Exponential models with differential equations — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: Core proportional growth/decay differential equations, general solution derivation via separation of variables, doubling time, half-life, and Newton's Law of Cooling for temperature change modeling.

You should already know: Separation of variables for separable differential equations. Derivative and antiderivative rules for exponential and logarithmic functions. Basic initial condition application.

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 AB 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 with differential equations are the most common applied differential equation type on the AP Calculus AB exam, accounting for 2-3% of total exam weight per the official CED. They appear in both multiple-choice (MCQ) and free-response (FRQ) sections, most often as contextual applied problems in FRQs. At their core, these models describe any quantity where the rate of change of the quantity is proportional to the current size of the quantity. This relationship matches a huge range of real-world phenomena: population growth, radioactive decay, drug concentration in the bloodstream, and temperature change, among others. Unlike algebraic exponential functions, exponential models are derived directly from the rate relationship described by a differential equation, so you must be able to move from the verbal description of the rate to the differential equation, then solve it to get the explicit function for the quantity. This topic relies heavily on separation of variables, which you should already be familiar with from earlier in the unit.

2. Core Proportional Growth and Decay Model

The fundamental assumption of any exponential model is: the rate of change of a quantity $y(t)$ is proportional to the current value of $y(t)$. Translating this statement to a differential equation gives: $$\frac{dy}{dt}=ky$$ where $k$ is the constant of proportionality. If $k>0$, the quantity is growing (rate is positive when $y$ is positive), and if $k<0$, the quantity is decaying (rate is negative when $y$ is positive). We can solve this separable differential equation to get a general solution:

1. Separate variables: $\frac{1}{y}dy=kdt$
2. Integrate both sides: $\ln |y|=kt+C$, where $C$ is the constant of integration
3. Exponentiate both sides to eliminate the logarithm: $|y|=e^{kt+C}=e^C{e}^{kt}$
4. Simplify by letting $A=\pm e^C$ (the $\pm$ accounts for the absolute value), so $y=A{e}^{kt}$
5. Apply the initial condition $y(0)=y_0$ (the quantity at $t=0$): $y_0=A{e}^0=A$, so the general solution becomes: $$y(t)=y_0{e}^{kt}$$

Worked Example

A culture of bacteria grows at a rate proportional to the current number of bacteria. At $t=0$ hours, there are 200 bacteria. At $t=3$ hours, there are 480 bacteria. Write an explicit formula for $y(t)$, the number of bacteria at time $t$.

1. We start with the core exponential model: $\frac{dy}{dt}=ky$, so general solution is $y(t)=y_0{e}^{kt}$. The initial condition gives $y_0=200$, so $y(t)=200e^{kt}$.
2. Substitute the second known point $t=3,y=480$: $480=200e^{3k}$. Divide both sides by 200 to get $2.4=e^{3k}$.
3. Take the natural logarithm of both sides: $\ln (2.4)=3k$, so solve for $k$: $k=\frac{\ln (2.4)}{3}\approx 0.2918$.
4. Substitute back to get the final formula: $y(t)=200e^{\left(\frac{\ln 2.4}{3}\right)t}$, or approximately $y(t)=200e^{0.2918t}$.

Exam tip: Always check the sign of $k$ after solving: if your problem describes decay, $k$ should be negative. If it describes growth, $k$ should be positive. This quick check catches 80% of common algebraic sign errors on the exam.

3. Doubling Time and Half-Life

Doubling time and half-life are special cases of exponential growth and decay that let you find $k$ directly without a second measurement, or find the time to reach a specific quantity. Doubling time ($T_{\text{double}}$) is the time required for a growing quantity to double in size. For any exponential growth model: $$2y_0=y_0{e}^{k{T}_{\text{double}}}⟹2=e^{k{T}_{\text{double}}}⟹\ln 2=k{T}_{\text{double}}$$ Rearranged, this gives $T_{\text{double}}=\frac{\ln 2}{k}$ or $k=\frac{\ln 2}{T_{\text{double}}}$ for growth.

