AP Calculus AB Quiz: Exponential Models With Differential Equations

What this quiz covers

This quiz focuses on Exponential Models With Differential Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $B(t)=C-0.07t$
2. $B(t)=Ce^{-0.07t}$
3. $B(t)=0.07t+C$
4. $B(t)=Ce^{0.07t}$ (correct answer)
5. $B(t)=C+e^{0.07t}$

Explanation: This problem involves solving an exponential differential equation representing compound interest growth. When we have $\frac{dB}{dt}=0.07B$, the savings balance grows at a rate proportional to its current amount with rate 0.07. This represents continuous compound interest where larger balances earn proportionally more interest. The differential equation $\frac{dy}{dt} = ky$ has the general solution $y = Ce^{kt}$, so with $k = 0.07$, we get $B(t) = Ce^{0.07t}$. Choice A would represent simple interest (linear growth), but continuous compounding requires exponential growth. To identify compound interest: when growth rate is proportional to current balance ($\frac{dB}{dt} = rB$), the solution is always exponential ($Ce^{rt}$).

Question 2

1. $y=Ce^{kt}$
2. $y=C-kt$
3. $y=C+e^{-kt}$
4. $y=kt+C$
5. $y=Ce^{-kt}$ (correct answer)

Explanation: This problem involves solving an exponential differential equation representing proportional decay. When we have $\frac{dy}{dt}=-ky$ with $k>0$, the quantity decreases at a rate proportional to its current amount. The negative sign with positive $k$ ensures decay behavior, as larger quantities experience proportionally larger decreases. The general solution to $\frac{dy}{dt} = -ky$ is $y = Ce^{-kt}$, representing exponential decay. Choice A would represent exponential growth since it lacks the negative exponent required for decay. To identify exponential decay: negative coefficients in proportional rate equations ($\frac{dy}{dt} = -ky$) always produce decay solutions with negative exponents ($Ce^{-kt}$).

Question 3

1. $y=0.5x+C$
2. $y=Ce^{-0.5x}$
3. $y=C+e^{0.5x}$
4. $y=Ce^{0.5x}$ (correct answer)
5. $y=C-0.5x$

Explanation: This problem involves solving an exponential differential equation representing compound interest growth. When we have $\frac{dy}{dx}=0.5y$, the balance grows at a rate proportional to its current value with rate 0.5. This represents continuous compound interest where larger balances earn proportionally more interest. The differential equation $\frac{dy}{dx} = ky$ has the general solution $y = Ce^{kx}$, so with $k = 0.5$, we get $y = Ce^{0.5x}$. Choice A represents simple interest (linear growth), but continuous compounding requires the exponential relationship. To distinguish compound from simple interest: proportional growth rates ($\frac{dy}{dx} = ry$) always indicate compound interest with exponential solutions ($Ce^{rx}$).

Question 4

1. $y=Ce^{(\ln 2)x}$
2. $y=Ce^{-(\ln 2)x}$ (correct answer)
3. $y=C+e^{-(\ln 2)x}$
4. $y=-(\ln 2)x+C$
5. $y=(\ln 2)x+C$

Explanation: This problem involves solving an exponential differential equation representing exponential decay with rate ln(2). When we have $\frac{dy}{dx}=-\ln(2)y$, the quantity decreases at a rate proportional to its current value with constant $-\ln(2)$. The negative constant indicates exponential decay, and the ln(2) factor relates to half-life calculations in exponential decay. The general solution to $\frac{dy}{dx} = ky$ is $y = Ce^{kx}$, so with $k = -\ln(2)$, we get $y = Ce^{-(\ln 2)x}$. Choice D represents linear decay, but the proportional relationship requires exponential behavior. To handle negative logarithmic constants: $-\ln(2)$ produces exponential decay solutions with the corresponding negative exponent.

Question 5

1. $y=Ce^{0.02x}$ (correct answer)
2. $y=Ce^{-0.02x}$
3. $y=0.02x+C$
4. $y=C+e^{0.02x}$
5. $y=C-0.02x$

Explanation: This problem involves solving an exponential differential equation where the rate of change is proportional to the current quantity. When we have $\frac{dy}{dx}=0.02y$, the quantity grows at a rate proportional to its current value with proportionality constant 0.02. This positive constant indicates exponential growth, though at a slow rate of 2% of the current value. The standard solution to $\frac{dy}{dx} = ky$ is $y = Ce^{kx}$, so with $k = 0.02$, we get $y = Ce^{0.02x}$. Choice C represents linear growth, which would occur if the rate of change were constant rather than proportional to the current quantity. To recognize small exponential growth: even small positive constants in proportional rate equations produce exponential growth, not linear growth.

