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P2_soln

# P2_soln - Comments:.2.11 Problem1(a (x,y,z G = 4 x 2 4 y 2...

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Prelim II – Solutions – Spring 2009 – Math 1920 Comments: The solution to problem 3 is directly from the homework solutions for 16.2.11 Problem 1 (a) – The normal to the hyperboloid is found by taking the gradient of G(x,y,z) G = (4 x 2 + 4 y 2 4 z 2 ) =< 8 x ,8 y , 2 z > At the point P(1,2,4) , G =< 8,16, 8 > The equation of the plane is G ( P P 0 ) = 0 So, 8( x 1) + 16( y 2) 8( z 4) = 0 (b) – The direction of max increase in z at the point P(1,2,4) is found by taking grad( z ). From our equation and the point, we find that z = 4 x 2 + 4 y 2 4 Therefore, z = 4 x 4 x 2 + 4 y 2 4 , 4 y 4 x 2 + 4 y 2 4 z P (1,2,4) = 1,2 (c) – The change in z at point P(1,2,4) is found by taking the directional derivative. D u z = z ˆ u = 1,2 0,1 = 2

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Problem 16.1.26 v = < 1 , 2 , t > ⇒ | v | = 5 + t M = R 2 0 3 5 + t 5 + t dt = 15 t + 3 2 t 2 2 0 = 36 M yz = R 2 0 3 t (5 + t ) dt = 15 2 t 2 + t 3 2 0 = 38 M xz = R 2 0 6 t (5 + t ) dt = 2 15 2 t 2 + t 3 2 0 = 76 M xy = R 2 0 2 t 3 / 2 (5 + t ) dt = 2 2 t 5 / 2 + 2 7 t 7 / 2 2 0 = 144 7 2 ¯ x = M yz M = 38 36 = 19 18 ¯ y = M xz M = 76 36 = 19 9 ¯ z = M xy M = 144 7 2 36 = 4 2 7 Problem 16.2.11 a) C : r = < t, t, t > v = < 1 , 1 , 1 > F = < 3 t 2 - 3 t, 3 t, 1 > W = R 1 0 F · v dt = R 1 0 3 t 2 + 1 dt = t 3 + t 1 0 = 2 b) C : r = < t, t 2 , t 4 > v = < 1 , 2 t, 4 t 3 > F = < 3 t 2 - 3 t, 3 t 4 , 1 > W = R 1 0 F · v dt = R 1 0 3 t 2 - 3 t + 6 t 5 + 4 t 3 dt = t 3 - 3 2 t 2 + t 6 + t 4 1 0 = 3 2 c) C 1 : r 1 =

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P2_soln - Comments:.2.11 Problem1(a (x,y,z G = 4 x 2 4 y 2...

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