How Do You Model Exponential Growth or Decay (Population; Radioactive Decay; Compound Interest)?

Use an exponential model when a quantity changes by a constant percentage or constant multiplicative factor over equal time intervals. The general forms are $y=ka^x+c$ or $y=ke^{rx}+c$, where $k$ is the initial amount and $c$ is the horizontal asymptote.

For IB Mathematics AI, begin by identifying the time unit and checking that the stated rate applies to that same unit. A yearly rate needs an exponent measured in years; a monthly rate needs months or an equivalent conversion. Then check the model's domain: in context, $t\geq0$, and population values may need rounding to whole individuals. Exponential models are most reliable within the observed time range because long-term extrapolation may become unrealistic outside that range.

Building the Model

Start with the initial value, convert the percentage change into a multiplier, and raise that multiplier to the number of time periods.

Here, $y_0$ is the initial value and $t$ is the number of time periods. If the model has a non-zero limiting value, include a vertical translation $c$.

Population Growth

A population of $5000$ grows by $3\%$ annually:

$$P(t)=5000(1.03)^t$$

After $10$ years, $P(10)=6719.58$, so the model predicts approximately $6720$ individuals.

Radioactive Decay

For an initial mass of $80\text{ g}$ with half-life $6$ years:

$$M(t)=80\left(\frac{1}{2}\right)^{t/6}$$

After $18$ years, three half-lives have passed, giving $M(18)=10\text{ g}$.

Compound Interest

An investment of \1000$at$4.8%$ per year, compounded monthly, is modelled by

$$A(t)=1000\left(1+\frac{0.048}{12}\right)^{12t}.$$

After $5$ years, A(5)=\1270.66$.

A common misconception is that exponential change adds the same amount each period. It does not: the percentage remains constant, so the numerical change becomes larger or smaller over time.

Exam Technique

This is SL 2.5, Modelling functions and may be tested on Paper 1 or Paper 2. Write the model before using your GDC, retain full precision, and interpret the result in context; for money, give the final answer to two decimal places.