. ⎜⎜ ⎟⎟ ⎜⎜ ⎜⎜ ⎟ ⎟ ⎜⎝ 1 ⎟⎠ ⎜⎝ 0 ⎟⎟⎠ ⎜⎝ 2 ⎟⎟⎠

Using (9.22) it is straightforward to show that (u(i ), u( j ) ) ≠ 0, (i ≠ j) and hence the eigenvectors are orthogonal. 10.9

(a) The eigenvalues are obtained from

1 −λ i = 0. −2i 1 −λ Expanding the determinant gives

λ 2 − 2λ −1 = 0 , with solutions λ1,2 = 1 ± 2 . For λ1 = 1 + 2 , the eigenvector u(1) = (x1 , x 2 )T is given from the solution of Au(1) = λ1u(1) . This gives

x 2 = −i 2x1 , and so the normalised form of u(1) is

u(1) =

1 ⎛⎜ 1 ⎜⎜ 3 ⎜⎝ −i 2

⎞⎟ ⎟⎟ . ⎟⎟ ⎠

In a similar way using λ2 = 1 − 2 , we find

u(2) =

1 ⎛⎜ 1 ⎜⎜ 3 ⎜⎝ i 2

⎞⎟ ⎟⎟ . ⎟⎟ ⎠

Using (9.22) we find that (u(1), u(2) ) = −1 3 and hence the eigenvectors are not orthogonal. (b) The eigenvalues of A are found from the solutions of

1 −λ 1 + i = (1 −λ)2 −(1 −i)(1 + i) = 0 , 1 −i 1 −λ

S10.5

i.e. λ1,2 = 1 ± 2 . For λ1 = 1 + 2 , the eigenvector u(1) = (x1 , x 2 )T is given from the solution of Au(1) = λ1u(1) . This gives

(1 + i)

x1 =

2

x2 ,

and so the normalized form of u(1) is

⎛ 1 ⎜⎜ (1 + i) ⎜ 2 ⎜⎜⎝ 1

u(1) =

⎞ 2 ⎟⎟⎟ . ⎟⎟ ⎠

In a similar way using λ2 = 1 − 2 , we find

u(2) =

⎛ 1 ⎜⎜ −(1 + i) ⎜ 2 ⎜⎜⎝ 1

⎞ 2 ⎟⎟⎟ . ⎟⎟ ⎠

To show that these eigenvectors are orthogonal, we calculate (u(i ), u( j ) ) for i, j = 1, 2 and show that (u(i ), u( j ) ) = δij . For example,

(u(1), u(2) ) =

1 2

( (1 −i)

⎛ 1 ⎜⎜ −(1 + i) ⎜ 2 ⎜⎜⎝ 1

)

2 1

The other relations follow in a similar way. 10.10

Using (10.19a,b), we have

ˆ(1) = u ˆ(1) = x giving

(xˆ

(1)

)

2

(1, 0, 1) ,

(xˆ

, u(2) = 2 ,

Hence

1

(

)

(1)

, u(3) = 2 .

)

ˆ(1), u(2) x ˆ(1) = (1, 1, −1) x(2) = u(2) − x so that

ˆ(2) = x and

(x

(2)

1 3

(1, 1, −1)

)

, u(3) = −1

Finally, this gives

S10.6

3.

⎞ 2 ⎟⎟⎟ = 0. ⎟⎟ ⎠

(

)

(

)

ˆ(1), u(3) x ˆ(1) − x ˆ(2), u(3) x ˆ(2) x(3) = u(3) − x = (−2 3, 4 3, 2 3), and hence

ˆ(3) = x

1 6

(−1, 2, 1) .

So the orthonormal set of vectors is

ˆ(1) = x

1 2

ˆ(2) = (1, 0, 1) , x

1 3

ˆ(3) = (1, 1, −1) , x

1 6

(−1, 2, 1).

10.11

det A = −40

A−1

⎛ 0.25 −0.25 0.50 0.25 0.00 −0.50 ⎞⎟⎟ ⎜⎜ ⎟ ⎜⎜ 1.90 0.50 0.20 0.50 −0.80 −2.80 ⎟⎟⎟ ⎜⎜ ⎟ 0.13 0.20 0.45 ⎟⎟ = ⎜⎜⎜ −0.48 −0.13 −0.05 ⎟ ⎜⎜ 2.43 0.38 0.15 0.63 −0.60 −3.35 ⎟⎟⎟ ⎟ ⎜⎜ 0.13 0.00 0.25 ⎟⎟⎟ ⎜⎜ 0.13 −0.13 −0.25 ⎟ ⎜⎝ −2.88 −0.13 −0.25 −0.88 1.00 4.25 ⎟⎠

Eigenvalues :

λ1 = 4.56, λ2 = −3.68, λ3 = −0.58 + 2.40i, λ4 = −0.58 − 2.40i , λ5 = 2.09,

λ6 = 0.19.

Eigenvectors:

x1 −0.25

x2

x3

x4

0.10 −0.54 + 0.00i −0.54 + 0.00i

x5

x6

0.36

− 0.06

−0.57 −0.18

0.18− 0.08i

0.18 + 0.08i

−0.83

− 0.48

−0.25 −0.13

−0.08 + 0.49i

−0.08− 0.49i

0.00

0.06 − 0.55

−0.64

0.50

0.44 − 0.04i

0.44 + 0.04i −0.29

−0.23 −0.77

0.29− 0.18i

0.29 + 0.18i

0.29

0.03

−0.33 + 0.09i

−0.33− 0.09i

0.11

0.68

−0.29

0.32

*10.12 The eigenvalues are found from

1 −λ 2 1 = (1 −λ)2(2 −λ)− 2(2 −λ) 0 2 −λ 0 2 1 1 −λ = (2 −λ)(λ 2 − 2λ −1) = 0,

S10.7

,

with solutions λ1 = 2 , λ2 = 1 + 2 and λ3 = 1 − 2 . The corresponding eigenvectors

u = (x1,x 2 )T are found from

(A −λ1, 2I )u(1, 2) = 0 . This gives the unormalised vectors

u

(1)

⎛ 1 ⎞⎟ ⎛ ⎛ −3 ⎞⎟ ⎜⎜ ⎜⎜ ⎟⎟ (3) ⎜⎜ 1 ⎟⎟ (2) = ⎜⎜ 1 ⎟⎟ , u = ⎜⎜⎜ 0 ⎟⎟⎟ , u = ⎜⎜⎜ 0 ⎜⎜ ⎟ ⎟⎟ ⎜⎜ ⎜⎜ ⎜⎝ −5 ⎟⎟⎟⎠ ⎝ 2 ⎟⎠ ⎝ − 2

⎞⎟ ⎟⎟ ⎟⎟ ⎟⎟ ⎟⎟ ⎠

and hence the matrix N is

⎛ −3 ⎜⎜ ⎜ N = ⎜⎜ 1 ⎜⎜ ⎜⎜⎝ −5

1 ⎞⎟⎟ ⎟⎟ 0 0 ⎟⎟ ⎟⎟ ⎟ 2 − 2 ⎟⎠

1

with N−1

⎛ 0 ⎞⎟ 1 0 ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ 1 2 (3 2 + 5) 2 2 1 2 2 =⎜ ⎟⎟. ⎜⎜ ⎟ ⎜⎜ 1 2 (3 2 − 5) 2 2 −1 2 2 ⎟⎟⎟ ⎝ ⎠

Finally, the diagonal matrix A′ is given by

A′ = N−1AN ⎛ 0 1 0 ⎜⎜ ⎜⎜ 1 2 2 = ⎜ 1 2 (3 2 + 5) 2 2 ⎜⎜ ⎜⎜ 1 2 (3 2 − 5) 2 2 −1 2 2 ⎝ ⎛ 1 2 1 ⎞⎟⎛⎜ −3 ⎜⎜ ⎟⎜⎜ ×⎜⎜ 0 2 0 ⎟⎟⎟⎜⎜ 1 ⎜⎜ ⎟⎜ ⎜⎝ 2 1 1 ⎟⎟⎠⎜⎜⎜ −5 ⎝

⎞⎟ ⎟⎟ ⎟⎟ ⎟⎟ ⎟⎟ ⎟⎟⎠ 1 ⎞⎟⎟ ⎛⎜ 2 0 0 ⎟⎟ ⎜⎜ 0 0 ⎟⎟ = ⎜⎜ 0 1 + 2 0 ⎟⎟ ⎜ ⎜ ⎟ 0 1− 2 2 − 2 ⎠⎟ ⎜⎝ 0

1

⎞⎟ ⎟⎟ ⎟⎟, ⎟⎟ ⎟⎟ ⎟⎠

and, as expected, the diagonal elements are the eigenvalues of A. *10.13 Using the co-ordinates shown, the coupled equations of motion are

x1 = −

k k k k (x1 − x 2 ); x2 = − (x 2 − x 3 )− (x 2 − x1 ); x3 = − (x 3 − x 2 ) . m µ µ m

Setting x i = ai cos(ωt) +bi sin(ωt) , these equations can be written in the matrix form

⎛ ⎞ ⎛ x ⎞⎟ ⎛ k m −k m ⎜⎜ 1 ⎟ 0 ⎞⎟⎟⎜⎜ x1 ⎟⎟ ⎜⎜ ⎟ ⎟ ⎟ ⎜ ⎜⎜ ⎟⎟⎜ x ⎟⎟ = ω 2 ⎜⎜ x ⎟⎟. ⎜⎜ 2 ⎟⎟ ⎜⎜ −k µ 2k µ −k µ ⎟⎟⎜⎜ 2 ⎟⎟ ⎟ ⎟⎟ ⎜⎜ ⎟⎜ ⎜⎜ −k m k m ⎟⎟⎠⎜⎜⎝ x 3 ⎟⎟⎟⎠ ⎜⎝ 0 ⎜⎝ x 3 ⎟⎟⎠ The eigenvalues may be found from the characteristic equation

S10.8

k m − ω2 −k µ

−k m

0

2k µ − ω

0

2

−k m

=0

−k µ k m−ω

2

and are ω 2 = 0, k m , (k m + 2k µ) . The corresponding eigenfunctions are then found by substituting these eigenvalues values back into the eigenvalue equation and solving. We find: (i) ω 2 = 0 gives x1 = x 2 = x 3 . This corresponds to a pure translation along the x axis without vibration. It is shown as (a) in the figure below. (ii) ω 2 = k m gives x1 = −x 3 , x 2 = 0 . This corresponds to the two outer masses moving in opposite directions while the middle one is stationary. This is shown as (b) in the figure below. (iii) ω 2 = k m + 2k µ gives x1 = x 3 , x 2 = −(2m µ)x1 . This corresponds to the two outer masses moving together and the inner mass moving in the opposite direction, so there is zero net momentum. This is shown as (c) in the figure below.

m

µ

m

m

µ

m

m

µ

m

*10.14 (a) The restoring force is 2T sin θ = 2Ty  , so that

m

d2y 2Ty , =− 2  dt

and hence d2y dt 2 = −2ω02y , where ω02 = T m , as required. (b) The forces on the three particles are

T T T y1 + (y 2 −y1 ) = (−2y1 + y 2 ),    T T T F2 = − (y 2 −y1 ) + (y 3 −y 2 ) = (y1 − 2y 2 + y 3 ),    T T T F3 = − (−y 2 + y 3 )− y 3 = (y 2 − 2y 3 ).    F1 = −

S10.9

The equations of motion take the form

d2y = −ω02Ay , dt 2 where y = (y1 ,y 2 ,y 3 )T and the matrix

⎛ 2 −1 0 ⎞⎟⎟ ⎜⎜ ⎟⎟ ⎜⎜ A = ⎜ −1 2 −1 ⎟⎟ . ⎟⎟ ⎜⎜ ⎜⎜⎝ 0 −1 2 ⎟⎟⎠ The eigenvalues of A are given by

2 −λ

−1

0

−1

2 −λ

−1

0

−1

2 −λ

= (2 −λ) 2 −λ −1 + −1 −1 0 2 −λ −1 2 −λ =(2 −λ)(λ 2 − 4λ + 2) = 0,

giving λ = λ2 ,λ2 ,λ3 = 2, 2 + 2 , 2 − 2 , respectively. The eigenvalues of ω02A are then ω 2 = λ ω02 , so that the frequencies of the normal modes are

ω1 = 2ω0 , ω2 = (2 + 2)1/2 ω0 , ω3 = (2 − 2)1/2 ω0 .

To find the normal modes, we find the matrix P that diagonalises A. Suitable normalised eigenvectors which are solutions of Ay = λi y are

⎛ 1 ⎜ 1 ⎜⎜ ⎜ − 2 2 ⎜⎜ ⎜⎝ 1

⎞⎟ ⎟⎟ ⎟⎟ ⎟⎟ ⎟⎟ ⎠

⎛ 1 ⎜ 1 ⎜⎜ ⎜⎜ 0 2 ⎜⎜ ⎜⎝ −1

⎞⎟ ⎟⎟ ⎟⎟ ⎟⎟ ⎟⎟ ⎠

⎛ 1 ⎞⎟ ⎜ ⎟⎟ 1 ⎜⎜ ⎜⎜ 2 ⎟⎟⎟ 2⎜ ⎜⎝ 1 ⎟⎟⎟⎠

so that

⎛ ⎜ 1 2 1 ⎜⎜⎜ P= ⎜ − 2 0 2 ⎜⎜ ⎜⎜ 1 − 2 ⎝

⎞ 1 ⎟⎟⎟ ⎟ 2 ⎟⎟⎟, ⎟⎟ 1 ⎟⎠

and the normal modes y′ are related to y by y = P y′ . Hence

y1 = 12 (y1′ + 2y 2′ + y 3′ ), y 2 = 12 (− 2y1′ + 2y 3′ ), y 3 = 12 (y1′ − 2y 2′ + y 3′ ).

S10.10

The normal modes are then characterised by

y1 : y 2 : y 3 = 1 : − 2 : 1

(y 2′ = y 3′ = 0),

y1 : y 2 : y 3 = 1 : 0 : −1

(y1′ = y 3′ = 0),

y1 : y 2 : y 3 = 1 : 2 : 1

(y1′ = y 2′ = 0),

as shown in the figure below.

*10.15 At equilibrium, the tensions are balanced by gravity, so we need only consider the change in tension caused by the further extension (or contraction) of the springs. Hence,

d2x1

3m m so that

dt 2 d2x 2 dt 2

= −4kx1 + k(x 2 − x1 ) = −5kx1 + kx 2 , = −k(x 2 − x1 ), d2x = −ω02Ax , 2 dt

where x = (x1 ,x 2 )T and

⎛ 5 3 −1 3 ⎜ A = ⎜⎜ ⎜⎜ −1 1 ⎝

⎞⎟ ⎟⎟ . ⎟⎟ ⎟⎠

We next introduce variables x ′ = P−1x , where P is a matrix such that

⎛ λ 0 ⎞⎟ ⎜ ⎟⎟ , P AP = ⎜⎜ 1 ⎜⎜ 0 λ ⎟⎟ 2 ⎟ ⎝ ⎠ −1

so that

d2x i′ dt

2

= −λi ω02x i′

i = 1, 2

with solutions

x i′ = Ai cos ωit + Bi sin ωit = Ai cos ωit ,

S10.11

where ωi2 = −λi ω02 and we have used the fact that the masses are stationary at t = 0 , so that dx i dt and dx i′ dt vanish at t = 0 . We then have x = P x ′ so that

x1 = x1′ 3 − x 2′ 3 ,

x 2 = −x1′ + x 2′ .

At t = 0 this gives

⎫ 0 = 53 A1 − 13 A2 ⎪ ⎪ ⎪ ⎬ ⇒ A1 = 1 4, A2 = 5 4 , 1 = −A1 + A2 ⎪ ⎪ ⎪ ⎭ so that we have at arbitrary t

x1(t) =

5 12

cos ω1t − 125 cos ω2t ,

x 2(t) = − 14 cos ω1t + 54 cos ω2t. *10.16 The equation of the surface may be written in the form xT A x = k , where

x = (x,y,z)T , k = −9 and A is the symmetric matrix

⎛ 11 8 −2 ⎞⎟ ⎜⎜ ⎟⎟ ⎜ A = ⎜⎜ 8 5 10 ⎟⎟⎟ . ⎟⎟ ⎜⎜ ⎜⎜⎝ −2 10 2 ⎟⎟⎠ First we find the eigenvalues of A from

11 −λ 8

8

−2

5 −λ

10

10

2 −λ

−2

= 0,

which expanding, gives

λ 3 −18λ 2 − 81λ + 1458 = (λ −18)(λ − 9)(λ + 9) = 0 , i.e. the eigenvalues are λ1 = 18, λ2 = 9, λ3 = −9 . The corresponding eigenvectors are found by solving the equation

Ax = λx and are

⎛ ⎞ 1 ⎜⎜⎜ −2 ⎟⎟⎟ (1) u = ⎜ −2 ⎟⎟ , u(2) = ⎟ 3 ⎜⎜ ⎜⎝ −1 ⎟⎟⎠

⎛ −2 ⎞⎟ ⎜ ⎟⎟ 1 ⎜⎜ ⎜⎜ 1 ⎟⎟⎟ , u(3) = ⎟ 3 ⎜⎜ ⎜⎝ 2 ⎟⎟⎠

S10.12

⎛ 1 ⎞⎟ ⎜ ⎟⎟ 1 ⎜⎜ ⎜⎜ −2 ⎟⎟⎟ , . ⎟ 3 ⎜⎜ ⎜⎝ 2 ⎟⎟⎠

The principal axes are therefore along the lines

2x + 2y + z = 0 , 2x −y − 2z = 0, x − 2y + 2z = 0, and in terms of new variables x ′, y ′, z ′ along these axes the equation of the surface is [cf. (10.35)]

(x ′)2 (y ′)2 (z ′)2 −(x ′)2 (y ′)2 (z ′)2 + + = − + = 1. (k λ1 ) (k λ2 ) (k λ3 ) (1 2) (1) (1)

(1)

Since two terms are negative, this is a two-sheet hyperboloid, as illustrated in Figure 10.5(b). The shortest distance between the sheets occurs at x ′ = y ′ = 0 , when z ′ = ±1 , so the shortest distance is 2. Note: If the principal axes are differently labeled (e.g. the x ′, y ′, z ′ axes are associated with λ2 , λ3 and λ1 , respectively, instead of λ1 , λ2 and λ3 ) the shortest distance is always along the axis associated with λ3 = −9 , i.e. the term with positive sign in (1). *10.17 (a) The characteristic equation is

1 −λ 0 0 2 0 2 −λ −1 = (1 −λ)(λ − 4λ + 3) 0 −1 2 −λ = (1 −λ)(λ −1)(λ − 3) = 0, with eigenvalues are λ1 = λ2 = 1 , λ3 = 3 . So from (10.35) it is an ellipsoid with semiaxes a = k 1/2, b = k 1/2, c = (k 3)1/2 . It is therefore an oblate spheroid, as shown in Fig 10.4(b). (b) In this case, the characteristic equation is

1 −λ 0 0 2 −λ −1 1

−1 = −(λ 3 − 4λ 2 + 3λ + 1) = 0, 1 1 −λ

so that the eigenvalues are determined by

f (λ) = λ 3 − 4λ 2 + 3λ + 1 = 0 . Since f (1) ≠ 0 and f (−1) ≠ 0 , this does not factorise in any obvious way. However, using the curve sketching method of Section 3.5, we have

f (x → ±∞) = ±∞ and

f (1) = 1,

f (0) = 1 , f (−1) = −7 ,

S10.13

and the turning points are at

df = 3λ 2 − 8λ + 3 = 0, dλ i.e. at (8 ± 28) 6 ≈ 0.45 and 2.21, where f (0.45) ≈ 1.6 and f (2.21) ≈ −1. In addition

f (3) = 1 . So f (λ) has the behaviour sketched in the figure below where the turning points are marked by crosses, and the other calculated points by solid circles. Hence there is one negative and two distinct positive eigenvalues, so that it is a one-sheet hyperboloid of the type shown in Figure 10.5(a).

*10.18 On diagonalising A with an orthogonal matrix P −1 = P T , we have from (10.33) and (10.34) 3

Q = xT Ax = x ′T Λx ′ = ∑ λi (x i′)2 ,

(1)

i=1

where, since an orthogonal transformation preserves the lengths of vectors, 3

∑ (x ′) i=1

2

i

=1 .

