Exponential Growth and Decay

Exponential Growth and Decay Models

By the end of this lesson, you will be able to set up, solve, and interpret exponential growth and decay models from differential equations.

The Basic Model

The DE $\frac{dP}{dt} = kP$ says: the rate of change of P is proportional to P itself.

Solution: $P(t) = P_0 e^{kt}$

where $P_0 = P(0)$ is the initial value and $k$ is the proportionality constant.

• $k > 0$: exponential growth
• $k < 0$: exponential decay

Applications

Population growth: $P(t) = P_0 e^{rt}$ where r is the per-capita growth rate.

Radioactive decay: $N(t) = N_0 e^{-\lambda t}$ where $\lambda > 0$ is the decay constant.

Newton's Law of Cooling: $T(t) - T_{\text{env}} = (T_0 - T_{\text{env}})e^{-kt}$ — the temperature difference decays exponentially.

Compound interest (continuously compounded): $A(t) = Pe^{rt}$

Half-Life and Doubling Time

Half-life $t_{1/2}$: the time for a quantity to halve.

$P_0/2 = P_0 e^{kt_{1/2}}$ → $e^{kt_{1/2}} = 1/2$ → $kt_{1/2} = -\ln 2$ → $t_{1/2} = \frac{\ln 2}{|k|}$

For radioactive carbon-14: $t_{1/2} = 5730$ years.

Doubling time $t_d$: the time for a quantity to double.

$t_d = \frac{\ln 2}{k}$

Rule of 70: $t_d \approx 70/r%$ — if r = 3%, doubling time ≈ 23 years.

Example 1: Bacterial Growth

A culture starts with 500 bacteria and grows at 8% per hour. When will there be 5000 bacteria?

$P(t) = 500e^{0.08t}$

$5000 = 500e^{0.08t}$

$10 = e^{0.08t}$

$\ln 10 = 0.08t$

$t = \frac{\ln 10}{0.08} \approx \frac{2.303}{0.08} \approx 28.8$ hours

Example 2: Carbon-14 Dating

A bone fragment contains 25% of its original carbon-14. The half-life of $^{14}$C is 5730 years. How old is the fragment?

$N(t) = N_0 e^{kt}$

From the half-life: $0.5 = e^{k \cdot 5730}$ → $k = \frac{\ln 0.5}{5730} = \frac{-\ln 2}{5730}$

$0.25 = e^{kt}$ → $\ln(0.25) = kt$ → $t = \frac{\ln(0.25)}{k} = \frac{-\ln 4}{-\ln 2 / 5730} = \frac{5730 \ln 4}{\ln 2} = 5730 \cdot 2 = 11460$ years