Math 31S. Rumbos Fall Solutions to Assignment #16


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1 Mth 31S. Rumbos Fll Solutions to Assignment #16 1. Logistic Growth 1. Suppose tht the growth of certin niml popultion is governed by the differentil eqution 1000 dn N dt = 100 N, (1) where N(t) denote the number of individuls in the popultion t time t. () Suppose there re 200 individuls in the popultion t time t = 0. Sketch the grph of N = N(t). Solution: The eqution in (1) describes logistic growth in popultion with intrinsic growth rte r = 100/1000 nd crrying cpcity K = 100. A sketch of the solution with initil popultion N(0) = 200 is shown in Figure 1. N 100 t Figure 1: Sketch of Solution to (1) with N o = 200 (b) Will there ever be more thn 200 individuls in the popultion? Will there ever be fewer thn 100 individuls? Explin your nswer. Solution: The sketch of the solution to (1) subject to the initil condition N(0) = 200 shows tht the popultion size will never be bove 200 or below Spred of virl infection 2. Let I(t) denote the totl number of people infected with virus. Assume tht I(t) grows ccording to logistic model. Suppose 1 Adpted from Problem 6 on pge 521 in Hughes Hllett et l, Clculus, Third Edition, Wiley, Adpted from Problem 7 on pge 521 in Hughes Hllett et l, Clculus, Third Edition, Wiley, 2002
2 Mth 31S. Rumbos Fll tht 10 people hve the virus originlly nd tht, in the erly stges of the infection the number of infected people doubles every 3 dys. It is lso estimted tht, in the long run 5000 people in given re will become infected. () Solve n pproprite logistic model to find formul for computing I(t), where t is the time from the initil infection mesured in weeks. Sketch the grph of I(t). Solution: The function I solves the logistic eqution di dt = ri(k I), (2) where r is the intrinsic growth rte of infection nd K is the limiting number of people who will become infected in the long run. Thus, K = (3) In order to estimte r, we pproximte the spred of the infection with n exponentil model with doubling time of 3 dys or 3/7 weeks. Thus, r = ln 2 3/7 = , (4) in units of 1/week. The solution to (2) subject to the initil condition I(0) = I o is given by I(t) = I o K, for t R. (5) I o + (K I o )e rt Substituting the vlues of I o = 10, nd K nd r given in (3) nd (4), respectively, into (5) yields the solution I(t) = 50000, for t R. (6) 10 + (4990)e t A sketch of the grph of the function in (6) is pictured in Figure 2. (b) Estimte the time when the rte of infected people begins to decrese. Solution: The rte of infection will begin to decrese when the number of infected people is hlf of the limiting vlue; nmely, when I(t) = 2500,
3 Mth 31S. Rumbos Fll I K t Figure 2: Sketch of function in (6) or, ccording to (6), when Solving the eqution in (7) yields t = = (7) 10 + (4990)e t 1 ln(499) = 3.84 weeks Thus, the rte of infection will begin to decrese in bout 3 weeks nd 5 dys nd 21 hours. 3. Non Logistic Growth 3. There re mny clsses of orgnisms whose birth rte is not proportionl to the popultion size. For exmple, suppose tht ech member of the popultion requires prtner for reproduction, nd ech member relies on chnce encounters for meeting mte. Assume tht the expected number of encounters is proportionl to the product of numbers of femle nd mle members in the popultion, nd tht these re eqully distributed; hence, the number of encounters will be proportionl to the squre of the size of the popultion. Use conservtion principle to derive the popultion model dn dt = N 2 bn, (8) 3 Adpted from Problem 12 on pge 39 in Brun, Differentil Equtions nd their Applictions, Fourth Edition, Springer Verlg, 1993
4 Mth 31S. Rumbos Fll where nd b re positive constnts. Explin your resoning. Solution: Begin with the conservtion principle dn dt In this cse we hve nd = Rte of individuls in Rte of individuls out. (9) Rte of individuls in = N 2, (10) Rte of individuls outbn, (11) where nd b re positive constnts of proportionlity. The eqution in (8) follows from (9) fter substituting (10) nd (11). 4. For the eqution in (8), () find the vlues of N for which the popultion size is not chnging; Solution: Rewrite the eqution in (8) s ( dn dt = N N b ). (12) We see from (12) tht dn dt = 0 when N = 0 or N = b. (b) find the rnge of positive vlues of N for which the popultion size is incresing, nd those for which it is decresing; Solution: We see from (12) tht dn dt > 0 for N > b, nd dn dt < 0 for N < b. This, the popultion size increses for N > b, nd decreses for N < b. (c) find rnges of positive vlues of N for which the grph of N = N(t) is concve up, nd those for which it is concve down; Solution: Differentite on both sides of (8) with respect to t to obtin d 2 N dt 2 = 2N dn dt bdn dt, (13)
5 Mth 31S. Rumbos Fll where we hve pplied the Chin Rule. The eqution in (13) cn be rewritten s ( d 2 N dt = 2 N b ) dn 2 2 dt. (14) Substituting the expression for dn dt d 2 N dt 2 = 22 N in (12) into (14) then yields ( N b ) ( N b ). (15) 2 In view of (15) we see tht, for positive vlues of N, the sign of d2 N is dt 2 determined by the signs of the two right most fctors in (15). The signs of these two fctors re displyed in Tble 1. The concvity of of the grph N b 2 N b b/2 b/ N (t) + + grph of N(t) concve up concve down concve up Tble 1: Concvity of the grph of N = N(t) of N = N(t) is lso displyed in Tble 1. From tht tble we get tht the grph of N = N(t) is concve up for nd concve down for 0 < N < b 2 b 2 < N < b. or N > b, (d) Sketch possible solutions. Solution: Putting together the informtion on concvity in Tble 1 nd the fct tht N(t) increses for N > b/ nd decreses for 0 < N < b/,
6 Mth 31S. Rumbos Fll N b t Figure 3: Possible Solutions to Logistic eqution we obtin the sketches of possible solutions to the eqution in (8) displyed in Figure For the eqution in (8), () use seprtion of vribles nd prtil frctions to find solution stisfying the initil condition N(0) = N o, for N o > 0. Solution: Seprte vrible in the eqution in (12) to obtin 1 N(N b/) dn = dt. (16) Use prtil frctions in the integrnd on the left hnd side to (16) nd integrte on the right hnd side to get to get { 1 } b N + 1 dn = t + c 1, (17) N b/ for some constnt c 1. Evlute the integrl on the left hnd side of (17) nd simplify to get ( ) N b/ ln = bt + c 2, (18) N for some constnt c 2. Next, tke the exponentil function on both sides of (18) to get N b/ = c 3 e bt, (19) N
7 Mth 31S. Rumbos Fll where we hve set c 3 = e c 2. Using the continuity of N nd of the exponentil function we deduce from (19) tht N b/ = c e 2t/b, (20) N for some constnt c. The eqution in (20) cn now be solved for N s function of t to get N(t) = b/. (21) 1 c ebt Next, use the initil condition N(0) = N o to obtin from (20) tht Substituting the vlue of c in (22) into (21) yields N(t) = c = N o b/ N o. (22) N o b/. (23) N o + (b/ N o ) ebt (b) Wht hppens to N(t) s t if N o > b/? Wht hppens if N o < b/? Why is b/ clled threshold vlue? Solution: We first consider the cse in which 0 < N o < b/. In this cse, the function in (23) is defined for ll vlues of t nd lim N(t) = 0, t since b > 0. On the other hnd, is N o > b/, then the function in (23) ceses to exist when (N o b/) e bt = N o. As t pproches tht time, N(t). Thus, depending on whether N o < b/ or N o > b/, the popultion will eventully go extinct or it will hve unlimited growth in finite time. Thus, b/ is the threshold popultion vlue which determines growth or extinction.
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