Page 140 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 6] PARTIAL DERIVATIVES 131

Solution:
ðaÞ z ¼ f ðx þ x; y þ yÞ f ðx; yÞ
2 2
¼fðx þ xÞ ð y þ yÞ 3ð y þ yÞg fx y 3yg
2
2
2
¼ 2xy x þðx 3Þ y þð xÞ y þ 2x x y þð xÞ y
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ðAÞ ðBÞ
The sum (A)is the principal part of z and is the differential of z, i.e., dz. Thus,
2
2
dz ¼ 2xy x þðx 3Þ y ¼ 2xy dx þðx 3Þ dy
ðbÞ
@z @z
2
Another method: dz ¼ dx þ dy ¼ 2xy dx þðx 3Þ dy
@x @y
ðcÞ z ¼ f ðx þ x; y þ yÞ f ðx; yÞ¼ f ð4 0:01; 3 þ 0:02Þ f ð4; 3Þ
2
2
¼fð3:99Þ ð3:02Þ 3ð3:02Þg fð4Þ ð3Þ 3ð3Þg ¼ 0:018702
2 3
dz ¼ 2xy dx þðx 3Þ dy ¼ 2ð4Þð3Þð 0:01Þþð4 3Þð0:02Þ¼ 0:02
Note that in this case z and dz are approximately equal, because x ¼ dx and y ¼ dy are
sufficiently small.
(d)We must find f ðx þ x; y þ yÞ when x þ x ¼ 5:12 and y ¼ y ¼ 6:85. We can accomplish this by
choosing x ¼ 5, x ¼ 0:12, y ¼ 7, y ¼ 0:15. Since x and y are small, we use the fact that
f ðx þ x; y þ yÞ¼ f ðx; yÞþ z is approximately equal to f ðx; yÞþ dz, i.e., z þ dz.
2
Now z ¼ f ðx; yÞ¼ f ð5; 7Þ¼ ð5Þ ð7Þ 3ð7Þ¼ 154
2
2
dz ¼ 2xy dx þðx 3Þ dy ¼ 2ð5Þð7Þð0:12Þþð5 3Þð 0:15Þ¼ 5:1:
Then the required value is 154 þ 5:1 ¼ 159:1 approximately. The value obtained by direct com-
putation is 159.01864.

2
2
3
2
2 y=x
6.16. (a) Let U ¼ x e . Find dU. (b) Show that ð3x y 2y Þ dx þðx 4xy þ 6y Þ dy can be
written as an exact differential of a function ðx; yÞ and find this function.
(a) Method 1:

@U 2 y=x y y=x @U 2 y=x 1
¼ x e þ 2xe ; ¼ x e
@x x 2 @y x
@U @U
Then dU ¼ dx þ dy ¼ð2xe y=x ye y=x Þ dx þ xe y=x dy
@x @y
Method 2:
2
2 y=x
2
dU ¼ x dðe y=x Þþ e y=x dðx Þ¼ x e dðy=xÞþ 2xe y=x dx

2 y=x xdy ydx y=x y=x y=x y=x
¼ x e þ 2xe dx ¼ð2xe ye Þ dx þ xe dy
x 2
(b) Method 1:
@ @
2
2
3
2
Suppose that ð3x y 2y Þ dx þðx 4xy þ 6y Þ dy ¼ d ¼ dx þ dy:
@x @y
@ 2 2 @ 3 2
Then (1) ¼ 3x y 2y ; (2) ¼ x 4xy þ 6y
@x @y
From (1), integrating with respect to x keeping y constant, we have
3 2
¼ x y ¼ 2xy þ FðyÞ