Exponential Growth and Decay
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 6.8: “Exponential Growth and Decay” |
| Direct link | https://openstax.org/books/calculus-volume-1/pages/6-8-exponential-growth-and-decay |
| Supplementary | OpenStax Calculus Volume 1, Section 3.9: “Derivatives of Exponential and Logarithmic Functions” |
| Supplementary link | https://openstax.org/books/calculus-volume-1/pages/3-9-derivatives-of-exponential-and-logarithmic-functions |
Both sources are free and openly licensed.
Key idea
There is one differential equation that a student of differential equations meets before any other, and it is the simplest interesting one: the rate of change of a quantity is proportional to the quantity itself. Written symbolically, \[ \frac{dy}{dt} = ky. \] This single equation is the prototype. Population growth under ideal conditions, radioactive decay, continuously compounded interest, and the cooling of a hot object all reduce to it or to a small variation of it.
The reason this equation is so central is that it has a clean, memorable solution, and you can almost guess it. The equation asks for a function whose derivative is a constant multiple of itself. The exponential function is the one function with that defining property, so the solution is $y = y_0 e^{kt}$. When $k$ is positive the quantity grows; when $k$ is negative it decays.
Securing this one model now pays off immediately. The study of differential equations treats $\frac{dy}{dt} = ky$ as the example that motivates every method: it is the first equation you model, the first you solve by separation, and the first whose direction field you read. Knowing the answer in advance lets you focus on the methods rather than the arithmetic.
Prerequisite Check
Before this lesson, make sure you can do all of the following:
If exponentials or logarithms are shaky, review The Chain Rule and a logarithm-rules lesson first.
Quick Reference
The growth-decay equation. A quantity $y(t)$ whose rate of change is proportional to its size satisfies \[ \frac{dy}{dt} = ky, \] where $k$ is the proportionality constant (the relative growth rate).
The solution. Every solution has the form \[ y(t) = y_0 e^{kt}, \] where $y_0 = y(0)$ is the initial amount.
- $k > 0$: exponential growth (the quantity increases without bound)
- $k < 0$: exponential decay (the quantity decreases toward zero)
Half-life and doubling time.
| Quantity | Formula | Meaning |
|---|---|---|
| Doubling time (growth, $k > 0$) | $T_2 = \dfrac{\ln 2}{k}$ | time for $y$ to double |
| Half-life (decay, $k < 0$) | $T_{1/2} = \dfrac{\ln 2}{\lvert k\rvert}$ | time for $y$ to halve |
Key Concepts
1. The Model: Rate Proportional to Amount
Many quantities change at a rate set by how much is currently present. A bacterial colony with twice the cells divides into twice as many new cells per minute. A radioactive sample with twice the atoms emits twice as many particles per second. The mathematical statement of “rate proportional to amount” is \[ \frac{dy}{dt} = ky. \] The constant $k$ is the relative growth rate: it is the rate of change per unit of the quantity, $\frac{1}{y}\frac{dy}{dt} = k$.
This is a differential equation because it relates an unknown function $y(t)$ to its own derivative. The task is to find the function.
2. Solving the Equation by Separation
The equation $\frac{dy}{dt} = ky$ is separable, so the variables can be moved to opposite sides and each side integrated. (This is a preview of the separation-of-variables method, treated in full in a later lesson.) Divide by $y$ and multiply by $dt$: \[ \frac{1}{y}\, dy = k\, dt. \] Integrate both sides: \[ \int \frac{1}{y}\, dy = \int k\, dt \quad\Longrightarrow\quad \ln\lvert y\rvert = kt + C_1. \] Exponentiate to undo the logarithm: \[ \lvert y\rvert = e^{kt + C_1} = e^{C_1}e^{kt}. \] Writing $C = \pm e^{C_1}$ absorbs the absolute value and the constant into a single constant, giving the general solution \[ y(t) = Ce^{kt}. \]
The constant $C$ is the initial amount. Set $t = 0$: $y(0) = Ce^{0} = C$. So $C = y_0$, the amount present at the start, and the solution is usually written $y(t) = y_0 e^{kt}$. The arbitrary constant from integration is exactly the initial condition.
