AP Calculus BC 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 BC.

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. $S(t)=3e^{0.7t}$ (correct answer)
2. $S(t)=0.7e^{3t}$
3. $S(t)=3e^{t/0.7}$
4. $S(t)=3(0.7)^t$
5. $S(t)=3+0.7t$

Explanation: This problem models exponential growth in biological systems through differential equations. The equation dS/dt = 0.7S with S(0) = 3 indicates tumor growth at a rate proportional to current size, characteristic of unrestricted exponential growth. Solving by separation of variables gives S(t) = 3e^(0.7t), where the positive coefficient creates exponential increase. The initial condition S(0) = 3 determines the coefficient in front of the exponential function. Choice D uses a base other than e, which doesn't arise naturally from the differential equation solution process. When modeling biological growth, recognize that dy/dt = ky with positive k always leads to solutions of the form y₀e^(kt).

Question 2

1. $3$ (correct answer)
2. $9\ln 3$
3. $9e^{1/\ln 3}$
4. $(\ln 3)e^{-9}$
5. $9-\ln 3$

Explanation: This problem involves exponential decay with a logarithmic rate constant. From dy/dt = -(ln 3)y with y(0) = 9, we get y(t) = 9e^(-(ln 3)t). To find y(1): y(1) = 9e^(-(ln 3)×1) = 9e^(-ln 3). Using the property e^(-ln a) = 1/a, we have e^(-ln 3) = 1/3, so y(1) = 9×(1/3) = 3. The logarithmic decay constant creates a specific pattern where the quantity is divided by 3 over unit time. Choice B incorrectly multiplies by ln 3 rather than recognizing the exponential-logarithm relationship. When decay constants involve logarithms, use properties like e^(-ln a) = 1/a to evaluate the expressions.

Question 3

1. $P(t)=90e^{-0.05t}$ (correct answer)
2. $P(t)=-0.05e^{90t}$
3. $P(t)=90e^{0.05t}$
4. $P(t)=0.05e^{-90t}$
5. $P(t)=90-0.05t$

Explanation: This problem models exponential decay in environmental cleanup. The differential equation dP/dt = -0.05P with P(0) = 90 represents pollutant removal at a rate proportional to current concentration, common in natural degradation processes. The solution P(t) = 90e^(-0.05t) shows exponential decrease from initial pollutant level 90, with coefficient -0.05 indicating moderate cleanup rate. This models realistic environmental remediation where pollutant concentration decreases exponentially over time. Choice C incorrectly uses a positive exponent, which would model pollutant accumulation rather than the expected removal in environmental cleanup processes. For pollution decay models, always verify that negative rate constants produce decreasing pollutant concentrations.

Question 4

1. $25e^{1}$ (correct answer)
2. $\frac{1}{5}e^{125}$
3. $25e^{25}$
4. $25\left(\frac{1}{5}\right)^5$
5. $25+1$

Explanation: This problem requires evaluating an exponential growth model with a fractional rate. From dy/dt = (1/5)y with y(0) = 25, we get y(t) = 25e^((1/5)t). To find y(5): y(5) = 25e^((1/5)×5) = 25e^1. The fractional growth rate 1/5 multiplied by time 5 gives the simple exponent 1, making the evaluation straightforward. This represents moderate growth over time period 5, reaching 25e times the original amount. Choice C incorrectly uses the time value as part of a larger exponent, which has no basis in the solution method for this differential equation. When working with fractional rates, carefully multiply by the time value to determine the final exponent.

Question 5

1. $8e^{-2}$ (correct answer)
2. $-0.5e^{32}$
3. $8e^{2}$
4. $0.5e^{-8}$
5. $8-2$

Explanation: This problem involves evaluating an exponential decay model for pharmacokinetics. From dA/dt = -0.5A with A(0) = 8, we get A(t) = 8e^(-0.5t). To find A(4): A(4) = 8e^(-0.5×4) = 8e^(-2). This represents drug elimination following first-order kinetics, where half of the remaining drug is eliminated per unit time proportional to the rate constant. The calculation shows the drug amount after 4 time units of elimination. Choice C uses a positive exponent, which would incorrectly model drug accumulation rather than elimination from the body. For pharmacokinetic decay problems, ensure the negative rate constant produces decreasing drug concentrations over time.

