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How do you model growth and decay over time and interpret the gradient and area of a real-life graph as a rate of change?

Model exponential growth and decay over repeated periods; and interpret the gradient of a graph as a rate of change and the area under a graph in real-life contexts (Higher tier).

A focused answer to the OCR GCSE Mathematics ratio content on growth, decay and rates of change, covering exponential growth and decay models and interpreting the gradient and area of real-life graphs as rates of change.

Generated by Claude Opus 4.811 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. Exponential growth and decay
  3. Gradient as a rate of change
  4. Area under a graph
  5. Why this matters

What this dot point is asking

OCR references R15 and R16 cover growth and decay over repeated periods (exponential change) and, at Higher tier, interpreting the gradient of a graph as a rate of change and the area under a graph in real-life contexts. This content ties the multiplier idea from percentages to graphs and to compound measures. It appears mainly at Higher tier and on the calculator paper, and gradient-as-rate questions are a reliable multi-mark feature that tests AO2 interpretation.

Exponential growth and decay

Growth and decay multiply by a fixed factor each period.

So a population of 2000020000 growing at 8%8\% a year reaches 20000×1.0852938720000 \times 1.08^5 \approx 29387 after 55 years, and a £1200012000 car depreciating at 20%20\% a year falls to 12000×0.84=12000 \times 0.8^4 = £4915.204915.20 after 44 years. Growth and decay are exponential, not linear, which is why a single percentage cannot just be multiplied by the number of years. Contexts include savings, populations, radioactive decay and depreciation.

Gradient as a rate of change

The gradient of a real-life graph measures how fast one quantity changes with another.

So a distance-time graph rising from 2020 m to 8080 m between 44 s and 1010 s has gradient 8020104=10\dfrac{80 - 20}{10 - 4} = 10 m/s, the speed. For a curved graph, the rate of change at a point is the gradient of the tangent at that point, which you estimate by drawing the tangent and finding its gradient. A negative gradient on a distance-time graph means returning towards the start.

A flat (horizontal) section of a distance-time graph means the object is stationary, because the distance is not changing. On a speed-time graph, a flat section means constant speed (zero acceleration), and a downward slope means the object is slowing down (deceleration). Being able to translate the shape of a graph into a word description of the motion, and the reverse, is a frequent OCR question that targets AO2 communication, so practise describing each segment in plain English.

Area under a graph

The area under certain graphs has a physical meaning.

Why this matters

Growth and decay model the real financial and natural world, and OCR sets them alongside compound interest because they share the multiplier idea. Gradient-as-rate and area-under-a-graph are the GCSE foundations of calculus, and they appear in physics-style questions where a graph must be read for speed, acceleration or distance. Stating what the gradient or area represents, in words, is exactly the AO2 communication OCR rewards.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20194 marksA car bought for £1800018000 depreciates by 15%15\% each year. Work out its value after 33 years, to the nearest pound. (Higher, Paper 4, calculator.)
Show worked answer →

Depreciation of 15%15\% per year is exponential decay with multiplier 0.850.85.

After 33 years the value is 18000×0.85318000 \times 0.85^3.

0.853=0.6141250.85^3 = 0.614125, so 18000×0.614125=11054.2518000 \times 0.614125 = 11054.25.

To the nearest pound, the car is worth £1105411054.

Markers award a mark for the multiplier 0.850.85, a mark for the power 33, a mark for the calculation, and a mark for £1105411054. Subtracting 15%15\% three times as 45%45\% off (giving £99009900) is the standard error.

OCR 20213 marksThe graph of a car's journey shows distance against time. Between t=10t = 10 s and t=30t = 30 s the line is straight, rising from 5050 m to 250250 m. Find the speed of the car over this interval. (Higher, Paper 4, calculator.)
Show worked answer →

On a distance-time graph, the gradient is the speed.

Gradient =change in distancechange in time=250503010=20020=10= \dfrac{\text{change in distance}}{\text{change in time}} = \dfrac{250 - 50}{30 - 10} = \dfrac{200}{20} = 10.

So the speed is 1010 m/s.

Markers give a mark for identifying the gradient as speed, a mark for the correct differences, and a mark for 1010 m/s. Reading single coordinates instead of the change, or dividing time by distance, are the usual slips.

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