Exponential Growth & Decay
Learn exponential growth and decay for NSW Year 12 Mathematics Advanced (NESA outcome MAV-12-06). When a quantity changes in proportion to its current size, as with bacteria or radioactive material, it follows an exponential curve.
You will learn to use the model \(\dfrac{dQ}{dt} = kQ\) with solution \(Q = Ae^{kt}\), show by substitution that it fits, find the initial value \(A\) and the growth constant \(k\) from data, and solve for times such as a half-life — the maths behind population and decay forecasts.
Theory
When the rate is proportional to the amount, growth is exponential: \(\dfrac{dQ}{dt}=kQ\) gives \(Q=Ae^{kt}\). This Year 12 Mathematics Advanced topic (MAV-12-06) models growth and decay.
\(\dfrac{dQ}{dt}=kQ\Rightarrow Q=Ae^{kt}\), where \(A\) is the initial value. \(k>0\) is growth; \(k<0\) is decay. Check: \(\dfrac{d}{dt}(Ae^{kt})=kAe^{kt}=kQ\).
Method
- Identify \(A\) (initial value) and \(k\).
- Positive \(k\) grows; negative \(k\) decays.
- Substitute a time to evaluate, or a value to solve.
| \(Q(0)\) | \(=\) | \(5e^{0}=5\) |
\(k=-0.1<0\), so decay.
| \(\dfrac{dP}{dt}\) | \(=\) | \(0.2\times 50=10\) |
| \(Q(2)\) | \(=\) | \(8e^{1}=8e\) |
Common pitfalls
Frequently asked questions
What is the exponential growth and decay law?
When the rate of change is proportional to the amount, dQ dt equals k Q, and the solution is Q equals A e to the k t.
What does A represent?
The initial value, the amount when t is zero.
How do you tell growth from decay?
k greater than zero gives growth; k less than zero gives decay toward zero.
How do you check Q = Ae^{kt} solves dQ/dt = kQ?
Differentiate: the derivative of A e to the k t is k A e to the k t, which equals k Q.