Spring 2006 - Gross' Class - Exam 2

Spring 2006 - Gross' Class - Exam 2 - Midterm #2 Solutions...

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Midterm #2 Solutions - Green Version Professor Gross Spring 2006 May 22, 2006 1. (a) If r ( t ) = cos(3 t ) i + ln(3 t + 1) j + e cos t k , what is r 0 ( t )? r 0 ( t ) = - 3 sin(3 t ) i + 3 3 t + 1 j + ( - sin t ) e cos t k (b) Reparameterize the curve r ( t ) = 3 sin t i + 4 t j + 3 cos t k with respect to arc length measured from the point where t = 0 in the direction of increasing t . First, we will compute r 0 ( t ): r 0 ( t ) = 3 cos t i + 4 j - 3 sin t k . Now, we use the formula for arc length: s ( t ) = Z t 0 | r 0 ( u ) | d u = Z t 0 | 3 cos u i + 4 j - 3 sin u k | d u = Z t 0 p (3 cos u ) 2 + 4 2 + (3 sin u ) 2 d u = Z t 0 p 9 cos 2 u + 16 + 9 sin 2 u d u = Z t 0 q 9(sin 2 u + cos 2 u ) + 16 d u = Z t 0 9 + 16 d u = Z t 0 9 + 16 d u = Z t 0 25 d u = Z t 0 5 d u = 5 | t 0 = 5 t So s = 5 t , which, solving for t gives us t ( s ) = s/ 5, so we plug this back in to the original equation for r ( t ), and get r ( t ( s )) = 3 sin s 5 i + 4 s 5 j + 3 cos s 5 k . 1
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2. (a) Find the velocity and position of a particle that has the given acceleration and the given initial velocity and positions: a ( t ) = 3 t j + k , v (0) = 2 i , r (0) = k . a ( t ) = 3 t j + k v ( t ) = Z a ( t ) d t = Z 3 t j + k d t = 3 t 2 2 j + t k + c = 3 t 2 2 j + t k + c Now, pluging in t = 0, we get that v (0) = 0 j + 0 k + c = c , but we also know that v (0) = 2 i , so that means that c = 2 i , so v ( t ) = 3 t 2 2 j + t k + 2 i = 2 i + 3 t 2 2 j + t k . Now, to get the position, we
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Spring 2006 - Gross' Class - Exam 2 - Midterm #2 Solutions...

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