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251sol4

# 251sol4 - 2 t A âˆš 1 4 t 2 = 4 t âˆš 1 4 t 2 a N = a Â N =...

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Assignment 4 Multivariate Calculus Math 251 (Fall 2010) A1. A1. The driver of a car follows the parabolic trajectory r ( t ) = ( t, t 2 ), for 2 t 2. Find the tangential and normal components a T and a N of the acceleration of the car (as functions of t ). Solution We compute the velocity, acceleration, the tangent vector and the normal vector. v ( t ) = r ( t ) = ( 1 , 2 t ) a ( t ) = v ( t ) = ( 0 , 2 ) T ( t ) = v ( t ) | v ( t ) | = ( 1 , 2 t ) 1 + 4 t 2 . We may use the chain and quotient rules to compute N ( t ). T ( t ) = (Bigg 1 2 8 t (1 + 4 t 2 ) 3 / 2 , 2 · 1 + 4 t 2 2 t · (8 t ) 1 2 (1 + 4 t 2 ) 1 / 2 1 + 4 t 2 )Bigg = (bigg 4 t (1 + 4 t 2 ) 3 / 2 , 2(1 + 4 t 2 ) 8 t 2 (1 + 4 t 2 ) 3 / 2 )bigg = (− 4 t, 2 ) (1 + 4 t 2 ) 3 / 2 | T ( t ) | = 2 4 t 2 + 1 (1 + 4 t 2 ) 3 / 2 = 2 1 + 4 t 2 N ( t ) = T ( t ) | T ( t ) | = (− 2 t, 1 ) 1 + 4 t 2 . Note that we could have computed N ( t ) much quicker by noting that it is just the rotation of T ( t ) by π/ 2 radians. Finally, we compute the components of a ( t ). a T = a · T = ( 0 , 2
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Unformatted text preview: , 2 t A âˆš 1 + 4 t 2 = 4 t âˆš 1 + 4 t 2 a N = a Â· N = a , 2 A Â· aâˆ’ 2 t, 1 A âˆš 1 + 4 t 2 = 2 âˆš 1 + 4 t 2 . Note that as t increases from 0 to 2, the tangential acceleration increases from 0 to 8 âˆš 65 â‰ˆ . 992, whereas the centripetal acceleration decreases from 2 down to 2 âˆš 65 â‰ˆ . 248. This corresponds to smoothly pressing the accelerator while smoothly straightening the steering wheel. The passengerâ€™s body would smoothly transform from being pressed against the door to being pressed against the seat-back. During this time interval, the carâ€™s speed increases from | v (0) | = 1 to | v (2) | = âˆš 65 â‰ˆ 8 . 06. a...
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