A15, R15 - Gradients and areas under graphs (Higher) | AQA GCSE Mathematics

A curve records more than its plotted heights. In A15 you will calculate or estimate gradients and areas, and interpret them in journeys and financial models. R15 connects a chord's gradient to an average rate and a tangent's gradient to the rate at one instant.

Average rates and chords (Higher)

A distance-time curve follows $s=t^2$ , with distance in metres and time in seconds. Between $1$ and $3$ seconds its average speed is $4\,\mathrm{m/s}$ , yet at $3$ seconds its speed is $6\,\mathrm{m/s}$ . Can both be correct? We need to separate a rate over an interval from a rate at one instant.

You already know

A straight line's gradient is vertical change divided by horizontal change. Speed is distance divided by time. Substituting into a function gives coordinates on its graph.

At $t=1$ , $s=1$ ; at $t=3$ , $s=9$ . The object covers $8$ metres in those $2$ seconds. Joining the endpoints with a straight line shows the steady rate that would produce the same overall change.

Chord and average rate of change

A chord is a straight line segment joining two points on a curve. Its gradient is the average rate of change of the vertical quantity over that input interval. The endpoints, rather than the curve's changing steepness between them, determine this average.

A distance-time curve may have changing speed throughout the interval. Its chord still gives a single average. On a balance-time graph, the same calculation gives average change in money per unit time.

Average rate from two points

For endpoints $(x_1,y_1)$ and $(x_2,y_2)$ , with $x_2\ne x_1$ ,

\text{average rate}=\frac{y_2-y_1}{x_2-x_1}

Use final minus initial in both differences. The units are vertical-axis units divided by horizontal-axis units.

Not on the formulae sheet: learn it.

You can obtain the endpoints from a graph, a table or an algebraic rule. The rate calculation is the same in all three forms. The particular model $s=t^2$ is given in the question; it is not a formula to memorise.

Worked example 1: the first interval

Question. For $s=t^2$ , work out the average speed between $t=1$ and $t=3$ . Non-calculator.

Plan. Calculate the two distances, then divide the distance change by the elapsed time.

\begin{aligned} s(1)&=1^2=1,\quad s(3)=3^2=9 &&\text{endpoint distances}\\ \text{change in distance}&=9-1=8\,\mathrm{m} &&\text{final minus initial}\\ \text{average speed}&=\frac{8}{3-1}=4\,\mathrm{m/s} &&\text{divide by elapsed time} \end{aligned}

Why this step? The question starts at $t=1$ , so neither the distance nor the time should be measured from zero.

Answer. $4\,\mathrm{m/s}$ , the gradient of the chord joining $(1,1)$ to $(3,9)$ .

Check. A steady speed of $4\,\mathrm{m/s}$ for $2$ seconds would cover the required $8$ metres.

Move only the interval later, keeping its duration at $2$ seconds. The curve is steeper there.

Worked example 2: a later interval

Question. For the same model, work out the average speed between $t=3$ and $t=5$ . Non-calculator.

Plan. Use the two new endpoints of the interval.

\begin{aligned} s(3)&=9,\quad s(5)=25 &&\text{substitute the endpoints}\\ \text{average speed}&=\frac{25-9}{5-3} &&\text{chord gradient}\\ &=\frac{16}{2}=8\,\mathrm{m/s} &&\text{evaluate} \end{aligned}

Why this step? An average depends on the chosen interval. A curve does not have one gradient that describes its whole length.

Answer. $8\,\mathrm{m/s}$ .

Check. This later chord is steeper than the first. Its average speed is greater, matching the increasing steepness of the distance-time curve.

[DIAGRAM: asset_name: Chords show average rates over different intervals; asset_slug: aqa8300_r09_chords; caption: Each chord represents the total change divided by the interval length; description: Two distance-time panels show the same blue quadratic s equals t squared. The first has a sage chord from 1 comma 1 to 3 comma 9, with a dashed right-angled triangle labelled rise 8 metres and run 2 seconds. The second has a sage chord from 3 comma 9 to 5 comma 25, with rise 16 metres and run 2 seconds.]
Each chord represents the total change divided by the interval length

If the table instead records a balance of £70 on day $2$ and £58 on day $5$ , the average change is $(58-70)/(5-2)=-4$ pounds per day. The negative sign means the balance falls, not that “negative days” have passed.

Watch out

Wrong: using $9/3=3$ for the average between $t=1$ and $t=3$ . That describes the interval from zero instead. Right: subtract both starting values, giving $(9-1)/(3-1)=4$ .

Choose the correct endpoints for another interval on the same kind of model.

Instantaneous rates and tangents (Higher)

For a rate at one instant, look at the curve's direction at that point. A long chord averages over changing directions. A tangent follows the local direction instead.

Tangent and instantaneous rate

A tangent is a straight line through a point on a smooth curve with the same local direction as the curve there. Its gradient gives the instantaneous rate of change at that input. On a distance-time graph it gives speed at that instant; on a velocity-time graph it gives acceleration.