3. Verifying the Solution Directly
You do not have to trust the derivation. Differentiate the proposed solution and check.
Example 1. Verify that $y(t) = y_0 e^{kt}$ solves $\dfrac{dy}{dt} = ky$.
Differentiate with the chain rule (the inside is $kt$, derivative $k$): \[ \frac{dy}{dt} = y_0 \cdot e^{kt} \cdot k = k\left(y_0 e^{kt}\right) = ky. \] The derivative equals $k$ times the function, so the equation holds. The initial value checks too: $y(0) = y_0 e^{0} = y_0$.
This is the same verification skill used throughout the course: propose a function, differentiate it, confirm it satisfies the equation.
4. Growth Versus Decay
The sign of $k$ decides the behavior.
- If $k > 0$, then $e^{kt}$ increases as $t$ increases. The quantity grows, and the larger it gets the faster it grows. This is exponential growth.
- If $k < 0$, then $e^{kt}$ decreases toward zero as $t$ increases. The quantity decays, never quite reaching zero. This is exponential decay.
Example 2. A bacterial culture starts with $500$ cells and grows at a relative rate of $k = 0.4$ per hour. Find the population after $3$ hours.
The model gives $P(t) = 500\, e^{0.4 t}$. At $t = 3$: \[ P(3) = 500\, e^{0.4 \cdot 3} = 500\, e^{1.2} \approx 500 \cdot 3.320 = 1660 \text{ cells.} \] (The answer is rounded to the nearest whole cell, since a population is a count.)
5. Finding the Rate Constant From Data
Often $k$ is not given. Instead you know the quantity at two times and must solve for $k$.
Example 3. A sample of $80$ mg of a radioactive isotope decays to $50$ mg after $6$ years. Find the decay constant $k$, then the amount remaining after $10$ years.
Start from $y(t) = 80\, e^{kt}$ (here $y_0 = 80$). Use the data point $y(6) = 50$: \[ 50 = 80\, e^{6k} \quad\Longrightarrow\quad e^{6k} = \frac{50}{80} = 0.625. \] Take the natural logarithm of both sides: \[ 6k = \ln(0.625) \approx -0.470 \quad\Longrightarrow\quad k \approx -0.0783 \text{ per year.} \] The constant is negative, confirming decay. Now find $y(10)$: \[ y(10) = 80\, e^{-0.0783 \cdot 10} = 80\, e^{-0.783} \approx 80 \cdot 0.457 = 36.6 \text{ mg.} \]
mislabeling the initial amount. The constant $y_0$ is the amount at $t = 0$, not the amount at the first measurement time given in the problem. In Example 3, $y_0 = 80$ because the sample begins at $80$ mg; the $50$ mg reading at year $6$ is the data point used to find $k$, not the value of $y_0$.
6. Half-Life and Doubling Time
The time for a quantity to halve (decay) or double (growth) does not depend on the starting amount. That is a signature feature of exponential change.
Doubling time ($k > 0$): solve $2y_0 = y_0 e^{kT_2}$, which gives $e^{kT_2} = 2$, so \[ T_2 = \frac{\ln 2}{k}. \]
Half-life ($k < 0$): solve $\tfrac{1}{2}y_0 = y_0 e^{kT_{1/2}}$, which gives $e^{kT_{1/2}} = \tfrac{1}{2}$, so \[ T_{1/2} = \frac{\ln(1/2)}{k} = \frac{-\ln 2}{k} = \frac{\ln 2}{\lvert k\rvert}. \]
Example 4. A radioactive substance has a half-life of $12$ years. Find its decay constant $k$.
