Ratio, proportion and rates of change

Gradients and rates of change

9 past-paper questions, easiest first. Try each one, then reveal the answer.

Foundation & Higher1 of 9 from Foundation papers

Revision guide by The Full Marks

Gradients and rates of change: what you need to know

The gradient of a graph tells you how fast one quantity changes compared with another, such as speed on a distance–time graph. Foundation students read gradients from straight lines; Higher students also estimate them on curves.

Foundation students interpret gradients of straight lines; chords, tangents and instantaneous rates on curves are Higher only.

Key facts

Gradient $= \dfrac{\text{change in } y}{\text{change in } x}$ , with the units ' $y$ units per $x$ unit'.
On a distance–time graph the gradient is speed; on a speed–time graph it is acceleration.
A negative gradient means the quantity is decreasing.
Higher: a chord between two points on a curve gives the average rate of change.
Higher: a tangent at a point gives the instantaneous rate of change at that moment.

Gradient of a straight line

Pick two points on the line far apart.
Work out the change in $y$ and the change in $x$ .
Divide, then say what it means in context with units.

Rate of change on a curve (Higher only)

Draw a tangent that just touches the curve at the point, with a ruler.
Pick two points far apart on your tangent.
Find its gradient; state the answer as an estimate with units.

Worked examples

Example 1 Foundation · grade 5

A straight-line graph shows the water in a tank falling from $600$ litres at $0$ minutes to $150$ litres at $15$ minutes. Find the gradient and explain what it means.

Show the solution

Change in volume $= 150 - 600 = -450$ litres
Change in time $= 15$ minutes
Gradient $= -450 \div 15 = -30$
The water flows out of the tank at $30$ litres per minute.

Example 2 Higher · grade 7

A distance–time curve starts at $(0, 0)$ and passes through $(4, 20)$ , where distance is in metres and time in seconds. A tangent drawn at $t = 4$ passes through $(1, 2)$ and $(7, 38)$ . (a) Estimate the speed at $t = 4$ . (b) Find the average speed over the first $4$ seconds.

Show the solution

(a) Gradient of tangent $= \frac{38 - 2}{7 - 1} = \frac{36}{6} = 6$
Speed at $t = 4$ is about $6$ m/s
(b) Chord from $(0, 0)$ to $(4, 20)$ : $20 \div 4 = 5$ m/s

Common mistakes

Dividing a single reading by the time (e.g. $20 \div 4$ ) when the question wants a tangent. That's an average, not the instantaneous rate.
Describing a gradient vaguely ('it goes down'). Give the rate with units.
Drawing a tangent that cuts through the curve. It should just touch at that point.

Exam tips

Draw your tangent long and pick far-apart points; answers within a range get the marks.
For 'what does the gradient represent', answer in context: 'the rate the water flows out, in litres per minute'.

Questions people ask

What does the gradient of a distance–time graph show?

The speed. A steeper line means faster, and a flat line means stopped.

How do you find the rate of change on a curved graph?

Draw a tangent at the point you want and work out its gradient. This is Higher tier.

Exam questions: Gradients and rates of change (9, easiest first)

Gradients and rates of change1 markGrade 5No calculatorOn both tiers

November 2024 · Paper 1 Foundation · Q24

Interpret the gradient of a volume–time graph in context.

Show answer

Answer	The rate at which the volume of water changes (litres leaving per second)	1 mark

Worked solution Checked against the mark scheme

The gradient is change in volume ÷ change in time.

Here $-20 \div 50 = -0.4$ , so water leaves the tank at 0.4 litres per second: the rate of change of volume.

Watch out: Saying 'negative correlation' or just 'the volume goes down'.

Gradients and rates of change3 marksGrade 7No calculatorHigher

November 2023 · Paper 1 Higher · Q15

Estimate the gradient of a curved distance-time graph at a given time by drawing a tangent.

Show answer

Answer	$25$ (accept 22 to 28)	3 marks

Worked solution Checked against the mark scheme

Draw a tangent touching the curve at $t = 3$ (where $d \approx 42$ ).

Pick two points on the tangent, for example $(2, 17)$ and $(4, 67)$ .

Gradient $\approx \frac{67 - 17}{4 - 2} = \frac{50}{2} = 25$ (metres per second)

Watch out: Using a chord from the origin (42 ÷ 3) instead of a tangent.

Gradients and rates of change4 marksGrade 7CalculatorHigher

June 2023 · Paper 3 Higher · Q17

Draw a tangent to a distance-time curve to estimate its gradient, and say what the gradient means.

Show answer

(a)	About 18 (accept 16.5 to 19.5)	3 marks
(b)	The speed (in m/s) at that time	1 mark

Worked solution Checked against the mark scheme

(a) Draw a straight line touching the curve at $t = 2.5$ (where the distance is about 21 m).

Pick two points on the tangent, e.g. $(1.4, 0)$ and $(4, 48)$ .

Gradient $= \dfrac{48 - 0}{4 - 1.4} \approx 18$

(b) Distance ÷ time is speed, so the gradient is the speed at 2.5 seconds, about 18 m/s.

Watch out: Using a chord between two points on the curve instead of a tangent, or reading the distance at 2.5 s.

Gradients and rates of change3 marksGrade 7CalculatorHigher

November 2021 · Paper 3 Higher · Q15

Estimate the instantaneous rate of flow from a volume–time curve by drawing a tangent and finding its gradient.

