The Fate of a Stocked Lake Population Under Fishing Pressure

  • Thread starter Shackleford
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In summary, the conversation discusses a lake that is stocked with walleye pike and how the population of pike is governed by the logistic equation. The logistic model is modified to account for fishing and the equilibrium points are found and classified as asymptotically stable and unstable. The fate of the fish population is discussed using qualitative analysis and it is determined that the population will decrease to zero if the initial population is 1000 fish and will increase to the stable equilibrium point of 8.8K fish if the initial population is 2000 fish. The stability of the equilibrium points is determined by testing the sign of P' in each interval.
  • #1
Shackleford
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Consider a lake that is stocked with walleye pike and that the population of pike is governed by the logistic equation

P' = 0.1P(1- P/10),

where time is measured in days and P in thousands of fish. Suppose that fishing is started in this lake and that 100 fish are removed each day.

(a) Modify the logistic model to account for the fishing.

P' = 0.1P(1- P/10) - 0.1

(b) Find and classify the equilibrium points for your model.

Equilibrium points: 1.12702 (asymptotically stable), 8.87298 (unstable)

(c) Use qualitative analysis to completely discuss the fate of the fish population with this model. In particular, if the initial fish population is 1000, what happens to the fish as time passes? What will happen to an initial population having 2000 fish?

With 1000 fish, the population will decrease to zero.

With 2000 fish, the population will increase to the 8.8K fish EQ point.


I hope I did this correctly.
 
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  • #2
I like your model and your EQ pts, but how did you decide stable vs. unstable?
 
  • #3
Billy Bob said:
I like your model and your EQ pts, but how did you decide stable vs. unstable?

Hey, Bob.

Well, I drew a phase line and tested the sign of P' in each of the three intervals. I just redid it and got a different answer.

Each of the intervals has a negative sign, so they are all unstable.
 
  • #4
Each of the intervals has a negative sign

That's still not what I got.
 
  • #5
Billy Bob said:
That's still not what I got.

Oops. The 1.2 < t < 8.8 interval is positive. The other two are negative. So, the 8.8 EQ point is stable. I think I had that originally.
 
  • #6
Shackleford said:
Oops. The 1.2 < t < 8.8 interval is positive. The other two are negative. So, the 8.8 EQ point is stable. I think I had that originally.

That's what I got.
 
  • #7
Billy Bob said:
That's what I got.

Great. So, my answer to (c) should be correct.
 

What is the Logistic Equation Problem?

The Logistic Equation Problem is a mathematical model used to describe the growth of a population that is limited by resources. It takes into account the population's initial size, the rate of growth, and the carrying capacity of the environment.

What is the formula for the Logistic Equation Problem?

The formula for the Logistic Equation Problem is: P(t) = K / (1 + A*e^(-rt)), where P(t) is the population at time t, K is the carrying capacity, A is the initial population, r is the growth rate, and e is the base of the natural logarithm.

How is the Logistic Equation Problem used in real life?

The Logistic Equation Problem is used in many fields, including biology, ecology, and economics. It can be used to model the growth of animal populations, the spread of diseases, and the demand for products or services.

What is the significance of the carrying capacity in the Logistic Equation Problem?

The carrying capacity represents the maximum number of individuals that an environment can support. In the Logistic Equation Problem, it acts as a limiting factor for population growth. Once the population reaches the carrying capacity, the growth rate decreases and the population stabilizes.

What are some limitations of the Logistic Equation Problem?

Some limitations of the Logistic Equation Problem include the assumption of continuous and unlimited resources, constant population growth rate, and lack of accounting for external factors such as predators or disease. It is also not applicable to all species and environments.

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