Logistics Equations

 

  1. Modeling Population

    The population of Tahoe was 15,000 in 1984 and now is 20,000.  If we use the model that the growth is proportional to the population, then

            dP/dt = kp 

    or 

            dp/p = kdt, 

            dp/p = kdt

            lnP = kt + c, P = Cekt  

    since 

            P(0) = 15000, C = 15,000

    so 

    P = 15000ekt  

    since 

            P(15) = 20000,

            20000 = 15000ek15  

    or 

            k = .019

    this model says that the population  of Tahoe in 200 years will be

    P(200) = 15000e.019(215) = 891,629 

    or almost one million people.

    We should hope that this is not a good model.

    Alternatively, we can assume that the population will never become greater than 50,000 people and that the rate of population growth is proportional to the product of the population and 50,000 minus the population.

            dP/dt = kP(50000 - P)

            dP/[P(50000 - P)] = kdt

            dP/P(50000 - P)   =   kdt

    Now use partial fractions:

            A/P + B/(50000 - P) => A(50000 - P) + BP = 1

            A = 1/50000, B = 1/50000

    Hence

            1/50000 [dP/P + dP/(50000 - P)] = kt + C

            ln(P) - ln(50000 - P) = at + b

            ln[P/(50000 - P) = at + b

            P(0) = 15,000 gives

            ln15/35 = b, b = ln3/7

            P(15) = 20000 gives

            ln2/3 = 15a + ln3/7 or

            a = (ln(14/9)/15

    Hence in 200 years we have

            ln[P/(50000 - P] = at + b

    or 

            P/(50000 - P) = eat+ b = M

            P = 50000M - MP, P = 50000M/(1 + M)

    Now when 

            t = 215, M =   241.2

    so that 

            P =     49794

    Which is a more reasonable estimation.


  2. Exercises:  

    1. On Monday the Ebola virus is caught by 3 people in an isolated community of 10,000 people.  By Thursday, 40 people have the virus.  How long will it be until half of the population have the virus.

    2. Water is pouring out of a lake at the rate of 10 gallons per second.  Water from an contaminated MTBE source of 3 mg per gallon is pouring into the lake at 12 gallons per second.  The lake initially is uncontaminated and contains 1,000,000 of water.  How long will it take until the lake has 1 mg per gallon of MTBE contamination?  Hint:  Rate = Rate In - Rate Out.

       

 

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