1
Solutions for Chapter 9
The Chemistry Maths Book
Erich Steiner
University of Exeter
Second Edition 2008
Solutions
Chapter 9. Functions of several variables
9.1 Concepts
9.2 Graphical representation
9.3 Partial differentiation
9.4 Stationary points
9.5 The total differential
9.6 Some differential properties
9.7 Exact differentials
9.8 Line integrals
9.9 Multiple integrals
9.10 The double integral
9.11 Change of variables
© E Steiner 2008
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Solutions for Chapter 9
Section 9.1
1.
Find the value of the function f ( x, y ) = 2 x 2 + 3 xy − y 2 + 2 x − 3 y + 4 for
(i)
( x, y ) = (0, 1)
(ii) ( x, y ) = (2, 0)
(i)
f (0,1) = 0 + 0 − 1 + 0 − 3 + 4 = 0
(iii) ( x, y ) = (3, 2)
(ii) f (2, 0) = 8 + 0 − 0 + 4 − 0 + 4 = 16
(iii) f (3, 2) = 18 + 18 − 4 + 6 − 6 + 4 = 36
2.
Find the value of f (r , θ , φ ) = r 2 sin 2 θ cos 2 φ + 2 cos 2 θ − r 3 sin 2θ sin φ for
(i)
(r , θ , φ ) = (1, π 2, 0) (ii) (r , θ , φ ) = (2, π 4, π 6)
(iii) (r , θ , φ ) = (0, π, π 3) .
(i)
f (1, π 2, 0) = 1× sin 2 π 2 × cos 2 0 + 2 cos 2 π 2 − 1× sin π× sin 0
= 1+ 0 − 0 = 1
(ii) f (2, π 4, π 6) = 22 × sin 2 π 4 × cos 2 π 6 + 2 cos 2 π 4 − 23 × sin π 2× sin π 6
1 3
1
1
3
= 4 × × + 2 × − 8 ×1× = −
2 4
2
2
2
(iii) f (0, π, π 3) = 0 + 2 cos 2 π − 0 = 2
Section 9.3
Find
∂z
∂z
and
for
∂x
∂y
∂z
= 4x ,
∂x
3.
z = 2 x2 − y 2 :
4.
z = x 2 + 2 y 2 − 3x + 2 y + 3 :
5.
z = e2 x +3 y :
6.
z = sin( x 2 − y 2 ) :
© E Steiner 2008
∂z
= −2 y
∂y
∂z
= 2x − 3 ,
∂x
∂z
= 2e 2 x + 3 y = 2 z ,
∂x
∂z
= 4y + 2
∂y
∂z
= 3e 2 x +3 y = 3z
∂y
∂z
= 2 x cos( x 2 − y 2 ) ,
∂x
∂z
= −2 y cos( x 2 − y 2 )
∂y
3
Solutions for Chapter 9
7.
2
z = e x cos xy : By the product rule,
2
2
∂z
∂
∂
= e x × cos xy + (e x ) × cos xy
∂x
∂x
∂x
2
2
= e x × (− y sin xy ) + (2 xe x ) × cos xy
= ⎡⎣ 2 x cos xy − y sin xy ⎤⎦ e x
2
2
2
∂z
∂
= e x × cos xy = − xe x sin xy
∂y
∂y
Find all the nonzero partial derivatives of
8.
z = x 2 − 3x 2 y + 4 xy 2 :
First derivatives:
Second :
Third:
∂z
∂z
= 2 x − 6 xy + 4 y 2 ,
= −3 x 2 + 8 xy
∂x
∂y
∂2 z
∂x 2
= 2 − 6 y,
∂3 z
∂y∂x 2
∂3 z
∂y 2 ∂x
9.
= −6 =
=8=
∂2 z
∂2 z
∂2 z
= −6 x + 8 y =
,
= 8x
∂y∂x
∂x∂y ∂y 2
∂3 z
∂3 z
= 2
∂x∂y∂x ∂x ∂y
∂3 z
∂3 z
=
∂y∂x∂y ∂x∂y 2
u = 3 x 2 + y 2 + 2 xy 3 :
u x = 6 x + 2 y 3 , u y = 2 y + 6 xy 2
u xx = 6, u yx = 6 y 2 = u xy , u yy = 2 + 12 xy
u yyx = 12 y = u yxy = u xyy
u yyy = 12 x
u xyyy = 12 = u yxyy = u yyxy = u yyyx
Find all the first and second partial derivatives of
10. z = 2 x 2 y + cos( x + y ) :
∂z
∂z
= 4 xy − sin( x + y ),
= 2 x 2 − sin ( x + y ),
∂x
∂y
∂2 z
∂x 2
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= 4 y − cos ( x + y ),
∂2 z
∂2 z
∂2 z
,
= 4 x − cos ( x + y ) =
= − cos ( x + y )
∂x∂y
∂y∂x ∂y 2
4
Solutions for Chapter 9
11. z = sin( x + y ) e x − y : By the product rule,
∂z
= ⎡sin( x + y ) + cos( x + y ) ⎤⎦ e x − y
∂x ⎣
∂z
= ⎡⎣ cos( x + y ) − sin( x + y ) ⎤⎦ e x − y
∂y
∂2 z
∂x
2
= 2 cos( x + y )e x − y ,
∂2 z
∂y
2
= −2 cos( x + y )e x − y
2
∂ z
∂2 z
= −2sin( x + y )e x − y =
∂x∂y
∂y∂x
Show that f xy = f yx for
12. f = x3 − 3 x 2 y + y 3 :
f yx = −6 x ⎫⎪
⎬ → f yx = f xy
f y = −3x 2 + 3 y 2 , f xy = −6 x ⎪⎭
f x = 3x 2 − 6 xy,
13. f = x 2 cos( y − x) : By the product rule,
f x = x 2 sin( y − x) + 2 x cos( y − x),
f y = − x 2 sin( y − x),
14. f =
xy
x2 + y 2
: By the quotient rule,
fx =
( x 2 + y 2 ) ( y ) − ( xy )(2 x)
fy =
f yx =
f xy =
Therefore
© E Steiner 2008
f yx = x 2 cos( y − x) − 2 x sin( y − x) ⎫⎪
⎬ → f yx = f xy
f xy = x 2 cos( y − x) − 2 x sin( y − x) ⎪⎭
( x 2 + y 2 )2
( x 2 + y 2 ) ( x ) − ( xy )(2 y )
( x 2 + y 2 )2
=
=
y3 − x2 y
( x 2 + y 2 )2
x3 − xy 2
( x 2 + y 2 )2
( x 2 + y 2 ) 2 × (3 y 2 − x 2 ) − ( y 3 − x 2 y ) × (4 y )( x 2 + y 2 )
( x 2 + y 2 )4
( x 2 + y 2 ) 2 × (3 x 2 − y 2 ) − ( x3 − xy 2 ) × (4 x)( x 2 + y 2 )
f yx = f xy
( x 2 + y 2 )4
=−
=−
( x4 + y 4 − 6 x2 y 2 )
( x 2 + y 2 )3
( x4 + y 4 − 6 x2 y 2 )
( x 2 + y 2 )3
5
Solutions for Chapter 9
Show that f xyz = f yzx = f zxy for
15. f = cos( x + 2 y + 3z ) :
Therefore
f x = − sin( x + 2 y + 3 z ),
f zx = −3cos( x + 2 y + 3 z ),
f yzx = 6sin( x + 2 y + 3 z )
f y = −2sin( x + 2 y + 3 z ),
f xy = −2 cos( x + 2 y + 3 z ),
f zxy = 6sin( x + 2 y + 3 z )
f z = −3sin( x + 2 y + 3 z ),
f yz = −6 cos( x + 2 y + 3 z ),
f xyz = 6sin( x + 2 y + 3 z )
f xyz = f yzx = f zxy
16. f = xy e yz
Therefore
f x = ye yz ,
f zx = y 2 e yz ,
f yzx = (2 y + y 2 z )e yz
f y = ( x + xyz )e yz ,
f xy = (1 + yz )e yz ,
f zxy = (2 y + y 2 z )e yz
f z = xy 2 e yz ,
f yz = (2 xy + xy 2 z )e yz ,
f xyz = (2 y + y 2 z )e yz
f xyz = f yzx = f zxy
17. If r = ( x 2 + y 2 + z 2 )1 2 find
∂r ∂r ∂r
, , .
∂x ∂y ∂z
Let
u = x 2 + y 2 + z 2 , r = u1 2 .
