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math 21a final solution5

# math 21a final solution5 - FIFTH PRACTICE FINAL Name Math...

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1/17/2007, FIFTH PRACTICE FINAL Math 21a, Fall 2007 Name: MWF 9 Chen-Yu Chi MWF 10 Oliver Knill MWF 10 Corina Tarnita MWF 11 Veronique Godin MWF 11 Stefan Hornet MWF 11 Jay Pottharst MWF 12 Chen-Yu Chi MWF 12 Ming-Tao Chuan TTH 10 Thomas Barnet-Lamb TTH 10 Rehana Patel TTH 11:30 Thomas Barnet-Lamb TTH 11:30 Thomas Lam Please mark the box to the left which lists your section. Do not detach pages from this exam packet or unstaple the packet. Show your work. Answers without reason- ing can not be given credit except for the True/False and multiple choice problems. Please write neatly. Do not use notes, books, calculators, comput- ers, or other electronic aids. Unspecified functions are assumed to be smooth and defined everywhere unless stated otherwise. You have 180 minutes time to complete your work. 1 20 2 10 3 10 4 10 5 10 6 10 7 10 8 10 9 10 10 10 11 10 12 10 13 10 14 10 Total: 150

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Problem 1) True/False questions (20 points) 1) T F For any two vectors vectorv and vectorw one has proj vectorv ( vectorv × vectorw ) = vector 0. Solution: The projection of a vector vectorv onto a vector vectorw which is perpendicular to vectorv is zero. 2) T F Any parameterized surface S is either the graph of a function f ( x, y ) or a surface of revolution. Solution: A counter example is an asymetric ellipsoid. It is neither a graph, nor a surface of revolution. 3) T F If the directional derivative D vectorv ( f ) of f into the direction of a unit vector vectorv is zero, then vectorv is perpendicular to the level curve of f . Solution: It is either tangent to the level curve or at a critical point. 4) T F The linearization L ( x, y ) of f ( x, y ) = 5 x 100 y at (0 , 0) satisfies L ( x, y ) = 5 x 100 y . Solution: The linearization of any linear function at (0 , 0) is the function itself. 5) T F If a parameterized curve vector r ( t ) intersects a surface { f = c } at a right angle, then at the point of intersection we have f ( vector r ( t )) × vector r ( t ) = 0. Solution: This is clear, once you know what the question means. The conditon f ( vector r ( t )) × vector r ( t ) = 0 means that the velocity vector vector r ( t ) is parallel to the gradient vector, which means that it is perpendicular to the level surface. 6) T F The curvature of the curve vector r ( t ) = ( cos(3 t 2 ) , sin(6 t 2 ) ) at the point vector r (1) is larger than the curvature of the curve vector r ( t ) = ( 2 cos(3 t ) , 2 sin(6 t ) ) at the point vector r (1).
Solution: While curvature is independent of the parameterization of the curve, the two circles have also a different radius. The second curve has twice the radius, so half the curvature. 7) T F At every point ( x, y, z ) on the hyperboloid x 2 y 2 + z 2 = 10, the vector ( x, y, z ) is normal to the hyperboloid. Solution: Indeed, it is the gradient of the level surface. And the gradient is perpendicular to the level surface.

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math 21a final solution5 - FIFTH PRACTICE FINAL Name Math...

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