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01spfin - Math 222 Final Exam May 14 2001 Answers I(50...

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Math 222 – Final Exam May 14 – 2001 Answers I. (50 points.) The motion of a particle in the plane is governed by the equations dx dt = x, dy dt = - x 2 . (i) Find parametric equations for its coordinates ( x,y ) at time t if ( x,y ) = (1 , - 4) when t = 0. (Hint: Find x first.) Answer: We solve the equation dx/dt = x by separation of variables: dx x = dt = ln( x ) = t + C = x = e t + C . When t = 0 we have x = 1 so 1 = e 0+ C so C = 0 so x = e t . Thus dy dt = - x 2 = - e 2 t so y = - e 2 t / 2 + C 1 . When t = 0 we have y = - 4 so - 4 = - e 0 / 2 + C 1 so C 1 = - 7 / 2 y = - e 2 t + 7 2 . (ii) Find the velocity vector v and the acceleration vector a . Answer: v = dx dt i + dy dt j = x i - x 2 j = e t i - e 2 t j and a = d 2 x dt 2 i + d 2 y dt 2 j = e t i - 2 e 2 t j . (The velocity vector v and acceleration vector a were defined in the lecture on parametric equa- tions. They are introduced in the text in Chapter 12.) II. (60 points.) The questions on this page involve the three points A ( - 1 , 0 , 2) , B (2 , 1 , - 1) , C (1 , - 2 , 2) . (i) Find the angle ABC .
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Answer: The angle θ is given by the formula cos θ = -→ BA · --→ BC ± ± ± -→ BA ± ± ± ± ± ± --→ BC ± ± ± Now -→ BA = ( - 1 - 2) i + (0 - 1) j + (2 - 2) k = - 3 i - j --→ BC = (1 - 2) i + ( - 2 - 1) j + (2 - ( - 1)) k = - i - 3 j + 3 k so -→ BA · --→ BC = ( - 3)( - 1) + ( - 1)( - 3) + (0)(3) = 6 , ± ± ± -→ BA ± ± ± = 9 + 1 + 0 = 10 , ± ± ± --→ BC ± ± ± = 1 + 9 + 9 = 19 and θ = cos - 1 (6 / 190). (ii) Find an equation for the plane containing the three points A , B , and C . Answer:
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01spfin - Math 222 Final Exam May 14 2001 Answers I(50...

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