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math 21a final solution6

# math 21a final solution6 - SIXTH PRACTICE EXAM Name Math...

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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

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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. Solution: It is incompressible because the divergence is zero but it is not conservative, because the curl is not identically zero. Already the third component is not. 10) T F The vector field vector F ( x , y , z ) = ( x , y , z ) is the curl of a vector field.

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