Then
∂r ∂r ∂u 1 −1 2
x
=
×
= u
× 2x =
r
∂x ∂u ∂x 2
Simularly,
∂r y
= ,
∂y r
∂r z
=
∂z r
18. If φ = f ( x − ct ) + g ( x + ct ) , where c is a constant, show that
Let
u = x + ct , v = x − ct
Then
φ = f (u ) + g (v ),
Therefore
© E Steiner 2008
∂ 2φ
∂x 2
=
1 ∂ 2φ
c 2 ∂t 2
∂ 2φ
∂x 2
=
1 ∂ 2φ
c 2 ∂t 2
∂φ ∂f ∂u ∂g ∂v ∂f ∂g
=
+
=
+ ,
∂x ∂u ∂x ∂v ∂x ∂u ∂v
∂ 2φ
∂φ ∂f ∂u ∂g ∂v
∂f
∂g
=
+
= c −c
∂t ∂u ∂t ∂v ∂t
∂u
∂v
∂ 2φ
∂x 2
∂t
2
=
.
∂2 f
∂u 2
= c2
+
∂2 f
∂u
2
∂2 g
∂v 2
+ c2
2
∂g
∂v 2
6
Solutions for Chapter 9
⎛ ∂y ⎞
19. If xyz + x 2 + y 2 + z 2 = 0 , find ⎜ ⎟ .
⎝ ∂x ⎠ z
We have
⎡
⎡ ⎛ ∂y ⎞ ⎤
⎛ ∂ ⎞ ⎡
⎛ ∂y ⎞ ⎤
2
2
2⎤
⎜ ⎟ ⎣ xyz + x + y + z ⎦ = ⎢ yz + xz⎜ ⎟ ⎥ + ⎣⎡ 2 x ⎦⎤ + ⎢ 2 y⎜ ⎟ ⎥ = 0
⎝ ∂x ⎠ z
⎝ ∂x ⎠ z ⎦⎥
⎣⎢
⎣⎢ ⎝ ∂x ⎠ z ⎥⎦
Therefore
2 x + yz
⎛ ∂y ⎞
⎛ ∂y ⎞
(2 y + xz ) ⎜ ⎟ + (2 x + yz ) = 0 → ⎜ ⎟ = −
∂
x
∂
x
y + xz
2
⎝ ⎠z
⎝ ⎠z
20. For the van der Waals equation
⎛
n2 a ⎞
⎜⎜ p + 2 ⎟⎟ (V − nb) − nRT = 0
V ⎠
⎝
⎛ ∂V ⎞
⎛ ∂V ⎞
⎛ ∂p ⎞
⎛ ∂p ⎞
Find (i) ⎜
⎟ , (iii) ⎜
⎟ , (ii) ⎜
⎟ , (iv) ⎜
⎟ .
⎝ ∂T ⎠ p, n
⎝ ∂T ⎠V , n
⎝ ∂V ⎠T , n
⎝ ∂p ⎠T , n
(i) We have
⎛∂
⎜
⎝ ∂T
⎧⎪⎛
⎫⎪
n2 a ⎞
⎞
+
(V − nb) − nRT ⎬
p
⎜
⎨
⎟
2 ⎟
⎜
⎟
V ⎠
⎠ p, n ⎪⎩⎝
⎪⎭
⎡⎛ 2n 2 a ⎞
⎛
n 2 a ⎞ ⎤ ⎛ ∂V ⎞
= ⎢⎜⎜ − 3 ⎟⎟ (V − nb) + ⎜⎜ p + 2 ⎟⎟ ⎥ ⎜
− nR
⎟
V ⎠ ⎥⎦ ⎝ ∂T ⎠ p, n
⎢⎣⎝ V ⎠
⎝
⎡
n 2 a 2n3 ab ⎤ ⎛ ∂V
= ⎢p− 2 +
⎥⎜
V
V 3 ⎥⎦ ⎝ ∂T
⎢⎣
Therefore
⎛ ∂V ⎞
= nR
⎜
⎟
⎝ ∂T ⎠ p, n
⎞
− nR = 0
⎟
⎠ p, n
⎛
n 2 a 2n3 ab ⎞
⎜⎜ p − 2 +
⎟
V
V 3 ⎟⎠
⎝
⎫⎪
⎛∂ ⎞
n2 a ⎞
⎪⎧⎛
⎨⎜⎜ p + 2 ⎟⎟ (V − nb) − nRT ⎬
⎜ ⎟
V ⎠
⎝ ∂p ⎠T , n ⎪⎩⎝
⎪⎭
(ii)
⎡
2n 2 a ⎛ ∂V
= ⎢1 − 3 ⎜
⎢
V ⎝ ∂p
⎣
⎛
⎞ ⎤
n 2 a ⎞ ⎛ ∂V
⎟ ⎥ (V − nb) + ⎜⎜ p + 2 ⎟⎟ ⎜
V ⎠ ⎝ ∂p
⎠T , n ⎥⎦
⎝
⎡
n 2 a 2n3 ab ⎤ ⎛ ∂V
= (V − nb) + ⎢ p − 2 +
⎥⎜
V
V 3 ⎥⎦ ⎝ ∂p
⎢⎣
Therefore
© E Steiner 2008
⎛ ∂V
⎜
⎝ ∂p
⎞
= − (V − nb)
⎟
⎠T , n
⎛
n 2 a 2n3 ab ⎞
⎜⎜ p − 2 +
⎟
V
V 3 ⎟⎠
⎝
⎞
=0
⎟
⎠T , n
⎞
⎟
⎠T , n
7
Solutions for Chapter 9
⎧⎪⎛
⎫⎪
n2 a ⎞
⎛∂ ⎞
⎨⎜⎜ p + 2 ⎟⎟ (V − nb) − nRT ⎬
⎜
⎟
V ⎠
⎝ ∂T ⎠V , n ⎩⎪⎝
⎭⎪
(iii)
⎛ ∂p ⎞
= (V − nb)⎜
− nR = 0
⎟
⎝ ∂T ⎠V , n
Therefore
nR
⎛ ∂p ⎞
=
⎜
⎟
⎝ ∂T ⎠V , n V − nb
⎛ ∂
⎜
⎝ ∂V
(iv)
⎫⎪
n2 a ⎞
⎪⎧⎛
⎞
+
p
(V − nb) − nRT ⎬
⎜
⎨
⎟
2 ⎟
⎜
⎟
V ⎠
⎠T , n ⎩⎪⎝
⎭⎪
⎛ ⎛ ∂p ⎞
⎛
n2 a ⎞
2n 2 a ⎞
= ⎜⎜
− 3 ⎟ (V − nb) + ⎜⎜ p + 2 ⎟⎟ = 0
⎟
⎜ ⎝ ∂V ⎠T , n V ⎟
V ⎠
⎝
⎝
⎠
Therefore
⎛ ∂p
⎜
⎝ ∂V
2n 2 a ( p + n 2 a V 2 )
⎞
=
−
⎟
V − nb
⎠T , n V 3
Section 9.4
Find the stationary points of the following functions:
21. f ( x, y ) = 3 − x 2 − xy − y 2 + 2 y
For a stationary value,
∂f
= −2 x − y = 0 and
∂x
Therefore
x = − 2 3, y = 4 3
∂f
= −x − 2 y + 2 = 0
∂y
and the single stationary point lies at
( x, y ) = (− 2 3, 4 3)
22. f ( x, y ) = x3 + y 2 − 3 x − 4 y + 2
For the stationary values,
∂f
= 3x 2 − 3 = 0 → x = ±1
∂x
∂f
= 2y − 4 = 0 → y = 2
∂y
There are two stationary points, at
( x, y ) = (1, 2) and (−1, 2)
© E Steiner 2008
8
Solutions for Chapter 9
23. 4 x 3 − 3 x 2 y + y 3 − 9 y
For the stationary values,
∂f
= 12 x 2 − 6 xy = 0 → 6 x(2 x − y ) = 0 → x = 0 or x = y 2
∂x
and
∂f
= −3x 2 + 3 y 2 − 9 = 0;
∂y
⎧x=0
→ 3y2 − 9 = 0 → y = ± 3
⎪
⎨
9 2
⎪ x = y 2 → y − 9 = 0 → y = ±2
⎩
4
There are four stationary points, at
( x, y ) = (0, 3), (0, − 3), (1, 2), and ( −1, −2)
Determine the nature of of the stationary points of the functions in Exercises 21 − 23:
24. 3 − x 2 − xy − y 2 + 2 y (Exercise 21)
We have
f x = −2 x − y, f y = − x − 2 y + 2
f xx = −2 < 0 ⎫
⎪
⎪
f yy = −2 < 0 ⎬ → f xx f yy − f xy 2 = 3 > 0
⎪
f xy = −1
⎪⎭
The stationary point is therefore a maximum.
