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Exam_solutions_2_ - MATH 241, 2nd Examination Solutions and...

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MATH 241, 2nd Examination Solutions and Grading Key Prof. Jonathan Rosenberg Friday, October 28, 2005 1. (25 points, divided as indicated) (a) (15 points) Find the tangent plane to the surface x sin y cos z + e xy + z 2 = 1 at the point (1 , 0 , 0) . Solution. Let f ( x, y, z ) = x sin y cos z + e xy + z 2 ; then f = (sin y cos z + ye xy , x cos y cos z + xe xy , x sin y sin z + 2 z ) . Thus f (1 , 0 , 0) = (0 , 2 , 0) , and this must be normal to the level surface of f through the point (1 , 0 , 0) . So the equation of the tangent plane is (0 , 2 , 0) · ( x 1 , y, z ) = 0 or 2 y = 0 or y = 0 . (b) (10 points) Is it possible to locally solve for z as a smooth function of x and y (in a neighborhood of (1 , 0) ), and if so, what are ∂z/∂x and ∂z/∂y when x = 1 and y = 0 ? Solution. No, it is not possible to locally solve for z as a smooth function of x and y (in a neighborhood of (1 , 0) ), since the tangent plane y = 0 is perpendicular to the x - y plane. (Or to phrase it another way, there are no points in the tangent plane over points with
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This note was uploaded on 09/09/2009 for the course MATH 241 taught by Professor Wolfe during the Spring '08 term at Maryland.

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Exam_solutions_2_ - MATH 241, 2nd Examination Solutions and...

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