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Unformatted text preview: MAC 2313, Fall 2010 — Midterm 2 Review Problems Solutions  1 1 Find a vector function that represents the curve of intersection of the cylinder x 2 + y 2 = 16 and the plane x + z = 5. Solution: Any vector function r ( t ) = ( 4cos t, 4sin t, z ( t ) ) will lie on the cylinder, so we just need to choose z ( t ) so that this function lies on the plane. The plane is given by x + z = 5, so we want 4cos t + z ( t ) = 5, which gives z ( t ) = 5 − 4cos t . The answer is then r ( t ) = ( 4cos t, 4sin t, 5 − 4cos t ) . 2 A particle moves with position function r ( t ) = t ln t i + t j + e − t k . Find the velocity, acceleration, and speed of the particle. Solution: We have v ( t ) = r ′ ( t ) = (ln t + 1) i + j − e − t k , a ( t ) = v ′ ( t ) = 1 t i + e − t k , Speed =  v ( t )  = radicalBig (ln t + 1) 2 + 1 + e − 2 t . 3 Compute the position vector for a particle which passes through the origin at time t = 0 and has velocity vector v ( t ) = 2 t i + sin t j + cos t k . Solution: Letting r ( t ) denote the position of the particle, we have r ( t ) = integraldisplay v ( t ) dt, = integraldisplay 2 t i + sin t j + cos t k dt, = t 2 i − cos t j + sin t k + C , where C is a constant vector. Since the particle passes through the origin at time t = 0, we want r (0) = , i.e., r (0) = − j + C = , so we choose C = j , which gives us r ( t ) = t 2 i + (1 − cos t ) j + sin t k . MAC 2313, Fall 2010 — Midterm 2 Review Problems Solutions  2 4 Find the arc length of the curve r ( t ) = cos 3 t j + sin 3 t k from t = 0 to t = 1. Solution: We have Arc Length = integraldisplay 1 Speed dt = integraldisplay 1  v ( t )  dt = integraldisplay 1 vextendsingle vextendsingle (− 3cos 2 t sin t, 3sin 2 t cos t ) vextendsingle vextendsingle dt = integraldisplay 1 radicalbig 9cos 4 t sin 2 t + 9sin 4 t cos 2 t dt = integraldisplay 1 3cos t sin t radicalbig cos 2 t + sin 2 t dt = 3 integraldisplay 1 cos t sin t dt. Now we make a usubstitution ( u can be set to either cos t or sin t ) and evaluate the integral to get 3sin 2 1 2 . 5 Consider the curve defined by r ( t ) = ( 4sin ct, 3 ct, 4cos ct ) . What value of c makes the arc length of the curve traced by r ( t ), for 0 ≤ t ≤ 1, equal to 10? Solution: We first compute the arc length for any value of c , and then we solve this to find which value of c makes the arc length 10. We have v ( t ) = r ′ ( t ) = ( 4 c cos ct, 3 c, − 4 c sin ct ) , MAC 2313, Fall 2010 — Midterm 2 Review Problems Solutions  3 so Arc length for 0 ≤ t ≤ 1 = integraldisplay 1 Speed dt, = integraldisplay 1  v ( t )  dt, = integraldisplay 1 radicalBig (4 c cos ct ) 2 + (3 c ) 2 + ( − 4 c sin ct ) 2 dt = integraldisplay 1 radicalbig 16 c 2 cos 2 ct + 9 c 2 + 16 c 2 sin 2 ct dt = integraldisplay 1 √ 25 c 2 dt = 5 c....
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This note was uploaded on 12/10/2011 for the course MAC 2313 taught by Professor Keeran during the Fall '08 term at University of Florida.
 Fall '08
 Keeran
 Calculus

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