Question 6

1. $k=-3$ (correct answer)
2. $k=-9$
3. $k=3$
4. $k=6$
5. $k=e^{-9}$

Explanation: This problem involves finding the decay constant from given conditions over time. From $\frac{dM}{dt} = kM$ with $M(0) = 6$ and $M(3) = 6e^{-9}$, we know $M(t) = 6e^{kt}$. Substituting the condition at t = 3: $6e^{k \times 3} = 6e^{-9}$, which simplifies to $e^{3k} = e^{-9}$. Therefore, $3k = -9$, giving $k = -3$. The negative value confirms this models a decay process where material decreases over time. Choice C gives $k = 3$, which would produce $M(3) = 6e^{9}$ instead of the required $6e^{-9}$, representing growth rather than decay. When finding decay constants, ensure the rate constant is negative to match decreasing exponential behavior.

Question 7

1. $7e^{kt}$ (correct answer)
2. $ke^{7t}$
3. $7e^{t/k}$
4. $7k^{t}$
5. $7+kt$

Explanation: This problem tests understanding of the general form of exponential models. The differential equation dy/dt = ky with y(0) = 7 represents the fundamental exponential growth or decay equation, depending on the sign of k. The general solution is always y(t) = y₀e^(kt), where y₀ is the initial condition. Since y₀ = 7, the solution must be y(t) = 7e^(kt). The constant k determines whether we have growth (k > 0) or decay (k < 0), but the form remains the same. Choice B incorrectly places k as a coefficient outside the exponential, which doesn't satisfy the original differential equation. For any differential equation of the form dy/dt = ky, the solution is always y(t) = y₀e^(kt) regardless of the specific value of k.

Question 8

1. $80e^{-0.4}$ (correct answer)
2. $-0.4e^{80}$
3. $80e^{0.4}$
4. $0.4e^{-80}$
5. $80-0.4$

Explanation: This problem involves evaluating an exponential decay model at a specific time point. The differential equation dM/dt = -0.4M with M(0) = 80 gives us the solution M(t) = 80e^(-0.4t), representing radioactive decay. To find M(1), we substitute t = 1: M(1) = 80e^(-0.4×1) = 80e^(-0.4). The negative exponent confirms this is a decay process, where the amount decreases over time. Choice C uses a positive exponent, which would incorrectly model growth instead of decay for a radioactive sample. For decay problems, always verify that the exponent includes the negative sign when evaluating at specific time points.

Question 9

1. $P(t)=200e^{0.3t}$ (correct answer)
2. $P(t)=0.3e^{200t}$
3. $P(t)=200e^{t/0.3}$
4. $P(t)=200(0.3)^t$
5. $P(t)=200+0.3t$

Explanation: This problem requires solving a first-order linear differential equation to find an exponential model. The equation $\frac{dP}{dt} = 0.3P$ indicates that the rate of change is proportional to the current population, which is the defining characteristic of exponential growth. When we separate variables and integrate, we get $\ln|P| = 0.3t + C$, which gives us $P = C e^{0.3t}$. Applying the initial condition $P(0) = 200$, we find $C = 200$, so $P(t) = 200 e^{0.3t}$. Choice D represents exponential decay with a base less than 1, which contradicts the positive growth rate. When solving differential equations of the form $\frac{dy}{dt} = ky$ with initial condition $y(0) = y_0$, always look for the solution $y(t) = y_0 e^{kt}$.

Question 10

1. $k=2$ (correct answer)
2. $k=6$
3. $k=\frac{1}{2}$
4. $k=12$
5. $k=e^6$

Explanation: This problem requires finding the growth constant in an exponential model using given boundary conditions. From $\frac{dN}{dt} = kN$ with $N(0) = 12$, we know $N(t) = 12e^{kt}$. Using the condition $N(3) = 12e^6$, we substitute: $12e^{k \times 3} = 12e^6$, which simplifies to $e^{3k} = e^6$. Taking natural logarithms of both sides gives $3k = 6$, so $k = 2$. The exponential form of the given condition directly reveals the relationship between k and time. Choice B would give k = 6, which fails because $3k$ would equal 18, not 6 as required. When finding growth constants, match the exponents in the exponential expressions to establish the relationship $kt = (\text{given exponent})$.

