math2E practice midterm1 with solution

math2E practice midterm1 with solution - MIDTERM SOLUTIONS...

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Unformatted text preview: MIDTERM: SOLUTIONS FOR PRACTICE PROBLEMS Section 13.5 21. F ( x, y, z ) = 3 x 2 z + 2 z 3 − 3 yz = 0. Then F x = 6 xz, F y = − 3 z, F z = 3 x 2 + 6 z 2 − 3 y, and therefore, ∂z ∂x = − F x F z = − 2 xz x 2 + 2 z 2 − y , ∂z ∂x = − F y F z = z x 2 + 2 z 2 − y . 25. Note that x θ = − r sin θ , and y θ = r cos θ . Therefore, by the Chain Rule, f θ = f x x θ + f y y θ = − f x r sin θ + f y r cos θ. Section 13.6 19. f ( x, y ) = x 3 yz 2 − 4 xy . Then ∇ f ( x, y, z ) = ( 3 x 2 yz 2 − 4 y, x 3 z 2 − 4 x, 2 x 3 yz ) , and ∇ f (1 , − 1 , 2) = (− 8 , , − 4 ) . We have to find the derivative of f at (1 , − 1 , 2) in the direction of v = ( 2 , , − 1 ) . The corresponding unit vector is u = v / bardbl v bardbl = 1 radicalbig 2 2 + 0 2 + ( − 1) 2 ( 3 , − 2 ) = ( 2 / √ 5 , , − 1 / √ 5 ) . Then D u f (1 , − 1 , 2) = ∇ f (1 , − 1 , 2) · u = (− 8 , , − 4 ) · ( 2 / √ 5 , , − 1 / √ 5 ) = − 12 √ 5 . 41. We need to find find the equations of the tangent plane and the normal line to the level surface F ( x, y, z ) = x 2 + y 3 − z = 0 at the point (1 , − 1 , 0). Then ∇ F ( x, y, z ) = ( 2 x, 3 y 2 , − 1 ) , hence n = ∇ F (1 , − 1 , 0) = ( 2 , 3 , − 1 ) is normal to our surface at the point (1 , − 1 , 0). The normal line to the surface passes through (1 , − 1 , 0) in the direction of n...
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This note was uploaded on 01/27/2010 for the course MATH 2E taught by Professor Wong during the Winter '08 term at UC Irvine.

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math2E practice midterm1 with solution - MIDTERM SOLUTIONS...

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