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Differential equations help, modelling population

  1. Jun 15, 2010 #1
    in a scientific study, the size , p of a population at t hours is being studied. Initially p = 560 and after 6 hours, p is found to be 1218. In a simple model of population growth, the rate of the population is taken to be proportional to the population at that time. Using this model predict.

    a) the size of the population 24 hours after the start of the experiment.
    b) how long it will take for the population to increase tenfold.

    I keep getting the wrong answer, here's my working any help appreciated.


    [tex] \frac{dp}{dt} = kt [/tex]

    [tex] \int \frac{1}{p} dp = \int k dt [/tex]

    [tex] ln (p) = kt +c [/tex]

    [tex] p = p_{0}e^{kt} [/tex] p_{0} = 560 , t= 0

    [tex] p = 560e^{kt} [/tex]

    [tex] 1218 = 560e^{k6} [/tex]

    [tex] k = ln(\frac{1218}{3360}) [/tex]

    so to find after 24 hours I would just put t = 42 into my formula, and get ;

    [tex] p = 560e^{ln\frac{1218}{3360}24} [/tex]

    But when I solve this I get 4872, and the correct answer is 12532

    What am I doing wrong?

    Thank you.
     
  2. jcsd
  3. Jun 15, 2010 #2

    rl.bhat

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    Homework Helper

    Your calculation of k value is wrong.

    [tex]6k = ln(\frac{1218}{560})[/tex]

    [tex]k = \frac{1}{6}ln(\frac{1218}{560})[/tex]

    Now find k.
     
  4. Jun 15, 2010 #3
    Thank you,

    But thats what I have dont, is that not the same as multiplying the denominator by 6?
     
  5. Jun 15, 2010 #4

    rl.bhat

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    Homework Helper

    No. Find the logarithm and divide it by 6 to get k.
     
  6. Jun 15, 2010 #5
    Okay, thank you.

    I get it now
     
  7. Jun 15, 2010 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Though not important there is a typo in your original equation. You clearly intended the equation to be
    [tex]\frac{dp}{dt}= kp[/tex]
    not "= kt".

     
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