y = (1 y)cos t Answer: The equation is not autonomous because of the cos t term.

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1 Math 211 Homework #4 Februar 9, = Answer: Note that = is autonomous, having form = f(). Solve the equation f()= 0 to find the equilibrium points. f()= = 0 = 1. Thus, (t) = 1 is an equilibrium solution, as shown in the following figure t Equilibrium solution of = = (1 )cos t Answer: The equation is not autonomous because of the cos t term.

2 f () ( 1,0) (1,0) Answer: The equilibrium points are =±1. Both are unstable. The equilibrium solutions are (t) = 1 and (t) = = 2 7 Answer: (i) In this case, f()= 2 7, whose graph is shown in the figure below f ()=2 7 (7/2,0) Graph of the right-hand side of = 2 7. (ii) The phase line is easil captured from the figure above and is shown in the figure below. 7 2 Phase line for = 2 7.

3 (iii) The phase line in the figure above indicates that solutions decrease if <7/2 and increase if >7/2. This allows us to easil construct the phase portrait shown in the t plane in the figure below. Note the unstable equilibrium solution, (t) = 7/2. (0,7/2) t Phase portrait for = = ( + 1)( 2 9) Answer: (i) In this case, f() = ( + 1)( 2 9) factors as f() = ( + 1)( 3)( + 3), whose graph is shown in the figure below. f ()=(+1)( 2 9) ( 3,0) ( 1,0) (3,0) Graph of the right-hand side of = ( + 1)( 2 9). (ii) The phase line is easil captured from the figure above, and is shown in the figure below.

4 3 1 3 Phase line for = ( + 1)( 2 9). (iii) The phase line in the figure below shows that solutions decrease if < 3, increase for 3 << 1, decrease if 1 <<3, and increase for >3. This allows us to easil construct the phase portrait shown in the t plane in the figure below. Note the unstable equilibrium solution, (t) = 3, the stable equilibrium solution, (t) = 1, and the unstable equilibrium solution, (t) = 3. (0,3) t (0, 1) (0, 3) Phase portrait for = ( + 1)( 2 9) A tank contains 100 gallons of pure water. A salt solution with concentration 3 lb/gal enters the tank at a rate of 2 gal/min. Solution drains from the tank at a rate of 2 gal/min. Use qualitative analsis to find the eventual concentration of the salt solution in the tank. Answer: Let x(t) represent the amount of salt in the tank at time t. The rate at which solution enters the tank is given b Rate In = 2 gal/min 3 lb/gal = 6 lb/min. The rate at which solution leaves the tank is Consequentl, Rate Out = 2 gal/min dx dt x 100 lb/gal = 1 50 x lb/min. = x.

5 Let c(t) represent the concentration of salt in the solution at time t. Thus, c(t) = x(t)/100 and 100c = x. 100c = (100c) c = c Let f(c) = 6/100 (1/50)c. Setting f(c) = 0 produces the equilibrium point c = 3, as shown in the figure below. f (c)= c (3,0) c Graph of f(c)= 6/100 (1/50)c The phase line on the c-axis in the figure above shows that c = 3 is a stable equilibrium point so a trajector with initial condition c(0) = 0 (the initial concentration of salt is zero) should approach the stable equilibrium solution c(t) = 3, as shown in the figure below,

6 c (0,3) (0,0) t Eventual salt concentration A biologist prepares a culture. After 1 da of growth, the biologist counts 1000 cells. After 2 das of growth, he counts Assuming a Malthusian model, what is the reproduction rate and how man cells were present initiall? Answer: The equation of the Malthusian model is P(t) = Ce rt. Appl the cell counts to solve for r and C. P(1) = 1000, so Ce r = 1000, i.e. r = ln(1000/c). Also, P(2) = 3000, so Ce 2r = 3000, i.e. r = 1 ln(3000/c). Setting these equal 2 and solving, one obtains C = 1000/3. Substituting C into r = ln(1000/c) gives that r = ln Suppose a population is growing according to the logistic equation, ( dp = rp 1 P ). dt K Prove that the rate at which the population is increasing is at its greatest when the population is at one-half of its carring capacit. Hint: Consider the second derivative of P. Answer: To find the minimum and maximum growth rate, use the first derivative test on the growth rate dp /dt. That is, set d 2 P/dt 2 = 0. Obtain d 2 ( P dt = 2 rp 1 2P ). k Thus, dp /dt reaches an extremum when P = 0 and when P = k/2. One wants to show that dp /dt is a maximum when P = k/2. So, one checks that d 2 P/dt 2 changes sign from positive to negative at P = k/2. When P = k/2, P = r(k/2)(1 (k/2)/k) = rk/4, which is positive. So an possible change in sign of d 2 P/dt 2 is in the third factor (1 2P/k). This factor does change sign from positive to negative at P = k/2, so d 2 P/dt 2 changes sign from positive to negative at P = k/2. Therefore dp /dt reaches a maximum at P = k/2.