If λm is the smallest eigenvalue, the smallest value of (1) occurs when (x m′ )2 = 1 and all the other terms vanish, so that Q ≥ λm as required. Finally, the condition for

Q > 0 for any non-vanishing vector is that all eigenvalues must be positive. *10.19

The equation may be written in the form xT Ax = k , where

⎛ 3 −4 ⎞⎟ ⎟⎟, A = ⎜⎜⎜ ⎜⎝ −4 −3 ⎟⎟⎠

⎛ x ⎞⎟ ⎟⎟, x = ⎜⎜⎜ ⎜⎝ y ⎟⎟⎠

and k = 5 . We now find the orthogonal transformation x ′ = P x such that

S10.14

⎛ λ 0 ⎞⎟ ⎜ ⎟⎟ ≡ Λ . P AP = ⎜⎜ 1 ⎜⎜ 0 λ ⎟⎟⎟ 2 ⎠ ⎝ −1

Using standard methods, the eigenvalues are λ1 = 5, λ2 = −5 and the corresponding orthogonal eigenvectors are

u

(1)

1 ⎛⎜ 2 ⎜⎜ = 5 ⎝⎜ −1

⎞⎟ ⎟⎟, ⎟⎟ ⎠

u

(2)

⎛ ⎞ 1 ⎜⎜ 1 ⎟⎟ ⎟⎟ , = ⎜ 5 ⎜⎜⎝ 2 ⎟⎟⎠

so that

P=

1 ⎛⎜ 2 1 ⎞⎟⎟ T −1 ⎜⎜ ⎟⎟ = P = P , ⎜ ⎟ 5 ⎝ −1 2 ⎠

and the equation xT Ax = x ′T Λx = 5 becomes

x ′2 − y ′2 = 1 , which is the equation of a hyperbola. The x ′ and y ′ axes lie along the u(1) and u(2) directions y = −x 2 , y = 2x , respectively, i.e. at right angles with the x ′ axis at

θ = tan−1(−1 2) = 0.42 rad ≈ 24 0 to the x axis, as shown in the figure below. The two branches are closest at (x ′, y ′) = (±1, 0) , where

⎛ ⎞⎛ ⎛ x ⎞⎟ ⎛ ⎞ ⎞ ⎜⎜ ⎟⎟ = 1 ⎜⎜ 2 1 ⎟⎟⎟⎜⎜ ±1 ⎟⎟⎟ = 1 ⎜⎜ ±2 ⎟⎟⎟ , ⎜⎜ ⎜⎜ y ⎟⎟ ⎜ ⎜ ⎟ ⎝ ⎠ 5 ⎝ −1 2 ⎟⎠⎜⎝ 0 ⎟⎠ 5 ⎜⎝ 1 ⎟⎟⎠ i.e. at (x,y) = (2

5 ,−1

5), (−2

5 ,1

5) , respectively.

S10.15

SOLUTIONS TO PROBLEMS 11 Solution 11.1 11.1

(a) On the line AB, y = 3x 2 −1 2 and so 3

3 1 1 ⎡ 3x 3 x 2 ⎤⎥ I a = ∫ x(3x −1)dx = ⎢⎢ − ⎥ = 11. 2 1 2 ⎢⎣ 3 2 ⎥⎦1

(b) On the line AC, y = −2x + 3 and so 0

I AC = ∫ 1

1

⎡ 2x 3 3x 2 ⎤ ⎥ = −5. x(−2x + 3)dx = ⎢⎢ − 2 ⎥⎥⎦ 0 6 ⎢⎣ 3

On the line CB, y = x 3 + 3 , and so 3

⎡ x 3 3x 2 ⎤ ⎛x ⎞ ⎥ = 33 . x ⎜⎜⎜ + 3⎟⎟⎟ dx = ⎢⎢ + ⎟⎠ ⎜⎝ 3 2 ⎥⎦⎥ 0 2 ⎢⎣ 9

3

ICB = ∫ 0

Thus I b = I AC + ICB = 33 2 − 5 6 = 47 3. 11.2

(a) As a function of x only 1

I =∫ 0

1

⎡x 3 x 4 ⎤ 7 (x + x )dx = ⎢⎢ + ⎥⎥ = . 4 ⎦⎥ 0 12 ⎢⎣ 3 2

3

(b) As a function of y only, we substitute x = y 1/3 and dx = 13 y −2/3 to give 1

1 ⎤ 1 1⎡ 3 7 I = ∫ (y 2/3 + y )y −2/3dy = ⎢y + y 4/3 ⎥ = . ⎢ ⎥ 3 0 3⎣ 4 ⎦ 0 12

11.3

For the upper semicircle, 0 ≤ y ≤ 1 , y = (1 − x 2 )1/2 and hence dy dx = −x (1 − x 2 )1/2 . Similarly,

for

the

lower

semicircle, −1 ≤ y ≤ 0 , y = −(1 − x 2 )1/2

2 1/2

dy dx = x (1 − x ) . Using these in (11.11a) gives 1/2

−1

I = ∫ ⎡⎢x + x(1 − x ) ⎣ 2

2 1/2

1

1

+∫

−1

⎛ x 2 ⎞⎟⎟ + (1 − x )⎤⎥ ⎜⎜⎜1 + ⎟ (−dx) ⎦ ⎜⎝ 1 − x 2 ⎟⎠ 2

1/2

2 ⎛ ⎞ ⎡x 2 − x(1 − x 2 )1/2 + (1 − x 2 )⎤ ⎜⎜1 + x ⎟⎟ dx = 2π, ⎢⎣ ⎥⎦ ⎜⎜ 2⎟ 1 − x ⎟⎠ ⎝

using the substitution x = sin θ .

S11.1

and

hence

11.4

(a) In plane polar co-ordinates:

x = cos θ, dx = −sin θ dθ, y = sin θ, dy = cos θ dθ . Then

(x + 2y)dx − 2x dy = (− 12 sin 2θ − 2)dθ and the line integral becomes 2π

⎞ ⌠ ⎛1 −⎮ ⎜⎜⎜ sin 2θ + 2⎟⎟⎟ dθ = −4π . ⎟⎠ ⌡ ⎜⎝ 2 0

(b) For two sections of the path x is a constant and hence dx = 0 , and for the other two sections y is a constant and hence dy = 0 . Then the line integral is 1

−1

−1

1

−2∫ dy + ∫ (x + 2)dx + 2∫ dy + ∫ (x − 2)dx −1

1

1

−1

1

−1

1

−1

= −2 ⎡⎢y ⎤⎥ + ⎡⎢x 2 2 + 2x ⎤⎥ + 2 ⎡⎢y ⎤⎥ + ⎡⎢x 2 2 − 2x ⎤⎥ ⎣ ⎦ −1 ⎣ ⎣ ⎦1 ⎦1 ⎣ ⎦ −1 = −16. 11.5

(a) In this case x = y = z and so 1

1

I = ∫ (x 2 + 2x + 1)dx = ⎡⎢x 3 3 + x 2 + x ⎤⎥ = 7 3 . ⎣ ⎦0 0

(b) If we write I = I 1 + I 2 + I 3 , where the subscripts refer to the three parts of the path, then for (0, 0, 0) → (1, 0, 0) , y = z = 0 and hence dy = dz = 0 , so I 1 = 0 . For

(1, 0, 0) → (1,1, 0) , x = 1, z = 0 and hence dx = dz = 0 , so 1

I 2 = ∫ 2dy = 2 . 0

For (1,1, 0) → (1,1,1) , x = 1, y = 1 and hence dy = dy = 0 , so 1

I 3 = ∫ dz = 1 . 0

Therefore I = I 1 + I 2 + I 3 = 3 . 11.6

The area is shown shaded in the diagram below.

S11.2

The limits on x are (0,1 −y) and the limits on y are (0,1) . Hence 1−y

1

1

1−y

I = ∫ dy ∫ (2 − x −y) dx = ∫ ⎡⎢(2 − x −y)−1 ⎤⎥ dy ⎣ ⎦0 −2

0 1

=∫ 0

11.7

0

0

⎛ ⎞ ⎜⎜1 − 1 ⎟⎟ dy = ⎡y + ln(2 −y)⎤1 = 1 − ln 2. ⎟ ⎢⎣ ⎦⎥ 0 ⎜⎜⎝ 2 −y ⎟⎠

Integrating with respect to x keeping y fixed gives x=2

⎡x 2 ⎤ ⎢ + 3xy ⎥ = 2 + 6y . (x + 3y)dx = ∫ ⎢2 ⎥ ⎢⎣ ⎥⎦ x=0 x=0 x=2

Then integrating with respect to y gives y=2

∫ (2 + 6y ) dy = ⎡⎢⎣2y + 3y

y=1

2

y=2

⎤ = 11. ⎥⎦ y=1

Repeating the exercise but reversing the order of integrations gives y=2

⎡ 3 2⎤ 9 ∫ (x + 3y)dy = ⎢⎢xy + 2 y ⎥⎥ = x + 2 ⎣ ⎦ y=1 y=1

y=2

and x=2

⎡ 2 ⎤ ⎛ ⎞ ⎜⎜x + 9 ⎟⎟ dx = ⎢ x + 9 x ⎥ = 11. ∫ ⎜⎜⎝ 2 ⎟⎟⎠ ⎢2 2 ⎥ ⎢⎣ x=0 ⎦⎥ x=0 x=2

The orders of integration can be interchanged without changing the limits in this case because the region of integration is rectangular. 11.8

For a given value of y in the range 0 ≤ y ≤ a , the value of x lies between the y-axis and the parabola x 2 = 4a 2 − 4ay , so the region of integration is the shaded area shown on the figure below.

S11.3

Thus the integration limits of x are 0 ≤ x ≤ 2a , and for any x in this range,

y max = a − x 2 4a , i.e. 0 ≤ y ≤ a − x 2 4a . Thus, a−x 2 4a

2a

I = ∫ dx

∫

0

11.9

f (x,y)dy .

0

(a) The region of integration is shown in figure (a) below. On reversing the order of integration, one obtains 1

I =∫ 0

y

2−y

2

1

2

dy dy y (2 −y)2 x dx + x dx = dy + ∫ y∫ ∫2 ∫ 2y dy = 2 ln 2 −1. y ∫0 1 0 0 1

(b) The region of integration is shown in figure (b) above. On reversing the order of integration, one obtains 1

1

∫ [y(2 −y)]

1/2

0

1

dy ∫ dx = ∫ (1 −y)[y(2 −y)]1/2 dy y

0

1

= 11.10

3/2 ⎤ 1⎡ 1 ⎢{y(2 −y)} ⎥ = . ⎥⎦ 0 3 3 ⎢⎣

Reversing the order of integration gives

S11.4

6

I =∫

2

2

∫

x(y 3 + 1)1/2 dy dx = ∫

0 x/3

0

2

3y

∫ x(y

3

+ 1)1/2 dx dy

0

2

x=3y

= ∫ ⎡⎢x(y 3 + 1)1/2 2⎤⎥ dy = ∫ ⎡⎢9y 2(y 3 + 1)1/2 2⎤⎥ dy ⎣ ⎦ x=0 ⎣ ⎦ 0

0

2

= ⎡⎢(y + 1) ⎤⎥ = 27. ⎣ ⎦0 3

11.11

3/2

If we set

P = e xy + cosx sin y; Q = e x + sin x cosy then

∂P ∂Q = e x cosx cosy = ∂y ∂x and so by Green’s theorem in the plane, I = 0 11.12

The integration path may be written

∫

C

where

B

C

=∫ +

∫

A

+

B

D

A

∫

+∫

C

,

D

along AB, y = 0 and 0 ≤ x ≤ 1, along BC, x = 1 and 0 ≤ y ≤ 1, along CD, y = 1 and x decreases from 1 to 0, along DA, x = 0 and y decreases from 1 to 0.

Therefore, 1

∫

C

0

(xy dx + x 2dy) = ∫ dy + ∫ x dx = 12 . 0

1

Using Green’s theorem in a plane with P(x,y) = xy and Q(x,y) = x 2 , the line integral may be written 1

∫

C

(xy dx + x 2dy) = ∫ 0

1

1

∫ (2x − x)dx dy = 2, 0

which verifies the theorem. 11.13

The path is shown in the figure below. If we define

P = e x cosy ⇒

∂P ∂Q = −e x sin y; Q = −e x sin y ⇒ = −e x sin y , ∂y ∂y

then the integral around the closed path ADBA is equal to the integral

S11.5

⌠⌠ ⎛⎜ ∂Q ∂P ⎞⎟ ⎟⎟ dx dy = 0 − ⎮⎮ ⎜⎜ ∂y ⎟⎠ ⌡⌡ ⎜⎝ ∂x area ABDA

and hence

∫

ADB

=−∫

.

B→A

In the latter integral, y = 0 and dy = 0 and so ln 2

∫

B→A

= ∫ e x cosy dx = ∫ e x dx = ⎡⎢e x ⎤⎥ =3 2. ⎣ ⎦ −ln 2

Hence

∫

= −3 2 .

ABD

11.14

In the notation of (11.23a), one easily shows that (11.23b) is satisfied, so that they are both perfect differentials. Using (11.23a) then implies

(a)

∂I ∂I ∂I = y +z, = x +z, = x +y, ∂x ∂y ∂z

so that

I = xy + xz + yz +c, with I B − I A = 11, and

(b)

∂I ∂I ∂I = xy 2 + z , = x 2y + 2, = x, ∂x ∂y ∂z

so that

I = 12 x 2y 2 + xz + 2y +c, with I B − I A = − 32 . 11.15

The line integral to be evaluated is

I = ∫ ⎡⎢(x 2 + y 2 )dx + 2xydy ⎤⎥ . ⎣ ⎦ (a) Using y = x 2 2

S11.6

2

I =∫ 0

2

⎡x 3 x5 ⎤ 32 (x + x 4 + x )dx = ⎢⎢ + ⎥⎥ = . 4 ⎥⎦ 0 3 ⎢⎣ 3 2

4

4

(b) We will write I = I 1 + I 2 , where I 1 is the integral along the straight line from

(0, 0) to (2, 0) , and I 2 is the integral along the straight line from (2, 0) to (2, 2) . In I 1 , y = 0 and so 2

I1 = ∫ 0

2

⎡x 3 ⎤ 8 x dx = ⎢⎢ ⎥⎥ = . ⎢⎣ 3 ⎥⎦ 0 3 2

In I 2 , x = 2 with dx = 0 , and so 2

2

I 2 = ∫ 4y dy = ⎡⎢2y 2 ⎤⎥ = 8, ⎣ ⎦0 0

so that I = 8 3 + 8 = 32 3 . (c) Using x = t 2 2 and y = t , gives 2

2 ⎤ ⎞ ⎡ t6 t4 t4 ⎤ ⌠ ⎡⎛⎜ t 4 32 2⎟ 3⎥ ⎢ I = ⎮ ⎢⎜⎜ + t ⎟⎟t + t ⎥ dt = ⎢⎢ + + ⎥⎥ = . ⎮ ⎜⎝ 4 ⎟ 4 ⎥⎦ 0 3 ⎠ ⎢⎣ 24 4 ⌡ ⎢⎣ ⎥⎦ 0

Thus the line integral is independent of the path, which is consistent with df being an exact differential.

11.16

We first substitute for x and y, and use the Jacobian

a cos θ cosφ −a sin θ sin φ

J=

b cos θ sin φ

b sin θ cosφ

= ab sin θ cos θ

to transform the surface element. Then the ranges of the new variables are found from

tan φ = ay bx ⇒ 0 ≤ φ ≤ π 2 and

x 2 a 2 + y 2 b 2 = sin 2 θ ⇒ 0 ≤ θ ≤ π 2 . Hence π/2 π/2

I=

∫∫ 0

(a 3 sin 3 θ cos3 φ)(b sin θ sin φ) ab sin θ cos θ dθdϕ (1 − sin 2 θ cos2 φ − sin 2 θ sin 2 φ)1/2

0 π/2

π/2

= a 4b 2 ∫ sin 5 θ dθ ∫ sin φ cos3φ dφ = 2a 4b 2 15. 0

0

S11.7

11.17

Using the curvilinear co-ordinates u1 = 2xy and u 2 = x 2 − y 2 , the area is specified by

0 ≤ u1 ≤ 2a 2 ;

− ∞ < u2 < ∞ .

To change variables in the integrand we use the Jacobian

J=

∂(u1,u 2 ) ∂(x,y)

=

∂u1

∂u1

∂x ∂u 2

∂y ∂u 2

∂x

∂y

=

2y 2x = −4(x 2 + y 2 ) . 2x −2y

Hence from (11.26), the integral is 2a 2

∞

I = ∫ du1 ∫ du 2 0

−∞

2xy(x 2 + y 2 ) exp ⎡⎢−(x 2 + y 2 )2 ⎤⎥ ⎣ ⎦ 4(x 2 + y 2 ) 2a 2

2a 2

∞ ⎤ 1 1⎡ 1 −u 2 −u 2 = ∫ u1e 1 du1 ∫ e 2 du 2 = ⎢− exp(−u12 )⎥ ⎥ 4 0 4 ⎢⎣ 2 ⎦0 −∞

=

π

π⎡ 1 − exp(−4a 4 )⎤⎥ , ⎢ ⎣ ⎦ 8

using the standard integral (11.27). 11.18

Using r = xi + yj + zk gives

r = uw cosφi + uw sin φj + 12 (u 2 − w 2 )k . Hence, by (11.29) we have

eu =

w cosφi + w sin φj + uk u2 + w2

, ew

u cosφi + u sin φj −wk u2 + w2

, eφ = −sin φi + cosφj ,

from which we see that

eu ⋅ ew = eu ⋅ e φ = ew ⋅ e φ = 0 i.e. they are orthogonal. From (11.30) we have

dr = hu eu du + hw ew dw + hφ eφ dφ , where

hu = hw = u 2 + w 2 , hφ = uw . Hence,

dr2 = (u 2 + w 2 )(du 2 + dw 2 ) + u 2w 2dφ 2 and

S11.8

dυ = huhwhφ du dw dφ = uw u 2 + w 2 du dw dφ . 11.19

From (11.29) and (11.30b), we have

hu eu = α sinh u cosw i + α cosh u sin w j , hw ew = −α cosh u sin w i + α sinh u cosw j , so that huhw eu ⋅ ew = 0 and hence eu ⋅ ew = 0 , since hu , hw > 0 . For u = constant , we have

1 = cos2 w + sin 2 w =

x2 y2 + , α2 cosh 2 u α2 sinh 2 u

which is the equation of an ellipse with semi-axes a = α cosh u , b = α sinh u . Alternatively, if w = constant , we have

1 = cosh 2 u − sinh 2 u =

x2 y2 − , α2 cos2 w α2 sin 2 w

which is the equation of a hyperbola. They are sketched in the positive quadrant, in the figure below, together with eu , ew at the point of intersection.

11.20

The region is covered by: (a) first keeping x and y fixed and integrating from z = x 2 to z = 2 ; then (b) by keeping x fixed and integrating from y = 0 to y = 3 ; and finally (c) integrating from x = 0 to x = 2 (where z = x 2 meets z = 2 ). The integral is therefore: x= 2 y=3 z=2

(a) I =

∫ ∫ ∫

(xz i + x j − 2y 2k)dz dy dx

x=0 y=0 z=x 2 x= 2 y=3

=

∫ ∫

x=0 y=0

x= 2 y=3

=

∫ ∫

z=2

⎡(xz 2 2)i + xz j − 2y 2zk ⎤ dy dx ⎣⎢ ⎦⎥ z=x 2

x=0 y=0

⎡(2x − x 5 2)i + (2x − x 3 )j − 2y 2(2 − x 2 )k ⎤ dy dx , ⎢⎣ ⎥⎦

then,

S11.9

x= 2

(b) I =

∫

x=0

x= 2

=

∫

x=0

y=3

⎡(2x − x 5 2)y i + (2x − x 3 )y j − 2 y 3(2 − x 2 )k ⎤ 3 ⎢⎣ ⎥⎦ y=0 dx ⎡(6x − 3x 5 2)i + (6x − 3x 3 ) j −18(2 − x 2 )k ⎤ dx , ⎣⎢ ⎦⎥

and finally, x= 2

(c) I = ⎡⎢(3x 2 − x 6 4)i + (3x 2 − 3x 4 4) j −18(2x − x 3 3)k ⎤⎥ ⎣ ⎦ x=0 = 4 i + 3 j − 24 2 k. 11.21

We will use spherical co-ordinates (r, θ,φ) , where from (11.39a) and (11.44),

x = r sin θ cosφ, y = r sin θ sin φ, z = r cos θ , dυ = r 2 sin θ dr dθ dφ , and the octant is defined by

0 ≤ r ≤ a , 0 ≤ θ ≤ π 2, 0 ≤ φ ≤ π 2 . The integral is then π/2

a

π/2

∫ dr ∫ dθ ∫ dφ [r 0

0

3

sin θ cos2 θ cosφ exp(r 2 a 2 )]r 2 sin θ .

