Page 141 - Schaum's Outline of Theory and Problems of Advanced Calculus
P. 141
132 PARTIAL DERIVATIVES [CHAP. 6
where FðyÞ is the ‘‘constant’’ of integration. Substituting this into (2)yields
3
3
2
3
x 4xy þ F ð yÞ¼ x 4xy þ 6y 2 from which F ð yÞ¼ 6y ; i.e., Fð yÞ¼ 2y þ c
0
0
3
2
3
Hence, the required function is ¼ x y 2xy þ 2y þ c, where c is an arbitrary constant.
Note that by Theorem 3, Page 122, the existence of such a function is guaranteed, since if
3
2
2
2
2
P ¼ 3x y 2y and Q ¼ x 4xy þ 6y ,then @P=@y ¼ 3x 4y ¼ @Q=@x identically. If @P=@y 6¼
@Q=@x this function would not exist and the given expression would not be an exact differential.
Method 2:
2
2
2
2
2
2
3
3
ð3x y 2y Þ dx þðx 4xy þ 6y Þ dy ¼ð3x ydx þ x dyÞ ð2y dx þ 4xy dyÞþ 6y dy
3 2 3 3 2 3
¼ dðx yÞ dð2xy Þþ dð2y Þ¼ dðx y 2xy þ 2y Þ
3 2 3
¼ dðx y 2xy þ 2y þ cÞ
2
3
3
Then the required function is x y 2xy þ 2y þ c.
This method, called the grouping method,is based on one’s ability to recognize exact differential
combinations and is less than Method 1. Naturally, before attempting to apply any method, one should
determine whether the given expression is an exact differential by using Theorem 3, Page 122. See
Theorem 4, Page 122.
DIFFERENTIATION OF COMPOSITE FUNCTIONS
6.17. Let z ¼ f ðx; yÞ and x ¼ ðtÞ, y ¼ ðtÞ where f ; ; are assumed differentiable. Prove
dz @z dx @z @y
dt ¼ @x dt þ @y dt
Using the results of Problem 6.14, we have
dz z @z x @z y x y @z dx @z dy
¼ lim ¼ lim þ 1 þ 2
dt t!0 t t!0 @x t þ @y t t t ¼ @x dt þ @y dt
x dx y dy
since as t ! 0we have x ! 0; y ! 0; 1 ! 0; 2 ! 0; ! ; ! :
t dt t dt
2
xy
6.18. If z ¼ e , x ¼ t cos t, y ¼ t sin t, computer dz=dt at t ¼ =2.
dz @z dx @z dy 2 xy 2 xy 2
¼ð y e Þð t sin t þ cos tÞþð2xye Þðt cos t þ sin tÞ:
dt ¼ @x dt þ @y dt
3
2
At t ¼ =2; x ¼ 0; y ¼ =2: Then dz ¼ð =4Þð =2Þþð0Þð1Þ¼ =8:
dt t¼ =2
2
3
Another method. Substitute x and y to obtain z ¼ e t sin t cos t and then differentiate.
6.19. If z ¼ f ðx; yÞ where x ¼ ðu; vÞ and y ¼ ðu; vÞ, prove that
@z @z @x @z @y @z @z @x @z @y
; :
@u @x @u @y @u @v @x @v @y @v
ðaÞ ¼ þ ðbÞ ¼ þ
(a)From Problem 6.14, assuming the differentiability of f ; ; ,we have
@z z @z x @z y x y @z @x @z @y
¼ lim ¼ lim þ 1 þ 2
@u u!0 u u!0 @x u þ @y u u u ¼ @x @u þ @y @u
(b) The result is proved as in (a)by replacing u by v and letting v ! 0.