123 Sample Exam 2 Solns

123 Sample Exam 2 Solns - MATH 23 Sample Second Exam...

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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, it’s 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 ) . (a) Find the directional derivative of f at the point (2 , π 8 ) in the direction of the vector a = < - 1 , 3 > = - i + 3 j. The gradient is < f x , f y > = < - 2 y sin(2 xy ) , - 2 x sin(2 xy ) >, which is < - π 4 , - 4 > . Notice | a | = 10 , so the unit vector is u = 1 10 < - 1 , 3 > .
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