25. x3 + y 2 − 3 x − 4 y + 2 (Exercise 22)
We have
f x = 3x 2 − 3, f y = 2 y − 4
f xx = 6 x, f yy = 2, f xy = 0
Therefore
( x, y ) = (1, 2) → f xx = 6 > 0, f yy = 2 > 0, f xx f yy − f xy 2 = 12 > 0
a minimum
( x, y ) = (−1, 2) → f xx = −6 < 0, f yy = 2 > 0, f xx f yy − f xy 2 = −12 < 0
a saddle point
© E Steiner 2008
9
Solutions for Chapter 9
26. 4 x 3 − 3x 2 y + y 3 − 9 y (Exercise 23)
f x = 12 x 2 − 6 xy, f y = −3 x 2 + 3 y 2 − 9
We have
f xx = 24 x − 6 y, f yy = 6 y, f xy = −6 x
Then
f xx f yy − f xy 2
type
0
−108 < 0
saddle point
−6 3 < 0
0
−108 < 0
saddle point
12 > 0
12 > 0
−6
108 > 0
minimum
−12 < 0
−12 < 0
6
108 > 0
maximum
x
y
f xx
f yy
0
3
−6 3 < 0
6 3<0
0
− 3
6 3>0
1
2
−1
−2
f xy
27. Find the stationary value of the function f = 2 x 2 + 3 y 2 + 6 z 2 subject to the constraint
x + y + z = 1,
(i) by using the constraint to eliminate z from the function, (ii) by the method of Lagrange
multipliers.
(i) We have z = 1 − x − y . Then
f ( x, y , z ) = 2 x 2 + 3 y 2 + 6 z 2
→ F ( x, y ) = 2 x 2 + 3 y 2 + 6(1 − x − y ) 2
= 8 x 2 + 9 y 2 − 12 x − 12 y + 12 xy + 6
Then , for a stationary value,
∂F
⎫
= 16 x − 12 + 12 y = 0 ⎪
1
1
1
∂x
⎪
→ x = , y = and z = 1 − x − y =
⎬
∂F
2
3
6
= 18 y − 12 + 12 x = 0 ⎪
⎪⎭
∂y
The function, a quadratic form in x, y and z, has the single stationary value f ( x, y, z ) = 1 at
( x, y, z ) = (1 2, 1 3, 1 6). This is a minimum value. Thus
f xx = 16 > 0, f yy = 18 > 0, f xy = 12,
f xx f yy − f xy 2 > 0
© E Steiner 2008
Solutions for Chapter 9
(ii) Let
Then
10
φ = (2 x 2 + 3 y 2 + 6 z 2 ) − λ ( x + y + z )
∂φ
⎫
= 4x − λ = 0 ⎪
∂x
⎪
∂φ
x
2x
1
⎪
= 6y − λ = 0 ⎬ → z = , y =
, and x + y + z = 1 → x =
∂y
3
3
2
⎪
⎪
∂φ
= 12 z − λ = 0 ⎪
∂z
⎭
and ( x, y, z ) = (1 2, 1 3, 1 6) , as in (i).
28. Find the maximum value of the function f = x 2 y 2 z 2 subject to the constraint x 2 + y 2 + z 2 = c 2 ,
(i) by using the constraint to eliminate z from the function, (ii) by the method of Lagrange
multipliers.
Let
u = x 2 , v = y 2 , w = z 2 , and a = c 2 .
Then
f = uvw subject to u + v + w = a .
(i) We have w = a − u − v . Then
f = uv (a − u − v ) = auv − u 2v − uv 2
and , for a stationary value,
∂f
∂f
= av − 2uv − v 2 = 0,
= au − u 2 − 2uv = 0
∂u
∂v
The possible solutions are
(u , v ) = (0, 0), (0, a), (a, 0) → f = 0 , the minimum
and
(u , v ) = (a 3, a 3) → w = a 3 → f = a 3 27 .
Therefore
f = c 6 27 at x 2 = y 2 = z 2 , the (local) maximum.
(ii) Let
Then
φ = uvw − λ (u + v + w ) .
∂φ
= vw − λ = 0
∂u
∂φ
= uw − λ = 0
∂v
∂φ
= uv − λ = 0
∂w
with nonzero solution u = v = w = a 3 , as in (i).
© E Steiner 2008
Solutions for Chapter 9
11
29. (i) Find the stationary points of the function f = ( x − 1) 2 + ( y − 2) 2 + ( z − 2)2 subject to the
constraint x 2 + y 2 + z 2 = 1 . (ii) Show that these lie at the shortest and longest distances of the point
(1, 2, 2) from the surface of the sphere x 2 + y 2 + z 2 = 1 .
(i) By the method of Lagrange multipliers, let
φ = ( x − 1)2 + ( y − 2) 2 + ( z − 2) 2 − λ ( x 2 + y 2 + z 2 )
For a stationary value,
⎫
∂φ
= 2( x − 1) − 2λ x = 0 → x(1 − λ ) = 1 ⎪
∂x
⎪
⎪⎪
∂φ
= 2( y − 2) − 2λ y = 0 → y (1 − λ ) = 2 ⎬ → y = z = 2 x
∂y
⎪
⎪
∂φ
= 2( z − 2) − 2λ z = 0 → z (1 − λ ) = 2 ⎪
⎪⎭
∂z
Therefore
x2 + y 2 + z 2 = 9 x2 = 1 → x = ±1 3 .
There are therefore two stationary points:
( x, y, z ) = (1 3, 2 3, 2 3) → f = (1 3 − 1) 2 + (2 3 − 2) 2 + (2 3 − 2) 2 = 4
( x, y, z ) = (−1 3, −2 3, −2 3) → f = (−1 3 − 1)2 + (−2 3 − 2) 2 + (−2 3 − 2) 2 = 16
(ii) Geometrically, the quantity
f = ( x − 1) 2 + ( y − 2) 2 + ( z − 2)2
is the distance of point (1, 2, 2) from a point ( x, y, z ) on the surface of sphere x 2 + y 2 + z 2 = 1
with radius r = 1 and centre at (0, 0, 0) . The stationary points are then the maximum and
minimum values of this distance. As illustrated in Figure 1, the line defined by y = z = 2 x
passes through the four points (−1 3, −2 3, −2 3), (0, 0, 0), (1 3, 2 3, 2 3) , and (1, 2, 2).
Figure 1
© E Steiner 2008
12
Solutions for Chapter 9
30. (i) Show that the problem of finding the stationary values of the function
E ( x, y, z ) = a ( x 2 + y 2 + z 2 ) + 2b( xy + yz )
subject to the constraint x 2 + y 2 + z 2 = 1 (a and b are constants) is equivalent to solving the
secular equations
(a − λ ) x + by
=0
bx + (a − λ ) y + bz
=0
by + (a − λ ) z = 0
These equations have solutions for three possible values of the Lagrangian multiplier:
λ1 = a, λ2 = a + 2b, λ3 = a − 2b
(ii) Find the stationary point corresponding to each value of λ (assume x is positive). (iii) Show
that the three stationary values of E are identical to the corresponding values of λ. (This is the
Hückel problem for the allyl radical, CH2CHCH2; see also Example 17.9).
(i) By the method of Lagrange multipliers, let
φ = E ( x, y , z ) − λ ( x 2 + y 2 + z 2 )
= (a − λ )( x 2 + y 2 + z 2 ) + 2b( xy + yz )
For stationary values
∂φ
∂φ
∂φ
= 2(a − λ ) x + 2by = 0,
= 2(a − λ ) y + 2bx + 2bz = 0,
= 2(a − λ ) z + 2by = 0
∂x
∂y
∂z
and the secular equations are
(a − λ ) x + by
=0
bx + (a − λ ) y + bz
=0
by + (a − λ ) z = 0
(ii ) For λ = a,
0 + by + 0 = 0 ⎫
⎪
bx + 0 + bz = 0 ⎬ → y = 0, z = − x
0 + by + 0 = 0 ⎪⎭
Then, given that x 2 + y 2 + z 2 = 1 , the corresponding stationary point is
1 ⎞
⎛ 1
( x, y , z ) = ⎜
, 0, −
⎟.
2⎠
⎝ 2
For λ = a + 2 b,
bx
0
and
© E Steiner 2008
= 0⎫
⎪⎪
− 2 by + bz = 0 ⎬ → y = 2 x, z = x
+ by − 2 bz = 0 ⎪⎪⎭
− 2 bx + by +
0
⎛1 1 1⎞
( x, y , z ) = ⎜ ,
, ⎟.
2 2⎠
⎝2
13
Solutions for Chapter 9
Similarly for λ = a − 2b ,
1 1⎞
⎛1
( x, y , z ) = ⎜ , −
, ⎟.
