Instantaneous Rate Of Change
Learn the instantaneous rate of change for Queensland Year 11 Mathematical Methods (QCAA). It is the rate at a single instant, equal to the gradient of the tangent — the value average rates approach as the interval shrinks.
You will learn to estimate this rate from shrinking intervals, find it exactly from first principles for power and polynomial functions, and read it as an instantaneous velocity — the heart of the derivative.
Every question with a fully worked solution.
- Instantaneous Rate Of Change - Video - Average versus instantaneous rates of change Watch
Theory
In Year 11 Mathematical Methods (QCAA, Unit 2), the instantaneous rate of change of \(y=f(x)\) at a point is the gradient of the tangent there — the limit that the average rates (secant gradients) approach as the interval shrinks to zero. This page shows the secants-to-tangent idea, the first-principles rule \(f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}\) for simple power and polynomial functions, and instantaneous velocity.
The instantaneous rate of change of \(y=f(x)\) at \(x=a\) is how fast \(y\) is changing at that exact moment. You cannot use two separated points, so instead you take average rates over intervals \([a,a+h]\) and let \(h\) shrink towards \(0\).
As the second point slides towards \((a,f(a))\), the secant lines rotate towards the tangent at that point. The instantaneous rate of change is therefore the gradient of the tangent, written \(f'(a)\).
Computing this limit for a general \(x\) is called finding the derivative from first principles: \(f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}\). For a position-time graph, the tangent gradient is the instantaneous velocity.
The instantaneous rate of change (the derivative) from first principles is
The quantity inside the limit, before letting \(h\to0\), is the average rate (secant gradient) over \([x,x+h]\):
How to find an instantaneous rate from first principles
- Set up: write the average rate over \([x,x+h]\), that is \(\dfrac{f(x+h)-f(x)}{h}\), expanding \(f(x+h)\).
- Simplify: cancel the leading terms in the numerator, then divide every remaining term by \(h\).
- Take the limit: let \(h\to0\); the terms still containing \(h\) vanish, leaving \(f'(x)\). Substitute the point if a value is needed.
Average rate over \([1,1+h]\) — expand \((1+h)^2\):
| \(\dfrac{f(1+h)-f(1)}{h}\) | \(=\) | \(\dfrac{(1+h)^2-1}{h}\) |
| \(=\) | \(\dfrac{1+2h+h^2-1}{h}\) | |
| \(=\) | \(\dfrac{2h+h^2}{h}\) | |
| \(=\) | \(2+h\) |
| \(h\) | 1 | 0.5 | 0.1 | →0 |
|---|---|---|---|---|
| \(2+h\) | 3 | 2.5 | 2.1 | 2 |
Let \(h\to0\) for the instantaneous rate:
| \(f'(1)\) | \(=\) | \(\lim_{h\to0}(2+h)\) |
| \(=\) | \(2\) |
Average rate \(=2+h\); instantaneous rate at \(x=1\) is \(2\).
Set up the difference quotient — expand \((x+h)^2\):
| \(\dfrac{f(x+h)-f(x)}{h}\) | \(=\) | \(\dfrac{(x+h)^2-x^2}{h}\) |
| \(=\) | \(\dfrac{x^2+2xh+h^2-x^2}{h}\) | |
| \(=\) | \(\dfrac{2xh+h^2}{h}\) | |
| \(=\) | \(2x+h\) |
Take the limit as \(h\to0\):
| \(f'(x)\) | \(=\) | \(\lim_{h\to0}(2x+h)\) |
| \(=\) | \(2x\) |
\(f'(x)=2x\); the instantaneous rate of change at any \(x\) is \(2x\).
Average velocity over \([t,t+h]\) — expand \(5(t+h)^2\):
| \(\dfrac{s(t+h)-s(t)}{h}\) | \(=\) | \(\dfrac{5(t+h)^2-5t^2}{h}\) |
| \(=\) | \(\dfrac{5t^2+10th+5h^2-5t^2}{h}\) | |
| \(=\) | \(\dfrac{10th+5h^2}{h}\) | |
| \(=\) | \(10t+5h\) |
Let \(h\to0\) for the instantaneous velocity:
| \(v\) | \(=\) | \(\lim_{h\to0}(10t+5h)\) |
| \(=\) | \(10t\) |
Substitute \(t=3\):
| \(v\) | \(=\) | \(10(3)\) |
| \(=\) | \(30\text{ m/s}\) |
Instantaneous velocity at \(t=3\) s is \(30\) m/s.
Difference quotient — expand \((x+h)^2-3(x+h)\):
| \(\dfrac{f(x+h)-f(x)}{h}\) | \(=\) | \(\dfrac{\bigl[(x+h)^2-3(x+h)\bigr]-\bigl[x^2-3x\bigr]}{h}\) |
| \(=\) | \(\dfrac{x^2+2xh+h^2-3x-3h-x^2+3x}{h}\) | |
| \(=\) | \(\dfrac{2xh+h^2-3h}{h}\) | |
| \(=\) | \(2x-3+h\) |
Take the limit as \(h\to0\):
| \(f'(x)\) | \(=\) | \(\lim_{h\to0}(2x-3+h)\) |
| \(=\) | \(2x-3\) |
Substitute \(x=4\):
| \(f'(4)\) | \(=\) | \(2(4)-3\) |
| \(=\) | \(5\) |
\(f'(x)=2x-3\); the instantaneous rate at \(x=4\) is \(5\).
Common pitfalls
Frequently asked questions
What is the instantaneous rate of change?
The rate of change of \(y=f(x)\) at a single point, equal to the gradient of the tangent there and written \(f'(a)\).
How is the instantaneous rate the limit of average rates?
As the interval \([a,a+h]\) shrinks (\(h\to0\)), the secant gradients approach the tangent gradient, so the average rates approach the instantaneous rate.
What does first principles mean?
Finding the derivative directly from the limit \(f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}\), rather than from a rule.
Why can't you just put h equals 0?
Because while \(h\) is in the denominator you would get \(\dfrac{0}{0}\). You must simplify and cancel the \(h\) first, then take the limit.
What is instantaneous velocity?
The instantaneous rate of change of position — the gradient of the tangent to the position-time graph at that instant.
How does the instantaneous rate differ from the average rate?
The average rate is a secant gradient over an interval; the instantaneous rate is a tangent gradient at one point, the limit of those secants.