0

This may be written as the product of three independent integrals I = I 1I 2I 3 , where π/2

a

I 1 = ∫ r exp(r a )]dr, I 2 = 5

2

2

0

π/2

∫ sin

2

2

θ cos θ dθ, I 3 =

0

∫ cosφ dφ . 0

On changing variables to t = r 2 , and using the given integral, we obtain 2

a ⎛t ⎞ 1 a 6 (e − 2) . I 1 = ∫ t 2 exp ⎜⎜⎜ 2 ⎟⎟⎟ dt = ⎜⎝a ⎟⎠ 2 0 2

Evaluation of the other integrals gives I 2 = π 16 and I 3 = 1 , and hence

I = I 1I 2I 3 = 11.22

πa 6 (e − 2) . 32

We use cylindrical polar co-ordinates (ρ,φ,z) with origin at the base of the hemisphere and the z-axis vertical, along the axis of the hemisphere. Since in the Cartesian system, the equation of the hemisphere is x 2 + y 2 + z 2 = R 2 , in cylindrical

S11.10

co-ordinates x 2 + y 2 = ρ 2 , so this becomes ρ 2 + z 2 = R 2 . The region occupied by the liquid is then

0 ≤ ρ ≤ (R 2 − z 2 )1/2 , 0 ≤ φ ≤ 2π, 0 ≤ z ≤ h . and it occupies the volume 2π

(R 2 −z 2 )1/2

h

V = ∫ dφ ∫ dz 0

0

∫

ρ dρ ,

0

where we have used dυ = ρ dρ dφ dz for the volume element in cylindrical coordinates. Evaluating the integrals in turn gives 2π

ρ=(R 2 −z 2 )1/2

⎡ ρ2 ⎤ ⎢ ⎥ ⎢2⎥ ⎢⎣ ⎥⎦ ρ=0

h

V = ∫ dφ ∫ 0

0

2π

=∫ 0

2π

h

dz = ∫ dφ ∫ 0

z=h

2π 1 ⎡⎢ 2 z 3 ⎤⎥ R z − ⎥ dφ = ∫ 2 ⎢⎢⎣ 3 ⎥⎦ z=0 0

0

1 2 R − z 2 dz 2

(

)

1 ⎛⎜ 2 h3 ⎞ ⎜⎜R h − ⎟⎟⎟ dφ 2 ⎜⎝ 3 ⎟⎠

⎛ h3 ⎞ = π ⎜⎜⎜R 2h − ⎟⎟⎟. ⎜⎝ 3 ⎟⎠

11.23 Referring to the figure below, for any values of x and y the range of z is

0 ≤ z ≤ (1 − x −y) , where (1 − x −y) lies on the plane P. The possible values of x and y are those lying on the face OAB, bounded by x = 0 and y = 0 , and the line x +b = 1 . In this case,

0 ≤ x ≤ 1 −y

and

0 ≤y ≤1.

Hence the integral is 1

1−y

1−x−y

I = ∫ dy ∫ dx 0

0

∫

ln(1 − x −y − z)dz.

0

To do the integral over z we use the given integral with a = 1 − x −y , n = 1 and

p = z , which gives 1

1−y

0 1

0 1−y

0

0

z=1−x−y

I = ∫ dy ∫ dx ⎡⎢−(1 − x −y − z)ln(1 − x −y − z) + (1 − x −y − z)⎤⎥ ⎣ ⎦ z=0 = ∫ dy ∫ dx ⎡⎢(1 − x −y)ln(1 − x −y)−(1 − x −y)⎤⎥ , ⎣ ⎦ where we have used the fact that

lim(x ln x) = 0 . x→0

S11.11

Next we use the given integral with a = 1 −y ,n = 2 and p = x , which gives 1

I = ∫ dy[− 12 (1 −y − x)2 ln(1 −y − x) + 14 (1 −y − x)2 − x(1 −y)− 12 x 2 ]x=1−y x=0 0 1

{

= ∫ dy − 12 (1 −y)2 + 12 (1 −y)2 ln(1 −y)− 14 (1 −y)2

}

0 1

= ∫ dy ⎡⎢ 12 (1 −y)2 ln(1 −y)− 43 (1 −y)2 ⎤⎥ . ⎣ ⎦ 0

Finally we use the given integral with a = 1, n = 3 and p = y , which gives 1

I = ⎡⎢− 16 (1 −y)3 ln(1 −y) + 181 (1 −y)3 − 43 (y −y 2 + 13 y 3 )⎤⎥ ⎣ ⎦0 = −11 36.

S11.12

SOLUTIONS TO PROBLEMS 12 12.1

The electric field is E = −∇φ = −2x i + 2y j and at (2, 1) E = −4i + 2j . Thus

E = 2 5 and its direction is −2i + j . At (–3, 2), −∇φ = 6i + 4 j . Thus the direction of most rapid decrease is

3i + 2j . The directional derivative is

ˆ , where at (1, 2) dφ ds = ∇φ ⋅ u ˆ= u

1 10

(3i − j) and ∇φ = 2i − 4 j .

Thus

dφ = ds 12.2

(a)

∇ψ =

1 10

(2i − 4 j)⋅(3i − j) = 10 .

∂ψ ∂ψ ∂ψ i+ j+ k = 2x i − 2yz j −y 2k , ∂x ∂y ∂z

which at (1, 1, 1) is ∇ψ = 2i − 2j − k .

ˆ is a unit vector in the ˆ , where n (b) The directional derivative is dψ ds = ∇ψ ⋅ n required direction. Thus, ˆ= n

1 6

(i − 2j + k)

and

dψ 1 5 = (2i − 2j − k)⋅(i − 2j + k) = . ds 6 6 (c) ∇ψ is perpendicular to the surface ψ = constant , so the normal line is

r = (1,1,1) + (2,−2,1)t . 12.3

(a) Using (12.10), gives curl A = y i −(3z 3 − 4xz)j . (b) First form SA = 2x 3yz 3i − x 2y 2z 2 j + 3x 3yz 4 k then use (12.10) to gives

curl(SA) = x 2z(3xz 3 + 2y 2 )i + 3x 2yz 2(2x − 3z 2 )j − 3xz 2(y 2 + x 2z)k . (c) curl curl A = ∇ ×(∇ ×A) , so using the result of (a) and then again using (12.10), gives curl curl A = (−4x + 9z 2 )i + (4z −1)k . (d) grad(A ⋅ curl A) = ∇[A ⋅(∇ ×A)] , so using the result of (a) gives

A ⋅(∇ ×A) = −2xyz 2 + 3yz 4 = f (x,y,z)

S12.1

and

∇[A ⋅(∇ ×A)] =

∂f ∂f ∂f i+ j+ k ∂x ∂y ∂z = −2yz 2i + z 2(3z 2 − 2x)j + 4yz(3z 2 − x)k.

(e) from (12.14a), curl gradS = 0 . 12.4

(a) If we define a vector field F by F ≡ ∇ ×(ψV) , then

i F = ∂ ∂x ψVx

j k ∂ ∂y ∂ ∂z , ψVy

ψVz

so that

⎛ ∂V ⎞ ⎡ ∇ ×(ψV)⎤ = ∂ (ψV ) − ∂ (ψV ) = ψ ⎜⎜ ∂Vz − y ⎟⎟⎟ +V ∂ψ −V ∂ψ ⎜⎜ z y z y ⎣⎢ ⎦⎥ x ∂y ∂z ∂z ⎟⎟⎠ ∂y ∂z ⎝ ∂y = ⎡⎢ ψ ∇ ×V ⎤⎥ − ⎡⎢ V × ∇ψ ⎤⎥ , ⎣ ⎦x ⎣ ⎦x and similarly for the other components, so that

curl(ψV) = ψ curl V − V× ∇ψ (b) Since αV = grad ψ and ∇ ×(∇ψ) = 0 by (12.14a), then

0 = ∇ ×(αV) = α curl V − V× ∇α . Hence

αV ⋅ curl V = V ⋅ V× ∇α = 0, since the triple product a ⋅ b×c = 0 if any pair of a, b, c are equal. 12.5

We use the two identities:

∇ ×(∇ ×a) = ∇(∇ ⋅ a)− ∇ 2a

(1)

which is (12.14c), and

∇ ×(φa) = φ(∇ ×a)− a ×(∇φ) , from Table 12.1 with φ = ∇ ⋅ a , i.e.,

∇ ×a(∇ ⋅ a) = (∇ ⋅ a)(∇ ×a)− a × ∇(∇ ⋅ a) . Using (1) and (2) in the given expression, gives

(∇ ⋅ a)(∇ ×a)− a × ∇(∇ ⋅ a) + a × ∇(∇ ⋅ a)− a × ∇ 2a + a × ∇ 2a = (∇ ⋅ a)(∇ ×a).

S12.2

(2)

12.6

In spherical polar co-ordinates [cf. Section 11.3.1]

x = r sin θ cosφ, y = r sin θ sin φ, z = r cos θ and

er = sin θ cosφ i + sin θ sin φ j + cos θ k, eθ = cos θ cosφ i + cos θ sin φ j − sin θ k, eφ = −sin φ i + cosφ j, (a) ψ = 2yz = r 2 sin 2θ sin φ . (b) Using the results above,

i ⋅ er = sin θ cosφ, i ⋅ eθ = cos θ cosφ, i ⋅ eφ = −sin φ so that

i = sin θ cosφ er + cos θ cosφ eθ − sin φ eφ since er , eθ , and eφ are orthogonal. Similarly

j = sin θ sin φ er + cos θ sin φ eθ + cosφ eφ and

k = cos θ er − sin θ eθ . Hence, collecting terms,

V = x j −y k = r sin θ sin φ(sin θ cosφ − cos θ)er + r sin θ sin φ(cos θ cosφ + sin θ)eθ + r sin θ cos2 φ eφ . (c) Using, in Cartesian co-ordinates,

∇ψ =

∂ψ ∂ψ ∂ψ i+ j+ k = 2z j + 2y k ∂x ∂y ∂z

and the expressions above for y, z, j and k in spherical polar co-ordinates, gives

∇ψ = 4r cos θ sin θ sin φ er + 2r sin φ(cos2 θ − sin 2 θ)eθ + 2r cos θ cosφeφ . (d) In Cartesian co-ordinates,

i ∇ ×V = ∂ ∂x 0

j k ∂ ∂y ∂ ∂z = −i + k x

−y

= (cos θ − sin θ cosφ)er −(sin θ + cos θ cosφ)eθ + sin φeφ

S12.3

12.7

Since dz = 0 around the circle, the integral reduces to

∫

C

V ⋅ dr = ∫ ⎡⎢(x 2 + y 2 )y dx −(x 2 + y 2 )x dy ⎤⎥. ⎣ ⎦

If we use plane polar co-ordinates in the xy-plane, around the circle we have r = a , so that

x = a cos θ, y = a sin θ, dx = −a sin θdθ, dy = a cos θdθ and 2π 4 2 2 4 ∫ V ⋅ dr = −a ∫ (sin θ + cos θ)dθ = −2πa . C

0

To verify this result, we can Green’s theorem in the plane, (11.20) in which case we evaluate the double integral over the interior R of the circle C of

∂ ∂x

⎡−(x 2 + y 2 )x ⎤ − ∂ ⎢⎣ ⎥⎦ ∂y

⎡−(x 2 + y 2 )y ⎤ = −4(x 2 + y 2 ) = −4r 2 . ⎢⎣ ⎥⎦

and so the integral is a

2π

−∫ dr ∫ (4r 2 )r dθ = −2πa 4 . 0

12.8

0

The work done is

W =∫  F ⋅ dr . Along the given curve,

x = 1 − cost, dx = sin t dt, y = sin t, dy = cost dt, z = t, dz = dt. Thus,

F ⋅ dr = (y + z)dx −(x + z)dy + (x + y)dz = (−t cost + t sin t + sin t − 2cost + 2)dt, and the work done along the curve is 2π

W = ∫ (−t cost + t sin t + sin t − 2cost + 2)dt = 2π . 0

Along the z-axis, x = y = 0 , dx = dy = 0 , and Fz = 0 , so that F ⋅ dr = 0 and so no work 12.9

is done. Hence the total work done is 2π .

In the xy-plane, z = 0 and hence dz = 0 , so using dr = dx i + dy j + dz k , gives

S12.4

∫

C

F ⋅ dr = ∫ ⎡⎢(x − 2y 2 )dx + (3x + 2y)dy ⎤⎥ , ⎣ ⎦

where we must remember that x and y are not independent variables along the curve specified. If we work in the two dimensional polar co-ordinates, the integration path C corresponds to

r(θ) = cos θ i + sin θ j ,

0≤θ≤π

where θ is the angle from the x-axis measured in the counter-clockwise direction. Hence

dx = −sin θ dθ ,

dy = cos θ dθ ,

and substituting in the integral gives 0

∫ C

F ⋅ dr = ∫ (sin θ cos θ + 2sin 3 θ + 3cos2θ)dθ) π

0

= ∫ ⎡⎢ 12 sin 2θ + 2sin θ − 2sin θ cos2 θ + 32 + 32 cos 2θ ⎤⎥ dθ ⎣ ⎦ π

0

= ⎡⎢− 14 cos 2θ − 2cos θ + 23 cos3 θ + 32 θ + 43 sin 2θ ⎤⎥ ⎣ ⎦π = −(9π + 16) 6. 12.10 (a) The work done is B

W = ∫ F ⋅ dr =W1 +W2 , A

where O

B

W1 = ∫ F ⋅ dr and W2 = ∫ F ⋅ dr A

O

Along AO,

F = (x 2 + y 2 )j and dr = −i dx ⇒ F ⋅ dr = 0 and W1 = 0. Along OB,

F = (x 2 + y 2 )j = y 2 j and dr = j dy ⇒ F ⋅ dr = y 2dy, and hence a

W2 = ∫ y 2 dy = 0

3

so that W =W1 +W2 = a 3 . (b) In this case,

S12.5

a3 , 3

F = a 2 j and r = a cos θ i + a sin θ j, with dr = a(−sin θ i + cos θ j)dθ . So the work done is π/2

B

W = ∫ F ⋅ dr = a

3

A

12.11

∫ cos θ dθ = a

3

sin (π 2) = a 3 .

0

The contour is shown in the figure below.

The work done is 2 W =∫  F ⋅ dr = ∫ (xy dx −y dy) .

If we set

P = xy,

∂P ∂Q = x, Q = −y 2, = 0, ∂y ∂x

then by Green’s theorem in the plane (11.20),

⌠⌠ ⎛ ∂Q ∂P ⎞⎟ ⎟⎟ dx dy = −∫∫ x dx dy W = ⎮⎮ ⎜⎜⎜ − ∂y ⎟⎠ ⌡⌡ ⎜⎝ ∂x area area

1

2 y

1

= − ∫ dy ∫ x dx = −∫ 2y dy = −1. y=0

12.12

x=0

0

To prove that V is a conservative field, it is simplest to show that curl V = 0 . For example, the x component is

∂Vz ∂y

−

∂Vy ∂z

= 2xy sinh(2xz)− 4xy cosh(xz)sinh(xz) = 0 ,

and likewise for the other components. To find the scalar potential, we have

∇φ =

∂φ ∂φ ∂φ i+ j+ k, ∂x ∂y ∂z

with

S12.6

∂φ ∂φ ∂φ = −y 2z sinh(2xz), = −2y cosh 2(xz), = −xy 2 sinh(2xz), ∂x ∂y ∂z and thus by inspection φ = −y 2 cosh 2(xz) +c , where c is a constant. 12.13

Taking the curl of F gives

curl F = (c + 1)i + (a − 4)j + (b − 2)k . For F to be conservative, a = 4, b = 2, c = −1 . Therefore ,

curl F = 0 ,

and

hence

we

must

choose

F = (x + 2y + 4z)i + (2x − 3y − z)j + (4x −y + 2z)k. , The scalar potential φ is then found from F = −∇φ , where

−

∂φ = x + 2y + 4z, ∂x

so that

−φ = 12 x 2 + 2yx + 4zx + f (y,z) , where f (y,z) is a function to be determined. Similarly for the y-component,

−φ = 2xy − 32 y 2 − zy + g(x,z) , and from the z-component,

−φ = 4xz −yz + z 2 + h(x,y). On comparing these three results, we find the general solution

−φ = 12 x 2 + 2yx + 4zx − 32 y 2 − zy + z 2 + d , where d is an arbitrary constant. 12.14

In cylindrical co-ordinates, we have

x = ρ cosφ, y = ρ sin φ, z = z , so that on the curved surface of the cylinder, ρ = 2 and one sees from Figure 11.14 that

ds = ρ dφ dz eρ = dφ dz(x i + y j) using (11.35). Hence

S12.7

A ⋅ ds = dφ dz(8xy + yz) = dφ dz(8ρ 2 sin φ cosφ + ρz sin φ) = dφ dz(32sin 2φ + 4z sin φ), and thus π 2

3

I = ∫ dφ ∫ dz(32sin 2φ + 4z sin φ) 0

0

π 2

π 2

= 96 ∫ sin 2φ dφ + 18 ∫ sin φ dφ = 114. 0

12.15

0

Consider the face corresponding to x = 1 . The vector ds is outward from this face and normal to it and so can be written dy dz i . Then

V ⋅ ds = (x 2i + 12 y 2 j + 12 z 2k)⋅ dy dz i = x 2dy dz , and since x = 1 , the contribution to the integral is , z=1 y−1

∫ ∫

dy dz = 1 .

z=0 y=0

In a similar way, on the opposite face with x = 0 , the corresponding integral is zero. If this is repeated for the three other pairs of faces, it is straightforward to show that one face of each pair contributes 12 and the other is zero. So, finally, the total surface integral from all three pairs is 2. 12.16

The relevant diagram is shown below.

In spherical co-ordinates,

x = a sin θ cosφ, y = a sin θ sin φ, z = a cos θ , where the hemisphere corresponds to the ranges 0 ≤ θ ≤ π 2, 0 ≤ φ ≤ 2π and

ds = hθhφ dθ dφ = a 2 sin θ dθ dφ using (12.39). Thus,

S12.8

y 2 + z 2 = a 2(sin 2 θ sin 2 φ + cos2 θ) and π 2

2π

I = a 4σ ∫ dφ ∫ [sin 2θ sin 2 φ + cos2 θ]sin θ dθ = 43 πσa 4 . 0

0

12.17 (a) If we take the axis through the centre to be the z-axis, and use spherical polar coordinates, the distance from a point (r, θ,φ) to the axis is r0 = r sin θ . Hence from (12.46) and (11.44),

ICM = ∫ Ω

3M ρr sin θr sin θ dr dθ dφ = 4πa 3 2

2

2

a

2π

π

∫ r dr ∫ dφ ∫ sin θ dθ, 4

0

3

0

0

where Ω is the volume of the sphere. The integral π

∫ 0

1

sin 3θ dθ = −∫ (1 − cos2θ)d cos θ = 4 3, −1

so that finally, ICM = 52 Ma 2 . (b) By the parallel axis theorem (12.57), the moment about a tangent to the sphere is

I = ICM + Ma 2 = 75 Ma 2 . 12.18

We use cylindrical polar co-ordinates with the origin at the centre of the base of the cylinder, the z-axis along the axis of the cylinder, and the y-axis along the axis of rotation. Then the distance r0 from a point (ρ,φ,z) to the axis of rotation is [cf. Figure 11.13(a)]

r02 = x 2 + z 2 = ρ 2 sin 2 φ + z 2 and hence by (12.46), the moment of inertia

I = ∫ ρ0[ρ 2 sin 2 φ + z 2 ]ρ dρ dφ dz, Ω

where ρ0 is the density and Ω is the volume of the cylinder. Hence, using

ρ0 = M πa 2d ,

M I= 2 πa d =

a

2π

d

∫ ρ dρ ∫ sin φ dφ ∫ 3

0

2

0

0

M dz + 2 πa d

Ma 2 Md 2 + . 4 3

S12.9

a

2π

d

∫ ρ dρ ∫ dφ ∫ z 0

0

0

2

dz

12.19

By the divergence theorem the integral is

∫∫∫ V

⎡∂ ⎤ ∂ ∂ (∇ ⋅ F)dυ = ∫∫∫ ⎢ (4xz) + (−y 2 ) + (yz)⎥ dυ ⎢ ∂x ⎥ ∂y ∂z ⎣ ⎦ V 1

= ∫∫∫ (4z − 2y)dυ = ∫ V

0

1

=

∫ 1

∫

∫ ∫ (4z − 2y) dx dy dz 0

0

dx ∫ dy[2z 2 − 2yz ]z=1 z=0 y=0

y=1

x=0

12.20

1

1

x=0

=

1

dx ⎡⎢2y −y 2 ⎤⎥ = 1. ⎣ ⎦ y=0

From Green’s second identity (12.52b), we have

∫∫∫ (φ ∇ φ 2

Ω

i

j

)

− φj ∇2φi dυ = ∫ ∫ (φi ∇φj − φj ∇φi ) ⋅ ds = 0 , S

using the boundary conditions φi = 0 , φj = 0 on S. Hence

(γ j − γi )∫∫∫ φi φj dυ = 0, Ω

so that

∫∫∫ 12.21

Ω

φ1φj dυ = 0 if γi ≠ γ j .

