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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 Eduqas GCSE Mathematics ratio content on growth, decay and rates of change, covering exponential growth and decay multipliers, the gradient of a graph as a rate, and the area under a graph in real-life contexts.

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. The gradient as a rate of change
  4. The area under a graph
  5. Linking the ideas

What this dot point is asking

The Eduqas ratio content asks you, at Higher tier, to model exponential growth and decay over repeated periods, and to interpret the gradient of a real-life graph as a rate of change and the area under a graph in context. Growth and decay extend compound interest to any repeated proportional change (populations, depreciation, radioactive decay), while the graph work links to physics through speed-time and distance-time graphs. Both are reliable Higher-tier tasks because they combine calculation with interpretation, exactly the AO2 and AO3 reasoning Eduqas weights at half the marks.

Exponential growth and decay

Repeated proportional change is exponential, not linear: the same multiplier is applied each period, so the change compounds.

So a savings account of 4000 pounds growing at 5%5\% per year reaches 4000×1.056=5360.384000 \times 1.05^6 = 5360.38 pounds after 66 years, while a substance decaying at 12%12\% per hour drops to (start)×0.88n\left(\text{start}\right) \times 0.88^n after nn hours. The defining feature is that the percentage applies to the current amount each time, so growth accelerates and decay slows, unlike a constant linear change.

The gradient as a rate of change

On a graph of one quantity against another, the gradient measures how fast one changes with respect to the other.

For a curved graph, the rate is changing, so to find the rate at a particular instant you draw a tangent to the curve at that point and measure its gradient. This is an estimate, because the tangent is drawn by eye, so Eduqas accepts answers within a range.

The area under a graph

The area between a graph and the horizontal axis often has a physical meaning.

For a curved speed-time graph, estimate the area by dividing it into strips (trapezia) and summing them, which gives an approximate distance.

Linking the ideas

Growth, decay and rates of change all describe how a quantity behaves over time, and they connect across the course: the multiplier method comes from percentage change, the powers echo the index laws, and the graph interpretation feeds directly into physics. Because these questions ask you to both calculate and explain (state the rate, interpret the area), setting out clear reasoning alongside the arithmetic is what secures full marks. A good habit is to label what each gradient or area represents in words, so a tangent gradient becomes "the speed at t=4t = 4 seconds is about 66 m/s" rather than a bare number, which is exactly the communication Eduqas mark schemes look for.

A further subtlety is that growth and decay never reach zero or infinity in finite steps: decay halves and halves again but never quite vanishes, which is why these models suit real quantities that fade gradually. Recognising the difference between this multiplicative behaviour and a simple linear fall is the conceptual point examiners test most often, so when a question describes a fixed percentage change "each year" or "each hour", reach for a power of the multiplier rather than a repeated subtraction.

Exam-style practice questions

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

Eduqas 20193 marksThe population of a colony of bacteria grows by 8 percent each hour. At the start there are 5000 bacteria. How many are there after 3 hours? Give your answer to the nearest whole number. (Higher, Component 2, calculator.)
Show worked answer →

An 8 percent increase each hour means a multiplier of 1.081.08, applied once per hour.

After 3 hours: 5000×1.083=5000×1.259712=6298.565000 \times 1.08^3 = 5000 \times 1.259712 = 6298.56.

To the nearest whole number, this is 6299 bacteria.

Markers award a mark for the multiplier 1.081.08, a mark for raising it to the power 3, and a mark for the rounded answer. Multiplying by 1.08×31.08 \times 3 (treating it as simple, linear growth) instead of 1.0831.08^3 is the standard error.

Eduqas 20224 marksA car's value depreciates by 15 percent each year. It was bought new for 18000 pounds. Work out its value after 4 years, to the nearest pound, and state what fraction of its original value remains. (Higher, Component 2, calculator.)
Show worked answer →

Depreciation of 15 percent per year means a multiplier of 0.850.85 each year.

After 4 years: 18000×0.854=18000×0.52200625=9396.1118000 \times 0.85^4 = 18000 \times 0.52200625 = 9396.11 pounds, so 9396 pounds to the nearest pound.

The fraction remaining is 0.854=0.5220.85^4 = 0.522\ldots, about 52%52\% of the original value.

Markers give marks for the multiplier 0.850.85, for the power of 4, for the rounded value, and for the fraction remaining. Subtracting 15 percent four times as a flat amount, rather than compounding, loses the structure marks.

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