Rates of Change with Differentiation
Learn rates of change with differentiation for NSW Year 12 Mathematics Advanced (NESA outcome MAV-12-06). When a quantity such as a volume, population or temperature changes over time, its derivative tells you how fast it is changing at any instant.
You will learn to differentiate a formula to find the instantaneous rate at a given time, read the sign to tell rising from falling, use the second derivative to see whether the rate is speeding up, and keep units such as litres per minute — the core of rates of change.
Theory
The rate of change of a quantity is its derivative with respect to time. This Year 12 Mathematics Advanced topic (MAV-12-06) finds and interprets \(\dfrac{dQ}{dt}\).
The instantaneous rate of change of \(Q\) is \(\dfrac{dQ}{dt}\); its sign shows increase (\(+\)) or decrease (\(-\)). Evaluate it at a time for the rate then. The second derivative is the rate of change of the rate.
Method
- Differentiate \(Q(t)\).
- Substitute the time for the instantaneous rate.
- Read the sign for increasing or decreasing.
| \(\dfrac{dV}{dt}\) | \(=\) | \(3t^{2}\) |
| \(\text{at }2\) | \(=\) | \(12\) |
| \(\dfrac{dN}{dt}\) | \(=\) | \(50e^{0.5t}\) |
| \(\text{at }0\) | \(=\) | \(50\) |
| \(\dfrac{dT}{dt}\) | \(=\) | \(-30e^{-t}\) |
| \(\text{at }0\) | \(=\) | \(-30\) |
| \(2t-6\) | \(=\) | \(0\Rightarrow t=3\) |
Common pitfalls
Frequently asked questions
What is a rate of change?
It is how fast a quantity changes, found as the derivative with respect to time, dQ dt.
How do you find an instantaneous rate?
Differentiate the quantity, then substitute the given time.
What does a negative rate mean?
The quantity is decreasing at that instant.
What is the second derivative in a rates problem?
It is the rate of change of the rate, such as acceleration when the quantity is displacement.