# 13-3 - Solutions to Homework Assignment 10 1 r t =< 2cos...

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Unformatted text preview: Solutions to Homework Assignment 10 1. r ( t ) = < 2cos t, 5 ,- 2sin t >, so L = Z 10- 10 p (2cos t ) 2 + 5 2 + (- 2sin t ) 2 dt = Z 10- 10 √ 29 dt = √ 29 t fl fl fl fl 10- 10 = √ 29(10- (- 10)) = 20 √ 29 . 3. r ( t ) = < √ 2 ,e t ,- e- t >, so L = Z 1 p 2 + e 2 t + e- 2 t dt = Z 1 p ( e t + e- t ) 2 dt = Z 1 ( e t + e- t ) dt = e t- e- t fl fl 1 = e- e- 1 . 9. r ( t ) = < 2 ,- 3 , 4 >, so we have s ( t ) = Z t p 2 2 + (- 3) 2 + 4 2 du = √ 29 u fl fl fl fl t = √ 29 t. Therefore, t = 1 √ 29 s. We get r ( t ( s )) = ¿ 2 s √ 29 , 1- 3 s √ 29 , 5 + 4 s √ 29 . Notice that with this, if s = 1 , the point on the curve (a line) is ¿ 2 √ 29 , 1- 3 √ 29 , 5 + 4 √ 29 . The distance of this along the curve from the point (0 , 1 , 5) (where we started) is p 4 / 29 + 9 / 29 + 16 / 29 = 1 since the curve is just a line, and this is the value of s. This is the point of the reparametrization: the arc length is exactly the value of the parameter. (Try this for s = 2 and s = 7.)= 7....
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13-3 - Solutions to Homework Assignment 10 1 r t =< 2cos...

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