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AP Calculus BC 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, T, in degrees Fahrenheit, of a cup of tea is modeled by the function T(m), where m is the time in minutes after the tea is poured. What are the units of T'(5)?

All Questions (7)

The temperature, T, in degrees Fahrenheit, of a cup of tea is modeled by the function T(m), where m is the time in minutes after the tea is poured. What are the units of T'(5)?

A) minutes

B) degrees Fahrenheit

C) minutes per degree Fahrenheit

D) degrees Fahrenheit per minute

Correct Answer: D

The derivative T'(m) represents the rate of change of temperature with respect to time. The units of the function's output (T, in degrees Fahrenheit) are divided by the units of the function's input (m, in minutes). Therefore, the units of T'(m) are degrees Fahrenheit per minute.

Let C(t) represent the concentration of a medication in a patient's bloodstream, in milligrams per liter (mg/L), t hours after the medication was administered. The statement C'(4) = -1.2 is interpreted as:

A) Four hours after administration, the concentration is 1.2 mg/L.

B) The concentration decreases by an average of 1.2 mg/L each hour during the first four hours.

C) Four hours after administration, the concentration is decreasing at a rate of 1.2 mg/L per hour.

D) The total decrease in concentration after four hours is 1.2 milligrams per liter.

Correct Answer: C

The derivative C'(t) represents the instantaneous rate of change of the concentration with respect to time. C'(4) = -1.2 means that at the specific moment t=4 hours, the concentration is changing at a rate of -1.2 mg/L per hour. The negative sign indicates that the concentration is decreasing.

The number of bacteria in a petri dish is modeled by the function N(t) = 500e^(0.2t), where t is the time in hours. At what rate is the number of bacteria changing at t = 10 hours?

A) 100e^2 bacteria per hour

B) 500e^2 bacteria per hour

C) 2500e^2 bacteria per hour

D) 5000e^2 bacteria per hour

Correct Answer: A

To find the rate of change, we need to find the derivative of N(t) and evaluate it at t = 10. The derivative is N'(t) = d/dt [500e^(0.2t)] = 500 * e^(0.2t) * 0.2 = 100e^(0.2t). Evaluating at t = 10 gives N'(10) = 100e^(0.2 * 10) = 100e^2 bacteria per hour.

The cost, C, in dollars, of producing x items is given by C(x). The term 'marginal cost' is defined as the rate of change of the cost with respect to the number of items produced. Which of the following represents the marginal cost when 50 items are produced?

A) C(50)

B) C(50) / 50

C) C'(50)

D) C(51) - C(50)

Correct Answer: C

The rate of change of a function is given by its derivative. The marginal cost is defined as the rate of change of the cost function C(x). Therefore, the marginal cost when 50 items are produced is the value of the derivative of the cost function evaluated at x = 50, which is C'(50).

Water is being pumped into a cylindrical tank. The volume of water, V, in cubic feet, is a differentiable function of time t, in minutes. The line defined by y = 1.5t + 8 is tangent to the graph of V(t) at the point where t = 4. What is the rate at which the volume of water in the tank is changing at t = 4 minutes?

A) 1.5 cubic feet per minute

B) 4 cubic feet per minute

C) 8 cubic feet per minute

D) 14 cubic feet per minute

Correct Answer: A

The instantaneous rate of change of a function at a point is equal to the slope of the tangent line to the function's graph at that point. The derivative V'(t) represents the rate of change of the volume. The problem states that the line y = 1.5t + 8 is tangent to the graph of V(t) at t = 4. The slope of this line is 1.5. Therefore, V'(4) = 1.5 cubic feet per minute.

The value of a certain investment, V(t), in dollars, is recorded at various times t, in years, as shown in the table below. Which of the following is the best approximation for the rate at which the investment's value is changing at t = 6 years? | t (years) | 0 | 3 | 6 | 9 | 12 | |-----------|------|------|------|------|------| | V(t) ($) | 1000 | 1250 | 1600 | 2100 | 2800 |

A) $100 per year

B) $116.67 per year

C) $141.67 per year

D) $166.67 per year

Correct Answer: C

The instantaneous rate of change at t=6 can be approximated by the average rate of change over a small interval containing t=6. The best approximation using the given data is the average rate of change over the interval from t=3 to t=9. This is calculated as (V(9) - V(3)) / (9 - 3) = (2100 - 1250) / 6 = 850 / 6 ≈ $141.67 per year.

The radius r of a spherical balloon is increasing at a constant rate of 2 centimeters per minute. At the instant when the radius of the balloon is 10 centimeters, what is the rate of change of the volume of the balloon? (The volume V of a sphere with radius r is V = (4/3)πr³).

A) 80π cm³/min

B) 400π cm³/min

C) 800π cm³/min

D) (4000/3)π cm³/min

Correct Answer: C

This is a related rates problem. We are given dV/dt and need to find dV/dt. First, differentiate the volume formula with respect to time t: dV/dt = d/dt[(4/3)πr³] = (4/3)π * 3r² * (dr/dt) = 4πr²(dr/dt). We are given that dr/dt = 2 cm/min and we want to find dV/dt when r = 10 cm. Substitute the given values into the derivative: dV/dt = 4π(10)²(2) = 4π(100)(2) = 800π cm³/min.