Question

It is more likely that the amount of fishing is governed by the current number of fish present, so instead of a constant number of fish being caught, the rate is proportional to the current number of fish present, with proportionality constant \(k\), as \(P^{\prime}=0.4 P\left(1-\frac{P}{10000}\right)-k P\)Solve this equation, assuming a value of \(k=0.05\) and an initial condition of 2000 fish.

Short answer

The population of fish over time is given by \( P(t) = \frac{10000}{1 + 3e^{-0.35t}} \).

Step-by-step solution

Understanding the Differential Equation

The given differential equation is \( P' = 0.4P(1 - \frac{P}{10000}) - kP \). It is a logistic differential equation with harvesting, where \( P \) denotes the population of fish, and \( k \) is a proportionality constant representing the rate at which fish are being caught.

Substitute the Value of k

Substitute \( k = 0.05 \) into the equation: \( P' = 0.4P(1 - \frac{P}{10000}) - 0.05P \). This simplifies to: \( P' = P(0.4 - 0.05)(1 - \frac{P}{10000}) \).

Simplify the Equation

Simplify the equation by reducing the terms inside the brackets: \( P' = P(0.35)(1 - \frac{P}{10000}) \), which is \( P' = 0.35P(1 - \frac{P}{10000}) \).

Set Up the Integrative Form

The simplified equation \( P' = 0.35P(1 - \frac{P}{10000}) \) is separable. Integrate to solve. Rewrite it as \( \frac{dP}{P(1 - \frac{P}{10000})} = 0.35 dt \).

Integrate Both Sides

Separate the variables to integrate both sides. Integrate the left: \( \int \frac{dP}{P(1 - \frac{P}{10000})} \), and the right: \( \int 0.35 dt \). Use partial fraction decomposition for the left side integration.

Integration Using Partial Fractions

The left side can be split into partial fractions: \( \frac{1}{P} + \frac{1}{10000-P} \). Integrate to get \( \ln|P| - \ln|10000-P| = 0.35t + C \), where \( C \) is an integration constant.

Solve for Integration Constant C

Use the initial condition \( P(0) = 2000 \) to determine \( C \). Substitute into the integrated equation: \( \ln|2000| - \ln|8000| = C \). Simplify to find \( C = \ln(\frac{2000}{8000}) = \ln(0.25) \).

Solve for P

Plug \( C = \ln(0.25) \) back into the equation: \( \ln|P| - \ln|10000-P| = 0.35t + \ln(0.25) \). Exponentiate both sides to solve for \( P(t) \). Simplify to get the solution for \( P(t) \): \( P(t) = \frac{10000}{1 + 3e^{-0.35t}} \).

Verify Solution

Verify the solution \( P(t) = \frac{10000}{1 + 3e^{-0.35t}} \) satisfies the initial condition by checking if \( P(0) = 2000 \). Substitute \( t=0 \) in the solution to confirm.

Key concepts

Fish Population Model

A fish population model is a mathematical representation used to predict and understand changes in fish population over time, considering various factors. In our context, this model takes the form of a logistic differential equation with a harvesting term. The logistics component accounts for the natural growth and limitations due to environmental factors, while the harvesting term represents human activities, such as fishing, affecting population levels.

The equation given in the exercise models how the fish population (\(P\)) grows, stabilizes, and potentially declines when fishing, governed by a rate \(k\), is introduced. The logistic model suits this scenario well as it predicts the population's carrying capacity, given by \(10,000\) fish in this case, and illustrates how population fluctuates around this value as fishing activity intensifies. By setting initial conditions and substituting values like \(k\), we can simulate realistic ecological changes and make insightful predictions.

Separable Differential Equations

Separable differential equations are a specific class of differential equations where variables can be rearranged so that each side of the equation depends only on one of the different variables. This separation allows us to perform integration conveniently.

To solve the fish population model given in the exercise, we transformed the differential equation into a separable form: \(\frac{dP}{P(1 - \frac{P}{10000})} = 0.35 dt\). By separating the variables, we positioned all terms involving \(P\) on one side and differential \(dt\) on the other, paving the way to integrate both sides separately.

This approach simplifies the solving process significantly because it reduces complex differential equations into manageable integrals, hence helping to determine behavior over time.

Partial Fraction Decomposition

Partial fraction decomposition is a method used in calculus to break down complex rational expressions into simpler fractions that are easy to integrate. This technique is particularly helpful when dealing with separable differential equations, such as in our fish population model.

In our problem, to integrate the left-hand side of the separated equation \( \int \frac{dP}{P(1 - \frac{P}{10000})} \), we decompose the fraction into simpler parts: \( \frac{1}{P} + \frac{1}{10000-P} \). Each of these partial fractions can be integrated individually using basic integration techniques.

By applying partial fraction decomposition, we simplified the integration process and obtained a clearer path to expressing the function that models population changes.

Initial Value Problem

An initial value problem is a type of differential equation that, aside from the differential equation itself, also includes an initial condition that specifies the value of the function at a particular point. This crucial piece of information enables us to find a specific solution to the differential equation.

In this case, the initial condition given was \(P(0) = 2000\), which indicates the fish population at the starting point. By using this initial condition, we determined the constant \(C\) during the integration process. After substituting \(C\) back into our general solution, we refined the solution to reflect the real-life scenario accurately.

Verifying this solution by checking it against the initial condition \(P(0) = 2000\) ensures the accuracy and relevance of the model, confirming that our solution works perfectly within initial constraints.