P1 4.1 - Gradients and rates of change | Edexcel A-Level International Mathematics

A straight line has one gradient, but a curve can be steep in one place and almost flat in another. Learn how a tangent gives a curve's gradient at one point, how derivative notation records that gradient, and how to interpret it as an instantaneous rate of change.

Secants approaching a tangent

How can a curve have a gradient when its steepness keeps changing? Begin with two points, where the familiar straight-line gradient still works, then bring those points closer together.

On $y=x^2$ , take $P=(1,1)$ and another point $Q$ . The segment joining them is a chord; the whole straight line through them is a secant. Its gradient describes the average change in $y$ per unit change in $x$ between the points.

Gradient between two points

For distinct points with $x_1\ne x_2$ ,

m=\frac{y_2-y_1}{x_2-x_1}.

Use the same order of subtraction in numerator and denominator.

Not in the formula booklet: learn it.

First keep $P$ fixed and choose a second point a short distance to its right.

Worked example 1: a chord gradient

Question. Find the gradient of the chord from $P=(1,1)$ to the point on $y=x^2$ where $x=1.5$ .

Plan. Find the second point's height, then divide its vertical change from $P$ by its horizontal change.

\begin{aligned} y_Q&=1.5^2=2.25 &&\text{point on the curve}\\ m_{PQ}&=\frac{2.25-1}{1.5-1} &&\text{change in height over change in }x\\ &=\frac{1.25}{0.5}=2.5 \end{aligned}

Answer. The chord gradient is $2.5$ .

Check. Moving $0.5$ units horizontally along this line changes its height by $2.5\times0.5=1.25$ , as required.

Change only how close $Q$ is to $P$ : reduce the horizontal separation from $0.5$ to $0.1$ .

Worked example 2: a closer second point

Question. Find the gradient of the chord from $P=(1,1)$ to the point on $y=x^2$ where $x=1.1$ .

Plan. Use the same gradient calculation with the closer point.

\begin{aligned} y_Q&=1.1^2=1.21\\ m_{PQ}&=\frac{1.21-1}{1.1-1}\\ &=\frac{0.21}{0.1}=2.1 \end{aligned}

Answer. The chord gradient is $2.1$ .

Check. The line through $P$ with this gradient gives $1+2.1(0.1)=1.21$ at $x=1.1$ , so it also passes through $Q$ .

The chord becomes a better description of the curve close to $P$ . The picture shows the line approaching a fixed position.

[DIAGRAM: asset_name: Chords approaching a tangent; asset_slug: chords-to-tangent; caption: As the second point approaches P on the parabola, the chord line approaches the tangent of gradient two.; description: Three equal-scale coordinate panels show y equals x squared near P at one comma one. Second points have x coordinates two, one point five and one point one, giving chord gradients three, two point five and two point one. The tangent y equals two x minus one is dashed in each panel.]
As the second point approaches P on the parabola, the chord line approaches the tangent of gradient two.

To see the exact limiting value, write the two points as $(a,a^2)$ and $(a+h,(a+h)^2)$ , where $h\ne0$ :

\frac{(a+h)^2-a^2}{h} =\frac{2ah+h^2}{h} =2a+h.

As $h$ approaches zero from either side, this gradient approaches $2a$ . At $a=1$ , the values $2.1,2.01,2.001$ approach $2$ from above; using $h=-0.1,-0.01,-0.001$ gives $1.9,1.99,1.999$ , approaching $2$ from below.

Derivative at a point

The derivative at a point is the gradient of the tangent there. When the chord gradients approach one finite value as the second point approaches from either side, that value is the tangent gradient.

A tangent describes the curve's direction locally. It need not meet the curve only once, and it can cross the curve elsewhere.

Watch out

Do not put $h=0$ into the original quotient: that gives $0/0$ , which is undefined. Simplify for $h\ne0$ , then consider the value approached as $h$ tends to zero.

Try a chord approaching from the other side.

Derivative notation

A derivative function records the tangent gradient at each allowed value of the input, rather than at just one point.

The calculation above worked for every real $a$ , so for $f(x)=x^2$ , the gradient function is $2x$ . At $x=3$ , the curve's height is $9$ , while its tangent gradient is $6$ . Those are different pieces of information.

Derivative notation

For $y=f(x)$ , write the derivative as

f'(x)\quad\text{or}\quad\frac{\mathrm{d}y}{\mathrm{d}x}.

Read $f'(x)$ as “f prime of x”, and $\mathrm{d}y/\mathrm{d}x$ as “the derivative of y with respect to x”. The operation of finding it is differentiation.