2 2⎠
⎝2
E = a ( x 2 + y 2 + z 2 ) + 2b( xy + yz ) = a + 2b( xy + yz )
(iii) We have
For λ = a ,
E = a + 2b(0 + 0) = a
For λ = a ± 2b ,
1
1 ⎞
⎛
E = a + 2b ⎜ ±
±
⎟ = a ± 2b
⎝ 2 2 2 2⎠
Section 9.5
Find the total differential df :
31. f ( x, y ) = x 2 + y 2 :
∂f
∂f
= 2 x,
= 2 y → df = 2 x dx + 2 y dy
∂x
∂y
32. f ( x, y ) = 3x 2 + sin( x − y ) :
∂f
∂f
= 6 x + cos( x − y ),
= − cos( x − y )
∂x
∂y
→ df = ⎡⎣6 x + cos( x − y ) ⎤⎦ dx − cos( x − y ) dy
∂f
∂f
1
= 3x 2 y 2 ,
= 2 x3 y +
y
∂x
∂y
33. f ( x, y ) = x3 y 2 + ln y :
⎡
1⎤
→ df = 3x 2 y 2 dx + ⎢ 2 x3 y + ⎥ dy
y⎦
⎣
34. f ( x, y, z ) =
1
2
2
( x + y + z 2 )1 2
1
Let
u = x2 + y 2 + z 2 , f =
Then, for example,
∂f ∂f ∂u − x
x
=
=
=− 2
2
∂x ∂u ∂x u 3 2
( x + y + z 2 )1 2
and
df = −
12
u
1
2
2
( x + y + z 2 )3 2
.
⎡⎣ x dx + y dy + z dz ⎤⎦
35. f (r , θ , φ ) = r sin θ sin φ : df = sin θ sin φ dr + r cos θ sin φ dθ + r sin θ cos φ dφ
© E Steiner 2008
14
Solutions for Chapter 9
36. Write down the total differential of the volume of a two-component system in terms of changes in
temperature T, pressure p, and amounts nA and nB of the components A and B. Use the full
notation with subscripts for constant variables.
⎛ ∂V
⎛ ∂V ⎞
dV = ⎜
dT + ⎜
⎟
⎝ ∂T ⎠ p, nA , nB
⎝ ∂p
⎛ ∂V
⎞
dp + ⎜
⎟
⎠T , nA , nB
⎝ ∂nA
⎞
⎛ ∂V ⎞
d nA + ⎜
d nB
⎟
⎟
⎠T , p, nB
⎝ ∂nB ⎠T , p, nA
Section 9.6
37. Given z = x 2 + 2 xy + 3 y 2 , where x = (1 + t )1 2 and y = (1 − t )1 2 , find
dz
dt
by (i) substitution, (ii) the chain rule (9.21).
z = x 2 + 2 xy + 3 y 2 → z = (1 + t ) + 2(1 + t )1 2 (1 − t )1 2 + 3(1 − t )
(i)
= 4 − 2t + 2(1 − t 2 )1 2
Then
dz
2t
= −2 −
dt
(1 − t 2 )1 2
= −2 −
(ii)
y x
( y 2 − x2 )
= −2 + −
xy
x y
⎛ 1 ⎞
dz ∂z dx ∂z dy
⎛ 1 ⎞
= × + ×
= (2 x + 2 y ) × ⎜ ⎟ + (2 x + 6 y ) × ⎜ − ⎟
dt ∂x dt ∂y dt
⎝ 2x ⎠
⎝ 2y ⎠
= −2 +
y x
as in (i)
−
x y
38. Given u = e x − y , where x = 2 cos t and y = 3t , use the chain rule to find
du ∂u dx ∂u dy
=
× + ×
= e x− y × (−2sin t ) + −e x − y × (3)
dt ∂x dt ∂y dt
(
= −(2sin t + 3) e 2cos t −3t
© E Steiner 2008
)
(
)
du
.
dt
15
Solutions for Chapter 9
39. Given u = ln( x + y + z ) , where x = a cos t , y = b sin t , and z = ct , find
du
.
dt
du ∂u dx ∂u dy ∂u dz
1
(−a sin t + b cos t + c)
=
× + × + × =
dt ∂x dt ∂y dt ∂z dt x + y + z
=
− a sin t + b cos t + c
a cos t + b sin t + ct
40. Given z = ln (2 x + 3 y ), x = a cos θ , y = a sin θ , use the chain rule to find
dz
.
dθ
2
3
dz ∂z dx ∂z dy
= ×
+ ×
=
× (−a sin θ ) +
× (a cos θ )
2x + 3y
dθ ∂x dθ ∂y dθ 2 x + 3 y
=
3cos θ − 2sin θ
2 cos θ + 3sin θ
41. Given f = sin(u + v ) , where v = cos u , (i) find
df
df
, (ii) if u = e −t , find
.
du
dt
df ∂f ∂f d v
=
+ ×
= cos(u + v ) + cos(u + v ) × (− sin u )
du ∂u ∂v d u
(i)
= ⎡⎣1 − sin u ⎤⎦ cos(u + v ) = ⎡⎣1 − sin u ⎤⎦ cos(u + cos u )
(ii) We have
Therefore
du
= −e−t = −u
dx
df df du
=
= −u ⎡⎣1 − sin u ⎤⎦ cos(u + cos u )
dt du dt
⎛ ∂y ⎞
42. If z = x5 y − sin y , find ⎜ ⎟ .
⎝ ∂x ⎠ z
We have
⎛ ∂z ⎞
⎛ ∂z ⎞
4
5
⎜ ⎟ = 5 x y, ⎜ ⎟ = x − cos y
⎝ ∂x ⎠ y
⎝ ∂y ⎠ x
Therefore
⎛ ∂y ⎞
⎛ ∂z ⎞
⎜ ⎟ = − ⎜ ∂x ⎟
∂
x
⎝ ⎠y
⎝ ⎠z
© E Steiner 2008
⎛ ∂z ⎞
−5 x 4 y
⎜ ⎟ = 5
⎝ ∂y ⎠ x x − cos y
16
Solutions for Chapter 9
⎛ ∂V
43. For the van der Waals gas, use the expressions for ⎜
⎝ ∂T
⎛ ∂V
⎞
and ⎜
⎟
⎠ p, n
⎝ ∂p
⎞
from Exercise 20
⎟
⎠T , n
⎛ ∂p ⎞
to find ⎜
⎟
⎝ ∂T ⎠V , n
⎛
n 2 a 2n3 ab ⎞ ⎛ ∂V
⎜⎜ p − 2 +
⎟, ⎜
V
V 3 ⎟⎠ ⎝ ∂p
⎝
We have
⎛ ∂V
⎜
⎝ ∂T
Therefore
⎛ ∂p ⎞
⎛ ∂V
= −⎜
⎜
⎟
⎝ ∂T
⎝ ∂T ⎠V , n
⎞
= nR
⎟
⎠ p, n
⎡
= − ⎢ nR
⎣⎢
=
⎞
⎟
⎠ p, n
⎞
= − (V − nb)
⎟
⎠T , n
⎛
n 2 a 2n3 ab ⎞
⎜⎜ p − 2 +
⎟
V
V 3 ⎟⎠
⎝
⎛ ∂V ⎞
⎜
⎟
⎝ ∂p ⎠T , n
⎛
n 2 a 2n3 ab ⎞ ⎤
⎜⎜ p − 2 +
⎟⎥
V
V 3 ⎠⎟ ⎦⎥
⎝
⎡
⎢ − (V − nb)
⎣⎢
⎛
n 2 a 2n3 ab ⎞ ⎤
⎜⎜ p − 2 +
⎟⎥
V
V 3 ⎠⎟ ⎥⎦
⎝
nR
V − nb
⎛ ∂z ⎞
44. If z = x 2 + y 2 and u = xy , find ⎜ ⎟ by (i) substitution, (ii) equation (9.30).
⎝ ∂y ⎠u
(i) We have
Therefore
(ii) We have
Therefore
x=
u
u2
→ z = 2 + y2
y
y
⎛ ∂z ⎞
2u 2
2 x2
2( y 2 − x 2 )
+ 2y =
⎜ ⎟ = − 3 + 2y = −
y
y
y
⎝ ∂y ⎠ u
⎛ ∂z ⎞
⎛ ∂x ⎞
u
x
⎛ ∂z ⎞
⎜ ⎟ = 2 x, ⎜ ∂y ⎟ = 2 y, ⎜ ∂y ⎟ = − 2 = −
y
y
⎝ ∂x ⎠ y
⎝ ⎠x
⎝ ⎠u
⎛ ∂z ⎞ ⎛ ∂z ⎞ ⎛ ∂x ⎞
⎛ ∂z ⎞
⎜ ⎟ = ⎜ ∂y ⎟ + ⎜ ∂x ⎟ ⎜ ⎟
⎝ ∂y ⎠ u ⎝ ⎠ x ⎝ ⎠ y⎝ ∂y ⎠ u
⎛ x ⎞ 2( y 2 − x 2 )
