Mathematics 317 Solutions to Midterm

# Mathematics 317 Solutions to Midterm - Mathematics 317...

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Mathematics 317 Solutions to Midterm #1 Fall 2009 1. (a) . ) ( T r r v v dt ds ds d t = = = & (b) Differentiating (a) with respect to t , one obtains , ) ( 2 2 N T T T T T T T v a v t v ds d v v dt ds ds d v v v v κ + = + = + = + = = & & & & & & from the first Frenet-Serret formula. (c) . 0 ) ( = × = r r r A r & t Suppose , ) ( ) ( α A t a t = for some constant vector α , then the particle’s path satisfies . 0 = α r Hence the motion is planar. 2. (a) . 2 sin cos k j i r t t t - = & Hence v = . 4 1 4 sin cos 2 2 2 2 t t t t s + = + + = = r & & Thus the distance travelled by the particle over the time interval π t 0 is given by the integral . 4 1 0 2 dt t + π (b) . 2 cos sin k j i r - - = t t & & Then . 2 ) 0 ( ) 0 ( ; 2 ) 0 ( , ) 0 ( j k r r k j r i r = × - = = & & & & & & Hence the curvature . 5 2 1 ) 0 ( ) 0 ( ) 0 ( ) 0 ( 2 3 = + = × = r r r & & & & κ (c) From 1(b), the tangential component of acceleration is given by 0 4 1 4 ) 4 1 ( 2 2 = + = + t t t dt d at t = 0, and the normal component of acceleration at t = 0 is given by . 5 ) 0 ( ) 0 ( 2 = v κ 3. (a) Note that the first equation describes a circle. Hence, use the parametrization , sin ) ( , cos ) ( t t y t t x = = and then from the second equation one obtains Subscribe to view the full document. Unformatted text preview: 2 ) ( ) ( 2 ) ( t t t y t x t z--=--= This yields . ) sin cos 2 ( sin cos ) ( k j i r t t t t t--+ + = (b) . ) sin (cos sin cot ; ) cos (sin cos sin k j i r k j i r t t t t t t t + +--=-+ +-= &amp; &amp; &amp; At the point (0,1,1). , 2 / = t and hence . ) 2 / ( , ) 2 / ( k j r k i r +-= +-= &amp; &amp; &amp; Consequently, . ) 2 / ( ) 2 / ( j i k r r---= × &amp; &amp; &amp; Hence the curvature at the given point is . 2 3 ) 2 ( 3 ) 2 / ( ) 2 / ( ) 2 / ( ) 2 / ( 2 / 3 2 / 1 3 3 = = × = r r r &amp; &amp; &amp; &amp; (c) Note that 2 = + + z y x defines a plane and hence the curve is planar. The curve is an ellipse since the intersection of a plane and a circular cylinder yields an ellipse (which could be a circle—not the case here)....
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