Calculus 3 Practice Exam 3

Calculus 3 Practice Exam 3 - so | T ± | = 3 5 Thus N = T...

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Math 209 Exam 3 Practice 1. Find parametric equations for the line containing the points P (1 , 2 , 4) and Q (3 , 0 , 7). 2. Find the equation of the plane containing the point (3 , - 1 , 4) with normal vector 2 , 5 , 6 . 3. Find Z 1 0 4 t, 5 , 12 t 3 fi dt For the remaining problems, let r ( t ) = 4 t, - 3 sin( t ) , 3 cos( t ) . 4. Find the length of the curve between t = 0 and t = π 2 . 5. Find the unit tangent vector T at t = π 2 . 6. Find the unit normal vector N at t = π 2 . 7. Find the unit binormal vector B at t = π 2 . 8. Find the curvature κ at t = π 2 . 9. Find the velocity and acceleration vectors at t = π 2 . 10. Find the tangential and normal components of the acceleration vector at t = π 2 Bonus: Evaluate Z 1 - 1 e t , e - t , 1 fi dt
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Math 209 Exam 3 Practice Solutions 1. -→ PQ = 2 , - 2 , 3 , so r = 1 , 2 , 4 + 2 , - 2 , 3 t = 1 + 2 t, 2 - 2 t, 4 + 3 t Thus, x = 1 + 2 t, y = 2 - 2 t z = 4 + 3 t . 2. 2( x - 3) + 5( y + 1) + 6( z - 4) = 0, which reduces to 2 x + 5 y + 6 z = 25 . 3. Z 1 0 4 t, 5 , 12 t 3 fi dt = £› 2 t 2 , 5 t, 3 t 4 fi/ 1 0 = 2 , 5 , 3 4. r = 4 , - 3 cos t, - 3 sin t , so | r | = 5, thus L = Z π/ 2 0 5 dt = 5 π/ 2. 5. T = r
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Unformatted text preview: , so | T ± | = 3 5 . Thus, N = T ± | T ± | = ± , sin t,-cos t ² = ± , 1 , ² . 7. B = T × N = ¿ 4 5 , ,-3 5 ± × ± , 1 , ² = ¿ 3 5 , , 4 5 ± 8. The curvature κ = | T ± | | r ± | = 3 25 9. The velocity v = r ± = ± 4 ,-3 cos t,-3 sin t ² = ± 4 , ,-3 ² . The acceleration a = r ±± = ± , 3 sin t,-3 cos t ² = ± , 3 , ² . 10. The tangential component of acceleration at t = π 2 is given by a T = a · T = ± , 3 , ² · ¿ 4 5 , ,-3 5 ± = 0 The normal component of acceleration at t = π 2 is given by a N = a · N = ± , 3 , ² · ± , 1 , ² = 3 Bonus: Z 1-1 › e t , e-t , 1 fi dt = £› e t ,-e-t , t fi/ 1-1 = › e-e-1 , e-e-1 , 2 fi . Page 2...
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