138_pdfsam_math 54 differential equation solutions odd

138_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Chapter 3 Therefore, the velocity is given by v (t) = 6 − Ke−2t . Setting v = 1 when t = 0, we find that 1=6−K ⇒ K = 5. Thus the equation for velocity v (t) is v (t) = 6 − 5e−2t . The limiting velocity of the sailboat under these conditions is found by letting time approach infinity: t→∞ lim v (t) = lim 6 − 5e−2t = 6 (m/sec). t→∞ To determine the equation of motion we will use the equation of velocity obtained previously and substitute dx/dt for v (t) to obtain dx = 6 − 5e−2t , dt Integrating this equation we obtain x(t) = 6t + Setting x = 0 when t = 0, we find 0=0+ 5 + C1 2 ⇒ 5 C1 = − . 2 5 −2t e + C1 . 2 x(0) = 0. Thus the equation of motion for the sailboat is given by x(t) = 6t + 5 −2t 5 e −. 2 2 23. In this problem, there are two forces acting on a boat: the wind force F1 and the water resistance force F2 . Since the proportionality constant in the water resistance force is different for the velocities below and above of a certain limit (5 m/sec for the boat A and 6 m/sec for the boat B), for each boat we have two differential equations. (Compare with Problem 7.) Let x1 (t) denote the distance passed by the boat A for the time t, v1 (t) := dx1 (t)/dt. Then the equation describing the motion of the boat A before it reaches the velocity 5 m/sec is dv m1 dt 134 (A) (A) (A) (A) = F1 + F2 = 650 − (A) b1 v1 ⇒ dv1 dt (A) = 65 4 (A) − v1 . 6 3 (3.12) ...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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