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3.12a · Rates of Change · Sub-skill
Calculate rates of change from straight-line graphs, and see how chords on a curve get closer to the gradient at a point.
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Build it up, step by step
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The rate of change of $y$ with respect to $x$ is $\dfrac{\text{change in } y}{\text{change in } x}$. Its units are '(units of $y$) per (unit of $x$)': kilometres per hour, pounds per month, degrees per minute.
On a straight-line graph the rate of change is constant: choose any two points and calculate rise ÷ run. A steeper line means a faster rate; a negative gradient means the quantity is falling.
On a curve the rate keeps changing. The average rate of change between two points is the gradient of the straight line joining them — a chord: $\dfrac{y_2 - y_1}{x_2 - x_1}$.
If you zoom in close to a point on a smooth curve it looks like a straight line. Taking a shorter and shorter chord gives gradients that settle towards one value — the gradient at that point, which is the gradient of the tangent.
For $y = x^2$, find the average rate of change between $x = 1$ and $x = 3$, then between $x = 1$ and $x = 1.1$.
$x = 1 \to y = 1$; $x = 3 \to y = 9$: gradient $= \dfrac{9 - 1}{3 - 1} = 4$.
$x = 1.1 \to y = 1.21$: gradient $= \dfrac{1.21 - 1}{0.1} = 2.1$.
The shorter chord gives a value closer to the gradient at $x = 1$, which is 2.
Test yourself
The temperature at 9 am is 12 °C, at 11 am it is 15 °C and at 1 pm it is 24 °C.
(a) Find the average rate of change of temperature between 9 am and 1 pm, in °C per hour.
(b) Find the average rate between 9 am and 11 am, and between 11 am and 1 pm.
(c) When was the temperature rising faster? [5 marks]
(a) $\dfrac{24 - 12}{4} = $ 3 °C per hour. (2 marks)
(b) 9–11 am: $\dfrac{15 - 12}{2} = $ 1.5 °C/h. 11 am–1 pm: $\dfrac{24 - 15}{2} = $ 4.5 °C/h. (2 marks)
(c) Between 11 am and 1 pm (4.5 > 1.5). (1 mark)
Practice
Five short questions. Work through them, then reveal the mark scheme to check.