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Quiz5_sol - NAME Spring 2011 MAC 2313 Quiz 5 UFID Section...

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NAME: Spring 2011, MAC 2313, Quiz 5 UFID: Section: 3124 1. Find the maximum rate of change of f ( x, y, z ) = tan( x +2 y +3 z ) at point ( - 5 , 1 , 1) and the direction in which it occurs. (2 points) Solution . First we find the partial derivatives of f ( x, y, z ): f x ( x, y, z ) = 1 cos 2 ( x + 2 y + 3 z ) , f y ( x, y, z ) = 2 cos 2 ( x + 2 y + 3 z ) , f z ( x, y, z ) = 3 cos 2 ( x + 2 y + 3 z ) . (1) Hence the gradient of f at point ( - 5 , 1 , 1) is f ( - 5 , 1 , 1) = h f x ( - 5 , 1 , 1) , f y ( - 5 , 1 , 1) , f z ( - 5 , 1 , 1) i = h 1 , 2 , 3 i . (2) The maximal rate of change is therefore k∇ f ( - 5 , 1 , 1) k = 1 2 + 2 2 + 3 2 = 14, and the direction at which the maximal rate of change occurs is v = f ( - 5 , 1 , 1) k∇ f ( - 5 , 1 , 1) k = 1 14 h 1 , 2 , 3 i . (3) 2. Find all points at which the direction of fastest change of the function f ( x, y ) = x 2 + y 2 - 2 x - 4 y is i + j . (2 points) Solution . The partial derivatives of f ( x, y ) are f x ( x, y ) = 2 x - 2 , f y ( x, y ) = 2 y - 4 . (4) A point that has the fastest change in the direction i + j must have a parallel gradient. Namely, f = h f x , f y i is a multiple of i + j . Hence, f x = f y , i.e. 2 x - 2 = 2 y - 4 . (5) Therefore, all points on y = x
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