Math 13 Spring 2010 Exam 1 February 23, 2010

Math 13 Spring 2010: Exam 1 February 23, 2010 Name: Instructions: Each problem is scored out of 8 points for a total of 32 points. You may not use any outside materials(eg. notes or calculators). You have 50 minutes to complete this exam. Remember to fully justify your answers. Score: 1 Problem 1. (a) Find an equation for the tangent line to ~r(t) = hcos t, 2 sin t, ti at (1, 0, 0). (b) Find the distance from (1, 4, 2) to the plane 2x − 6y + z = 3. (c) Find the angle of intersection of the two planes 2x + y − 3z = −1 and x + 2y − z = 4. Proof. (a) The point occurs at t = 0. We have ~r 0 (t) = h− sin t, 2 cos t, 1i and hence ~r 0 (0) = h0, 2, 1i, which is the direction of the line. We have the line h1, 2t, ti . (b) We use the distance formula |ax0 + by0 + cz0 + d| √ a2 + b2 + c2 |2 − 24 + 2 − 3| = √ 4 + 36 + 1 23 =√ . 41 d= (c) We know the angle between two planes is the angle between their normal vectors. The two normal vectors are ~n1 = h2, 1, −3i and ~n2 = h1, 2, −1i . We compute the angle between them as cos θ = So we have ~n1 · ~n2 2+2+3 7 = √ √ =√ . |~n1 | |~n2 | 14 6 84 7 θ = cos−1 √ . 84 2 Problem 2. (a) Find the vector function that represents the curve of intersection of the paraboloid z = 4x2 +y 2 and the parabolic cylinder y = x2 . Find the point(s) where this curve intersects the plane x + 2y = 3? (b) Suppose you start at the point (0, 0, 3) and move 5π units along the curve ~r(t) = h3 sin t, 4t, 3 cos ti. Where are you now? Proof. (a) We parameterize the cylinder as x = t, y = t2 . Then we use the paraboloid to get ® ­ ~r(t) = t, t2 , 4t2 + t4 . To find the point(s) of intersection we solve t + 2t2 − 3 = (t − 1)(2t + 3) = 0 to get t = 1, t = − 32 to get the two points ~r(1) = h1, 1, 5i µ ¶ ¿ À 3 3 9 81 and ~r − = − , ,9 + . 2 2 4 16 (b) We parameterize with respect to arc length. We compute ~r 0 (t) = h3 cos t, 4, −3 sin ti p √ |~r 0 (t)| = 9 cos2 t + 16 + 9 sin2 t = 25 = 5. The point (0, 0, 3) occurs at t = 0 so we have Z t s(t) = |~r 0 (t)| dt 0 Z = t 5dt = 5t 0 So we have t = 5s . So the parametrization with respect to arc length is ¿ À s 4 s r(s) = 3 sin , s, 3 cos ; 5 5 5 So we have r(5π) = h0, 4π, −3i . 3 ­ ® Problem 3. Given the curve ~r(t) = t2 − 1, 2t + 3, t2 − 4t (a) Find the curvature at (3, 7, −4). (b) Find an equation for the normal plane at (3, 7, −4). Proof. (a) The point occurs at t = 2. We compute ~r 0 (t) = h2t, 2, 2t − 4i ~r 0 (2) = h4, 2, 0i √ |~r 0 (2)| = 20 ~r 00 (t) = h2, 0, 2i We compute h4, 2, 0i × h2, 0, 2i = h4, −8, 4i = 4 h1, −2, −1i So we have κ= |4 h1, −2, −1i| 3 |h4, 2, 0i| = √ √ 4 6 3 √ = √ . 20 20 5 10 (b) The normal plane has normal vector T~ . So compute 1 T~ (t) = p 4t2 + 4 + (2t − 4)2 h2t, 2, 2t − 4i 1 T~ (2) = √ h4, 2, 0i 20 which is in the direction h2, 1, 0i. So we have the plane 2(x − 3) + (y − 7) = 0. 4 Problem 4. Given ~r(t) = 2t2 î + tĵ + k̂. (a) Find the tangential and normal components of acceleration as functions of t. (b) What is the minimum speed and for what t does the minimum speed occur? Proof. (a) We have r0 (t) = h4t, 1, 0i r00 (t) = h4, 0, 0i so we have r0 · r00 16t =√ 0 |r | 16t2 + 1 0 00 |0, 0, −4| 4 |r × r | = =√ =√ . 0 2 |r | 16t + 1 16t2 + 1 aT = aN (b) We compute speed as v = |~r 0 (t)| = p 16t2 + 1. This is a max/min problem so we solve for the critical points: d ~v (t) = aT . dt aT = √ 16t =0 16t2 + 1 to get t = 0. This gives a minimum speed of 1 and is clearly a minimum since v contains only t2 . 5