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251sol4 - 2 t A √ 1 4 t 2 = 4 t √ 1 4 t 2 a N = a 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 √ 1 + 4 t 2 = 4 t √ 1 + 4 t 2 a N = a · N = a , 2 A · a− 2 t, 1 A √ 1 + 4 t 2 = 2 √ 1 + 4 t 2 . Note that as t increases from 0 to 2, the tangential acceleration increases from 0 to 8 √ 65 ≈ . 992, whereas the centripetal acceleration decreases from 2 down to 2 √ 65 ≈ . 248. This corresponds to smoothly pressing the accelerator while smoothly straightening the steering wheel. The passenger’s body would smoothly transform from being pressed against the door to being pressed against the seat-back. During this time interval, the car’s speed increases from | v (0) | = 1 to | v (2) | = √ 65 ≈ 8 . 06. a...
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