16----yellowexam2sol[1]

16----yellowexam2sol[1] - yellow MATH 32A Exam 2 November...

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Sheet1 Page 1 yellow MATH 32A Exam 2 November 29, 2007 LASTNAME FIRST NAME IDNO. YourTA: To receive credit, you must write your answer in the space provided. DO NOT WRITE BELOW THIS LINE 1 (25 pts) 5 (25 pts) 2 (25 pts) 6 (25 pts) 3 (25 pts) 7 (25 pts) 4 (25 pts) 8 (25 pts) TOTAL T PROBLEM 1 (25 Points) A particle travels counterclockwise around the circle of rad ius 12 centered at the origin in the xy-plane. At a certain moment, theparticleislocated at(12,0). At this moment, Answer: Solution: The acceleration vector has the form a = v 0T+ v2N The curvature of a circle of radius 12 is =1/12. At the location (12,0), the unit tangent vector is vertical and the unit normal vector points along the negative x-axis: T = h0,1i, N = h-1,0i It is given that 2 0 2 v =4 m/s,v =5 m/s Therefore, 1 14 a =5T+42N =5h0,1i+42h-1,0i = h- ,5i 12 12 3 1 PROBLEM 2 (25 Points) the particlehs speed is 4 m/s and is increasing at a rate of 5 m/s2 . What is the particlehs acceleration vector?
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Sheet1 Page 2 Calculate the curvature of r(t)= ht2,t-1,ti at t =1. Answer: Solution: Use the formula ||r00(t) r0(t)|| = ||r0(t)||3 We have -2 r 0(t)= h2t -t,1i 00(t)-3 r = h2,2t,0i 00(t) -2k rr 0(t)= -2t-3i+2j+6t At t =1 we have r 0 = h2,-1,1i r r = -2i+2j+6k and v p ||-2i+2j+6k|| 4+4+36 2 11 = Err:520 63/2 ||h2,-1,1i||3 (4+1+1)3/2 | PROBLEM 3 (25 Points) A particle moves along the path described by 31 r(t)= h1t, t2i
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This note was uploaded on 04/21/2008 for the course MATH 31A taught by Professor Jonathanrogawski during the Fall '07 term at UCLA.

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16----yellowexam2sol[1] - yellow MATH 32A Exam 2 November...

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