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AP Calculus BC Practice Quiz: Modeling Situations 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 7 questions to check your progress.

Question 1 of 7

The rate of growth of a bacterial population, P, is directly proportional to the number of bacteria present at time t. Which differential equation models this relationship, where k is a positive constant?

All Questions (7)

The rate of growth of a bacterial population, P, is directly proportional to the number of bacteria present at time t. Which differential equation models this relationship, where k is a positive constant?

A) dP/dt = kP

B) dP/dt = k/P

C) dP/dt = P + k

D) dP/dt = kt

Correct Answer: A

The phrase 'the rate of growth of a bacterial population' translates to the derivative dP/dt. The phrase 'is directly proportional to' translates to '= k *' for some constant k. 'The number of bacteria present' is P. Combining these pieces gives the differential equation dP/dt = kP.

The temperature, T, of a cup of coffee is cooling in a room with a constant ambient temperature of 70°F. The rate of change of the coffee's temperature is proportional to the difference between its temperature and the ambient temperature. Which differential equation describes this situation, where k is a positive constant?

A) dT/dt = k(T + 70)

B) dT/dt = -k(T - 70)

C) dT/dt = 70k

D) dT/dt = kT

Correct Answer: B

The 'rate of change of the coffee's temperature' is dT/dt. 'is proportional to' means it equals a constant times some quantity. The 'difference between its temperature and the ambient temperature' is (T - 70). Since the coffee is cooling, its temperature T is decreasing, so dT/dt must be negative when T > 70. The expression -k(T - 70) correctly models this, as T - 70 is positive, and the term is multiplied by a negative constant -k.

A tank contains a volume V of water. Water flows into the tank at a constant rate of 5 gallons per minute, and water drains from the tank at a rate proportional to the volume of water in the tank. Which differential equation models the rate of change of the volume of water in the tank, where k is a positive constant?

A) dV/dt = 5 - kV

B) dV/dt = k(5 - V)

C) dV/dt = 5 + kV

D) dV/dt = 5V - k

Correct Answer: A

The rate of change of the volume, dV/dt, is the rate of water flowing in minus the rate of water flowing out. The 'rate in' is a constant 5. The 'rate out' is 'proportional to the volume, V,' which can be written as kV. Therefore, the net rate of change is dV/dt = (rate in) - (rate out) = 5 - kV.

In a school with a total of 1000 students, the rate at which a rumor spreads is proportional to the product of the number of students who have heard the rumor, S, and the number of students who have not yet heard the rumor. Which differential equation models this relationship, where k is a positive constant?

A) dS/dt = kS

B) dS/dt = k(1000 - S)

C) dS/dt = kS(1000 - S)

D) dS/dt = kS(1000)

Correct Answer: C

The 'rate at which a rumor spreads' is dS/dt. 'is proportional to' translates to '= k *'. The statement requires the 'product of the number of students who have heard the rumor, S,' and 'the number of students who have not yet heard the rumor.' The number who have not heard it is the total (1000) minus those who have (S), which is (1000 - S). The product is S(1000 - S). Combining these parts gives dS/dt = kS(1000 - S).

The acceleration of an object is defined as the rate of change of its velocity, v, with respect to time, t. According to a certain physical model, the acceleration of an object is inversely proportional to the square of its velocity. Which differential equation represents this model, where k is a constant?

A) dv/dt = kv^2

B) dv/dt = k/v^2

C) dv/dt = k/v

D) d^2v/dt^2 = k/v^2

Correct Answer: B

Acceleration is the 'rate of change of its velocity, v,' which is the derivative dv/dt. The phrase 'is inversely proportional to' translates to '= k /' for some constant k. 'the square of its velocity' is v^2. Putting these pieces together results in the differential equation dv/dt = k/v^2.

The rate of change of a fish population, P, in a lake is modeled by two factors. The population increases at a rate proportional to the current population. However, due to fishing, the population decreases at a constant rate of 200 fish per year. Which differential equation describes this situation, where k is a positive constant?

A) dP/dt = kP + 200

B) dP/dt = k(P - 200)

C) dP/dt = kP - 200

D) dP/dt = 200 - kP

Correct Answer: C

The overall 'rate of change of a fish population' is dP/dt. This rate is the sum of the rates of increase and decrease. The increase is 'proportional to the current population,' which is kP. The decrease is 'at a constant rate of 200,' which is represented by -200. Combining these gives the net rate of change: dP/dt = kP - 200.

The amount of a radioactive substance, A, in grams, decreases at a rate proportional to the amount present. Which differential equation models the amount of the substance at time t?

A) dA/dt = kA, where k > 0

B) dA/dt = -kA, where k > 0

C) dA/dt = k/A, where k > 0

D) dA/dt = k - A, where k > 0

Correct Answer: B

The 'rate of change of the amount' is dA/dt. The substance 'decreases,' which means the rate of change dA/dt must be negative. The rate is 'proportional to the amount present,' A. This can be written as dA/dt = cA for some constant of proportionality c. Since A is always positive (it's an amount) and dA/dt must be negative (it's decreasing), the constant c must be negative. Option B, dA/dt = -kA where k > 0, correctly represents this, as -k is a negative constant. Option A would model growth, not decay.