Exponential models with differential equations — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: The general exponential growth/decay differential equation, separation of variables solution, Newton’s Law of Cooling, half-life calculations, continuous growth models, and interpreting solutions in real contextual problems.

You should already know: Separation of variables for separable differential equations. Basic derivative and integral rules for exponential functions. Evaluating initial conditions to find constants of integration.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Exponential models with differential equations?

Exponential models describe quantities that change at a rate proportional to the current size of the quantity, and they are one of the most frequently tested differential equation models on the AP Calculus BC exam. This topic makes up roughly 10-15% of Unit 7 (Differential Equations) exam weight, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often as part of a contextual application question that also tests other skills like interpretation or integration. A synonym sometimes used for this topic is proportional growth/decay models, since the core relationship is that rate of change is proportional to current amount. Unlike linear models where rate is constant, exponential models have changing rates that match many real-world phenomena: population growth with unlimited resources, radioactive decay, cooling of objects, continuous compounding of interest, and drug concentration in the bloodstream. On the AP exam, you will typically be given either the differential equation relationship, or a description of the proportionality, and asked to solve for the function, find a value at a specific time, or interpret the constant or rate in context.

2. Core Exponential Growth and Decay Model

The core relationship that defines all basic exponential models is that the rate of change of a quantity $y(t)$ with respect to time $t$ is proportional to the current value of $y$. Mathematically, this translates to the first-order separable differential equation: $$\frac{dy}{dt}=ky$$ where $k$ is the constant of proportionality. If $k>0$, the model describes exponential growth (quantity increases over time); if $k<0$, it describes exponential decay (quantity decreases over time).

We can solve this equation via separation of variables to get the general solution. Separate variables to get $\frac{1}{y}dy=kdt$, then integrate both sides: $\ln |y|=kt+C_1$, where $C_1$ is the constant of integration. Exponentiate both sides to get $|y|=e^{kt+C_1}=e^{C_1}{e}^{kt}$. Let $C=\pm e^{C_1}$, so $y=C{e}^{kt}$. Applying the initial condition $y(0)=y_0$ (the initial amount at $t=0$) gives $C=y_0$, so the final general solution is: $$y(t)=y_0{e}^{kt}$$

Worked Example

A population of bacteria increases at a rate proportional to its current size. At time $t=0$ hours, the population is 2000. At time $t=2$ hours, the population is 3000. What is the population at $t=6$ hours?

1. From the problem description, we use the core exponential model: $\frac{dy}{dt}=ky$, so the general solution is $y(t)=y_0{e}^{kt}$.
2. Substitute the initial condition $y(0)=2000$ to get $y_0=2000$, so $y(t)=2000e^{kt}$.
3. Use the given point at $t=2$ to solve for $k$: $3000=2000e^{2k}⟹\frac{3}{2}=e^{2k}⟹\ln \left(\frac{3}{2}\right)=2k⟹k=\frac{1}{2}\ln \left(\frac{3}{2}\right)$.
4. Evaluate $y(6)$ using logarithm rules to simplify: $y(6)=2000e^{6k}=2000e^{3\ln (3/2)}=2000{\left(\frac{3}{2}\right)}^3=2000\cdot \frac{27}{8}=6750$.

The population at $t=6$ hours is 6750.

Exam tip: Simplify $e^{kt}$ using logarithm exponent rules before plugging in a calculator; this avoids rounding errors and often gives an exact integer answer, which is what AP exam questions almost always expect.

3. Half-Life and Doubling Time

Half-life (for decay) and doubling time (for growth) are common special cases of exponential models where you are given the time to halve or double the quantity, instead of a second point to solve for $k$. For a half-life $T_{1/2}$, by definition half of the original quantity remains at $t=T_{1/2}$, so $y(T_{1/2})=\frac{1}{2}{y}_0$. Substitute into the general solution to get $\frac{1}{2}{y}_0=y_0{e}^{k{T}_{1/2}}$. Cancel $y_0$ and take the natural log of both sides: $\ln \left(\frac{1}{2}\right)=k{T}_{1/2}⟹-\ln 2=k{T}_{1/2}$, so $k=-\frac{\ln 2}{T_{1/2}}$.

