AP Calculus AB Exam Question - Logistic Growth

Consider a population of bacteria that follows the logistic growth model. The population initially starts with 100 bacteria and has a carrying capacity of 1000. The growth rate is defined as:

$$\frac{dP}{dt} = kP\left(1-\frac{P}{1000}\right)$$

a) Find the general solution to this differential equation.

b) If the initial population is 100 bacteria, find the specific solution to this differential equation.

c) Determine the limiting population as $t$ approaches infinity, and interpret this value in the context of the problem.

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Solution:

a) General Solution

To find the general solution to the given logistic growth model, we need to solve the differential equation:

$$\frac{dP}{dt} = kP\left(1-\frac{P}{1000}\right)$$

Rearranging the equation, we get:

$$\frac{1}{P\left(1-\frac{P}{1000}\right)} dP = k dt$$

Now, we can integrate both sides:

$$\int \frac{1}{P\left(1-\frac{P}{1000}\right)} dP = \int k dt$$

Using partial fraction decomposition, we can write the left-hand side integral as:

$$\int \left(\frac{1}{P} + \frac{1000}{1000-P}\right) dP$$

Now we can integrate both terms separately:

$$\ln|P| - \ln|1000-P| = kt + C$$

where $C$ is the constant of integration.

Rearranging the equation, we get:

$$\ln\left|\frac{P}{1000-P}\right| = kt + C_1$$

where $C_1 = C - \ln(1000)$.

Taking the exponential of both sides:

$$\left|\frac{P}{1000-P}\right| = e^{kt+C_1}$$

Since $e^C$ is a positive constant, we can remove the absolute values:

$$\frac{P}{1000-P} = Ce^{kt}$$

where $C = \pm e^{C_1}$.

Finally, we can solve this equation for $P$:

$$P = \frac{1000Ce^{kt}}{1 + Ce^{kt}}$$

This is the general solution to the given logistic growth model.

b) Specific Solution

To find the specific solution for the given differential equation with an initial population of 100 bacteria, we can use the initial condition $P(0) = 100$.

Plugging this value into the general solution:

$$100 = \frac{1000Ce^{k(0)}}{1 + Ce^{k(0)}}$$

Simplifying the equation:

$$100 + 100Ce^{k(0)} = 1000Ce^{k(0)}$$

Dividing both sides by 100:

$$1 + Ce^{k(0)} = 10Ce^{k(0)}$$

Subtracting $Ce^{k(0)}$ from both sides:

$$1 = 9Ce^{k(0)}$$

Dividing by 9:

$$C = \frac{1}{9e^{k(0)}}$$

Substituting this value of $C$ back into the general solution:

$$P = \frac{1000\left(\frac{1}{9e^{k(0)}}\right)e^{kt}}{1 + \left(\frac{1}{9e^{k(0)}}\right)e^{kt}}$$

Simplifying further:

$$P = \frac{1000e^{(k-k(0))t}}{1 + \frac{e^{(k-k(0))t}}{9e^{k(0)}}}$$

This is the specific solution to the given logistic growth model with an initial population of 100 bacteria.

c) Limiting Population

To determine the limiting population as $t$ approaches infinity, we need to find the asymptote of the specific solution. We can notice that as $t$ approaches infinity, the exponential terms $e^{(k-k(0))t}$ and $e^{k(0)t}$ grow indefinitely.

Therefore, the denominator term $1 + \frac{e^{(k-k(0))t}}{9e^{k(0)}}$ tends to $1$, since the exponential term in the denominator grows much faster than the constant term 9.

Thus, the limiting population as $t$ approaches infinity is given by:

$$P_{\text{limit}} = 1000$$

This means that the population will eventually stabilize at 1000 bacteria, which is the carrying capacity.