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spring02prelim2sol

spring02prelim2sol - Spring 2002 Prelim 2 Solutions 1a...

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Spring 2002, Prelim 2 Solutions 1a. Since a ( t ) = d v ( t ) dt , v ( t ) = Z a ( t ) dt = •Z 1 - 2 t 2 dt i + •Z (cos πt ) dt j + •Z (sin πt ) dt k , = t + 2 t i + 1 π sin πt j - 1 π cos πt k + C 1 . The constant vector, C 1 , is computed from the initial velocity vector: v (1) = 1 + 2 1 i + 1 π sin π j + 1 π cos π k + C 1 , = 3 i + 1 π k + C 1 = i + 1 π k , C 1 = - 2 i , v ( t ) = t + 2 t - 2 i + 1 π sin πt j - 1 π cos πt k . Since v ( t ) = d r ( t ) dt , r ( t ) = Z v ( t ) = •Z t + 2 t - 2 dt i + •Z 1 π sin πt dt j - •Z 1 π cos πt dt k , = 1 2 t 2 + 2 ln t - 2 t i - 1 π 2 cos πt j - 1 π 2 sin πt k + C 2 . The constant vector, C 2 , is computed from the initial position vector: r (1) = 1 2 + 2 ln 1 - 2 i - 1 π 2 cos π j - 1 π 2 sin π k + C 2 , = - 3 2 i + 1 π 2 j + C 2 = 4 i + 2 j + k , C 2 = 11 2 i + 2 - 1 π 2 j + k , r ( t ) = 1 2 t 2 + 2 ln t - 2 t + 11 2 i - 1 π 2 cos πt + 1 π 2 - 2 j - 1 π 2 sin πt - 1 k . 1b. To find when the speed is the slowest, find the time when | v | 2 = v · v has a minimum.
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