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Year 11 Methods (Unit 1 & 2) Rates Of Change

Recognising Relationships

20 practice questions 1 video lesson Theory + worked examples

Learn to recognise rates of change from graphs and tables for Queensland Year 11 Mathematical Methods (QCAA). A rate tells you how quickly one quantity changes as another does.

You will learn to spot increasing and decreasing intervals, tell a constant rate from a changing one, find where the rate is greatest, least or zero, and calculate an average rate of change — the intuition behind calculus in the QCAA course.

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 2), before any differentiation rules, you learn to recognise relationships from a graph or table: read the sign of the rate of change (rising, falling or flat), find where a quantity is increasing or decreasing, locate where the rate is greatest, least or zero, and tell a constant (straight-line) rate from a changing (curved) one. This page shows how to read the gradient of a graph as a rate of change.

The rate of change of a quantity tells you how fast it is changing as another quantity varies. On a graph of \(y\) against \(x\), the rate of change is the gradient: where the graph rises the rate is positive, where it falls the rate is negative, and where it is horizontal the rate is zero.

A constant rate produces a straight-line graph — the steepness is the same everywhere. A changing rate produces a curve: the graph gets steeper where the rate grows and flatter where it shrinks. The point where a curve is steepest has the greatest rate of change; a turning point (a peak or trough) has a rate of change of zero.

The average rate of change between two points is the gradient of the straight line joining them — the secant — found as change in \(y\) over change in \(x\).

Rising is positive, falling is negative, flat is zero. Read the direction and the steepness of the graph, and describe the rate in the words and units of the context.
Sign of the rate of change on a curveA curve rising to a peak then falling: positive rate on the rise, zero rate at the peak, negative rate on the fall. x y rising peak falling
Positive rate on the rise (gold), zero rate at the peak (grey), negative rate on the fall (red).
Constant rate is the gradient of a straight lineA straight distance-time line; the constant speed equals rise over run, sixty metres over four seconds. x y run 4 s rise 60 m
A straight line has a constant rate: the gradient is \(\dfrac{\text{rise}}{\text{run}}=\dfrac{60}{4}=15\) m/s.

The rate of change read from a graph is a gradient. Over an interval from \(x=a\) to \(x=b\), the average rate of change is

\[\text{average rate}=\dfrac{\text{change in }y}{\text{change in }x}=\dfrac{f(b)-f(a)}{b-a}\]
f(b)-f(a)b-a

For a straight-line graph this gradient is the same over every interval — the rate is constant:

\[\text{constant rate}=\dfrac{\text{rise}}{\text{run}}\]
riserun
Same gradient everywhere ⇔ straight line ⇔ constant rate. A curve whose gradient changes has a changing rate; it is steepest where the rate is greatest.

How to read a rate of change from a graph

  1. Direction: decide whether the graph is rising (positive rate), falling (negative rate) or horizontal (zero rate).
  2. Steepness: compare how steep the graph is — the steepest part has the greatest rate; a straight line has a constant rate.
  3. Describe: state the rate with its sign and, in a practical context, its units and meaning (for example metres per second).
Example 1 — Sign of the rate
A walker's distance from home \(d\) (metres) is graphed against time \(t\) (minutes). Describe the rate of change of \(d\) on each stage.
Solution

Read the direction of the graph on each stage.

Stage 1 — the graph rises from \((0,0)\) to \((8,400)\):

\(\text{rate}\)\(=\)\(\dfrac{400-0}{8-0}\)
\(=\)\(+50\text{ m/min (positive)}\)

Stage 2 — the graph is horizontal from \((8,400)\) to \((12,400)\):

\(\text{rate}\)\(=\)\(\dfrac{400-400}{12-8}\)
\(=\)\(0\text{ m/min (zero)}\)

Stage 3 — the graph falls from \((12,400)\) to \((20,200)\):

\(\text{rate}\)\(=\)\(\dfrac{200-400}{20-12}\)
\(=\)\(-25\text{ m/min (negative)}\)

Walking away at \(+50\) m/min, resting at \(0\) m/min, then returning at \(-25\) m/min.

Distance from home over timePiecewise graph: rising to four hundred metres, flat, then falling to two hundred metres. x y
+50,0,-25
Example 2 — Where the rate is greatest
Points \(P(1,1)\), \(Q(2,4)\) and \(R(3,9)\) lie on the curve \(y=x^2\). At which point is the rate of change greatest?
Solution

The rate of change is the steepness of the curve. On \(y=x^2\) the curve gets steeper as \(x\) increases, so compare the short intervals just before each point.

