exam 1 solutions - smallest that happens at the times t = 0...

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p r r 18.02 Exam 1 Solutions Problem 1. a) P = (1 , 0 , 0), Q = (0 , 2 , 0) and R = (0 , 0 , 3). Therefore --! = ˆ ı - QR = - + 3 k ˆ . QP and --! --! QR · --! QP h 1 , - 2 , 0 i · h 0 , - 2 , 3 i = 4 p 1 2 + 2 2 p 2 2 + 3 2 b) cos = p 65 = QP --! QP --! Problem 2. a) --! PR = h- 1 , 0 , 3 i . PQ = h- 1 , 2 , 0 i , -! --! PQ -! PR = ˆ ı ˆ k ˆ - 1 2 0 - 1 0 3 = ı + + 2 k ˆ . 1 = 1 6 2 + 3 2 + 2 2 = 1 p 49 = 7 --! PQ -! PR Then area ( Δ ) = . 2 2 2 2 b) A normal to the plane is given by - N = PQ PR h 6 , 3 , 2 i . Hence the equation has the form ! --! -! = 6 x + 3 y + 2 z = d . Since P is on the plane d = 6 1 + 3 1 + 2 1 = 11. In conclusion the equation of the · · · plane is 6 x + 3 y + 2 z = 11 . c) The line is parallel to h 2 - 1 , 2 - 2 , 0 - 3 i = h 1 , 0 , - 3 i . ! = 6 - 6 = Since - 0, the line is N · h 1 , 0 , - 3 i parallel to the plane. Problem 3. a) -! AB OA = h 10 t, 0 i and --! = h cos t, sin t i , hence B r --! OB = -! AB OA + --! = h 10 t + cos t, sin t i . O A The rear bumper is reached at time t = and the position of B is (10 - 1 , 0). b) - V ! = h 10 - sin t, cos t i , thus 2 2 | -! V | 2 = (10 - sin t ) 2 + cos t = 100 - 20 sin t + sin 2 t + cos t = 101 - 20 sin t. The speed is then given by p 101 - 20 sin t . The speed is smallest when sin t is largest i.e. sin t = 1. It occurs when t = / 2. At this time, the position of the bug is (5 , 1). The speed is largest when
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Unformatted text preview: smallest; that happens at the times t = 0 or ⇡ for which the position is then (0 , 0) and (10 ⇡-1 , 0). Problem 4. a) | M | =-12. b) 5, b = 7.-a = 2 3 2 x 1 1 4 t/ 12 + 1 1 3 3 2 3 2 3 4 5 = 4 4 5 5 = 4 5 c)-5 7-8 7-5 4 7 t/ 12-2-5 t/ 12 + 1 t y 12 z d ~ r 1 7 5 d) dt = 12 , 12 ,-12 . Problem 5. a)-N ·-r ( t ) = 6 , where-N ! ! ! = h 4 ,-3 ,-2 i . b) We di ↵ erentiate-N ·-r ( t ) = 6: ! ! d d d d-! N-N ·-r ( t ) +-N ! ! ! ·-r ( t ) +-! N-r ( t ) ! ! and hence-! N ? d r ( t ). ! dt--r ( t ) =-! ! r ( t ) ! ·-= 0 = · · dt dt dt dt...
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