Rates of Change with Differentiation
Rates of change use the derivative \(\dfrac{dy}{dx}\) to measure how one quantity changes with respect to another at an instant.
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
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.