Question 6

1. $y=C+e^{-0.12x}$
2. $y=Ce^{0.12x}$
3. $y=C-0.12x$
4. $y=Ce^{-0.12x}$ (correct answer)
5. $y=0.12x+C$

Explanation: This problem involves solving an exponential differential equation representing exponential decay. When we have $\frac{dy}{dx}=-0.12y$, the quantity decreases at a rate proportional to its current value with constant -0.12. The negative proportionality constant indicates exponential decay at 12% of the current value per unit change. The general solution to $\frac{dy}{dx} = ky$ is $y = Ce^{kx}$, so with $k = -0.12$, we get $y = Ce^{-0.12x}$. Choice C represents linear decay, but the proportional relationship requires exponential rather than linear behavior. To recognize percentage-based exponential decay: negative decimal constants like -0.12 produce exponential decay solutions, representing 12% proportional decrease per unit.

Question 7

1. $T(t)=T_0+kt$
2. $T(t)=T_0e^{kt}$ (correct answer)
3. $T(t)=T_0e^{-kt}$
4. $T(t)=T_0e^{kt}+k$
5. $T(t)=kt+T_0e^{kt}$

Explanation: This question addresses exponential decay in temperature, though the notation might seem counterintuitive at first. The equation dT/dt = kT with k < 0 means temperature changes at a rate proportional to its current value, but the negative k ensures cooling. The general solution is T(t) = Ce^(kt), and with initial condition T(0) = T₀, we get T(t) = T₀e^(kt). Since k is negative, we can write this as T₀e^(-|k|t), showing explicit decay. Choice C (T₀e^(-kt)) would actually represent growth since k < 0 makes -k positive, contradicting the cooling behavior. The key insight: in dy/dt = ky, the sign of k determines growth (k > 0) or decay (k < 0), but the solution form y = Ce^(kt) remains the same.

Question 8

1. $y=3x+C$
2. $y=C+e^{3x}$
3. $y=Ce^{3x}$ (correct answer)
4. $y=Ce^{-3x}$
5. $y=C-3x$

Explanation: This problem involves solving an exponential differential equation where the rate of change is proportional to the current quantity. When we have $\frac{dy}{dx}=3y$, the quantity $y$ changes at a rate proportional to its current value with proportionality constant 3. This positive constant indicates exponential growth, as larger values of $y$ produce proportionally larger rates of increase. The standard solution to $\frac{dy}{dx} = ky$ is $y = Ce^{kx}$, so with $k = 3$, we get $y = Ce^{3x}$. Choice A represents linear growth, which would occur if the rate were constant rather than proportional to the current value. To recognize exponential vs linear behavior: proportional rates always yield exponential solutions, while constant rates yield linear solutions.

Question 9

1. $N(t)=kt+C$
2. $N(t)=C+e^{kt}$
3. $N(t)=Ce^{-kt}$
4. $N(t)=Ce^{kt}$ (correct answer)
5. $N(t)=Ck^t$

Explanation: This problem presents the fundamental exponential growth differential equation dN/dt = kN with k > 0. When the growth rate is proportional to the current population size, we get exponential behavior described by N(t) = Ce^(kt). The positive constant k ensures the population grows over time, as e^(kt) increases when k > 0 and t increases. This proportional relationship is what distinguishes exponential growth from other types—the larger the population, the faster it grows. Choice E (N(t) = Ck^t) might seem plausible but represents a different exponential base (k instead of e), which doesn't solve our differential equation. Remember: differential equations of the form dy/dt = ky always have solutions y = Ce^(kt), where e is the natural exponential base.

Question 10

1. $P(t)=0.07t+C$
2. $P(t)=Ce^{0.07t}$ (correct answer)
3. $P(t)=Ce^{-0.07t}$
4. $P(t)=Ce^{0.07t}+0.07$
5. $P(t)=0.07t+Ce^{0.07t}$

Explanation: This problem explicitly states that growth rate is a percentage of current size, the defining characteristic of exponential growth. The equation dP/dt = 0.07P means the population grows at 7% of its current value per unit time. The general solution to this exponential differential equation is P(t) = Ce^(0.07t), where C represents the initial population size. This captures how larger populations grow faster in absolute terms while maintaining the same relative growth rate. Choice A (0.07t + C) would mean the population grows by a constant 0.07 units per time period, not 7% of its current size. The key recognition pattern: when a rate is described as a percentage or proportion of the current value, you have an exponential differential equation.

