RaySix

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at November 3rd 2023, 2:42:46 am.

AP Calculus AB Exam Question:

A population of bacteria follows a logistic growth model given by the equation:

\frac{{dP}}{{dt}} = kP \left(1 - \frac{{P}}{{M}}\right)

Where P is the population size at time t, k is a constant representing the growth rate, and M is the carrying capacity of the population.

Answer:

To solve the logistic growth equation for P(t), we need to separate the variables and integrate both sides. Let's start by rearranging the equation:

\frac{{dP}}{{P(1 - \frac{{P}}{{M}})}} = kdt

Now, we can separate the variables by multiplying both sides by the denominator:

\int \frac{{dP}}{{P(1 - \frac{{P}}{{M}})}} = \int kdt

Using partial fraction decomposition, we can rewrite the left side of the equation as:

\int \left( \frac{1}{P} + \frac{\frac{1}{M}}{1 - \frac{P}{M}} \right) dP

Simplifying further:

\int \frac{1}{P} dP + \int \frac{\frac{1}{M}}{1 - \frac{P}{M}} dP = \int kdt

Integrating each term separately:

\ln|P| - \ln|M - P| = kt + C

Using logarithm properties, we can simplify further:

\ln\left(\frac{P}{M - P}\right) = kt + C

Now, we need to eliminate the logarithm. We can do this by exponentiating both sides:

\frac{P}{M - P} = e^{kt + C}

Since e^C is just a constant, we can rewrite it as K (another constant):

\frac{P}{M - P} = Ke^{kt}

Cross-multiplying:

P = K(M - P)e^{kt}

Expanding and rearranging the equation:

P = KMe^{kt} - KP e^{kt}

Finally, factoring out P on the right side:

P = \frac{KMe^{kt}}{1 + Ke^{kt}}

Hence, the solution to the logistic growth equation for P(t) is:

\boxed{ P(t) = \frac{KMe^{kt}}{1 + Ke^{kt}} }