Biologists stocked a lake with 300 fish and estimated the environment's capacity to be 6200. The number of fish tripled in the first year. Assume that the size of the fish population satisfies the logistic equation dP/dt=kP(1-P/K), where t is measured in years. Find an expression for the size of the population by solving the equation using the separation of variables. Note that you must find a value for k, which can not be entered as a variable.

We are given a logistic population model described by the equation:

\[
\frac{dP}{dt} = kP\left(1 - \frac{P}{K}\right)
\]

where:
- \( P(t) \) is the population at time \( t \),
- \( k \) is the growth rate constant,
- \( K = 6200 \) is the carrying capacity of the environment,
- \( P(0) = 300 \) is the initial population,
- \( P(1) = 900 \) (since the population tripled after the first year).

Our goal is to solve this equation using separation of variables and find the value of \( k \).

### Step 1: Separate variables
We start by rearranging the equation:

\[
\frac{dP}{P\left(1 - \frac{P}{K}\right)} = k \, dt
\]

Now, we need to simplify the left-hand side. We perform partial fraction decomposition:

\[
\frac{1}{P\left(1 - \frac{P}{K}\right)} = \frac{A}{P} + \frac{B}{K - P}
\]

Multiplying both sides by \( P(K - P) \) gives:

\[
1 = A(K - P) + B P
\]

Now, expand this equation:

\[
1 = A K - A P + B P
\]

Equating the coefficients of \( P \) and the constant terms:

- Constant term: \( A K = 1 \) → \( A = \frac{1}{K} \)
- Coefficient of \( P \): \( -A + B = 0 \) → \( B = A = \frac{1}{K} \)

So, we have the partial fractions:

\[
\frac{1}{P\left(1 - \frac{P}{K}\right)} = \frac{1}{K} \left( \frac{1}{P} + \frac{1}{K - P} \right)
\]

### Step 2: Integrate both sides
Now, substitute the partial fractions back into the original equation:

\[
\frac{1}{K} \left( \frac{1}{P} + \frac{1}{K - P} \right) \, dP = k \, dt
\]

Integrating both sides:

\[
\frac{1}{K} \left( \ln|P| - \ln|K - P| \right) = k t + C
\]

This simplifies to:

\[
\frac{1}{K} \ln \left| \frac{P}{K - P} \right| = k t + C
\]

Multiplying through by \( K \) gives:

\[
\ln \left| \frac{P}{K - P} \right| = kK t + C_1
\]

Exponentiating both sides:

\[
\frac{P}{K - P} = e^{kK t + C_1}
\]

Let \( C_2 = e^{C_1} \), so:

\[
\frac{P}{K - P} = C_2 e^{kK t}
\]

### Step 3: Solve for \( P \)
Now, solve for \( P \):

\[
P = \frac{C_2 e^{kK t} (K - P)}{1} \quad \text{Rearrange to get:}


Explore how to solve a logistic growth population problem using separation of variables. This calculus-based problem involves solving the differential equation dP/dt = kP(1 - P/K) for fish population growth in a lake, with an initial population of 300 fish and an environment capacity of 6200. Suitable for university-level students studying calculus and differential equations.

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