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Ready-to-teach lesson. Includes a video tutorial, printable worksheets and Higher GCSE exam-style practice with worked solutions.
This lesson covers instantaneous rates of change for Higher GCSE graphs. Students draw a tangent at a point, calculate its gradient, and interpret that rate in context.
Students can already find the gradient of a straight line, because that gradient stays the same. On a curve it changes, so the straight-line method only works after they draw a tangent.
I teach this before the additional practice, once the tangent method is secure. The lesson pack supplies the slides, and the revision lesson is for the exam run.
Watch the tangent being drawn first. The gradient of that line is the rate at that moment. Try the worksheet below after the video.
A chord between two points gives an average rate of change. A tangent touches the curve at one point, so its gradient is the rate at that instant.
Gradient equals change in y divided by change in x. The units come from the two axes, so metres divided by seconds gives metres per second.
On a distance-time graph that gradient is speed. On a speed-time or velocity-time graph it is acceleration. The axes decide the meaning, not the shape of the curve.
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Each grid shows y = x in blue. The red line is the graph to name. Steepness and direction decide the match before anyone calculates.
Prompts / Questions to consider
The curve is no longer a straight line, so the starter method stops working. Students now draw a tangent where the question names the time.
The temperature falls quickly at first, then the curve flattens as the food gets closer to the freezer temperature. The rate of cooling is not constant.
Ask for the rate at which the temperature is decreasing, not the gradient written as a negative number. The rate of decrease is positive.
Prompts / Questions to consider
The plenary keeps the tangent method and adds the meaning of the gradient. By 45 minutes the curve is almost level.
The tea starts near 75°C and falls towards about 18°C. That resting value matches the temperature of the room around the cup.
The gradient at 10 minutes is negative, so the temperature is falling. The size of that gradient is the rate of cooling, in °C per minute.
Prompts / Questions to consider
More able: On y = x/2 + 2/x, check the tangent at x = 1 with dy/dx = 1/2 – 2/x². Both methods give -1.5.
Less able: Give two points already marked on the tangent. Their job is the division, then naming the units.
| Phase | Focus | Time |
|---|---|---|
| Starter | Match red lines to equations using y = x | 8 min |
| Development | Draw tangents on the freezer graph at 5 and 14 minutes | 15 min |
| Check | Mini-whiteboards: gradient, units, rate of decrease | 8 min |
| Main / stretch | Worksheet questions 1 to 4, then the calculus check | 15 min |
| Plenary | Tea curve at 10 minutes and the resting temperature | 8 min |
Use this checklist after the lesson or as a revision self-check.
Try these Higher GCSE-style questions on instantaneous rates of change, then reveal the solutions. Select an image to view it full screen.
Use the distance-time graph to estimate the speed of the car at 5 seconds.
At t = 5 the curve is at about 32 m. Draw a tangent at (5, 32).
A workable tangent passes through about (3, 0) and (7, 64). Gradient = 64 ÷ 4 = 16.
Speed is about 16 m/s. An answer from about 14 m/s to 20 m/s is reasonable.
Elsie’s graph shows distance, in metres, against time. Estimate the gradient at t = 4, say what it represents, and explain why it is an estimate.
a) At t = 4 the distance is about 8 m. Draw a tangent at (4, 8) through about (0, 3.2) and (5, 9.2).
Gradient = 6 ÷ 5 = 1.2. An answer from about 0.8 to 1.6 is reasonable.
b) The axes are distance and time, so the gradient is Elsie’s speed at 4 seconds, about 1.2 m/s.
c) The tangent is drawn by eye and the coordinates are read from the graph, so the gradient is an estimate.
Estimate the gradient of the parachutist’s velocity-time graph after 3 seconds, then interpret it.
a) At t = 3 the velocity is about 29 m/s. A tangent at (3, 29) passes through about (0, 5) and (6, 53).
Gradient = 48 ÷ 6 = 8. An answer from about 6 to 10 is reasonable.
b) The axes are velocity and time, so this is the acceleration at 3 seconds, about 8 m/s².
The curve is y = x/2 + 2/x for 0 < x ≤ 8. Solve x/2 + 2/x = 3 from the graph, then estimate the gradient at x = 1.
a) Draw y = 3. It meets the curve at about x = 0.8 and x = 5.2. Solving gives x = 3 ± √5, which is 0.76 and 5.24.
b) At x = 1, y = 2.5. A tangent through about (0, 4) and (2, 1) has gradient -3 ÷ 2 = -1.5.
Check: dy/dx = 1/2 – 2/x², so at x = 1 the gradient is 0.5 – 2 = -1.5 exactly.
Watch a grade 9 example of estimating a rate of change from a curve. Pause before the working and sketch the tangent yourself.
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It is the gradient of the tangent to the curve at that point. A chord between two points gives an average rate instead.
Draw a tangent at the point and choose two clear points on that line. Divide the change in y by the change in x.
Speed. Distance divided by time gives metres per second. It is not the acceleration.
Acceleration. A positive gradient means the velocity is increasing. A zero gradient means the velocity is steady.
You draw the line by eye and read the coordinates from the graph. A small change in the tangent changes the gradient.
Next, use the additional practice so students meet a new curve and have to choose the tangent themselves. Keep the revision lesson on gradients of curves for the run-up to the exam.
Try this tomorrow. On the tea graph, ask only for the gradient at 10 minutes and what the negative sign means. See who states a rate of cooling in °C per minute.
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