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Math 31S. Rumbos Fall 2011 1 Solutions to Assignment #16 1. Logistic Growth1 . Suppose that the growth of a certain animal population is governed by the diο¬erential equation 1000 ππ = 100 β π, π ππ‘ (1) where π (π‘) denote the number of individuals in the population at time π‘. (a) Suppose there are 200 individuals in the population at time π‘ = 0. Sketch the graph of π = π (π‘). Solution: The equation in (1) describes logistic growth in a population with intrinsic growth rate π = 100/1000 and carrying capacity πΎ = 100. A sketch of the solution with initial population π (0) = 200 is shown in Figure 1. β‘ π 100 π‘ Figure 1: Sketch of Solution to (1) with ππ = 200 (b) Will there ever be more than 200 individuals in the population? Will there ever be fewer than 100 individuals? Explain your answer. Solution: The sketch of the solution to (1) subject to the initial condition π (0) = 200 shows that the population size will never be above 200 or below 100. β‘ 1 Adapted from Problem 6 on page 521 in HughesβHallett et al, Calculus, Third Edition, Wiley, 2002 Math 31S. Rumbos Fall 2011 2 2. Spread of a viral infection2 . Let πΌ(π‘) denote the total number of people infected with a virus. Assume that πΌ(π‘) grows according to a logistic model. Suppose that 10 people have the virus originally and that, in the early stages of the infection the number of infected people doubles every 3 days. It is also estimated that, in the long run 5000 people in a given area will become infected. (a) Solve an appropriate logistic model to ο¬nd a formula for computing πΌ(π‘), where π‘ is the time from the initial infection measured in weeks. Sketch the graph of πΌ(π‘). Solution: The function πΌ solves the logistic equation ππΌ = ππΌ(πΎ β πΌ), ππ‘ (2) where π is the intrinsic growth rate of infection and πΎ is the limiting number of people who will become infected in the long run. Thus, πΎ= Λ 5000. (3) In order to estimate π, we approximate the spread of the infection with an exponential model with doubling time of 3 days or 3/7 weeks. Thus, π= Λ ln 2 = Λ 1.6173, 3/7 (4) in units of 1/week. The solution to (2) subject to the initial condition πΌ(0) = πΌπ is given by πΌ(π‘) = πΌπ πΎ , πΌπ + (πΎ β πΌπ )πβππ‘ for π‘ β β. (5) Substituting the values of πΌπ = 10, and πΎ and π given in (3) and (4), respectively, into (5) yields the solution πΌ(π‘) = 50000 , 10 + (4990)πβ1.6173π‘ for π‘ β β. A sketch of the graph of the function in (6) is pictured in Figure 2. 2 (6) β‘ Adapted from Problem 7 on page 521 in HughesβHallett et al, Calculus, Third Edition, Wiley, 2002 Math 31S. Rumbos Fall 2011 3 πΌ πΎ π‘ Figure 2: Sketch of function in (6) (b) Estimate the time when the rate of infected people begins to decrease. Solution: The rate of infection will begin to decrease when the number of infected people is half of the limiting value; namely, when πΌ(π‘) = 2500, or, according to (6), when 50000 = 2500. 10 + (4990)πβ1.6173π‘ (7) Solving the equation in (7) yields π‘= Λ 1 ln(499) = Λ 3.84 weeks. 1.6173 Thus, the rate of infection will begin to decrease in about 3 weeks and 5 days and 21 hours. β‘ 3. NonβLogistic Growth3 . There are many classes of organisms whose birth rate is not proportional to the population size. For example, suppose that each member of the population requires a partner for reproduction, and each member relies on chance encounters for meeting a mate. Assume that the expected number of encounters is proportional to the product of numbers of female and male members in the population, and that these are equally distributed; hence, 3 Adapted from Problem 12 on page 39 in Braun, Diο¬erential Equations and their Applications, Fourth Edition, SpringerβVerlag, 1993 Math 31S. Rumbos Fall 2011 4 the number of encounters will be proportional to the square of the size of the population. Use a conservation principle to derive the population model ππ = ππ 2 β ππ, ππ‘ (8) where π and π are positive constants. Explain your reasoning. Solution: Begin with the conservation principle