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# notes001 (45) - v t = j v t j = q(3 t 2(4 t 2 = 5 t The...

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MATH 2203 °Quiz 3 (Version 2) Solution September 24, 2010 NAME_____________________________ The motion of a particle is given by x = 3 2 t 2 y = 2 t 2 t > 0 . Find ° the position function: r ( t ) ° the velocity: v ( t ) ( 4 points if you get to here ) ° the speed: v ( t ) = j v ( t ) j ° the acceleration: a ( t ) ( 10 points if you get to here ) ° the unit tangent vector: T ( t ) ° the curvature: ° ( t ) ( 16 points if you get to here ) ° the tangential component of acceleration: a T ( t ) ° the normal component of acceleration: a N ( t ) ( 20 points if you get to here ) Solution: The vector°valued function for the position of the particle is r ( t ) = 3 2 t 2 i ± 2 t 2 j . The velocity is v ( t ) = d r dt = 3 t i + 4 t j 1

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and the speed is v ( t ) =
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Unformatted text preview: v ( t ) = j v ( t ) j = q (3 t ) 2 + (4 t ) 2 = 5 t . The acceleration is a ( t ) = d v dt = 3 i + 4 j . The unit tangent vector is T ( t ) = 1 v ( t ) v ( t ) = 1 5 t (3 t i + 4 t j ) = 3 5 i + 4 5 j . Also d T dt = and & & & & d T dt & & & & = 0 so the curvature is & ( t ) = 1 v ( t ) & & & & d T dt & & & & = 1 5 t (0) = 0 . The tangential component of acceleration is a T ( t ) = dv dt = d dt (5 t ) = 5 and the normal component of acceleration is a N ( t ) = & ( t ) ( v ( t )) 2 = (0) (5 t ) 2 = 0 . 2...
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notes001 (45) - v t = j v t j = q(3 t 2(4 t 2 = 5 t The...

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