Test3_sol - NAME: Spring 2011, MAC 2313, Test 3 4/7/2011,...

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NAME: Spring 2011, MAC 2313, Test 3 UFID: 4/7/2011, Section: 3124 1. Use the Chain Rule to find ∂z/∂s and ∂z/∂t for z = tan( u/v ), u = 2 s + 3 t , v = 3 s - 2 t . (10 points) Solution . By the Chain Rule, we have ∂z ∂s = ∂z ∂u · ∂u ∂s + ∂z ∂v · ∂v ∂s = sec 2 ± u v ² · 1 v · 2 + sec 2 ± u v ² · - u v 2 · 3 = 2 v - 3 u v 2 · sec 2 ± u v ² = - 13 t (3 s - 2 t ) 2 sec 2 ³ 2 s + 3 t 3 s - 2 t ´ , ∂z ∂t = ∂z ∂u · ∂u ∂t + ∂z ∂v · ∂v ∂t = sec 2 ± u v ² · 1 v · 3 + sec 2 ± u v ² · - u v 2 · ( - 2) = 3 v + 2 u v 2 · sec 2 ± u v ² = 13 s (3 s - 2 t ) 2 sec 2 ³ 2 s + 3 t 3 s - 2 t ´ . 2. Find the directional derivative of the function f ( x,y,z ) = xyz at (3 , 2 , 6) in the direction of v = h- 1 , - 2 , 2 i . (10 points) Solution . First we calculate f x = yz 2 xyz , f y = xz 2 xyz , f z = xy 2 xyz . Then we know that f x (3 , 2 , 6) = 2 · 6 2 36 = 1 , f y (3 , 2 , 6) = 3 · 6 2 36 = 3 2 , f z (3 , 2 , 6) = 3 · 2 2 36 = 1 2 . Also, there is
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Test3_sol - NAME: Spring 2011, MAC 2313, Test 3 4/7/2011,...

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