AP Calculus BC Practice Quiz: Logistic Models with Differential Equations (BC ONLY)

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 11 questions to check your progress.

Question 1 of 11

A population, $P$, grows according to the logistic model. The rate of change of the population is jointly proportional to the current population size and the difference between the carrying capacity, $M$, and the current population size. Which differential equation describes this relationship?

A)$\frac{dP}{dt} = kP(M-P)$B)$\frac{dP}{dt} = k(M-P)$C)$\frac{dP}{dt} = kP$D)$\frac{dP}{dt} = kP(P-M)$

All Questions (11)

A) $\frac{dP}{dt} = kP(M-P)$

B) $\frac{dP}{dt} = k(M-P)$

C) $\frac{dP}{dt} = kP$

D) $\frac{dP}{dt} = kP(P-M)$

Correct Answer: A

The phrase "jointly proportional to" means the rate is a constant, $k$, times the product of the two quantities mentioned. The two quantities are "the current population size" ($P$) and "the difference between the carrying capacity, $M$, and the current population size" ($M-P$). This translates directly to the equation $\frac{dP}{dt} = kP(M-P)$. [cite: 2807]

The number of students at a school who have heard a rumor, $R(t)$, is modeled by the logistic differential equation $\frac{dR}{dt} = 0.02R(500-R)$. What is the maximum number of students who will eventually hear the rumor?

A) 2

B) 250

C) 500

D) 502

Correct Answer: C

The logistic differential equation is given in the form $\frac{dy}{dt} = ky(a-y)$, where $a$ is the carrying capacity, or the limiting value. In the given equation, $\frac{dR}{dt} = 0.02R(500-R)$, the carrying capacity is $a=500$. This represents the maximum number of students who will hear the rumor. [cite: 2809]

A population of bears, $B(t)$, in a national park is modeled by the logistic differential equation $\frac{dB}{dt} = 0.01B(200-B)$. At what population size is the bear population growing the fastest?

A) 50

B) 100

C) 150

D) 200

Correct Answer: B

For a logistic model $\frac{dy}{dt} = ky(a-y)$, the rate of change is fastest when the population is at half the carrying capacity, i.e., when $y = a/2$. In this model, the carrying capacity is $a=200$. Therefore, the population is growing fastest when the number of bears is $200/2 = 100$. [cite: 2810]

The population of fish in a lake, $F(t)$, is modeled by $\frac{dF}{dt} = 0.04F(1000-F)$, where $t$ is in years. If the initial population is $F(0) = 1200$, which of the following statements is true?

A) The fish population will increase and approach 1200.

B) The fish population will decrease and approach 1000.

C) The fish population will increase and approach infinity.

D) The fish population will decrease and approach 0.

Correct Answer: B

The carrying capacity of the lake is 1000 fish. The initial population is 1200, which is above the carrying capacity. When $F > 1000$, the term $(1000-F)$ is negative, making $\frac{dF}{dt}$ negative. A negative rate of change means the population is decreasing. The population will decrease until it approaches the carrying capacity of 1000. [cite: 2806, 2808]

The growth of a yeast culture is described by the differential equation $\frac{dy}{dt} = 0.1y(1 - \frac{y}{800})$, where $y$ is the mass of the yeast in grams and $t$ is in hours. If the initial mass is 50 grams, what is the limit of the mass of the yeast as $t$ approaches infinity?

A) 50

B) 400

C) 800

D) Infinity

Correct Answer: C

This logistic equation is in the form $\frac{dy}{dt} = ry(1 - \frac{y}{L})$, where $L$ is the carrying capacity. By comparing the given equation to this form, we can identify the carrying capacity as $L=800$. The carrying capacity is the limiting value of the population as $t \to \infty$. [cite: 2809]

The spread of a flu virus in a community is modeled by $\frac{dP}{dt} = 0.003P(2500-P)$, where $P(t)$ is the number of people infected at time $t$ in days. The number of infected people is increasing fastest when the number of infected people is:

A) 3

B) 1250

C) 2500

D) 7500

Correct Answer: B

The logistic model shows that the rate of change, $\frac{dP}{dt}$, is at its maximum when the population $P$ is half of the carrying capacity. The carrying capacity, $a$, in the equation $\frac{dP}{dt} = 0.003P(2500-P)$ is 2500. Therefore, the population is increasing fastest when $P = 2500 / 2 = 1250$. [cite: 2810]

Let $P(t)$ represent the number of people in a town who have a certain new phone. The rate at which people get the phone is given by the logistic differential equation $\frac{dP}{dt} = 0.01P(4000-P)$. If $P(0) = 500$, which of the following statements is true?

