1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    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

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Population model: dP/dt = k sqrtP
  1. Population Model (Replies: 2)

  2. Population model (Replies: 1)