Steepness near \(P\), on \([0.5,\,1]\):

\(\text{gradient}\)\(=\)\(\dfrac{1-0.25}{1-0.5}\)
\(=\)\(1.5\)

Steepness near \(Q\), on \([1.5,\,2]\):

\(\text{gradient}\)\(=\)\(\dfrac{4-2.25}{2-1.5}\)
\(=\)\(3.5\)

Steepness near \(R\), on \([2.5,\,3]\):

\(\text{gradient}\)\(=\)\(\dfrac{9-6.25}{3-2.5}\)
\(=\)\(5.5\)

The curve is steepest at \(R(3,9)\), so the rate of change is greatest there.

A curve getting steeperParabola with three points P, Q, R; the curve is steeper further to the right. x y P Q R
R(3,9)
Example 3 — Average rate from a table
A tank drains as shown. Find the average rate of change of volume \(V\) (litres) over \([0,2]\) and over \([4,6]\) minutes, and say when it drains fastest.
Solution
\(t\) (min)0246
\(V\) (L)40302420

Average rate over \([0,2]\):

\(\dfrac{\Delta V}{\Delta t}\)\(=\)\(\dfrac{30-40}{2-0}\)
\(=\)\(\dfrac{-10}{2}\)
\(=\)\(-5\text{ L/min}\)

Average rate over \([4,6]\):

\(\dfrac{\Delta V}{\Delta t}\)\(=\)\(\dfrac{20-24}{6-4}\)
\(=\)\(\dfrac{-4}{2}\)
\(=\)\(-2\text{ L/min}\)

Both rates are negative (draining). \(|-5|>|-2|\), so the tank empties faster at the start — it is decreasing at a decreasing rate.

\([0,2]\): \(-5\) L/min; \([4,6]\): \(-2\) L/min; it drains fastest in the first two minutes.

Petrol draining over timeA decreasing curve that flattens: the tank drains quickly at first, then more slowly. x y
-5,-2
Example 4 — Rate before and after a peak
A ball's height is \(h=-5t^2+20t\) (metres), \(t\) in seconds. Describe how the rate of change of height behaves, and where it is zero.
Solution

Track the direction of the height as time increases. The graph is a parabola opening downwards with a peak.

Average rate on the way up, \([0,2]\):

\(\text{rate}\)\(=\)\(\dfrac{h(2)-h(0)}{2-0}\)
\(=\)\(\dfrac{20-0}{2}\)
\(=\)\(+10\text{ m/s (rising)}\)

Average rate on the way down, \([2,4]\):

\(\text{rate}\)\(=\)\(\dfrac{h(4)-h(2)}{4-2}\)
\(=\)\(\dfrac{0-20}{2}\)
\(=\)\(-10\text{ m/s (falling)}\)

The height rises to a peak of \(20\) m at \(t=2\) s, where the graph is momentarily horizontal, then falls.

The rate is positive before \(t=2\) s, zero at the peak \((2,20)\), and negative after.

Height of a ball over timeA parabola rising to a peak height of twenty metres at two seconds, then falling. x y peak
(2,20)

Common pitfalls

Confusing the value with the rate. A quantity can be large while its rate of change is small (a high, flat graph) — the rate is the steepness, not the height.
Reading a falling graph as a positive rate. If the graph goes down as you move right, the rate of change is negative, even though the numbers may still be positive.
Thinking a curve has a constant rate. Only a straight line has a constant rate. On a curve the rate changes from point to point; it is greatest where the curve is steepest.
Forgetting the units. In a practical context, always give the rate its units and meaning, such as \(50\) metres per minute.

Frequently asked questions

How do you tell the sign of a rate of change from a graph?

If the graph rises as you move right the rate is positive; if it falls the rate is negative; if it is horizontal the rate is zero.

Where is the rate of change greatest on a curve?

Where the curve is steepest. The rate of change is the gradient, so the steepest part of the graph has the greatest rate.

What does a straight-line graph tell you about the rate?

That the rate is constant — the same over every interval — because a straight line has the same gradient everywhere.

What is the average rate of change between two points?

The gradient of the straight line (the secant) joining them: \(\dfrac{f(b)-f(a)}{b-a}\), the change in \(y\) divided by the change in \(x\).

What does a rate of change of zero mean?

The quantity is momentarily not changing. On a graph this is a horizontal part, such as a peak, a trough or a flat section.

What is the difference between increasing at an increasing rate and at a decreasing rate?

Both are rising, so the rate is positive. Increasing at an increasing rate gets steeper (rate growing); increasing at a decreasing rate flattens out (rate shrinking).