From $T_{1/2} = \frac{\ln 2}{\lvert k\rvert}$, \[ \lvert k\rvert = \frac{\ln 2}{12} \approx \frac{0.693}{12} \approx 0.0578, \qquad k \approx -0.0578 \text{ per year.} \] The sign is negative because the substance decays.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Treating a later reading as $y_0$ | using $50$ as $y_0$ in Example 3 | $y_0$ is the amount at $t = 0$ ($80$ mg) |
| Sign of $k$ for decay | reporting $k > 0$ for a decaying sample | Decay requires $k < 0$ |
| Dropping the chain-rule factor when verifying | $\frac{d}{dt}e^{kt} = e^{kt}$ | $\frac{d}{dt}e^{kt} = ke^{kt}$ |
| Half-life depends on starting amount | scaling $T_{1/2}$ with $y_0$ | Half-life is independent of $y_0$ |
| Forgetting the logarithm step | solving $e^{6k}=0.625$ by division | Take $\ln$ of both sides |
Leveled Practice
Level 1: Direct Application
Problem 1. A population follows $P(t) = 200\, e^{0.05 t}$ with $t$ in years. What is the population at $t = 0$, and is it growing or decaying?
Show answer
At $t = 0$: $P(0) = 200\, e^{0} = 200$. Since $k = 0.05 > 0$, the population is growing.
Problem 2. Write the growth-decay differential equation whose solution is $y(t) = 7e^{-3t}$.
Show answer
Differentiate: $\frac{dy}{dt} = 7\cdot(-3)e^{-3t} = -3\left(7e^{-3t}\right) = -3y$. The differential equation is \[ \frac{dy}{dt} = -3y, \qquad y(0) = 7. \]
Problem 3. A quantity grows at relative rate $k = 0.1$ per hour from an initial amount of $40$. Find the amount after $5$ hours.
Show answer
\[ y(5) = 40\, e^{0.1 \cdot 5} = 40\, e^{0.5} \approx 40 \cdot 1.6487 \approx 65.9. \]
Level 2: Finding the Constant
Problem 4. A culture doubles every $3$ hours. Find its relative growth rate $k$.
Show answer
From $T_2 = \frac{\ln 2}{k}$, \[ k = \frac{\ln 2}{T_2} = \frac{\ln 2}{3} \approx \frac{0.693}{3} \approx 0.231 \text{ per hour.} \]
Problem 5. A $100$ mg sample decays to $90$ mg in $2$ years. Find $k$ and the half-life.
Show answer
From $90 = 100\, e^{2k}$: $e^{2k} = 0.9$, so $2k = \ln(0.9) \approx -0.1054$, giving $k \approx -0.0527$ per year.
Half-life: $T_{1/2} = \frac{\ln 2}{\lvert k\rvert} \approx \frac{0.693}{0.0527} \approx 13.2$ years.
Problem 6. The element carbon-14 has a half-life of about $5730$ years. Find its decay constant.
Show answer
\[ \lvert k\rvert = \frac{\ln 2}{5730} \approx \frac{0.693}{5730} \approx 1.21 \times 10^{-4} \text{ per year}, \qquad k \approx -1.21 \times 10^{-4}. \]
Level 3: Applications and Connections
Problem 7. A bank account earns interest compounded continuously at an annual rate of $4\%$, so the balance $B$ satisfies $\frac{dB}{dt} = 0.04\,B$. If the initial deposit is $\$2000$, how long until the balance reaches $\$3000$?
Show answer
The solution is $B(t) = 2000\, e^{0.04 t}$. Set $B(t) = 3000$: \[ 3000 = 2000\, e^{0.04 t} \implies e^{0.04 t} = 1.5 \implies 0.04 t = \ln 1.5 \approx 0.405. \] \[ t \approx \frac{0.405}{0.04} \approx 10.1 \text{ years.} \]
Problem 8. A bone fragment contains $30\%$ of its original carbon-14. Using the decay constant $k \approx -1.21 \times 10^{-4}$ per year, estimate its age.
Show answer
The fraction remaining is $\frac{y}{y_0} = e^{kt} = 0.30$. Take logarithms: \[ kt = \ln(0.30) \approx -1.204 \implies t = \frac{-1.204}{-1.21 \times 10^{-4}} \approx 9950 \text{ years.} \] The fragment is roughly $9900$ to $10000$ years old.