Question 11

1. $y(t)=16e^{-3t/4}$ (correct answer)
2. $y(t)=\frac{3}{4}e^{-16t}$
3. $y(t)=16e^{3t/4}$
4. $y(t)=16\left(-\frac{3}{4}\right)^t$
5. $y(t)=16-\frac{3}{4}t$

Explanation: This problem involves exponential decay with a fractional coefficient. The differential equation $dy/dt = -(3/4)y$ with $y(0) = 16$ represents quantity decrease at a fractional rate. The solution $y(t) = 16e^{-(3/4)t}$ shows exponential decay where the fractional coefficient creates the exponent $-3t/4$. The negative sign ensures decreasing quantity over time, while fraction $3/4$ determines the decay speed. Choice C incorrectly uses a positive exponent, which would model quantity increase contrary to the negative rate in the differential equation. For decay problems with fractional rates, ensure the negative sign appears in the exponential function to represent the decreasing behavior correctly.

Question 12

1. $\frac{12}{e}$ (correct answer)
2. $-\frac{1}{6}e^{72}$
3. $12e$
4. $\frac{1}{6}e^{-12}$
5. $12-1$

Explanation: This problem requires evaluating an exponential decay model with a fractional rate. From dy/dt = -(1/6)y with y(0) = 12, we get y(t) = 12e^(-(1/6)t). To find y(6): y(6) = 12e^(-(1/6)×6) = 12e^(-1) = 12/e. The fractional decay rate -1/6 multiplied by time 6 gives the exponent -1, and e^(-1) = 1/e. This demonstrates how fractional rates over longer time periods can produce simple exponential values. Choice C uses a positive exponent, which would incorrectly model growth rather than decay as specified by the negative rate constant. For fractional decay rates, ensure proper multiplication with time to determine the final negative exponent.

Question 13

1. $A(t)=50e^{-0.2t}$ (correct answer)
2. $A(t)=-0.2e^{50t}$
3. $A(t)=50e^{0.2t}$
4. $A(t)=0.2e^{-50t}$
5. $A(t)=50-0.2t$

Explanation: This problem demonstrates exponential decay modeling through differential equations. The equation dA/dt = -0.2A shows that the medicine amount decreases at a rate proportional to the current amount, with the negative coefficient indicating decay. Separating variables gives dA/A = -0.2dt, and integration yields ln|A| = -0.2t + C, so A = Ce^(-0.2t). Using the initial condition A(0) = 50, we determine C = 50, giving us A(t) = 50e^(-0.2t). Choice C incorrectly uses a positive exponent, which would represent growth rather than decay. For exponential models, the sign of the coefficient in the differential equation directly determines whether we have growth (positive) or decay (negative).

Question 14

1. $S(t)=150e^{0.02t}$ (correct answer)
2. $S(t)=0.02e^{150t}$
3. $S(t)=150e^{t/0.02}$
4. $S(t)=150(0.02)^t$
5. $S(t)=150+0.02t$

Explanation: This problem applies exponential modeling to chemical or physical processes. The differential equation dS/dt = 0.02S with S(0) = 150 represents substance increase at a small positive rate, possibly chemical synthesis or accumulation. The solution S(t) = 150e^(0.02t) shows slow exponential growth from the initial amount of 150 units. The small positive coefficient 0.02 creates gradual increase over time rather than rapid growth. Choice D uses (0.02)^t as the base, which doesn't arise from differential equation solutions and would represent a very different mathematical relationship. For substance accumulation models, use the natural exponential function with the rate constant as the exponent coefficient.

Question 15

1. $10e^{2}$ (correct answer)
2. $10e^{8}$
3. $\frac{1}{2}e^{40}$
4. $10e^{0.5+4}$
5. $10+\frac{1}{2}(4)$

Explanation: This bacterial growth problem demonstrates exponential increase with a fractional growth rate. The equation dB/dt = B/2 has solution B(t) = Ce^(t/2), where k = 1/2. With B(0) = 10, we get B(t) = 10e^(t/2). At t = 4, B(4) = 10e^(4/2) = 10e^2. Choice B (10e^8) incorrectly multiplies 1/2 × 4 × 4 instead of just 1/2 × 4, possibly confusing the calculation. For exponential models dy/dt = ky, always multiply k by t directly in the exponent, without any additional operations.

