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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\textbf{average rate}; a tangent at one point gives the instantaneous rate\textbf{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)
1

Gradient as a rate

Gradient =change in ychange in x= \dfrac{\text{change in } y}{\text{change in } x}, with units "yy-units per xx-unit".

A distance–time graph rising 120 km in 1.5 hours has gradient 8080: 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.

2

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.

Example 1The graph of y=t2y = t^{2} is drawn. A tangent at t=5t = 5 passes through (3,5)(3, 5) and (7,45)(7, 45). Estimate the rate of change at t=5t = 5, and find the average rate from t=1t = 1 to t=4t = 4.
1Tangent gradient: 45−57−3=404=10\dfrac{45 - 5}{7 - 3} = \dfrac{40}{4} = 10.
2Chord from (1,1)(1, 1) to (4,16)(4, 16): 16−14−1=5\dfrac{16 - 1}{4 - 1} = 5.
AnswerInstantaneous rate at t=5t = 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=t2y = t^2 between t=2t = 2 and t=6t = 6.

Show worked solution
Solution
  1. 1At t=2t = 2, y=4y = 4; at t=6t = 6, y=36y = 36.
  2. 2Average rate =36−46−2=324=8= \frac{36 - 4}{6 - 2} = \frac{32}{4} = 8.

Answer: 88

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.

3

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.

Example 1A cyclist accelerates from rest to 12 m/s in 6 seconds, then rides at 12 m/s for 10 seconds. How far do they travel?
1Triangle: 12×6×12=36\dfrac12 \times 6 \times 12 = 36 m.
2Rectangle: 10×12=12010 \times 12 = 120 m.
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?

Show hint

The area under the speed–time graph is a trapezium.

Show worked solution
Solution
  1. 1Distance = area under the graph =12(4+10)×6= \frac{1}{2}(4 + 10) \times 6.
  2. 2=42= 42 m.

Answer: 4242 m

4

Estimating with trapezia

Split a curved area into vertical strips of equal width. Each strip is approximately a trapezium with area 12(a+b)h\dfrac12(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.

Example 1A curve passes through (0,0)(0, 0), (1,3)(1, 3), (2,5)(2, 5), (3,6)(3, 6) and (4,6.5)(4, 6.5). Estimate the area under it from x=0x = 0 to x=4x = 4 using four strips.
1Strips: 12(0+3)+12(3+5)+12(5+6)+12(6+6.5)\tfrac12(0 + 3) + \tfrac12(3 + 5) + \tfrac12(5 + 6) + \tfrac12(6 + 6.5).
2=1.5+4+5.5+6.25= 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)(0, 2), (2,6)(2, 6) and (4,8)(4, 8). Using two trapezia of width 2, estimate the area under the curve from x=0x = 0 to x=4x = 4.

Show worked solution
Solution
  1. 1First strip: 12(2+6)×2=8\frac{1}{2}(2 + 6) \times 2 = 8.
  2. 2Second strip: 12(6+8)×2=14\frac{1}{2}(6 + 8) \times 2 = 14. Total 2222.

Answer: 2222 square units

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=ΔyΔx\text{gradient} = \dfrac{\Delta y}{\Delta x}
Rate of change.
area of trapezium=12(a+b)h\text{area of trapezium} = \tfrac12(a + b)h
Estimating area under a curve.
area under v–t=distance\text{area under } v\text{–}t = \text{distance}
Interpreting area.

Watch out for these

Reading the yy-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 12\tfrac12 in the trapezium formula.
Each strip is 12(a+b)h\tfrac12(a + b)h.

Stretch yourself

A car’s velocity is recorded every 2 seconds: at t=0,2,4,6,8t = 0, 2, 4, 6, 8 it is 0,6,10,12,130, 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.

Reveal solution
SolutionArea ≈2×12[(0+6)+(6+10)+(10+12)+(12+13)]=6+16+22+25=69\approx 2 \times \tfrac12[(0 + 6) + (6 + 10) + (10 + 12) + (12 + 13)] = 6 + 16 + 22 + 25 = 69 m. The curve bends downwards (its gradient decreases), so the straight trapezium tops lie below the curve: 69 m is an underestimate. ■\blacksquare

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

    Answer: 88 m/s

Answer on paper first, then mark it.
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.

  1. (a)Work out the acceleration in the first 8 seconds.[2]
    Show mark scheme (a)

    Answer: 2.52.5 m/s²

  2. (b)Work out the total distance travelled.[3]
    Show mark scheme (b)

    Answer: 370370 m

Answer on paper first, then mark it.
Question 3Grade 83 marks
  1. A graph shows the depth of water in a tank (cm) against time (minutes). The tangent to the curve at t=3t = 3 passes through (1,2)(1, 2) and (5,14)(5, 14). Estimate the rate at which the depth is increasing at t=3t = 3, and give its units.[3]
    Show mark scheme

    Answer: 33 cm/min

Answer on paper first, then mark it.

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

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