{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

samsoln23x210 - MATH 23 Sample Second Exam Solutions was...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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) .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}