= 2 y + 2x × ⎜ − ⎟ =
y
⎝ y⎠
© E Steiner 2008
17
Solutions for Chapter 9
⎛ ∂z ⎞
45. Given z = x sin y and u = x 2 + 2 xy + 3 y 2 , find ⎜ ⎟ .
⎝ ∂y ⎠ u
We have
⎛ ∂z ⎞
⎛ ∂x ⎞
⎛ ∂u ⎞
⎛ ∂z ⎞
⎜ ⎟ = sin y, ⎜ ∂y ⎟ = x cos y, ⎜ ∂y ⎟ = − ⎜ ∂ ⎟
∂
x
⎝ ⎠y
⎝ ⎠x
⎝ ⎠u
⎝ y ⎠x
Therefore
⎛ ∂z ⎞ ⎛ ∂z ⎞ ⎛ ∂x ⎞
⎛ ∂z ⎞
⎜ ⎟ = ⎜ ∂y ⎟ + ⎜ ∂x ⎟ ⎜ ⎟
⎝ ∂y ⎠u ⎝ ⎠ x ⎝ ⎠ y⎝ ∂y ⎠u
2x + 6 y
⎛ ∂u ⎞
⎜ ⎟ =−
∂
x
2
x + 2y
⎝ ⎠y
⎛ x + 3y ⎞
= x cos y − ⎜
⎟ sin y
⎝ x+ y ⎠
46. If U = U (V , T ) and p = p (V , T ) are functions of V and T and if H = U + pV , show that
⎡⎛ ∂U ⎞
⎛ ∂H ⎞
⎛ ∂U ⎞
⎜
⎟ −⎜
⎟ = ⎢⎜
⎟ +
⎝ ∂T ⎠ p ⎝ ∂T ⎠V ⎣⎢⎝ ∂V ⎠T
We have
⎛ ∂H
⎜
⎝ ∂T
⎞ ⎛ ∂U
⎟ =⎜
⎠ p ⎝ ∂T
⎞
⎟ +
⎠p
⎤ ⎛ ∂V ⎞
p⎥ ⎜
⎟ .
⎦⎥ ⎝ ∂T ⎠ p
⎛ ∂V
p⎜
⎝ ∂T
⎞
⎟
⎠p
and, by equation (9.30),
⎛ ∂U ⎞
⎛ ∂U ⎞
⎛ ∂U ⎞ ⎛ ∂V ⎞
⎜
⎟ =⎜
⎟ +⎜
⎟ ⎜
⎟
⎝ ∂T ⎠ p ⎝ ∂T ⎠V ⎝ ∂V ⎠T ⎝ ∂T ⎠ p
Therefore
⎛ ∂H
⎜
⎝ ∂T
⎡⎛ ∂U
⎞
⎟ = ⎢⎜
⎠ p ⎢⎣⎝ ∂T
⎞
⎛ ∂U
⎟ +⎜
⎠V ⎝ ∂V
⎞ ⎛ ∂V
⎟ ⎜
⎠T ⎝ ∂T
⎡⎛ ∂U ⎞
⎛ ∂H ⎞
⎛ ∂U ⎞
⎜
⎟ −⎜
⎟ = ⎢⎜
⎟ +
⎝ ∂T ⎠ p ⎝ ∂T ⎠V ⎣⎢⎝ ∂V ⎠T
© E Steiner 2008
⎞ ⎤
⎟ ⎥+
⎠ p ⎥⎦
⎛ ∂V
p⎜
⎝ ∂T
⎤ ⎛ ∂V ⎞
p⎥ ⎜
⎟
⎦⎥ ⎝ ∂T ⎠ p
⎞
⎟
⎠p
18
Solutions for Chapter 9
47. Given x = au + bv and y = bu − av , where a and b are constants, (i) if f is a function of x and y,
⎛ ∂f ⎞
⎛ ∂f ⎞
⎛ ∂f ⎞
⎛ ∂f ⎞
⎛ ∂f ⎞
express ⎜ ⎟ and ⎜ ⎟ in terms of ⎜ ⎟ and ⎜ ⎟ , (ii) if f = x 2 + y 2 , find ⎜ ⎟
⎝ ∂u ⎠v
⎝ ∂v ⎠u
⎝ ∂u ⎠v
⎝ ∂x ⎠ y
⎝ ∂y ⎠ x
⎛ ∂f ⎞
and ⎜ ⎟ in terms of u and v .
⎝ ∂v ⎠u
(i) We have
⎛ ∂f ⎞ ⎛ ∂f ⎞ ⎛ ∂x ⎞ ⎛ ∂f ⎞ ⎛ ∂y ⎞
⎜ ⎟ = ⎜ ⎟ ⎜ ⎟ +⎜ ⎟ ⎜ ⎟
⎝ ∂u ⎠v ⎝ ∂x ⎠ y⎝ ∂u ⎠v ⎝ ∂y ⎠ x⎝ ∂u ⎠v
Then
⎛ ∂f ⎞
⎛ ∂f ⎞
⎛ ∂f ⎞
⎜ ⎟ = a⎜ ⎟ + b⎜ ⎟
⎝ ∂u ⎠v
⎝ ∂x ⎠ y
⎝ ∂y ⎠ x
Similarly,
⎛ ∂f ⎞
⎛ ∂f ⎞
⎛ ∂f ⎞
⎜ ⎟ = b⎜ ⎟ − a⎜ ⎟
∂
v
∂
x
⎝ ⎠u
⎝ ⎠y
⎝ ∂y ⎠ x
⎛ ∂f ⎞
⎛ ∂f ⎞
(ii) If f = x 2 + y 2 then ⎜ ⎟ = 2 x, ⎜ ⎟ = 2 y
⎝ ∂x ⎠ y
⎝ ∂y ⎠ x
Therefore
⎛ ∂f ⎞
⎜ ⎟ = 2ax + 2by = 2a(au + bv ) + 2b(bu − av )
⎝ ∂u ⎠v
= 2(a 2 + b 2 )u
Similarly,
© E Steiner 2008
⎛ ∂f ⎞
2
2
⎜ ⎟ = 2(a + b )v
⎝ ∂v ⎠u
19
Solutions for Chapter 9
48. Given u = x n + y n and v = x n − y n , where n is a constant,
1 ⎛ ∂y ⎞ ⎛ ∂v ⎞
⎛ ∂x ⎞ ⎛ ∂u ⎞
(i) show that ⎜ ⎟ ⎜ ⎟ = = ⎜ ⎟ ⎜ ⎟ .
⎝ ∂u ⎠v ⎝ ∂x ⎠ y 2 ⎝ ∂v ⎠u⎝ ∂y ⎠ x
⎛ ∂f ⎞
⎛ ∂f ⎞
(ii) If f is a function of x and y, express ⎜ ⎟ and ⎜ ⎟ in terms of
∂
x
⎝ ⎠y
⎝ ∂y ⎠ x
⎛ ∂f ⎞
⎛ ∂f ⎞
⎜ ⎟ and ⎜ ⎟ .
∂
u
⎝ ⎠v
⎝ ∂v ⎠ u
⎛ ∂f ⎞
⎛ ∂f ⎞
Hence, (iii) if f = u 2 − v 2 , find ⎜ ⎟ and ⎜ ⎟ in terms of x and y.
