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math237_a4

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

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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

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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) . In order to obtain the equation of the tangent line to the level curve f ( x, y ) = 3 at the point (2 , 1) we will make use of the orthogonality between f (2 , 1) and this tangent line. There are two ways of getting this equation:
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