Exponential models with differential equations

AP Calculus BC· AP Calculus BC CED — Differential Equations· 14 min read

1. Core Exponential Growth and Decay Model★★☆☆☆⏱ 4 min

Exponential models describe quantities that change at a rate proportional to the current size of the quantity. They are one of the most frequently tested differential equation models on the AP Calculus BC exam, appearing in both multiple-choice and free-response questions. The core relationship is that the rate of change of $y (t)$ with respect to time $t$ is proportional to the current value of $y$ .

📘 Definition

Exponential Growth/Decay Differential Equation

\frac{dy}{dt} = ky

First-order separable differential equation where the rate of change of a quantity is proportional to the current quantity. $k > 0$ indicates growth, $k < 0$ indicates decay.

Example:

Unlimited population growth, radioactive decay, continuous compound interest

🔬 Derivation

Goal:

Derive the general solution for the core exponential model

Starting from:

\frac{dy}{dt} = ky

1
Separate variables to group terms with $y$ and $t$ :
2
$\frac{1}{y} dy = k dt$
3
Integrate both sides of the equation:
4
$\ln|y| = kt + C_1$
5
Exponentiate both sides to eliminate the natural logarithm:
6
$|y| = e^{kt + C_1} = e^{C_1}e^{kt}$
7
Let $C = \pm e^{C_{1}}$ to absorb the sign and constant, resulting in $y = C e^{k t}$ . Apply the initial condition $y (0) = y_{0}$ (initial amount at $t = 0$ ) to find $C = y_{0}$ .

Result:

The general solution for the exponential model is $y (t) = y_{0} e^{k t}$ , where $y_{0}$ is the initial quantity at $t = 0$ .

📐 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
Use the core exponential model, with general solution:
2
$y(t) = y_0 e^{kt}$
3
Substitute the initial condition $y (0) = 2000$ to get $y_{0} = 2000$ :
4
$y(t) = 2000 e^{kt}$
5
Use the given point at $t = 2$ to solve for $k$ :
6
$3000 = 2000 e^{2k} \implies \frac{3}{2} = e^{2k} \implies \ln\left(\frac{3}{2}\right) = 2k \implies k = \frac{1}{2}\ln\left(\frac{3}{2}\right)$
7
Evaluate $y (6)$ , simplifying with logarithm rules to avoid rounding error:
8
$y(6) = 2000 e^{6k} = 2000 e^{3\ln(3/2)} = 2000 \left(\frac{3}{2}\right)^3 = 2000 \cdot \frac{27}{8} = 6750$

Exam tip:

Simplify $e^{k t}$ 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.

2. Half-Life and Doubling Time★★☆☆☆⏱ 3 min

Half-life (for decay) and doubling time (for growth) are common special cases of exponential models. Instead of giving a second point to solve for $k$ , you are given the time it takes for the quantity to halve or double. These relationships can be derived quickly on the exam if you forget them, but memorizing them saves time.

Half-life (decay): $k = - \frac{l n 2}{T _{1/2}}$ , where $T_{1/2}$ is the time to halve the quantity
Doubling time (growth): $k = \frac{l n 2}{T _{2}}$ , where $T_{2}$ is the time to double the quantity

📐 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, with initial amount $y_{0} = 100$ g:
2
$y(t) = 100 e^{kt}$
3
Use the half-life to solve for $k$ :
4
$k = -\frac{\ln 2}{15} \approx -0.0462$
5
Substitute $t = 40$ and simplify using exponent rules:
6
$y(40) = 100 e^{(-\ln 2 / 15)(40)} = 100 (2)^{-8/3} \approx 15.75$
7
Round to the nearest whole gram, giving 16g.

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.

3. Newton's Law of Cooling★★★☆☆⏱ 4 min

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.

📘 Definition

Newton's Law of Cooling

\frac{dT}{dt} = k(T - T_s)

Differential equation for temperature change of an object relative to constant ambient temperature. $k$ is always negative, so temperature approaches ambient temperature over time.

Example:

Cooling of hot food, warming of a cold drink in a room

Solving this via separation of variables gives the general solution, where $T_{0}$ is the initial temperature of the object at $t = 0$ :

T(t) = T_s + (T_0 - T_s)e^{kt}

📐 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: $T_{s} = 70° F$ , $T_{0} = 180° F$ , substitute into the general solution:
2
$T(t) = 70 + (180 - 70)e^{kt} = 70 + 110e^{kt}$
3
Use $T (5) = 150° F$ to solve for $k$ :
4
$150 = 70 + 110e^{5k} \implies \frac{8}{11} = e^{5k} \implies k = \frac{1}{5}\ln\left(\frac{8}{11}\right) \approx -0.0645$
5
Evaluate at $t = 15$ , simplifying with exponent rules:
6
$T(15) = 70 + 110e^{15k} = 70 + 110\left(\frac{8}{11}\right)^3 \approx 112.3°F$
7
Round to the nearest whole degree, giving 112°F.

