math 21a final solution6

math 21a final solution6 - 1 17 2007 SIXTH PRACTICE EXAM...

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Unformatted text preview: 1 / 17 / 2007, SIXTH PRACTICE EXAM 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 reasoning can not be given credit except for the True / False and multiple choice problems. • Please write neatly. • Do not use notes, books, calculators, computers, 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 Problem 1) True / False questions (20 points). No justifications are needed. 1) T F The angle between the vectors ( 1 , 2 , 3 ) and ( 2 , − 1 , − 1 ) is π 6 . Solution: The dot product is negative. It is an obtuse angle and can not be 30 ◦ . 2) T F ( 4 , 6 , 8 ) is a normal vector for the plane − 2 x − 3 y − 4 z = 5. Solution: Yes, it is parallel to the gradient vector (− 2 , − 3 , − 4 ) of the plane. 3) T F The plane tangent to the graph of f ( x , y ) = x 2 + y 2 at (3 , 4 , 25) is 6 x + 8 y = 50. Solution: The equation describes a cylinder, not a plane. We are in space, not in the plane. 4) T F The directional derivative of a function f in the direction of ∇ f can never be negative. Solution: D ∇ f f = ∇ f · ∇ f ≥ 0. 5) T F The surface parameterized by ( sin u , cos v , u 2 + v 2 ) , 0 ≤ u , v ≤ 1 has the same surface area as the surface parameterized by ( sin u 2 , cos v 2 , u 4 + v 4 ) , 0 ≤ u , v ≤ 1. Solution: Yes, surface area does not depend on the parametrization. 6) T F By the chain rule, integraltext b a ∇ f ( vector r ( t )) · vector r ′ ( t ) dt = integraltext b a d dt f ( vector r ( t )) dt = f ( vector r ( b )) − f ( vector r ( a )). Solution: Yes, this is the proof of the fundamental theorem of line integrals. 7) T F The function u ( x , y ) = x 2 + y 2 is a solution of the wave equation u xx = u yy . Solution: Yes, u xx = 2 and u yy = 2. 8) T F Let C be the unit circle parametrized counter-clockwise. If vector F ( x , y ) is a vector field and integraltext C vector F · d vector r = 0, then vector F is a gradient vector field. Solution: It is not su ffi cient to check the closed loop property for one closed path only. It has to be true for all closed paths. 9) T F The vector field vector F ( x , y , z ) = ( y 2 − z 2 , z 2 − x 2 , x 2 − y 2 ) is conservative....
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math 21a final solution6 - 1 17 2007 SIXTH PRACTICE EXAM...

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