Half-life ($T_{\text{half}}$) is the time required for a decaying quantity to decrease to half its original size. For any exponential decay model: $$\frac{1}{2}{y}_0=y_0{e}^{k{T}_{\text{half}}}⟹\ln \left(\frac{1}{2}\right)=k{T}_{\text{half}}⟹-\ln 2=k{T}_{\text{half}}$$ Rearranged, this gives $T_{\text{half}}=-\frac{\ln 2}{k}$ or $k=-\frac{\ln 2}{T_{\text{half}}}$ for decay. Since $k$ is negative for decay, $T_{\text{half}}$ is always positive, which matches its physical meaning.

Worked Example

Radioactive Carbon-14 has a half-life of 5730 years. A fossil fragment has 12% of its original Carbon-14 remaining. How old is the fossil, to the nearest 100 years?

1. Start with the general decay model $y(t)=y_0{e}^{kt}$. Using the half-life formula, $k=-\frac{\ln 2}{5730}\approx -1.21\times 10^{-4}$ per year.
2. We know that 12% of the original Carbon-14 remains, so $y(t)=0.12y_0$. Substitute into the model: $0.12y_0=y_0{e}^{kt}$. Cancel $y_0$ (non-zero) to get $0.12=e^{kt}$.
3. Take the natural logarithm of both sides: $\ln (0.12)=kt$. Solve for $t$: $t=\frac{\ln (0.12)}{k}$.
4. Substitute $k$: $t=\frac{\ln (0.12)}{-\frac{\ln 2}{5730}}=5730\cdot \frac{\ln (0.12)}{-\ln 2}\approx 5730\cdot \frac{-2.120}{-0.693}\approx 17500$ years.

Exam tip: If the question asks for an exact form, leave your answer in terms of natural logarithms instead of approximating. Only approximate when the question explicitly asks for a numerical value.

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 a constant ambient (surrounding) temperature. The core assumption is: the rate of change of the object's temperature is proportional to the difference between the object's temperature and the ambient temperature. Translating this to a differential equation: $$\frac{dT}{dt}=k(T-T_s)$$ where $T(t)$ is the object temperature at time $t$, $T_s$ is the constant ambient temperature, and $k$ is a negative constant of proportionality. 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 model works for both cooling (hot object in cool environment) and warming (cool object in warm environment): if $T_0>T_s$, $T(t)$ decreases toward $T_s$; if $T_0<T_s$, $T(t)$ increases toward $T_s$.

Worked Example

A hot cup of tea at 95°C is placed in a 20°C room. After 5 minutes, the temperature of the tea is 70°C. What is the temperature of the tea after 10 minutes?

1. We have $T_s=20^{\circ }\text{C}$, $T_0=95^{\circ }\text{C}$, so substitute into the general solution: $T(t)=20+(95-20)e^{kt}=20+75e^{kt}$.
2. Use the known point $t=5,T=70$: $70=20+75e^{5k}⟹50=75e^{5k}⟹\frac{2}{3}=e^{5k}$.
3. To find $T(10)$, note that $e^{10k}=(e^{5k}{)}^2={\left(\frac{2}{3}\right)}^2=\frac{4}{9}$.
4. Substitute back: $T(10)=20+75\left(\frac{4}{9}\right)=20+\frac{100}{3}\approx 53.3^{\circ }\text{C}$.

Exam tip: Keep intermediate values in exact form (like $e^{5k}=2/3$) instead of approximating $k$ early. This avoids rounding error that can cost you points on FRQ.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Writing Newton's Law of Cooling as $\frac{dT}{dt}=kT$ instead of $\frac{dT}{dt}=k(T-T_s)$. Why: Students confuse Newton's Law with the basic exponential growth/decay model and forget the rate depends on the temperature difference, not the temperature itself. Correct move: When you see a temperature problem, always write the $(T-T_s)$ term immediately after writing $dT/dt$ to avoid this error.
• Wrong move: Using the doubling time formula $k=\ln 2/T$ for half-life problems, resulting in a positive $k$ for decay. Why: Students mix up the formulas for growth and decay and fail to check the sign of $k$. Correct move: After calculating $k$ for any decay problem, confirm $k$ is negative; if it is positive, add a negative sign to your result.
• Wrong move: Omitting the $+T_s$ term in the Newton's Law general solution, writing $T(t)=(T_0-T_s)e^{kt}$. Why: Students forget to move $T_s$ to the left side of the equation after integrating. Correct move: After integrating and simplifying, always isolate $T$ before applying the initial condition to make sure you do not miss the $T_s$ term.
• Wrong move: Canceling $y_0$ when $y_0=0$, leading to an undefined solution for the model. Why: Exponential models assume non-zero initial quantity, but students forget the trivial solution for zero initial quantity. Correct move: If the initial quantity is zero, the quantity stays zero for all time; write $y(t)=0$ as your solution.
• Wrong move: Setting $A=e^C$ (positive only) after integrating $\ln |y|$, leading to an incorrect sign for $y$ when the initial quantity is negative. Why: Students forget the absolute value requires allowing $A$ to be negative. Correct move: After solving for $A$ using the initial condition, confirm the sign of $A$ matches the sign of the initial quantity.