Problem 9. Explain why the growth-decay equation $\frac{dy}{dt} = ky$ cannot, by itself, model a population over a very long time.
Show answer
When $k > 0$, the solution $y_0 e^{kt}$ grows without bound, so the model predicts an infinite population. Real populations are limited by food, space, and competition. As resources run short, the relative growth rate falls instead of staying constant. The logistic equation $\frac{dP}{dt} = kP\left(1 - \frac{P}{M}\right)$ corrects this by reducing the rate as $P$ approaches a carrying capacity $M$. The exponential model is the small-population approximation of the logistic model, accurate only while the population is far below its limit.
Mastery Checklist
You have mastered this skill when you can do all of the following without referring to notes:
Mental Model
Think of exponential change as interest paid on whatever you currently have.
If a quantity earns a fixed percentage of itself each instant, then a larger pile earns more, which makes the pile larger, which makes it earn even more. That self-reinforcing loop is what $\frac{dy}{dt} = ky$ encodes: the rate is always proportional to the current size, never fixed. The result is the exponential curve, which steepens as it climbs (growth, $k > 0$) or flattens as it falls toward zero (decay, $k < 0$).
The half-life and doubling time capture the most striking property of this loop: the time to change by a fixed factor is always the same, no matter where you start. A sample takes the same number of years to fall from $80$ mg to $40$ mg as from $40$ mg to $20$ mg. This is why carbon dating works with a single half-life number, and why the model is so compact.
Holding this picture frames the rest of the course. The exponential model is the honest answer when nothing limits the growth. Every later refinement, especially the logistic equation, is a correction added when the loop runs into a ceiling.
Connections
Within MATH347 (Differential Equations)
- Modeling with differential equations: “Rate proportional to amount” is the first modeling assumption taught, and it produces this equation directly.
- Separable equations: $\frac{dy}{dt} = ky$ is the cleanest separable equation, solved by the same separate-and-integrate steps used on harder ones.
- The logistic equation: The logistic model $\frac{dP}{dt} = kP(1 - P/M)$ starts from exponential growth and adds a slowing factor. Understanding the unlimited case is the prerequisite for understanding the limited one.
- Direction fields: The slope field of $\frac{dy}{dt} = ky$ shows the family of exponential curves at a glance, with the equilibrium line $y = 0$.
Toward Later Coursework
- Newton’s Law of Cooling: The equation $\frac{dT}{dt} = k(T - T_s)$ for a body cooling toward surrounding temperature $T_s$ is a shifted version of this model; substituting $u = T - T_s$ turns it back into $\frac{du}{dt} = ku$.
- Mixing problems: Many first-order applications approach a limiting value exponentially, governed by the same mathematics.
Audience Notes
For students who find math intimidating: There is really one formula to hold: $y = y_0 e^{kt}$. The starting amount is $y_0$, and $k$ controls how fast (positive grows, negative decays). Most problems are filling in two of the unknowns and solving for the third, which always comes down to taking a logarithm.
For students interested in proof: That the exponential is the unique solution of $\frac{dy}{dt} = ky$ with a given initial value is a uniqueness statement guaranteed by the existence-and-uniqueness theorem for first-order equations. The separation argument constructs the solution; the theorem says there is no other.
For students interested in careers: This model underlies compound interest in finance, drug-clearance rates in pharmacology (where half-life sets the dosing schedule), and radiometric dating in geology and archaeology. It is also the linearization every epidemic curve follows in its earliest, unlimited phase.
For gifted and curious students: The relative growth rate $\frac{1}{y}\frac{dy}{dt}$ being constant is what singles out the exponential among all functions. If instead the relative rate declines linearly with $y$, you get the logistic curve; if it varies periodically, you get oscillating models. Asking “what controls the relative rate” is the doorway from this one model to the whole landscape of population dynamics.