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3c-spring2011-exam_2_solutions

3c-spring2011-exam_2_solutions - Math 3C Exam#2 Solutions...

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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
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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 = - 1
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