Page 152 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 6] PARTIAL DERIVATIVES 143

which simplifies to
2
2
2
2
2
2
@ V 2 @ V 2 sin cos @V 2 sin cos @ V sin @V sin @ V
¼ cos
@x 2 @ 2 þ 2 @ @ @ þ @ þ 2 @ 2 ð7Þ
Similarly,
2
2
2
2
2
@ V ¼ sin 2 2 sin cos @V 2 sin cos @ V cos @V cos @ V
2 @ V
@y 2 @ 2 2 @ þ @ @ þ @ þ 2 @ 2 ð8Þ
2
2
2
2
@ V @ V @ V 1 @V 1 @ V
Adding ð7Þ and ð8Þ we find, as required, þ ¼ þ þ ¼ 0:
2
@x 2 @y 2 @ 2 @ @ 2
6.43. (a)If x ¼ f ðu; vÞ and y ¼ gðu; vÞ, where u ¼ ðr; sÞ and v ¼ ðr; sÞ, prove that @ðx; yÞ ¼
@ðr; sÞ
.
@ðx; yÞ @ðu; vÞ
@ðu; vÞ @ðr; sÞ
(b) Prove that @ðx; yÞ @ðu; vÞ ¼ 1 provided @ðx; yÞ 6¼ 0, and interpret geometrically.
@ðu; vÞ @ðx; yÞ @ðu; vÞ

x r x s x u u r þ x v v r x u u s þ x v v s
@ðx; yÞ

ðaÞ ¼ ¼

y r y s y u u r þ y v v r y u u s þ y v v s
@ðr; sÞ

x u x v u r u s

@ðx; yÞ @ðu; vÞ

¼ ¼

y u y v v r v s @ðu; vÞ @ðr; sÞ
using a theorem on multiplication of determinants (see Problem 6.108). We have assumed here, of
course, the existence of the partial derivatives involved.
Place r ¼ x; s ¼ y in the result of ðaÞ: Then @ðx; yÞ @ðu; vÞ @ðx; yÞ ¼ 1:
ðbÞ ¼
@ðu; vÞ @ðx; yÞ @ðx; yÞ
The equations x ¼ f ðu; vÞ, y ¼ gðu; vÞ defines a transformation between points ðx; yÞ in the xy plane
and points ðu; vÞ in the uv plane. The inverse transformation is given by u ¼ ðx; yÞ, v ¼ ðx; yÞ. The
result obtained states that the Jacobians of these transformations are reciprocals of each other.
6.44. Show that Fðxy; z 2xÞ¼ 0 satisfies under suitable conditions the equation
xð@z=@xÞ yð@z=@yÞ¼ 2x.What are these conditions?
Let u ¼ xy, v ¼ z 2x. Then Fðu; vÞ¼ 0 and
dF ¼ F u du þ F v dv ¼ F u ðxdy þ ydxÞþ F v ðdz 2 dxÞ¼ 0
ð1Þ
Taking z as dependent variable and x and y as independent variables, we have dz ¼ z x dx þ z y dy. Then
substituting in (1), we find
ð yF u þ F v z x 2Þ dx þðxF u þ F v z y Þ dy ¼ 0
Hence, we have, since x and y are independent,
yF u þ F v z x 2 ¼ 0 xF u þ F v z y ¼ 0
ð2Þ ð3Þ
Solve for F u in (3) and substitute in (2). Then we obtain the required result xz x yz y ¼ 2x upon dividing by
F v (supposed not equal to zero).
The result will certainly be valid if we assume that Fðu; vÞ is continuously differentiable and that F v 6¼ 0.