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Growth problem

  1. Feb 11, 2007 #1
    Here is the first question:
    A population P obeys the logistic model. It satisfies the equation
    [​IMG]

    Assume that P(0)=3. Find P(66)

    First I multiplied both sides by dt and integrated, giving:
    P=6/700Pt(7-P)+c
    If P(0)=3 then c=3
    P=6/700Pt(7-P)+3

    Then I divided everything by P and had
    1=6/700t(7/P-1)+3/P

    Now to find P(66)
    1=6/700*66(7/P-1)+3/P
    1=396/700(7/P-1)+3/P
    1=2772/700P-396/700+3/P
    4872/700P=1096/700
    P=4.445

    That's not right, what am I missing?
    Thanks.
     
  2. jcsd
  3. Feb 11, 2007 #2

    cristo

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    Your first equation is not visible. Perhaps you could rewrite it?
     
  4. Feb 11, 2007 #3

    ranger

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    You have to accept some sort of web certificate to view the equation. It seems to be located on some university's website. Heres the equation that I get:

    [tex]\frac{dP}{dt} = \frac {6}{700}P(7-P)[/tex]
     
  5. Feb 11, 2007 #4

    cristo

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    Ahh, ok, thanks for that, ranger. I must have clicked no automatically!

    You have this equation: [tex]\frac{dP}{dt} = \frac {6}{700}P(7-P)[/tex]. You cannot simply multiply by dt and integrate, since you have not integrated the terms including P wrt P! You must rearrange the equation to give: [tex]\int \frac{dP}{P(7-P)}=\int\frac{6}{700}dt +C[/tex]

    Do you know how to solve this?
     
  6. Feb 11, 2007 #5

    ranger

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    You did put +C accidentally right?
     
    Last edited: Feb 11, 2007
  7. Feb 11, 2007 #6
    Yeah, I can solve that. Didn't think to seperate variables for some retarded reason. Thanks for the help.
     
  8. Feb 11, 2007 #7

    cristo

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    Yea, I guess I haven't really integrated anything yet, so strictly the constant doesn't appear until the next line!

    You're welcome!
     
  9. Feb 11, 2007 #8
    OK I lied, I'm still not getting the right answer.

    dP/P(7-P)=6/700dt
    1/7log(P)-1/7log(7-P)=6t/700+c
    log(P)-log(7-P)=6t/100+c
    Using e
    P-7+P=e^(6t/100)+c
    P=(e^(6t/100)+7)/2+c
    P(0)=3 so c=-1
    Subbing 66 for t, I get 28.729

    Still not right, what am I doing wrong now?
    Thanks again.
     
  10. Feb 11, 2007 #9

    cristo

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    The - sign in red should be a +
    What you've done here is wrong. You must collect the logarithmic terms before you can take the exponential of both sides.
     
  11. Feb 11, 2007 #10
    You lost me with the collecting. Can you give me another example?
     
  12. Feb 11, 2007 #11

    cristo

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    Well, have you come across the general rule: log(a)+log(b)=log(ab) ?
     
  13. Feb 11, 2007 #12
    If I had I'd forgotten it. I should be able to solve from here (again). Thanks again for helping.
     
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