Page 138 - Schaum's Outline of Theory and Problems of Advanced Calculus
P. 138
CHAP. 6] PARTIAL DERIVATIVES 129
@ @ 3 xy 2 2 xy 2 2 2 2 xy 2
ðx y þ e Þ¼ 3x y þ e y ¼ 3x y þ y e
@x @x
ðaÞ x ¼ ¼
@ @ 3 xy 2 3 xy 2 3 xy 2
ðx y þ e Þ¼ x þ e 2xy ¼ x þ 2xy e
@y @y
ðbÞ y ¼ ¼
2
@ @ @ @ 2 2 xy 2 2 xy 2 2 4 xy 2
ðcÞ xx ¼ 2 ¼ ¼ ð3x y þ y e Þ¼ 6xy þ y ðe y Þ¼ 6xy þ y e
@x @x @x @x
2
@ @ 3 xy 2 @ xy 2 xy @
2
ðx þ 2xy e ðe Þþ e
ðdÞ yy ¼ 2 ¼ Þ¼ 0 þ 2xy ð2xyÞ
@y @y @y @y
2 2 xy
¼ 2xy e xy 2 2xy þ e xy 2 2x ¼ 4x y e 2 þ 2xe xy 2
2
@ @ @ @ 2 2 xy 2 2 2 xy 2 xy 2
ð3x y þ y e Þ¼ 3x þ y e 2xy þ e 2y
ðeÞ xy ¼ ¼ ¼
@y @x @y @x @y
2 xy 2
3 xy
2
¼ 3x þ 2xy e þ 2ye
2
@ @ @ @ 3 xy 2 2 xy 2 2 xy 2
ðx þ 2xy e Þ¼ 3x þ 2xy e y þ e 2y
@x @y @x @y @x
ð f Þ yx ¼ ¼ ¼
2 xy 2
3 xy
2
¼ 3x þ 2xy e þ 2ye
Note that xy ¼ yx in this case. This is because the second partial derivatives exist and are
continuous for all ðx; yÞ in a region r. When this is not true we may have xy 6¼ yx (see Problem 6.41,
for example).
2 2 2 1=2
6.11. Show that Uðx; y; zÞ¼ ðx þ y þ z Þ satisfies Laplace’s partial differential equation
2 2 2
@ U @ U @ U
¼ 0.
þ
þ
@x 2 @y 2 @z 2
We assume here that ðx; y; zÞ 6¼ð0; 0; 0Þ. Then
@U
1 2 2 2 3=2 2 2 2 3=2
2
@x ¼ ðx þ y þ z Þ 2x ¼ xðx þ y þ z Þ
2
@ U @ 2 2 2 3=2 3 2 2 2 5=2 2 2 2 3=2
2
@x 2 ¼ @x ½ xðx þ y þ z Þ ¼ð xÞ½ ðx þ y þ z Þ 2xþðx þ y þ z Þ ð 1Þ
2
2
2
2
2
3x 2 ðx þ y þ z Þ 2x y z 2
2 2 2 5=2 2 2 2 5=2 2 2 2 5=2
¼ ¼
ðx þ y þ z Þ ðx þ y þ z Þ ðx þ y þ z Þ
2
2
2
2
2
2
@ U 2y x z 2 @ U 2z x y 2
Similarly ¼ ; ¼ :
@y 2 2 2 2 5=2 @x 2 2 2 2 5=2
ðx þ y þ z Þ ðx þ y þ z Þ
2
2
2
@ U @ U @ U
Adding, þ þ ¼ 0:
@x 2 @y 2 @z 2
2
y @ z
2 1
6.12. If z ¼ x tan , find at ð1; 1Þ.
x @x @y
@z 2 1 @ 2 x 2 1 x 3
y
2
2
2
@y ¼ x 1 þð y=xÞ @y x ¼ x x þ y 2 x ¼ x þ y 2