9.12 Exercises 291
Section 9.6
37.Givenz 1 = 1 x
2
1 + 12 xy 1 + 13 y
2
, wherex 1 = 1 (1 1 + 1 t)
122
andy 1 = 1 (1 1 − 1 t)
122
, find by
(i)substitution,(ii)the chain rule (9.21).
38.Givenu 1 = 1 e
x−y, wherex 1 = 121 cos 1 tandy 1 = 13 t, use the chain rule to find.
39.Givenu 1 = 1 ln(x 1 + 1 y 1 + 1 z), wherex 1 = 1 a 1 cos 1 t,y 1 = 1 b 1 sin 1 t,andz 1 = 1 ct, find.
40.Givenz 1 = 1 ln(2x 1 + 13 y),x 1 = 1 a 1 cos 1 θ,y 1 = 1 a 1 sin 1 θ,use the chain rule to find.
41.Givenf 1 = 1 sin(u 1 + 1 v), wherev 1 = 1 cos 1 u, (i)find , (ii)ifu 1 = 1 e
−t, find.
42.Ifz 1 = 1 x
5
y 1 − 1 sin 1 y, find.
43.For the van der Waals gas, use the expressions for and from Exercise 20
to find
44.Ifz 1 = 1 x
2
1 + 1 y
2
andu 1 = 1 xy, find by (i)substitution, (ii)equation (9.30).
45.Givenz 1 = 1 x 1 sin 1 yandu 1 = 1 x
2
1 + 12 xy 1 + 13 y
2
, find.
46.IfU 1 = 1 U(V,T)andp 1 = 1 p(V, T)are functions of Vand Tand ifH 1 = 1 U 1 + 1 pV, show that
.
47.Givenx 1 = 1 au 1 + 1 bvandy 1 = 1 bu 1 − 1 av, where aand bare constants, (i)iffis a function of xand
y, express and in terms of and , (ii)iff 1 = 1 x
2
1 + 1 y
2
, find
and in terms of uand v.
48.Givenu 1 = 1 x
n1 + 1 y
nandv 1 = 1 x
n1 − 1 y
n, where nis a constant,
(i)show that. (ii)Iffis a function of xand y, express
and in terms of and. Hence, (iii)iff 1 = 1 u
2
1 − 1 v
2
, find
and in terms of xand y.
49.Ifx 1 = 1 au 1 + 1 bv, y 1 = 1 bu 1 − 1 av, andfis a function of xand y(see Exercise 47(i)), show that
(i) , (ii).
∂
∂∂
=
∂
∂
−
∂
∂
+−
∂
∂
22
2
2
2
22
2
f
u
ab
f
x
f
y
ba
f
v
()
xxy∂
∂
∂
∂
∂
=+
∂
∂
∂
∂
2
2
2
2
22
2
2
2
2
f
u
f
ab
f
x
f
v y
()
x
f
y
∂
∂
y
f
x
∂
∂
u
∂f
∂
v
v
∂
∂
f
u
x
f
y
∂
∂
y
f
x
∂
∂
v
v
∂ v
∂
∂
∂
==
∂
∂
∂
∂
x
u
u
x
y
y
y ux
1
2
u
∂f
∂
v
v
∂
∂
f
u
x
f
y
∂
∂
y
f
x
∂
∂
u
∂f
∂
v
v
∂
∂
f
u
pV T
H
T
U
T
U
V
p
∂
∂
−
∂
∂
=
∂
∂
∂
∂
p
V
T
u
z
y
∂
∂
u
z
y
∂
∂
Vn
p
T
,
∂
∂
Tn
V
p
,
∂
∂
p
V
T
,n
∂
∂
z
y
x
∂
∂
df
dt
df
du
dz
dθ
du
dt
du
dt
dz
dt