Exponential Growth and Decay
The equation \(dy/dt = ky\) says a quantity changes at a rate proportional to its current size. Its solution is \(y = y_0 e^{kt}\) — exponential growth when \(k > 0\), decay when \(k < 0\) — and the single constant \(k\) sets everything, including how long doubling or halving takes.
By the end you'll be able to solve \(dy/dt = ky\), read growth vs. decay off the sign of \(k\), and compute doubling time or half-life with \(\ln 2 / k\).
Predict: double k from 0.050 to 0.100 — does the doubling time halve? And what happens the moment k crosses below zero? Slide to check.
This is \(y = y_0 e^{kt}\). Drag k to change the rate — its sign flips the curve between growth and decay — and drag \(y_0\) to change the starting amount. The dashed marker shows where \(y\) reaches double (growth) or half (decay) of \(y_0\) — watch whether it moves when you change \(y_0\). Hover or focus the curve and points for exact readouts.
y(t) = 100·e^(0.050t) — growth · doubling time = ln 2 / 0.050 ≈ 13.86
\(dy/dt = ky\) is the simplest and most common growth model — the more of \(y\) you have, the faster it changes — and its solution \(y = y_0 e^{kt}\) makes doubling time and half-life a one-line computation: \(\ln 2 / k\).
Where the solution comes from
You already have the tool for this: \(dy/dt = ky\) is separable. Divide by \(y\), integrate both sides:
$$\int \frac{dy}{y} = \int k\, dt \;\Longrightarrow\; \ln|y| = kt + C \;\Longrightarrow\; y = y_0 e^{kt},$$
where \(y_0 = y(0)\) is the initial amount. If \(k > 0\) the exponent grows and you get exponential growth; if \(k < 0\) it shrinks and you get exponential decay. The sign of one constant flips the whole behavior — exactly what you saw sliding \(k\) through zero above.
Half-life and doubling time fall out of k
Ask when a growing quantity doubles: \(y_0 e^{kt} = 2y_0\). The \(y_0\) cancels, leaving \(e^{kt} = 2\), so
$$t_{\text{double}} = \frac{\ln 2}{k}.$$
The identical algebra with \(e^{kt} = \tfrac{1}{2}\) (or writing decay as \(y = y_0e^{-kt}\) with \(k > 0\)) gives the same formula for the half-life: \(t_{\text{half}} = \ln 2 / k\). Notice what's not in the formula: \(y_0\). A population growing at rate \(k\) takes the same time to double whether it starts at 10 or 10 million — that's why the dashed marker above never moves sideways when you slide \(y_0\).
The same equation shows up wherever change is proportional to amount: bacterial colonies and populations (\(k > 0\)), radioactive samples like carbon-14 (\(k < 0\)), and continuously compounded interest, where a principal \(P\) grows as \(\text{Balance} = Pe^{rt}\). Radiocarbon dating runs the decay model backwards: carbon-14's half-life of 5,730 years fixes \(k = \ln 2 / 5730 \approx 0.000121\) per year, and measuring how much carbon-14 remains in a fossil tells you how long it has been decaying.
Visually, \(dy/dt = ky\) means the curve's steepness is proportional to its height: growth curves accelerate forever, decay curves flatten toward zero without touching it. In equal time steps of \(\ln 2/k\), the curve's height multiplies by the same factor of 2 (or ½) every step — that repeated-doubling rhythm is the signature of exponentials, and it's what the marker in the demo pins down.
A colony grows as \(f(t) = 200e^{0.02t}\), with \(t\) in minutes.
After 300 minutes: \(f(300) = 200e^{0.02(300)} = 200e^{6} \approx
80{,}686\) bacteria.
Time to reach 100,000: solve \(200e^{0.02t} = 100{,}000\), so
\(e^{0.02t} = 500\) and \(t = \dfrac{\ln 500}{0.02} \approx 310.73\) minutes.
Doubling time: \(t_{\text{double}} = \dfrac{\ln 2}{0.02} \approx
34.66\) minutes — consistent with the colony roughly doubling from \(\approx
80{,}686\) at \(t = 300\) toward \(\approx 161{,}000\) by around \(t \approx 335\).
A radioactive sample decays as \(dy/dt = -0.10\,y\), with \(t\) in days, starting from \(y_0 = 500\) grams. Write the solution: \(y = 500e^{-0.10t}\). Now the half-life: solve \(e^{-0.10t} = \tfrac{1}{2}\), so \(t_{\text{half}} = \dfrac{\ln 2}{0.10} = \) ____ days.
Reveal the answer
\(t_{\text{half}} = \ln 2 / 0.10 \approx\) 6.93 days. The 500 grams never entered the computation — half-life depends only on the rate constant. Set \(k = -0.10\) on the slider above and check the dashed marker against this number.
More info — why "rate proportional to amount" forces an exponential
Another way to see it: \(dy/dt = ky\) says the relative growth rate \(\frac{1}{y}\frac{dy}{dt}\) is a constant \(k\). But \(\frac{1}{y}\frac{dy}{dt}\) is exactly the derivative of \(\ln y\) — so \(\ln y\) rises along a straight line of slope \(k\), and \(y\) itself is the exponential of a line: \(y_0e^{kt}\). This is the same separation-of-variables computation from the previous lesson, just read geometrically: exponential growth is a straight line in log scale. The OpenStax section linked under Dive deeper works this derivation in full alongside the population and radiocarbon applications.
Check your understanding
A quantity satisfies \(dy/dt = 0.05y\). How long does it take to double?
Two populations grow with the same rate constant \(k > 0\). One starts at 10 individuals, the other at 10 million. Which doubles first?
To solve \(dy/dt = ky\) by separation of variables you write \(\int \frac{dy}{y} = \int k\,dt\). What general solution does this produce?
Carbon-14 decays as \(y = y_0 e^{-0.000121\,t}\) (\(t\) in years). What is its half-life?
Recap
- \(dy/dt = ky\) — rate of change proportional to current amount — has solution \(y = y_0 e^{kt}\), found by separation of variables.
- \(k > 0\) gives exponential growth; \(k < 0\) gives exponential decay.
- Doubling time and half-life are both \(\ln 2 / k\) (using the decay rate's magnitude) — they depend only on \(k\), never on \(y_0\).
- The same equation models populations, continuously compounded interest (\(Pe^{rt}\)), and radioactive decay — carbon-14's 5,730-year half-life is what powers radiocarbon dating.
Dive deeper
- OpenStax Calculus Volume 1, §6.8 Exponential Growth and Decay Derive the half-life/doubling-time formula ln(2)/k and apply it to population and radiocarbon dating.
Sources
- Exponential growth and decay