On a straight-line graph the rate of change is the same everywhere. On a curved graph it is different at every point — a car speeds up and slows down, a cup of tea cools quickly then slowly.
• Average rate of change between two points → find the gradient of the chord joining them.
• Instantaneous rate of change at one particular moment → find the gradient of the tangent at that point.
A chord is a straight line joining two points on the curve. Its gradient tells you the average rate of change over that interval.
- Read off the coordinates of the two points from the graph (or use the equation).
- Subtract to find the change in $y$ and the change in $x$.
- Divide.
- State the units and interpret the answer in context.
A ball's distance fallen is modelled by $d = 5t^2$, where $d$ is in metres and $t$ in seconds. Find the average speed between $t = 1$ and $t = 3$.
Answer: $20$ m/s
The ball is not travelling at $20$ m/s at any particular instant in that interval — that is just the average.
Using $d = 5t^2$ again, find the average speed over intervals that close in on $t = 2$.
| Interval | Change in $d$ | Change in $t$ | Average speed |
|---|---|---|---|
| $t = 2$ to $t = 3$ | $45 - 20 = 25$ | $1$ | $25$ m/s |
| $t = 2$ to $t = 2.5$ | $31.25 - 20 = 11.25$ | $0.5$ | $22.5$ m/s |
| $t = 2$ to $t = 2.1$ | $22.05 - 20 = 2.05$ | $0.1$ | $20.5$ m/s |
| $t = 2$ to $t = 2.01$ | $20.2005 - 20 = 0.2005$ | $0.01$ | $20.05$ m/s |
The values are closing in on $20$ m/s. That is the instantaneous speed at $t = 2$ — and it is exactly what the tangent at $t = 2$ would give you.
A tangent is a straight line that just touches the curve at one point, matching its steepness there.
- Mark the point on the curve where the rate is wanted.
- Lay a ruler so that it touches the curve at that point only, with equal gaps between the ruler and the curve on both sides.
- Draw a long line — the longer the tangent, the more accurate your reading.
- Pick two points on the tangent (not on the curve) that are easy to read.
- Find the gradient of the tangent and give the units.
On a distance–time graph, a tangent drawn at $t = 4$ s passes through the points $(2,\ 10)$ and $(6,\ 34)$, with distance in metres. Estimate the speed at $t = 4$.
The speed at $t = 4$ s is about $6$ m/s.
The word "estimate" is used because the tangent was drawn by eye.
A cup of coffee cools. A tangent to the temperature–time curve at $t = 10$ minutes passes through $(4,\ 78)$ and $(16,\ 42)$, with temperature in °C. Find the rate of cooling at $t = 10$.
The coffee is cooling at $3$ °C per minute at that moment.
The negative sign is important: it shows the temperature is falling.
| Shape | Gradient | Meaning on a distance–time graph |
|---|---|---|
| Rising, getting steeper | Positive and increasing | Speeding up |
| Rising, getting shallower | Positive and decreasing | Slowing down (still moving forward) |
| Horizontal at a peak | Zero | Momentarily stationary |
| Falling, getting steeper | Negative, getting more negative | Returning, and speeding up |
| Falling, getting shallower | Negative, approaching zero | Returning, and slowing down |
The graph of the depth of water in a draining bath is a curve that falls steeply at first and then more gently. Explain how the rate of draining changes, without doing any calculations.
The rate of change is decreasing in magnitude throughout.
The gradient of a graph gives a rate. The area underneath gives a total.
For a curve you cannot use one simple shape, so you estimate the area by splitting it into strips — usually trapeziums.
A car's speed in m/s is recorded every $2$ seconds:
| Time (s) | $0$ | $2$ | $4$ | $6$ |
|---|---|---|---|---|
| Speed (m/s) | $0$ | $7$ | $12$ | $15$ |
Estimate the distance travelled in the first $6$ seconds.
Chord
Joins two points; its gradient is the average rate of change.
Tangent
Touches at one point; its gradient is the instantaneous rate.
Drawing a tangent
Long line, equal gaps either side, then use points on the line.
Sign
Negative gradient means the quantity is decreasing.
Steepness
Steeper tangent → faster rate of change at that moment.
Area
Area under a speed–time graph $=$ distance travelled.
Trapezium rule
Each strip: $\tfrac{1}{2}(a+b)h$; add them all up.
Estimate wording
Say "estimate" — hand-drawn tangents and strips are approximate.
A curve passes through $(2,\ 7)$ and $(8,\ 43)$. Find the average rate of change between these points.
▶ Show solution
Gradient of the chord $= \dfrac{43 - 7}{8 - 2} = \dfrac{36}{6} = 6$
A tangent to a distance–time curve at $t = 5$ s passes through $(1,\ 6)$ and $(9,\ 46)$, with distance in metres. Estimate the speed at $t = 5$ s.
▶ Show solution
Gradient $= \dfrac{46 - 6}{9 - 1} = \dfrac{40}{8} = 5$
Speed $\approx 5$ m/s
The height of a plant is modelled by $h = t^2 + 3t$, where $h$ is in cm and $t$ in weeks. Find the average growth rate between weeks $2$ and $6$.
