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Models for Population Growth

  1. Jul 23, 2017 #1
    1. The problem statement, all variables and given/known data
    The Pacific halibut fishery has been modeled by the differential equation.

    [itex]\displaystyle\dfrac{dy}{dt}=ky\left(1-\dfrac{y}{M} \right)[/itex]

    where y(t) is the biomass (the total mass of the members of the population) in kilograms at time t (measured in years), the carrying capacity is estimated to be [itex]M = 7\times 10^7 kg[/itex], and [itex]k=0.78[/itex] per year.


    (a) If [itex]y(0)= 2\times 10^7 kg[/itex], find the biomass a year later. (Round your answer to two decimal places.)
    (b) How long will it take for the biomass to reach [itex]4\times 10^7 kg[/itex]? (Round your answer to two decimal places.)


    2. Relevant equations

    [itex]\displaystyle P= \dfrac{K}{1+Ce^{-kt}}[/itex]

    3. The attempt at a solution

    K = carrying capacity [itex]\implies 7\times 10^7 kg[/itex]
    k = [itex]0.78[/itex] per year

    At time 0, biomass is [itex]2\times 10^7 kg[/itex] [itex]\implies [/itex][itex]y(0)= 2\times 10^7 kg[/itex]

    C = the difference between the carrying capacity and the initial capacity subtracted by 1.

    [itex]C= \dfrac{7\times 10^7}{2 \times 10^7}-1=\dfrac{5}{2}[/itex]



    [itex]P=\dfrac{7 \times 10^7}{1+\dfrac{5}{2}e^{-0.78\cdot1}}= 70000001.15 kg[/itex]

    I'm trying to solve for the biomass (P) after 1 year. This answer doesn't seem correct. Am I using the wrong number for variable t? Or I'm not solving for P right away?
     
  2. jcsd
  3. Jul 23, 2017 #2

    mfb

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    The two sides here are not equal. The denominator is larger than 1, the fraction cannot be larger than the numerator.
     
  4. Jul 23, 2017 #3

    Ray Vickson

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    Is ##P## the same as ##y##? I assume so.

    Your final equation is incorrect; it should be
    $$ \frac{7 \times 10^7}{1 +\displaystyle \frac{5}{2} e^{-0.78 \times 1}} = 4 \times 10^7$$
     
  5. Jul 23, 2017 #4
    Thanks, guys, I got the right answer now. Major parentheses mistake for the calculator.

    I can solve b, but for a.) I got [itex]3.26\times10^7[/itex]
     
  6. Jul 24, 2017 #5

    Ray Vickson

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    Why the "but"? Your answer for (a) is correct.
     
  7. Jul 24, 2017 #6
    Sorry for my bad English. For (a) I got [itex]3.26 \times 10^7[/itex]
     
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