⎝ ∂x ⎠ y
⎝ ∂y ⎠ x
(i) We have
Also
Therefore
(ii)
⎛ ∂u ⎞
n −1 ⎛ ∂v ⎞
n −1
⎜ ⎟ = nx , ⎜ ⎟ = −ny
∂
∂
⎝ x ⎠y
⎝ y ⎠x
xn =
1
1
⎛∂ ⎞
⎛ ∂x ⎞
(u + v ) → ⎜ ⎟ x n = nx n −1 ⎜ ⎟ =
2
⎝ ∂u ⎠v
⎝ ∂u ⎠v 2
yn =
1
1
⎛∂ ⎞
⎛ ∂y ⎞
(u − v ) → ⎜ ⎟ y n = ny n −1 ⎜ ⎟ = −
2
2
⎝ ∂v ⎠u
⎝ ∂v ⎠u
1⎫
⎛ ∂u ⎞ ⎛ ∂x ⎞
⎜ ⎟ ⎜ ⎟ = ⎪
⎝ ∂x ⎠ y⎝ ∂u ⎠v 2 ⎪
1 ⎛ ∂y ⎞ ⎛ ∂v ⎞
⎪
⎛ ∂x ⎞ ⎛ ∂u ⎞
⎬ → ⎜ ⎟ ⎜ ⎟ = =⎜ ⎟ ⎜ ⎟
⎝ ∂u ⎠v ⎝ ∂x ⎠ y 2 ⎝ ∂v ⎠u ⎝ ∂y ⎠ x
⎛ ∂v ⎞ ⎛ ∂y ⎞
1⎪
⎜ ⎟ ⎜ ⎟ = ⎪
⎝ ∂y ⎠ x⎝ ∂v ⎠u 2 ⎪⎭
⎛ ∂f ⎞
⎛ ∂f ⎞ ⎛ ∂u ⎞ ⎛ ∂f ⎞ ⎛ ∂v ⎞
⎜ ⎟ = ⎜ ⎟ ⎜ ⎟ +⎜ ⎟ ⎜ ⎟
⎝ ∂x ⎠ y ⎝ ∂u ⎠v ⎝ ∂x ⎠ y ⎝ ∂v ⎠u ⎝ ∂x ⎠ y
⎡⎛ ∂f ⎞ ⎛ ∂f ⎞ ⎤
= nx n −1 ⎢⎜ ⎟ + ⎜ ⎟ ⎥
⎢⎣⎝ ∂u ⎠v ⎝ ∂v ⎠u ⎥⎦
Similarly,
(iii) We have
Therefore
⎡
⎛ ∂f ⎞
⎛ ∂f ⎞ ⎤
n −1 ⎛ ∂f ⎞
⎜ ⎟ = ny ⎢⎜ ⎟ − ⎜ ⎟ ⎥
⎝ ∂y ⎠ x
⎣⎢⎝ ∂u ⎠v ⎝ ∂v ⎠u ⎦⎥
f = u2 − v 2 →
∂f
∂f
= 2u,
= −2v
∂u
∂v
⎡
⎛ ∂f ⎞
⎛ ∂f ⎞ ⎤
n −1 ⎛ ∂f ⎞
n −1
n −1 n
⎜ ⎟ = nx ⎢⎜ ⎟ + ⎜ ⎟ ⎥ = nx ⎡⎣ 2u − 2v ⎤⎦ = 4nx y
x
u
v
∂
∂
∂
⎝ ⎠y
⎣⎢⎝ ⎠v ⎝ ⎠u ⎦⎥
⎡
⎛ ∂f ⎞
⎛ ∂f ⎞ ⎤
n −1 ⎛ ∂f ⎞
n −1
n n −1
⎜ ⎟ = ny ⎢⎜ ⎟ − ⎜ ⎟ ⎥ = ny ⎡⎣ 2u + 2v ⎤⎦ = 4nx y
⎢⎣⎝ ∂u ⎠v ⎝ ∂v ⎠u ⎥⎦
⎝ ∂y ⎠ x
© E Steiner 2008
20
Solutions for Chapter 9
49. If x = au + bv , y = bu − av , and f is a function of x and y (see Exercise 45(i)), show that
(i)
∂2 f
∂u
2
+
⎛ ∂2 f ∂2 f
= (a 2 + b 2 ) ⎜⎜ 2 + 2
∂v
∂y
⎝ ∂x
∂2 f
2
⎞
⎛ ∂2 f ∂2 f
∂2 f
= ab ⎜⎜ 2 − 2
⎟⎟ , (ii)
∂u∂v
∂y
⎝ ∂x
⎠
2
⎞
2
2 ∂ f
.
⎟⎟ + (b − a )
∂x∂y
⎠
From Exercise 47,
⎛ ∂f ⎞
⎛ ∂f ⎞
⎛ ∂f ⎞
⎜ ⎟ = a⎜ ⎟ + b⎜ ⎟ ,
⎝ ∂u ⎠v
⎝ ∂x ⎠ y ⎝ ∂y ⎠ x
(i) We have
∂2 f
∂u
2
=
∂ ⎛ ∂f ⎞ ∂ ⎛ ∂f
∂f ⎞
⎜a +b ⎟
⎜ ⎟=
∂u ⎝ ∂u ⎠ ∂u ⎝ ∂x
∂y ⎠
=
∂ ⎛ ∂f
∂f ⎞ ∂x ∂ ⎛ ∂f
∂f ⎞ ∂y
⎜ a + b ⎟× + ⎜ a + b ⎟×
∂x ⎝ ∂x
∂y ⎠ ∂u ∂y ⎝ ∂x
∂y ⎠ ∂u
= a2
Similarly,
Therefore
(ii) We have
∂2 f
∂v 2
∂2 f
∂u
2
= b2
+
∂2 f
∂x 2
− ab
∂x 2
+ ba
∂2 f
∂2 f
∂2 f
+ ab
+ b2 2
∂x∂y
∂y∂x
∂y
∂2 f
∂2 f
∂2 f
− ba
+ a2 2
∂x∂y
∂y∂x
∂y
2
∂2 f
2
2 ⎛∂ f
=
a
+
b
+
(
)
⎜
⎜ ∂x 2 ∂y 2
∂v 2
⎝
⎞
⎟⎟
⎠
⎛ ∂f ⎞ ⎞
∂2 f
∂ ⎛ ∂f ⎞ ∂ ⎛ ⎛ ∂f ⎞
⎜ b⎜ ⎟ − a⎜ ⎟ ⎟
=
⎜ ⎟=
∂u∂v ∂u ⎝ ∂v ⎠ ∂u ⎜ ⎝ ∂x ⎠ y ⎝ ∂y ⎠ x ⎟
⎝
⎠
∂ ⎛ ∂f
∂f ⎞ ∂x ∂ ⎛ ∂f
∂f ⎞ ∂y
⎜b − a ⎟× + ⎜b − a ⎟×
∂x ⎝ ∂x
∂y ⎠ ∂u ∂y ⎝ ∂x
∂y ⎠ ∂u
= ba
© E Steiner 2008
∂2 f
∂2 f
=
Therefore
⎛ ∂f ⎞
⎛ ∂f ⎞
⎛ ∂f ⎞
⎜ ⎟ = b⎜ ⎟ − a ⎜ ⎟
⎝ ∂v ⎠ u
⎝ ∂x ⎠ y ⎝ ∂y ⎠ x
∂2 f
∂x 2
⎛ ∂2 f ∂2 f
∂2 f
= ab ⎜⎜ 2 − 2
∂u∂v
∂y
⎝ ∂x
− a2
∂2 f
∂2 f
∂2 f
+ b2
− ab 2
∂x∂y
∂y∂x
∂y
2
⎞
2
2 ∂ f
⎟⎟ + (b − a )
∂y∂x
⎠
21
Solutions for Chapter 9
Show that the following functions of position in a plane satisfy Laplace’s equation:
50. f ( x, y ) = x5 − 10 x3 y 2 + 5 xy 4
∂f
∂2 f
= 5 x 4 − 30 x 2 y 2 + 5 y 4 ,
= 20 x3 − 60 xy 2
∂x
∂x 2
∂f
∂2 f
= −20 x3 y + 20 xy 3 ,
= −20 x3 + 60 xy 2
∂y
∂y 2
Therefore
∇2 f =
∂2 f
∂x 2
+
∂2 f
∂y 2
=0
B⎞
⎛
51. f (r , θ ) = ⎜ Ar + ⎟ sin θ
r⎠
⎝
We have
∇2 =
∂2
∂r 2
+
1 ∂
1 ∂2
+ 2
r ∂r r ∂θ 2
∂f ⎛
B⎞
∂2 f 2B
∂2 f
B⎞
⎛
sin
,
θ
= ⎜ A − 2 ⎟ sin θ ,
=
= − ⎜ Ar + ⎟ sin θ
r⎠
∂r ⎝
r ⎠
r3
∂r 2
∂θ 2
⎝
Therefore
∇2 f =
∂2 f
∂r
2
+
1 ∂f
1 ∂2 f
+ 2
r ∂r r ∂θ 2
B⎞
B⎞
1 ⎛
⎛ 2B
⎞ 1⎛
= ⎜ 3 sin θ ⎟ + ⎜ A − 2 ⎟ sin θ − 2 ⎜ Ar + ⎟ sin θ = 0
r
r⎠
r ⎠
r ⎝
⎝r
⎠
⎝
52. f (r , θ ) = r n cos nθ , n = 1, 2, 3, …
∂f
∂2 f
= nr n −1 sin nθ ,
= n(n − 1)r n − 2 sin nθ ,
∂r
∂r 2
∂f
∂2 f
= − nr n cos θ ,
= −n 2 r n sin θ
∂θ
∂θ 2
Therefore
∇2 f =
∂2 f
∂r 2
+
1 ∂f
1 ∂2 f
+ 2
r ∂r r ∂θ 2
= n(n − 1)r n − 2 sin θ + nr n − 2 sin θ − n 2 r n − 2 sin θ = 0
© E Steiner 2008
22
Solutions for Chapter 9
Section 9.7
Test for exactness:
53. F dx + G dy = (4 x + 3 y ) dx + (3 x + 8 y ) dy
We have
∂F
∂G
= 3,
= 3 → (4 x + 3 y ) dx + (3 x + 8 y ) dy is exact
∂y
∂x
54. F dx + G dy = (6 x + 5 y + 7) dx + (4 x + 10 y + 8) dy
∂F
∂G
= 5,
= 4 → (6 x + 5 y + 7) dx + (4 x + 10 y + 8) dy is not exact
∂y
∂x
55. F dx + G dy = y cos x dx + sin x dy
∂F
∂G
= cos x,
= cos x → y cos x dx + sin x dy is exact
∂y
∂x
⎛ ∂S ⎞
⎛ ∂V
56. Given the total differential dG = − S dT + V dp , show that ⎜ ⎟ = − ⎜
⎝ ∂T
⎝ ∂p ⎠T
The total differential of G = G (T , p ) is
⎛ ∂G ⎞
⎛ ∂G ⎞
dG = ⎜
⎟ dp
⎟ dT + ⎜
⎝ ∂T ⎠ p
⎝ ∂p ⎠T
Therefore
⎛ ∂G ⎞
⎛ ∂G ⎞
−S = ⎜
⎟
⎟ , V =⎜
⎝ ∂T ⎠ p
⎝ ∂p ⎠T
⎛ ∂S ⎞
⎛ ∂ ⎞ ⎛ ∂G ⎞
∂ 2G
−⎜ ⎟ = ⎜ ⎟ ⎜
=
,
⎟
⎝ ∂p ⎠T ⎝ ∂p ⎠T ⎝ ∂T ⎠ p ∂p∂T
∂ 2G
∂ 2G
⎛ ∂V ⎞
⎛ ∂ ⎞ ⎛ ∂G ⎞
=
⎟ =
⎜
⎟ =⎜
⎟ ⎜
⎝ ∂T ⎠ p ⎝ ∂T ⎠ p⎝ ∂p ⎠T ∂T ∂p ∂p∂T
and
© E Steiner 2008
⎛ ∂S ⎞
⎛ ∂V
⎜ ⎟ = −⎜
⎝ ∂T
⎝ ∂p ⎠T
⎞
⎟
⎠p
⎞
⎟ .