From Table 12.1, we have

∇ ⋅(ψa) = ψ(∇ ⋅ a) + a ⋅ ∇ ψ, which gives the desired identity on substituting a = ∇ψ . Integrating this identity over the volume Ω , and setting ∇2ψ = 0 , gives

∫

Ω

(∇ψ)2 dυ = ∫ ∇ ⋅(ψ ∇ψ)dυ = ∫ ψ ∇ψ ⋅ ds Ω

S

using the divergence theorem. Since ψ = 0 on S, this gives

∫

Ω

(∇ψ)2dυ = 0

and hence ∇ψ = 0 in Ω since (∇ψ)2 ≥ 0 . Therefore ψ is constant in Ω and must be zero because ψ = 0 on S. 12.22

By (12.58), div g = −4πρG . Hence by the divergence theorem

S12.10

∫∫ g ⋅ ds = ∫∫∫ S

Ω

div g dυ = −4πG ∫∫∫ ρ dυ, Ω

where ρ is the density of the material, i.e. the total flux through the surface is −4πG times the mass enclosed. In this case, the gravitational field is obviously spherically symmetric, g(r) = g(r)ˆr , where r = r . Hence, if S is a sphere of radius r, we have

∫∫ g ⋅ ds = 4πr F(r) 2

,

S

so that

g(r) = −

G r2

∫∫∫

Ω

ρ dυ ,

where the integral extends over the volume of the enclosed sphere and gives the enclosed mass. We then have: (a) r > b , i.e. whole mass enclosed, and g(r) = −

GM ˆr , r2

(b) r < a , i.e. no mass enclosed, and g(r) = 0 , (c) a < r < b , i.e. mass enclosed is

M(r 3 −a 3 ) GM ⎛⎜ r 3 −a 3 ⎞⎟⎟ and so g(r) = − ⎜ ⎟ ˆr . r 2 ⎜⎜⎝b 3 −a 3 ⎟⎠ (b 3 −a 3 )

The field vanishes for r < a , so the potential φ , where g = −grad φ , is constant in this region. To find it, we need to calculate the work done in bringing a unit mass from r = ∞ to r = a along a radius, a

a

φ(r ≤ a) = −∫ g(r)⋅ dr = − ∫ g(r)dr ∞ b

∞

a

GM ⌠ GM = ⎮ 2 dr + 3 ∫ ⌡ r (b −a 3 ) b ∞

3⎞ ⎛ ⎜⎜r − a ⎟⎟ dr ⎟ ⎜⎜ r 2 ⎟⎠ ⎝

=−

GM GM ⎛⎜a 2 b 2 a 3 ⎞⎟⎟ 2 + 3 − + a − ⎜ ⎟ b 2 b ⎟⎠ (b −a 3 ) ⎜⎜⎝ 2

=−

3GM(b + a) . 2(b 2 + ab + a 2 )

As (b −a) → 0 , V → −GM a , which is the result for a thin spherical shell, as expected. 12.23

(a)

∂ ∂ ∂ (φE x ) + (φEy ) + (φE z ) ∂x ∂y ∂z ∂φ ∂φ ∂φ = φ∇ ⋅ E + E x + Ey + Ez ∂x ∂y ∂z = φ∇ ⋅ E + E ⋅ ∇φ.

∇ ⋅(φE) =

S12.11

(b) At very large distances R → ∞ , the charge distribution can be approximated by a point charge at the origin, so that

E=

Q ˆr , 4πε0R 2

φ=

Q . 4πε0R

(c)

∫∫∫

Ω1

ρφ dυ = ∫∫∫ ρφ dυ Ω2

since ρ = 0 outside Ω1 . Hence, using Poisson’s equation (12.53)

∫∫∫

Ω1

ρφ dυ = ε0 ∫∫∫ φ div E dυ Ω2

= ε0 ∫∫∫ div(φE) dυ − ε0 ∫∫∫ E ⋅ ∇φ dυ Ω2

Ω2

using the identity derived in (a). Then, using the divergence theorem in the first integral and E = −∇φ in the second integral, gives

∫∫∫

Ω1

ρφ dυ = ε0 ∫∫∫ E2dυ + ε0 ∫ ∫ φE ⋅ ds , Ω2

S

where S is the surface of the larger sphere, and

ds = R 2 sin θ dθ dφ er = R 2 sin θ dθ dφˆr in spherical polar co-ordinates by (12.39). Substituting this, together with the results of part (b), gives

∫∫∫

Ω1

ρφdυ = ε0 ∫∫∫ E2 dυ + Ω2

Q2 16π 2ε0R 2

R

∫ 0

π

2π

0

0

dr ∫ dθ ∫ dφ.

The integral is 4πR , giving the desired result with c = Q 2 (4πε0 ) . 12.24

(a) Taking the divergence of the fourth Maxwell equation, and using the identity (12.46), gives

0 = div curl H = div j +

∂ div D, ∂t

which together with the first Maxwell equation gives the equation of continuity

div j +

∂ρ =0 , ∂t

so that charge is conserved.

S12.12

(b) Taking the divergence of both sides of Ohm’s law and using the first Maxwell equation, gives

div j = σ div E =

σ σρ div D = , ε ε

which, together with the continuity equation, gives

∂ρ σ =− ρ , ∂t ε with the solution

ρ(r,t) = ρ0 (r)e −σt/ε with the given boundary condition. This result is known as the ‘decay of free charge’ in a conductor. The charge does not actually decay of course, but migrates to the boundary of the conductor, where there is a discontinuity across the surface, so that Maxwell’s equations, in the form given do not apply on the surface. 12.25

The boundary C of S is a circle in the xy plane of unit radius and centred at the origin. This can be written as

x = cos θ, y = sin θ, z = 0.

0 ≤ θ ≤ 2π .

Then

∫ A ⋅ dr = ∫ [(2x −y)dx −yz dy −y z dz ] 2

C

2

C 2π

= ∫ (2cos θ − sin θ)(−sin θ)dθ = π. 0

Using Stokes’ Theorem, the integral becomes

∫∫ (∇ ×A)⋅ nˆ ds = ∫∫ (k ⋅ nˆ) ds , S

S

ˆ ds is just the projection of a surface on evaluating the curl using (12.10). But k ⋅ n element ds onto the xy plane. Hence the integral becomes

∫∫ (k ⋅ nˆ) ds = ∫∫ dx dy , S

R

where R is the projection of S on the xy-plane, so that we finally obtain y=+ (1−x 2 )

x=+1

∫

x=−1

dx

∫

1

dy = 4 ∫ dx 2

y=− (1−x )

0

(1−x 2 )

∫

dy = π,

o

in agreement with the value obtained by direct evaluation of the line integral.

S12.13

12.26

Using Stokes’ theorem, we have

∫∫

S

⎡ ∇ ×(φc)⎤ ⋅ ds = φc ⋅ dr . ∫C ⎢⎣ ⎥⎦

Expanding the integrand on the left-hand side gives

∇ ×(φc) = ∇φ ×c + φ∇ ×c = ∇φ ×c , since c is a constant vector. The triple scalar product on the left-hand side can therefore be written

⎡ ∇ ×(φc)⎤ ⋅ ds = (∇φ ×c)⋅ ds = c ⋅(ds× ∇φ) . ⎢⎣ ⎥⎦ Substituting this into the expression from Stokes’ theorem gives

c ⋅ ∫∫ ds× ∇φ = c ⋅ ∫  φ dr S

C

and since c is an arbitrary vector , we obtain the desired result. 12.27

The work done W is given by

W =∫  F ⋅ dr C

where C is the circuit shown in the figure below. (a) On the circle,

r = a cos θ i + a sin θ j dr = −a sin θ i + a cos θ j F = a 2(sin 2 θ i + cos2 θ j) Therefore 2π

∫

C

F ⋅ dr = −a

3

2π

∫ sin θ dθ + a ∫ cos θ dθ = 0. 3

3

0

3

0

(b) Using Stokes’ theorem,

W =∫  F ⋅ dr = ∫∫ curl F ⋅ ds , C

S

ˆ and curl F = 2(x −y)k ˆ , so that where ds = dx dy k W = 2∫∫ (x − y)dx dy = 0 S

since the area is symmetric in x and y.

S12.14

(c) Since curl F ≠ 0 , the field F is not conservative. For an arbitrary loop, W ≠ 0 .

S12.15

SOLUTIONS TO PROBLEMS 13 13.1

The function is even and hence has a cosine series ∞

f (x) = 12 a 0 + ∑ an cos(nx) , n=1

where, by (13.10), π ⎡ 2π/3 ⎤ 2⎢ 2π ⎥. an = ⎢ ∫ x cos(nx)dx + cos(nx)dx ⎥ ∫ π⎢ 0 3 2π/3 ⎥⎦ ⎣

Hence

a0 =

2 ⎡⎢ 1 4π 2 2π 2 ⎤⎥ 8π , + = π ⎢⎢⎣ 2 9 9 ⎥⎦⎥ 9

and for n ≠ 0 , 2π/3

π

2 ⌠ d 4 an = sin(nx)dx + ∫ cos(nx)dx ⎮x πn ⌡ dx 3 2π/3 0

2 ⎡⎢ ⎛⎜ 2nπ ⎞⎟ ⎤⎥ ⎟⎟ −1 = cos ⎜ πn 2 ⎢⎢⎣ ⎜⎜⎝ 3 ⎟⎠ ⎥⎥⎦

.

on combining the result of the two integrals. So, using (2.37a), the Fourier series is

f (x) = 13.2

4π 4 ∞ sin 2(nπ 3) − ∑ cos(nx). 9 π n=1 n2

Since f (x) is odd, the coefficients an = 0 , and the coefficients bn are given by π

bn =

π

2 1 f (x)sin(nx)dx = ∫ sin(nx)dx ∫ π 0 π π/2 π ⎛ nπ ⎞⎟⎤ 1 ⎡⎢ cos(nx) ⎤⎥ 1 ⎡⎢ n ⎜⎜ ⎟⎥ . = − =− (−1) − cos ⎜⎜⎝ 2 ⎟⎟⎠⎥ π ⎢⎣ n ⎥⎦ π/2 nπ ⎢⎢⎣ ⎥⎦

For n odd,

bn = −

1 1 (−1)n = , nπ nπ

and for n even, with n = 2p ,

⎧ ⎪ 0 ⎪ ⎪ 1 ⎡ ⎪ ⎤ bn = − 1 − cos(px)⎥ = ⎨ ⎦ ⎪ − 2 nπ ⎣⎢ ⎪ ⎪ ⎪ ⎩ nπ Thus,

S13.1

p even p odd

.

f (x) =

⎪⎧⎪ 1 ⎪⎫⎪⎤⎥ 1 ⎡⎢ 1 1 sin x + sin 3x +− 2 sin 2x + sin 6x + ⎨ ⎬⎥ ⎪⎪⎩ 2 ⎪⎪⎭⎥ π ⎢⎢⎣ 3 6 ⎦ 1 ∞ 1⎡ ∑ sin nx − sin 2nx ⎤⎦⎥ . π n odd n ⎣⎢

=

When x = π 2 , the Fourier series will sum to the mean value of f (x) on either side of the discontinuity, i.e.

1 2

(0 + 12 ) = 14 . But setting x = π 2 in the series gives ⎞ 1 ⎛⎜ π 1 3π 1 5π + sin +⎟⎟⎟ , ⎜⎜sin + sin ⎟⎠ π ⎜⎝ 2 3 2 5 2

so

1 − 13 + 15 − 71 + = π 4 . 13.3

The function is even and so has a cosine series with π

π 1 2 ⎡ 2π 2x 3 x 5 ⎤⎥ 14 a 0 = ∫ x 2(2π 2 − x 2 )dx = ⎢⎢ − ⎥ = π4. π −π π ⎢⎣ 3 5 ⎥⎦ 0 15

For n ≠ 0, π

π

2 −2 an = ∫ (2π 2x 2 − x 4 )cos(nx)dx = (4π 2x − 4x 3 )sin(nx)dx π 0 nπ ∫0 π

−2 24 = 2 ∫ (4π 2 −12x 2 )cos(nx)dx = nπ 0 πn 2

π

∫x

2

,

cos(nx)dx ,

0

where we have left out terms that vanish at each step. Integrating by parts twice more (cf. Example 13.4) then gives

(−1)n n4 n=1 ∞

an = 48∑ and so, finally

x 2(2π 2 − x 2 ) =

13.4

∞ 7 4 (−1)n π + 48∑ 4 cos(nx) . 15 n n=1

The function is even so only the coefficients an are non-zero. These are given by π

π 1 1 ⎡ sin{(µ + n)x} sin{(µ −n)x} ⎤⎥ , an = ∫ cos(µx)cos(nx)dx = ⎢ + ⎥ π −π π ⎢⎣ µ +n µ −n ⎦0

where we have used the trigonometric relation (2.36b) to give

cos(µx)cos(nx) = 12 [cos(µ + n)x + cos(µ −n)x ] .

S13.2

Evaluating the limits and using (2.36a), then gives

an =

⎛ 1 1 1 ⎞⎟ (−1)n+1 2µ ⎟= sin(µπ)cos(nπ)⎜⎜⎜ − sin(µπ) . ⎜⎝ n + µ n − µ ⎟⎟⎠ π π n 2 − µ2

So,

cos(µx) =

∞ ⎡ 1 ⎤ 2µ (−1)n+1 sin(µπ) ⎢⎢ 2 + ∑ 2 cos(nπ)⎥⎥ . 2 π n=1 n − µ ⎢⎣ 2µ ⎥⎦

To deduce an expansion for cot(µπ) , we set x = π in the expansion for cos(µx) and divide by sin(µπ) to give

cot(µπ) =

∞ 2µ ⎡⎢ 1 1 ⎤⎥ . − ∑ π ⎢⎣ 2µ 2 n=1 n 2 − µ 2 ⎥⎦

Finally, setting µ = 1 3 , leads directly to ∞

∑ 9n n=1

1 1 π 3 = − , 18 −1 2

2

as required. 13.5

(a) This does not satisfy the Dirichlet conditions because the integral (13.13) does not converge. (b) This does not satisfy the Dirichlet conditions because arcsin x is multivalued. (c) This does satisfy the Dirichlet conditions, with just a finite discontinuity at x = π + 2kπ , where k is an integer. Because sinh(−x) = sinh(x) , we have a sine series ∞

f (x) = sinh x = ∑ bn sin(nx)dx , n=1

where π

bn =

π

1 2 sinh x sin(nx)dx = ∫ sinh x sin(nx)dx . ∫ π −π π 0

Integrating by parts twice gives π

π

(1 + n 2 )∫ sinh x sin(nx)dx = −n ⎡⎢sinh x cos(nx)⎤⎥ = (−1)n+1n sinh π, ⎣ ⎦0 0

so that

bn =

2sinh π (−1)n+1n , π 1 + n2

giving

S13.3

sinh x =

⎤ 2sinh π ⎡⎢ sin x 2sin 2x 3sin 3x − + −⎥ , ⎢ 2 ⎥ π 5 10 ⎣ ⎦

−π < x < π

(d) This does not satisfy the Dirichlet conditions because x 2e −1/x has an infinite discontinuity at x = 0 . To see this, note that

ln f (x) = ln x 2 −1 x , so as x → 0+ , ln f → −∞ and f → 0, but x → 0− , ln f → +∞ and f → ∞ . 13.6

The Fourier coefficients are π

a0 =

2π

1 1 4 x 2 dx + ∫ π 2 dx = π 2 , ∫ π 0 π π 3

and for n ≠ 0 , π

an =

2π

π

1 1 1 x 2 cosnx dx + ∫ π 2 cosnx dx = ∫ x 2 cosnx dx, ∫ π 0 π π π 0

since the second integral vanishes. Integrating by parts twice then gives (cf. Example 13.4)

an =

2(−1)n . n2

Similarly

bn =

2 ⎡ π (−1)n −1⎤⎥ − . 3 ⎣⎢ ⎦ n πn

Hence the Fourier series is

f (x) =

∞ ∞ ⎧ ⎫ ⎪⎪ 2 ⎡ 2 2 (−1)n cosnx n ⎤ − π ⎪⎪⎬ sin nx . π + 2∑ + (−1) −1 ⎨ ∑ ⎦⎥ n ⎪ ⎪⎩ πn 3 ⎣⎢ 3 n2 n=1 n=1 ⎪ ⎪⎭

At x = π the series has to converge to π 2 , so setting x = π in the series gives

π2 = 13.7

∞ 2 2 1 π + 2∑ 2 3 n=1 n

⇒

1 π2 = . ∑ 2 6 n=1 n ∞

We use the general expression

f (x) =

⎡ ⎛ nπx ⎞⎟ ⎛ nπx ⎞⎟⎤ ⎟⎟ +bn sin ⎜⎜ ⎟⎥ + ∑ ⎢⎢an cos ⎜⎜⎜ ⎜⎜⎝ L ⎟⎟⎠⎥ , ⎜⎝ L ⎟⎠ 2 n=1 ⎢⎣ ⎥⎦

a0

where, by (13.21a),

S13.4

2L

1 x dx = 2L L ∫o

ao = and for n ≠ 0 , 2L

2L

0

0

⎛ nπx ⎞⎟ ⎛ nπx ⎞⎟ 1⌠ 1⌠ ⎟⎟ dx and bn = ⎮ x sin ⎜⎜ ⎟ an = ⎮ x cos ⎜⎜⎜ ⎜⎜⎝ L ⎟⎟⎠ dx . ⎜⎝ L ⎟⎠ L⌡ L⌡ Both integrals can be evaluated by parts and give 2L

2l

⎡ 1 ⎛ nπx ⎞⎟⎤ 1 ⌠ ⎛⎜ nπx ⎞⎟ L ⎟⎟⎥ − ⎟⎟ dx = 2 2 an = ⎢⎢ x sin ⎜⎜⎜ ⎮ sin ⎜⎜ ⎥ ⎟ ⎟ ⎜⎝ L ⎠⎥ ⎜ nπ ⎢⎣ nπ ⎦ 0 nπ ⌡ ⎝ L ⎠ 0

2L

⎡ ⎛ nπx ⎞⎤ ⎟⎟⎥ ⎢cos ⎜⎜ ⎢ ⎜⎜⎝ L ⎟⎟⎠⎥ = 0, ⎢⎣ ⎥⎦ 0

and 2L

2L

⎡ −1 ⎛ nπx ⎞⎟⎤ ⎛ nπx ⎞⎟ 1 ⌠ 2L L ⎜⎜ ⎟⎟⎥ + ⎟⎟ dx = − bn = ⎢⎢ x cos ⎜⎜⎜ cos + 2 2 ⎮ ⎥ ⎜ ⎟ ⎟ ⎜ ⎜ nπ ⌡ nπ n π ⎝ L ⎠⎥⎦ 0 ⎝ L ⎠ ⎢⎣ nπ 0

2L

⎡ ⎛ nπx ⎞⎤ 2L ⎟⎟⎥ ⎢sin ⎜⎜ ⎢ ⎜⎜⎝ L ⎟⎟⎠⎥ = − nπ . ⎢⎣ ⎥⎦ 0

So, finally,

f (x) = L − 13.8

⎛ nπx ⎞⎟ 2L ∞ 1 ⎟. sin ⎜⎜⎜ ∑ ⎜⎝ L ⎟⎟⎠ π n=1 n

Extend the function to the region 1 < x < 0 to produce an even function as shown below.