Show answer

Answer	About $0.9$ litres per second (accept $0.7$ to $1.1$ )	3 marks

Worked solution Checked against the mark scheme

Draw a tangent to the curve at $t = 12$ (where the volume is about 17 litres).

Pick two points far apart on the tangent, e.g. the tangent drops about 28 litres over 30 seconds.

Gradient $\approx 28 \div 30 \approx 0.93$ , so water flows out at about $0.9$ litres per second.

Watch out: Dividing the volume at t = 12 by 12 (an average rate) instead of drawing a tangent.

Gradients and rates of change3 marksGrade 7CalculatorHigher

November 2019 · Paper 3 Higher · Q19

Draw a tangent to a distance–time curve to estimate the speed at a given time.

Show answer

Answer	About $3.4$ m/s (3.05 to 3.7)	3 marks

Worked solution Checked against the mark scheme

Draw a tangent touching the curve at $t = 6$ (where the distance is about 10 m).

Pick two points on the tangent, e.g. it rises about 20 m over about 5.8 s.

Speed $= 20 \div 5.8 \approx 3.4$ m/s

Watch out: Dividing 10 by 6 (average speed from the start) instead of using a tangent.

Gradients and rates of change3 marksGrade 7CalculatorHigher

June 2018 · Paper 2 Higher · Q14

Estimate the speed at a given time by drawing a tangent to a distance-time curve.

Show answer

Answer	About $15$ m/s (11 to 19 accepted)	3 marks

Worked solution Checked against the mark scheme

Draw a tangent to the curve at $t = 5$ .

Pick two points far apart on the tangent and find rise ÷ run, e.g. a rise of about 60 m over 4 s.

Speed $\approx 60 \div 4 = 15$ m/s

Watch out: Dividing distance by time at $t = 5$ (about 30 ÷ 5) instead of using a tangent.

Gradients and rates of change4 marksGrade 7CalculatorHigher

June 2018 · Paper 3 Higher · Q15

Estimate the distance travelled from a speed–time curve using four strips, and say whether the estimate is too high or too low.

Show answer

(a)	About 498 m (488 to 507)	3 marks
(b)	Underestimate: the curve bends above the straight tops of the trapezia, so area under the curve is missed	1 mark

Worked solution Checked against the mark scheme

(a) Read speeds at 0, 5, 10, 15, 20 s: 0, 22, 28, 32, 35 m/s.

Each strip is a trapezium of width 5:

$2.5(0+22) + 2.5(22+28) + 2.5(28+32) + 2.5(32+35)$

$= 55 + 125 + 150 + 167.5 = 497.5$ m

(b) The curve bulges above the straight lines joining the points, so the trapezia leave out some area. It is an underestimate.

Watch out: Forgetting that the first strip starts at a speed of 0, or using a strip width of 4 instead of 5.

Gradients and rates of change4 marksGrade 8CalculatorHigher

November 2022 · Paper 2 Higher · Q21

Draw a tangent to a curved volume–time graph to estimate its gradient, then interpret what the gradient means.

Show answer

(a)	About $1.06$	3 marks
(b)	The rate at which the volume of water increases, in litres per second, at 17.5 seconds	1 mark

Worked solution Checked against the mark scheme

(a) Draw a tangent touching the curve at $t = 17.5$ .

Pick two points on it far apart and work out rise ÷ run, e.g. $\frac{18.5}{17.5} \approx 1.06$

(b) It is the rate the container fills: about 1.06 litres per second at $t = 17.5$ .

Watch out: Using a chord from the origin, 10.5 ÷ 17.5, which scores nothing.

Gradients and rates of change5 marksGrade 8CalculatorHigher

Specimen · Paper 2 Higher · Q20

Estimate an acceleration from a tangent to a velocity-time curve and the distance fallen using three trapezium strips.

Show answer

(a)	Between 3 and 4 m/s²	2 marks
(b)	452 m	3 marks

Worked solution Checked against the mark scheme

(a) Draw a tangent at $t = 6$ and find rise ÷ run, e.g. from about (2, 32) to (10, 60): $28 \div 8 = 3.5$ m/s².

(b) Strips of width 4: $v(0) = 0,\ v(4) = 35,\ v(8) = 51,\ v(12) = 54$

$\frac{4}{2}(0 + 35) = 70$

$\frac{4}{2}(35 + 51) = 172$

$\frac{4}{2}(51 + 54) = 210$

Total $= 70 + 172 + 210 = 452$ m

Watch out: In (a), using a chord from the origin to (6, 46) instead of a tangent.

Also tested in these questions

Their main topic is different, but part of each one tests this topic.

June 2024 · Paper 2 Higher · Q14: Estimate a car's deceleration from a tangent on a velocity-time graph and the distance travelled using three trapezium strips.
June 2022 · Paper 1 Higher · Q14: Estimate the gradient of a curved speed-time graph with a tangent and say what the area under it means.
November 2019 · Paper 2 Higher · Q21: Use the area under a speed–time graph to find when half the distance has been covered, then compare accelerations.
June 2019 · Paper 2 Higher · Q14: Estimate the gradient of a speed–time curve at a point, interpret it, and estimate distance travelled using four trapezium strips.
June 2018 · Paper 3 Higher · Q12: Find the gradient of a volume–time line graph and interpret its gradient and intercept.
Specimen · Paper 1 Higher · Q16: Match four sketch graphs to direct, inverse, square and inverse-square proportion.
Specimen · Paper 2 Higher · Q10: Find the gradient of a straight-line depth-time graph and say what it means.