Rates of Change: Gradients and Areas Under Graphs
GCSE Higher doesn’t use calculus. Instead you estimate rates of change by drawing chords and tangents, and you estimate areas under curves with trapezia. Then you say what your answer means, which is often where the last mark is.
1Average and instantaneous rates of change
| Rate | How to find it | On the graph |
|---|---|---|
| Average rate of change | Gradient of the chord joining two points | A straight line between two points on the curve |
| Instantaneous rate of change | Gradient of the tangent at a point | A line touching the curve at one point |
Worked example
Using the graph, estimate (a) the average acceleration over the first 8 seconds (b) the acceleration at t = 4.
- (a) Chord from (0, 0) to (8, 16): gradient = 16 ÷ 8 = 2 m/s².
- (b) The tangent at (4, 12) passes through (0, 4) and (8, 20): gradient = (20 − 4) ÷ 8 = 2 m/s².
2What the gradient and area mean
| Graph | Gradient means | Area under graph means |
|---|---|---|
| Distance–time | Speed | — |
| Velocity–time | Acceleration | Distance travelled |
| Cost–number | Cost per item | — |
| Rate of flow–time | Change in rate | Total volume |
Key ideaUnits help: the gradient has units “y-units per x-unit” (m per s → m/s). The area has units “y-units × x-units” (m/s × s → m).
3Estimating the area under a curve
Split the area into vertical strips of equal width and treat each strip as a trapezium: area = ½(a + b)h.
Worked example
Estimate the distance travelled in the first 8 seconds using 4 strips of equal width.
- Velocities at t = 0, 2, 4, 6, 8: 0, 7, 12, 15, 16.
- Area ≈ ½(0 + 7)×2 + ½(7 + 12)×2 + ½(12 + 15)×2 + ½(15 + 16)×2
- = 7 + 19 + 27 + 31 = 84.
Under or over?The curve bends downwards (concave), so each trapezium lies below the curve. The estimate is an underestimate. If the curve bends upwards, the trapezia over-estimate.
Exam tipMore, narrower strips give a better estimate.
4Graphs made of straight lines
For straight-line sections the area is exact: split it into rectangles, triangles and trapezia.
Worked example
A car accelerates from 0 to 12 m/s in 3 s, then travels at 12 m/s for another 3 s. How far does it travel?
- Triangle: ½ × 3 × 12 = 18.
- Rectangle: 3 × 12 = 36.
- Total = 54.
Common mistakes
- Drawing a tangent that crosses the curve, or reading coordinates from too short a line.
- Mixing up chord (average rate) and tangent (instantaneous rate).
- Not saying whether a trapezium estimate is an over- or underestimate, with a reason.
- Forgetting units, or giving the wrong meaning (for example, calling the area under a velocity–time graph a speed).
★Key Vocabulary Recap
- Chord
- A straight line joining two points on a curve.
- Tangent
- A line touching a curve at one point.
- Rate of change
- How fast one quantity changes compared with another; the gradient.
- Trapezium rule
- Estimating area using trapezium-shaped strips.
Specification reference (Pearson Edexcel GCSE (9–1) Mathematics 1MA1: A15, R14, R15)
- A15: calculate or estimate gradients of graphs and areas under graphs (including quadratic and other non-linear graphs), and interpret results in cases such as distance–time and velocity–time graphs.
- R14: interpret the gradient of a straight-line graph as a rate of change.
- R15: interpret the gradient at a point on a curve as the instantaneous rate of change; apply the concepts of average and instantaneous rate of change.
Next stepReady to practise? Try the 8 exam-style questions, then check your working against the worked solutions.
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Practice questions
Rates of Change: Gradients and Areas Under Graphs — Practice Questions
8 exam-style questions (18 marks), building from warm-up to grade 8–9. Show full working: Edexcel gives method marks even when the final answer is wrong.
Higher · Grades 6–7
- (2 marks)Grade 6A car accelerates steadily from rest to 12 m/s in 3 seconds, then travels at 12 m/s for 3 seconds. Find the total distance travelled.
