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AP Calculus AB Exam: Population Growth

Created by @nathanedwards

at October 31st 2023, 8:08:53 pm.

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#population-growth

#differential-equation

#ap-calculus-ab-exam

AP Calculus AB Exam Question:

The population of a certain species is growing at a rate proportional to both its current population and the difference between its population and a carrying capacity. The initial population is 200, and after 5 years, the population has reached 500.

a) Find the differential equation that models the growth of the population.

b) Determine the value of the constant of proportionality.

c) Find the carrying capacity for this population.

d) Use the differential equation to find the population at t = 10 years.

Answer:

a) Let P(t) denote the population of the species at time t. According to the problem, the population is growing at a rate proportional to both its current population and the difference between its population and a carrying capacity. We can set up the following differential equation to represent the growth of the population:

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

where C represents the carrying capacity and k is a constant of proportionality.

b) To determine the value of the constant of proportionality, we use the initial condition that the population is 200 when t = 0:

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

\frac{{dP}}{{dt}} = k(200) \left( C - 200 \right)

Substituting the initial population value and evaluating at t = 0, we get:

k(200) \left( C - 200 \right) = \frac{{dP}}{{dt}} \Bigg|_{t=0}

Since $\frac{{dP}}{{dt}} \Bigg|_{t=0} = \frac{{dP}}{{dt}} \Bigg|_{t=0}$ , we can solve for k:

k(200) \left( C - 200 \right) = k(200) \left( C - P \right) \Bigg|_{t=0}

Simplifying, we have:

0 = k(200) \left( C - 200 \right)

Since $C$ and $P$ are different at t = 0, we can conclude that the term $C - P \right) = C - 200 = 0, leading to \( C = 200$ .

Therefore, the differential equation becomes:

\frac{{dP}}{{dt}} = 200k(200 - P)

c) Now, we need to find the carrying capacity, which represents the population at which the growth stops. From the given information, we know that the population has reached 500 after 5 years. Using this information and the initial condition, we can solve for the constant of proportionality k:

\frac{{dP}}{{dt}} = 200k(200 - P)

\int \frac{{1}}{{(200 - P)}} \, dP = \int 200k \, dt

- \ln|(200 - P)| = 200kt + C_1

Applying the initial condition $P = 200$ when $t = 0$ , we have:

- \ln|(200 - 200)| = 200k(0) + C_1

0 = C_1

Therefore, the equation becomes:

- \ln|(200 - P)| = 200kt

Now, we use the additional information that the population is 500 after 5 years:

- \ln|(200 - 500)| = 200k(5)

\ln|300| = -1000k

-1000k = \ln|300|

k = \frac{{\ln|300|}}{{-1000}}

So, the carrying capacity of the population is 200, and the constant of proportionality k is approximately -0.1308.

d) Now, using the differential equation, we can find the population at $t = 10$ :

\frac{{dP}}{{dt}} = 200k(200 - P)

Separating variables and integrating, we have:

\int \frac{{1}}{{(200 - P)}} \, dP = \int 200k \, dt

- \ln|(200 - P)| = 200kt + C_2

Applying the initial condition $P = 200$ when $t = 0$ , we have:

- \ln|(200 - 200)| = 200k(0) + C_2

0 = C_2

Therefore, the equation becomes:

- \ln|(200 - P)| = 200kt

Evaluating at $t = 10$ :

- \ln|(200 - P)| = 200k(10)

Using the previously calculated value of $k$ :

- \ln|(200 - P)| = 200 \left(\frac{{\ln|300|}}{{-1000}}\right) (10)

- \ln|(200 - P)| = 20 \ln|300|

Taking the exponential of both sides:

|200 - P| = e^{-20 \ln|300|}

Since the population cannot be negative, we can drop the absolute value signs:

200 - P = e^{-20 \ln|300|}

P = 200 - e^{-20 \ln|300|}

Evaluating this expression, we find that the population at $t = 10$ years is approximately 430.23.