S6 - Solution to Set 6 1 Find dz/dt(a z z = 2x = 2y x y dx 1 dy = t1/2 = 2e2t dt 2 dt dz = xt1/2 4ye2t dt z z = cos x y sin xy = x sin xy x y dx dy

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Solution to Set 6 1. Find dz/dt (a) ∂z ∂x =2 x ∂y = 2 y dx dt = 1 2 t - 1 / 2 dy dt e 2 t dz dt = xt - 1 / 2 4 ye 2 t (b) =cos x y sin xy = x sin xy dx dt t dy dt =0 dz dt =(cos x y sin xy )2 t 2. (a) = 1 y 2 = 2 x y 3 ∂u =1 ∂v = 1 = 1 y 2 2 x y 3 = 1 y 2 + 2 x y 3 (b) =4 xye x 2 y x 2 e x 2 y = 1 2 r v u = 1 2 r u v = 1 u 2 xye x 2 y r v u 2 x 2 e x 2 y 1 u 2 xye x 2 y r u v 3. f x = y 2 cos xy 2 f y xy cos xy 2 f x ( 1 π )= π 2 cos π = π 2 f y ( 1 π 2 unit vector along ±u is 1 5 ( ± i 2 ± j ) D a f ( 1 π 1 5 ( π 2 +4) 4. The component of ± f n along the ± i direction is ∂f n /∂x = nf n - 1 ∂f/∂x , and other components are similar (by replacing x with y or z ). Therefore, ± f n = nf n - 1 ( f x ± i + f y ± j + f z ± k nf n - 1 ± f . 5. (a) Let f ( x, y )=sin( πxy ), the graph of the equation is the level curve f ( x, y 3 2 . f x = πy cos( πxy ) f y = πx cos( πxy ) f x ( 1 6 , 2) = 2 π cos π 3 = πf y ( 1 6 , 2) = π 12 A normal vector is π ± i + π 12 ± j . (b) The graph is a level curve the function f ( x, y e x 2 y . f x xye x 2 y f y = x 2 e x 2 y f x (1 , ln 3) = 6 ln 3 f y (1 , ln 3) = 3 A normal vector is 6 ln3 ± i +3 ± j . 1
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6. (a) f x = y 1 f y =1+ x f x (0 , 2) = 1 f y (0 , 2) = 1 A normal to the graph is
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This note was uploaded on 02/27/2010 for the course MATH MATH101 taught by Professor Chan during the Fall '09 term at HKUST.

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S6 - Solution to Set 6 1 Find dz/dt(a z z = 2x = 2y x y dx 1 dy = t1/2 = 2e2t dt 2 dt dz = xt1/2 4ye2t dt z z = cos x y sin xy = x sin xy x y dx dy

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