⎠p
23
Solutions for Chapter 9
Section 9.8
57. Evaluate the line integral
∫ ⎡⎣ xydx + 2 ydy ⎤⎦ on the line y = 2x from x = 0 to x = 2 .
C
∫
C
⎡⎣ xydx + 2 ydy ⎤⎦ =
∫
2
0
2
⎡ 2 x 2 + 8 x ⎤ dx = ⎡ 2 x3 + 4 x 2 ⎤ = 64
⎢3
⎥
⎣
⎦
⎣
⎦0 3
58. When the path of integration is given in parametric form x = x(t ) , y = y (t ) from t = tA to t = tB ,
the line integral can be evaluated as
∫
C
⎡⎣ F dx + G dy ⎤⎦ =
Evaluate
∫
tB
tA
⎡ dx
dy ⎤
⎢ F dt + G dt ⎥ dt .
⎣
⎦
∫ ⎡⎣(x + 2 y)dx + ( y + x)dy ⎤⎦ on the curve with parametric equations x = t , y = t
2
2
2
C
from A(0, 0) to B(1, 1) .
∫
1
∫ ⎡⎣(t + 2t )+ (t + t)× 2t ⎤⎦ dt
⎡ t 5t ⎤
= ⎡ 2t + 5t ⎤ dt = ⎢ +
⎥ =2
⎣
⎦
∫
⎢⎣ 3 3 ⎥⎦
⎡( x 2 + 2 y ) dx + ( y 2 + x) dy ⎤ =
⎣
⎦
C
2
2
5
2
4
0
1
6
3
0
59. Evaluate
1
0
∫ ⎡⎣ xydx + 2 ydy ⎤⎦ on the curve y = x
2
from x = 0 to x = 2 (see Exercise 57).
C
∫ ⎡⎣ xydx + 2 ydy ⎤⎦ = ∫
C
60. Evaluate the line integral
2
0
2
⎡ 4⎤
⎡ x3 + 4 x3 ⎤ dx = ⎢ 5 x ⎥ = 20
⎣
⎦
⎢⎣ 4 ⎥⎦ 0
∫ ⎡⎣(x + 2 y)dx + ( y + x)dy ⎤⎦ on the curve with parametric equations
2
2
C
x = t 2 , y = t from A(0, 0) to B(1, 1) (see Exercise 58).
∫
1
∫ ⎡⎣(t + 2t)× 2t + (t + t ) ⎤⎦ dt
⎡t
⎤
7
= ⎡ 2t + 6t ⎤ dt = ⎢ + 2t ⎥ =
∫ ⎣ ⎦ ⎣⎢ 3 ⎦⎥ 3
⎡( x 2 + 2 y ) dx + ( y 2 + x) dy ⎤ =
⎣
⎦
C
2
2
0
1
0
© E Steiner 2008
4
5
2
6
1
3
0
24
Solutions for Chapter 9
61. The total differential of entropy as a function of T and p is (Example 9.27)
⎛ ∂S
dS = ⎜
⎝ ∂T
⎛ ∂S ⎞
⎞
⎟ dT + ⎜ ⎟ dp
⎠p
⎝ ∂p ⎠T
⎛ ∂S
Given that, ⎜
⎝ ∂T
C p 5R
⎛ ∂S ⎞
⎞
⎛ ∂V
=
and ⎜ ⎟ = − ⎜
⎟ =
2T
∂
p
⎠p T
⎝ ∂T
⎝ ⎠T
R
⎞
⎟ = − for 1 mole of ideal gas, show that
p
⎠p
the (reversible) heat absorbed by the ideal gas round the closed path shown in Figure 2 is equal to
the work done by the gas; that is,
∫v T dS =∫v pdV (see Example 9.25).
We have
⎛ ∂S
dS = ⎜
⎝ ∂T
=
⎛ ∂S ⎞
⎞
⎟ dT + ⎜ ⎟ dp
⎠p
⎝ ∂p ⎠T
R
5R
dT − dp
2T
p
The line integral round the closed path is the sum of
the lines integrals along the four sides:
∫v T dS =∫
C1
Then
T dS +
∫
T dS +
C2
∫
T dS +
C3
C1 ( constant p = p1 ) :
∫
T dS =
∫
T dS =
∫
T dS =
∫
T dS =
C1
C2 ( constant T = T2 ) :
C3
C4 ( constant T = T1 ) :
∫
T2
Figure 2
5R
⎛ 5R ⎞
(T2 − T1 )
⎜
⎟ dT =
2
2
⎝
⎠
∫
p2
p1
∫
T1
∫
p1
T2
C2
T dS
C4
T1
C2
C3 ( constant p = p2 ) :
∫
p2
⎛ RT2 ⎞
p2
⎜−
⎟ dT = − RT2 ln
p ⎠
p1
⎝
5R
⎛ 5R ⎞
(T1 − T2 )
⎜
⎟ dT =
2
⎝ 2 ⎠
⎛ RT1 ⎞
p1
⎜−
⎟ dp = − RT1 ln
p
p
⎝
⎠
2
Therefore
∫
v T dS =
p
p
p
5R
5R
(T2 − T1 ) − RT2 ln 2 +
(T1 − T2 ) − RT1 ln 1 = − R (T2 − T1 ) ln 2
2
2
p1
p2
p1
and this is equal to the work
v∫ pdV
demonstrated in Example 9.25.
© E Steiner 2008
done by (one mole of) the gas round the closed path, as
25
Solutions for Chapter 9
62. (i) Show that F dx + Gdy for F = 9 x 2 + 4 y 2 + 4 xy and G = 8 xy + 2 x 2 + 3 y 2 is an exact
differential. (ii) By choosing an appropriate path, evaluate
∫ [F dx + G dy] from ( x, y) = (0, 0)
C
to (1, 2) .
(iii) Show that the result in (ii) is consistent with the differential as the total differential of
z ( x, y ) = 3 x3 + 4 xy 2 + 2 x 2 y + y 3 .
⎛ ∂F ⎞
⎛ ∂G ⎞
⎜
⎟ = 8 y + 4x = ⎜
⎟
y
∂
⎝ ∂x ⎠ y
⎝
⎠x
(i)
(ii) For the path shown in Figure 3,
∫ [F dx + G dy] = ∫
F dx +
C1
C
∫
G dy
C2
Figure 3
where
1
C1 :
∫
( F ) y =0 dx =
C2 :
∫
(G )x =1 dx =
C1
C2
∫ 9x dx = 3
2
0
∫
2
0
Therefore
∫ [F dx + G dy] = 31
C
(iii) If
then
z ( x, y ) = 3 x3 + 4 xy 2 + 2 x 2 y + y 3
⎛ ∂z ⎞
2
2
⎜ ⎟ = 9 x + 4 y + 4 xy = F
⎝ ∂x ⎠ y
⎛ ∂z ⎞
2
2
⎜ ⎟ = 8 xy + 2 x + 3 y = G
⎝ ∂y ⎠ x
and
© E Steiner 2008
2
(8 y + 2 + 3 y 2 ) dy = ⎡ 4 y 2 + 2 y + y 3 ⎤ = 28
⎣
⎦0
⎛ ∂z ⎞
⎛ ∂z ⎞
dz = ⎜ ⎟ dx + ⎜ ⎟ dy = F dx + G dy
⎝ ∂x ⎠ y
⎝ ∂y ⎠ x
26
Solutions for Chapter 9
63. Evaluate
∫
C
⎡ 2 xy dx + ( x 2 − y 2 ) dy ⎤ on the circle with parametric equations x = cos θ ,
⎣
⎦
y = sin θ ,
(i) from A(1, 0) to B(0, 1) and
(ii) around a complete circle ( θ = 0 → 2π ).
