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test1 rev sol

# test1 rev sol - Math 136 Answers to Review Problems for...

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Math 136, Answers to Review Problems for Test 1 1. We calculate D f using the chain rule and the fact that F ( x, y, f ( x, y )) = 0. We define G ( x, y ) = x y f ( x, y ) and note that D g = 1 0 0 1 ∂f ∂x ∂f ∂y . We know G is differentiable since its coordinate functions are C 1 (the first two coordinate functions are C , and the third, f , is C 1 since F and ( x 0 , y 0 , z 0 ) satisfy the Implicit Function Theorem). Using the chain rule on the equation F ( x, y, f ( x, y )) = 0, we get 0 = D 0 = D ( F ( x, y, f ( x, y ) ) = D ( F G ) = D F · D G where · is matrix multiplication. Now, using the fact that D F ( x, y, z ) = ( ∂F ∂x , ∂F ∂y , ∂F ∂z ) we see (1) 0 = ∂F ∂x + ∂F ∂z ∂f ∂x , ∂F ∂y + ∂f ∂y ∂F ∂z . We can solve (1) for ∂f ∂x and ∂f ∂y and this give ∂f ∂x = - ∂F ∂x ∂F ∂z and ∂f ∂y = - ∂F ∂y ∂F ∂z evaluated at ( x 0 , y 0 , f ( x 0 , y 0 )). Here we used that ∂F ∂z ( x 0 , y 0 , z 0 ) = 0 (and z 0 = f ( x 0 , y 0 )). 2. The function F in this problem satisfies the hypotheses of the Implicit Function Theo- rem, so we can assert there is an open neighborhood U of x 0 and an open neighborhood V of y 0 and a C 1 function f : U V such that F ( x, f ( x )) = F ( x 0 , y 0 ) for all x U . (a) So, as there are an infinite number of points in U , and for each x U , F ( x, f ( x )) = F ( x 0 , y 0 ), there are an infinite number of points that satisfy F ( x, y ) = F ( x 0 , y 0 ). (b) No by (a). (c) This might be proven in class, but if not, here is the proof. Let G be the auxiliary function in the Implicit Function Theorem, G ( x, y ) = ( x, F ( x, y )). In the proof

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