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.
Derivative of $e^x$: What is $\frac{d}{dx}[e^{3x}]$? → If unsure, review Derivative of Exponential Functions
Logarithm rules: Solve $e^{2t} = 5$ for $t$. → If unsure, review Natural Logarithm
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
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, which 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 outFinding k From Two Data Points
The key procedure: Given $y(t_1) = y_1$ and $y(t_2) = y_2$, find $k$.
- Write: $y_1 = y_0 e^{kt_1}$ and $y_2 = y_0 e^{kt_2}$
- Divide: $\frac{y_2}{y_1} = \frac{y_0 e^{kt_2}}{y_0 e^{kt_1}} = e^{k(t_2 - t_1)}$
- Take ln: $\ln\left(\frac{y_2}{y_1}\right) = k(t_2 - t_1)$
- 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
Which of the following describes a quantity that changes at a rate proportional to its size? Select all that apply.
(A) A tank drains at a constant rate of 5 gallons per minute.
(B) A bacterial colony doubles every 3 hours.
(C) A car travels at a constant speed of 60 mph.
(D) The value of a savings account increases by 4% each year (compounded continuously).
(E) A radioactive sample loses half its mass every 100 years.
A population of insects grows at a rate proportional to its size. If there are 800 insects initially and the relative growth rate is 0.15 per day, write a formula for the population $P(t)$ after $t$ days. Then find the population after 1 week.
A colony of bacteria has 1,200 cells at noon and 4,800 cells at 3:00 PM the same day. Assuming exponential growth:
(a) Find the relative growth rate $k$.
(b) Find a formula for the population $P(t)$, where $t$ is hours after noon.
(c) What was the population at 1:00 PM?
(d) When will the population reach 20,000 cells?
A research lab is studying a bacterial culture. They know the population was 500 cells at 8:00 AM and 32,000 cells at 2:00 PM. Unfortunately, they didn’t record the initial measurement time.
(a) Find the relative growth rate.
(b) What was the population at 10:00 AM?
(c) When did the population first reach 1,000 cells?
(d) At what time would the culture have started from a single cell, according to this model?
Prove that if $y(t)$ satisfies the differential equation $\frac{dy}{dt} = ky$ with initial condition $y(0) = y_0$, then $y(t) = y_0 e^{kt}$ is the only solution.
Hint: Consider the function $g(t) = y(t) \cdot e^{-kt}$ and show that $g'(t) = 0$.
CCI-Style Conceptual Questions
Two populations grow exponentially. Population A has $k = 0.05$ per year. Population B has $k = 0.10$ per year. Both start with 1000 individuals.
Which statement is TRUE?
(A) Population B is always exactly twice as large as Population A.
(B) Population B grows at twice the rate (individuals per year) as Population A.
(C) Population B’s relative growth rate is twice that of Population A.
(D) After one year, Population B will be twice as large as Population A.
Common Misconceptions
the growth constant $k$ in $y(t) = y_0 e^{kt}$ directly equals the percentage change in $y$ per unit time.
This is the rate-as-fixed-number error. The constant $k$ is the relative instantaneous rate, meaning $\frac{dy/dt}{y} = k$. Over a finite time interval, the actual factor of change is $e^{k \Delta t}$, not $1 + k\Delta t$. For $k = 0.10$ and $\Delta t = 1$ year, the quantity grows by a factor of $e^{0.10} \approx 1.105$, a 10.5% increase, not exactly 10%. The distinction matters whenever precise predictions or comparisons are required.
Mastery Checklist
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 |
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 |
Last updated: 2026-01-22