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AP Calculus AB Practice Quiz: Rates of Change in Applied Contexts Other Than Motion

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

Test your understanding with short quizzes. This quiz has 7 questions to check your progress.

Question 1 of 7

The temperature of a pot of water, T, in degrees Celsius, is a function of time, t, in minutes, given by T(t). What is the best interpretation of T'(10) = 5?

All Questions (7)

The temperature of a pot of water, T, in degrees Celsius, is a function of time, t, in minutes, given by T(t). What is the best interpretation of T'(10) = 5?

A) The temperature of the water is 5 degrees Celsius at 10 minutes.

B) The temperature of the water increases by 5 degrees Celsius every 10 minutes.

C) After 10 minutes, the temperature of the water is increasing at a rate of 5 degrees Celsius per minute.

D) The average rate of change of the temperature over the first 10 minutes is 5 degrees Celsius per minute.

Correct Answer: C

The derivative of a function at a point represents the instantaneous rate of change at that point. T'(10) represents the rate of change of the temperature T with respect to time t at the specific instant t = 10 minutes. Therefore, T'(10) = 5 means that at t = 10 minutes, the temperature is increasing at a rate of 5 degrees Celsius per minute.

The number of bacteria in a petri dish is given by the function P(t), where t is measured in hours. The statement P'(4) = 200 is given. Which of the following is the best interpretation of this statement?

A) At 4 hours, the number of bacteria is 200.

B) The number of bacteria increases by 200 in the first 4 hours.

C) The average rate of increase of bacteria over the first 4 hours is 200 bacteria per hour.

D) At the 4-hour mark, the number of bacteria is increasing at a rate of 200 bacteria per hour.

Correct Answer: D

The derivative, P'(t), represents the instantaneous rate of change of the population of bacteria with respect to time. P'(4) = 200 means that at the precise moment t = 4 hours, the population is growing at a rate of 200 bacteria per hour. It does not describe the total population or an average rate over an interval.

The volume of a spherical balloon is increasing. The volume, V, in cubic centimeters, is given by V = (4/3)πr³, where r is the radius in centimeters. At the instant when the radius is 10 cm, the radius is increasing at a rate of 0.5 cm/sec. At what rate is the volume increasing at this instant?

A) 20π cm³/sec

B) 50π cm³/sec

C) 200π cm³/sec

D) 400π cm³/sec

Correct Answer: C

This is a related rates problem. We are given dV/dt and need to find it. First, differentiate V with respect to time t: dV/dt = d/dt[(4/3)πr³] = 4πr²(dr/dt). We are given r = 10 cm and dr/dt = 0.5 cm/sec. Plugging these values in: dV/dt = 4π(10)²(0.5) = 4π(100)(0.5) = 200π cm³/sec. This shows how the derivative is used to solve problems involving rates of change.

Let C(x) be the cost, in dollars, to produce x widgets. What are the units of C'(x)?

A) dollars

B) widgets

C) dollars per widget

D) widgets per dollar

Correct Answer: C

The derivative C'(x) represents the rate of change of cost with respect to the number of widgets produced. In Leibniz notation, this is dC/dx. The units of the derivative are the units of the dependent variable (C, in dollars) divided by the units of the independent variable (x, in widgets). Therefore, the units are dollars per widget. This is often referred to as the marginal cost.

Water is draining from a conical tank. The volume of water in the tank, V, in cubic feet, is given by V(t) = (1/3)(12 - t)³, where t is the time in hours and 0 ≤ t ≤ 12. At what rate is the volume of water changing at t = 4 hours?

A) -192 cubic feet per hour

B) -64 cubic feet per hour

C) 64 cubic feet per hour

D) 192 cubic feet per hour

Correct Answer: B

The rate of change of the volume is the derivative of the volume function, V'(t). To find the derivative of V(t) = (1/3)(12 - t)³, we use the power rule and the chain rule. V'(t) = (1/3) * 3(12 - t)² * d/dt(12 - t) = (12 - t)² * (-1) = -(12 - t)². To find the rate at t = 4 hours, we substitute t = 4 into the derivative: V'(4) = -(12 - 4)² = -(8)² = -64. The negative sign indicates that the volume is decreasing, which is consistent with the tank draining. The rate is -64 cubic feet per hour.

The amount of a certain medication in a patient's bloodstream, A, in milligrams (mg), is measured at various times, t, in hours. The data is recorded in the table below. Using the data, what is the best approximation for A'(3)? | t (hours) | 0 | 2 | 4 | 6 | |---|---|---|---|---| | A(t) (mg) | 0 | 150 | 180 | 160 |

A) 15 mg/hr

B) 30 mg/hr

C) 60 mg/hr

D) 75 mg/hr

Correct Answer: A

The derivative A'(3) can be approximated by the average rate of change over a small interval containing t=3. The best interval available from the data is from t=2 to t=4. The average rate of change is the slope of the secant line: [A(4) - A(2)] / (4 - 2) = (180 - 150) / 2 = 30 / 2 = 15. Therefore, the best approximation for the instantaneous rate of change at t=3 is 15 mg/hr.

The graph of y = P(t) shows the population of a town, in thousands, over a 10-year period. At which of the labeled points is the population growing at the fastest rate?

A) Point A

B) Point B

C) Point C

D) Point D

Correct Answer: B

The rate of growth of the population is represented by the derivative, P'(t), which is the slope of the tangent line to the graph of P(t). To find where the population is growing fastest, we need to find the point on the graph with the steepest positive slope. By visually inspecting the graph, the tangent line at Point B is steeper than at any other labeled point, indicating the highest instantaneous rate of change (fastest growth). At Point A, the slope is positive but less steep than at B. At Point C, the slope is zero (peak population). At Point D, the slope is negative (population is decreasing).