Average & Instantaneous Rates of Change
The fastest 100m sprint ever was 9.58 seconds. But the runner was not running at 10.44 m/s for the entire race. His speed varied: slow at the start, explosive in the middle, then a slight fade at the end. In this lesson, you will learn how to measure average speed over an interval and glimpse how calculus will let us find his exact speed at any single instant.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
A car travels 120 km in 2 hours, so its average speed is 60 km/h. Does this mean the speedometer showed exactly 60 km/h at every moment? What do you think happens to the accuracy of the average speed as we measure it over shorter and shorter time intervals?
There are only two moves in this entire lesson. Lock them into muscle memory and the rest is just calculation.
The average rate of change measures overall steepness between two points on a curve. It is the gradient of the secant line joining those points.
The instantaneous rate of change is what happens when those two points get infinitely close together. The secant becomes a tangent and we capture the exact rate at a single instant.
Key facts
- The formula for average rate of change
- The difference quotient and what it represents
- That average rate equals the gradient of a secant
Concepts
- Why average speed smooths out variation over an interval
- How shrinking the interval improves the approximation of instantaneous speed
- The connection between rate of change and real-world speed
Skills
- Calculate average rate of change from a table, graph or equation
- Interpret average rate of change in real-world contexts
- Estimate instantaneous rate of change using small intervals
When a quantity changes, we often want to know how fast it changes. If a function moves from $x = a$ to $x = b$, the average rate of change is:
This is exactly the same calculation as finding the gradient of a straight line through the two points $(a, f(a))$ and $(b, f(b))$. On a curve, this line is called a secant — it cuts through the curve at two points.
Average rate = slope of secant (grey dashed). Instantaneous rate = slope of tangent (red) at a single point.
From speed to any rate
The same idea applies to any changing quantity:
- Speed: change in distance over change in time
- Growth rate: change in population over change in time
- Flow rate: change in volume over change in time
- Cost per unit: change in cost over change in quantity
The geometric idea
Geometrically, as the second point on the curve moves closer to the first, the secant line pivots and approaches a limiting position: the tangent line at that point. The gradient of this tangent is the instantaneous rate of change.
Average rate of change = $\dfrac{f(b)-f(a)}{b-a}$ = gradient of the secant through $(a, f(a))$ and $(b, f(b))$; Secant line: a straight line crossing the curve at two distinct points
Pause — copy the average rate of change formula $\dfrac{f(b)-f(a)}{b-a}$ and its geometric meaning (gradient of the secant line) into your book.
Quick check: True or false — the average rate of change over an interval equals the gradient of the tangent line at the midpoint of that interval.
Worked examples · 3 in a row, reveal as you go
The temperature of a chemical solution is recorded over time:
| Time (h) | 0 | 2 | 4 | 6 | 8 |
|---|---|---|---|---|---|
| Temp (°C) | 15 | 22 | 28 | 24 | 18 |
Find the average rate of change of temperature over the first 4 hours.
Find the average rate of change of $f(x) = x^2$ over the interval $[1, 3]$.
A runner's position after $t$ seconds is $s(t) = t^2$ metres. Estimate the instantaneous speed at $t = 2$ by calculating average speeds over $[2, 2.1]$ and $[2, 2.01]$.
Quick check: For $f(x) = x^2$, the average rate of change over $[2, 5]$ is:
Common errors · the 3 traps that cost marks
Odd one out: Three of the following are examples of a rate of change. Which one is NOT?
Quick-fire practice · 5 problems
Find the average rate of change of $f(x) = 3x + 2$ from $x = 0$ to $x = 4$.
Find the average rate of change of $f(x) = x^2$ from $x = 2$ to $x = 5$.
A runner's distance from the start is recorded:
| Time (s) | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| Distance (m) | 0 | 3 | 8 | 15 |
Find the average speed over the first 3 seconds.
Estimate the instantaneous rate of change of $f(x) = x^2$ at $x = 2$ using $h = 0.1$.
Explain the difference between average rate of change and instantaneous rate of change.
Fill the blanks: drag each token into the matching blank.
The gradient of the ___ line gives the ___ rate of change. As the interval shrinks, this approaches the ___ rate, which equals the gradient of the ___.
Earlier you were asked: Does an average speed of 60 km/h mean the speedometer showed exactly 60 km/h at every moment? No. Average speed is calculated over a whole interval and smooths out all the variation. The speedometer shows instantaneous speed, which can be faster or slower than the average at any given moment. As we measure over shorter and shorter intervals, the average speed gets closer and closer to the instantaneous speed.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. The temperature of a liquid is given by $T(t) = 20 + 5t - t^2$, where $t$ is in minutes. Find the average rate of change of temperature from $t = 1$ to $t = 4$. Show all working. 3 MARKS
Q2. A particle's position is $s(t) = t^3$ metres after $t$ seconds. Estimate the instantaneous velocity at $t = 2$ by using intervals of width $0.1$ and $0.01$. Show all working. 4 MARKS
Q3. A car's speedometer shows instantaneous speed. Explain why the average speed over a trip can be different from the instantaneous speed at any moment. 3 MARKS
📖 Comprehensive answers (click to reveal)
Drill 1: $\frac{f(4)-f(0)}{4-0} = \frac{14-2}{4} = 3$
Drill 2: $\frac{f(5)-f(2)}{5-2} = \frac{25-4}{3} = 7$
Drill 3: $\frac{15-0}{3-0} = 5$ m/s
Drill 4: $\frac{f(2.1)-f(2)}{0.1} = \frac{4.41-4}{0.1} = 4.1$
Drill 5: Average rate is calculated over an interval and smooths out all variation. Instantaneous rate is the exact rate at a single point, found by shrinking the interval to zero.
Q1 (3 marks): $T(1) = 20 + 5(1) - (1)^2 = 24$ [1]. $T(4) = 20 + 5(4) - (4)^2 = 24$ [1]. Average rate $= \frac{24-24}{4-1} = 0$ °C/min [1].
Q2 (4 marks): Over $[2, 2.1]$: $s(2.1) = 9.261$, $s(2) = 8$, rate $= \frac{9.261-8}{0.1} = 12.61$ m/s [1]. Over $[2, 2.01]$: $s(2.01) = 8.120601$, rate $= \frac{8.120601-8}{0.01} = 12.0601$ m/s [1]. As the interval shrinks, the values approach 12 [1]. Estimated instantaneous velocity at $t = 2$ is approximately $12$ m/s [1].
Q3 (3 marks): Average speed is total distance divided by total time over the whole trip [1]. Instantaneous speed is the exact speed at a single moment in time [1]. They differ because a car speeds up, slows down, and stops during a trip, so the average smooths out these changes while the instantaneous reading captures only one moment [1].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering rate-of-change questions. Lighter alternative to the boss.
Mark lesson as complete
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