A graph carries a rate in its steepness. A chord between two points gives the average rate; a tangent at one point gives the instantaneous rate.
You can answer "how fast" from a picture with no formula at all. Two points and a ruler give the average rate; one point and a well-drawn tangent give the speed at that instant.
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Recall, your gut answer first
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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Recall, your gut answer first
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A distance-time graph rises steeply, then flattens, then rises steeply again. Describe what the object is doing in each part, using the word speed. Where is it fastest?
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Steepness is the rate
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Steepness is the rate
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The average rate of change between two points is the gradient of the chord joining them, $\dfrac{y_2 - y_1}{x_2 - x_1}$. The instantaneous rate at a point is the gradient of the tangent there. Same idea, two points versus one.
average rate $= \dfrac{y_2 - y_1}{x_2 - x_1}$ (chord) instantaneous rate $=$ gradient of the tangent
Read the units off the axes
A gradient on a distance-time graph is metres per second. On a cost-quantity graph it is dollars per item. The units come from the axes, vertical over horizontal.
A tangent touches, it does not cross
Draw the tangent so it just grazes the curve at the point, then use two points far apart on the tangent to find its gradient. Using points on the curve gives a chord instead.
Sign carries meaning
A negative gradient means the quantity is decreasing. On a distance-time graph that is motion back towards the start; on a cooling graph it is temperature falling.
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What you'll master
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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What you'll master
Know
Key facts
The average rate of change over an interval is the gradient of the chord across that interval.
The instantaneous rate of change at a point is the gradient of the tangent at that point.
The gradient carries units, read as the vertical axis unit per horizontal axis unit.
A negative gradient means the quantity is decreasing.
Understand
Concepts
Why a chord gives an average and a tangent gives an instant.
Why a tangent gradient must be estimated from two points on the tangent, not on the curve.
Why the same graph can have very different rates at different points.
Can do
Skills
Find an average rate of change from two points on a graph.
Draw a tangent and estimate an instantaneous rate from it.
Interpret a rate in the context of the situation, with units.
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Key terms
Average rate of changeThe change in the vertical quantity divided by the change in the horizontal one, over an interval. Like this: from $(1, 4)$ to $(5, 20)$ the average rate is $\dfrac{20 - 4}{5 - 1} = 4$.
ChordThe straight line joining two points on a curve. Like this: joining $(1, 4)$ and $(5, 20)$ on a curve gives a chord of gradient $4$.
Instantaneous rate of changeThe rate at one exact moment, given by the gradient of the tangent there. Like this: the speed shown on a car speedometer at one instant.
TangentA straight line touching a curve at one point and matching its steepness there. Like this: the tangent to a distance-time graph at $t = 3$ gives the speed at $t = 3$.
GradientRise divided by run, carrying the units of the two axes. Like this: $\dfrac{60 \text{ m}}{4 \text{ s}} = 15$ metres per second.
Interpreting a rateSaying what the number means in the situation, with units. Like this: a gradient of $15$ means the car is travelling at $15$ metres per second.
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Average rate of change, the gradient of a chord
Work through the core explanation before applying it.
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Average rate of change, the gradient of a chord
core concept
Pick two points on the graph and join them. That straight line is a chord, and its gradient is the average rate of change between those two points.
On a distance-time graph running from $(2, 30)$ to $(6, 90)$: the gradient is $\dfrac{90 - 30}{6 - 2} = \dfrac{60}{4} = 15$, so the average speed over that interval is $15$ metres per second.
It is an average because it ignores everything between the two points. The object may have sped up, slowed down, or stopped; the chord only reports the overall effect.
Units come from the axes. Vertical unit per horizontal unit, always. Metres over seconds gives metres per second; dollars over items gives dollars per item.
Quick check: a graph passes through $(1, 5)$ and $(4, 20)$. What is the average rate of change between them?
The average rate of change between two points is the gradient of the chord joining them, $\dfrac{y_2 - y_1}{x_2 - x_1}$. It reports the overall change and says nothing about what happened in between.
Pause, copy the chord definition, the worked $15$ metres per second, and the units rule, into your book.
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Instantaneous rate, the gradient of a tangent
core concept
We just saw how two points give an average. That raises a question: what if you want the rate at one exact moment? This card answers it → use the tangent at that point instead of a chord.
A tangent touches the curve at one point and matches its steepness there. Its gradient is the instantaneous rate of change at that point.
To estimate it: draw the tangent carefully, then pick two points that lie on the tangent and are far apart, and find the gradient between them. Far apart matters, because a small run magnifies any reading error.
If the tangent at $t = 4$ passes through $(2, 20)$ and $(8, 80)$, the instantaneous rate is $\dfrac{80 - 20}{8 - 2} = 10$ units per second.
Read the two points off the tangent, not the curve. Points taken from the curve give a chord gradient, which is the average over that interval and not the instantaneous rate you were asked for.
Which of these does NOT give an instantaneous rate of change?
The instantaneous rate at a point is the gradient of the tangent there. Estimate it from two widely separated points that lie on the tangent, never from points on the curve.
Pause, copy the tangent definition, the two-points-on-the-tangent method, and the worked $10$ units per second, into your book.