▶ Show solution
At $t = 2$: $h = 4 + 6 = 10$ cm.
At $t = 6$: $h = 36 + 18 = 54$ cm.
Average rate $= \dfrac{54 - 10}{6 - 2} = \dfrac{44}{4} = 11$ cm per week.
A tangent to a temperature–time curve at $t = 8$ minutes passes through $(2,\ 84)$ and $(14,\ 30)$, with temperature in °C. Find the rate of change and describe what it means.
▶ Show solution
Gradient $= \dfrac{30 - 84}{14 - 2} = \dfrac{-54}{12} = -4.5$
The temperature is falling at $4.5$ °C per minute at $t = 8$ minutes.
For the curve $y = x^2$, calculate the gradient of the chord from $x = 3$ to (a) $x = 4$, (b) $x = 3.5$, (c) $x = 3.1$. What value are these closing in on?
▶ Show solution
(a) $\dfrac{16 - 9}{4 - 3} = 7$
(b) $\dfrac{12.25 - 9}{0.5} = \dfrac{3.25}{0.5} = 6.5$
(c) $\dfrac{9.61 - 9}{0.1} = \dfrac{0.61}{0.1} = 6.1$
The values are approaching $6$ — the gradient of the tangent at $x = 3$.
Estimate the distance travelled in the first $8$ seconds from this speed table, using trapezium strips.
| Time (s) | $0$ | $2$ | $4$ | $6$ | $8$ |
|---|---|---|---|---|---|
| Speed (m/s) | $0$ | $5$ | $9$ | $12$ | $14$ |
▶ Show solution
$\tfrac{1}{2}(0+5)\times 2 = 5$
$\tfrac{1}{2}(5+9)\times 2 = 14$
$\tfrac{1}{2}(9+12)\times 2 = 21$
$\tfrac{1}{2}(12+14)\times 2 = 26$
Total $\approx 5 + 14 + 21 + 26 = \mathbf{66}$ m
A distance–time graph curves upwards, becoming steeper. At $t = 2$ the tangent has gradient $3$; at $t = 6$ it has gradient $11$. Describe the motion and find the average acceleration between these times.
▶ Show solution
The gradient is the speed, so the object speeds up from $3$ m/s to $11$ m/s — it is accelerating.
Average acceleration $= \dfrac{11 - 3}{6 - 2} = \dfrac{8}{4} = 2\text{ m/s}^2$
Water is poured into a container. The graph of depth against time is a curve that rises steeply at first and then more gradually. Explain what this tells you about the shape of the container.
▶ Show solution
A steep gradient means the depth rises quickly, which happens where the container is narrow.
A shallow gradient means the depth rises slowly, which happens where the container is wide.
So the container is narrow at the bottom and gets wider towards the top — a cone or bowl shape standing point-down.
A rocket's height is modelled by $h = 40t - 5t^2$ metres, where $t$ is in seconds.
(a) Find the average velocity between $t = 1$ and $t = 3$. (b) Find the average velocity between $t = 5$ and $t = 7$. (c) Explain the sign of your answer to (b).
▶ Show solution
(a) $h(1) = 40 - 5 = 35$ m; $h(3) = 120 - 45 = 75$ m.
Average velocity $= \dfrac{75 - 35}{2} = 20$ m/s.
(b) $h(5) = 200 - 125 = 75$ m; $h(7) = 280 - 245 = 35$ m.
Average velocity $= \dfrac{35 - 75}{2} = -20$ m/s.
(c) The negative sign shows the rocket is falling over that interval — it has passed its highest point (at $t = 4$) and is coming back down.
A speed–time graph for a car over $10$ seconds gives these readings.
| Time (s) | $0$ | $2.5$ | $5$ | $7.5$ | $10$ |
|---|---|---|---|---|---|
| Speed (m/s) | $20$ | $15$ | $11$ | $8$ | $6$ |
(a) Estimate the distance travelled. (b) Is your estimate too big or too small? Explain. (c) Estimate the deceleration at $t = 5$ s using the readings either side.
▶ Show solution
(a) Strip width $h = 2.5$.
$\tfrac{1}{2}(20+15)(2.5) = 43.75$
$\tfrac{1}{2}(15+11)(2.5) = 32.5$
$\tfrac{1}{2}(11+8)(2.5) = 23.75$
$\tfrac{1}{2}(8+6)(2.5) = 17.5$
Total $\approx 43.75 + 32.5 + 23.75 + 17.5 = \mathbf{117.5}$ m
(b) The speed is falling and the curve is concave up (the drops get smaller: $5$, $4$, $3$, $2$), so the straight strip tops lie above the curve. The estimate is therefore an over-estimate.
(c) Use the chord from $t = 2.5$ to $t = 7.5$ as an approximation to the tangent at $t = 5$:
Gradient $= \dfrac{8 - 15}{7.5 - 2.5} = \dfrac{-7}{5} = -1.4$
The car is decelerating at about $1.4\text{ m/s}^2$.