Winter 2007 - Takeda's Class - Exam 2 (Version 1)

# Winter 2007 - Takeda's Class - Exam 2 (Version 1) - Math...

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Math 20C, Midterm 2 Solutions February 26, 2007 Name : PID : TA : Sec. No : Sec. Time : This exam consists of 6 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 one 4-by-6 index card, both sides. Score 1 10 2 10 3 10 4 10 5 10 Total 50 1

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1. (a) Let f ( x, y ) = y 2 cos( xy ) + 2 - ln y . Find f x . f x = - y 3 sin( xy ) (b) For z = y x 2 , sketch several level curves. 2
2. (a) Find the arc length of the curve r ( t ) = 2 sin t i +5 t j +2 cos t k for 0 t 1 Since r ( t ) = 2 cos t i + 5 j - 2 sin t k , we have | r ( t ) | = 4 cos 2 t + 25 + 4 sin 2 t = 29 . So the arc length is 1 0 29 dt = 29 . (b) Consider a particle moving in space with the following position vector. x ( t ) = e t i + 2 t j + t 2 k . Find the acceleration a ( t ). a ( t ) = x ( t ) = e t i + 2 k . 3

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3. Show that the following limit does not exist. lim ( x,y ) (0 , 0) sin 2 x x 2 + y Along the path x = 0, the limit is lim ( x,y ) (0 , 0) sin 2 x x 2 + y = lim y 0 0 0 + y = 0 . But along the path y = 0, the limit is lim ( x,y ) (0 , 0) sin 2 x x 2 + y = lim x 0 sin 2 x x 2 = lim x 0 sin x x 2 = 1 . So the two limits do not match up and hence the limit does not exist. 4
4. Let f ( x, y ) = x 2 +sin( xy ). Find all unit vectors u = a, b so that D u f (1 , 0) = 1, i.e. the directional derivative of

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