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Unformatted text preview: Math 3C — Exam #2 Solutions Laney College, Spring 2011 Fred Bourgoin 1. A contour diagram for z = f ( x,y ) is shown below. (12 points) (a) Is z an increasing or decreasing function of x ? Of y ? Solution. z is an increasing function of x because, as you move in the positive x-direction (to the right), the function increases. z is a decreasing function of y . (b) Is f x positive or negative? How about f y ? Solution. f x ( x,y ) > 0 and f y ( x,y ) < ) for the same reasons as above. (c) Is f xx positive or negative? How about f yy ? Solution. f xx ( x,y ) > 0 because f is increasing faster as x increases ( f is concave up in the x-direction). f yy ( x,y ) > 0 because f is decreasing faster as y increases ( f is concave down in the y-direction). (d) Sketch the direction of vector ∇ f at points P and Q . Solution. See the diagram. vector ∇ f should be perpendicular to the level curve to which it corresponds and pointing in the direction of increasing values of f . (e) Is vector ∇ f longer at P or at Q ? How do you know? Solution vector ∇ f is longer at P because the contours are closer together at P than at Q . 2. Find the points on the hyperboloid x 2 +4 y 2- z 2 = 4 where the tangent plane is parallel to the plane 2 x + 2 y + z = 5. (12 points) Solution. Let ( x,y,z ) be any point on the hyperboloid, which can be thought 1 of as one level surface of g ( x,y,z ) = x 2 + 4 y 2- z 2 . Then the gradient vector ∇ f ( x,y,z ) = 2 x vector i + 8 y vector j- 2 z vector k is perpendicular to the tangent plane at the point ( x,y,z ). We need vector ∇ f ( x,y,z ) to be parallel to 2 vector i + 2 vector j + vector k , a vector normal to the given plane; that is, 2 x = 2 k , 8 y = 2 k , and- 2 z = k for some scalar k . This means that y = 1 4 x and z...
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This note was uploaded on 04/19/2011 for the course MTH 110 taught by Professor Helenius during the Spring '08 term at Grand Valley State University.
- Spring '08