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s6 - Solution to Set 6 1 Find dz/dt(a z z = 2x = 2y x y dx...

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Solution to Set 6 1. Find dz/dt (a) ∂z ∂x = 2 x ∂z ∂y = 2 y dx dt = 1 2 t - 1 / 2 dy dt = 2 e 2 t dz dt = xt - 1 / 2 4 ye 2 t (b) ∂z ∂x = cos x y sin xy ∂z ∂y = x sin xy dx dt = 2 t dy dt = 0 dz dt = (cos x y sin xy ) 2 t 2. (a) ∂z ∂x = 1 y 2 ∂z ∂y = 2 x y 3 ∂x ∂u = 1 ∂x ∂v = 1 ∂y ∂u = 1 ∂y ∂v = 1 ∂z ∂u = 1 y 2 2 x y 3 ∂z ∂v = 1 y 2 + 2 x y 3 (b) ∂z ∂x = 4 xye x 2 y ∂z ∂y = 2 x 2 e x 2 y ∂x ∂u = 1 2 v u ∂x ∂v = 1 2 u v ∂y ∂u = 1 u 2 ∂y ∂v = 0 ∂z ∂u = 2 xye x 2 y v u 2 x 2 e x 2 y 1 u 2 ∂z ∂v = 2 xye x 2 y u v 3. f x = y 2 cos xy 2 f y = 2 xy cos xy 2 f x ( 1 π , π ) = π 2 cos π = π 2 f y ( 1 π , π ) = 2 unit vector along u is 1 5 ( i 2 j ) D a f ( 1 π , 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 = π f y ( 1 6 , 2) = π 12 A normal vector is πi + π
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