# Population Dynamics

#### alane1994

##### Active member
Suppose that the population of a species of fish in a certain lake is hypothesized to grow according to a logistic model of population growth with growth rate $$r=0.3$$ and carrying capacity $$k=3000$$. Assume initially there are 2500 fish of the species in the lake. Determine the correct differential equation for each of the scenarios below.

(a) Each year 150 fish are harvested from the lake.
(b) Each year 25% of the fish are harvested from the lake.
(c) What is a maximum (within 50) safe fixed amount to harvest each year in order to assure that there will always be some fish in the lake.

Now I am looking and re-reading in my book, but I am rather confused by this. I am reading in the section entitled "Autonomous Equations and Population Dynamics".

#### MarkFL

Staff member
If $P$ is the population at time $t$, $r$ is the growth rate, and $k$ is the carrying capacity, then we may model the population with the IVP:

$$\displaystyle \frac{dP}{dt}=rP\left(1-\frac{P}{k} \right)$$ where $$\displaystyle P(0)=P_0$$

This model does not account for any harvesting. So what you need to do is introduce a term that accounts for the harvesting in each case. Beginning with part (a), what should you add to the ODE to account for 150 fish being harvested each year?

#### alane1994

##### Active member
Would it be a simple subtraction of 150?

#### MarkFL

Staff member
Would it be a simple subtraction of 150?
Yes, it is as simple as that.

#### alane1994

##### Active member
Would we subtract $$0.25 P_0$$?

On a side note, I asked this on another site that I am rather active on.
OpenStudy
The guy literally copied what you said, verbatim.

Funny enough he didn't even format it properly so that the $$\LaTeX$$ would display....

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#### MarkFL

Staff member
You are subtracting 25% of the initial population each year, you want to subtract 25% of the current population instead...

When you said someone copied what I said, I thought perhaps they gave you the same equation, but I see they did in fact simply copy and paste my response.

#### alane1994

##### Active member
How would I format that so that it would be subtracting 0.25 of the population of the specific years?

$$0.25 P$$ perhaps?

And would you put that into the equation where the P normally would be?

$$\displaystyle \frac{dP}{dt} = rP(1-\frac{0.25P}{k})$$

Well, I have to go now. I will be back on later, and if not, definitely tomorrow!

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#### MarkFL

Staff member
How would I format that so that it would be subtracting 0.25 of the population of the specific years?

$$0.25 P$$ perhaps?

And would you put that into the equation where the P normally would be?

$$\displaystyle \frac{dP}{dt} = rP(1-\frac{0.25P}{k})$$
Yes, the term to add to the original ODE would be $$\displaystyle -\frac{1}{4}P$$.

If you replace $P$ as you did with $0.25P$ you are changing the natural carrying capacity...you are in effect making it 4 times as large.

For part (c) we want to analyze the stationary solutions. Suppose we define the annual harvest to be $$\displaystyle 50n$$ where $$\displaystyle n\in\mathbb{N}$$, then we want to look at the roots of:

$$\displaystyle \frac{3}{10}P-\frac{1}{10000}P^2-50n=0$$

$$\displaystyle P^2-3000P+500000n=0$$

If the discriminant is negative, then there are no stationary solutions, and extinction is imminent.

You see, we have the time rate of change of the population being equal to a quadratic, which opens downward. If there are no real roots, then the population must always be on the decline, since $$\displaystyle \frac{dP}{dt}<0$$ for all $P$.

#### alane1994

##### Active member
Thank you very much for your help and time, I really appreciate it! I remember back to when we briefly covered this earlier in the semester... didn't quite get it then either