AP Calculus BC 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
All Questions (14)
A) dP/dt = kP
B) dP/dt = k/P
C) dP/dt = kt
D) d²P/dt² = 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) y = kt + y₀
B) y = y₀e^(kt)
C) y = e^(kt) + y₀
D) y = y₀e^(-kt)
Correct Answer: B
The solution to the exponential growth and decay model, dy/dt = ky, with an initial condition y = y₀ when t = 0, is of the form y = y₀e^(kt). This is the standard solution for this type of differential equation. [cite: 2802]
A) A(t) = 100e^(0.05t)
B) A(t) = -0.05t + 100
C) A(t) = 100e^(-0.05t)
D) A(t) = e^(-0.05t) + 100
Correct Answer: C
The differential equation is of the form dy/dt = ky, where k = -0.05. The initial condition is y₀ = 100. The particular solution is y = y₀e^(kt). Substituting the given values, we get A(t) = 100e^(-0.05t). [cite: 2801, 2802]
A) It is the initial population.
B) It is the carrying capacity of the environment.
C) It is the time it takes for the population to double.
D) It is the relative (or proportional) growth rate of the population.
Correct Answer: D
The equation dP/dt = kP states that the rate of change of the population (dP/dt) is proportional to the population size (P). The constant of proportionality, k, represents the relative growth rate. For example, if k=0.02, it means the population is growing at a rate of 2% of its current size per unit of time. [cite: 2798, 2800]
A) 50e²
B) 50e
C) 50 + 2e
D) e² + 50
Correct Answer: A
The differential equation is dN/dt = kN with k=0.2 and the initial condition N(0) = 50. The solution is N(t) = N₀e^(kt), which becomes N(t) = 50e^(0.2t). To find the number of cells after 10 hours, we calculate N(10) = 50e^(0.2 * 10) = 50e². [cite: 2801, 2802]
A) dP/dt = (ln(2)/3)P
B) dP/dt = (3/ln(2))P
C) dP/dt = (2/3)P
D) dP/dt = (ln(3)/2)P
Correct Answer: A
The model is P(t) = P₀e^(kt). We are given that P(3) = 2P₀. So, 2P₀ = P₀e^(k*3). Dividing by P₀ gives 2 = e^(3k). Taking the natural logarithm of both sides gives ln(2) = 3k, so k = ln(2)/3. The differential equation is dP/dt = kP, which is dP/dt = (ln(2)/3)P. [cite: 2800, 2802]
A) 40e
B) 40e²
C) 40e⁻¹
D) 40e⁻²
Correct Answer: D
This is an exponential decay model applied to motion. The solution is of the form v(t) = v₀e^(kt). Here, v₀ = 40 and k = -0.5. So, v(t) = 40e^(-0.5t). We need to find v(4), which is v(4) = 40e^(-0.5 * 4) = 40e⁻². [cite: 2799, 2801, 2802]
A) t = 6
B) t = 8
C) t = 10
D) t = 12
Correct Answer: B
The model is y(t) = y₀e^(kt). We have y₀ = 120. So, y(t) = 120e^(kt). We use y(4) = 30 to find k: 30 = 120e^(4k) -> 1/4 = e^(4k) -> ln(1/4) = 4k -> k = ln(1/4)/4. Now we solve for t when y(t) = 7.5: 7.5 = 120e^(kt) -> 7.5/120 = e^(kt) -> 1/16 = e^(kt). Since 1/16 = (1/4)², we have (1/4)² = (e^(4k))^(t/4) = e^(kt). This means t/4 = 2, so t = 8. [cite: 2801, 2802]
A) y = Ce^(5t)
B) y = 5t + C
C) y = e^(5t) + C
D) y = 5Ce^t
Correct Answer: A
The differential equation dy/dt = ky has the general solution y = Ce^(kt), where C is an arbitrary constant representing the initial value. In this case, k=5, so the general solution is y = Ce^(5t). [cite: 2801, 2802]
A) V(t) = 1000e^((ln(1.5)/5)t)
B) V(t) = 1000e^((ln(0.5)/5)t)
C) V(t) = 1000(1.5)^(t/5)
D) V(t) = 1500e^((ln(1.5)/5)t)
Correct Answer: A
The model is V(t) = V₀e^(kt) with V₀ = 1000. We are given V(5) = 1500. So, 1500 = 1000e^(k*5). This simplifies to 1.5 = e^(5k). Taking the natural log, we get ln(1.5) = 5k, so k = ln(1.5)/5. Substituting this k back into the general form gives V(t) = 1000e^((ln(1.5)/5)t). Note that option C is equivalent, but option A is written in the standard y=y₀e^(kt) form. [cite: 2801, 2802]
A) The value of a car depreciating at a rate proportional to its current value.
B) The number of people in a town growing at a constant rate of 500 people per year.
C) The temperature of a hot object cooling in a room, where the rate of cooling is proportional to the temperature difference (Newton's Law of Cooling, a variation of the model).
D) The mass of a radioactive element decaying at a rate proportional to its mass.
Correct Answer: B
The model dy/dt = ky describes a rate of change that is proportional to the current amount. A constant rate of change, like 500 people per year, is modeled by dy/dt = 500, which is not proportional to y. Options A, C, and D all describe rates of change proportional to a quantity. [cite: 2798, 2800]
A) 2.5
B) 5
C) 1.25
D) 0.5
Correct Answer: A
We have y(t) = y₀e^(kt). We are given two points: 10 = y₀e^(2k) and 40 = y₀e^(4k). Divide the second equation by the first: (40/10) = (y₀e^(4k))/(y₀e^(2k)), which simplifies to 4 = e^(2k). Taking the natural log, ln(4) = 2k, so k = ln(4)/2 = ln(2). Now substitute this back into the first equation: 10 = y₀e^(2*ln(2)) = y₀e^(ln(4)) = 4y₀. Solving for y₀ gives y₀ = 10/4 = 2.5. [cite: 2801, 2802]
A) 0 grams
B) 5 grams
C) 8 grams
D) 10 grams
Correct Answer: C
The model is A(t) = A₀e^(kt) with A₀ = 50. We use A(10) = 20 to find k: 20 = 50e^(10k) -> 2/5 = e^(10k). We need to find A(20) = 50e^(20k) = 50(e^(10k))². Since we know e^(10k) = 2/5, we can substitute this value: A(20) = 50 * (2/5)² = 50 * (4/25) = 2 * 4 = 8 grams. [cite: 2801, 2802]
A) dA/dt = -(ln(2)/5730)A
B) dA/dt = -(5730/ln(2))A
C) dA/dt = -5730A
D) dA/dt = -(ln(0.5) * 5730)A
Correct Answer: A
The model is A(t) = A₀e^(kt). The half-life information means A(5730) = 0.5A₀. So, 0.5A₀ = A₀e^(k*5730). This simplifies to 0.5 = e^(5730k). Taking the natural log gives ln(0.5) = 5730k. Since ln(0.5) = ln(1/2) = -ln(2), we have k = -ln(2)/5730. The differential equation is dA/dt = kA, so dA/dt = -(ln(2)/5730)A. [cite: 2800, 2802]