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Unformatted text preview: Math 263 Test 1 Solutions 1. (a) Find the vector equation for the line tangent to the curve r ( t ) = h t,e 1- t , ln t i at the point r (1). (b) Consider the function f ( x,y ) = xe xy . Find an equation for the tangent plane of the surface z = f ( x,y ) at the point (1 , , 1). (c) Find the point P of intersection of the line l ( t ) = h 1 , 1 , i + t h 1 2 ,- 1 , 1 i and the plane E given by the equation x + y- z = 0 . Solution: (a) First of all we note that r (1) = h 1 , 1 , i . Next we calculate r ( t ) = 1 2 t i- e 1- t j + 1 t k and therefore we get r (1) = h 1 2 ,- 1 , 1 i . Combining all this we get that the equation for the tangent line is given by l ( t ) = h 1 , 1 , i + t h 1 2 ,- 1 , 1 i . (b) The partial derivatives of f are given by f x ( x,y ) = e xy (1 + xy ) and f y ( x,y ) = x 2 e xy . Hence we have f x (1 , 0) = 1 and f y (1 , 0) = 1 . Therefore the equation for the tangent plane is z- 1 = f x (1 , 0)( x- 1) + f y (1 , 0)( y- 0) = x- 1 + y or x + y- z = 0 ....
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This note was uploaded on 02/03/2010 for the course MATH MATH 263 taught by Professor Tobiaslahm during the Fall '09 term at The University of British Columbia.
- Fall '09