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

Which of the following is the solution to the differential equation $\frac{dy}{dt}=-0.2y$ with the initial condition $y(0)=10$? A) $y=10e^{0.2t}$ B) $y=10e^{-0.2t}$ C) $y=-0.2e^{10t}$ D) $y=e^{-0.2t}+10$

Worked Solution: For the core exponential differential equation $\frac{dy}{dt}=ky$, the general solution with initial condition $y(0)=y_0$ is $y(t)=y_0{e}^{kt}$. Here, $k=-0.2$ and $y_0=10$. Substituting these values gives $y(t)=10e^{-0.2t}$. Option A has the wrong sign on $k$, option C swaps $k$ and $y_0$, and option D adds $y_0$ instead of multiplying it. The correct answer is B.

Question 2 (Free Response)

A population of deer in a national park grows at a rate proportional to the current population size. At $t=0$ years, there are 1200 deer. Two years later, there are 1500 deer. (a) Write an explicit formula for $P(t)$, the deer population at time $t$. (b) What is the deer population after 10 years, rounded to the nearest whole number? (c) At what time will the population reach 3000 deer, rounded to two decimal places?

Worked Solution: (a) For exponential growth, $P(t)=P_0{e}^{kt}$, so $P_0=1200$. Substitute $t=2,P=1500$: $$1500=1200e^{2k}⟹\frac{5}{4}=e^{2k}⟹k=\frac{\ln (5/4)}{2}\approx 0.1116$$ The explicit formula is $P(t)=1200e^{\left(\frac{\ln (5/4)}{2}\right)t}$.

(b) At $t=10$: $$P(10)=1200e^{5\ln (5/4)}=1200{\left(\frac{5}{4}\right)}^5\approx 1200(3.0518)\approx 3662$$ The population after 10 years is 3662 deer.

(c) Set $P(t)=3000$: $$3000=1200e^{kt}⟹2.5=e^{kt}⟹t=\frac{\ln (2.5)}{k}=\frac{2\ln (2.5)}{\ln (1.25)}\approx 8.22$$ The population reaches 3000 deer at approximately 8.22 years.

Question 3 (Application / Real-World Style)

A hospital technician pulls a vial of insulin out of a 4°C refrigerator and places it on a counter in a 22°C room. After 15 minutes, the temperature of the vial is 10°C. What is the temperature of the vial after 30 minutes, to the nearest degree Celsius? Include units in your answer.

Worked Solution: Use Newton's Law of Cooling (warming): $T_s=22^{\circ }\text{C}$, $T_0=4^{\circ }\text{C}$. General solution: $$T(t)=22+(4-22)e^{kt}=22-18e^{kt}$$ At $t=15$, $T=10$: $10=22-18e^{15k}⟹\frac{2}{3}=e^{15k}$. For $t=30$: $$T(30)=22-18(e^{15k}{)}^2=22-18{\left(\frac{2}{3}\right)}^2=22-8=14$$ After 30 minutes on the counter, the insulin vial has a temperature of $14^{\circ }\text{C}$, which is still cool enough for safe clinical use.

7. Quick Reference Cheatsheet

8. What's Next

Exponential models are the foundational applied differential equation for AP Calculus AB, and they prepare you for the next core topic in Unit 7: logistic differential equation models, which add a carrying capacity to population growth to account for limited resources. Without mastering how to translate a verbal rate description to a differential equation, solve it via separation of variables, and interpret the result in context, you will not be able to correctly set up or solve logistic models, which frequently appear on FRQ sections of the exam. Beyond Unit 7, exponential models connect to integration applications and improper integrals, and they are a core tool for any real-world application of calculus to dynamic systems.

Follow-on topics: Logistic differential equations Separable differential equations Slope fields Euler's method