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2) Compute the arc length of K. A Space curve. D Follow the standard method. √ I 1) As r(0) = (0, 2, 0), and √ r (t) = (1 + cos t, − 2 sin t, 1 − cos t), r (0) = (2, 0, 0), it follows that a parametric description of the tangent of K at (0, √ √ x(u) = (0, 2, 0) + (2u, 0, 0) = (2u, 2, 0), u ∈ R. 4 Figure 49: The curve K and its tangent at (0, √ 2, 0). 2) The arc length of K is 1 −1 r (t) dt 1 = −1 1 = −1 (1+cos t)2 +2 sin2 t+(1−cos t)2 dt 2 + 2 cos2 t + 2 sin2 t dt = 1 −1 √ 4 dt = 2 · 2 = 4. Please click the advert what‘s missing in this equation?

D Find ds and then compute. I 1) As s (t) = r (t) and r (t) = a sin t2 , cos t2 , f˚ as s (t) = a sin2 (t2 ) + cos2 (t2 ) = a, we get s(t) = at and t(s) = 1 s. a The parametric description with the arc length is x = r(t) = a t 0 sin u2 du , t 0 cos u2 du =a s a 0 sin u2 du , s a 0 cos u2 du . 5 –1 Figure 44: The clothoid for a = 1 and s ∈ [−4, 4]. 2) From r (t) = a sin t2 , cos t2 ∼ a sin t2 , cos t2 , 0 , and r (t) = 2ta cos t2 , − sin t2 ∼ 2ta cos t2 , − sin t2 , 0 , follows that 2ta cos t2 −2ta sin t2 0 0 0 1 {ez × r (t)} · r (t) = a sin t 2 a cos t 2 =− 2ta cos t2 −2ta sin t2 a sin t2 a cos t2 0 = −2ta2 .

5. 4 that K ds = √ 1 {13 13 − 9}. 27 Alternatively, Y (x) = 1 K ds = = 1+ 0 4 t+ 9 2 −1 x 3 , thus 3 4 −2 x 3 dx = 9 3 2 1 = 0 13 9 1 0 3 2 1 x− 3 − 4 9 3 1 4 4 dx = t + dt 9 2 0 9 √ √ 8 1 13 13 − = {13 13 − 8}. = 27 27 27 2 x3 + 3 2 wanted: ambitious people Please click the advert At NNE Pharmaplan we need ambitious people to help us achieve the challenging goals which have been laid down for the company. Kim Visby is an example of one of our many ambitious co-workers. Besides being a manager in the Manufacturing IT department, Kim performs triathlon at a professional level.

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