Calc 3 Quiz 3 - 5 / 2) s 2 . Furthermore, ( t ) = 1 / (5 t...

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Instructor: Dr. Yuli B. Rudyak MAC 2313, Calculus 3 with Analytic Geometry, Fall 2011 QUIZ 3, OCTOBER 5, 2011 Problem 1. Find a vector function that represents the curve of inter- section of the cylinder x 2 + y 2 = 4 and the surface z = xy . Solution: For the cylinder we have x = 2 cos t,y = 2 sin t . Now, z = xy = 4 cos t sin t . Problem 2. Find the parametric equation for the tangent line to the curve x = t 2 - 1, y = t 2 + 1, z = t + 1. Solution: Take any point t 0 . The tangent vector at the point is h 2 t 0 , 2 t 0 , 1) i . So, the equation of the tangent line is: x = t 2 0 - 1 + 2 t 0 t , y = t 2 0 + 1 + 2 t 0 t , z = t 2 0 + t . Problem 3. Find the length of the curve r ( t ) = t 2 i + 2 t j + ln t k , 1 t e . Solution: We have r 0 ( t ) = h 2 t,t, 1 /t i . So L = Z e 1 1 + t 2 t dt = e 2 . Problem 4. Find the arc length function of the curve r ( t ) = h t 2 , sin t - t cos t, cos t + t sin t i , t [0 ,s ]. Find the curvature of the curve at t = 5. Solution: We have | r 0 ( t ) | = ( 5 / 2) t . So, L = Z s 0 = 5 tdt = (
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Unformatted text preview: 5 / 2) s 2 . Furthermore, ( t ) = 1 / (5 t ). Problem 5. Find the equation of the normal plane of the curve x = t,y = t 2 ,z = t 3 at the point (1 , 1 , 1 , ). Solution: We have r (1) = h 1 , 2 , 3 i is the normal vector to the normal plane. So, the equation of the normal plane is: ( x-1)+2( y 1 )+3( z-1) = 0, or x + 2 y + 3 z = 6. Problem 6. Find the tangential and normal components of the accel-eration vector r ( t ) = t i + t 2 j + 3 t k . 1 Solution: We have a T = | r ( t ) r 00 ( t ) | | r ( t )] = 4 t 4 t 2 + 10 . Furthermore, a N = | r ( t ) r 00 ( t ) | | r ( t )] = 2 10 4 t 2 + 10 ....
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This note was uploaded on 10/20/2011 for the course MAC 2313 taught by Professor Keeran during the Fall '08 term at University of Florida.

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Calc 3 Quiz 3 - 5 / 2) s 2 . Furthermore, ( t ) = 1 / (5 t...

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