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Having difficulty setting up a Diff Eq for this situation

  1. Jul 6, 2004 #1
    Suppose the population of mosquitos in a certain area increases at a rate proportional to the current population... in the absense of other factors, the population doubles each week. There are 200,000 mosquitos in the area initially, and predators (birds, etc) eat 20,000 mosquitos /day. Determine the populatoin of mosquitos in the area at any time t.

    So I know that for this proportional rate--- we can set up a differential equation of :

    dP/dt = rP - (The number eaten)

    so dP/dt = 1*P - 140000

    I use r =1 due to the fact that the population is doubling--- so increasing by 100 percent, and i am using 140000 because thats how many mosquitos are eaten in a week--- I am trying to get everything so t is in weeks.

    So simplified

    [tex]dP/dt = P-140000[/tex]

    and I simply solve for P using seperable equations, however the book gives the answer as

    P = 201977.31 - 1977.31(e^(ln2)t) for 0 < t < 6.6745 (in weeks)

    I obviously did not get this solution from the one above--- for what should I be looking to solve this properly and get a consistent answer?

    Thanks for your helP!!
     
  2. jcsd
  3. Jul 6, 2004 #2

    AKG

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    [tex]dP/dt = P - 140000[/tex]

    [tex]\int _{P_0} ^{P(t)} \frac{dP}{P - 140000} = \int _0 ^t dt[/tex]

    [tex]\ln [P(t) - 140000] - \ln (P_0 - 140000) = t[/tex]

    [tex]\ln [P(t) - 140000] = t + \ln (60000)[/tex]

    [tex]P(t) = 60000e^t + 140000[/tex]

    Actually, there may be a reason why this is wrong. In a week, it is not as though the population doubles, then 140,000 mosquitoes are eaten. Mosquitoes are constantly being eaten, and then doubling. If we could just say the mosquitoes double, then so many are eaten, then we should be able to say that so many are eaten, and then they double. This will give a different result. So your (our) approach had it so that they double, and then instantly at the end of the week, so many are eaten. So either this approach is wrong (and I'm not sure of the right approach) or we had it right, and the book's answer was wrong.
     
  4. Jul 6, 2004 #3
    Yeah I understand what you are considering--- but it would seem that if they wanted us to even consider it, that they wouldve told us that as the week goes by whether the growth rate of the mosquitos is linear or nonlinear... etc. Interesting....
     
  5. Jul 6, 2004 #4

    AKG

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    You're right. My best guess now is to take their solution, and reverse engineer it and see if you can figure out what equation in the form "dP/dt = ..." they started out with, and maybe you can figure out how they got it with the given numbers. For one, I can't understand why they'd have [itex]e^{t(\ln 2)}[/itex] and not just [itex]2e^t[/itex]. You can rewrite their equation:

    [tex]P = 200000 + 1977.31(1 - 2e^t)[/tex]

    ... if I understand what you wrote correctly. So, what is the signficance of 1977.31? Does it relate to 140,000 in anyway? Is ln(1977.31) a signficant number? If you're sure the book is right, it might not be a waste of time to figure out how they got their equation.

    Wait a minute. Please expand the rewritten version of the formula, and confirm that it is the equation the book's given. Then use that formula to evaluate P(0). If my equation is a correct interpreation of the book, [itex]P(0) \neq 200,000[/itex] according to that formula.
     
  6. Jul 7, 2004 #5
    Hmm their equation does indeed satisfy the IC--- the equation i wrote in the first message is exactly what they had in the book--- i double cheked--- interesting...
     
    Last edited: Jul 7, 2004
  7. Jul 7, 2004 #6
    No...The solution of your book is correct. The rate you include in your equation is wrong. Since the mosquitoes double each week the rate "r" in your equation should be ln(2), not 1.

    To "feel" the correctness of the solution, just note that 201977.31=140000/ln(2).
     
  8. Jul 7, 2004 #7
    btw e^(t*ln(2)) isnt the same as 2e^t i realized--- that may have been another error we made whoops!
     
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