Here is the distinction for the square function:

Notation	Meaning	Value
$f(x)$	Height of the curve at a general input	$x^2$
$f(3)$	Height when $x=3$	$9$
$f'(x)$	Tangent gradient at a general input	$2x$
$f'(3)$	Tangent gradient when $x=3$	$6$

A prime changes the function being evaluated. You substitute into $f$ to find a height and into $f'$ to find a gradient. The notation $\left.\mathrm{d}y/\mathrm{d}x\right|_{x=3}=6$ says the same as $f'(3)=6$ .

A positive derivative means the curve rises as $x$ increases locally; a negative derivative means it falls. A zero derivative gives a horizontal tangent. It does not by itself tell you everything about the curve's shape.

Key idea

$f(a)$ locates the point's height; $f'(a)$ describes the curve's direction at that point. Neither value can generally replace the other.

Use the prime to choose which information is relevant.

Instantaneous rates and units

When the axes represent quantities, a tangent gradient measures how fast one quantity is changing with respect to the other at that instant.

If a distance changes from $9\,\mathrm{m}$ to $16\,\mathrm{m}$ between $3\,\mathrm{s}$ and $4\,\mathrm{s}$ , its average rate is $(16-9)/(4-3)=7\,\mathrm{m}\,\mathrm{s}^{-1}$ . This uses a whole interval. The derivative gives the rate at one instant by taking the limiting gradient.

Instantaneous rate of change

If $y$ depends on $x$ , then $\mathrm{d}y/\mathrm{d}x$ is the instantaneous rate of change of $y$ with respect to $x$ . Its units are the units of $y$ divided by the units of $x$ .

The same graph calculation can therefore describe a journey or a filling tank. Keep the variable and its units attached to the rate.

Worked example 3: average and instantaneous rates

Question. A moving object's distance $s$ metres after $t$ seconds is $s=t^2$ , for $t\ge0$ . Its derivative is $\mathrm{d}s/\mathrm{d}t=2t$ . Find its instantaneous rate at $t=3$ and its average rate from $t=3$ to $t=4$ .

Plan. Evaluate the derivative for the instantaneous rate, and use a difference quotient for the average rate.

\begin{aligned} \left.\frac{\mathrm{d}s}{\mathrm{d}t}\right|_{t=3}&=2(3)=6\\ \text{average rate}&=\frac{4^2-3^2}{4-3}=7 \end{aligned}

Answer. The instantaneous rate is $6\,\mathrm{m}\,\mathrm{s}^{-1}$ ; the average rate is $7\,\mathrm{m}\,\mathrm{s}^{-1}$ .

Check. The object covers $7\,\mathrm{m}$ in the one-second interval. Its rate increases from $6$ to $8\,\mathrm{m}\,\mathrm{s}^{-1}$ , so an average of $7$ fits.

Change only the interval's end from $4$ to $3.1$ . The instantaneous rate at $3$ stays fixed while the average rate moves closer to it.

Worked example 4: a shorter time interval

Question. For the same model $s=t^2$ , with $\mathrm{d}s/\mathrm{d}t=2t$ , find the instantaneous rate at $t=3$ and the average rate from $t=3$ to $t=3.1$ .

Plan. Keep the tangent rate, then calculate the gradient over the shorter interval.

\begin{aligned} \left.\frac{\mathrm{d}s}{\mathrm{d}t}\right|_{t=3}&=6\\ \text{average rate}&=\frac{3.1^2-3^2}{3.1-3} =\frac{0.61}{0.1}=6.1 \end{aligned}

Answer. The instantaneous rate is $6\,\mathrm{m}\,\mathrm{s}^{-1}$ ; the average rate is $6.1\,\mathrm{m}\,\mathrm{s}^{-1}$ .

Check. An average of $6.1$ over $0.1\,\mathrm{s}$ predicts $0.61\,\mathrm{m}$ , matching the change in $s$ .

A rate can be negative. For example, if $\mathrm{d}V/\mathrm{d}t=-4\,\mathrm{litres}\,\mathrm{min}^{-1}$ , a tank's volume is decreasing at $4$ litres per minute. The negative sign describes direction of change, not a negative volume.

Watch out

An average rate needs two input values. A derivative evaluated at one input gives an instantaneous rate. Always divide the dependent quantity's units by the input quantity's units.

Our opening problem now has a precise answer: a curve has a gradient at a point because nearby chord gradients can approach one tangent gradient.

Quick Recap

A chord gradient is an average change; its limiting value gives the tangent gradient.
$f'(x)$ and $\mathrm{d}y/\mathrm{d}x$ both denote the derivative when $y=f(x)$ .
Evaluate the derivative at the required input to obtain an instantaneous rate.
A rate's units are output units divided by input units; its sign tells you the direction of change.

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P1 4.1 - Gradients and rates of change

Secants approaching a tangent

Derivative notation

Instantaneous rates and units

Quick Recap

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