pract_mt_2_solns

pract_mt_2_solns - MATH 32A/1: Practice Midterm 2 Solutions...

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Unformatted text preview: MATH 32A/1: Practice Midterm 2 Solutions Dr. Frederick Park Note: This Exam is Slightly Longer than the Actual Midterm 1. Find the length of the curve r ( t ) = < 2 t 3 / 2 , cos2 t, sin2 t > , 0 t 1 Solution: Will be done in review session as I want to review curvature. There have been lots of questions on it! 2. Reparametrize the curve r ( t ) = < e t ,e t sin t,e t cos t > with respect to arc length measured from the point (1 , , 1) in the direction of increasing t . Solution: The point (1 , , 1) corresponds to t = 0. We also note that r ( t ) = e t < 1 , sin t, cos t > . Now ds dt = | r ( t ) | . So r ( t ) = e t h , cos t,- sin t i + e t h 1 , sin t, cos t i = e t h 1 , cos t +sin t, cos t- sin t i . Hence, | r ( t ) | = 3 e t . Thus, s ( t ) = R t | r ( u ) | du = R t 3 e u du = 3( e t- 1). Thus s ( t ) = s = 3( e t- 1) implying that t = ln( s/ 3 + 1). Therefore, r ( t ( s )) = ( s/ 3 + 1) h 1 , sin(ln( s/ 3 + 1)) , cos(ln( s/ 3 + 1)) i . 3. Find the curvature of the curve y = x 4 at the point (1,1). Solution: Due to visual nature, problem will be done in Review. Also, see problem 1. 4. Find an equation of the osculating plane of the curve defined by the position function r ( t ) = < sin2 t,t, cos2 t > at the point (0 ,, 1). Is this plane perpen- dicular to the normal plane? Solution: Clearly the point (0 ,, 1) corresponds to t = . The Osculat- ing plane is a familiar friend from class (see your notes) and contains both the Tangent T and Normal N vectors. The Binormal vector B = T N is perpendicular to both the Tangent and Normal vectors and hence will define the direction of the Osculating plane. If this doesnt make sense, re-read the previous 2 sentences twice....
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pract_mt_2_solns - MATH 32A/1: Practice Midterm 2 Solutions...

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