Spring 2007 - Linshaw's Class - Quiz 3

Spring 2007 - Linshaw's Class - Quiz 3 - 1 − 1 − 1 0...

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Name: PID: TA: Sec. No: Sec. Time: Math 20C. Quiz 3 May 11, 2007 1. Find the directional derivative of f ( x, y ) = 2 x 2 y at the point (3 , 4) in the direction of the vector 4 i 3 j . Solution: We have f ( x, y ) = 4 x y i + x 2 y j , so f (3 , 4) = 24 i + 9 2 j . The unit vector u in the direction of 4 i 3 j is 4 5 i 3 5 j . Hence D u f (3 , 4) = f (3 , 4) · u = (24) p 4 5 P p 9 2 Pp 3 5 P = 33 2 , which may be left unsimpli±ed. 2. Let f ( x, y ) = 3 x x 3 2 y 2 + y 4 . Find all critical points of f , and use the second derivative test to classify each critical point as a local minimum, local maximum, or saddle point. Solution: We have f x ( x, y ) = 3 3 x 2 and f y ( x, y ) = 4 y + 4 y 3 . Setting f x ( x, y ) = 0 yields 3(1 x 2 ) = 0 so x = 1 or x = 1. Setting f y ( x, y ) = 0 yields 4 y (1 y 2 ) = 0, or y = 0, y = 1, or y = 1. Thus the critical points are: (1 , 1) , (1 , 0) , (1 , 1) , (
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Unformatted text preview: 1 , − 1) , ( − 1 , 0) , ( − 1 , 1) . We have f xx ( x, y ) = − 6 x , f xy ( x, y ) = 0, and f yy ( x, y ) = − 4 + 12 y 2 . Hence D ( x, y ) = − 6 x ( − 4 + 12 y 2 ). D (1 , − 1) = − 48, so (1 , − 1) is a saddle point. D (1 , 0) = 24 and f xx (1 , 0) = − 6, so (1 , 0) is a local maximum. D (1 , 1) = − 48, so (1 , 1) is a saddle point. D ( − 1 , − 1) = 48 and f xx ( − 1 , − 1) = 6, so ( − 1 , − 1) is a local minimum. D ( − 1 , 0) = − 24, so ( − 1 , 0) is a saddle point. D ( − 1 , 1) = 48 and f xx ( − 1 , 1) = 6, so ( − 1 , 1) is a local minimum....
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This note was uploaded on 04/29/2008 for the course MATH 20C taught by Professor Helton during the Spring '08 term at UCSD.

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