Quiz2_sol

# Quiz2_sol - 2 1 ²² i 2 j ± s √ 2 1 ² cos ± log ± s...

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NAME: Spring 2011, MAC 2313, Quiz 2 UFID: Section: 3124 1. Find the parametric equations for the tangent line to the curve with given parametric equation x = e - t cos t,y = e - t sin t,z = e - t at point (1 , 0 , 1). (2 points) Solution. . At point (1 , 0 , 1), t = 0. Also, there are x 0 ( t ) = - e - t cos t - e - t sin t = - e - t (cos t + sin t ) y 0 ( t ) = e - t cos t - e - t sin t = e - t (cos t - sin t ) z 0 ( t ) = - e - t (1) Therefore r 0 (0) = h x 0 (0) ,y 0 (0) ,z 0 (0) i = h- 1 , 1 , - 1 i . Hence the parametric equations for the tangent line (passing through point (1 , 0 , 1) with direction h- 1 , 1 , - 1 i ) is x = 1 - t,y = t,z = 1 - t . 2. Reparametrize the curve r ( t ) = e 2 t cos 2 t i +2 j + e 2 t sin 2 t k with respect to arc length measured from the point where t = 0 in the direction of increasing t . (2 points) Solution. . We have r 0 ( t ) = 2 e 2 t h cos t - sin t, 0 , cos t + sin t i k r 0 ( t ) k = 2 e 2 t ( (cos t - sin t ) 2 + (cos t + sin t ) 2 ) 1 / 2 = 2 2 e 2 t . s ( t ) = Z t 0 k r 0 ( τ ) k = Z t 0 2 2 e 2 τ = 2( e 2 t - 1) (2) Therefore, e 2 t = s/ 2 + 1 and t = 1 2 log( s/ 2 + 1). Hence r ( s ) = ± s 2 + 1 ² cos ± log ± s
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Unformatted text preview: 2 + 1 ²² i +2 j + ± s √ 2 + 1 ² cos ± log ± s √ 2 + 1 ²² k (3) 3. Find the curvature of r ( t ) = h e t cos t,e t sin t,t i at the point (1 , , 0). (2 points) Solution. . At point (1 , , 0), t = 0. Also, there is First we have r ( t ) = h e t (cos t-sin t ) ,e t (cos t + sin t ) , 1 i r 00 ( t ) = h-2 sin te t , 2 cos te t , i . So r (0) = h 1 , 1 , 1 i r 00 (0) = h , 2 , i . Therefore r (0) × r 00 (0) = h-2 , ,-2 i and hence k r (0) × r 00 (0) k = 2 √ 2 also k r (0) k = √ 3 . Therefore k (0) = k r (0) × r 00 (0) k k r (0) k 3 = 2 √ 2 3 √ 3 = 2 √ 6 9 (4)...
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