AP Calculus AB Practice Quiz: Exponential Models with Differential Equations

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

Test your understanding with short quizzes. This quiz has 14 questions to check your progress.

Question 1 of 14

The rate of growth of a population of rabbits, P, is directly proportional to the size of the population at any time t. Which of the following differential equations correctly models this relationship?

A)dP/dt = kPB)dP/dt = k/PC)dP/dt = kP^2D)d^2P/dt^2 = kP

All Questions (14)

A) dP/dt = kP

B) dP/dt = k/P

C) dP/dt = kP^2

D) d^2P/dt^2 = kP

Correct Answer: A

The statement 'The rate of change of a quantity is proportional to the size of the quantity' is modeled by the differential equation dy/dt = ky. In this context, the quantity is the population P, so the equation is dP/dt = kP. [cite: 2800]

A quantity y follows the exponential decay model dy/dt = -0.05y. If the initial amount of the quantity at t=0 is y_0 = 200, what is the particular solution for y(t)?

A) y = 200e^(-0.05t)

B) y = -0.05e^(200t)

C) y = 200e^(0.05t)

D) y = e^(-0.05t) + 200

Correct Answer: A

The differential equation dy/dt = ky with the initial condition y=y_0 at t=0 has the solution y = y_0 * e^(kt). Here, k = -0.05 and y_0 = 200. Substituting these values gives the particular solution y = 200e^(-0.05t). [cite: 2802, 2801]

The number of bacteria in a culture grows at a rate proportional to the number present. If the initial number of bacteria is 1000 and it doubles in 3 hours, which equation represents the number of bacteria, N, at time t hours?

A) N(t) = 1000e^((ln(2)/3)t)

B) N(t) = 1000e^((3/ln(2))t)

C) N(t) = 1000e^(2t/3)

D) N(t) = 1000 + (1000/3)t

Correct Answer: A

The model is N(t) = N_0 * e^(kt), where N_0 = 1000. We are given that N(3) = 2000. So, 2000 = 1000e^(k*3), which simplifies to 2 = e^(3k). Taking the natural logarithm of both sides gives ln(2) = 3k, so k = ln(2)/3. The particular solution is N(t) = 1000e^((ln(2)/3)t). [cite: 2801, 2802]

In the exponential growth model dP/dt = kP, where P represents the population of a city and t is time in years, what does the constant 'k' represent?

A) The initial population of the city.

B) The time it takes for the population to double.

C) The constant of proportionality, representing the relative growth rate.

D) The final population of the city.

Correct Answer: C

In the differential equation dy/dt = ky, 'k' is the constant of proportionality that relates the rate of change (dP/dt) to the quantity itself (P). It represents the relative (or per-capita) rate of growth. [cite: 2798, 2800]

A radioactive substance decays at a rate proportional to the amount present. If its half-life is 50 years, what is the value of the decay constant k in the model dy/dt = ky?

A) k = -ln(2)/50

B) k = ln(2)/50

C) k = -50/ln(2)

D) k = 50 * ln(2)

Correct Answer: A

The solution is y = y_0 * e^(kt). Half-life means that when t=50, y = 0.5*y_0. So, 0.5*y_0 = y_0 * e^(k*50). Dividing by y_0 gives 0.5 = e^(50k). Taking the natural logarithm, ln(0.5) = 50k. Since ln(0.5) = ln(1/2) = -ln(2), we have -ln(2) = 50k, which means k = -ln(2)/50. The constant k must be negative for decay. [cite: 2799, 2801]

What is the general solution to the differential equation dy/dt = 7y?

A) y = 7t + C

B) y = Ce^(7t)

C) y = 7e^(Ct)

D) y = (7/2)y^2 + C

Correct Answer: B

The differential equation dy/dt = ky has the general solution y = y_0 * e^(kt). In this case, k=7 and y_0 is an arbitrary constant, often denoted as C. Therefore, the general solution is y = Ce^(7t). [cite: 2802, 2801]

The rate of change of the amount of a medicine in the bloodstream, M, is proportional to the amount present. An initial dose of 100 mg is administered. After 4 hours, 60 mg remain. How much medicine will be present after 8 hours?

A) 20 mg

B) 30 mg

C) 36 mg

D) 40 mg

Correct Answer: C

The model is M(t) = M_0 * e^(kt) with M_0 = 100. We have M(4) = 60, so 60 = 100e^(4k). This gives e^(4k) = 0.6. We want to find M(8) = 100e^(8k). We can write e^(8k) as (e^(4k))^2. Therefore, M(8) = 100 * (0.6)^2 = 100 * 0.36 = 36 mg. [cite: 2801, 2799]

The velocity v of a particle moving along a line is described by the differential equation dv/dt = -0.2v. If the initial velocity v(0) is 50 m/s, what is the velocity at t=5 seconds?