A) The number of people with the phone is increasing, and the rate of increase is also increasing.

B) The number of people with the phone is increasing, but the rate of increase is decreasing.

C) The number of people with the phone is decreasing, and the rate of decrease is increasing.

D) The number of people with the phone is decreasing, and the rate of decrease is decreasing.

Correct Answer: A

The carrying capacity is $a=4000$. The population grows fastest at $a/2 = 2000$. The initial population is $P(0)=500$. Since $0 < 500 < 4000$, the term $(4000-P)$ is positive, so $\frac{dP}{dt}$ is positive, meaning the population is increasing. The graph of $\frac{dP}{dt}$ versus $P$ is a downward-opening parabola with its vertex at $P=2000$. Since the initial population $P=500$ is between 0 and 2000, it is on the interval where the rate of growth, $\frac{dP}{dt}$, is increasing. [cite: 2808, 2810]

The rate of spread of a new software application is jointly proportional to the number of users who have the application, $U$, and the number of potential users who do not yet have it. If the total number of potential users is 1,000,000, which differential equation models this situation?

A) $\frac{dU}{dt} = kU(1,000,000)$

B) $\frac{dU}{dt} = k(1,000,000 - U)$

C) $\frac{dU}{dt} = kU(1,000,000 - U)$

D) $\frac{dU}{dt} = kU + (1,000,000 - U)$

Correct Answer: C

The carrying capacity is the total number of potential users, which is $a = 1,000,000$. The number of users who have the application is $U$. The number of potential users who do not have it is the difference between the carrying capacity and the current number of users, which is $(1,000,000 - U)$. The statement "jointly proportional to" means the rate, $\frac{dU}{dt}$, is a constant $k$ times the product of these two quantities: $\frac{dU}{dt} = kU(1,000,000 - U)$. [cite: 2806, 2807]

Consider the logistic differential equation $\frac{dy}{dt} = 0.2y(10-y)$. If the initial condition is $y(0)=12$, what is $\lim_{t \to \infty} y(t)$?

A) 0

B) 10

C) 12

D) Infinity

Correct Answer: B

The carrying capacity for this logistic model is $a=10$. The initial value is $y(0)=12$, which is above the carrying capacity. For any initial value $y(0) > 0$, the solution to the logistic equation will approach the carrying capacity as $t \to \infty$. Even though the population starts above the carrying capacity and will decrease, it will still approach the stable equilibrium at $y=10$. [cite: 2808, 2809]

The population of a species of rabbits, $R(t)$, is modeled by the logistic equation $\frac{dR}{dt} = 0.08R(50-R)$. What is the maximum growth rate of the rabbit population?

A) 25

B) 50

C) 100

D) 200

Correct Answer: B

The growth rate, $\frac{dR}{dt}$, is maximized when the population is at half the carrying capacity. The carrying capacity is $a=50$. The population is growing fastest when $R = 50/2 = 25$. To find the maximum rate, substitute $R=25$ back into the differential equation: $\frac{dR}{dt} = 0.08(25)(50-25) = 0.08(25)(25) = 2(25) = 50$. [cite: 2810]

A rumor spreads through a school with a total of 1200 students. The rate of spread is jointly proportional to the number of students who have heard the rumor, $S$, and the number of students who have not. At the moment when 200 students have heard the rumor, it is spreading at a rate of 40 students per hour. At what number of students is the rumor spreading the fastest?

A) 200

B) 600

C) 1000

D) 1200

Correct Answer: B

The problem describes a logistic model. The carrying capacity, $a$, is the total number of students, which is 1200. The differential equation is $\frac{dS}{dt} = kS(1200-S)$. The information about the rate at a specific time (40 students/hour when S=200) could be used to find $k$, but it is not needed to answer the question. The rumor spreads fastest when the number of students who have heard it is half the carrying capacity. Therefore, the rate is maximum when $S = a/2 = 1200/2 = 600$. [cite: 2807, 2809, 2810]