When the rate of change depends on how much you already have
2026-02-26
You should know
What multiplication means, how powers like 2^3 work, and how constant rate was described in the previous chapter.
You will learn
Why some systems grow by multiplying instead of adding, how exponential curves behave, and how logarithms help us work backwards from growth.
Why this matters
Population growth, epidemics, compounding, and many diffusion processes make sense only when we notice that the rate depends on the amount already present.
If this gets hard, focus on…
The contrast between adding the same amount each step and multiplying by the same factor each step. That one distinction unlocks the whole chapter.
In 1950, the population of Lagos, Nigeria, was around 290,000. By 1975 it had grown to roughly 2 million. By 2000, 8 million. Today it exceeds 15 million and is one of the fastest-growing cities on Earth. The pattern — roughly doubling every 15 years — cannot be described with a straight line. The rate of growth is not a constant in absolute terms; it is constant as a fraction of what is already there.
This is the defining feature of exponential growth: the more there is, the faster it grows. It governs population increase under unconstrained conditions, the compounding of investment returns, radioactive decay running in reverse, and the early stage of every epidemic. The differential equation at its heart — dN/dt = rN — is the simplest self-referential relationship in mathematics. Solving it produces the exponential function; inverting it produces the logarithm. Between them, these two functions describe more of the natural and human world than almost any others, and the time it takes to understand them thoroughly repays itself many times over in the models that follow.
What happens when the rate of growth is proportional to the current amount?
A city grows by 2% per year. The population next year depends on the population this year. A forest fire spreads faster when the burning perimeter is longer. Radioactive decay happens faster when more atoms remain.
This is different from linear growth. In linear growth, you add the same amount each period (100 people per year, 5 meters per day). In exponential growth, you multiply by the same factor (1.02× per year, 2× per day).
The mathematical question: How do we model this? What does the solution look like? And how do we work backwards — if we know the final size, how do we find the time it took to get there?
Linear growth: - Add a constant amount each period - y = mx + b - Graph is a straight line - Doubling time gets longer as you grow (because you’re adding the same amount to a larger base)
Exponential growth: - Multiply by a constant factor each period - y = y_0 e^{rt} (we’ll derive this) - Graph is a curve that gets steeper - Doubling time is constant (the same percentage growth always takes the same time)
Example:
| Year | Linear (+100/yr) | Exponential (×1.1/yr) |
|---|---|---|
| 0 | 1000 | 1000 |
| 1 | 1100 | 1100 |
| 2 | 1200 | 1210 |
| 3 | 1300 | 1331 |
| 10 | 2000 | 2594 |
| 20 | 3000 | 6727 |
After 20 years, the exponential model is more than twice the linear model, even though they started the same.
The defining property of exponential growth:
The rate of change is proportional to the current amount.
In symbols:
\frac{dN}{dt} = rN
Where: - N is the quantity (population, mass, money) - t is time - r is the growth rate (a constant, units: 1/time)
This is a differential equation — an equation involving a derivative. It says: “the bigger N gets, the faster it grows.”
We want to find N(t) — a function that satisfies this equation.
Separation of variables:
\frac{dN}{dt} = rN
Divide both sides by N:
\frac{1}{N} \frac{dN}{dt} = r
Multiply both sides by dt:
\frac{1}{N} dN = r \, dt
Integrate both sides:
\int \frac{1}{N} dN = \int r \, dt
\ln N = rt + C
Where C is the constant of integration.
Exponentiate both sides:
N = e^{rt + C} = e^C \cdot e^{rt}
Let N_0 = e^C (the initial population at t = 0):
N(t) = N_0 e^{rt}
This is the solution. Exponential growth.
Example: If r = 0.03 per year, the population grows by 3% per year.
We could write exponential growth as:
N(t) = N_0 \cdot 2^{kt}
or
N(t) = N_0 \cdot 10^{mt}
These are all equivalent. We use e because it makes the calculus clean. The derivative of e^{rt} is r e^{rt}. The derivative of $2^{kt}$ involves an extra factor of \ln 2.
The number e \approx 2.71828 is the natural base — the one where the exponential function is its own derivative (up to a constant).
If you’re given a growth rate as “doubles every 5 years,” you can find r:
N(t) = N_0 \cdot 2^{t/5}
Rewrite using $2 = e^{}$:
N(t) = N_0 \cdot \left(e^{\ln 2}\right)^{t/5} = N_0 e^{(\ln 2 / 5) t}
So:
r = \frac{\ln 2}{T_{\text{double}}} = \frac{0.693}{5} = 0.139 \text{ per year}
If y = e^x, then x = \ln y.
The logarithm undoes the exponential. It answers the question: “What power do I raise e to in order to get y?”
Examples: - e^2 \approx 7.39, so \ln(7.39) = 2 - e^0 = 1, so \ln(1) = 0 - e^{-1} \approx 0.368, so \ln(0.368) = -1
\ln(ab) = \ln a + \ln b
\ln(a/b) = \ln a - \ln b
\ln(a^b) = b \ln a
These make logarithms powerful tools for multiplication → addition conversions.
Problem: A population starts at N_0 = 1000 and grows at r = 0.05 per year. When does it reach N = 5000?