For doubling time $T_2$ (growth), by definition $y(T_2)=2y_0$, so following the same steps gives $k=\frac{\ln 2}{T_2}$. These relations can be derived quickly on the exam if you forget them, but memorizing them saves time.

Worked Example

A 100g sample of a radioactive isotope has a half-life of 15 years. How much of the original isotope remains after 40 years, to the nearest gram?

1. Start with the general decay solution: $y(t)=y_0{e}^{kt}$, with initial amount $y_0=100$g.
2. Use the half-life to solve for $k$: $T_{1/2}=15$, so $k=-\frac{\ln 2}{15}\approx -0.0462$.
3. Substitute $t=40$ and simplify: $y(40)=100e^{(-\ln 2/15)(40)}=100(2{)}^{-40/15}=100(2{)}^{-8/3}\approx 100(0.1575)\approx 15.75$.
4. Round to the nearest whole gram: 16g.

16 grams of the original isotope remain after 40 years.

Exam tip: After solving for k in any decay problem, confirm k is negative; if you get a positive k, you forgot the negative sign from $\ln (1/2)$ and need to correct your work before proceeding.

4. Newton's Law of Cooling

Newton's Law of Cooling is a modified exponential model that describes the temperature change of an object relative to the constant temperature of its surroundings (called the ambient temperature). Unlike basic exponential models where rate is proportional to current amount, here the rate of change of the object's temperature is proportional to the difference between the object's temperature and the ambient temperature. The differential equation is: $$\frac{dT}{dt}=k(T-T_s)$$ where $T(t)$ is the object's temperature at time $t$, $T_s$ is the constant ambient temperature, and $k$ is always a negative constant (if the object is hotter than its surroundings, it cools, so $\frac{dT}{dt}$ is negative, which matches a positive $(T-T_s)$ times negative $k$).

Solving this via separation of variables gives the general solution: $$T(t)=T_s+(T_0-T_s)e^{kt}$$ where $T_0$ is the initial temperature of the object at $t=0$. This matches intuition: as $t\to \infty$, the exponential term approaches 0, so $T(t)\to T_s$, which matches real-world behavior.

Worked Example

A hot cup of coffee has an initial temperature of 180°F. It is placed in a room with constant temperature 70°F. After 5 minutes, the coffee's temperature is 150°F. What is the temperature of the coffee after 15 minutes, to the nearest degree?

1. Identify known values: $T_s=70°F$, $T_0=180°F$, so the general solution is $T(t)=70+(180-70)e^{kt}=70+110e^{kt}$.
2. Use $T(5)=150°F$ to solve for $k$: $150=70+110e^{5k}⟹80=110e^{5k}⟹\frac{8}{11}=e^{5k}⟹k=\frac{1}{5}\ln \left(\frac{8}{11}\right)\approx -0.0645$.
3. Evaluate at $t=15$ and simplify: $T(15)=70+110e^{15k}=70+110{\left(\frac{8}{11}\right)}^3=70+110\left(\frac{512}{1331}\right)\approx 70+42.3=112.3°F$.

The temperature of the coffee after 15 minutes is approximately 112°F.

Exam tip: Never forget the shifted $T_s$ term; a common mistake is using the basic exponential solution $T(t)=T_0{e}^{kt}$, which incorrectly predicts the object's temperature will approach 0 instead of the ambient temperature.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Writing Newton's Law of Cooling with a positive $k$ and concluding the object's temperature will grow indefinitely far from the ambient temperature. Why: Students mix up sign conventions for proportionality. Correct move: Always check the limit as $t\to \infty$: $T(t)$ should approach $T_s$, not infinity or zero, to confirm your sign for $k$ is correct.
• Wrong move: Getting a positive $k$ for a half-life decay problem after deriving it from $T_{1/2}$. Why: Students forget that $\ln (1/2)=-\ln 2$ and drop the negative sign. Correct move: Confirm $k$ is negative for all decay problems before substituting into the solution function.
• Wrong move: Rounding $k$ to two decimal places early in the problem, leading to a final answer that is outside the AP exam's acceptable error range. Why: Students rush to calculate a decimal value for $k$ instead of simplifying with logarithm rules. Correct move: Simplify $e^{kt}$ to a power of a constant before calculating a final decimal value.
• Wrong move: Omitting the constant of integration $+C$ after integrating $\frac{1}{y}dy=kdt$, leading to the solution $y=e^{kt}$ with no constant term. Why: Students rush the separation of variables step. Correct move: Add the constant of integration immediately after integrating both sides, before exponentiating or rearranging.
• Wrong move: Stating that $k=0.03$ means 3% growth per year for a continuous exponential model. Why: Students confuse discrete annual growth with continuous growth. Correct move: For continuous models, $k$ is the proportionality constant; to get percentage growth per unit time, multiply $k$ by 100.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