Then p = 2 , bn = 0 , and the coefficients an are given by 12

a 0 = 2∫ dx = 1 0

and for n ≥ 1 , 1

an = 2∫ f (x)cos(nπx)dx = 0

12 2 ⎡ 2 sin(nπx)⎤⎥ = sin (nπ 2) . ⎢ ⎣ ⎦ 0 nπ nπ

Thus,

S13.5

f (x) =

⎤ 1 2 ⎡⎢ cos(3πx) cos(5πx) + cos(πx)− + −⎥ ⎥ 2 π ⎢⎣ 3 5 ⎦ =

13.9

1 2 ∞ (−1)n+1 + ∑ cos[(2n −1)πx ]. 2 π n=1 (2n −1)

(a) For −π < x < π , integrating the given series gives ∞ x3 π 2x (−1)n =c + + 4∑ 3 sin nx , 3 3 n n=1

where the integration constant c = 0 for consistency at x = 0 . Integrating again then gives ∞ x4 π 2x 2 (−1)n+1 = c′ + + 4∑ cosnx , 12 6 n4 n=1

where c ′ is an integration constant to be determined. Substituting the original series for x 2 and multiplying by 12 gives ∞ ⎛ π2 1⎞ x 4 = a 0 + 48∑ (−1)n ⎜⎜⎜ 2 − 4 ⎟⎟⎟ cosnx , ⎜⎝ 6n n ⎟⎠ n=1

where π

2 1 π4 4 a 0 = 12c ′ + π 4 = x dx = . 3 2π ∫ 5 −π So finally

x4 =

∞ ⎛ π2 π4 1⎞ + 48∑ (−1)n ⎜⎜⎜ 2 − 4 ⎟⎟⎟ cosnx ⎜⎝ 6n 5 n ⎟⎠ n=1

and setting x = π and cosnπ = (−1)n , gives ∞

⎛ π2

∑ ⎜⎜⎜⎜⎝ 6n n=1

2

−

1 ⎞⎟⎟ π 4 , ⎟= n 4 ⎟⎠ 60

as required. (b) Differentiating the series for x 4 gives ∞ ⎛ π2 1⎞ x 3 = 12∑ (−1)n+1 ⎜⎜⎜ − 3 ⎟⎟⎟ sin nx, ⎜⎝ 6n n ⎟⎠ n=1

f (x) in [−π, π] since f (x) = x ′ in [−π, π] f (x + 2π) = f (x) is a continuous function at all x. Differentiating again gives

which is a valid series for

S13.6

and

∞ ⎛ π2 1⎞ x 2 = 4∑ (−1)n+1 ⎜⎜⎜ − 2 ⎟⎟⎟ cosnx , ⎜⎝ 6 n ⎟⎠ n=1

which differs from the original series and is not a valid series since the function f (x) = x 3 in [−π, π] with f (x + 2π) = f (x) has jump discontinuities at x = (2n + 1)π . 13.10

The series given in Example 13.4 has a sum over terms proportional to n −2 . This can be converted to a sum over terms proportional to n −1 by differentiating the series to give

⎛ πnx ⎞⎟ (−1)n+1 ⎟ sin ⎜⎜⎜ ⎜⎝ 2 ⎟⎟⎠ πn n=1 ∞

2x = 8∑

Then setting x = 1 and using the result

⎧ 0 ⎛ nπ ⎞ ⎪ ⎪ sin ⎜⎜⎜ ⎟⎟⎟ = ⎨ ⎜⎝ 2 ⎟⎠ ⎪ (−1)(n−1)/2 ⎪ ⎪ ⎩

for n even for n odd

,

gives ∞ (−1)(n−1)/2 (−1)n = 4∑ . n n odd n=0 (2n + 1) ∞

π = 4∑ 13.11

The Fourier coefficients an are given by 2

2 2 a 0 = ∫ (x −1)2dx = ⎡⎢x 3 3 − x 2 + x ⎤⎥ = ⎣ ⎦0 3 0

and for n ≥ 1 , 2

2

an = ∫ (x −1)2 cos(nπx)dx = ∫ (x 2 − 2x + 1) cos(nπx)dx . 0

0

The integrals may be done by parts and give 2

∫ 0

2

⎡ x 2 sin(nπx) 2x cos(nπx) 2sin(nπx) ⎤ ⎥ = 4 , x cos(nπx)dx = ⎢⎢ + − 2 3 nπ (nπ) (nπ) ⎥⎥⎦ 0 (nπ)2 ⎢⎣ 2

2

⎡ 2x sin(nπx) 2cos(nπx) ⎤ ⎥ = 0, −2∫ x cos(nπx)dx = ⎢⎢− − 2 ⎥ nπ (nπ) ⎣ ⎦0 0 2

and 2

∫ 0

2

⎡ sin(nπx) ⎤ ⎥ = 0. cos(nπx)dx = ⎢ ⎢ nπ ⎥ ⎣ ⎦0

Likewise, the coefficients bn are given by

S13.7

2

bn = ∫ (x −1)2 sin(nπx)dx 0

and again, integrating by parts, without details, gives 2

∫

2

−2∫ x sin(nπx)dx = 4 nπ ,

x 2 sin(nπx)dx = −4 nπ,

0

0

and 2

∫ sin(nπx)dx = 0

.

0

So, finally,

f (x) =

1 4 ∞ cos(nπx) + 2∑ . 3 π n=1 n2

The coefficients an in the above sum are proportional to n −2 . To get a sum over terms proportional to n −4 we can use Parseval’s theorem, p/2

1 1 2 1 ∞ ⎡ 2 2 f (x)dx = a + ∑ a +bn2 ⎤⎥ . ∫ ⎦ p −p/2 4 0 2 n=1 ⎢⎣ n Firstly, 2

2

1 1 1 f 2(x)dx = ∫ (x 4 − 4x 3 + 6x 2 − 4x + 1)dx = , ∫ 2 0 2 0 5 so that

1 1 1 16 1 = + , 4 ∑ 4 5 9 2π n n i.e.

1

∑n n

13.12

4

=

π4 . 90

From the Fourier series for cos(µx) we can identify

a0 2

=

sin(µπ) µπ

and

an =

2µ (−1)n+1 . sin(µπ) 2 π (n − µ 2 )

Then Parseval’s theorem (13.31) gives 2

π 1 1 ⎡⎢ 2µπ + sin(2µπ) ⎤⎥ ⎛⎜a 0 ⎞⎟⎟ 1 ∞ 2 2 ⎜ cos (µx)dx = = + ⎟ ∑a , ⎥ ⎜⎜⎝ 2 ⎟⎟⎠ 2π ∫ π ⎢⎣ 4µ 2 n=1 n ⎦ −π

where the right-hand side is

S13.8

∞ sin 2(µπ) 2µ 2 1 2 . + sin (µπ) ∑ 2 2 2 2 2 2 µπ π n=1 (n − µ )

Equating these two and rearranging gives

⎡ 4µ 2 ⎤ 1 2 ⎢ ⎥ = sin (µπ) ∑ 2 2 2 ⎢ π ⎥ n=1 (n − µ ) ⎢⎣ ⎥⎦ ∞

−1

2 ⎡ ⎤ ⎢ π + sin(2µπ) − 2sin (µπ) ⎥ . 2 ⎢ ⎥ 2µ µπ ⎢⎣ ⎥⎦

Finally, setting µ = 1 2 gives the desired result. 13.13

The complex expansion is ∞

f (x) = ∑ cne inx , −∞

where π

π 2 ⎤ ⎡ x3 1 2 ⎢ + x ⎥ = 1 π 2, c0 = (x + x)dx = ⎢ 6π 4π ⎥ 2π ∫ ⎢⎣ ⎥⎦ −π 3 −π

and for n ≠ 0 , π

cn =

1 f (x)e −inx dx . 2π ∫ −π

These integrals my be evaluated using the relation given, or by parts, and give π

π

⎡ e −inx ⎤ ⎡ −inx ⎤ 2 2 ⎥ +⎢ e ⎥ cn = ⎢⎢ (n x − 2inx − 2) (−1 −inx) 3 2 ⎥ ⎢ ⎥ ⎢⎣ 2π(in) ⎥⎦ −π ⎢⎣ 2π(in) ⎥⎦ −π ⎡2 i⎤ = (−1)n ⎢ 2 + ⎥ . ⎢n n ⎥⎦ ⎣ So, finally

f (x) =

∞ ⎛2 π2 i⎞ + ∑ (−1)n ⎜⎜⎜ 2 + ⎟⎟⎟e inx . ⎜⎝ n 3 n=−∞ n ⎟⎠ n≠0

13.14

The complex coefficients are given by (13.36), 0

cn =

π

1 1 (−1)e −inx dx + (+1)e −inx dx ∫ ∫ 2π −π 2π 0 0

π

1 ⎡⎢ −1 −inx ⎤⎥ 1 ⎡⎢ 1 −inx ⎤⎥ 1 ⎡ = e + e = 1 −(−1)n ⎤⎥ . ⎢ ⎥ ⎢ ⎥ ⎣⎢ ⎦ 2π ⎣ −in 2π −in nπi ⎦ −π ⎣ ⎦0 Thus

⎧⎪ 2 (nπi) n odd ⎪ cn = ⎨ ⎪⎪ 0 n even ⎪⎩ and the complex Fourier series is

S13.9

,

f (x) =

2 ∞ 1 inx ∑ e . iπ n =-∞ n n odd

13.15

The function is

⎧ ⎪ 0 ⎪ ⎪ ⎪ f (x) = ⎪ ⎨ −2x − 2a ⎪ ⎪ ⎪ −2x + 2a ⎪ ⎪ ⎩

x >a .

−a < x < 0 a >x >0

Its Fourier transform is ∞

f (x) =

∫ g(k)e

ikx

dk

−∞

where ∞ 0 a 1 1 ⎡⎢ −ikx g(k) = ∫ f (x)e dx = 2π ⎢⎢ ∫ + ∫ 2π −∞ 0 ⎣ −a

⎤ ⎥ f (x)e −ikx dx. ⎥ ⎥⎦

The integrals needed are 0 −ikx ∫ e dx = − −a

a −ikx ∫ e dx = 0

1 ik

⎡1 −e ika ⎤ , ⎢⎣ ⎥⎦

1⎡ 1 −e −ika ⎤⎥ , ⎢ ⎣ ⎦ ik

0

∫ xe

−ikx

dx = −

−a

a

∫ xe

−ikx

dx =

0

1 (ik)2

⎡1 −e ika (1 −ika)⎤ , ⎢⎣ ⎥⎦

1 ⎡ 1 −e −ika (1 + ika)⎤⎥ , 2 ⎢⎣ ⎦ (ik)

so that 0

I 1 = ∫ (−2x − 2a)e −ikx dx = − −a

2 ⎡ 1 + ika −e ika ⎤⎥ , ⎦ k 2 ⎣⎢

and a

I 2 = ∫ (−2x + 2a)e −ikx dx = 0

2 ⎡ 1 −ika −e −ika ⎤⎥ . 2 ⎣⎢ ⎦ k

Then

g(k) =

1 2[ka − sin(ka)] I1 + I 2 ) = ( 2π iπk 2

and hence ∞

−2i ⌠ [ka − sin(ka)] ikx f (x) = e dk . ⎮ π ⌡ k2 −∞

3.16

From (13.44b) we have

S13.10

∞

1 g(k) = ∫ f−(x)(coskx −i sin kx)dx 2π −∞ ∞

∞

i i =− f−(x)sin(kx)dx = − ∫ f−(x)sin(kx)dx , ∫ 2π −∞ π 0 since the cosine tern vanishes by symmetry. From this one sees that g(k) = −g(−k) , so that (13.44a) gives ∞

f−(x) =

∫ g(k)(coskx + i sin kx)dk

−∞ ∞

∞

= i ∫ g(k)sin(kx)dk = 2i ∫ g(k)sin(kx)dk. −∞

0

Defining gs (k) ≡ ig(k) then gives ∞

gs (k) =

1 f (x)sin(kx)dx π ∫0 −

and ∞

f−(x) = 2∫ gs (k) sin(kx)dk 0

as required. 13.17

The Fourier sine transform is ∞

g(k) =

1 e −x sin(kx)dx . π ∫0

Integrating by parts twice gives

(k 2 + 1)g(k) =

∞ 1⎡ k −ke −x cos(kx)⎤⎥ = . ⎢ ⎣ ⎦ 0 π π

The inverse sine transform is then ∞

∞

2⌠ k f (x) = 2∫ g(k)sin(kx)dk = ⎮ 2 sin(kx)dk = e −x , ⌡ π k +1 0 0

and hence ∞

π ⌠ x sin(mx) dx = e −m , ⎮ ⌡ x2 +1 2

m >0 .

0

13.18

The function f (x) is even and hence the cosine transform

S13.11

∞

g(k) =

3

1 1 f (x)cos(kx)dx = ∫ (1 − x 2 )cos(kx)dx . ∫ π 0 π 0

Since 3

∫ cos(kx)dx = 0

sin 3k , k

then integrating by parts twice gives

g(k)= −

2 πk 3

⎡3k cos(3k) + (4k 2 −1)sin(3k)⎤ . ⎢⎣ ⎥⎦

It follows from the inverse transform ∞

f (x) = 2∫ g(k)cos(kx)dk 0

∞

4⌠ 1 =− ⎮ 3 π⌡k

⎡3k cos(k) + (4k 2 −1)sin(3k)⎤ cos(kx)dk ⎢⎣ ⎥⎦

0

that ∞

⌠ 1 ⎮ 3 ⌡t 0

13.19

⎡3t cos(3t) + (4t 2 −1)sin(3t)⎤ cost dt = − πf (1) = 0. ⎢⎣ ⎥⎦ 4

(

)

Set f (x) = exp(−αx 2 ) and g(k) = (2 πα)−1 exp −k 2 4α . (a) f (x −a) = exp[−α(x −a)2 ], so that by (13.55a) −ika −k 2 /4α

e e F ⎡⎢ exp −α(x −a)2 ⎤⎥ = e −ika g(k) = ⎣ ⎦ 2 πα

{

}

(b) f ′(x) = −2αx exp(−αx 2 ), so that 2 ik F ⎡⎢−2αx exp(−αx 2 )⎤⎥ = e −k /4α ⎣ ⎦ 2 πα

by (13.53a), and hence 2 −ik F ⎡⎢x exp(−αx 2 )⎤⎥ = e −k /4α . 3 1/2 ⎣ ⎦ 4(πα )

(c) We have, ∞

2 2 1 1 e −αx e −ikx dx = e −k /4α . ∫ 2π −∞ 2 πα

S13.12

.

Differentiating both sides with respect to α gives

−

∞ 2 1 1 ⎛⎜ 1 −3/2 1 −5/2 2 ⎞⎟ −k 2 /4α x 2 e −αx e −ikx dx = + α k ⎟⎟e , ⎜⎜− α ∫ ⎟⎠ 2π −∞ 4 2 π ⎜⎝ 2

so that

F ⎡⎢x 2 e −αx ⎣ 13.20

2

⎤ = 1 2α−3/2 −k 2α−5/2 e −k 2 /4α . ⎥⎦ 8 π

(

)

For real f (x) we have, from (13.49a) and (13.51a), g(k) = g *(−k) , while for antismmetric

f (x) we have g(−k) = −g(k) . Hence for real, antisymmetric f (x) , we have g(k) = −g *(k) , so that g(k) must be pure imaginary. (a) Writing the trigonometric functions in terms of exponentials, gives

cos 2x = 12 ⎡⎢e 2ix +e −2ix ⎤⎥ and sin 3x = ⎣ ⎦

1 2i

⎡e 3ix −e −3ix ⎤ , ⎢⎣ ⎥⎦

so that

sin(3x)cos(2x) =

1 4i

⎡e 5ix +e ix −e −ix −e −5ix ⎤ . ⎣⎢ ⎦⎥

But F ⎡⎢e iqx ⎤⎥ = δ(k −q) , and so ⎣ ⎦

F ⎡⎢cos(2x)sin(3x)⎤⎥ = ⎣ ⎦

1 4i

⎡δ(k − 5) + δ(k −1)− δ(k + 1)− δ(k + 5)⎤ . ⎣⎢ ⎦⎥

(b) Using (13.62b) gives

δ(x 2 −a 2 ) =

1 ⎡ δ(x −a)− δ(x + a)⎤⎥ . ⎦ 2a ⎣⎢

Also, ∞

1 e ika −ikx F ⎡⎢δ(x ± a)⎤⎥ = δ(x ± a)e dx = . ⎣ ⎦ 2π ∫ 2π −∞ So finally,

1 ⎡ −ika i F ⎡⎢δ(x 2 −a 2 )⎤⎥ = e −e ika ⎤⎥ = − sin ka , ⎣ ⎦ 4aπ ⎢⎣ ⎦ 2πa *13.21 The convolution of T(x) with itself (called auto-convolution) is

⎧⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ∞ ⎪ C(x) = ∫ T(x ′)T(x − x ′)dx ′ = ⎪⎨ ⎪⎪ −∞ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎩

x+1/2

∫

T(x ′)dx ′ = 0

x >1

dx ′ = 1 − x

0 0. The boundary condition y(0) = 0 gives A = 0 , while y(1) = 0 gives either B = 0 , or sin k = 0 , i.e.

k = λ −1 = nπ , ,

n = ±1, ± 2, …

Hence λ −1 = n 2π 2 , so that the eigenvalues are

λ = λn = 1 + n 2π 2 and the corresponding eigenfunctions are therefore

yn (x) = An sin(nπx). 15.10

This is an Euler equation. It is transformed into a linear equation with constant coefficients by the substitution x = e t , as shown in Section 14.3. Then the differentials in (1) become

dy dy dt dy = = e −t = e −ty ′, dx dt dx dt and

d2y d ⎛⎜ dy ⎞⎟ d ⎛⎜ 1 dy ⎞⎟ d ⎛ dy ⎞⎟ ⎟⎟ = e −t ⎜⎜e −t ⎟⎟ = ⎜⎜ ⎟⎟ = ⎜⎜ 2 dx ⎜⎝ dx ⎟⎠ dx ⎜⎝ x dt ⎟⎠ dt ⎜⎜⎝ dt ⎟⎠ dx = e −2t

d2y dy −e −2t = e −2ty ′′ −e −2ty ′. 2 dt dt

S15.8

where the primed notation now refers to differentiating with respect to t. Substituting these expressions into (1) gives

y ′′ + y ′ + ( 14 + λn )y = 0

(2)

which is the form of (14.24) with a = b = 1 , c = ( 14 + λn ) . Now since b 2 − 4ac < 0 , the general solution of (2) is, from (14.28c)

y(t) = e αt (A sin βt + B cos βt), where α = −b 2a = −1 2 , β = 4ac −b 2 2a = λn

and A and B are arbitrary

constants. Thus,

y(t) = e −t/2(A sin λn t + B cos λn t). But the boundary conditions are

y(x = 1) = y(x = e) = 0 , i.e. y(t = 0) = y(t = 1) = 0 , λn = nπ , or λn2 = n 2π 2 . Converting back to the

and imposing these gives B = 0 and x variable we have the eigenfunctions

yn (x) = Ax −1/2 sin(nπ ln x) . Finally to normalise yn (x) we form the integral e

2 ⌠ sin (nπ ln x) 1 = A2 ⎮ dx , ⌡ x 1

which may be evaluated by changing the variable to z = ln x , giving e

2 ⎡ ln x sin(2nπ ln x) ⎤ ⎥ =A . 1=A ⎢ − ⎢ 2 ⎥ 4nπ 2 ⎣ ⎦1 2

So finally

yn (x) = 2x −1/2 sin(nπ ln x) . *15.11 Substituting (15.35) for Ql (x) into the left and right-hand sides of (1) and comparing the coefficients of powers of x one sees that the recurrence relation is satisfied. Equation (1) is identical with the recurrence relation (15.39a) for Legendre polynomials. Hence, since from Example 15.8,

S15.9

Q0 (x) =

1 ⎛⎜1 + x ⎞⎟ P0 (x) ⎛⎜1 + x ⎞⎟ ⎟⎟ = ⎟, ln ⎜ ln ⎜⎜ ⎜⎝ 1 − x ⎟⎟⎠ 2 ⎜⎜⎝ 1 − x ⎟⎠ 2

and

Q1(x) =

P (x) ⎛1 + x ⎞⎟ x ⎛⎜1 + x ⎞⎟ ⎟⎟ −1 = 1 ln ⎜⎜ ⎟ ln ⎜⎜ ⎜⎜⎝ 1 − x ⎟⎟⎠ −1, 2 ⎜⎝ 1 − x ⎟⎠ 2

this implies

Ql (x) =

⎛1 + x ⎞⎟ ⎟ −q (x) , ln ⎜⎜⎜ ⎜⎝ 1 − x ⎟⎟⎠ l 2

Pl (x)

as required, where q 0 (x) = 0 , q1(x) = 1 , and

lql (x) = (2l −1)xql−1(x)−(l −1)ql−2(x) , so that

q 2(x) = 32 xq1(x)− 12 q 0 (x) = 32 x , q 3(x) = 53 xq 2(x)− 23 q1(x) = 52 x 2 − 23 , q 4 (x) = 74 xq 3(x)− 43 q 2(x) = 15.12

35 8

55 x 3 − 24 x.