(iii) Confirm that the differential 2 xy dx + ( x 2 − y 2 ) dy is exact.
We have
x = cos θ , dx = − sin θ dθ
y = sin θ , dy = cos θ dθ
Then
∫ ⎡⎣2xy dx + (x − y ) dy ⎤⎦ = ∫ ⎡⎣(2 cosθ sin θ × (− sin θ ) + (cos θ − sin θ ) × cosθ ⎤⎦ dθ
= ⎡⎣cos 2θ cos θ − sin 2θ sin θ ⎤⎦ dθ
∫
= cos 3θ dθ
∫
2
2
2
2
C
The paths of integration are anticlockwise, as illustrated in Figure 4.
(i) θ = 0 → π 2 :
∫
π2
∫
2π
0
(ii) θ = 0 → 2π
0
π 2
⎡1
⎤
1
cos3θ dθ = ⎢ sin 3θ ⎥ = −
3
3
⎣
⎦0
2π
⎡1
⎤
cos3θ dθ = ⎢ sin 3θ ⎥ = 0
3
⎣
⎦0
Figure 4
(iii) The zero result of integration round the closed loop in (ii) suggests that the differential is
exact. Thus,
if
2 xy dx + ( x 2 − y 2 ) dy = F dx + G dy
then
∂F
∂G
= 2x =
∂y
∂x
© E Steiner 2008
27
Solutions for Chapter 9
Section 9.9
3
64. Evaluate the integral
2
∫ ∫ ( x y + xy ) dxdy and show that the result is independent of the order
0
2
2
1
of integration.
3
2
3
⎡
⎢
⎣⎢
∫ ∫ (x y + xy ) dxdy = ∫ ∫
0
2
2
1
0
=
∫
3
0
2
1
⎤
( x y + xy ) dx ⎥ dy =
⎦⎥
2
2
∫
3⎡ 3
0
2
x y x2 y2 ⎤
+
⎢
⎥ dy
2 ⎥⎦
⎣⎢ 3
1
3
⎡7
⎡7 2 1 3 ⎤
3 2⎤
⎢ 3 y + 2 y ⎥ dy = ⎢ 6 y + 2 y ⎥ = 24
⎣
⎦
⎣
⎦0
Interchanging the order of integration,
2
3
2
⎡
⎢
⎣⎢
∫ ∫ (x y + xy )dy dx = ∫ ∫
1
2
2
0
1
=
∫
2
1
3
0
⎤
( x y + xy ) dy ⎥ dx =
⎦⎥
2
2
∫
2⎡ 2 2
1
2
⎡9 2
⎤
⎡3 3 9 2⎤
⎢ 2 x + 9 x ⎥ dx = ⎢ 2 x + 2 x ⎥ = 24
⎣
⎦
⎣
⎦1
and the integration is independent of order.
π
65. Evaluate the integral
∫∫
0
π
∫∫
0
R
R
e − r cos 2 θ sin θ drdθ
0
e − r cos 2 θ sin θ dr dθ =
0
∫
π
cos 2 θ sin θ dθ ×
0
∫
0
π
R
⎡ cos3 θ ⎤
−r
= ⎢−
⎥ × ⎡ −e ⎤
⎣
⎦
0
3 ⎦⎥
⎣⎢
0
=
© E Steiner 2008
2
(1 − e − R )
3
R
3
x y
xy 3 ⎤
+
⎢
⎥ dx
3 ⎥⎦
⎣⎢ 2
0
e − r dr
28
Solutions for Chapter 9
Section 9.10
Evaluate the integral and sketch the region of integration:
2
66.
∫∫
0
2x
( x 2 + y 2 ) dydx =
2
∫ ∫
0
x
=
∫
2
∫
2
0
=
⎡
⎢
⎣⎢
0
2x
x
⎤
( x 2 + y 2 ) dy ⎥ dx
⎦⎥
2x
⎡ 2
y3 ⎤
⎢ x y + ⎥ dx
3 ⎥⎦
⎣⎢
x
10 x 3
40
dx =
3
3
Figure 5
The region of integration, shown in Figure 5, lies between x = 0 and x = 2 and, from the limits of
integration for y, between the lines y = x and y = 2 x .
∫∫
a
67.
0
a2 − x2
xy dydx =
0
0
=
⎡
x⎢
⎢
⎣
∫ ∫
a
2
∫
a
0
0
a2 − x2
⎤
y 2 dy ⎥ dx
⎥
⎦
⎡1
⎤
x ⎢ (a 2 − x 2 )3 2 ⎥ dx
⎣3
⎦
Let u = a 2 − x 2 , du = −2 x dx .
Then u = a 2 when x = 0 , and u = 0 when x = a .
∫∫
a
Therefore
0
0
a2 − x2
xy 2 dydx = −
1
6
∫
0
u 3 2 du =
a
a5
15
Figure 6
The region of integration, shown in Figure 6, lies between x = 0 and x = a , and between y = 0
and y = a 2 − x 2 . Comparison with equation (9.59) shows that the integration is over that quarter
of the circle of radius a that lies in the first quadrant.
© E Steiner 2008
29
Solutions for Chapter 9
68. (i) Show from a sketch of the region of integration that
(2 − y ) 2
2
∫∫
1
∫∫
x3 dx dy =
2 y −4
0
0
2− 2 x
x3 dy dx +
0
∫∫
−4
0
( x + 4) 2
x3 dy dx ,
0
(ii) evaluate the integral.
(i) Figure 7
The region of integration lies between y = 0 and y = 2 and, from the limits of integration for x,
between the lines x = 2 y − 4 and x = (2 − y ) 2 . Interchange of the order of integration gives
Figure 7.
1
(ii)
∫∫
0
2−2 x
x3 dy dx =
0
0
∫∫
−4
1
∫ ∫
0
( x + 4) 2
x3 dy dx =
0
2
∫∫
Therefore
0
⎡
x3 ⎢
⎣⎢
2− 2 x
0
⎡
x3 ⎢
−4
⎢⎣
0
∫ ∫
(2 − y ) 2
⎤
dy ⎥ dx =
⎦⎥
( x + 4) 2
0
x3 dx dy = −
2 y −4
1
∫ x (2 − 2x) dx = 10
1
3
0
⎤
1
dy ⎥ dx =
2
⎥⎦
∫
0
−4
x3 ( x + 4) dx = −
256
10
51
2
Section 9.11
Transform to polar coordinates and evaluate:
1
69.
∫∫
0
1− x 2
( x 2 + 2 xy ) dydx
0
Integration is that illustrated in Figure 6, with a = 1 .
1
Then
∫∫
0
1− x 2
( x 2 + 2 xy ) dydx =
1
∫ ∫
r =0
0
=
∫
1
∫
1
π 2
θ =0
r 3 dr ×
0
=
0
© E Steiner 2008
⎡ r 2 cos 2 θ + 2r 2 cos θ sin θ ⎤ rdrdθ
⎣
⎦
∫
π2
∫
π2
0
r 3 dr ×
0
⎡cos 2 θ + 2 cos θ sin θ ⎤ dθ
⎣
⎦
⎡1
⎤
1⎛π ⎞
⎢ 2 (1 + cos 2θ ) + sin 2θ ⎥ dθ = 4 ⎜ 4 + 1⎟
⎝
⎠
⎣
⎦
30
Solutions for Chapter 9
70.
∞
∞
−∞
−∞
∫ ∫
e −2( x
2
+ y 2 )1 2
( x 2 + y 2 )3 dxdy
The integration is over the whole plane. Therefore
∞
∞
−∞
−∞
∫ ∫
e −2( x
2
+ y 2 )1 2
( x 2 + y 2 )3 dxdy =
∞
∫ ∫
π
r =0 θ =0
=
∫
∞
e −2 r r 7 dr
∞
∫ ∫
0
∞
e−( x
2
∫
π
dθ =
θ =0
r =0
71.
e−2 r r 6 × r dr dθ
7!
8
2
× 2π =
+ y2 ) 2
x dx dy
0
The integration is over one quarter of the xy−plane. Therefore
∞
∫ ∫
0
∞
e−( x
2
+ y2 ) 2
x dx dy =
0
=
© E Steiner 2008
∞
1
4
∫ ∫
1
4
∞
2π
2
e− r r 2 cos 2 θ × r dr dθ
r =0 θ =0
∫
0
2
e − r r 3 dr ×
∫
2π
0
cos 2 θ dθ =
1 1
π
× × π=
8
4 2
315π
8