A quantity $y(t)$ decays according to the differential equation $\frac{dy}{dt}=-0.02y$. If the initial quantity is 100, what is the half-life of the quantity, to the nearest tenth of a unit? (A) 13.9 (B) 24.3 (C) 34.7 (D) 69.3

Worked Solution: For any exponential decay model with half-life $T_{1/2}$, we use the relation $\ln (1/2)=k{T}_{1/2}$. Rearranging gives $T_{1/2}=\frac{-\ln 2}{k}$. Substitute $k=-0.02$: $T_{1/2}=\frac{-0.6931}{-0.02}=34.655$, which rounds to 34.7. The correct answer is (C).

Question 2 (Free Response)

A population of deer in a protected forest grows at a rate proportional to the current number of deer. At time $t=0$ years, the population is 200. After 4 years, the population is 280. (a) Write the explicit function $P(t)$ that gives the deer population at time $t$. (b) What is the population after 10 years, to the nearest whole number? (c) At what time $t$ will the population reach 1000 deer, to the nearest tenth of a year?

Worked Solution: (a) For exponential growth, the general solution is $P(t)=P_0{e}^{kt}$. Substitute $P_0=200$ and use $P(4)=280$ to solve for $k$: $280=200e^{4k}⟹1.4=e^{4k}⟹k=\frac{\ln 1.4}{4}$. The explicit function is: $$\boxed{P(t)=200e^{\left(\frac{\ln 1.4}{4}\right)t}}$$

(b) Evaluate at $t=10$: $P(10)=200(1.4{)}^{10/4}=200(1.4{)}^{2.5}\approx 200(2.208)\approx 442$. $$\boxed{442\text{ deer}}$$

(c) Set $P(t)=1000$ and solve for $t$: $1000=200e^{(\ln 1.4/4)t}⟹\ln 5=\left(\frac{\ln 1.4}{4}\right)t⟹t=\frac{4\ln 5}{\ln 1.4}\approx 19.1$. $$\boxed{19.1\text{ years}}$$

Question 3 (Application / Real-World Style)

A veterinarian administers a 200mg dose of a drug to a dog. The drug is eliminated from the dog's bloodstream at a rate proportional to the amount of the drug present. After 3 hours, 120mg of the drug remains. A blood test can detect the drug if there is at least 10mg present in the bloodstream. How many hours after injection will the drug no longer be detectable? Round your answer to the nearest whole hour, and interpret your result in context.

Worked Solution: Let $A(t)$ be the amount of drug in mg at time $t$ hours. The proportional elimination rate gives $\frac{dA}{dt}=kA$, with general solution $A(t)=200e^{kt}$. Use $A(3)=120$ to solve for $k$: $120=200e^{3k}⟹0.6=e^{3k}⟹k=\frac{\ln 0.6}{3}\approx -0.1703$. Set $A(t)=10$ (the detection threshold) and solve for $t$: $10=200e^{-0.1703t}⟹0.05=e^{-0.1703t}⟹\ln (0.05)=-0.1703t⟹t\approx 18$ hours.

Interpretation: The drug will no longer be detectable in the dog's bloodstream approximately 18 hours after the initial injection.

7. Quick Reference Cheatsheet

8. What's Next

Exponential models are the foundation for more complex differential equation models that you will study next in Unit 7. Immediately after mastering exponential models, you will move on to logistic growth models, which account for carrying capacity and limited resources in population growth, a topic that frequently appears in AP Calculus BC FRQs. Without a solid understanding of how exponential models are derived from differential equations and how to solve for constants of proportionality, you will struggle to separate variables and interpret logistic models, which share many solution properties with exponential models. Exponential models also connect to integration of exponential functions and parametric differential equations later in the course.

• Logistic models with differential equations
• Separation of variables
• Euler's method for approximate solutions