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Exponential Growth and Decay Model

Why Does Everything Seem to Grow (or Shrink) Exponentially?

Bacteria doubling every hour. A savings account earning interest. A radioactive sample losing mass. A population of rabbits exploding across a continent. What do these seemingly unrelated phenomena have in common?

They all share a fundamental property: the rate of change is proportional to the current amount. When you have more bacteria, they reproduce faster. When you have more money, you earn more interest. When you have more radioactive atoms, more decay per second.

This simple relationship—"the more you have, the faster it changes"—leads to one of the most important differential equations in science.

Before You Start

Quick Self-Check (click to expand)

Can you answer these questions? If not, review the linked prerequisite first.

1. Derivative of $e^x$: What is $\frac{d}{dx}[e^{3x}]$?

1. Logarithm rules: Solve $e^{2t} = 5$ for $t$.

Prerequisite Map

Quick Reference

Property	Value
Section	§3.8 Exponential Growth and Decay
Difficulty	Intermediate
Time	~20 minutes

Key Concepts

The Fundamental Differential Equation

If a quantity $y(t)$ changes at a rate proportional to its current value, then:

$$\boxed{\frac{dy}{dt} = ky}$$

where $k$ is called the proportionality constant or growth/decay constant.

• If $k > 0$: growth (the quantity increases)
• If $k < 0$: decay (the quantity decreases)

The Solution

The only functions satisfying $\frac{dy}{dt} = ky$ are exponential functions:

$$\boxed{y(t) = y_0 e^{kt}}$$

where $y_0 = y(0)$ is the initial value (the value at time $t = 0$).

Why This Works

Let's verify that $y(t) = y_0 e^{kt}$ actually satisfies $\frac{dy}{dt} = ky$:

$$\frac{dy}{dt} = \frac{d}{dt}\left[y_0 e^{kt}\right] = y_0 \cdot k e^{kt} = k \cdot (y_0 e^{kt}) = ky \quad \checkmark$$

The exponential function is special because its derivative is a constant multiple of itself.

The Relative Growth Rate

The quantity $\frac{dy/dt}{y} = k$ is called the relative growth rate—it tells you the rate of change per unit of the quantity itself.

Example	Meaning of $k = 0.03$
Population	Growing at 3% per year
Investment	Earning 3% interest (continuously compounded)
Radioactive mass	Losing 3% per year (if $k = -0.03$)

Visualizing Growth vs Decay

Growth (k > 0):                    Decay (k < 0):
    y                                  y
    |         ___                      |\
    |       _/                         | \___
    |     _/                           |     \___
    |   _/                             |         \___
    |__/                               |             \___
    +---------------→ t               +---------------→ t
    Curve gets steeper               Curve flattens out

Finding k From Two Data Points

The key procedure: Given $y(t_1) = y_1$ and $y(t_2) = y_2$, find $k$.

1. Write: $y_1 = y_0 e^{kt_1}$ and $y_2 = y_0 e^{kt_2}$
2. Divide: $\frac{y_2}{y_1} = \frac{y_0 e^{kt_2}}{y_0 e^{kt_1}} = e^{k(t_2 - t_1)}$
3. Take ln: $\ln\left(\frac{y_2}{y_1}\right) = k(t_2 - t_1)$
4. Solve: $k = \frac{1}{t_2 - t_1} \ln\left(\frac{y_2}{y_1}\right)$

$$\boxed{k = \frac{\ln(y_2/y_1)}{t_2 - t_1}}$$

Domain and Validity

Condition	Requirement	Why
$y_0$	Must be positive	Exponential functions output positive values only
$y(t)$	Always positive	$e^{kt} > 0$ for all $t$
$k$	Can be any real number	Positive = growth, negative = decay
$t$	Usually $t \geq 0$	Time typically starts from initial measurement

Common Pitfalls

Mistake	Why It's Wrong	Correct Approach
Confusing $k$ with percentage	$k = 0.03$ means 3% relative rate, but after 1 year it's $e^{0.03} \approx 1.0305$, not exactly 1.03	Use the formula; don't assume $k$ = actual percentage increase
Forgetting $y_0$ in the solution	Writing $y(t) = e^{kt}$	Always include initial value: $y(t) = y_0 e^{kt}$
Using wrong time units	Mixing hours and days when finding $k$	Keep units consistent throughout
Taking ln incorrectly	$\ln(y_2/y_1) \neq \ln y_2 / \ln y_1$	The log of a quotient is $\ln y_2 - \ln y_1$
Assuming decay means $y$ becomes 0	$e^{kt} \to 0$ as $t \to \infty$ but never equals 0	Exponential decay approaches but never reaches 0

Practice Problems

CCI-Style Conceptual Questions

Mastery Checklist

• [ ] I can recognize when a situation involves exponential growth or decay
• [ ] I can write the differential equation $\frac{dy}{dt} = ky$ for a given scenario
• [ ] I can state the general solution $y(t) = y_0 e^{kt}$ and explain each part
• [ ] I can find $k$ given two data points using logarithms
• [ ] I can use the model to predict future values or find when a value is reached
• [ ] I understand why $k$ is called the "relative growth rate"
• [ ] I can explain why exponential functions (and only exponentials) solve $\frac{dy}{dt} = ky$

Mental Model

The Snowball Effect:

Think of a snowball rolling downhill. The bigger it gets, the more snow it picks up with each rotation—and picking up more snow makes it even bigger, which lets it pick up even more snow. This self-reinforcing cycle is exponential growth.

For decay, imagine a leaky balloon: the more air inside, the harder it pushes out through the leak. As air escapes, there's less pressure, so it escapes more slowly. This self-limiting process is exponential decay.

Still Confused?

If you're struggling with specific parts of this skill, here's where to look:

Struggling with...	Review this
Why the derivative of $e^{kt}$ has that form	Derivative of Exponential Functions
Solving equations with $e$ and $\ln$	Natural Logarithm Properties
The concept of rate of change	Rates of Change
Differential equations in general	Return to §9.1 Introduction to Differential Equations

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What This Unlocks

After mastering this skill, you're ready for:

• Half-Life and Radioactive Decay — applying exponential decay to physics
• Newton's Law of Cooling — a shifted exponential model
• Continuously Compounded Interest — financial applications

Previous	Up	Next
—	Chapter 6 Index	Half-Life

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Last updated: 2026-01-22