Question 11

1. $A(t)=Ce^{-0.2t}$ (correct answer)
2. $A(t)=C-0.2t$
3. $A(t)=Ce^{0.2t}$
4. $A(t)=Ce^{-0.2t}+0.2$
5. $A(t)=-0.2t+C e^{-0.2t}$

Explanation: This question involves solving an exponential differential equation where the medication amount decreases proportionally to its current value. The equation dA/dt = -0.2A indicates exponential decay because the rate is negative and proportional to A itself. Solving this differential equation yields A(t) = Ce^(-0.2t), where C is determined by initial conditions. The negative coefficient (-0.2) in the differential equation becomes the exponent in the exponential function, maintaining the decay behavior. Choice B (C - 0.2t) would represent linear decay with constant rate -0.2, not the proportional decay described by our differential equation. When you see dy/dt = -ky (with k > 0), the solution is always y = Ce^(-kt), representing exponential decay.

Question 12

1. $y=1.5t+C$
2. $y=Ce^{-1.5t}$
3. $y=C+e^{1.5t}$
4. $y=C-1.5t$
5. $y=Ce^{1.5t}$ (correct answer)

Explanation: This problem involves solving an exponential differential equation representing exponential growth. When we have $\frac{dy}{dt}=1.5y$, the quantity grows at a rate proportional to its current value with proportionality constant 1.5. This positive constant indicates exponential growth, as larger quantities produce proportionally larger increases at 1.5 times the current amount. The standard solution to $\frac{dy}{dt} = ky$ is $y = Ce^{kt}$, so with $k = 1.5$, we get $y = Ce^{1.5t}$. Choice A represents linear growth, which would occur if the growth rate were constant rather than proportional to the current value. To identify exponential growth: positive constants in proportional rate equations ($\frac{dy}{dt} = ky$) always produce exponential growth solutions ($Ce^{kt}$).

Question 13

1. $P(t)=Ce^{-0.03t}$ (correct answer)
2. $P(t)=Ce^{0.03t}$
3. $P(t)=C+e^{-0.03t}$
4. $P(t)=-0.03t+C$
5. $P(t)=0.03t+C$

Explanation: This problem involves solving an exponential differential equation representing population decline. When we have $\frac{dP}{dt}=-0.03P$, the population decreases at a rate proportional to its current size with constant -0.03. The negative proportionality constant indicates exponential decay, as declining populations have proportionally fewer individuals to sustain reproduction. The standard solution to $\frac{dy}{dt} = ky$ is $y = Ce^{kt}$, so with $k = -0.03$, we get $P(t) = Ce^{-0.03t}$. Choice B would represent exponential growth rather than the required population decline. To recognize population decline: negative constants in proportional growth equations ($\frac{dP}{dt} = kP$) always indicate exponential decay solutions.

Question 14

1. $y=C+e^{x}$
2. $y=x+C$
3. $y=Ce^{x}$ (correct answer)
4. $y=Ce^{-x}$
5. $y=C-x$

Explanation: This problem involves solving the fundamental exponential differential equation. When we have $\frac{dy}{dx}=y$, the quantity grows at a rate equal to its current value, which is the most basic form of exponential growth. This means the derivative equals the function itself, a defining property of the exponential function. The standard solution to $\frac{dy}{dx} = ky$ is $y = Ce^{kx}$, and with $k = 1$, we get $y = Ce^{x}$. Choice A represents addition of an exponential term rather than multiplication by an exponential factor. To recognize the natural exponential: when $\frac{dy}{dx} = y$, the solution is always $Ce^{x}$, the natural exponential function family.

Question 15

1. $y(t)=Ce^{0.9t}$
2. $y(t)=Ce^{-0.9t}$ (correct answer)
3. $y(t)=C+e^{-0.9t}$
4. $y(t)=-0.9t+C$
5. $y(t)=0.9t+C$

Explanation: This problem involves solving an exponential differential equation representing exponential decay. When we have $\frac{dy}{dt}=-0.9y$, the quantity decreases at a rate proportional to its current value with constant -0.9. The negative proportionality constant indicates exponential decay at 90% of the current value per unit time. The general solution to $\frac{dy}{dt} = ky$ is $y = Ce^{kt}$, so with $k = -0.9$, we get $y = Ce^{-0.9t}$. Choice D represents linear decay, but the proportional relationship requires exponential rather than linear behavior. To identify moderate exponential decay: negative constants with absolute values less than 1 still produce exponential decay, just at rates slower than natural decay.

