Page 144 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 6] PARTIAL DERIVATIVES 135
2
@ U
6.28. If x ¼ 2r s and y ¼ r þ 2s, find in terms of derivatives with respect to r and s.
@y @x
Solving x ¼ 2r s, y ¼ r þ 2s for r and s: r ¼ð2x þ yÞ=5, s ¼ð2y xÞ=5.
Then @r=@x ¼ 2=5, @s=@x ¼ 1=5, @r=@y ¼ 1=5, @s=@y ¼ 2=5. Hence we have
@U @U @r @U @s 2 @U 1 @U
@x ¼ @r @x þ @s @x ¼ 5 @r 5 @s
2

@ U @ @U @ 2 @U 1 @U @r @ 2 @U 1 @U @s

@y @x ¼ @y @x ¼ @r 5 @r 5 @s @y þ @s 5 @r 5 @s @y
! !
2
2
2
2
2 @ U 1 @ U 2 @ U 1 @ U
1
2
¼ 2 þ 2
5 @r 5 @r @s 5 5 @s @r 5 @s 5
!
2
2
2
1 @ U @ U @ U
2 þ 3 2
25 @r @r @s @s
¼ 2 2
assuming U has continuous second partial derivatives.
IMPLICIT FUNCTIONS AND JACOBIANS
3 5 2 3 2
6.29. If U ¼ x y, find dU=dt if (1) x þ y ¼ t; ð2) x þ y ¼ t .
Equations (1) and (2)define x and y as (implicit) functions of t. Then differentiating with respect to t,
we have
2
4
5x ðdx=dtÞþ dy=t ¼ 1 2xðdx=dtÞþ 3y ðdy=dtÞ¼ 2t
ð3Þ ð4Þ
Solving (3) and (4) simultaneously for dx=dt and dy=dt,
4
1 1 5x 1
2 4
dx 2t 3y 2 3y 2t dy 2x 2t 10x t 2x
4 2
dt ¼ 5x 4 1 ¼ 15x y 2x ; dt ¼ 5x 4 1 ¼ 15x y 2x
4 2

2x 3y 2 2x 3y 2
4
2
dU @U dx @U dy 3y 2t ! 10x t 2x !
2
3
Then ¼ þ ¼ð3x yÞ þðx Þ :
4 2
4 2
dt @x dt @y dt 15x y 2x 15x y 2x
6.30. If Fðx; y; zÞ¼ 0 defines z as an implicit function of x and y in a region r of the xy plane, prove
that (a) @z=@x ¼ F x =F z and (b) @z=@y ¼ F y =F z , where F z 6¼ 0.
@z @z
Since z is a function of x and y, dy.
dz ¼
@x dx þ @y
@F @F @F @F @F @z @F @F @z
Then dy ¼ 0.
@x dx þ @y dy þ @z dz ¼ @x þ @z @x dx þ @y þ @z @y
dF ¼
Since x and y are independent, we have
@F @F @z @F @F @z
¼ 0 ¼ 0
@x @z @x @y @z @y
ð1Þ þ ð2Þ þ
from which the required results are obtained. If desired, equations (1) and (2)can be written directly.
6.31. If Fðx; y; u; vÞ¼ 0 and Gðx; y; u; vÞ¼ 0, find (a) @u=@x; ðbÞ @u=@y; ðcÞ @v=@x; ðdÞ @v=@y.
The two equations in general define the dependent variables u and v as (implicit) functions of the
independent variables x and y. Using the subscript notation, we have