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Unformatted text preview: MATH 23 Sample Second Exam Solutions was: April, 2003 NAME : Dodson, B. Section 110 313 (Last, First) 1. ( 10 points ) Sketch the level curves . . . . (directly from homework.) 2. ( 10 points ) Let f ( x, y ) = ye xy . Find each of the following. (directly from homework.) 3. ( 10 points ) Find an equation of the tangent plane to the surface z = x 2 3 y 2 at the point ( 3 , 2 , 3) . For this problem, and the other tangent plane problem, its useful to have the correct formula. Here the surface is a graph, z = f ( x, y ) , for which the normal vector of the tangent plane is given by < f x , f y , 1 > . We compute f x = 2 x, f y = 6 y. In this case, the point on the surface has the form ( a, b, f ( a, b )) , and the partial derivatives are to be evaluated at ( a, b ) = ( 3 , 2) . So the normal is < 6 , 12 , 1 > and a point on the plane is the point of tangency, so an equation for the plane is 6( x + 3) 12( y 2) ( z + 3) = 0 . In the other case (from the other exam problems, #2), for the level surface xyz 2 + y 3 = 3 at the point (2 , 1 , 1) , we have F ( x, y, z ) = c, and the normal is < F x , F y , F z > = < yz 2 , xz 2 + 3 y 2 , 2 xyz > = < 1 , 2 + 3 , 2 > = < 1 , 5 , 2 >, with partials evaluated at (2 , 1 , 1) . Here the equation is ( x 2) + 5( y 1) 2( z + 1) = 0 . 4. ( 15 points ) Let f be the function f ( x, y ) = cos(2 xy ) ....
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 Spring '06
 YUKICH
 Math

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