hutchings-final-sol[1]

hutchings-final-sol[1] - Answers to Hutchings’ 03 Final...

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Unformatted text preview: Answers to Hutchings’ 03 Final Note that you will be expected to provide more detail in your solutions than what is presented below. 1. Let x ( t ) and y ( t ) be the x and y components of r , respectively. Now d dt f ( x ( t ) , y ( t )) = f x ( x ( t ) , y ( t )) x ( t ) + f y ( x ( t ) , y ( t )) y ( t ) Since f x ( x, y ) = ye xy , and f y ( x, y ) = xe xy , d dt f ( x ( t ) , y ( t )) | t =0 = 2 e 1 · 2 (3) + 1 e 1 · 2 (4) = 10 e 2 2. (a) A normal vector to the surface is the gradient of f ( x, y, z ) = x 2 + y 2 + z 2 . ∇ f = h 2 x, 2 y, 2 z i , so a normal at (3 , 2 , 1) is h 6 , 4 , 2 i . (b) A normal vector to the surface at (2 , 1 , 3) is the gradient of f ( x, y, z ) =- x 2 + y 2 + z at the point: h- 4 , 2 , 1 i . Thus, the tangent plane is- 4( x- 2) + 2( y- 1) + 1( z- 3) = 0 3. The only critical point of f is (2 , 0), which lies outside the constraint. We use the method of Lagrange multiplies to check the boundary g ( x, y ) = x 2 + 2 y 2 = 1. From the equation ∇ f = λ ∇ g , and the boundary equation, we get the following three equations to solve: 2 x- 4 = λ (2 x ) (1) 2 y = λ 4 y (2) x 2 + 2 y 2 = 1 (3) From (2), we have that λ = 1 2 , or y = 0. If y = 0, then x = ± 1 by (3). If λ = 1 2 , then x = 4 by (1), so 2 y 2 =- 15, which is a contradiciton. Thus, we only have (1 , 0) and (...
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This note was uploaded on 03/09/2010 for the course PHYSICS 7A taught by Professor Lanzara during the Fall '08 term at Berkeley.

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hutchings-final-sol[1] - Answers to Hutchings’ 03 Final...

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