Exam tip:

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

4. AP Style Practice Worked Examples★★★☆☆⏱ 3 min

📐 Worked Example

A quantity $y (t)$ decays according to the differential equation $\frac{d y}{d t} = - 0.02 y$ . 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

1
Use the half-life relation $T_{1/2} = \frac{- l n 2}{k}$ , where $k = - 0.02$ :
2
$T_{1/2} = \frac{-\ln 2}{-0.02} = 34.655$
3
Round to the nearest tenth, so the correct answer is (C) 34.7.

📐 Worked Example

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?

1
Part (a): Start with general exponential growth solution, substitute $P_{0} = 200$ :
2
$P(t) = 200 e^{kt}$
3
Use $P (4) = 280$ to solve for $k$ :
4
$280 = 200 e^{4k} \implies 1.4 = e^{4k} \implies k = \frac{\ln 1.4}{4}$
5
The explicit function is $P (t) = 200 e^{(\frac{l n 1.4}{4}) t}$
6
Part (b): Evaluate at $t = 10$ , simplifying with exponent rules:
7
$P(10) = 200 (1.4)^{10/4} = 200 (1.4)^{2.5} \approx 442$
8
Part (c): Set $P (t) = 1000$ and solve for $t$ :
9
$1000 = 200 e^{(\ln 1.4 / 4)t} \implies \ln 5 = \frac{\ln 1.4}{4} t \implies t = \frac{4 \ln 5}{\ln 1.4} \approx 19.1$

📐 Worked Example

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.

1
Let $A (t)$ be the amount of drug in mg at time $t$ hours. The proportional elimination rate gives the differential equation $\frac{d A}{d t} = k A$ , with general solution $A (t) = 200 e^{k t}$ .
2
Use $A (3) = 120$ to solve for $k$ :
3
$120 = 200 e^{3k} \implies 0.6 = e^{3k} \implies k = \frac{\ln 0.6}{3} \approx -0.1703$
4
Set $A (t) = 10$ (the detection threshold) and solve for $t$ :
5
$10 = 200 e^{-0.1703 t} \implies 0.05 = e^{-0.1703 t} \implies t \approx 18$
6
Interpretation: The drug will no longer be detectable in the dog's bloodstream approximately 18 hours after the initial injection.

5. Common Pitfalls

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 $e^{k t}$ to a power of a constant before calculating a final decimal value.

Wrong move:

Omitting the constant of integration $+ C$ after integrating $\frac{1}{y} d y = k d t$ , leading to the solution $y = e^{k t}$ 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; the percentage growth per unit time is $k \times 100$ .

6. Quick Reference Cheatsheet

Category	Formula	Notes
Core Exponential Model DE	$\frac{d y}{d t} = k y$	Applies when rate of change of $y$ is proportional to current $y$ ; $k > 0$ = growth, $k < 0$ = decay
General Solution	$y (t) = y_{0} e^{k t}$	$y_{0}$ = value of $y$ at $t = 0$ ; derived via separation of variables
Half-Life (Decay)	$T_{1/2} = \frac{l n 2}{∣ k ∣}$	Time to reduce quantity by half; $k = - \frac{l n 2}{T _{1/2}}$
Doubling Time (Growth)	$T_{2} = \frac{l n 2}{k}$	Time to double a growing quantity; $k > 0$ for all growth
Newton's Law of Cooling DE	$\frac{d T}{d t} = k (T - T_{s})$	$T_{s}$ = constant ambient temperature; $k < 0$ for cooling
Newton's Law of Cooling Solution	$T (t) = T_{s} + (T_{0} - T_{s}) e^{k t}$	$T_{0}$ = initial object temperature; as $t \to \infty$ , $T (t) \to T_{s}$
Solve for $k$ with known point	$ln (\frac{y ( t _{1} )}{y _{0}}) = k t_{1}$	Use after substituting a known point $(t_{1}, y (t_{1}))$ into the general solution

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · BC MCQ
Half-life calculation problem
2022 · BC FRQ
Newton's Law of Cooling application

What's Next

Exponential models are the foundation for more complex differential equation models you will study next in Unit 7 of AP Calculus BC. 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.