ππ = Rate of individuals in β Rate of individuals out. ππ‘ (9) In this case we have Rate of individuals in = ππ 2 , (10) Rate of individuals outππ, (11) and where π and π are positive constants of proportionality. The equation in (8) follows from (9) after substituting (10) and (11). β‘ 4. For the equation in (8), (a) ο¬nd the values of π for which the population size is not changing; Solution: Rewrite the equation in (8) as ( ) π ππ = ππ π β . ππ‘ π (12) ππ π = 0 when π = 0 or π = . β‘ ππ‘ π (b) ο¬nd the range of positive values of π for which the population size is increasing, and those for which it is decreasing; ππ π ππ Solution: We see from (12) that > 0 for π > , and < 0 for ππ‘ π ππ‘ π π π < . This, the population size increases for π > , and decreases for π π π π< . β‘ π We see from (12) that Math 31S. Rumbos Fall 2011 5 (c) ο¬nd ranges of positive values of π for which the graph of π = π (π‘) is concave up, and those for which it is concave down; Solution: Diο¬erentiate on both sides of (8) with respect to π‘ to obtain ππ π2 π ππ βπ , (13) = 2ππ 2 ππ‘ ππ‘ ππ‘ where we have applied the Chain Rule. The equation in (13) can be rewritten as ) ( π2 π π ππ . (14) = 2π π β ππ‘2 2π ππ‘ ππ Substituting the expression for in (12) into (14) then yields ππ‘ ( )( ) π2 π π π 2 = 2π π π β πβ . (15) ππ‘2 2π π π2 π is ππ‘2 determined by the signs of the two rightβmost factors in (15). The signs of these two factors are displayed in Table 1. The concavity of of the graph In view of (15) we see that, for positive values of π , the sign of πβ π 2π β + + πβ π π β β + 0 π/2π π/π β²β² π (π‘) + β + graph of π (π‘) concaveβup concaveβdown concaveβup Table 1: Concavity of the graph of π = π (π‘) of π = π (π‘) is also displayed in Table 1. From that table we get that the graph of π = π (π‘) is concave up for 0<π < and concave down for π 2π or π π> , π π π <π < . 2π π β‘ Math 31S. Rumbos Fall 2011 6 (d) Sketch possible solutions. Solution: Putting together the information on concavity in Table 1 and the fact that π (π‘) increases for π > π/π and decreases for 0 < π < π/π, we obtain the sketches of possible solutions to the equation in (8) displayed in Figure 3. π π π π‘ Figure 3: Possible Solutions to Logistic equation β‘ 5. For the equation in (8), (a) use separation of variables and partial fractions to ο¬nd a solution satisfying the initial condition π (0) = ππ , for ππ > 0. Solution: Separate variable in the equation in (12) to obtain β« β« 1 ππ = π ππ‘. (16) π (π β π/π) Use partial fractions in the integrand on the leftβhand side to (16) and integrate on the rightβhand side to get to get } β« { π 1 1 β + ππ = ππ‘ + π1 , (17) π π π β π/π for some constant π1 . Evaluate the integral on the leftβhand side of (17) and simplify to get ) ( β£π β π/πβ£ = ππ‘ + π2 , (18) ln β£π β£ Math 31S. Rumbos Fall 2011 7 for some constant π2 . Next, take the exponential function on both sides of (18) to get β£π β π/πβ£ = π3 πππ‘ , (19) β£π β£ where we have set π3 = ππ2 . Using the continuity of π and of the exponential function we deduce from (19) that π β π/π 2 = π ππ π‘/π , (20) π for some constant π. The equation in (20) can now be solved for π as a function of π‘ to get π/π π (π‘) = . (21) 1 β π πππ‘ Next, use the initial condition π (0) = ππ to obtain from (20) that π= ππ β π/π . ππ (22) Substituting the value of π in (22) into (21) yields π (π‘) = ππ π/π . ππ + (π/π β ππ ) πππ‘ (23) β‘ (b) What happens to π (π‘) as π‘ β β if ππ > π/π? What happens if ππ < π/π? Why is π/π called a threshold value? Solution: We ο¬rst consider the case in which 0 < ππ < π/π. In this case, the function in (23) is deο¬ned for all values of π‘ and lim π (π‘) = 0, π‘ββ since π > 0. On the other hand, is ππ > π/π, then the function in (23) ceases to exist when (ππ β π/π) πππ‘ = ππ . As π‘ approaches that time, π (π‘) β β. Thus, depending on whether ππ < π/π or ππ > π/π, the population will eventually go extinct or it will have unlimited growth in a ο¬nite time. Thus, π/π is the threshold population value which determines growth or extinction. β‘