Applications: Growth, Decay and Population

Many real quantities change at a rate proportional to their current size. If $N = N(t)$ is such a quantity at time $t$, this law is written as the differential equation

where $k$ is the constant of proportionality: $k > 0$ describes growth (population, bacteria, continuously-compounded money) and $k < 0$ describes decay (radioactive material, depreciation of an asset).

Solving the law. This is variables-separable:

where $N_0 = N(0)$ is the initial value (taking $e^{c_1} = N_0$). This exponential law $N = N_0 e^{kt}$ is the backbone of every growth/decay application.

How the numbers are used. Two pieces of data are typically given — the initial value $N_0$ and one later reading — which fix $N_0$ and $k$ (or, more cleverly, fix the combination $e^{k\cdot(\text{unit time})}$). A very common shortcut: if the quantity is multiplied by a factor $r$ over a time span $T$ (e.g. “doubles in 25 years” means $r = 2$), then over a span $mT$ it is multiplied by $r^{m}$ — because $e^{k\cdot mT} = \left(e^{kT}\right)^{m} = r^{m}$. This often avoids computing $k$ explicitly.

Worked illustration. A city's population grows at a rate proportional to itself and doubles every 25 years. If today's population is $1$ lakh, what will it be after $50$ years?

With $N = N_0 e^{kt}$ and $N_0 = 1$ lakh: doubling in 25 years gives $e^{25k} = 2$. After 50 years,

$N = N_0\,e^{50k} = N_0\left(e^{25k}\right)^{2} = 1\times 2^{2} = 4 \text{ lakh}.$ …