Pm1aSolns - Math 265(Butler Practice Midterm I A(Solutions 1 A particle moves through three dimensional space with velocity v(t = sec2 t 2 sec t

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Math 265 (Butler) Practice Midterm I — A (Solutions) 1. A particle moves through three dimensional space with velocity v ( t ) = h sec 2 t, 2 sec t tan t, tan 2 t i . At time t = 0 the particle is at h 0 , 1 , 2 i , find the position function of the particle for - π/ 4 t π/ 4. If r ( t ) is the position function then r 0 ( t ) = v ( t ). So taking antiderivatives we have r ( t ) = Z v ( t ) dt = ±Z sec 2 tdt, Z 2 sec t tan Z tan 2 tdt ² = ± tan t + C, 2 sec t + D, Z (sec 2 t - 1) dt ² = ± tan t + C, 2 sec t + D, tan t - t + E ² . Two of the three integrals are straightforward. The last one is the trickiest but this follows by relating tan 2 t (something which we cannot directly inte- grate) to sec 2 t (something which is easy to integrate). Now all that is left is to determine the constants C,D,E . We have r (0) = h C, 2 + D,E i = h 0 , 1 , 2 i giving the constants we need. So we have that the position function of the particle is r ( t ) = ± tan t, 2 sec t - 1 , tan t - t + 2 ² . (The condition for - π/ 4 t π/ 4 is not needed directly, it only is used to guarantee that we stay away from the vertical asymptotes (solutions cannot be pushed past vertical asymptotes, something you will learn in your future math classes).)
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2. Find the point on the plane x +2 y = 5+3 z which is closest to the point (4 , 4 , - 7).
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This note was uploaded on 02/06/2012 for the course MATH 265 taught by Professor Gregorac during the Spring '08 term at Iowa State.

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Pm1aSolns - Math 265(Butler Practice Midterm I A(Solutions 1 A particle moves through three dimensional space with velocity v(t = sec2 t 2 sec t

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