Suppose a second series ∞

f (x) = ∑ ck′Pk (x), k=0

ck ≠ ck′

(3)

exists. Subtracting (3) from (1), we have ∞

0 = ∑ (ck − ck′ )Pk (x). k=0

Then multiplying by Pl (x) , for any l, and integrating, gives ∞

1

k=0

−1

0 = ∑ (ck − ck′ )∫ Pk (x)Pl (x) dx =

2(ck − ck′ ) 2l + 1

,

where we have used the orthogonality condition (15.30). Hence ck = ck′ for all l, in contradiction to assumption (3), so no second series exists. Substituting (1) into (3) and using the orthogonality condition (15.30), gives 1

1

2 ∞ ∞ ∞ 2 ⌠⎡ ⎤ dx = ∑ ∑ c c P (x)P (x)dx = ∑ 2ck . f (x) ⎮ ⎣⎢ k l∫ k l ⎦⎥ ⌡ k=0 l=0 k=0 2k + 1 −1 −1

*15.13 Differentiating the generating function partially with respect to h gives

S15.10

∞ ∂G(x,h) (x −h) = = lPl (x)h l−1 . ∑ ∂h (1 − 2xh + h 2 )3/2 l=0

Using the expression for G, this becomes

(1 − 2xh + h 2 )∑ lPl (x)h l−1 = (x − h)∑ Pl (x)h l . l

l

Equating the coefficients of h l on each side gives

(l + 1)Pl+1 − 2lxPl + (l −1)Pl−1 + Pl−1 − xPl = 0 , and hence

(2l + 1)xPl = (l + 1)Pl+1 + lPl−1 as required. *15.14 From (15.41a) the contributions of the three charges to V(r) are, in an obvious notation,

Vp (e, r) =

e 4π ε0r

l

⎛ p ⎞⎟ ⎜⎜ ⎟ P (cos θ), ∑ ⎜⎝ r ⎟⎟⎠ l l=0 ⎜

e V−p (e, r) = 4π ε0r

∞

l

⎛ p ⎞⎟ ⎜⎜ ⎟ P [cos(π − θ)], ∑ ⎜⎝ r ⎟⎟⎠ l l=0 ⎜ ∞

and

V0 (−2e, r) = −

2e , 4πε0r

where θ is the angle between r and p. Then, since cos(π − θ) = −cos θ and

Pl (−cos θ) = (−1)l Pl (cos θ) , adding the three terms gives

e V= 4πε0r

l

⎛ p ⎞⎟ ⎜⎜ ⎟ P (cos θ) ⎡1 + (−1)l ⎤ = 2e ∑ ⎜⎝ r ⎟⎟⎠ l ⎣⎢ ⎦⎥ 4πε r l=1 ⎜ 0 ∞

2n

⎛ p ⎞⎟ ⎜⎜ ⎟ P (cos θ) ∑ ⎜⎝ r ⎟⎟⎠ 2n n=1 ⎜ ∞

and if r  p , then

V→

2ep 2 ep 2 P (cos θ) = (3cos2 θ −1). 4πε0r 3 2 4πε0r 3

*15.15 Since Pk (−x) = (−1)k Pk (x) , we have

P2n+1(0) = 0 and P2n′ (0) = 0. Using this, the recurrence relation (15.39a) gives

2nP2n (0) = −(2n −1)P2n−2(0)

S15.11

so that

P2n (0) = −

(2n −1) (2n −1)(2n − 3) P2n−2(0) = P2n−4 (0) 2n 2n(2n − 2)

==

(−1)n (2n −1)!! (−1)n (2n −1)!! P0 (0) = , (2n)!! (2n)!!

as required, since P0 (0) = 1 . Similarly, from the recurrence relation (15.39b), we obtain n

(−1) (2n + 1)!! P2n+1′(0) = (2n + 1)P2n (0) = , (2n)!! using the previous result for P2n (0) . *15.16 Equation (15.50a) gives l/2 (−1)k (2l − 2k)!x l−2k ≡ ck x l−2k . ∑ l k=0 2 k !(l − 2k)!(l −k)! k=0 l/2

Pl (x) = ∑ On comparing this with

l

Pl (x) = ∑ anx n n=0

we see that an = ck and an+2 = ck−1 , where n = l − 2k . Hence

an+2 an

=

ck−1 ck

=

(−1)(2l − 2k + 2)(2l − 2k + 1) , k (l − 2k + 2)(l − 2k + 1)(l −k + 1) −1

and since 2k = l −n , this gives

an+2 an

=

(−1)(l −n)(l + n + 1) , (n + 2)(n + 1)

which is (15.26) and defines Pl (x) for even l (with a 2n+1 = 0 ) up to a normalisation factor. At x = 0 , the only term to survive in (15.50a) is the one with 2k = l , so that

Pl (0) =

(−1)l/2l ! , 2l (l 2)!(l 2)!

even l.

Setting l = 2n , gives

P2n (0) =

(−1)n (2n)! (−1)n (2n −1)!! = , (2n)!! 22n n !n !

S15.12

which is the result obtained in Problem 15.12, and confirms the normalisation. *15.17 From (15.33) and Rodriques’ formula, one obtains 1

cn =

1

n (2n + 1) (2n + 1) 1 n n d 2 n x P (x)dx = x ∫ n ∫ dx n (x −1) dx. n 2 2 2 n ! −1 −1

The proceeding as in Example 15.10, but with l = n , one obtains 1

∫

1

xn

−1

dn (x 2 −1)n dx = (−1)n n ! ∫ (x 2 −1)n dx dx n−1 −1 1

= n ! ∫ (1 + x)n (1 − x)n dx . −1

Sbstituting x = 2t −1, dx = 2dt , 1 + x = 2t , 1 − x = 2(1 −t) , one obtains 1

1

∫ (1 + x) (1 − x) dx = 2 ∫ t n

n

2n+1

−1

n

(1 −t)n dt = 22n+1

0

n !n ! (2n + 1)!

using the given integral. Hence

cn = 15.18

(2n + 1) 1 n !n ! 2n n !n ! 2n+1 n !2 = . 2 (2n + 1)! (2n)! 2n n !

Substituting z = kx in the equation, we obtain, after a little algebra the equation

z2

d2y(z) dy(z) +z + (z 2 −1)y(z) = 0 , 2 dz dz

which is Bessel’s equation for order ν = 1 . Therefore the only solution that satisfies y(z = kx) = 0 at x = 0 is y = AJ 1(kz) . The boundary condition y = 0 at x = 1 , then gives J 1(k) = 0 , which from the table at the end of Section 15.4.1 is satisfied by

k = 3.8317, 7.0156, 10.1735, 13.3237 plus higher values. *15.19 These relations follow directly from the recurrence relations (15.71). In particular, at a maximum of J n (x) , the derivative J n′ (x) = 0 , so that (15.71d) gives

J n−1(x) = J n+1(x), while if J n (x) = 0 , (15.71c) gives

S15.13

J n−1(x) = −J n+1(x). 15.20

For ν = 2 and x = 2 , the series (15.65) becomes

(−1)n . n=0 n !(n + 2)! ∞

J 2(2) = ∑

Since this is an alternating series, (5.62) guarantees that the error is less than the magnitude of the first term that is omitted. Evaluating the contributions in turn to six decimal places gives

n=0

0.500000

n=4

0.000058

n = 1 −0.166667 n = 5 −0.000002 ≈ 2×10−6 n=2 0.020833 n = 6 3.4 ×10−8 n = 3 −0.001389 Hence to achieve an error of less than 10−5 , needs 5 terms and gives

J 2(2) = 0.35283 to 5 decimal places. Adding one extra term with n = 5 , and calculating with sufficient accuracy, would reduce the error to 3.4 ×10−8 . 15.21

(a) From (15.68),

N ν (x) ≡

J ν (x)cos(νπ)−J −v (x) sin(νπ)

,

where for integer m, this is interpreted as

⎡ J (x)cos(νπ)−J (x) ⎤ −ν ⎥. N m (x) = lim ⎢⎢ ν ⎥ ν→m sin(νπ) ⎢⎣ ⎥⎦ For ν = 0 , the numerator and denominator both vanish, but by l’Hôpital’s rule,

⎡ J ′ (x)cos(νπ)− π sin(νπ)J (x)− J ′ (x) ⎤ ν −ν ⎥, N 0 (x) = lim ⎢⎢ ν ⎥ ν→0 π cos(νπ) ⎢⎣ ⎥⎦ where the derivatives are with respect to ν . From (15.65) and (15.62a), ν

ν+2

⎛x ⎞ 1 ⎛⎜ x ⎞⎟ J ν (x) = ⎜⎜ ⎟⎟ +O ⎜⎜⎜ ⎟⎟⎟ ⎜⎝ 2 ⎟⎠ νΓ(ν) ⎜⎝ 2 ⎟⎠ which using (1) is

S15.14

,

(2)

ν

ν+2

⎛x ⎞ ⎛x ⎞ J ν (x) = ⎡⎢1 + γν +O(ν 2 )⎤⎥ ⎜⎜⎜ ⎟⎟⎟ +O ⎜⎜⎜ ⎟⎟⎟ ⎣ ⎦ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠

,

(3a)

and similarly −ν

2−ν

⎛x ⎞ ⎛x ⎞ J −ν (x) = ⎡⎢1 − γν +O(ν 2 )⎤⎥ ⎜⎜⎜ ⎟⎟⎟ +O ⎜⎜⎜ ⎟⎟⎟ ⎣ ⎦ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠

.

(3b)

To find the derivatives of these functions, we set y = (x 2)ν , so that ln y = ν ln(x 2) and d ln y dν = (1 y)dy dν = ln(x 2) . Hence ν

ν

⎛ x ⎞⎟ ⎛ x ⎞⎟ d ⎛⎜ x ⎞⎟ ⎜⎜ ⎟⎟ = ⎜⎜⎜ ⎟⎟ ln ⎜⎜⎜ ⎟⎟ , ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ dx ⎜⎝ 2 ⎟⎠

(4a)

and similarly −ν

d ⎛⎜ x ⎞⎟ ⎜ ⎟⎟ dx ⎜⎜⎝ 2 ⎟⎠

−ν

⎛x ⎞ ⎛x ⎞ = −⎜⎜⎜ ⎟⎟⎟ ln ⎜⎜⎜ ⎟⎟⎟ . ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠

(4b)

The derivatives of the Bessel functions are then ν

ν

⎛x ⎞ ⎛x ⎞ ⎛x ⎞ J ν′ (x) = (1 + γν)⎜⎜⎜ ⎟⎟⎟ ln ⎜⎜⎜ ⎟⎟⎟ + γ ⎜⎜⎜ ⎟⎟⎟ , ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ and −ν

−ν

⎛x ⎞ ⎛x ⎞ ⎛x ⎞ ′ (x) = −(1 − γν)⎜⎜ ⎟⎟⎟ ln ⎜⎜ ⎟⎟⎟ − γ ⎜⎜ ⎟⎟⎟ . J −ν ⎜⎜⎝ 2 ⎟⎠ ⎜⎜⎝ 2 ⎟⎠ ⎜⎜⎝ 2 ⎟⎠ So finally, using these expressions in (2) and taking the limit ν → 0 , gives

N 0 (x) =

2⎡ ln x − ln 2 + γ ⎤⎥ +O(x 2 ). ⎦ π ⎢⎣

(b) As in part (a) above, from (15.69),

N 1(x) ≡ lim ν→1

J ν (x)cos(νπ)−J −v (x) sin(νπ)

,

which may be written

⎡ J (x)cos(νπ)−J (x) ⎤ −1−ε ⎥ N 1(x) = lim ⎢⎢ 1+ε ⎥ ε→0 sin(νπ) ⎢⎣ ⎥⎦ where ν = 1 + ε . Both numerator and denominator vanish in the limit ε → 0 , but using l’Hôpital’s rule, we have

S15.15

⎡ J ′ (x)cos(νπ)− π sin(νπ)J (x)− J ′ (x) ⎤ 1+ε −1−ε ⎥ , N 1(x) = lim ⎢⎢ 1+ε ⎥ ε→0 π cos(νπ) ⎢⎣ ⎥⎦

(5)

where the derivatives are now with respect to ε . Proceeding as in part (a), and using (15.65) and (15.62a), we have 1+ε

⎛ x ⎞⎟ 1 ⎜⎜ ⎟ J 1+ε (x) = ⎟ ε(1 + ε)Γ(ε) ⎜⎜⎝ 2 ⎟⎠

3+ε

⎛x ⎞ +O ⎜⎜⎜ ⎟⎟⎟ ⎜⎝ 2 ⎟⎠

1+ε

⎛x ⎞ = ⎡⎢1 + (γ −1)ε +O(ε2 )⎤⎥ ⎜⎜⎜ ⎟⎟⎟ ⎣ ⎦ ⎜⎝ 2 ⎟⎠

3+ε

⎛x ⎞ +O ⎜⎜⎜ ⎟⎟⎟ ⎜⎝ 2 ⎟⎠

,

where we have used the expansion (1) for εΓ(ε) . Similarly, −1−ε

1 ⎛⎜ x ⎞⎟ J −1−ε (x) = ⎜ ⎟⎟ Γ(−ε) ⎜⎜⎝ 2 ⎟⎠

1−ε

⎛ x ⎞⎟ 1 ⎜⎜ ⎟ − ⎟ (−ε)Γ(−ε) ⎜⎜⎝ 2 ⎟⎠ −1−ε

⎛x ⎞ = ⎡⎢−ε +O(ε2 )⎤⎥ ⎜⎜⎜ ⎟⎟⎟ ⎣ ⎦ ⎜⎝ 2 ⎟⎠

3−ε

⎛x ⎞ +O ⎜⎜⎜ ⎟⎟⎟ ⎜⎝ 2 ⎟⎠

1−ε

⎛x ⎞ − ⎡⎢1 − γε +O(ε2 )⎤⎥ ⎜⎜⎜ ⎟⎟⎟ ⎣ ⎦ ⎜⎝ 2 ⎟⎠

3−ε

⎛x ⎞ +O ⎜⎜⎜ ⎟⎟⎟ ⎜⎝ 2 ⎟⎠

.

Hence 1+ε

⎛x ⎞ ′ (x) = (γ −1)⎜⎜ ⎟⎟⎟ J 1+ε ⎜⎜⎝ 2 ⎟⎠

1+ε

⎛x ⎞ + ⎜⎜⎜ ⎟⎟⎟ ⎜⎝ 2 ⎟⎠

⎛x ⎞ ln ⎜⎜⎜ ⎟⎟⎟, ⎜⎝ 2 ⎟⎠

and −1−ε

⎛x ⎞ ′ (x) = −⎜⎜ ⎟⎟⎟ J −1−ε ⎜⎜⎝ 2 ⎟⎠

1−ε

⎛x ⎞ + γ ⎜⎜⎜ ⎟⎟⎟ ⎜⎝ 2 ⎟⎠

1−ε

⎛x ⎞ + ⎜⎜⎜ ⎟⎟⎟ ⎜⎝ 2 ⎟⎠

⎛x ⎞ ln ⎜⎜⎜ ⎟⎟⎟ , ⎜⎝ 2 ⎟⎠

where we have neglected terms which obviously vanish in the limit ε → 0 . Substituting into (5) gives

⎛x ⎞ ⎛ x ⎞ ⎛ x ⎞⎤ 1 ⎡⎢ ⎛⎜ 2 ⎞⎟ ⎟⎟ + (2γ −1)⎜⎜ ⎟⎟⎟ + 2 ⎜⎜ ⎟⎟⎟ ln ⎜⎜ ⎟⎟⎟⎥ +O(x 2 ), − ⎜ ⎜⎜⎝ 2 ⎟⎠ ⎜⎜⎝ 2 ⎟⎠ ⎜⎜⎝ 2 ⎟⎠⎥ π ⎢⎢⎣ ⎜⎜⎝ x ⎟⎠ ⎥⎦

N 1(x) = as required. *15.22 We need to prove that

2n+1/2

⎛ x ⎞⎟ (−1)n ⎜⎜ ⎟ J 1/2(x) = ∑ ⎜⎝ 2 ⎟⎟⎠ n=0 Γ(n + 1)Γ(n + 3 2) ⎜ ∞

is equal to 1/2

1/2

⎛ 2 ⎞⎟ ⎛ ⎞ ⎜⎜ ⎟ sin x = ⎜⎜ 2 ⎟⎟ ⎜⎜⎝ πx ⎟⎟⎠ ⎜⎜⎝ πx ⎟⎟⎠ From (1) we have

S15.16

(−1)n x 2n+1 . n=0 (2n + 1)! ∞

∑

(1)

1/2

⎛ 2⎞ J 1/2(x) = ⎜⎜⎜ ⎟⎟⎟ ⎜⎝ x ⎟⎠

2n+1

⎛ x ⎞⎟ (−1)n ⎜⎜ ⎟ ∑ ⎜⎜⎝ 2 ⎟⎟⎠ Γ(n + 1)Γ(n + 3 2) n=0 ∞

where Γ(n + 1) = n ! , and

Γ(n + 3 2) = (n + 1 2)Γ(n + 1 2) = [(n + 1 2)(n −1 2)1 2]Γ(1 2) =

[(2n + 1)(2n −1)1] π , 2n+1

so that 1/2

⎛ 2⎞ J 1/2(x) = ⎜⎜⎜ ⎟⎟⎟ ⎜⎝ πx ⎟⎠

(−1)n x 2n+1 . ∑ n n=0 2 n ![(2n + 1)(2n −1)1] ∞

Now

(2n + 1)(2n −1)1 =

(2n + 1)! (2n + 1)! = (2n)(2n − 2)2 2n n !

and hence 1/2

⎛ 2⎞ J 1/2(x) = ⎜⎜⎜ ⎟⎟⎟ ⎜⎝ πx ⎟⎠

1/2

(−1)n x 2n+1 ⎛⎜ 2 ⎞⎟ = ⎜⎜ ⎟⎟ sin x ∑ ⎜⎝ πx ⎟⎠ n=0 (2n + 1)! ∞

as required. Using the recurrence relation (15.71a), 1/2

⎛ 2⎞ J −1/2(x) = ⎜⎜⎜ ⎟⎟⎟ cosx . ⎜⎝ πx ⎟⎠ Similarly, using (15.71b),

−x −1/2J 3/2(x) =

d ⎡ −1/2 x J 1/2(x)⎤⎥ , ⎦ dx ⎣⎢

giving 1/2

⎛ 2 ⎞ ⎡ sin x ⎤ J 3/2(x) = ⎜⎜⎜ ⎟⎟⎟ ⎢ − cosx ⎥ . ⎥ ⎜⎝ πx ⎟⎠ ⎢⎣ x ⎦

S15.17

SOLUTIONS TO PROBLEMS 16 16.1

We use the method of separation of variables. Thus, writing

u(x,y,z) = X(x)Y (y)Z(z) and dividing by u gives

X ′′ Y ′′ Z ′′ + + + k2 = 0 X Y Z which leads to the three ODEs

X ′′ + kx2X = 0, Y ′′ + ky2Y = 0, Z ′′ + kz2Z = 0, where kx , ky , kz are three arbitrary constants subject to the constraint

k 2 = kx2 + ky2 + kz2 .