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Reading and interpreting a practical graph
core concept
We just saw how to get a number out of a graph. That raises a question: what does the number mean? This card answers it → the gradient is a rate in the situation, with units and a sign.
A flat section has gradient $0$: the quantity is not changing. On a distance-time graph that is an object at rest.
A negative gradient means the quantity is decreasing. On a cooling curve it is temperature falling; on a distance-time graph it is movement back towards the start.
Steeper means faster. Comparing two points on the same graph, the one with the steeper tangent has the greater instantaneous rate, whatever the units happen to be.
An answer without units is unfinished. "The rate is $15$" earns less than "the car is travelling at $15$ metres per second". Interpretation is usually its own mark.
Fill the blank: a horizontal section of a distance-time graph means the speed is .
Gradient $0$ means no change, a negative gradient means the quantity is decreasing, and a steeper tangent means a faster rate. Always state a rate with its units and say what it means in context.
Pause, copy the three cases (zero, negative, steeper) and the note that interpretation carries its own mark, into your book.
Worked examples · 3 in a row, reveal as you go
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Work examples end to end
Follow the reasoning through complete worked solutions.
PROBLEM 1 · AVERAGE RATE FROM A CHORD
A distance-time graph passes through $(2, 30)$ and $(6, 90)$, with distance in metres and time in seconds. Find the average speed over this interval.
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Average rate $= \dfrac{y_2 - y_1}{x_2 - x_1}$
The gradient of the chord.
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$= \dfrac{90 - 30}{6 - 2} = \dfrac{60}{4}$
Substitute the two points.
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$= 15$ metres per second
Units are metres over seconds, from the axes.
PROBLEM 2 · INSTANTANEOUS RATE FROM A TANGENT
The tangent to a curve at $t = 4$ passes through the points $(2, 20)$ and $(8, 80)$. Find the instantaneous rate of change at $t = 4$.
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Both $(2, 20)$ and $(8, 80)$ lie on the tangent, so use them directly
A tank drains so that its volume-time graph falls from $(0, 500)$ to $(10, 0)$, volume in litres and time in minutes. Find the average rate of change and say what it means.
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Average rate $= \dfrac{0 - 500}{10 - 0}$
Take the points in the same order top and bottom.
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$= -50$ litres per minute
Negative because the volume is decreasing.
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The tank empties at an average of $50$ litres per minute
The sign is carried by the word "empties", so it is not repeated.
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Quick-fire practice
Work through the core explanation before applying it.
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Quick-fire practice
+10 XP
A graph passes through $(1, 8)$ and $(5, 24)$. Find the average rate of change.
A tangent passes through $(0, 6)$ and $(4, 18)$. Find the instantaneous rate.
A distance-time graph is horizontal between $t = 3$ and $t = 7$. What is the speed there?
A graph falls from $(0, 90)$ to $(6, 30)$. Find the average rate of change and say what the sign means.
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Revisit the steep, flat, steep graph
Run the quick drill and copy the summary into your book.
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Revisit the steep, flat, steep graph
At the start you described an object whose distance-time graph rose steeply, flattened, then rose steeply again. Name the speed in the flat section, and explain how you could compare the two steep sections without measuring anything precisely.
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Multiple choice
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Multiple choice
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Short answer
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Short answer
ApplyBand 33 marks
Q1. A distance-time graph passes through $(1, 12)$ and $(5, 44)$, with distance in metres and time in seconds. Find the average speed over this interval, and explain why it may not equal the speed at $t = 3$. (3 marks)
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ApplyBand 43 marks
Q2. The tangent to a curve at $x = 5$ passes through $(1, 10)$ and $(9, 42)$. Find the instantaneous rate of change at $x = 5$, and state one thing that would make this estimate less reliable. (3 marks)
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UnderstandBand 42 marks
Q3. Explain the difference between the gradient of a chord and the gradient of a tangent, and say what each one measures. (2 marks)
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📖 Comprehensive answers (click to reveal)
Practice 1: $\dfrac{24 - 8}{5 - 1} = 4$. Practice 2: $\dfrac{18 - 6}{4 - 0} = 3$. Practice 3: $0$, the object is at rest. Practice 4: $\dfrac{30 - 90}{6 - 0} = -10$; the negative sign means the quantity is decreasing.
Q1 (3 marks): Average speed $= \dfrac{44 - 12}{5 - 1}$ [1]. $= \dfrac{32}{4} = 8$ metres per second [1]. This is an average across the whole interval, so the speed at $t = 3$ may be higher or lower; only the gradient of the tangent at $t = 3$ gives that [1].
Q2 (3 marks): Both points lie on the tangent, so gradient $= \dfrac{42 - 10}{9 - 1}$ [1]. $= \dfrac{32}{8} = 4$ units per unit of $x$ [1]. The estimate depends on how accurately the tangent was drawn, so a tangent drawn slightly off, or two points taken close together, would make it less reliable [1].
Q3 (2 marks): A chord joins two points on the curve, so its gradient is the average rate of change across that whole interval [1]. A tangent touches at one point, so its gradient is the instantaneous rate of change at that single point [1].
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Review and finish
Take the module quiz if you are ready, then mark the lesson complete.
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Boss battle · Rate Reader
earn bronze · silver · gold
Read average and instantaneous rates off graphs and interpret them in context. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.