Exponential Growth & Decay
Exponential growth and decay model quantities whose rate is proportional to their size, \(\dfrac{dN}{dt}=kN\), giving \(N=N_0e^{kt}\).
Part of the NSW Year 12 Mathematics Advanced course, in the Calculus area of study (Applications of calculus focus area) of the 2024 syllabus. Work through practice questions with fully worked solutions and video lessons, or scroll down for the theory summary and worked examples.
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.