7 In The Biolog of Population Growth, published in 1925, the biologist Ramond Pearl reported the data shown in Table 3 for the growth of a population of fruit flies. Table 1 The growth of another population of fruit flies. Da Number of flies Da Number of flies Da Number of flies (a) Notice that data were not collected sstematicall. However, data were collected on das 9 and 18. Use the method of Example 1.18 to estimate the natural reproductive rate and the carring capacit for a logistic model. Answer: Use the following three data points to estimate r and k: Da Number of Flies P P P Set h = 9. Equation (1.17) gives r = 1 ( ) 163(52 6) 9 ln = (163 52) Then using this in equation (1.16), one obtains K = (6)(52)(1 e 0.269(9) ) = e 0.269(9) The plot of the logistic curve with these parameters is shown as the dashed curve in Figure 1. (b) Answer: This part was not assigned since the nonlinear least squares program required to do it is not readil available. However, we will present the results because the comparison of the two results is quite striking. To compute the parameters we want to find values of the parameters P 0, r, and K which minimise 11 [ ( )] P(tj ) 2 log, j=1 where t j and P j are the data from Table 1, and P(t)is the logistic function in equation with parameters P 0, r, and K. To do this we write a MATLAB m-file ex313b.m P j

8 with the contents function = ex313b(u) K = u(1); P0 = u(2); r = u(3); t=[ ]; N=[ ]; pp = K*P0./(P0 + (K - P0)*exp(-r*t)); = log(pp./n); Then we execute the commands >> u = lsqnonlin( ex313b,[320,6,0.3]); Optimization terminated successfull: Relative function value changing b less than OPTIONS.TolFun >> K=u(1) K = >> P0 = u(2) P0 = >> r=u(3) r = (c) The plot of the logistic curve with these parameters is shown as the solid curve in Figure 1. The MATLAB command lsqnonlin is not part of the standard MATLAB. It is part of the Optimization Toolbox, which must be purchased separatel. This means that while the command should be available on an Rice owned computer, it will probabl not be available on our home computer using the Student Edition of MATLAB. Is the logistic model a good one for this data? Answer: As can be seen in Figure 1, the logistic model with the parameters from part (a) poorl approximates the data for short times, and at long times it is terrible.

9 However, when we use all of the data, as we did in part (b), we get a logistic curve that fits reasonabl well. 400 Number of flies Time in das Figure A biologist develops a culture that obes the modified logistic equation ( P = 0.38p 1 P ) h(t), (1) 1000 where the harvesting is defined b the piecewise function { 200, if t<3, h(t) = 0, otherwise. (2) (a) Use our numerical solver to plot solution trajectories for initial bacterial populations ranging between 0 and You ll note that in some cases, the population recovers, but in others, the bacterial count goes to zero. Determine experimentall the critical initial population that separates these two behaviors. Answer: In Figure 2, the nine curves are given b initial conditions P(0) = 300, P(0) = 400, P(0) = 410, P(0) = 414, P(0) = 415, P(0) = 420, P(0) = 600, P(0) = 800, P(0) = Note that the solution curve with initial condition P(0) = 415 never reaches zero, although the curve with initial condition P(0) = 414 does. Thus, the critical population is between 414 and 415.

10 p p = (0.38) p (1 p/1000) h h = 200 (t<3) t Figure 2. (b) Use an analtic method to determine the exact value of the critical initial population found in part (a). Justif our answer. Answer: The critical initial population will be the initial condition for a solution curve which has P = 0 when t = 3. Writing the differential equation for t<3gives dp = 0.38P P dt Separate variables. dp P P 200 = dt. Integrating this, one obtains t + C = ( ) P tan Using conditions P(3) = 0, one obtains 3 + C = ( ) tan or C = Now setting t = 0 and solving for P = P 0, one obtains P 0 = On the da of his birth, Jason s grandmother pledges to make available $50,000 on his eighteenth birthda for his college education. She negotiates an account paing 6.25% annual interest, compounded continuousl, with no initial deposit, but agrees to deposit a fixed amount each ear. What annual deposit should be made to reach her goal? Answer: Let P(t) represent the balance in the account t ears after Jason s da of birth. Let r represent the annual rate, d the annual deposit. Since no initial investment is required, P = rp + d, P(0) = 0.

11 This equation is linear, with integrating factor e rt. Consequentl, ( e rt P ) = de rt, e rt P = d r e rt + C, P = d r + Cert. Use P(0) = 0 to produce C = d/r and P(t) = d r ( e rt 1 ). Because P(18) = 50000, = d ( e r(18) 1 ), r d = 50000r e 18r 1, d = 50000(0.0625) e 18(0.0625) 1, d $1, Don and Heidi would like to bu a home. The ve examined their budget and determined that the can afford monthl paments of $1000. If the annual interest is 7.25% and the term of the loan is 30 ears, what amount can the afford to borrow? Answer: Let P(t)represent the loan balance after t ears. Let r represent the annual interest rate, w the annual pament, and P 0 the amount of the loan. Then P = rp w, P(0) = P 0. The equation is linear with integrating factor e rt. Consequentl, ( e rt P ) = we rt, e rt P = w r e rt + C, P = w r + Cert. Use P(0) = P 0 to produce C = P 0 w/r and P(t) = w ( r + P 0 w ) e rt. r

12 Now, $1,000 per month makes $12,000 per ear, so w = Furthermore, the term of the loan is 30 ears, so P(30) = 0 and 0 = w ( r + P 0 w ) e r(30), r we 30r w = P 0 re 30r, P 0 = w ( e 30r 1 ), re 30r P 0 = ( e 30(0.0725) 1 ), e 20(0.0725) P 0 $146, 713. M3.18. Here are the MATLAB commands that will produce the figure. x=0:0.01:4*pi; hold on plot(x, sin(x), - ) plot(x, cos(x), : ) legend( = sin(x), = cos(x) ) xlabel( x ) label( ) hold off M3.19. Here are the MATLAB commands that will produce the figure. x=-3:0.01:3; hold on plot(x, sin(x), - ) plot(x, x - (x.^3)/6, -- ) plot(x, x - (x.^3)/6 + (x.^5)/120, : ) legend( = sin(x), = x - x^3/6, = x - x^3/6 + x^5/120 ) title( Talor polnomial approximations to = sin x. ) xlabel( x-axis ) label( -axis ) hold off The polnomials given are the third- and fifth-order Talor polnomials for the sine function.