Thomas Calculus 13th [Solutions]

Chapter 14 Additional and Advanced Exercises 1077
3. Substitution of u u( x ) and v v( x ) in g( u, v ) gives g( u( x), v( x )) which is a function of the independent
v dg g g
v v
variable x. Then, g( u, v) f ( t) dt du dv f ( t) dt du f ( t)
dt dv
u dx u dx v dx u u dx v u dx
u
v
f ( t) dt du f ( t) dt dv f ( u( x)) du f ( v( x)) dv f ( v( x)) dv f ( u( x))
du
u v dx v u dx dx dx dx dx
2
df d f 2 df 2
4. Applying the chain rules, f r r r
x f xx
. Similarly,
dr x 2 x dr 2
dr
x
2
d f 2 df 2
f r r
zz . Moreover,
2 z dr 2
dr
z
2 2
2
r x r y z y
; r
x
x y z
y
x y z
x y z
2 2 2 2 3
x 2 2 2
2 2 2
2
d f
2
df 2
f r r
yy and
2 y dr 2
dr
y
2 2 2
r x z
y
x y z
2 3
2 2 2
;
and
r
z
2 2
2
z
r x y
2 2 2
x y z z
x y z
2 3
2 2 2
.
Next, fxx f yy fzz
0
2 2 2 2 2 2
2 2
d f x df y z d f y df x z
dr x y z dr
dr x y z dr
x y z x y z
2 2 2 2 3 2 2 2 2 3
2 2 2 2 2 2
d
dr
2 2
2 2 2 2
d f z df x y d f 2 df d f 2 df
0 0 0
2 2 2 2 3 2 2 2 2
2
dr x y z dr dr r dr
2 2 2
dr x y z
dr
x y z
f 2 f , where
r
df df 2 dr
2
f ln f 2ln r ln C f Cr , or
dr f r
df 2
Cr f ( r)
C b a b for some constants a and b (setting a C )
dr r r
n
5. (a) Let u tx, v ty , and w f ( u, v) f ( u( t, x), v( t, y)) f ( tx, ty) t f ( x, y ), where t, x, and y are
n 1
independent variables. Then nt f ( x , y ) w w u w v x w y w
t u t v t u v . Now, w w u w v
x u x v x
w ( ) w (0) w w 1 w .
u t v t u u t x
Likewise, w w u w v w (0) w ( t
y u y v y u v
)
w 1 w
n 1 y
. Therefore, nt f
v t y
( x , y ) x w y w x w w
u v t x t y . When t 1, u x, v y,
and w f f
( x , y ) w
x x
and w f ( , ) f f
nf x y x y
y x x y
, as claimed.
1
(b) From part (a), nt n f ( x , y ) x w w
u
y v . Differentiating with respect to t again we obtain
n 2
2 2 2 2
2
2 2
2
2
( n 1) t n f ( x , y ) x w u w v w v w v w 2 w w .
2 t x v u t y u v t y 2 t x xy 2 u v
y Also from
2
u v u v
part (a),
2 2 2 2 2
w w 2
t w t w u t w v t w , w w t w
2 x x x u 2 x v u x 2 2
x u u y y y y v
2 2 2
2
t w u t w v t w , and
u v y 2 y
2
v
v
2 2 2 2
1 w w , 1 w w , and
2 2 2 2 2 2
t x u t y v
2 2 2 2
w w w w u w v 2
t t t t w
u
y x y x y u 2 y v u y v u
2 2
1 w w
2
t y x v u
2 2 2 2 2
n 2
2
( n 1) t n f ( x xy
y
, y ) x w w w
2 2 2 y x
for t 0.
2 2
t x t t y
When t 1, w f ( x, y ) and we have
2 2 2
2 f f 2 f
n( n 1) f ( x, y) x 2xy y as claimed.
2 x y
2
x
y
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