Model exponential growth and decay using $y = A \cdot r^t$ or $y = A e^{kt}$, including problems involving population growth, radioactive decay, depreciation and continuous compound interest

What this dot point is asking

QCAA wants you to model real-world growth and decay scenarios with exponential functions, choosing between $y = A r^t$ (discrete or growth factor form) and $y = A e^{kt}$ (continuous form), and to solve for any of the parameters $A$, $r$, $k$ or $t$ from worded conditions.

The two standard forms

Discrete growth factor form. $y = A r^t$ where:

• $A$ is the initial value at $t = 0$.
• $r$ is the per-time-unit multiplier.
• $r > 1$ for growth, $0 < r < 1$ for decay.
• $t$ is time in the units that match $r$.

Continuous form. $y = A e^{kt}$ where:

• $A$ is initial value.
• $k$ is the continuous growth rate ($k > 0$ growth, $k < 0$ decay).
• Used when growth is compounded continuously (continuously compounded interest, radioactive decay in mathematically clean form).

The two forms convert via $r = e^k$ or $k = \ln r$.

Building the model from a scenario

Identify $A$ from the initial value. Identify $r$ or $k$ from a single additional condition (typically "after $T$ time units the value is $V$").

For percentage growth at rate $p\%$ per period: $r = 1 + p/100$. For percentage decay: $r = 1 - p/100$ (assuming $p < 100$).

For half-life $T_{1/2}$: $r^{T_{1/2}} = 1/2$, so $r = (1/2)^{1/T_{1/2}}$.

For doubling time $T_d$: $r^{T_d} = 2$, so $r = 2^{1/T_d}$.

Continuous compound interest

$A(t) = P e^{rt}$ where $P$ is the principal, $r$ is the annual interest rate (as a decimal, continuously compounded), $t$ is time in years.

For discrete annual compounding the model is $A = P(1 + r)^t$; for $n$ times per year it is $A = P(1 + r/n)^{nt}$. As $n \to \infty$, this approaches $A = P e^{rt}$.

Choosing between the two forms

The growth-factor form $y = A r^t$ is natural when a problem gives a per-period multiplier, a percentage change, a half-life or a doubling time, because $r$ can be written down directly. The continuous form $y = A e^{kt}$ is natural for continuously compounded interest and for problems already expressed with $e$, and it is the form that connects to calculus, since $\tfrac{d}{dt}Ae^{kt} = kAe^{kt}$ shows the rate of change is proportional to the current amount. The two forms describe the same curve whenever $r = e^k$, so you can convert freely with $k = \ln r$.

Solving for the unknown parameter

Most worded problems supply the initial value $A$ plus one further data point, and the remaining parameter is found by substitution. If the extra information is a half-life or doubling time, use $r^{T} = \tfrac{1}{2}$ or $r^{T} = 2$ to get $r$. If it is a value at a known time, substitute and solve, taking a logarithm when the unknown is in the exponent. Solving for the time to reach a target value is the most common final part, and it always reduces to isolating the exponential and applying a logarithm.

How this appears in assessment

IA1 typically asks you to build a discrete model from a scenario and predict a value at a stated time. In the external assessment, Paper 1 may test identifying a growth factor or a doubling time, while Paper 2 sets a multi-part contextual problem: build the model, predict a value, then solve for the time a target is met. In Year 12 the same models are differentiated to find instantaneous rates of change, so a secure grasp here pays off later.