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Homework 4 Solutions

Homework 4 Solutions - WEEK 4 HOMEWORK SOLUTIONS 1 a Our...

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WEEK 4 HOMEWORK SOLUTIONS 1 a. Our curve is parameterized by r ( t ) = ( x ( t ) , y ( t )) for t R . To calculate curvature using the formula (1) κ ( t ) = | v ( t ) × a ( t ) | | v ( t ) | 3 , we must first write our parameterization as a three dimensional vector, then compute the associated velocity and acceleration. We can stick the plane that this curve is stuck in at any place along the z-axis, meaning we should just set the z component equal to the constant c . We then get; r ( t ) = ( x ( t ) , y ( t ) , c ) v ( t ) = ( x ( t ) , y ( t ) , 0) a ( t ) = ( x ( t ) , y ( t ) , 0) . To compute v ( t ) × a ( t ) it is best to first recognize that the only nonzero component will be ˆ k . We find that v ( t ) × a ( t ) = ( x ( t ) y ( t ) x ( t ) y ( t )) ˆ k | v ( t ) × a ( t ) | = | x ( t ) y ( t ) x ( t ) y ( t ) | . Input this into 1 we get κ ( t ) = | x ( t ) y ( t ) x ( t ) y ( t ) | ( x ( t ) 2 + y ( t ) 2 ) ( 3 / 2) , b. Define the function K ( t ) as K ( t ) = N ( t 0 ) · ( r ( t 0 ) r ( t )) | r ( t 0 ) r ( t ) | 2 . If we let θ be the angle between the two vectors in the denomiator we may writ this expression in terms of the respective magnitudes and θ . Writing this out and using 1
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2 WEEK 4 HOMEWORK SOLUTIONS the fact that N ( t ) is a unit vector we get that K ( t ) = | N ( t 0 ) || r ( t 0 ) r ( t ) | cos( θ ) | r ( t 0 ) r ( t ) | 2 = cos( θ ) | r ( t 0 ) r ( t ) | c. Now, we have oriented our coordinate system and set our initial time such that t 0 = 0, r (0) = (0 , 0), and N (0) = (0 , 1).
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