(1)

The solution for X(x) is

X(x) = Ak sin(kx x) + Bk cos(kx x) , x

x

but the boundary condition implies X(0) = X(L) = 0 , so that

Bk = 0 for all k, and kx L = nx π, x

where nx is a positive integer. The solution for X(x) s may thus be written

X(x) = Ak sin(kx x) = An sin (nx πx L ) . x

x

Similarly,

Y (y) = Bn sin (ny πy L ) and Z(z) = Bn sin (nz πz L ) , y

z

where ny and nz are also positive integers. So finally, the possible solutions are

⎛ n πx ⎞⎟ ⎛⎜ n πy ⎞⎟ ⎛ n πz ⎞⎟ u(x,y,z) = An n n sin ⎜⎜⎜ x ⎟⎟ sin ⎜⎜ y ⎟⎟⎟ sin ⎜⎜⎜ z ⎟⎟ x y z ⎜⎝ L ⎟⎟⎠ ⎜⎝ L ⎟⎠ ⎜⎝ L ⎟⎟⎠ where k 2 = (nx2 + ny2 + nz2 )π 2 L2 . 16.2

Substituting u = X(x)T(t) and dividing by u gives

S16.1

−

 2 X ′′ T′ . = i 2m X T

Then separating variables and denoting the separation constant by E gives

iT ′ = ET, (1)

X ′′ +

and

2mE X = 0, (2) . 2

The general solution of (1) is T = A exp(−iEt ) , so that

u(x,t) = X(x)exp(−iEt ) as required. Equation (2) may be written X ′′ + k 2X = 0, where E = k 2 2 2m , with the general solution

X(x) = B cos(kx) +C sin(kx) , where kL = 2πn in order to satisfy the periodic boundary condition. Hence k = 2πn L and 2

⎛ 2πn ⎞⎟  2 n 2h 2 ⎟⎟ En = ⎜⎜⎜ = . ⎜⎝ L ⎟⎠ 2m 2mL2 16.3

We seek a solution of the Laplace equation (16.5) in two dimensions, subject the boundary conditions

⎧⎪ 0 ⎪⎪ ⎪⎪ 0 u(x,y) = ⎪⎨ ⎪⎪ 0 ⎪⎪ 2 ⎪⎪ u 0 x (a − x) a ⎩

x = 0, 0 ≤ y ≤ b x = a, 0 ≤ y ≤b y = 0, 0 ≤ x ≤ a

.

y = b, 0 ≤ x ≤ a

Using the method of separation of variables (16.8), we write

u(x,y) = X(x)Y (y) , so that the Laplace equation becomes the two ODEs

1 d 2X = −k 2 , X dx 2

1 d2Y = k2 , Y dy 2

where k 2 is any real number. The only non-trivial solution to the first of these equations is

S16.2

X(x) = A cos(kx) + B sin(kx) , where A, B are arbitrary constants. Using the first boundary condition gives A = 0 , and the second implies sin(ka) = 0 for non-trivial solutions, i.e.

k = nπ a ,

n = 1, 2,…

so that

X n (x) = Bn sin(nπx a) . The solution of the equation for Y is

Y (y) = C cosh(ky) + D sinh(ky) = C cosh(nπy a) + D sinh(nπy a) , where C, D are arbitrary constants. The third boundary condition gives C = 0 , and so

Yn (y) = Dn sinh(nπy a) , and hence

un (x,y) = X n (x)Yn (y) = BnDn sin(nπx a)sinh(nπy a),

n = 1, 2, …

The complete solution is then, by the superposition principle, ∞

∞

n=1

n=1

u(x,y) = ∑ cnun (x,y) = ∑ cn sin(nπx a)sinh(nπy a), where the cn are a set of arbitrary constants. Finally, we need to impose the fourth boundary condition, which gives ∞

u(x,b) = u 0 x 2(a − x) a = ∑ cn sin(nπx a)sinh(nπb a),

0 ≤x ≤a

n=1

To find the coefficients, we multiply both sides by sin(mπx a) and integrate over the interval (0,a) . Using the orthogonality property of the sine function over this interval, [cf (12.36)] a

⎛ nπx ⎞⎟ ⎛ mπx ⎞⎟ a ⎟⎟ sin ⎜⎜ ⎟⎟ dx = δnm , ⎜ ⎟ ⎟ ⎜ a ⎠ ⎝ a ⎠ 2

∫ sin ⎜⎜⎜⎜⎝ 0

gives

⎛ nπb ⎞⎟ 2u ⎟= 0 cn sinh ⎜⎜⎜ ⎜⎝ a ⎟⎟⎠ a 2

a

⎛ nπx ⎞⎟ ⎟⎟ dx . a ⎟⎠

∫ x (a − x)sin ⎜⎜⎜⎜⎝ 2

0

Then, using the given integral, gives

S16.3

cn = −4u 0a 2

[1 + 2(−1)n ] 1 3 3 sinh(nπb a) nπ

and so the complete solution is

u(x,y) = − 16.4

4u 0a 2 π3

[1 + 2(−1)n ] sin(nπx a)sinh(nπy a) . sinh(nπb a) n3 n=1 ∞

∑

In Cartesian co-ordinates, the two-dimensional wave equation is

∂2u ∂2u 1 ∂2u + = ∂x 2 ∂y 2 υ 2 ∂t 2 and we seek solutions for 0 ≤ x ≤ a , 0 ≤ y ≤ b , subject to the boundary conditions

u(0,y) = u(a,y) = 0,

0 ≤ y ≤b

and

u(x, 0) = u(x,b) = 0,

0 ≤x ≤a .

Using the method of separation of variables, we can write

u(x,y,t) = X(x)Y (y)T(t) and proceeding as in Section 16.2.1, this leads to the three ODEs

d2T d 2X d2Y 2 2 2 = −υ α T , = −β x , = −γ 2x , 2 2 2 dt dx dy with α2 = β 2 + γ 2 . The solutions of the equations in the two spatial variables that satisfy the boundary conditions are

X(x) = sin(nπx a),

n = 1, 2,…

Y (y) = sin(mπy b),

m = 1, 2,…

and

and so α2 is given by 1/2

αnm

⎛n 2 m 2 ⎞ = π ⎜⎜⎜ 2 + 2 ⎟⎟⎟ . ⎜⎝ a b ⎟⎠

The solution for the time variable may therefore be written

Tnm (t) = anm cos(ωnmt) +bnm sin(ωnmt) , where

S16.4

1/2

ωnm = υαnm

⎛n 2 m 2 ⎞ = υπ ⎜⎜⎜ 2 + 2 ⎟⎟⎟ . b ⎟⎠ ⎝⎜ a

The complete solution is therefore ∞

u(x,y,t) = ∑ unm (x,y,t) , n,m

where

unm (x,y,t) = [anm cos(ωnmt) +bnm sin(ωnmt)]sin(πnx a)sin(πmy b) , as required. To find the constants we need to impose the further boundary condition

u(x,y, 0) = sin(xπ a)sin(yπ b) . This gives ∞

∞

u(x,y, 0) = ∑ ∑ anm sin (nπx a ) sin (mπy b ) = sin(xπ a)sin(yπ b) , n=1 m=1

so that a11 = 1 and anm = 0 otherwise. Similarly, bnm = 0 for all n, m because the plate is released from rest. So, for these boundary conditions, the motion of the plate is given by 1/2 ⎡⎛ 2 ⎤ ⎛ πx ⎞⎟ ⎛ πy ⎞⎟ ⎢⎜a +b 2 ⎞⎟⎟ ⎥ ⎜ ⎜ u(x,y,t) = sin ⎜⎜ ⎟⎟ sin ⎜⎜ ⎟⎟ cos ⎢⎜⎜ 2 2 ⎟ υπt ⎥ . ⎟ ⎜⎝ a ⎟⎠ ⎜⎝ b ⎟⎠ ⎢⎜⎝ a b ⎠ ⎥ ⎢⎣ ⎥⎦

16.5

We need to solve the one-dimensional diffusion equation,

∂ 2 u(x,t) 1 ∂u(x,t) , = κ ∂t ∂x 2 subject the the boundary conditions

u(0,t) = u(L,t) = 0 and

u(x, 0) = u 0x(L − x) . Following the argument of Section 16.2.3, the solution is given by [cf (16.34)] ∞

u(x,t) = u 0 ∑ Bn sin(nπx L)exp(−κn 2π 2t L2 ) , n=1

where the coefficients Bn are given by [cf (16.35)]

S16.5

Bn =

L L ⎛ nπx ⎞⎟ ⎛ nπx ⎞⎟ 2 2 ⎜⎜ ⎟ ⎟ dx, u(x, 0)sin dx = x(L − x)sin ⎜⎜⎜ ⎟ ∫ ∫ ⎜ ⎜⎝ L ⎟⎠ ⎜⎝ L ⎟⎟⎠ L 0 L 0

L ⎛ nπx ⎞⎟ ⎛ nπx ⎞⎟ 2 ⎟⎟ dx − ∫ x 2 sin ⎜⎜ ⎟ = 2∫ x sin ⎜⎜⎜ ⎜⎜⎝ L ⎟⎟⎠ dx, ⎜⎝ L ⎟⎠ L L

0

.

n = 1, 2,…

0

Both integrals may be done by parts. For the first integral, we have L

⎡ Lx ⎛ nπx ⎞⎟ ⎛ nπx ⎞⎟ ⎛ nπx ⎞⎟⎤ L2 ⎜⎜ ⎜⎜ ⎟⎟ dx = 2 ⎢− ⎟⎟ + ⎟⎥ 2∫ x sin ⎜⎜⎜ cos sin ⎢ nπ ⎜⎜⎝ L ⎟⎠ (nπ)2 ⎜⎜⎝ L ⎟⎟⎠⎥ ⎜⎝ L ⎟⎠ ⎢⎣ ⎥⎦ 0 0 2 L = −2 (−1)n , nπ L

and for the second integral, L

L

L ⎡ 2x 2 ⎛ nπx ⎞⎟ ⎛ nπx ⎞⎟⎤ ⎛ nπx ⎞⎟ 2 4 ⌠ ⎜⎜ ⎟⎟ dx = ⎢ ⎟⎟⎥ − ⎟ dx, − ∫ x 2 sin ⎜⎜⎜ cos x cos ⎜⎜⎜ ⎮ ⎢ ⎥ ⎜ ⎜⎝ L ⎟⎠ ⎜⎝ L ⎟⎠⎥ ⎜⎝ L ⎟⎟⎠ L 0 ⎢⎣ nπ ⎦ 0 nπ ⌡ 0

which, integrating by parts a second time becomes

−

L ⎛ nπx ⎞⎟ 2 ⎟ dx x 2 sin ⎜⎜⎜ ∫ ⎜⎝ L ⎟⎟⎠ L 0 L

⎡ 2x 2 ⎛ nπx ⎞⎟ 4Lx ⎛ nπx ⎞⎟ 4L2 ⎛ nπx ⎞⎟⎤ ⎜⎜ ⎜⎜ ⎟⎟− ⎟ ⎟⎥ = ⎢⎢ cos ⎜⎜⎜ sin − cos 2 ⎜⎜⎝ L ⎟⎟⎠ (nπ)3 ⎜⎜⎝ L ⎟⎟⎠⎥ , ⎟ ⎜ nπ L (nπ) ⎝ ⎠ ⎢⎣ ⎥⎦ 0 2 2 2 2L 4L 4L = (−1)n − (−1)n + , n = 1, 2,… 3 nπ (nπ) (nπ)3 Adding the two integrals gives

Bn =

4L2 (nπ)3

⎧ ⎪ 8L2 (nπ)3 n odd ⎡1 −(−1)n ⎤ = ⎪ . ⎨ ⎢⎣ ⎥⎦ ⎪ 0 n even ⎪ ⎪ ⎩

Because of this last relation, , finally we may write

u(x,t) = 16.6

8u 0L2 π3

sin[(2n −1)πx L]exp[−κ(2n −1)2 π 2t L2 ] . (2n −1)3 n=1 ∞

∑

We solve the Laplace equation in two dimensions expressed in Cartesian co-ordinates, i.e.,

∂ 2 u(x,y) ∂ 2 u(x,y) + = 0, ∂x 2 ∂y 2

S16.6

with the boundary conditions given. Using the method of separation of variables with a separation constant −α gives

d2X(x) = −αX(x). dx 2 If α < 0 , we set α = −k 2 and the solution is

X(x) = Ae kx + Be −kx , where A and B are constants. Imposing boundary condition u(0,y) = u(d,y) = 0 implies there is only the trivial solution A = B = 0 , so we take α > 0 and with α = k 2 , the solution is

X(x) = A cos(kx) + B sin(kx). Imposing the same boundary condition at x = 0 implies A = 0 and at x = d that

⎡ ⎛ nπ ⎞ ⎤ X(x) = Bn sin ⎢⎢sin ⎜⎜⎜ ⎟⎟⎟ x ⎥⎥ , ⎢⎣ ⎜⎝ d ⎟⎠ ⎥⎦ provided B ≠ 0 . In a similar way the solution for Y (y) is

Y (y) = C ne πny/d + Dne −πny/d . But imposing the boundary condition u(x,y) → 0 as y → ∞ implies C n = 0 , and so the full solution is

⎛ nπx ⎞⎟ ⎟, un (x,y) = Ene −nπy/d sin ⎜⎜⎜ ⎜⎝ d ⎟⎟⎠ where En ≡ BnDn . This is a particular solution and the general solution will be the linear superposition ∞ ∞ ⎛ nπx ⎞⎟ ⎟, u(x,y) = ∑ un (x,y) = ∑ Ene −nπy/d sin ⎜⎜⎜ ⎜⎝ d ⎟⎟⎠ n=0 n=1

provided the series converges and every term is twice differentiable. The final step is to impose the boundary condition at y = 0 , so that ∞ ⎛ nπx ⎞⎟ ⎟. u(x, 0) = f (x) = ∑ En sin ⎜⎜⎜ ⎜⎝ d ⎟⎟⎠ n=1

S16.7

This is a Fourier sine series and so from (13.12b), the coefficients En may be found from d

⎛ nπx ⎞⎟ 2⌠ ⎟ dx En = ⎮ f (x)sin ⎜⎜⎜ ⎜⎝ d ⎟⎟⎠ d⌡ 0

for a given f (x) . 16.7

We need to solve the two-dimensional Laplace equation subject to the boundary condition on the circle. Using the method of separation of variables, we showed in Section 16.3.1 that the general solution in plane polar co-ordinates is given by (16.42), ∞

u(r, θ) = C ln r + D + ∑ (C mr m + Dmr −m )(Am cosmθ + Bm sin mθ), m=1

where C , D , Am , Bm ,C m ,Dm are all constants. However, since u is finite at the origin, it follows that C = 0 and Dm = 0 , so that ∞

u(r, θ) = D + ∑ (C mr m )(Am cosmθ + Bm sin mθ). m=1

The boundary condition gives ∞

u(R, θ) = f (θ) = D + ∑ Rm (Am′ cosmθ + Bm′ sin mθ), m=1

where we have defined Am′ ≡ C mAm and Bm′ ≡ C mBm . To find Am′ , multiply by cos(nθ) divided by pi and integrate over θ from −π to π using the orthogonality conditions (13.6). This gives π ⎧⎪ 2D n = 0 1 an ≡ ∫ f (θ)cos(nθ)dθ = ⎪⎨ n ⎪⎪ R An′ n ≠ 0 π −π ⎪⎩

Likewise, multiplying by sin(nθ) π and integrating gives π

1 bn ≡ ∫ f (θ)sin(nθ)dθ = RnBn′ . π −π Then substituting, gives ∞

u(r, θ) = D + ∑ r n (An′ cosnθ + Bn′ sin nθ) n=1

n

⎛r ⎞ = + ∑ ⎜⎜⎜ ⎟⎟⎟ (an cosnθ +bn sin nθ). 2 n=1 ⎜⎝ R ⎟⎠ a0

∞

S16.8

16.8

Since u is a spherically symmetric potential, it only depends on the radial co-ordinate r, and so we only need the term in ∇2 that acts on r. Hence the Laplace equation becomes (cf 16.51)

∇2u(r) =

d ⎛⎜ 2 du ⎞⎟ ⎟⎟ = 0 . ⎜r dr ⎜⎜⎝ dr ⎟⎠

Integrating twice gives

u(r) =

a +b , r

where a and b are constants. But u(r) → 0 as r → ∞ and so b = 0 and u(r) = a r . 16.9

The problem involves a sphere and has cylindrical symmetry about the z-axis. We therefore use spherical polar co-ordinates but assume the solution is independent of the azimuthal angle φ , so that u = u(r, cos θ) . The general solution of Laplace’s equation (16.65) then reduces to ∞

∞

Bl

l=0

l=0

r l+1

u(r, θ) = ∑ Alr l Pl (cos θ) + ∑ where the An

and Bl

Pl (cos θ).

(1)

are undetermined constants and Pl (cos θ) are Legendre

polynomials. We next fix the constants by using appropriate boundary conditions. Far from the sphere, the electric field will be unaffected by its presence, so that E → −E 0ˆz as r → ∞ , and hence, using E = −∇u ,

u(r, θ) → −E 0z +c = −E 0rP1(cos θ) +c , where c is a constant to be determined and P1(cos θ) = cos θ . On comparing to (1), this gives

A0 = c , A1 = −E 0 , and Al = 0 for l > 1 so that the solution reduces to ∞

Bl

l=0

r l+1

u(r, θ) = c − E 0rP1(cos θ) + ∑

Pl (cos θ).

Setting u(a, θ) = 0 on the surface of the sphere then gives

B0 = −ac , B1 = E 0a 3 and Bl = 0 for l > 1 , so that (2) becomes

S16.9

(2)

⎛ a3 ⎞ ⎛ a⎞ u(r, θ) = c ⎜⎜⎜1 − ⎟⎟⎟ − E 0 ⎜⎜⎜1 − 3 ⎟⎟⎟r cos θ . ⎜⎝ ⎜⎝ r ⎟⎠ r ⎟⎠

(3)

Finally, at large distances, the first term in (3) generates a Coulomb field corresponding to a net charge (4πε0 )cq on the sphere. For an uncharged sphere, we must set c = 0 , to give finally

⎛ a3 ⎞ u(r, θ) = −E 0 ⎜⎜⎜1 − 3 ⎟⎟⎟r cos θ ⎜⎝ r ⎟⎠ as the desired potential. 16.10

Using the form of ∇2 in spherical polar co-ordinates as given in Table 12.1, then substituting u = R(r)Θ(θ)Φ(φ) and dividing throughout by u, one obtains

1 d ⎛⎜ 2 dR ⎞⎟ 1 d ⎛⎜ dΘ ⎞⎟ ⎟⎟ + 2 ⎟⎟ ⎜⎜r ⎜⎜sin θ dθ ⎟⎠ r R dr ⎜⎝ dr ⎟⎠ r Θsin θ dθ ⎜⎝ 2

+

1 d 2Φ g(θ) h(φ) + f (r) + 2 + 2 2 = 0. 2 2 2 r Φ sin θ dφ r r sin θ

Then, multiplying by r 2 and separating out the terms in r, one obtains

d ⎛⎜ 2 dR ⎞⎟ ⎟⎟ + r 2 f (r)R = kR ⎜⎜r ⎜ dr ⎝ dr ⎟⎠

(1)

and

1 d ⎛⎜ dΘ ⎞⎟ 1 d 2Φ h(φ) ⎟⎟ + + g(θ) + +k = 0 , ⎜⎜sin θ 2 2 Θsin θ dθ ⎜⎝ dθ ⎟⎠ Φ sin θ dφ sin 2 θ where k is an arbitrary constant. Multiplying by sin 2 θ then gives

sin θ d ⎛⎜ dΘ ⎞⎟ 1 d2Φ ⎟⎟ + + g(θ)sin 2 θ + h(φ) + k sin 2 θ = 0 ⎜⎜sin θ Θ dθ ⎜⎝ dθ ⎟⎠ Φ dφ 2 and separating variables again one obtains

1 d 2Φ + h(φ)Φ = LΦ Φ dφ 2

(2)

and

sin θ

d ⎛⎜ dΘ ⎞⎟ ⎟⎟ + g(θ)Θsin 2 θ + kΘsin 2 θ + LΘ = 0 , ⎜⎜sin θ ⎜ dθ ⎝ dθ ⎟⎠

(3)

where L is again an arbitrary separation constant. Hence the equation is completely separable, with R(r), Θ(θ) and Φ(φ) satisfying the ODEs (1), (2) and (3), respectively.

