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which simpli es to @2 V @2 V 2 sin  cos  @V 2 sin  cos  @2 V sin2  @V sin2  @2 V cos2  2 2 @  @ @  @ @x2 @ 2  @2 Similarly, @2 V @2 V 2 sin  cos  @V 2 sin  cos  @2 V cos2  @V cos2  @2 V sin2  2 2 2 @  @ @  @ @y @  2 @2 @2 V @2 V @2 V 1 @V 1 @2 V 2 2 2 0: 2  @  @2 @x @y @ 8 7
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Adding 7 and 8 we find, as required,
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@ x; y 6.43. (a) If x f u; v and y g u; v , where u  r; s and v r; s , prove that @ r; s @ x; y @ u; v . @ u; v @ r; s (b) Prove that
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 @ x; y  xr  @ r; s  yr
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@ x; y @ u; v @ x; y 1 provided 6 0, and interpret geometrically. @ u; v @ x; y @ u; v
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  xs   xu ur xv vr   ys   yu ur yv vr    xu xv  ur    y y  v
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u v r
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 xu us xv vs   yu us yv vs   us  @ x; y @ u; v  vs  @ u; v @ r; s
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using a theorem on multiplication of determinants (see Problem 6.108). course, the existence of the partial derivatives involved. b Place r x; s y in the result of a : Then @ x; y @ u; v @ x; y 1: @ u; v @ x; y @ x; y
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We have assumed here, of
The equations x f u; v , y g u; v de nes a transformation between points x; y in the xy plane and points u; v in the uv plane. The inverse transformation is given by u  x; y , v x; y . The result obtained states that the Jacobians of these transformations are reciprocals of each other.
6.44. Show that F xy; z 2x 0 satis es under suitable x @z=@x y @z=@y 2x. What are these conditions
Let u xy, v z 2x. 1 Then F u; v 0 and
conditions
equation
dF Fu du Fv dv Fu x dy y dx Fv dz 2 dx 0
Taking z as dependent variable and x and y as independent variables, we have dz zx dx zy dy. Then substituting in (1), we nd yFu Fv zx 2 dx xFu Fv zy dy 0 Hence, we have, since x and y are independent, 2 yFu Fv zx 2 0 3 xFu Fv zy 0
Solve for Fu in (3) and substitute in (2). Then we obtain the required result xzx yzy 2x upon dividing by Fv (supposed not equal to zero). The result will certainly be valid if we assume that F u; v is continuously di erentiable and that Fv 6 0.
PARTIAL DERIVATIVES
[CHAP. 6
Supplementary Problems
FUNCTIONS AND GRAPHS 6.45. If f x; y 2x y , nd 1 xy b (a) f 1; 3 ; c b f 2 h; 3 f 2; 3 ; h c f x y; xy .
Ans. a 1 ; 4 6.46.
11 ; 5 3h 5
2x 2y xy 1 x2 y xy2
If g x; y; z x2 yz 3xy, nd (a) g 1; 2; 2 ; b g x 1; y 1; z2 ; c g xy; xz; x y . Ans. a 1; b x2 x 2 yz2 z2 3xy 3y; c x2 y2 x2 z xyz 3x2 yz Give the domain of de nition for which each of the following functional rules are de ned and real, and indicate this domain graphically.   1 2x y : a f x; y 2 ; b f x; y ln x y ; c f x; y sin 1 2 1 x y x y Ans: a x2 y2 6 1; b x y > 0;   2x y  c  x y@1
s x y z 1 is de ned and real (b) Indi(a) What is the domain of de nition for which f x; y; z x2 y2 z2 1 cate this domain graphically. Ans. (a) x y z @ 1; x2 y2 z2 < 1 and x y z A 1; x2 y2 z2 > 1
three-dimensional space represented by each of the following. g x2 y2 2y; x2 z2 y2 ; 2 2 2 x y z 16; h z x y; x2 4y2 4z2 36; i y2 4z; j x2 y2 z2 4x 6y 2z 2 0: Ans. a plane, (b) paraboloid of revolution, (c) hyperbolic paraboloid, (d) right circular cone, (e) sphere, ( f ) hyperboloid of two sheets, (g) right circular cylinder, (h) plane, (i) parabolic cylinder, ( j) sphere, center at 2; 3; 1 and radius 4. Construct a graph of the region bounded by x2 y2 a2 and x2 z2 a2 , where a is a constant. Describe graphically the set of points x; y; z such that: (a) x2 y2 z2 1; x2 y2 z2 ; b x2 y2 < z < x y. The level curves for a function z f x; y are curves in the xy plane de ned by f x; y c, where c is any constant. They provide a way of representing the function graphically. Similarly, the level surfaces of w f x; y; z are the surfaces in a rectangular xyz) coordinate system de ned by f x; y; z c, where c is any constant. Describe and graph the level curves and surfaces for each of the following functions: (a) f x; y ln x2 y2 1 ; b f x; y 4xy; c f x; y tan 1 y= x 1 ; d f x; y x2=3 2=3 2 2 2 y ; e f x; y; z x 4y 16z ; f sin x z = 1 y :
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