Solution: (a) Before opening the parachute, the differential equation is given by: dv dt. = v. v(0) = 0

Transcription

1 Math 2250 Lab 4 Name/Unid: 1. (25 points) A man bails out of an airplane at the altitute of 12,000 ft, falls freely for 20 s, then opens his parachute. Assuming linear air resistance ρv ft/s 2, taking ρ=0.2 without the parachute and ρ= 2 with the parachute. (a) Write down the ODE that describes the velocity v before he opens the parachute. (b) Find the velocity at time 3 s and the velocity when he just opens the parachute. (c) Find the height when he opens the parachute. (d) Write down the ODE describing the velocity after he opens the parachute. (e) Write down the equation for the height depending of time. (f) How long will it take him to reach the ground (from the moment he bails out of the plane)? Solution: (a) Before opening the parachute, the differential equation is given by: dv dt = v v(0) = 0 (b) Solving the ODE by integrating factor and using the initial condition, we get: v(t) = 160(e 0.2t 1) Then, we can evaluate v(t) at time t=3 and t=20 (20s is when the man opens the parachute) v(3) = ft/s, v(20) = ft/s (c) For finding the height, integrate v(t) so we get: y(t) = 800e 0.2t 160t + c Using the initial condition y(0)=12,000 ft. c= 12,800. Thus He opens the parachute at t=20s y(t) = 800e 0.2t 160t + 12, 800

2 y(20) = ft (d) After opening the parachute, the differential equation change to: dv = v dt v(0) = (e) Solving again by integrating factor and using the initial condition, we get: Integrating v(t) for getting the height, v(t) = 16e 2t y(t) = 8e 0.2t t + c Using the initial condition y(0) = , the equation for the height is: y(t) = 8e 0.2t 16t (f) To reach the ground means y = 0 Thus solving the last equation for t, y = 0 when t = s The total of time descending is t = = s. Page 2

3 2. (35 points) For the problem 1, apply the following methods to the initial value problem. Use step size h=1 on the time interval [0,3]. You have to compute this by hand (using calculator, not programming) (a) Euler Method. (b) Improved Euler Method. (c) Runge Kutta Method. (d) Compare these solutions with the exact solution (part b, problem 1) and comment your results. Solution: h=1, time interval [0,3], v(0) = 0 (a) Euler Method, v n+1 = v n + h f(t n, v n ) v 1 = 32 v 2 = 57.6 v 3 = (b) Improved Euler. n = 0 k 1 = f(t 0, v 0 ) = f(v 0 ) = 32 u 1 = v 0 + h k 1 = 32 k 2 = f(u 1 ) = 25.6 v 1 = v h(k 1 + k 2 ) = n = 1 k 1 = f(v 1 ) = u 2 = v 1 + h k 1 = k 2 = f(u 2 ) = v 2 = v h(k 1 + k 2 ) = n = 2 k 1 = f(v 2 ) = u 3 = v 2 + h k 1 = k 2 = f(u 3 ) = v 3 = v h(k 1 + k 2 ) = Page 3

4 (c) R K Method k 1 = f(x n, y n ) k 2 = f(x n + h 2, y n + h 2 k 1) k 3 = f(x n + h 2, y n + h 2 k 2) k 4 = f(x n+1, y n + hk 3 ) v n+1 = v n + h 6 (k 1 + k 2 + k 3 + k 4 ) n = 0 k 1 = 32 k 2 = k 3 = k 4 = v 1 = n = 1 k 1 = k 2 = k 3 = k 4 = v 1 = n = 2 k 1 = k 2 = k 3 = k 4 = v 1 = Page 4

5 (d) Results: We can compare the three methods with the answer on problem 1, part b), the exact solution is v(3)= Which Euler gives us v(3)= with error e=5.8899, Improved Euler v(3)= with error e=0.409 and Runge Kutta is v(3)= with error e= So clearly the best approximation is given by Runge Kutta method, being Euler Method the least accurate. Page 5

6 3. (40 points) If we drop a package from a helicopter with a parachute attached, the wind resistance provided by the parachute is not really linear, assume for a certain type of parachute the acceleration equation is modeled by (Assume velocity in ft/s) v (t) = v 0.15v 1.5 v(0) = 0 (a) What is the terminal velocity in this model? Notice that the slope function v 0.15v 1.5 is decreasing function of v for v > 0 that its value is 32 when v=0, and that its limit as v is. Thus the slope function has exactly one root and you can find it with the Matlab (or other software) solve command. (b) This is a differential equation which does NOT have an elementary solution. Use numerics.m to estimate the solution values at t=2 and t=5 seconds with the methods Euler, Improved Euler and Runge Kutta with step sizes h=0.2 and h=0.02. Write the approximate solution for each method with each h and each time t=2 and t=5. Plot your result and comment about the accuracy of all the methods. Which of the values of h you think is better, why? (c) Numerical Observations: What does the asymptote on the plot mean? What happen if you change the initial condition to v(0)= 5 and v(0)= 40? Plot these two different initial value problem for all the methods on the same plot.(you can use h=0.2 and work on the time interval [0,6]) What difference do you see on the plots of those initial value problems? What do you think is the reason of the difference on the plots? Solution: (a) The terminal velocity is found by calculating the root of the right hand side (acceleration equals to zero), using a software (Matlab) which is giving you ft/s (b) Plot of the methods: Euler, Improved Euler and RK for h=0.2 Page 6

7 35 30 Euler Heun RK y x For h=0.2 the values of v on t=2 are the following: Euler v(2)= Improved Euler v(2)= RK v(2)= For h=0.2 the values of v on t=5 are the following: Euler v(5)= Improved Euler v(5)= RK v(5)= Plot of the methods: Euler, Improved Euler and RK for h= Euler Heun RK y x Page 7

8 For h=0.02 the values of v on t=2 are the following: Euler v(2)= Improved Euler v(2)= RK v(2)= For h=0.2 the values of v on t=5 are the following: Euler v(5)= Improved Euler v(5)= RK v(5)= We can see that on our second graph the three methods give us similar numerical solutions. So h=0.02 is a better option than h=0.2 Page 8