Question 16

1. $N(t)=Ce^{-2t}$
2. $N(t)=C+e^{2t}$
3. $N(t)=2t+C$
4. $N(t)=Ce^{2t}$ (correct answer)
5. $N(t)=C-2t$

Explanation: This problem involves solving an exponential differential equation representing rapid population growth. When we have $\frac{dN}{dt}=2N$, the colony grows at a rate proportional to its current size with proportionality constant 2. This represents exponential growth where each individual contributes to reproduction, making larger populations grow proportionally faster at double the rate. The standard solution to $\frac{dy}{dt} = ky$ is $y = Ce^{kt}$, so with $k = 2$, we get $N(t) = Ce^{2t}$. Choice A would represent exponential decay rather than the required growth. To recognize exponential population growth: positive constants in proportional rate equations always produce exponential growth solutions with positive exponents.

Question 17

1. $y=C+kx$
2. $y=Ce^{kx}$ (correct answer)
3. $y=Ce^{-kx}$
4. $y=C+e^{kx}$
5. $y=C-kx$

Explanation: This problem involves solving an exponential differential equation where the rate of change is proportional to the current quantity. When we have $\frac{dy}{dx}=ky$ where $k<0$, the quantity changes at a rate proportional to its current value with a negative proportionality constant. Since $k<0$, this represents exponential decay, as the quantity decreases proportionally to its current size. The general solution to $\frac{dy}{dx} = ky$ is $y = Ce^{kx}$, and since $k<0$, we can write this as $y = Ce^{kx}$ where the exponent $kx$ is negative. Choice C represents linear decay, but proportional rates always require exponential solutions. To recognize decay with negative constants: $k<0$ in $\frac{dy}{dx} = ky$ produces exponential decay solutions $Ce^{kx}$ with negative exponents.

Question 18

1. $y=Ce^{-(3/2)x}$ (correct answer)
2. $y=Ce^{(3/2)x}$
3. $y=-(3/2)x+C$
4. $y=C+e^{-(3/2)x}$
5. $y=(3/2)x+C$

Explanation: This problem involves solving an exponential differential equation where the rate of change is proportional to the current quantity. When we have $\frac{dy}{dx}=-\frac{3}{2}y$, the quantity decreases at a rate proportional to its current value with constant $-\frac{3}{2}$. The negative fractional constant indicates exponential decay at a rate of 1.5 times the current value. The general solution to $\frac{dy}{dx} = ky$ is $y = Ce^{kx}$, so with $k = -\frac{3}{2}$, we get $y = Ce^{-(3/2)x}$. Choice C represents linear decay, but the proportional relationship requires exponential rather than linear behavior. To handle fractional decay constants: negative fractions in proportional rate equations still produce exponential decay solutions with the same fractional exponent.

Question 19

1. $m(t)=Ce^{-2t}$ (correct answer)
2. $m(t)=Ce^{2t}$
3. $m(t)=C+e^{-2t}$
4. $m(t)=-2t+C$
5. $m(t)=2t+C$

Explanation: This problem involves solving an exponential differential equation representing exponential decay. When we have $\frac{dm}{dt}=-2m$, the mass decreases at a rate proportional to its current amount with proportionality constant -2. The negative constant indicates decay, as larger masses lose material proportionally faster. The general solution to $\frac{dy}{dt} = ky$ is $y = Ce^{kt}$, so with $k = -2$, we get $m(t) = Ce^{-2t}$. Choice B would represent exponential growth rather than decay since it has a positive exponent. To distinguish exponential growth from decay: negative coefficients in $\frac{dy}{dt} = ky$ produce solutions with negative exponents ($Ce^{-|k|t}$).

Question 20

1. $y=Ce^{5x}$
2. $y=Ce^{-5x}$ (correct answer)
3. $y=-5x+C$
4. $y=C+e^{-5x}$
5. $y=5x+C$

Explanation: This problem involves solving an exponential differential equation where the rate of change is proportional to the current quantity. When we have $\frac{dy}{dx}=-5y$, the quantity decreases at a rate proportional to its current value with proportionality constant -5. The negative constant indicates exponential decay, as the quantity shrinks proportionally to its current size. The general solution to $\frac{dy}{dx} = ky$ is $y = Ce^{kx}$, so with $k = -5$, we get $y = Ce^{-5x}$. Choice C represents linear decay, but the proportional relationship requires exponential behavior rather than constant decrease. To identify decay patterns: negative constants in proportional rate equations ($\frac{dy}{dx} = -ky$) always produce exponential decay solutions ($Ce^{-kx}$).