{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern