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Figure 4.1Figure 4.2 LR Series circuitFigure 4.3 RC Series CircuitFigure 4.4Newton’s Law of Cooling/WarmingPopulation Growth and decayMixture of Two Salt SolutionsSurvivability with AIDSDraining a TankPursuit ProblemHarvesting of Renewable Natural ResourcesChapter 4Applications of First-order Differential Equations to Real WorldSystems4.1 Cooling/Warming Law4.2 Population Growth and Decay4.3 Radio-Active Decay and Carbon Dating 4.4 Mixture of Two Salt Solutions 4.5 Series Circuits4.6 Survivability with AIDS 4.7 Draining a tank 4.8 Economics and Finance 4.9 Mathematics Police Women4.10 Drug Distribution in Human Body 4.11 A Pursuit Problem 4.12 Harvesting of Renewable Natural Resources 4.13 Exercises In Section 1.4 we have seen that real world problems can be representedby first-order differential equations. In chapter 2 we have discussed few methods to solve first orderdifferential equations. We solve in this chapter first-order differential equationsmodeling phenomena of cooling, population growth, radioactive decay, mixture ofsalt solutions, series circuits, survivability with AIDS, draining a tank, economics78and finance, drug distribution, pursuit problem and harvesting of renewablenatural resources.4.1 Cooling/Warming lawWe have seen in Section 1.4 that the mathematical formulation ofNewton’s empirical law of cooling of an object in given by the linear first-orderdifferential equation (1.17))mTα(TdtdTThis is a separable differential equation. We have αdt)mT(TdTor ln|T-Tm |=t+c1or T(t) = Tm+c2et(4.1)Example 4.1: When a chicken is removed from an oven, its temperature ismeasured at 3000F. Three minutes later its temperature is 200o F. How long will ittake for the chicken to cool off to a room temperature of 70oF.Solution: In (4.1) we put Tm = 70 and T=300 at for t=0.T(0)=300=70+c2e.0This gives c2=230For t=3, T(3)=200Now we put t=3, T(3)=200 and c2=230 in (4.1) then 200=70 + 230 e.3or 2301303αe 79or 2313ln3 or 19018.02313ln31Thus T(t)=70+230 e-0.19018t(4.2)We observe that (4.2) furnishes no finite solution to T(t)=70 since limit T(t) =70. t ¥The temperature variation is shown graphically in Figure 4.1. We observethat the limiting temperature is 700F.Figure 4.14.2 Population Growth and DecayWe have seen in section 1.4.1 that the differential equation )()(tkNdttdNwhere N(t) denotes population at time t and k is a constant ofproportionality, serves as a model for population growth and decay of insects,animals and human population at certain places and duration. Solution of this equation is N(t)=Cekt, where C is the constant of integration: kdttNtdN)()(80Integrating both sides we getlnN(t)=kt+ln Cor ktCtN)(lnor N(t)=CektC can be determined if N(t) is given at certain time. Example 4.2: The population of a community is known to increase at a rateproportional to the number of people present at a time t. If the population hasdoubled in 6 years, how long it will take to triple?Solution : Let N(t) denote the population at time t. Let N(0) denote the initialpopulation (population at t=0).)(tkNdtdNSolution is N(t)=Aekt , where A=N(0)Ae6k=N(6) =2N(0) = 2Aor e6k=2 or k = 61ln 2Find t when N(t)=3A=3N(0)or N(0) ekt=3N(0)or te)2(ln613 or ln 3= 6)2(ln tor t= 2ln3ln69.6 years (approximately 9 years 6 months)81Example 4.3 Let population of country be decreasing at the rate proportional toits population. If the population has decreased to 25% in 10 years, how long will ittake to be half?Solution: This phenomenon can be modeled by )t(kNdtdNIts solution is N(t)=N(0) ekt, whereN(0) in the initial population For t=10, N(10)=41N(0)41N(0) = N(0) e10kor e10k=41or k=101ln 41Set N(t)=41N(0))0(N21e)0(Nt41ln101or t=41ln10121ln 8.3 years approximately. Example 4.4 Let N(t) be the population at time t and Let N0 denote the initialpopulation, that is, N(0)=N0.Find the solution of the model 822)()( tbNtaNdtdNwith initial condition N(0)=NoSolution: This is a separable differential equation, and its solution is tdsds)s(bN)s(aN)s(dNt02t0bNaBNA)bNa(N1bNaN12To find A and B, observe that )()()()(bNaNNbABAabNaNBNbNaAbNaBNATherefore, Aa+(B-bA)N=1. Since this equation is true for all values of N,we see that Aa=1 and B-bA=0. Consequently, A=a1, B=b/a, and )bsa(sdsNN0 = ds)bsabs1(a1NNobNaobNaoNNalnln1|bNabNa|NNlna1ooThus at = lnbNabNaNN00It can be verified that )t(bNabNao is always positive for 0<t<∞. Hence83at = ln bNabNoaNNoTaking exponentials of both sides of this equation gives eat=bNaobNaoNNN0(a-bN)eat = (a-bN0)NBringing all terms involving N to the left-hand side of this equation, we seethat [a-bNo + bN0eat] N(t) = aN0eator N(t)=ateobNobNaateoaN4.3 Radio-active Decay and Carbon Dating As discussed in Section 1.4.2. a radioactive substance decomposes at arate proportional to its mass. This rate is called the decay rate. If m(t) representsthe mass of a substance at any time, then the decay rate dtdmis proportional tom(t). Let us recall that the half-life of a substance is the amount of time for it todecay to one-half of its initial mass. Example 4.5. A radioactive isotope has an initial mass 200mg, which two yearslater is 50mg. Find the expression for the amount of the isotope remaining at anytime. What is its half-life?Solution: Let m be the mass of the isotope remaining after t years, and let -k bethe constant of proportionality. Then the rate of decomposition is modeled by dtdm= - km,84where minus sign indicates that the mass is decreasing. It is a separableequation. Separating the variables, integrating, and adding a constant in the formlnc, we get lnm+lnc = - ktSimplifying, lnmc = - kt (4.3)or mc = e-ktor m = c1e-kt, where c1=c1To find c1, recall that m =200 when t=0. Putting these values of mand t in (4.3) we get200 = c1 e-ko = c1.1or c1=200and m = 200e-kt (4.4)The value of k may now be determined from (4.4) by substituting t=2,m=150.150 = 200 e-2kor 432 keor –2k=ln 43This gives 2134ln21k (0.2877)= 0.1438 0.14The mass of the isotope remaining after t years is then given by 85m(t) =200e -.1438tThe half-life th is the time corresponding to m=100mg. Thus 100 = 200 e-0.14thor 21= e-0.14thor

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