S16.10

16.11

On substituting (2) into (1) and using (16.51), one obtains, in spherical polar coordinates

−

 2 ⎧⎪⎪ 1 d ⎡⎢ 2 d ⎛⎜ R ⎞⎟⎤⎥ ⎫⎪⎪ V(r) r RYlm ⎜⎜ ⎟⎟⎥ ⎬Ylm + ⎨ 2 ⎢ 2m ⎪⎪r dr ⎢⎣ dr ⎜⎝ r ⎟⎠⎥⎦ ⎪⎪ r ⎩ ⎭ 2 ⎪ ⎧ ⎡ ∂Ylm ⎤⎥ ⎪⎫⎪ R  2 ⎡⎢ 1 ∂ 2Ylm ⎤⎥ R  ⎪ 1 ∂ ⎢ R − sin θ = E Ylm . ⎨ 2 ⎬ − ⎢ 2 2 2 ⎥ ⎢ ⎥ ⎪ ⎪ 2m ⎪r sin θ ∂θ ⎢⎣ ∂θ ⎥⎦ ⎪ r 2m ⎢⎣ r sin θ ∂φ ⎥⎦ r r ⎩ ⎭

On substituting [cf. (16.67a)]

Ylm (θ,φ) = clmPl m (cos θ)e imφ , where Pl m (cos θ) is an associated Legendre polynomial and clm is a constant, and then evaluating the derivatives with respect to r and φ , we obtain

−

 2 ⎡⎢ 1 d2R ⎤⎥ m R P + V(r)Pl m 2 ⎥ l ⎢ 2m ⎢⎣ r dr ⎥⎦ r dPl m ⎤⎥ ⎪⎫⎪ R  2 ⎡⎢ m 2 ⎤⎥ m R  2 ⎪⎧⎪ 1 d ⎡⎢ R − sin θ Pl = E Pl m , ⎨ 2 ⎬ + ⎢ ⎥ 2 2 ⎢ ⎥ 2m ⎪⎪r sin θ dθ ⎢⎣ dθ ⎥⎦ ⎪⎪ r 2m ⎢⎣ r sin θ ⎥⎦ r r ⎪⎩ ⎪⎭

where we have cancelled out a factor of clme imφ . Finally, the Pl m are solutions of the associated Legendre equation, which can be written [cf. (16.56), (16.60)]

1 d ⎛⎜ dΘ ⎞⎟ m2 ⎟⎟ − 2 Θ = −l(l + 1)Θ . ⎜⎜sin θ sin θ dθ ⎜⎝ dθ ⎟⎠ sin θ Using this, the previous equation becomes

−

 2 ⎛⎜ 1 d2R ⎞⎟⎟ m VR m  2 l(l + 1) R m R Pl + Pl + Pl = E Pl m , ⎜⎜ 2 ⎟ 2 ⎟ 2m ⎜⎝ r dr ⎠ r 2m r r r

and hence

−

 2 d2R ⎡⎢ l(l + 1) 2 ⎤⎥ + V(r) + R = ER, 2m dr 2 ⎢⎢⎣ 2mr 2 ⎥⎥⎦

where the second term inside the brackets is called the centrifugal term. 16.12

The diffusion equation is

∇2u(r,t) =

1 ∂u(r,t) κ ∂t

S16.11

(1)

where ∇2 in spherical polar co-ordinates is given in Table 12.1. Since u does not depend on angular variables, ∇2 reduces to

∇2 =

1 ∂ ⎛⎜ 2 ∂ ⎞⎟ ⎟⎟ ⎜r r 2 ∂r ⎜⎜⎝ ∂r ⎟⎠ (1)

and

⎛ r 2 ⎞⎟⎤ ⎛ r 2 ⎞⎟⎤ 1 ∂ ⎡⎢ 3 1 ∂ ⎡⎢ 3 ⎜⎜− ⎜⎜− ⎟⎟⎥ = − 1 ⎟⎥ −r exp r exp ⎜⎜ 4κt ⎟⎥ ⎜⎜ 4κt ⎟⎟⎥ r 2 ∂r ⎢⎢⎣ 2κt 5/2 2κr 2t 5/2 ∂r ⎢⎢⎣ ⎝ ⎠⎥⎦ ⎝ ⎠⎥⎦ . ⎞⎟ ⎛ r 2 ⎞⎟ 1 ⎛⎜ r 2 ⎟⎟. = − 3⎟⎟ exp ⎜⎜⎜− ⎜ ⎟⎠ ⎜⎝ 4κt ⎟⎠ 2κt 5/2 ⎜⎝⎜ 2κt

∇2u =

The right-hand side of (1) is

⎛ r 2 ⎞⎟⎤ 1 ⎡ 1 r 2 ⎛ r 2 ⎞⎟ 1 ∂ ⎡⎢ 1 3 ⎤⎥ ⎜⎜− ⎜⎜− ⎟⎟⎥ = ⎢ ⎟ exp − exp ⎜⎜ 4κt ⎟⎥ κ ⎢ t 3/2 4κt 2 2t 5/2 ⎥ ⎜⎜ 4κt ⎟⎟ κ ∂t ⎢⎢⎣ t 3/2 ⎝ ⎠⎥⎦ ⎝ ⎠ ⎢⎣ ⎥⎦ 2 ⎞ ⎞⎟ ⎛ 1 ⎛⎜ r 2 ⎟⎟ exp ⎜⎜− r ⎟⎟⎟. = − 3 ⎜ ⎜⎜ 4κt ⎟ ⎟⎠ 2κt 5/2 ⎜⎝⎜ 2κt ⎝ ⎠ Hence u is a solution of the diffusion equation. 16.13

Since by symmetry there is no dependence on the azimuthal angle φ , we seek solutions of the Laplace equation ∇2u(r, θ) = 0 , using spherical polar co-ordinates. If u is finite at r = 0 , this leads to a general solution [cf. (16.65)] ∞

u(r, θ) = ∑ clr l Pl (µ) ,

(1)

l=0

where Pl are Legendre polynomials and µ ≡ cos θ . To find the coefficients cl , multiply by Pm (µ) and integrate over µ from –1 to +1. Using the orthogonality properties of the Legendre polynomials (15.30), we have 1

∫

−1

1

∞

u(r, θ)Pm (µ)dµ = ∑ clr l ∫ Pl (µ)Pm (µ)dµ l=0

−1

∞

= ∑ clr l l=0

2δlm 2m + 1

=

2cmr m 2m + 1

.

Thus, 1

cm =

2m + 1 ∫ u(r, θ)Pm (µ)dµ . 2r m −1

Setting r = R , and using the boundary conditions, this may be written

S16.12

cm =

0 1 ⎤ 2m + 1 ⎡⎢ ⎥. T P (µ)dµ +T P (µ)dµ ⎢ L∫ m ⎥ U∫ m m 2R ⎢ −1 ⎥⎦ 0 ⎣

The integrals may be evaluated using the explicit forms for the Legendre polynomials given in (15.28). Ignoring the integration constants, we find,

∫ P (µ)dµ = µ, ∫ P (µ)dµ = (µ 0 2

1 2

∫ P (µ)dµ = µ , ∫ P (µ)dµ = (5µ − 6µ ), 1 2

1

3

− µ),

3

2

4

1 8

2

and so

1 3 c0 = (TU +TL ), c1 = (T −TL ), 2 4R U 7 c2 = 0, c3 = − (T −TL ). 16R 3 U Finally, substituting into (1) gives the required result 3

1 3⎛r ⎞ 7 ⎛r ⎞ u(r, θ) = (TU +TL ) + ⎜⎜⎜ ⎟⎟⎟(TU −TL )P1(cos θ)− ⎜⎜⎜ ⎟⎟⎟ (TU −TL )P3(cos) +  2 4 ⎜⎝ R ⎟⎠ 16 ⎜⎝ R ⎟⎠ 16.14

In cylindrical polar co-ordinates, the Helmholtz equation is

1 ∂ ⎛⎜ ∂u ⎞⎟ 1 ∂ 2 u ∂ 2 u ⎟⎟ + + + k 2u = 0, ⎜ρ ρ ∂ρ ⎜⎜⎝ ∂ρ ⎟⎠ ρ 2 ∂φ 2 ∂z 2 which, on substituting u = Ρ(ρ)Φ(φ)Z(z) and dividing by u, becomes

1 d ⎛⎜ dΡ ⎞⎟ 1 1 d2Φ 1 d2Z ⎟⎟ + + + k 2 = 0. ⎜⎜ρ 2 2 2 ⎟ ⎜ Ρρ dρ ⎝ dρ ⎠ Φ ρ dφ Z dz Separating off the z-dependent term gives

1 d2Z = a2 Z dz 2

(1)

1 d ⎛⎜ dΡ ⎞⎟ 1 1 d2Φ ⎟⎟ + + (k 2 + a 2 ) = 0, ⎜ρ Ρρ dρ ⎜⎜⎝ dρ ⎟⎠ Φ ρ 2 dφ 2

(2)

and

where we have called the separation constant −a 2 . The solution of (1) is

Z = Ae− az + Beaz , where C and D are constants. The condition that Z → 0 as z → ∞ then requires a to be real and B to be zero, so that

S16.13

Z = Ce −az

(3)

if we choose a > 0 . Multiplying (2) by r 2 gives

r d ⎛⎜ dR ⎞⎟ 1 d2Θ ⎟⎟ + + p 2r 2 = 0, ⎜r R dr ⎜⎜⎝ dr ⎟⎠ Θ dθ 2 where p 2 = k 2 + a 2 . Separating variables again gives

d 2Φ = −m 2Φ 2 dφ

(4)

and

ρ2

d 2Ρ dΡ +ρ + (p 2ρ 2 −m 2 )Ρ = 0 . 2 dρ dρ

(5)

The general solution of (4) is

Φ = E cos(mφ) + F sin(mφ) ,

(6)

where E and F are constants, and m = 0,1, 2,… to ensure that Φ is single-valued, i.e. for Φ(φ) = Φ(φ + 2π) . On substituting x = pr , (5) becomes

x2

d2R dR +x + (x 2 − m 2 )R = 0 . 2 dx dx

This is Bessel’s equation (15.53a), and the only solutions that are finite at x = pr = 0 are

R(ρ) = J m (pρ) ,

(7)

Where J m are Bessel functions of the first kind. Combining (3), (6) and (7), we find that the separable solutions are

u(ρ,φ,z) = J m (pρ)e −az (A cosmφ + B sin mφ) , where a > 0, p = (k 2 + a 2 )1/2 , m is an integer, and the constants A = CE and B = CF . 16.15

The general solution of the Laplace equation in cylindrical co-ordinates is given by a linear combination of the separable solutions (16.77a,b). In our case, u must be finite as z → ∞ , which implies that the coefficients Ak = 0 ; and the first boundary condition implies that only terms proportional to sin θ are non-zero. Thus setting m = 1 in (16.77b), we have solutions of the form

S16.14

u(ρ,φ,z) = ckJ 1(kρ)e −kz sin φ,

(1)

for any k where ck ≡ Bk D1 . In addition, the second boundary condition requires that

J 1(k) = 0 , and so the general solution is ∞

u(ρ,φ,z) = ∑ cmJ 1(km ρ)e

−kmz

sin φ,

m=1

where km are the zeroes of J 1 . To find the coefficients, we set z = 0 in (1), multiply by

ρJ 1(kn ρ) and integrate over r. Using the boundary condition (1), this gives 1

∫ 0

∞

1

m=1

0

ρ 2J 1(kn ρ)dρ = ∑ cm ∫ ρJ 1(km ρ)J 1(kn ρ)dρ.

The right-hand side may be found using the orthogonality condition (15.74) and gives 2 2 cn ⎡ c J 1′(kn )⎤⎥ = n ⎡⎢J 2(kn )⎤⎥ , ⎢ ⎣ ⎦ ⎣ ⎦ 2 2

where we have used (15.72b) and the fact that J 1(kn ) = 0 . Thus,

cn =

1

2

∫ r J (k r)dr 2

2

⎡J (k )⎤ ⎢⎣ 2 n ⎥⎦

1

n

.

0

The integral may be evaluated using (15.71a). Thus,

1 d ⎡ 2 ρ J (k ρ)⎤ = ρ 2J 1(kn ρ) kn dρ ⎣⎢ 2 n ⎦⎥ and integrating gives 1

1 J 2(kn ) 1 ⎡ 2 ⎤ ⎢⎣ρ J 2(kn ρ)⎥⎦ 0 = k .

∫ ρ J (k ρ)dρ = k 2

1

n

n

0

n

So,

cn =

2 , knJ 2(kn )

and hence the final result is

2 −k z J 1(kn ρ)e n sin φ, n=1 knJ 2 (kn ) ∞

u(ρ,φ,z) = ∑ as required.

S16.15

16.16

At z = 1 and r = 1 2 , and using u 0 = 50 , we have

un =

100 −k J 0 (kn 2)e n knJ 1(kn )

(1)

and kn are the zeros of J 0 . Using values of the zeros from Table 15.1, we may construct the following table (using, for example an Excel spreadsheet).

n

kn

J 1(kn )

J 0 (kn 2)

exp(−kn )

un

1

2.4048

0.51915

0.66994

0.09028

4.84471

2

5.5201 −0.34026 −0.16841

0.00401

0.03591

3

8.6537

0.27145

−0.35628

0.00017

−0.00265

4 11.7915 −0.23246

0.12078

0.00001

−0.00003

So finally, u = 4.878 to three decimal places. *16.17 In both cases, we can use d’Alembert’s solution (16.85), where α and β are given by (16.83). In this way, we obtain: (a)

u(x,t) = 12 exp[−(x − υt)2 ]+ 12 exp[−(x + υt)2 ] ,

i.e. two waves peaked at x = ±υt travelling in opposite directions. x+υt

(b)

1 u(x,t) = q exp(−q 2 )dq . ∫ 2υ x−υt

On changing variables to z = q 2 , one easily shows that

∫ q exp(−q )dq = − 2

1 2

exp(−q 2 ) +c ,

so that

u(x,t) = −

2 1 ⎡ −(x+υt )2 −e −(x−υt ) ⎤⎥ . ⎢⎣e ⎦ 4υ

Again we have two waves centred at x = ±υt travelling in opposite directions, but with opposite magnitudes. *16.18 Using the notation of (16.86) we have: (a) A = 2, B = 52 , C = 2 , so that AC ≠ B 2 and we have to solve (16.90), i.e

2 + 5λi + 2λi2 = 0 , which has solutions λi = − 12 or 2. So by (16.88)

S16.16

u = f (x − 12 y) + g(x + 2y), where f, g are arbitrary functions. (b) A = 9, B = −3, C = 1 , so AC = B 2 and by (16.93)

u = f (x + 3y) + xg(x + 3y) . (c) A = 1, B = −2, C = 5 , so that AC ≠ B 2 and we must solve (16.90), i.e.

1 − 4λi + 5λi2 = 0 , which has solutions λi = (4 ± 3i) 5 . So by (16.88),

u = f ⎡⎢x + (4 + 3i)y 5⎤⎥ + g ⎡⎢x + (4 − 3i)y 5⎤⎥ , ⎣ ⎦ ⎣ ⎦ where f, g are arbitrary functions. (d) A = 1, B = 1, C = 0 , so that AC ≠ B 2 and (16.90) becomes

1 + 2λi = 0 , with a single solution λi = − 12 . Hence f (x − 12 y) is a solution. Also, since both terms in the equation contain a partial derivative with respect to x, g(y) is a solution for any function g. The general solution is therefore

u = f (x −y 2) + g(y) . *16.19 In the notation of (16.86), A = B = C = 1 , so that AC = B 2 and the solution is given by (16.93), i.e.,

u(x,y) = f (x −y) + xg(x −y) , where f and g are arbitrary functions. Applying the boundary conditions, we have u(0,y) = f (−y) = y 2 = (−y)2 , so that f (x −y) = (x −y)2 and

u(x,y) = (x −y)2 + xg(x −y). Similarly, at y = 0 ,

u(x, 0) = x 2 + xg(x) = sin x , so that

S16.17

g(x) =

sin x −x x

and hence

⎡ sin(x −y) ⎤ u(x,y) = (x −y)2 + x ⎢ −(x −y)⎥ ⎢ x −y ⎥ ⎣ ⎦ x sin(x −y) = y(y − x) + . x −y *16.20 The temperature distribution is given by the one-dimensional diffusion equation

∂ 2 u(x,t) 1 ∂u(x,t) = , κ ∂t ∂x 2 where x ≥ 0, t > 0 and κ is a constant. The boundary conditions in this case are

u(x, 0) = 0, x > 0

and

u(0,t) = u 0 , t > 0 ,

and in addition we have to impose the physically reasonable condition that u(x,t) → 0 as x → ∞ for all t. Taking the Laplace transform of the diffusion equation with respect to time gives

⎡ ∂ 2 u ⎤ 1 ⎡ ∂u ⎤ L ⎢⎢ 2 ⎥⎥ = L ⎢ ⎥ , ⎢ ⎥ ⎢⎣ ∂x ⎥⎦ κ ⎣ ∂t ⎦ which, using (14.45a) may be written

∂2 1 1 F(x, p) = ⎡⎢−u(x, 0) + pF(x, p)⎤⎥ = ⎡⎢ pF(x, p)⎤⎥ 2 ⎣ ⎦ ⎣ ⎦ κ κ ∂x where F(x, p) ≡ L[u(x,t)] . Using the boundary condition at t = 0 , leads to the solution

F(x, p) = c1(p)e x

p κ

+c2(p)e −x

p κ

,

where c1,2 are functions of p. The condition that u(x,t) → 0 as x → ∞ for all t implies that F(x, p) → 0 also and hence c1(p) = 0 . Imposing the boundary condition at x = 0 gives c2(p) = F(0, p) , where

F(0, p) = L ⎡⎢u(0, p)⎤⎥ = u 0 p , ⎣ ⎦ using Table 14.1. Combining these results gives

F(x, p) =

⎛ p ⎞⎟⎟ ⎜ exp ⎜⎜−x ⎟ , ⎜⎜⎝ p κ ⎟⎟⎠

u0

S16.18

and hence

⎡ exp(−x p κ) ⎤ ⎥. u(x,t) = u 0L−1 ⎢⎢ ⎥ p ⎢⎣ ⎥⎦ 16.21

Taking the Laplace transform of the PDE with respect to t using (14.45a) gives

∂ 2 F(x, p) = −u(x, 0) + pF(x, p) = pF(x, p) , ∂x 2 where the last step follows from the boundary condition (1). This has the solution

F(x, p) = A cosh(xp1/2 ) + B sinh(xp1/2 ) , where A and B are constants. Next we find F(0, p) from the definition of the Laplace transform, which gives ∞

F(0, p) = ∫ u(0,t)e −pt dt = 0

u0 p

,

and likewise ∞

∂F(x, p) = ∂x x=a ∫0

⎛ ∂u ⎞⎟ ⎜⎜ ⎟ ⎜⎜⎝ ∂x ⎟⎟⎠

e −pt dt = 0 , x=a

where we have used the boundary conditions (2) and (3). These two relations may be used to find the constants A and B from the general solution. Thus, A = u 0 p and

B =−

u 0 sinh(ap1/2 ) . p cosh(ap1/2 )

Substituting these into the general solution gives

F(x, p) =

⎤ u 0 ⎡⎢ sinh(ap1/2 ) 1/2 1/2 ⎥ cosh(xp )− sinh(xp ) ⎥ , p ⎢⎢⎣ cosh(ap1/2 ) ⎥⎦

which using (2.65b) may be written

F(x, p) =

u 0 cosh[(x −a)p1/2 ] . p cosh(ap1/2 )

So, finally

⎡ cosh{(x −a)p1/2 } ⎤ ⎥ , u(x,t) = u 0 L−1 ⎢⎢ 1/2 ⎥ p cosh(ap ) ⎢⎣ ⎥⎦ as required. *16.22 Taking the Laplace transform of the PDE

S16.19

∂2u ∂2u = 2 ∂x 2 ∂t with respect to t using (14.45b) gives

∂ 2 F(x, p) ∂u , = p 2F(x, p)− pu(x, 0)− 2 ∂t t=0 ∂x which using (1) becomes

∂ 2 F(x, p) = p 2F(x, p) , 2 ∂x with solution

F(x, p) = Ae px + Be −px , where A and B are constants. Next we find F(0, p) from the definition of the Laplace transform, which gives ∞

F(0, p) = ∫ u(0,t)e −pt dt = 0 , 0

where we have used (2), and likewise ∞

∂F(x, p) = ∂x x=a ∫0

⎛ ∂u ⎞⎟ ⎜⎜ ⎟ ⎜⎜⎝ ∂x ⎟⎟⎠

e −pt dt = x=a

f , p

where again we have used the boundary conditions (2). These two relations may be used to find the constants A and B from the general solution. Thus,

A+B = 0

Ape pa − Bpe −pa = f p

and

and solving gives

A = −B =

f . 2p cosh(pa) 2

Substituting these into the general solution gives

⎡ sinh(px) ⎤ ⎥ , F(x, p) = f ⎢⎢ 2 ⎥ p cosh(pa) ⎣ ⎦ and finally,

⎡ sinh(px) ⎤ ⎥ , u(x,t) = f L−1 ⎢⎢ 2 ⎥ p cosh(pa) ⎣ ⎦ as required.

S16.20

S16.21