notes001 (44) - The acceleration is a t = d v dt =& 4...

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MATH 2203 °Quiz 3 (Version 1) Solution September 24, 2010 NAME_____________________________ The motion of a particle is given by x = sin (2 t ) y = ° cos (2 t ) . 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 ) = sin (2 t ) i ° cos (2 t ) j . The velocity is v ( t ) = d r dt = 2 cos (2 t ) i + 2 sin (2 t ) j and the speed is v ( t ) = j v ( t ) j = q (2 cos (2 t )) 2 + (2 sin (2 t )) 2 = 2 . 1
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The acceleration is
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Unformatted text preview: The acceleration is a ( t ) = d v dt = & 4 sin (2 t ) i + 4 cos (2 t ) j . The unit tangent vector is T ( t ) = 1 v ( t ) v ( t ) = 1 2 (2 cos (2 t ) i + 2 sin (2 t ) j ) = cos (2 t ) i + sin (2 t ) j . Also d T dt = & 2 sin (2 t ) i + 2 cos (2 t ) j and & & & & d T dt & & & & = q ( & 2 sin (2 t )) 2 + (2 cos (2 t )) 2 = 2 so the curvature is & ( t ) = 1 v ( t ) & & & & d T dt & & & & = 1 2 (2) = 1 . The tangential component of acceleration is a T ( t ) = dv dt = d dt (2) = 0 and the normal component of acceleration is a N ( t ) = & ( t ) ( v ( t )) 2 = (1) (2) 2 = 4 . 2...
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