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Using the results of Problem 6.14, we have & ' dz z @z x @z y x y @z dx @z dy lim lim 1 2 dt t!0 t t!0 @x t @y t t t @x dt @y dt since as t ! 0 we have x ! 0; y ! 0; 1 ! 0; 2 ! 0; x dx y dy ! ; ! : t dt t dt
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6.18. If z exy , x t cos t, y t sin t, computer dz=dt at t =2.
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2 2 dz @z dx @z dy y2 exy t sin t cos t 2xyexy t cos t sin t : dt @x dt @y dt
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At t =2; x 0; y =2: Another method.
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 dz   2 =4 =2 0 1 3 =8: dt t =2
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Substitute x and y to obtain z et
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sin2 t cos t
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and then di erentiate.
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6.19. If z f x; y where x  u; v and y u; v , prove that a @z @z @x @z @y ; @u @x @u @y @u b @z @z @x @z @y : @v @x @v @y @v
(a) From Problem 6.14, assuming the di erentiability of f ; ; , we have & ' @z z @z x @z y x y @z @x @z @y lim lim 1 2 @u u!0 u u!0 @x u @y u u u @x @u @y @u (b) The result is proved as in (a) by replacing u by v and letting v ! 0.
CHAP. 6]
PARTIAL DERIVATIVES
6.20. Prove that dz
@z @z dx dy even if x and y are dependent variables. @x @y
1 dx xu du xv dv xw dw 2 dy yu du yv dv yw dw
Suppose x and y depend on three variables u; v; w, for example. Then
Thus,
zx dx zy dy zx xu zy yu du zx xv zy yv dv zx xw zy yw dw zu du zv dv zw dw dz
using obvious generalizations of Problem 6.19.
6.21. If T x3 xy y3 , x  cos , y  sin , nd
@T @T @x @T @ @x @ @y @T @T @x @T @ @x @ @y
(a) @T=@, (b) @T=@.
@y 3x2 y cos  3y2 x sin  @ @y 3x2 y  sin  3y2 x  cos  @
This may also be worked by direct substitution of x and y in T.
6.22. If U z sin y=x where x 3r2 2s, y 4r 2s3 , z 2r2 3s2 , nd
a @U @U @x @U @y @U @z @r @x @r @y @r @z @r &  ' &  y  '  y y 1 y z cos 2 6r z cos 4 sin 4r x x x x x 6ryz y 4z y y 2 cos cos 4r sin x x x x x @U @U @x @U @y @U @z @s @x @s @y @s @z @s &  ' &  y  '  y y 1 y z cos 2 2 z cos 6s2 sin 6s x x x x x 2yz y 6s2 z y y cos 6s sin cos 2 x x x x x
(a) @U=@r;
b @U=@s.
6.23. If x  cos , y  sin , show that
 2  2  2   @V @V @V 1 @V 2 2 . @x @y @  @
1 2
Using the subscript notation for partial derivatives, we have V Vx x Vy y Vx cos  Vy sin  V Vx x Vy y Vx  sin  Vy  cos  Dividing both sides of (2) by , we have 1 V Vx sin  Vy cos    Then from (1) and (3), we have
2 V
1 2 2 2 V Vx cos  Vy sin  2 Vx sin  Vy cos  2 Vx Vy 2
6.24. Show that z f x2 y , where f is di erentiable, satis es x @z=@x 2y @z=@y .
Let x2 y u. Then z f u . Thus
PARTIAL DERIVATIVES @z @z @u f 0 u 2xy; @x @u @x Then x @z f 0 u 2x2 y; @x 2y @z @z @u f 0 u x2 @y @u @y
[CHAP. 6
@z @z @z f 0 u 2x2 y and so x 2y : @y @x @y
Another method: We have Also, Then dz f 0 x2 y d x2 y f 0 x2 y 2xy dx x2 dy : dz @z @z dx dy: @x @y @z x3 f 0 x2 y . @y x @z @z 2y . @x @y
@z 2xy f 0 x2 y ; @x
Elimination of f 0 x2 y yields
6.25. If for all values of the parameter  and for some constant p, F x; y p F x; y identically, where F is assumed di erentiable, prove that x @F=@x y @F=@y pF.
Let x u, y v. Then F u; v p F x; y The derivative with respect to  of the left side of (1) is @F @F @u @F dv @F @F x y @ @u @ @v @ @u @v The derivative with respect to  of the right side of (1) is pp 1 F. @F @F y pp 1 F x @u @v Letting  1 in (2), so that u x; v y, we have x @F=@x y @F=@y pF. Then 2 1
6.26. If F x; y x4 y2 sin 1 y=x, show that x @F=@x y @F=@y 6F.
Since F x; y x 4 y 2 sin 1 y=x 6 x4 y2 sin 1 y=x 6 F x; y , the result follows from Problem 6.25 with p 6. It can of course also be shown by direct di erentiation.
6.27. Prove that Y f x at g x at satis es @2 Y=@t2 a2 @2 Y=@x2 , where f and g are assumed to be at least twice di erentiable and a is any constant.
Let u x at; v x at so that Y f u g v . @Y @Y @u @Y @v a f 0 u ag 0 v ; @t @u @t @v @t Then if f 0 u  df =du, g 0 v  dg=dv, @Y @Y @u @Y @v f 0 u g 0 v @x @x @x @v @x
By further di erentiation, using the notation f 00 u  d 2 f =du2 , g 00 v  d 2 g=dv2 , we have 1 @2 Y @Yt @Yt @u @Yt @v @ @ fa f 0 u a g 0 v g a fa f 0 u a g 0 v g a @t @u @t @v @t @u @v @t2 a2 f 00 u a2 g 00 v @2 Y @Yx @Yx @u @Yx @v @ @ f f 0 u g 0 v g f f 0 u g 0 v g @x @u @x @v @x @u @v @x2 f 00 u g 00 v Then from (1) and (2), @2 Y=@t2 a2 @2 Y=@x2 .
CHAP. 6]
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