Question 16

1. $80e^{-32}$
2. $80e^{-2}$ (correct answer)
3. $80e^{2}$
4. $-\frac{1}{4}e^{80t}$
5. $80-\frac{1}{4}(8)$

Explanation: This problem involves exponential decay modeled by a differential equation. The equation dA/dt = -A/4 has the general solution A(t) = Ce^(-t/4), where the negative coefficient indicates decay. With initial condition A(0) = 80, we get C = 80, so A(t) = 80e^(-t/4). To find A(8), we substitute: A(8) = 80e^(-8/4) = 80e^(-2). Choice A incorrectly uses -32 in the exponent by multiplying 8 × 4 instead of dividing. Remember that for dy/dt = ky, the solution is y = Ce^(kt), where k appears directly in the exponent, not modified by multiplication.

Question 17

1. $k=\ln(2)$
2. $k=\frac{\ln(2)}{2}$ (correct answer)
3. $k=\frac{2}{\ln(2)}$
4. $k=\frac{\ln(60)}{30}$
5. $k=\frac{60}{30}$

Explanation: This problem requires finding the growth constant k from given population data. The general solution to dN/dt = kN is N(t) = Ce^(kt), and with N(0) = 30, we get N(t) = 30e^(kt). Using N(2) = 60, we have 60 = 30e^(2k), which simplifies to e^(2k) = 2. Taking natural logarithm of both sides: 2k = ln(2), so k = ln(2)/2. Choice A (k = ln(2)) forgets to divide by 2, missing that the exponent contains 2k, not just k. When finding growth constants from doubling conditions, remember to account for the time variable in the exponent.

Question 18

1. $S(t)=0.1e^{500t}$
2. $S(t)=500e^{0.1+t}$
3. $S(t)=500e^{0.1t}$ (correct answer)
4. $S(t)=500e^{t/0.1}$
5. $S(t)=500+0.1t$

Explanation: This savings account problem models continuous compound interest through exponential growth. The equation dS/dt = 0.1S represents 10% continuous growth rate, with solution S(t) = Ce^(0.1t). Using S(0) = 500 gives C = 500, so S(t) = 500e^(0.1t). Choice D (S(t) = 500e^(t/0.1)) incorrectly inverts the growth rate, which would give S(t) = 500e^(10t), representing 1000% growth. When the differential equation has form dy/dt = ky, the solution is always y = Ce^(kt) with k in the numerator of the exponent, never inverted.

Question 19

1. $-1$ (correct answer)
2. $-3$
3. $1$
4. $3$
5. $-\frac{1}{3}$

Explanation: This problem requires finding the decay constant k from given information about an exponential model. Since dm/dt = km, the general solution is m(t) = Ce^(kt). Using m(0) = 12, we get C = 12, so m(t) = 12e^(kt). We're told m(3) = 12e^(-3), so substituting: 12e^(3k) = 12e^(-3). Dividing by 12 gives e^(3k) = e^(-3), which means 3k = -3, so k = -1. Choice B incorrectly identifies k = -3, perhaps confusing the exponent -3 with the value of k. To find k in exponential models, use two known values to set up an equation involving e^(kt).

Question 20

1. $Q(t)=2e^{-3t}$ (correct answer)
2. $Q(t)=-3e^{2t}$
3. $Q(t)=2e^{3t}$
4. $Q(t)=3e^{-2t}$
5. $Q(t)=2-3t$

Explanation: This problem models exponential decay in electrical systems. The differential equation dQ/dt = -3Q with Q(0) = 2 represents capacitor discharge, where charge decreases rapidly due to the large coefficient -3. The solution Q(t) = 2e^(-3t) shows exponential decrease from initial charge 2, with the coefficient -3 creating rapid discharge. This models realistic capacitor behavior in RC circuits where charge decays exponentially. Choice C incorrectly uses a positive exponent, which would model charge accumulation rather than the expected discharge in a capacitor system. For electrical discharge problems, ensure negative rate constants produce decreasing exponential functions representing charge loss.