Rates of Change
A rate of change tells you how quickly one quantity changes compared with another. On a graph, this idea is represented by gradient.
Foundation & Higher, with a substantial Higher section. You should already understand straight-line gradient; see Linear Graphs.
Change per unit
Rate of change = change in output ÷ change in input. The units come from the quantities involved.
If temperature rises from 14°C to 26°C in 3 hours, average rate = 12 ÷ 3 = 4°C per hour.
Gradient as a rate
For a straight-line graph, gradient = change in y ÷ change in x. A positive gradient means y increases as x increases; a negative gradient means y decreases.
Always attach contextual units, such as litres per minute or pounds per hour.
Here the tank starts with 20 L. Between (2,40) and (6,80), gradient = (80−40)/(6−2) = 10 L/min. The starting volume changes the intercept, not the rate. The horizontal side of the dashed triangle is 4 min and the vertical side is 40 L.
Using tables
Rates can be estimated from nearby table values. If a tank contains 120 L at 4 min and 168 L at 6 min, average fill rate over that interval is (168−120)÷(6−4)=24 L/min.
Worked example — compare rates from a table
| Time (min) | 0 | 2 | 5 | 9 |
|---|---|---|---|---|
| Volume (L) | 50 | 86 | 140 | 188 |
From 0 to 2 minutes, average rate = (86−50) ÷ 2 = 18 L/min.
From 5 to 9 minutes, average rate = (188−140) ÷ 4 = 12 L/min.
The volume increases across both intervals, but its average rate is smaller in the later interval. A table alone does not show whether the rate was constant, or briefly changed direction, between the recorded times.
Higher — average rate using a chord
For a curved graph, the gradient between two points gives an average rate of change over that interval. The straight line joining the two points is a chord.
Use two clear points on the chord, not two arbitrary points on the curve.
Higher worked example — average rate on a curve
For the plotted model V = t² + 10, V(2) = 14 L and V(6) = 46 L. The dashed chord joins these two points. Its gradient is (46−14)/(6−2) = 8 L/min. That is the average rate over 2 to 6 minutes, even though the curve gets steeper within the interval.
Higher — instantaneous rate using a tangent
The instantaneous rate at a point is estimated by drawing a tangent to the curve at that point and finding the tangent’s gradient.
Place a ruler through the chosen point and rotate it until its edge follows the curve’s local direction. Judge the fit on both sides close to the point, then draw the line. A tangent can meet the curve again elsewhere; what matters is its direction at the chosen point. Select two well-separated, readable points on the drawn tangent and use a large gradient triangle.
Higher worked example — estimate a tangent gradient
At t = 4 min, the curve has V = 26 L. Draw the dashed tangent through that point following the curve’s direction. On this drawing it passes approximately through (2,10) and (6,42).
Gradient ≈ (42−10)/(6−2) = 32/4 = 8 L/min. The tank’s instantaneous filling rate at 4 minutes is therefore about 8 litres per minute. The green triangle measures the tangent’s rise and run; its endpoints are not both on the curve.
This is a graphical estimate. Small differences are expected when a tangent or its coordinates are read from a printed graph. No differentiation is needed.
Interpreting the sign and size
A larger positive gradient means faster increase. A more negative gradient means faster decrease. A gradient near zero means the quantity is changing slowly at that point.
Apply rates to real-life graphs
Apply these methods to distance–time and velocity–time situations in Real-Life and Travel Graphs. Then explore Areas Under Graphs and Accumulated Change (Higher). These GCSE applications use graphs and geometric areas; calculus is not required.
Think Like a Mathematician
Why does a tangent estimate an instantaneous rate?
Over a very short interval, a smooth curve behaves more and more like its tangent. The tangent gradient captures the local direction of change.
Common Mistakes
Forgetting units: gradient is a rate with compound units.
Using curve points to calculate a tangent gradient: use points on the tangent line.
Confusing average and instantaneous rates: chords describe intervals; tangents describe a point.
Practice — rates of change
1. A quantity rises from 20 to 50 in 6 seconds. Find average rate of change.
(50−20)÷6=5 units/s.
2. Water volume falls from 900 L to 660 L in 8 min. Find average rate.
(660−900)÷8=−30 L/min.
3. A straight-line graph passes through (2,7) and (8,25). Find gradient.
(25−7)÷(8−2)=18÷6=3.
4. Explain what a gradient of −4°C/min means.
Temperature is decreasing by 4°C each minute.
5. A tank has 140 L at 3 min and 230 L at 6 min. Find average fill rate.
(230−140)÷3=30 L/min.
6. Higher: for the curve V = t² + 10, find the average rate from t = 0 to t = 4 min. State which line on a graph represents it.
V(0) = 10 L and V(4) = 26 L. Average rate = (26−10)/4 = 4 L/min, the gradient of the chord joining (0,10) to (4,26).
7. Higher: what does a tangent gradient represent at a point on a curve?
An estimate of the instantaneous rate of change at that point.
8. Higher: a tangent to a volume–time curve passes through (4,18) and (10,42), with time in seconds and volume in litres. Estimate the rate.
Gradient ≈ (42−18)/(10−4) = 24/6 = 4 L/s. Use points on the tangent, not arbitrary points on the curve.
9. Higher: a tangent has gradient −2.5 litres/s. Interpret the sign.
The volume is decreasing at about 2.5 litres each second.
10. Higher: use the curve V = t² + 10 in the diagrams. Draw a tangent at t = 6 min and estimate the rate. Why use well-separated points?
A reasonable tangent through (6,46) passes close to (4,22) and (8,70). Gradient ≈ (70−22)/(8−4) = 12 L/min. A larger triangle reduces the effect of small plotting or reading errors. Allow a sensible graphical estimate.
Frequently Asked Questions
Does a flat graph mean the quantity is zero?
No. A horizontal graph has zero rate of change, but its value can be non-zero.
Are tangent rates Foundation or Higher?
Instantaneous rates and chord/tangent methods for curves are Higher. Basic change per unit and straight-line gradients are shared; a tangent estimate does not require differentiation.
Continue Learning
Previous: Speed, Density and Pressure
Next: Real-Life and Travel Graphs
Next extension — Higher: Areas Under Graphs and Accumulated Change