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First since r 0 0 3 0 the signed distance along the

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First, since r (0) = <0, 3, 0>, the (signed) distance along the curve from the point given by r (0) on the graph to that given by r (t) is . s ϕ ( t ) t 0 r ( u ) du t 0 <1, 6sin(2 u ),6cos(2 u )> du t 0 37 1/2 du 37 1/2 ( t 0) 37 1/2 t . Solving for t in terms of s yields t 37 1/2 s so that ϕ 1 ( t ) 37 1/2 t . Thus, R ( s ) r ( ϕ 1 ( s )) r (37 1/2 s ) < s 37 1/2 , 3cos 2 s 37 1/2 , 3sin 2 s 37 1/2 >. Evidently, R (s) = r ( ϕ -1 (s)), or equivalently, r (t) = R ( ϕ (t)). ______________________________________________________________________ 4. (10 pts.) A particle moves smoothly in such a way that at a particular time t = 0, we have v (0) = < 1 , 2 > and a (0) = < 3 , 0 >. If we write a (0) in terms of T (0) and N (0), then a (0) = a T (0) T (0) + a N (0) N (0), where (a) T (0) = , (b) a T (0) = , (c) a N (0) = , and (d) N (0) = . (e) Also,
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TEST2/MAC2313 Page 3 of 5 ______________________________________________________________________ 5. (10 pts.) (a) Find the limit. lim t 1 3 t 2 , ln( t ) t 2 1 ,sin( π 2 t ) 3, 1 2 ,1 The only silliness is the middle limit. Either recognize a loggy derivative limit as part of the mess or use l’Hopital’s Rule there. (b) Find parametric equations for the line tangent to the graph of r ( t ) = (2 - ln( t )) i + t 2 j at the point where t 0 = 1. Since r ( t ) = < - t -1 , 2 t >, r (1) = < 2 - ln(1) , 1 > = < 2 , 1 >, and r (1) = < -1 , 2 >. Consequently, an appropriate set of parametric equations is given by x = 2 - t and y = 1 + 2t. ______________________________________________________________________ 6. (10 pts.) Let . f ( x , y ) y x 2 (a) Obtain an equation for the level curve for f that passes through the point (-1,2). Since f (-1,2) = 2 - (-1) 2 = 1, an equation for the level curve through (-1,2) is 1 = y - x 2 , an equation a garden variety parabola.
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