Matlab 20D, Week 2 Exercise 2.2 (a) : DField drew lines of solutions of points on the plot that I d click on. b)

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1 Matlab 20D, Week 2 Exercise 2.2 (a) In this exercise, we are going to use DFIELD to plot the direction field of (4). This can be done by entering the following values into the DFIELD setup window. In the boxes under the words "The differential equation" enter y' = (exp(-x) - y)*(exp(-x) y). In the independent variable box enter x. Now click in a few places on the plot. What are the lines which DFIELD has drawn? Also, using Options -> Keyboard input from the display window, enter the values x=2 and y=3 and click Compute. : DField drew lines of solutions of points on the plot that I d click on. (b) Considering how complicated differential equation (4) appears to be, what do you think is the utility of plotting direction fields? : To give us a sense of what the solution looks like, what the solutions are at several points, and/or find the general solution through the graph. Exercise 2.3 Plot the direction field of (5). Suppose that the experiment also reveals that the initial value is about (0,1). Click on points near (0,1) on the graph itself. Using these plots, think about what would happen if the initial value in the problem were not exactly (0,1)? Would this greatly affect what the solution looks like for this differential equation? : If the initial value in the problem were not exactly (0,1), that would not be a big problem, since solutions near the point resemble that of (0,1) in concavity and trend.

2 Exercise 2.4 Plot the direction field of (6). Use Options -> Keyboard input to plot the solution passing though (1,1) and then click on a few points near it. If the initial value were not exactly (1,1), how would does this affect the solution? : Solutions near (1,1) diverge, greatly affecting what the solution looks like. Exercise 2.5 Enter the above equation into DFIELD. In the Parameters & expressions section enter A = 1 and k = 2. In the section below that, change the minimum value of t to 0 (since we are not interested in negative values of t here). Plot the direction field and include it in your Word document. Plot the direction fields for different values of A and click on the direction field to plot some solutions for each of these values. What property do you think A represents in real life? : At differing values of A, the stabilization line moves correspondingly (i.e. A=1, stabilization line, where all solutions converge, is at y=1; A=2 is at y=2, and so forth). A is probably the constant representing the temperature of the environment.

3 Exercise 2.6 Let us try to figure out how long it will take to defrost a frozen chicken breast in the fridge, which keeps a constant temperature of 41 F. The skinless boneless chicken breast has been in the freezer so its temperature is uniform at -6 F. We'll suppose k = 0.4, based on the properties of the chicken. (a) Recall that an initial value problem consists of a differential equation along with an initial condition. Write out the initial value problem which we must solve here. : Here the initial temperature of the chicken is -6 degf: dy/dt=k*(41-y) with initial conditions, y(0)=-6. (b) What do you think the value of A should be? : A is the ambient temperature, which is 41 degf. (c) Let us consider the chicken breast fully defrosted when the temperature at its center reaches 39 F. How long does it take to defrost a chicken breast under the above conditions? A rough estimate from a direction field plot is sufficient. : Solution: 41-47*exp(-(2*t)/5). The chicken reaches 39 degf in ~8 time units. (d) How much time would be saved if the delicious chicken breast were thawed on the kitchen counter instead, given that room-temperature is around 69 F? : Solution: 69-75*exp(-(2*t)/5). It appears to take around 2.3 time units to reach 39 degf. Exercise 2.7 Use PPLANE to plot the direction field of (9). Click at some points on the diagram to see the solution curve through those points. Now try changing the values of x' and y' from 2 and -3 to other constant values. How does this change the direction field? : Changing the constant values of x and y changes the slope of the direction field. Exercise 2.8 Set the minimum values for both x and y to -10, and the maximum values to 10. Run PPLANE: (10). Click on some points on the diagram to see the solution curve through those points.

4 Exercise 2.9 Plot the solution to the differential equation y' = y^3 + x^2 using DFIELD. Now apply PPLANE to the system: x' =1 x(0)=0 y' = y^3 + x^2 Take a few different initial values for y(0). What do you notice? Dfield Plot: PPlane Plot: : Looks to me like they are very similar. There are subtle differences though, likely due to PPlane treating x and y as both dependent variables and accounting for the solution s trajectory C.

5 Exercise 2.10 (a) Enter the system (11) above into PPLANE, setting the parameter values a = b = c = d = 1. Plot the vector field for the system. Where in the x-y plane are the physically possible solutions (remember - x and y represent populations!)? : Although it is an unusual graph, the physically possible solutions lie in x,y>/=0. (b) Click on a few different places in the graph to view some solutions. (c) The predator-prey system has the following behaviors: A. The populations of foxes and rabbits are both relatively small. B. The small number of foxes allows the rabbit population to increase C. The increased number of rabbits allows the number of foxes to increase D. The increase in the fox population causes the rabbit population to decrease. E. The decreased supply of rabbits causes the fox population to decrease, returning to behavior A. On your plot in your Word document mark one of the solutions (that is, one loop) with the behaviors A, B, C, D and E where they occur on that solution.