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Winter 2007 - Takeda's Class - Final Exam

Winter 2007 - Takeda's Class - Final Exam - Math 20C Final...

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Math 20C, Final Exam March 21, 2007 Name : PID : TA : Sec. No : Sec. Time : This exam consists of 11 pages including this front page. Ground Rules 1. No calculator is allowed. 2. Show your work for every problem. A correct answer without any justification will receive no credit. 3. You may use two 4-by-6 index cards, both sides. 4. You have two hours for this exam. Score 1 10 2 10 3 6 4 10 5 12 6 10 7 10 8 10 9 10 10 12 Total 100 1

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1. (a) Let f ( x, y ) = xe y 2 . What is f xy ? f x = e y 2 , and so f xy = 2 ye y 2 . (b) Compute the double integral π 0 x 0 sin y dy dx π 0 x 0 sin y dy dx = π 0 - cos y y = x y =0 dx = π 0 ( - cos x + 1) dx = ( - sin x + x ) π 0 = π. 2
2. (a) Let f ( x, y, z ) = cos x + e x y 2 z . Find a symmetric equation of the line which goes through the origin and which is parallel to f (0 , 2 , - 1) Since f ( x, y, z ) = ( - sin x + e x y 2 z ) i + 2 e x yz j + e x y 2 k . we have f (0 , 2 , - 1) = - 4 i - 4 j + 4 k . Hence any vector which is parallel to - 4 i - 4 j + 4 k can be a direction vector of the line. So let us choose i + j - k . Hence the symmetric equation of the line is x = y = z - 1 . (b) Find the velocity vector of a particle that has the following acceleration and the initial velocity. a ( t ) = k , v (0) = i - j . v ( t ) = a ( t ) dt = t k + C 1 i + C 2 j + C 3 k , where C 1 , C 2 , C 3 are constants. Now v (0) = 0 k + C 1 i + C 2 j + C 3 k = i - j .

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