What is the solution for calculating population growth with mice?

[patrickmoloney]

Homework Statement

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 $M_0$ mice, has per-capita birth rate of $8 \frac{1}{week}$ and a per-capita death rate of $2\frac{1}{week}$. Also, 20 mouse traps are set each fortnight and they are always filled.

(a)Write down the word equation for the mice population $M(t)$
(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 $M(t)$ at any time $t$ in terms of the initial population $M_0$
(d) Find the equilibrium solution $M_e$
(e)Find the long-term solution with the dependence on $M_0$. What happens when $M_0 = M_e$
(f)Find the time interval on which $M(t)\ge 0$ with dependence on $M_0$
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.

Homework Equations

The Attempt at a Solution

(a) $$\Bigg( \text{Rate of change} \text{ of number of mice} \Bigg) = \Big(\text{Rate of Births}\Big)-\Big(\text{Normal rate of Reaths}\Big) - \Big(\text{Rate of deaths by mousetraps}\Big)$$

(b) $$\dfrac{dM}{dt} = 16M - 4M - 20$$ $$M(0) = M_0$$

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

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

(d) I think this has to do when the rate of change population stops changing. When $\frac{dM}{dt}=0$ 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.

 

[scottdave]

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.

 

[patrickmoloney]

I think that's what they want for long-term behavior.

What do you need to do to find long-term behaviour?, I've never hear of it before thanks for your reply :)

 

[Ray Vickson]

Homework Statement

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 $M_0$ mice, has per-capita birth rate of $8 \frac{1}{week}$ and a per-capita death rate of $2\frac{1}{week}$. Also, 20 mouse traps are set each fortnight and they are always filled.

(a)Write down the word equation for the mice population $M(t)$
(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 $M(t)$ at any time $t$ in terms of the initial population $M_0$
(d) Find the equilibrium solution $M_e$
(e)Find the long-term solution with the dependence on $M_0$. What happens when $M_0 = M_e$
(f)Find the time interval on which $M(t)\ge 0$ with dependence on $M_0$
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.

Homework Equations

The Attempt at a Solution

(a) $$\Bigg( \text{Rate of change} \text{ of number of mice} \Bigg) = \Big(\text{Rate of Births}\Big)-\Big(\text{Normal rate of Reaths}\Big) - \Big(\text{Rate of deaths by mousetraps}\Big)$$

(b) $$\dfrac{dM}{dt} = 16M - 4M - 20$$ $$M(0) = M_0$$

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

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

(d) I think this has to do when the rate of change population stops changing. When $\frac{dM}{dt}=0$ 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.

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.)

 

[patrickmoloney]

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!

Does that mean it should be $$\frac{dM}{dt}= 4M + M -20$$ I thought I needed to double the per-capita to get the right numbers.

 

[patrickmoloney]

"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.)

If the equilibrium solution is to find the $\lim_{t\to\infty}$ then what does long term solution want me to find? I read equilibrium is when $\frac{dM}{dt }=0$ Thanks for your reply.

 

[Ray Vickson]

Does that mean it should be $$\frac{dM}{dt}= 4M + M -20$$ I thought I needed to double the per-capita to get the right numbers.

Sorry: I did not notice that you doubled the rates. Your equation/solution are OK.

 

[Ray Vickson]

If the equilibrium solution is to find the $\lim_{t\to\infty}$ then what does long term solution want me to find? I read equilibrium is when $\frac{dM}{dt }=0$ Thanks for your reply.

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.

 

[patrickmoloney]

So for part (d) I just found $$M_e = \frac{5}{3}$$ I found this by setting $\frac{dM}{dt}=0$. I'm trying to solve part (e) by $$\lim_{t\to\infty} (M_0-\frac{5}{3})e^{12t}+\frac{5}{3}$$ The $M_0$ is really throwing me off however.

 

[Ray Vickson]

So for part (d) I just found $$M_e = \frac{5}{3}$$ I found this by setting $\frac{dM}{dt}=0$. I'm trying to solve part (e) by $$\lim_{t\to\infty} (M_0-\frac{5}{3})e^{12t}+\frac{5}{3}$$ The $M_0$ is really throwing me off however.

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.

 

[patrickmoloney]

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.

That makes so much sense. cause of what's in the parenthesis $M_0 -\frac{5}{3}$. 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?

 

[Ray Vickson]

That makes so much sense. cause of what's in the parenthesis $M_0 -\frac{5}{3}$. 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?

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.

 

[patrickmoloney]

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.

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?

 

[patrickmoloney]

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.

I got it! thanks for all your help. I found $$t = \frac{1}{12}\log\Bigg(\frac{\frac{5}{3}}{\frac{5}{3}-M_0}\Bigg)$$