A) 50e

B) 50/e

C) 50e^2

D) 50/e^2

Correct Answer: B

This is an application of exponential decay to motion. The solution is v(t) = v_0 * e^(kt), with v_0 = 50 and k = -0.2. So, v(t) = 50e^(-0.2t). We need to find v(5), which is 50e^(-0.2*5) = 50e^(-1) = 50/e. [cite: 2799, 2801]

For the exponential decay model represented by dy/dt = ky, which of the following must be true about the constant of proportionality k?

A) k > 0

B) k < 0

C) k = 0

D) k can be any real number

Correct Answer: B

In the model dy/dt = ky, if the quantity y is decaying or decreasing, its rate of change dy/dt must be negative. Since y itself is a positive quantity, the constant k must be negative to make the product ky negative. [cite: 2798, 2800]

A particular solution to a differential equation is y = 10e^(0.5t). Which of the following differential equations and initial conditions does this function solve?

A) dy/dt = 5y, with y(0) = 10

B) dy/dt = 0.5y, with y(0) = 10

C) dy/dt = 10y, with y(0) = 0.5

D) dy/dt = 0.5y, with y(0) = 0.5

Correct Answer: B

The solution is of the form y = y_0 * e^(kt). By comparing y = 10e^(0.5t) to the general form, we can identify the initial condition y_0 = 10 and the constant k = 0.5. The differential equation is therefore dy/dt = ky, which is dy/dt = 0.5y. The initial condition is y at t=0, which is y(0) = 10e^(0.5*0) = 10e^0 = 10. [cite: 2801, 2802]

The value of an investment grows at a rate proportional to its current value. If an initial investment of $5000 grows to $7000 in 4 years, how many years will it take for the initial investment to triple?

A) 4 * ln(3) / ln(1.4)

B) 4 * ln(1.4) / ln(3)

C) ln(3) / (4 * ln(1.4))

D) 12

Correct Answer: A

The model is V(t) = 5000e^(kt). We know V(4) = 7000, so 7000 = 5000e^(4k), which means 1.4 = e^(4k). Solving for k gives k = ln(1.4)/4. We want to find t when V(t) = 15000. So, 15000 = 5000e^(kt), which means 3 = e^(kt). Taking the log, ln(3) = kt. Substituting k, we get ln(3) = (ln(1.4)/4) * t. Solving for t gives t = 4 * ln(3) / ln(1.4). [cite: 2801, 2799]

If dy/dt = ky and y(0) = 100, and y(2) = 400, what is the value of k?

A) ln(2)

B) ln(4)

C) 2

D) ln(200)

Correct Answer: A

The solution is y(t) = y_0 * e^(kt). With y(0) = 100, the solution is y(t) = 100e^(kt). We are given y(2) = 400. So, 400 = 100e^(k*2). Dividing by 100 gives 4 = e^(2k). Taking the natural logarithm of both sides, ln(4) = 2k. Since ln(4) = ln(2^2) = 2ln(2), we have 2ln(2) = 2k, so k = ln(2). [cite: 2801]

Two populations, A and B, grow exponentially. Population A starts with 100 individuals and follows the model dA/dt = 0.05A. Population B starts with 200 individuals and follows the model dB/dt = 0.03B. At what time t will the two populations be equal?

A) t = 50 * ln(2)

B) t = 50 * ln(0.5)

C) t = 2 * ln(50)

D) t = ln(2) / 0.08

Correct Answer: A

The particular solution for population A is A(t) = 100e^(0.05t). The particular solution for population B is B(t) = 200e^(0.03t). We set them equal: 100e^(0.05t) = 200e^(0.03t). Divide by 100: e^(0.05t) = 2e^(0.03t). Divide by e^(0.03t): e^(0.05t - 0.03t) = 2, which simplifies to e^(0.02t) = 2. Taking the natural logarithm: 0.02t = ln(2). Solving for t: t = ln(2)/0.02 = 50 * ln(2). [cite: 2801, 2802]

The rate of decay of a substance is proportional to the amount present. If the amount of the substance is 80 grams at t=3 and 20 grams at t=6, what was the initial amount of the substance at t=0?

A) 160 grams

B) 240 grams

C) 320 grams

D) 640 grams

Correct Answer: C

The model is y(t) = y_0 * e^(kt). We have two points: y(3) = 80 and y(6) = 20. So, 80 = y_0 * e^(3k) and 20 = y_0 * e^(6k). Dividing the second equation by the first gives 20/80 = (y_0 * e^(6k)) / (y_0 * e^(3k)), which simplifies to 1/4 = e^(3k). Now use the first equation: 80 = y_0 * e^(3k). Substitute e^(3k) = 1/4: 80 = y_0 * (1/4). Solving for y_0 gives y_0 = 80 * 4 = 320 grams. [cite: 2801]