Model:
N(t) = 1000 e^{0.05t}
Solve for t when N = 5000:
$5000 = 1000 e^{0.05t}$
Divide both sides by 1000:
$5 = e^{0.05t}$
Take the natural logarithm of both sides:
\ln 5 = \ln(e^{0.05t}) = 0.05t
t = \frac{\ln 5}{0.05} = \frac{1.609}{0.05} = 32.2 \text{ years}
The population reaches 5000 after 32.2 years.
Problem: A radioactive isotope has a half-life of 12 years. You start with 80 grams.
(a) Decay model
Half-life T_{1/2} = 12 years means the mass multiplies by 0.5 every 12 years.
The decay rate:
r = \frac{\ln(0.5)}{T_{1/2}} = \frac{-0.693}{12} = -0.0578 \text{ per year}
(Negative because it’s decay.)
Model:
M(t) = 80 e^{-0.0578t}
Check: at t = 12,
M(12) = 80 e^{-0.0578 \times 12} = 80 e^{-0.693} = 80 \times 0.5 = 40 \text{ g} ✓
(b) Mass after 30 years
M(30) = 80 e^{-0.0578 \times 30} = 80 e^{-1.734} = 80 \times 0.177 = 14.1 \text{ g}
(c) When does M = 10 g?
$10 = 80 e^{-0.0578t}$
$0.125 = e^{-0.0578t}$
\ln(0.125) = -0.0578t
-2.079 = -0.0578t
t = \frac{2.079}{0.0578} = 36.0 \text{ years}
Only 10 grams remain after 36 years.
Below is an interactive model comparing analytical (exact formula) vs. numerical (step-by-step simulation) solutions.
<label>
Initial value ($N_0$):
<input type="range" id="n0-slider" min="10" max="200" step="10" value="100">
<span id="n0-value">100</span>
</label>
<label>
Growth rate ($r$ per year):
<input type="range" id="r-slider" min="-0.2" max="0.2" step="0.01" value="0.05">
<span id="r-value">0.05</span>
</label>
<label>
Time step (Δt):
<input type="range" id="dt-slider" min="0.1" max="2" step="0.1" value="0.5">
<span id="dt-value">0.5</span> years
</label><p><strong>Analytical solution:</strong> $N(t) = N_0 e^{rt}$ (smooth curve)</p>
<p><strong>Numerical solution:</strong> $N_{i+1} = N_i + r N_i \Delta t$ (stepped)</p>
<p>As Δt → 0, the numerical solution converges to the analytical one.</p>Try this: - Set r = 0.1 (10% growth). Watch the curve steepen. - Set r = -0.1 (decay). The curve drops toward zero but never quite reaches it. - Reduce Δt to 0.1. The numerical solution gets closer to the analytical curve. - Set r = 0 (no growth). Both solutions are flat lines.
Key insight: The analytical solution is exact. The numerical solution is an approximation, but it converges to the exact answer as the time step shrinks.
For exponential growth, the doubling time is:
T_{\text{double}} = \frac{\ln 2}{r} \approx \frac{0.693}{r}
For exponential decay, the half-life is:
T_{1/2} = \frac{\ln 2}{|r|} \approx \frac{0.693}{|r|}
Example: If a population grows at r = 0.035 per year (3.5% annual growth):
T_{\text{double}} = \frac{0.693}{0.035} = 19.8 \text{ years}
The population doubles roughly every 20 years.
A useful approximation:
T_{\text{double}} \approx \frac{70}{\text{percent growth rate}}
If a city grows at 3% per year:
T_{\text{double}} \approx \frac{70}{3} = 23.3 \text{ years}
(Exact: \frac{69.3}{3} = 23.1 years.)
The rule of 70 works because \ln 2 \approx 0.693 \approx 70/100.
Because the rate depends on the size, exponential growth is self-reinforcing. A 5% increase when you’re at 100 is small. A 5% increase when you’re at 10,000 is enormous.
This is why: - Pandemics spread rapidly before intervention - Compound interest outpaces simple interest - Invasive species are hard to eradicate once established - Climate feedback loops accelerate warming
In the real world, nothing grows exponentially forever. Resources run out. Space fills up. Predators respond. Exponential growth is a local model — valid over short timescales before constraints kick in.
The next model introduces logistic growth, where the growth rate itself slows as the population approaches a limit.
If someone says “the population doubles every 10 years,” that is not the growth rate. The growth rate is:
r = \frac{\ln 2}{10} = 0.0693 \text{ per year}
Don’t confuse the two.
You cannot take the logarithm of a negative number (in the real number system). If you’re solving e^{rt} = -5, there is no solution. Exponentials are always positive.
In the numerical simulation, if Δt is too large, the approximation breaks down. The error compounds at each step, and the numerical solution diverges from the analytical one.
Rule of thumb: Use Δt small enough that the solution doesn’t change much when you halve it again.
e^a \cdot e^b = e^{a+b}
\frac{e^a}{e^b} = e^{a-b}
(e^a)^b = e^{ab}
e^0 = 1
e^{-a} = \frac{1}{e^a}
\ln(e^x) = x
e^{\ln x} = x
\ln(ab) = \ln a + \ln b
\ln(a/b) = \ln a - \ln b
\ln(a^b) = b \ln a
\ln 1 = 0
If you have \log_{10} x (common logarithm) and need \ln x (natural logarithm):
\ln x = \ln 10 \cdot \log_{10} x \approx 2.303 \cdot \log_{10} x