Page 151 - Schaum's Outline of Theory and Problems of Advanced Calculus

142 PARTIAL DERIVATIVES [CHAP. 6

Then
0
f xx ð0; 0Þ¼ lim f x ðh; 0Þ f x ð0; 0Þ ¼ lim ¼ 0
ðcÞ
h!0 h h!0 h
0
f y ð0; kÞ f y ð0; 0Þ
f yy ð0; 0Þ¼ lim ¼ lim ¼ 0
ðdÞ
k!0 k k!0 k
k
f xy ð0; 0Þ¼ lim f x ð0; kÞ f x ð0; 0Þ ¼ lim ¼ 1
ðeÞ
k!0 k k!0 k
h
f yx ð0; 0Þ¼ lim f y ðh; 0Þ f y ð0; 0Þ ¼ lim ¼ 1
ð f Þ
h!0 h h!0 h
Note that f xy 6¼ f yx at ð0; 0Þ. See Problem 6.13.
2
2
@ V @ V
6.42. Show that under the transformation x ¼ cos , y ¼ sin the equation þ ¼ 0 becomes
2 2 @x 2 @y 2
@ V 1 @V 1 @ V
¼ 0.
2
@ 2 @ @ 2
þ
þ
We have
@V @V @ @V @ @V @V @ @V @
ð1Þ ¼ þ ð2Þ ¼ þ
@x @ @x @ @x @y @ @y @ @y
Differentiate x ¼ cos , y ¼ sin with respect to x, remembering that and are functions of x
and y
@ @ @ @
1 ¼ sin þ cos ; 0 ¼ cos þ sin
@x @x @x @x
Solving simultaneously,
@ @ sin
¼ cos ; ¼ ð3Þ
@x @x
Similarly, differentiate with respect to y. Then
@ @ @ @
0 ¼ sin þ cos ; 1 ¼ cos þ sin
@y @y @y @y
Solving simultaneously,
@ @ cos
¼ sin ;
@y @y ¼ ð4Þ
Then from (1) and (2),
@V @V sin @V @V @V cos @V
¼ cos ¼ sin
@x @ @ @y @ @
ð5Þ ð6Þ þ
Hence
2

@ V @ @V @ @V @ @ @V @
@x 2 ¼ @x @x ¼ @ @x @x þ @ @x @x
@ @V sin @V @ @ @V sin @V @

cos cos
@ @ @ @x @ @ @ @x
¼ þ
!
2
2
@ V sin @V sin @ V
¼ cos þ ðcos Þ
@ 2 2 @ @ @
!
2
2
@V @ V cos @V sin @ V sin
þ sin þ cos
@ @ @ @ @ 2