Exponential models with differential equations

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

1. Core Concept: Exponential Models★★☆☆☆⏱ 3 min

Exponential models are the most common applied differential equation type on the AP Calculus AB exam, accounting for 2-3% of total exam weight, appearing in both multiple-choice and free-response sections. These models describe quantities whose rate of change is proportional to their current size, matching many real-world phenomena from population growth to radioactive decay.

📘 Definition

Exponential Differential Equation Model

A dynamic model where the rate of change of a quantity is directly proportional to the current value of the quantity, written as a first-order separable differential equation.

Example:

Bacterial growth, radioactive decay, drug concentration, temperature change

2. Core Proportional Growth and Decay Model★★☆☆☆⏱ 4 min

The fundamental relationship for any exponential model translates to a simple differential equation, where $k$ is the constant of proportionality:

\frac{dy}{dt} = ky

If $k > 0$ , the quantity grows exponentially; if $k < 0$ , the quantity decays exponentially. We solve this via separation of variables to get the general solution with initial condition $y (0) = y_{0}$ :

🔬 Derivation

Goal:

Derive the general solution for $\frac{d y}{d t} = k y$ with $y (0) = y_{0}$

1
Separate variables
$\frac{1}{y} dy = k dt$
2
Integrate both sides
$\ln|y| = kt + C$
3
Exponentiate to eliminate the logarithm
$|y| = e^{kt+C} = e^C e^{kt}$
4
Simplify, letting $A = \pm e^{C}$ to account for absolute value
$y = A e^{kt}$
5
Apply initial condition $y (0) = y_{0}$
$y_0 = A e^0 \implies A = y_0$

Result:

The final solution for a basic exponential model is:

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
Apply the initial condition to the general solution
$y(t) = y_0 e^{kt}, \quad y_0 = 200 \implies y(t) = 200 e^{kt}$
2
Substitute the known point $t = 3, y = 480$
$480 = 200 e^{3k} \implies 2.4 = e^{3k}$
3
Solve for $k$
$\ln(2.4) = 3k \implies k = \frac{\ln(2.4)}{3} \approx 0.2918$
4
Write the final formula
$y(t) = 200 e^{\left(\frac{\ln 2.4}{3}\right)t}$

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

Doubling time (for growth) and half-life (for decay) are special cases that let you find $k$ directly without a second measurement, or calculate time to reach a specific quantity.

For exponential growth with doubling time $T_{double}$ (time to double the initial quantity):

T_{\text{double}} = \frac{\ln 2}{k}, \quad k = \frac{\ln 2}{T_{\text{double}}}

For exponential decay with half-life $T_{half}$ (time to reduce the initial quantity by half):

T_{\text{half}} = -\frac{\ln 2}{k}, \quad k = -\frac{\ln 2}{T_{\text{half}}}

Since $k$ is negative for decay, $T_{half}$ is always positive, matching 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
Calculate $k$ from the given half-life
$k = -\frac{\ln 2}{5730} \approx -1.21 \times 10^{-4}$
2
Set up the equation for 12% remaining mass
$0.12 y_0 = y_0 e^{kt} \implies 0.12 = e^{kt}$
3
Take natural logs and solve for $t$
$\ln(0.12) = kt \implies t = \frac{\ln(0.12)}{k}$
4
Substitute $k$ and calculate
$t = 5730 \cdot \frac{\ln(0.12)}{-\ln 2} \approx 17500$

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

Newton's Law of Cooling is a modified exponential model that describes temperature change of an object relative to a constant ambient (surrounding) temperature. The rate of change of the object's temperature is proportional to the difference between the object's temperature and the ambient temperature.

\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 $k < 0$ always. Solving via separation of variables gives the general solution for initial temperature $T (0) = T_{0}$ :

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

This model works for both cooling (hot object in a cool room) and warming (cool object in a warm room): the object's temperature always approaches the ambient temperature over time.

📐 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
Substitute known values into the general solution
$T(t) = 20 + (95 - 20)e^{kt} = 20 + 75 e^{kt}$
2
Use the known point to find $e^{5 k}$
$70 = 20 + 75 e^{5k} \implies \frac{2}{3} = e^{5k}$
3
Simplify $e^{10 k}$ without approximating $k$
$e^{10k} = (e^{5k})^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9}$
4
Calculate $T (10)$
$T(10) = 20 + 75\left(\frac{4}{9}\right) \approx 53.3^\circ\text{C}$

5. Common Pitfalls

Wrong move:

Writing Newton's Law of Cooling as $\frac{d T}{d t} = k T$ instead of $\frac{d T}{d t} = k (T - T_{s})$

Why:

Confusing Newton's Law with the basic exponential growth/decay model; rate depends on temperature difference, not absolute temperature

Correct move:

When working on a temperature problem, always write the $(T - T_{s})$ term immediately after $\frac{d T}{d t}$

Wrong move:

Using the doubling time formula $k = \frac{l n 2}{T}$ for half-life problems, resulting in a positive $k$ for decay

Why:

Mixing up growth and decay formulas and forgetting to check the sign of $k$

Correct move:

After calculating $k$ for any decay problem, confirm it is negative; add a negative sign if it is positive

Wrong move:

Omitting the $+ T_{s}$ term in the Newton's Law general solution, writing $T (t) = (T_{0} - T_{s}) e^{k t}$

Why:

Forgetting to fully isolate $T$ after integrating, leaving the constant ambient temperature term on the wrong side

Correct move:

After integrating and simplifying, always fully isolate $T$ before applying the initial condition

Wrong move:

Canceling $y_{0}$ when $y_{0} = 0$ , leading to an undefined model

Why:

Forgetting that exponential models assume non-zero initial quantity; zero initial quantity is a trivial special case

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 $A = e^{C}$ (only positive) after integrating $ln ∣ y ∣$ , leading to incorrect sign for $y$

Why:

Forgetting the absolute value in $ln ∣ y ∣$ requires allowing $A$ to be negative to match the sign of the initial quantity

Correct move:

After solving for $A$ with the initial condition, confirm the sign of $A$ matches the sign of the initial quantity

6. Quick Reference Cheatsheet

Category	Formula	Notes
Core Exponential Differential Equation	$\frac{d y}{d t} = k y$	Rate proportional to current quantity; $k > 0$ = growth, $k < 0$ = decay
General Core Solution	$y (t) = y_{0} e^{k t}$	$y_{0} = y (0)$ is initial quantity at $t = 0$
Doubling Time (Growth)	$T_{double} = \frac{l n 2}{k}, k = \frac{l n 2}{T _{double}}$	Only for positive $k$ (exponential growth)
Half-Life (Decay)	$T_{half} = - \frac{l n 2}{k}, k = - \frac{l n 2}{T _{half}}$	Only for negative $k$ (exponential decay); $T_{half}$ is always positive
Newton's Law Differential Equation	$\frac{d T}{d t} = k (T - T_{s})$	$T_{s}$ = constant ambient temperature; $k$ is always negative
Newton's Law General Solution	$T (t) = T_{s} + (T_{0} - T_{s}) e^{k t}$	$T_{0}$ = initial object temperature at $t = 0$
Solving for $k$	$k = \frac{l n ( y _{1} / y _{0} )}{t _{1}}$	Used when given $y (t_{1}) = y_{1}$ for any exponential model

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 · MCQ
Half-life decay calculation
2022 · FRQ
Newton's Law of Cooling problem
2021 · MCQ
Exponential growth model solution

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.