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(1) Construct the differential equation forQ(t).(2) Solve the differential equation.(3) Predict the long term quantity of the salt in the reservoir.Solution: (1)Rate of change of total quantity of salt = Rate In - Rate Out;Rate of change of total quantity of salt =dQdt;Rate In = salt IN per day = salt water volume IN per day×concentration of salt waterIN;Rate Out = salt OUT per day = salt water volume OUT per day×concentration ofsalt water OUT.We havedQdt= 0.1×50-1×Q100,⇒dQdt=500-Q100,Q(0) = 0.(2)Q(t) = 500(1-e-0.01t).(3)limt→∞Q(t) = 500.23
7.4 Exponential Growth and DecayIf the population growthdPdtis proportional to the population sizeP(t), thendPdt=kP.The solution isP(t) =P(0)ekt.k >0→growth,k <0→decay.•Absolute growth rate = rate of change of population =dP/dt,•Relative growth rate = percent change per unit time = (dP/dt)/P,•Doubling time (D) of exponential growth = the time required for it to double:P(t) =P(0)2t/D.•Half life (H) of exponential growth = the time required for it to be half:P(t) =P(0)12t/H.•Newtons Law of Cooling states that the rate of change of the temperature of a coolingbody is proportional to the difference between its temperature T and the temperatureof its surrounding mediumTs. The model is:dTdt=-k(T-Ts),wherekis a constant.Example.A bacteria culture growth at a rate proportional to its size. After 2 hours thereare 40 bacteria and after 4 hours the count is 120. Find an expression for the populationafter t hours.Solution: We measure the timetin hours. LetP(t) be the population atthours, thenwe havedPdt=kP.The solution of the equation isP(t) =P(0)ekt.24
Note thatP(2) = 40 andP(4) = 120, we obtain40 =P(0)e2k,120 =P(0)e4k.These imply thatP(0) =403ande2k= 3,ork= ln 3/2.We thus haveP(t) =4033t/2=403√3t=403e(ln 3/2)t.Example.The half-life of Sodium-24 is 15 hours. Suppose you have 100 grams of Sodium-24. How many grams remaining after 27 minutes (keep three decimals)?Solution: Assumem(t) be the amount afterthours. Thenm(t) =m(0)(12)t/H,wherem(0) = 100,H= 15 hours. Note that 27 minutes = 27/60=0.45 hours. Thusm(0.45) = 100(12)0.45/15= 100(12)0.03= 97.942g.Example.Use Newton’s Law of Cooling to determine the time of death of a healthy man.•He died in his room some time before noon;•At noon, his body temperature was found to be 70 degrees;•His body cooled another 5 degrees in 1 hour after noon;•The room temperature was a constant 60 degrees;•Normal temperature of people’s body is 98.6 degree.Solution: LetT(t) be the temperature of the body at timet. Taking noon ast= 0, wehaveT(0) = 70. Note thatTs= 60, we havedTdt=-k(T-60),this equation is separable. We obtaindTT-60=-kdt,⇒T(t) = 60 +Ce-kt.25
FromT(0) = 70 we get 70 = 60 +C,C= 10 andT(t) = 60 + 10e-kt.At 1:00pm, his body temperature is 70-5 = 65. HenceT(1) = 65.65 = 60 + 10e-k(1),⇒5 = 10e-k,⇒k= ln 2,⇒T(t) = 60 + 10e-tln 2.Thus98.6 = 60 + 10e-tln 2,⇒t=-ln 3.86/ln 2 =-1.95,which means 1 hour and 0.95(60) minutes before noon, or 1 hour and 57 minutes beforenoon. Or the time of death is 10:03AM.