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 growth with mice

  1. Mar 19, 2017 #1
    1. The problem statement, all variables and given/known data
    Hey guys I'm struggling to find much information of modelling single species population dynamics that relates to this question. A question like this is going to be coming up in my final exam and I need to be able to solve it. I'm struggling to even know where to start. I'm completely new to modelling populations. help would be much appreciated.

    A population, initially consisting of [itex]M_0[/itex] mice, has per-capita birth rate of [itex]8 \frac{1}{week}[/itex] and a per-capita death rate of [itex]2\frac{1}{week}[/itex]. Also, 20 mouse traps are set each fortnight and they are always filled.

    (a)Write down the word equation for the mice population [itex]M(t)[/itex]
    (b) Write the differential rate equation for the number of mice.
    (c) Solve the differential rate equation to obtain the formula for the mice population [itex]M(t)[/itex] at any time [itex]t[/itex] in terms of the initial population [itex]M_0[/itex]
    (d) Find the equilibrium solution [itex]M_e[/itex]
    (e)Find the long-term solution with the dependence on [itex]M_0[/itex]. What happens when [itex]M_0 = M_e[/itex]
    (f)Find the time interval on which [itex]M(t)\ge 0[/itex] with dependence on [itex]M_0[/itex]
    No idea what's being asked here. I can't seem to find any information. It doesn't help that there is no decent books about this online that can be found easily.

    thanks for all the help.

    2. Relevant equations


    3. The attempt at a solution
    (a) [tex]\Bigg(
    \text{Rate of change}
    \text{ of number of mice}
    \Bigg) = \Big(\text{Rate of Births}\Big)-\Big(\text{Normal rate of Reaths}\Big) - \Big(\text{Rate of deaths by mousetraps}\Big)
    [/tex]

    (b) [tex]\dfrac{dM}{dt} = 16M - 4M - 20[/tex] [tex]M(0) = M_0[/tex]

    (c) The differential equation I've got is[tex]\frac{dM}{dt}= 12M -20[/tex]. Using separation of variables we can find [itex]M(t)[/itex]
    [tex]M(t) = \dfrac{e^{12t+12c}+20}{12}[/tex] The question is asking for it in terms of initial population [itex]M_0[/itex]. We know that [itex]M(0)= M_0[/itex]. Therefore
    [tex]M_0 = \dfrac{e^{12c}+20}{12}[/tex]

    We know that [itex]e^{c_1}= C[/itex]. Then we can find [itex]M(t)[/itex] in terms of [itex]M_0[/itex]
    Which is [tex]M(t) = \dfrac{(12M_0 -20)e^{12t}+20}{12}[/tex]
    Is this what they mean by "in terms of the intial population [itex]M_0[/itex]"?

    (d) I think this has to do when the rate of change population stops changing. When [itex]\frac{dM}{dt}=0[/itex] I don't know how to find the solutions though.

    (e) I'm not sure how to find the long term behaviour, what does the question want me to find?

    (f) i also don't know what they are asking here since it's probably related to part (e).

    Thanks for all the help. I'm really struggling with this problem. been at it for about 4 hours now and there is nothing online that is even like this question.
     
    Last edited: Mar 19, 2017
  2. jcsd
  3. Mar 19, 2017 #2

    scottdave

    User Avatar
    Homework Helper
    Gold Member

    Your workings out to determine that M0 = (e^(12c) + 20)/12 looks correct to me. Note that since c is an arbitrary constant, then e^(12c) is also a constant. I will let k = 12*e^12c, so we have M0 = k + (20/12) = k + 5/3. Note that the solution to your differential equation is now M(t) = k*e^(12t) + (5/3). Then you can substitute k = M0-(5/3) into this to get an equation based on M0.

    Yes the equilibrium situation is when there is no change over time (dM/dt = ). Since you know that dM/dt = 12M - 20, just set this equal to zero and you have M = 20/12 or 1 2/3 mice. You cannot have a fractional mouse, but since each unit of time is 2 weeks, the number would oscillate between 1 and 2 mice; I think that's what they want for long-term behavior.
     
  4. Mar 19, 2017 #3
    What do you need to do to find long-term behaviour?, I've never hear of it before thanks for your reply :)
     
  5. Mar 19, 2017 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Your units are not correct: the birth and (normal) death rates are "per week", but the trapping deaths are "per two weeks"--a fortnight is two weeks!

    "Equilibrium solution" means the solution that will persist over the long run. If you look at your formula for ##M(t)## in terms of ##M_0## and ##t##, what happens to it as ##t \to \infty?## In order to have finite behavior for large ##t## you need some condition on the value of ##M_0##. (However, in my opinion, the question is rather poorly worded, so it is no wonder you are confused.)
     
  6. Mar 19, 2017 #5
    Does that mean it should be [tex]\frac{dM}{dt}= 4M + M -20 [/tex] I thought I needed to double the per-capita to get the right numbers.
     
  7. Mar 19, 2017 #6
    If the equilibrium solution is to find the [itex]\lim_{t\to\infty} [/itex] then what does long term solution want me to find? I read equilibrium is when [itex]\frac{dM}{dt }=0[/itex] Thanks for your reply.
     
  8. Mar 19, 2017 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Sorry: I did not notice that you doubled the rates. Your equation/solution are OK.
     
  9. Mar 19, 2017 #8

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Sometimes "equilibrium" means limiting behavior, but sometimes it means ##dM/dt=0##; it all depends on how the system behaves. However, your system is simple enough that both meanings are the same.
     
  10. Mar 19, 2017 #9
    So for part (d) I just found [tex]M_e = \frac{5}{3}[/tex] I found this by setting [itex]\frac{dM}{dt}=0[/itex]. I'm trying to solve part (e) by [tex]\lim_{t\to\infty} (M_0-\frac{5}{3})e^{12t}+\frac{5}{3}[/tex] The [itex]M_0[/itex] is really throwing me off however.
     
  11. Mar 19, 2017 #10

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    If ##M_0 > 5/3## we have ##M(t) \to +\infty## as ##t \to \infty##---population explosion. If ##M_0 < 5/3## the population ##M(t)## falls to zero at some finite time ##t## and then stays there---the differential equation no longer applies after that. That is the case of population extinction. For the very special case of ##M_0 = 5/3## the population remains finite and non-zero for all time---in fact, remains constant.
     
  12. Mar 19, 2017 #11
    That makes so much sense. cause of what's in the parenthesis [itex]M_0 -\frac{5}{3}[/itex]. Is that all the question means when it want long tern solution? I thought it was asking for graphs or something. Thanks.

    is there a way to find how long the mice population is greater than or equal to zero?
     
  13. Mar 19, 2017 #12

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    If ##M_0 < 5/3##, the population reaches zero when ##(M_0 - 5/3) e^{12 t} + 5/3 = 0.## Solve for ##t## using logarithms.
     
  14. Mar 19, 2017 #13
    unfortunately I'm getting the natural log of a negative number which of course if undefined. is there a way around getting an undefined answer?
     
  15. Mar 19, 2017 #14
    I got it! thanks for all your help. I found [tex]t = \frac{1}{12}\log\Bigg(\frac{\frac{5}{3}}{\frac{5}{3}-M_0}\Bigg)[/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Population growth with mice
  1. Population Growth (Replies: 6)

Loading...