R14–R15, A15Ratio, Proportion & Rates of ChangeFoundation & HigherGrades 5–8

Rates of Change

A speedometer shows how fast a car is going right now; a journey time shows how fast it went on average. Both are rates of change, and on a graph both are gradients. This lesson reads rates from straight-line and curved graphs, and uses the area under a graph to find a total.

45 min AQA GCSE Maths

The big picture

The gradient of a graph is "change in the vertical quantity per unit of the horizontal quantity". On a distance–time graph that is speed; on a velocity–time graph it is acceleration; on a graph of water depth against time it is the rate of filling.

When the graph is curved the gradient changes. A chord between two points gives an $average rate$ ; a tangent at one point gives the $instantaneous rate$ . Going the other way, the area under a velocity–time graph gives distance travelled, estimated with trapezia when the graph is curved.

What you'll learn

Interpret the gradient of a straight-line graph as a rate of change
Find an average rate of change using a chord
Estimate an instantaneous rate of change using a tangent (Higher)
Estimate the area under a graph and interpret it (Higher)

Gradient as a rate

Gradient $= \frac{change in y}{change in x}$ , with units " $y$ -units per $x$ -unit".

A distance–time graph rising 120 km in 1.5 hours has gradient $80$ : a speed of 80 km/h.

A velocity–time graph rising from 0 to 12 m/s in 6 s has gradient 2: an acceleration of 2 m/s².

Tip — Include units with every gradient in context. "2" alone rarely gets the mark; "2 m/s²" does.

Average and instantaneous rates

Average rate between two times: draw the chord joining those two points and find its gradient.

Instantaneous rate at one time: draw a tangent touching the curve at that point, choose two points far apart on the tangent, and find its gradient.

1Tangent gradient:

\frac{45 - 5}{7 - 3} = \frac{40}{4} = 10

2Chord from

(1, 1)

(4, 16)

\frac{16 - 1}{4 - 1} = 5

AnswerInstantaneous rate at

t = 5

: about 10. Average rate from 1 to 4: 5.

Your turnSame method, new numbers — try it before moving on.

Find the average rate of change of $y = t^{2}$ between $t = 2$ and $t = 6$ .

Your answer

Why this works

A tangent is only an estimate because it is drawn by eye. Mark schemes accept a range of answers, provided the line genuinely touches the curve at the right point.

Area under a graph

On a velocity–time graph, area under the graph $=$ distance travelled.

For straight-line sections, split into triangles, rectangles and trapezia.

1Triangle:

\frac{1}{2} \times 6 \times 12 = 36

2Rectangle:

10 \times 12 = 120

Answer156 m

Your turnSame method, new numbers — try it before moving on.

A car speeds up steadily from 4 m/s to 10 m/s over 6 seconds. How far does it travel in that time, in metres?

Your answer

Estimating with trapezia

Split a curved area into vertical strips of equal width. Each strip is approximately a trapezium with area $\frac{1}{2} (a + b) h$ .

If the curve bends so the straight tops lie below it, the estimate is too small; if they lie above it, the estimate is too large.

1Strips:

\frac{1}{2} (0 + 3) + \frac{1}{2} (3 + 5) + \frac{1}{2} (5 + 6) + \frac{1}{2} (6 + 6.5)

= 1.5 + 4 + 5.5 + 6.25

Answer17.25 square units (an underestimate, since the curve bends above the trapezium tops)

Your turnSame method, new numbers — try it before moving on.

A curve passes through $(0, 2)$ , $(2, 6)$ and $(4, 8)$ . Using two trapezia of width 2, estimate the area under the curve from $x = 0$ to $x = 4$ .

Your answer

Tip — Say whether your estimate is an over- or underestimate and give the reason — AQA often asks.

Think like an examiner

Draw tangents and chords accurately with a ruler.
Show the two points used to calculate a gradient.
Interpret gradients and areas using the units on the axes.

Remember these

gradient = \frac{Δ y}{Δ x}

Rate of change.

area of trapezium = \frac{1}{2} (a + b) h

Estimating area under a curve.

area under v - t = distance

Interpreting area.

Watch out for these

Reading the

y

-value instead of the gradient for a rate.

A rate is a change per unit, found from a gradient.

Using a chord for an instantaneous rate.

Use a tangent at that single point.

Using the gridline count instead of axis values.

Read values from the scales.

Forgetting the

\frac{1}{2}

in the trapezium formula.

Each strip is

\frac{1}{2} (a + b) h

Stretch yourself

A car’s velocity is recorded every 2 seconds: at $t = 0, 2, 4, 6, 8$ it is $0, 6, 10, 12, 13$ m/s. The graph is a smooth curve getting less steep. Estimate the distance travelled in 8 seconds using four trapezia, and say whether this is an overestimate or underestimate.

Hint — Each strip has width 2.

Questions students ask

What does a negative gradient mean?

The quantity is decreasing — on a velocity–time graph, the object is decelerating.

Does a horizontal line on a distance–time graph mean moving at constant speed?

No — the distance is not changing, so the object is stationary.

Will more strips give a better estimate?

Yes, narrower trapezia follow the curve more closely.

Key takeaways

Gradient is a rate of change, with units.
Chord: average rate. Tangent: instantaneous rate.
Area under a velocity–time graph gives distance.
Estimate curved areas with trapezia and judge over or under.

Exam-style questions

Practise it the way it is examined

3 original questions · 10 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 52 marks

A distance–time graph is a straight line from $(0, 0)$ to $(30, 240)$ , with time in seconds and distance in metres. Work out the speed.[2]

Question 2Grade 75 marks

A train accelerates uniformly from rest to 20 m/s in 8 seconds, travels at 20 m/s for 12 seconds, then decelerates uniformly to rest in 5 seconds.

(a)Work out the acceleration in the first 8 seconds.[2]
(b)Work out the total distance travelled.[3]

Question 3Grade 83 marks

A graph shows the depth of water in a tank (cm) against time (minutes). The tangent to the curve at $t = 3$ passes through $(1, 2)$ and $(5, 14)$ . Estimate the rate at which the depth is increasing at $t = 3$ , and give its units.[3]

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

0 / 0 first-time correct Streak 0Best at this level: 0

Generating a question…

Test yourself

Ready to practise Rates of Change? Pick a mode and earn XP & Dobloons.

Next step

Practise this topic

Real past-paper questions on Rates of Change, marked mark-by-mark. How you do feeds straight into your weak-topic list, so your revision keeps targeting what actually needs work.

Practise this topic

Rates of Change

Gradient as a rate

Average and instantaneous rates

Area under a graph

Estimating with trapezia

Practise it the way it is examined

Skill drill

Practise this topic

Related topics

Rates of Change

Gradient as a rate

Average and instantaneous rates

Area under a graph

Estimating with trapezia

Practise it the way it is examined

Skill drill

Practise this topic

Related topics