A gradient measures a rate of change: how fast one quantity changes compared with another. The average rate of change between two points is the gradient of the chord joining them. The rate at a single point is the gradient of the tangent there, found by drawing a tangent or, in the International GCSE, by differentiating. On a distance–time graph the gradient is speed; on a velocity–time graph it is acceleration, and the area underneath is distance.
Rates of change
- Average rate of change = change in y ÷ change in x (gradient of a chord)
- Instantaneous rate: gradient of the tangent at that point
- Distance–time graph: gradient = speed
- Velocity–time graph: gradient = acceleration; area = distance
Average rate of change
The average rate of change of y between two x-values is the gradient of the chord, the straight line joining the two points on the curve. For y = x² between x = 3 and x = 5, y changes from 9 to 25, so the average rate is 16 ÷ 2 = 8.
Instantaneous rate of change
The rate of change at one point is the gradient of the tangent there. On a GCSE paper, draw the tangent carefully with a ruler, choose two points far apart on it, and work out change in y ÷ change in x. Your answer is an estimate, so the mark scheme accepts a range of values.
Exact gradients by differentiating (iGCSE)
In the International GCSE, you can find the exact gradient by differentiating. For y = x² + 3x, dy/dx = 2x + 3, so at x = 4 the gradient is 11.
Distance–time and velocity–time graphs
On a distance–time graph, the gradient is the speed: 200 m in 25 s is 8 m/s. On a velocity–time graph, the gradient is the acceleration: from 10 m/s to 30 m/s in 8 s is 20 ÷ 8 = 2.5 m/s². A negative gradient means slowing down, called deceleration.
Area under a velocity–time graph
The area under a velocity–time graph gives the distance travelled. Split the area into rectangles, triangles and trapezia. A car accelerating steadily from rest to 12 m/s in 4 s travels the area of a triangle: ½ × 4 × 12 = 24 m. For curved graphs, estimate the area with strips.
Worked examples
Example 1. Find the average rate of change of y = x² between x = 3 and x = 5.
- At x = 3, y = 9; at x = 5, y = 25.
- (25 − 9) ÷ (5 − 3) = 16 ÷ 2.
Answer: 8.
Example 2. Find the exact gradient of y = x² + 3x at x = 4, by differentiating.
- dy/dx = 2x + 3.
- At x = 4: 8 + 3.
Answer: 11.
Example 3. A cyclist travels 200 m in 25 s at a constant speed. Find the speed.
- Speed is the gradient of the distance–time graph.
- 200 ÷ 25.
Answer: 8 m/s.
Example 4. A car's velocity rises steadily from 10 m/s to 30 m/s in 8 s. Find its acceleration.
- Gradient of the velocity–time graph: (30 − 10) ÷ 8.
Answer: 2.5 m/s².
Example 5. A car accelerates steadily from rest to 12 m/s in 4 s. How far does it travel in that time?
- Distance is the area under the velocity–time graph, a triangle.
- ½ × 4 × 12.
Answer: 24 m.
Common mistakes
- Using one point instead of two. A gradient needs a change in y and a change in x.
- Confusing a chord with a tangent. A chord joins two points; a tangent touches at one.
- Reading speed from a velocity–time graph's gradient. On a velocity–time graph the gradient is acceleration.
- Forgetting units. Speed is in m/s, acceleration in m/s².
- Drawing a tangent that crosses the curve. A tangent should just touch the curve at the point.
How rates of change are examined
Gradients of curves and rates of change are Higher tier content: on Edexcel GCSE (1MA1) papers you estimate them by drawing tangents and chords, and on the International GCSE (4MA1) you can also find them by differentiation. Questions ask you to:
- work out an average rate of change between two points
- estimate the gradient of a curve at a point by drawing a tangent
- interpret the gradient of a distance–time or velocity–time graph
- work out a distance from the area under a velocity–time graph
The GCSE algebra revision engine gives practice on gradients, and the linear graphs worksheet covers gradients of straight lines.
This connects straight-line gradients with differentiation and speed, distance and time.
Frequently asked questions
What is the average rate of change?
The gradient of the chord joining two points on a curve: change in y divided by change in x.
How do you estimate the gradient of a curve?
Draw a tangent at the point, then find the gradient of the tangent using two points on it.
What does the area under a velocity–time graph represent?
The distance travelled.