- (2 marks)Grade 6The height of a ball is h = 30t − 5t² metres. Find the average rate of change of height between t = 1 and t = 3, and state its units.
- (2 marks)Grade 7The table shows the velocity of a cyclist.Find the average acceleration between t = 10 and t = 20.
t (s) 0 10 20 30 40 v (m/s) 0 8 12 14 15 - (3 marks)Grade 7Using the table in Question 3, use 4 strips of equal width to estimate the distance travelled in the first 40 seconds.
Higher · Grades 8–9
- (1 mark)Grade 8The cyclist’s velocity graph from Question 3 curves downwards (it is concave). Is your answer to Question 4 an overestimate or an underestimate? Explain.
- (3 marks)Grade 8Using the graph of v = 4t − 0.25t² in the notes, the tangent at t = 4 passes through (0, 4) and (8, 20). Estimate the acceleration at t = 4 and explain what it means.
- (2 marks)Grade 8On a distance–time graph, the chord from (0, 0) to (30, 14) is drawn. What does its gradient represent, and what is its value?
- (3 marks)Grade 9Water flows into a tank. The rate of flow (litres per minute) is shown by a curve. Explain what the area under the curve between t = 0 and t = 10 represents, and give its units.
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Worked solutions
Show worked solutions
Rates of Change: Gradients and Areas Under Graphs — Worked Solutions
Full worked solutions to all 8 questions (18 marks), set out the way an Edexcel examiner wants to see them.
Higher · Grades 6–7
- (2 marks)Grade 6A car accelerates steadily from rest to 12 m/s in 3 seconds, then travels at 12 m/s for 3 seconds. Find the total distance travelled.Solution
- Area of triangle = ½ × 3 × 12 = 18.
- Area of rectangle = 3 × 12 = 36.
54 m - (2 marks)Grade 6The height of a ball is h = 30t − 5t² metres. Find the average rate of change of height between t = 1 and t = 3, and state its units.Solution
- h(1) = 25 and h(3) = 45.
- Average rate = (45 − 25) ÷ (3 − 1) = 10.
10 m/s - (2 marks)Grade 7The table shows the velocity of a cyclist.Find the average acceleration between t = 10 and t = 20.
t (s) 0 10 20 30 40 v (m/s) 0 8 12 14 15 Solution- (12 − 8) ÷ (20 − 10) = 4 ÷ 10.
0.4 m/s² - (3 marks)Grade 7Using the table in Question 3, use 4 strips of equal width to estimate the distance travelled in the first 40 seconds.Solution
- Width 10. Area ≈ ½(0 + 8)10 + ½(8 + 12)10 + ½(12 + 14)10 + ½(14 + 15)10
- = 40 + 100 + 130 + 145.
≈ 415 m
Higher · Grades 8–9
- (1 mark)Grade 8The cyclist’s velocity graph from Question 3 curves downwards (it is concave). Is your answer to Question 4 an overestimate or an underestimate? Explain.Solution
- The straight tops of the trapezia lie below a curve that bends downwards, so some area is missed.
Underestimate - (3 marks)Grade 8Using the graph of v = 4t − 0.25t² in the notes, the tangent at t = 4 passes through (0, 4) and (8, 20). Estimate the acceleration at t = 4 and explain what it means.Solution
- Gradient = (20 − 4) ÷ (8 − 0) = 2.
- At 4 seconds the velocity is increasing at 2 m/s every second.
2 m/s² - (2 marks)Grade 8On a distance–time graph, the chord from (0, 0) to (30, 14) is drawn. What does its gradient represent, and what is its value?Solution
- The gradient of a chord on a distance–time graph is the average speed.
- 14 ÷ 30 = 715 ≈ 0.467.
Average speed ≈ 0.467 units - (3 marks)Grade 9Water flows into a tank. The rate of flow (litres per minute) is shown by a curve. Explain what the area under the curve between t = 0 and t = 10 represents, and give its units.Solution
- Rate (litres/min) × time (min) = litres.
- The area is the total volume of water that flowed in during the first 10 minutes.
Total volume, in litres
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