math237_a4 - SOLUTIONS FOR ASSIGNMENT 4 Problem Set 3, A6...

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Unformatted text preview: SOLUTIONS FOR ASSIGNMENT 4 Problem Set 3, A6 iii): If F ( x, y, z ) = f parenleftBigg y z x , z x y , x y z parenrightBigg , show that x F x + y F y + z F z = 0 . What assumption do you need to make about f ? Assuming that F has continuous partial derivatives, the Chain Rule gives: F x = D 1 f x parenleftbigg y z x parenrightbigg + D 2 f x parenleftBigg z x y parenrightBigg + D 3 f x parenleftbigg x y z parenrightbigg = y z x 2 D 1 f 1 y D 2 f + 1 z D 3 f, (1) F y = D 1 f y parenleftbigg y z x parenrightbigg + D 2 f y parenleftBigg z x y parenrightBigg + D 3 f y parenleftbigg x y z parenrightbigg = 1 x D 1 f z x y 2 D 2 f 1 z D 3 f, (2) F z = D 1 f z parenleftbigg y z x parenrightbigg + D 2 f z parenleftBigg z x y parenrightBigg + D 3 f z parenleftbigg x y z parenrightbigg = 1 x D 1 f + 1 y D 2 f x y z 2 D 3 f. (3) By relations (1), (2) and (3) we get: x F x + y F y + z F z = parenleftbigg y z x + y x z x parenrightbigg D 1 f + parenleftBigg x y z x y + z y parenrightBigg D 2 f + parenleftbigg x z y z x y z parenrightbigg D 3 f = 0 , i.e. exactly what we were asked to show. 1 Problem Set 3, A12: Let f ( x, y ) = 2 xy y 2 . Use the gradient vector to find the equation of the tangent line of the curve f ( x, y ) = 3 at the point (2 , 1) . Sketch the curve and the tangent line. Since f ( x, y ) = 2 xy y 2 , then f ( x, y ) = (2 y, 2 x 2 y ) and hence f (2 , 1) = (2 , 2) ....
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This note was uploaded on 07/28/2009 for the course MATH MATH137 taught by Professor Oancea during the Spring '08 term at Waterloo.

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math237_a4 - SOLUTIONS FOR ASSIGNMENT 4 Problem Set 3, A6...

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