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Population model: dP/dt = k sqrtP

  1. Jan 26, 2007 #1
    1. The problem statement, all variables and given/known data

    Suppose that when a certain lake is stocked with a population P of fish, the birth and death rates [tex]\alpha[/tex] and [tex]\beta[/tex] are inversely proportional to [tex]\sqrt{P}[/tex] , so that

    [tex] \frac{dP}{dt} = k \sqrt{p} [/tex]

    Find P(t) if P(0)=C

    3. The attempt at a solution

    [tex] \frac{dP}{\sqrt{p}} = k dt [/tex]


    [tex] 2 \sqrt{P} = k t + A [/tex]

    Sub 0 for t C for X

    [tex]A=2 \sqrt{C}[/tex]

    sub back in

    [tex] 2 \sqrt{P} = 2t+4 \sqrt{c} [/tex]

    [tex] P = (2 t k+4 \sqrt{C})^2 [/tex]
    Last edited: Jan 26, 2007
  2. jcsd
  3. Jan 27, 2007 #2
    [tex] 2 \sqrt{P} = 2t+4 \sqrt{c} [/tex]
    [tex] P = (2 t k+4 \sqrt{C})^2 [/tex]

    There are a few small problems with this, mostly just technical. Such as the second to last line is missing a k, and the we substitute C for P; there is no X.

    However, not so trivial is where did the coefficient of 2 for [tex] \sqrt{P}[/tex] go? How did the 2 in 2kt show up randomly? Your mistakes are algebraic and occur after your line "sub back in"

    Thus your solutions should be

    [tex] P(t) = \displaystyle\left( \frac{1}{2} kt + \sqrt(C) \right) ^2 [/tex]
  4. Jan 27, 2007 #3
    I ended up solving it correctly

    I ended up solving it correctly (2 min before it was due)

    the answer is

    [tex] \frac{(kt+2\sqrt{C})^2}{4} [/tex]
  5. Jan 27, 2007 #4


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    Science Advisor

    You were fine up to here.

    No, you just said [itex]2\sqrt{P}= kt+ A[/itex]. If, now, [itex]A= 2\sqrt{C}[/itex], then [itex]2\sqrt{P}= kt+ 2\sqrt{C}[/itex].
    Square both sides: [itex]4 P= (kt+ 2\sqrt{C})^2[/itex]
    [tex]P= \frac{kt+ 2\sqrt{C})^2}{4}[/tex]
    which is what you say you eventually got

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