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Winter 2007 - Takeda's Class - Practice Final Exam

Winter 2007 - Takeda's Class - Practice Final Exam - Math...

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Math 20C, Practice Final Exam Solutions March 18, 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 10 4 10 5 10 6 10 7 10 8 10 9 10 10 10 Total 100 1
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1. (a) Let f ( x, y ) = x y - ln | x + y 2 | . What is f x ? f x = yx y - 1 - 1 x + y 2 . (b) Compute the double integral 1 0 x 0 e y dy dx 1 0 x 0 e y dy dx = 1 0 e y y = x y =0 dx = 1 0 e x - 1 dx = ( e x - x ) 1 0 = e - 1 - 1 = e - 2 . 2
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2. (a) Let f ( x, y ) = xe y . Find the directional derivative of f at (1 , 0) in the direction of u = 3 5 , - 4 5 . First note that f ( x, y ) = e y i + xe y j . Hence D u ( f )(1 , 0) = f (1 , 0) · u = 1 , 1 · 3 5 , - 4 5 = 3 5 - 4 5 = - 1 5 . (b) Let a = 3 i - 1 j + 2 k and b = - i + 3 k . Compute a × b . a × b = i j k 3 - 1 2 - 1 0 3 = - 3 i - 11 j - k . 3
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3. Let S be the tangent plane to the surface defined by z = x 2 - 3 y 3 at (2 , 1 , 1). Find the symmetric equation of the line which is perpendicular to S and goes through the point (1 , 2 , 3). Let f ( x, y ) = x 2 - 3 y 3 . Then f x = 2 x and f y = - 9 y 2 . Then the equa- tion of S is z - 1 = f x (2 , 1)( x - 2) + f y (2 , 1)( y - 1) = 4( x - 2) - 9( y - 1) . Hence the normal vector of S is 4 , - 9 , - 1 . So the symmetric equation of the line is x - 1 4 = y - 2 - 9 = z - 3 - 1 . 4
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4. Compute the double integral 1 0 2 2 x e y 2 dy dx. We need to change the order of integration. 1 0 2 2 x e y 2 dy dx = 2 0 y/ 2 0 e y 2 dxdy = 2 0 xe y 2 x = y/ 2 x =0 dy = 2 0 y 2 e y 2 dy = 1 4 e y 2 2 0 = 1 4 e 4 - 1 4 .
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