To draw one, place a ruler through the required point and turn it until the line matches the curve's local direction. Extend it well either side. A tangent may cross a curve elsewhere; “touches the whole graph exactly once” is not a general definition.

Choose two widely separated, readable points on the tangent line. They do not both need to lie on the curve. Use a large right-angled gradient triangle to reduce the effect of reading errors.

Estimating a rate at one point

\text{instantaneous rate}\approx \frac{\text{vertical change along the tangent}}{\text{horizontal change along the tangent}}

The estimate depends on drawing and reading the tangent accurately. A horizontal tangent gives rate zero.

Not on the formulae sheet: learn it.

The first tangent below is drawn at $(3,9)$ on $s=t^2$ . The marked triangle uses $(2,3)$ and $(4,15)$ , points on the tangent, not points on the curve.

[DIAGRAM: asset_name: Tangent gradients for distance and money; asset_slug: aqa8300_r09_tangents; caption: Read two points on the tangent to estimate the rate at the marked point; description: The left panel shows s equals t squared, its tangent s equals 6t minus 9 at 3 comma 9, and tangent points 2 comma 3 and 4 comma 15. A dashed gradient triangle shows rise 12 and run 2. The right panel shows C equals 100 minus t squared, its tangent C equals 109 minus 6t at 3 comma 91, and tangent points 2 comma 97 and 4 comma 85, with a fall of 12 and run of 2.]
Read two points on the tangent to estimate the rate at the marked point

Worked example 3: speed at an instant

Question. A tangent to $s=t^2$ at $t=3$ passes through $(2,3)$ and $(4,15)$ . Estimate the speed at $t=3$ , with time in seconds and distance in metres. Non-calculator.

Plan. Use the tangent's two marked points to find its gradient.

\begin{aligned} \text{rise}&=15-3=12\,\mathrm{m} &&\text{tangent's vertical change}\\ \text{run}&=4-2=2\,\mathrm{s} &&\text{tangent's horizontal change}\\ \text{speed}&\approx12\div2=6\,\mathrm{m/s} &&\text{rise divided by run} \end{aligned}

Why this step? Using two points on the curve would give a chord and an average rate over an interval, rather than the rate at $t=3$ .

Answer. Approximately $6\,\mathrm{m/s}$ .

Check. The nearby chord from $t=2.9$ to $t=3.1$ has gradient $(3.1^2-2.9^2)/(3.1-2.9)=6$ , agreeing with the tangent.

Now use a falling graph. Only the sign of the local rate changes in the calculation; financial axes give the result a different meaning.

Worked example 4: a falling balance

Question. A balance-time curve is $C=100-t^2$ , with $C$ in pounds and $t$ in weeks. A tangent at $t=3$ passes through $(2,97)$ and $(4,85)$ . Estimate the instantaneous rate and interpret it. Non-calculator.

Plan. Calculate the signed tangent gradient, then attach the units from the axes.

\begin{aligned} \text{change in balance}&=85-97=-12 &&\text{final minus initial}\\ \text{change in time}&=4-2=2\,\mathrm{weeks} &&\text{horizontal interval}\\ \text{instantaneous rate}&\approx\frac{-12}{2}=-6 &&\text{pounds per week} \end{aligned}

Why this step? Keep the negative sign: the balance decreases as time increases. The units are pounds per week, rather than pounds alone.

Answer. Approximately $-6$ pounds per week. At that instant, the balance is falling at about £6 per week.

Check. The tangent's midpoint is $(3,91)$ , and $100-3^2=91$ , so the line passes through the required point.

The opening speeds are both correct. The average $4\,\mathrm{m/s}$ describes $1$ to $3$ seconds; $6\,\mathrm{m/s}$ describes just the instant at $3$ seconds. Here is a neat coincidence for this quadratic: the chord from $1$ to $5$ has gradient $(25-1)/4=6$ , equal to the tangent gradient at the midpoint time $3$ . Different questions can share an answer without meaning the same thing.

Watch out

Wrong: using $(2,4)$ and $(4,16)$ simply because they lie on $s=t^2$ . Those points define a chord. Right: draw the tangent at the stated time first, then use points on that line. When asked to show working, leave the tangent and gradient triangle visible.

Apply the same signed calculation to a new financial tangent.

Areas under straight sections (Higher)

Travelling at $6\,\mathrm{m/s}$ for $3$ seconds covers $18$ metres. On a speed-time graph that is a rectangle with height $6$ and width $3$ . Its area has the right units too: $\mathrm{m/s}\times\mathrm{s}=\mathrm{m}$ .

You already know

Rectangle area is width times height; triangle area is half base times perpendicular height. These two area rules are not on the formulae sheet: learn them. A straight sloping graph section encloses a triangle or trapezium with the time axis.

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A15, R15 - Gradients and areas under graphs (Higher)

Average rates and chords (Higher)

Instantaneous rates and tangents (Higher)

Areas under straight sections (Higher)

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