7.1 The derivative as a gradient and a rate of change — Practise with help

Work each question closed book first. Sketch the chord or the tangent before you calculate, and write down what the gradient means in the context.

Warm-up: remember

Find the gradient of the straight line through $(1, 3)$ and $(4, 15)$ .
For $f(x) = x^2 - 3x$ , work out $f(4)$ .
Make $h$ the subject of $4 = 8h$ .

Answers

Gradient $= \dfrac{15 - 3}{4 - 1} = \dfrac{12}{3} = 4$ .
$f(4) = 4^2 - 3 \times 4 = 16 - 12 = 4$ .
Divide both sides by $8$ : $h = \dfrac{4}{8} = 0.5$ .

If question 1 was hard, revise the gradient of a straight line in Chapter 3.

Practise with help

1. The curve $y = x^2 + 1$ passes through $P(1, 2)$ . Use chords from $P$ to $(1 + h, (1 + h)^2 + 1)$ to estimate the gradient at $P$ . Take $h = 1$ , $h = 0.5$ and $h = 0.1$ .

Hint

The gradient of each chord is $\dfrac{f(1 + h) - f(1)}{h}$ . Work out $f(1 + h)$ for each value of $h$ , then subtract $2$ and divide by $h$ .

Next step

For $h = 1$ the chord runs to $x = 2$ , where $y = 5$ . Its gradient is $\dfrac{5 - 2}{1}$ . Repeat for $h = 0.5$ and $h = 0.1$ .

Full working

$h$	$x = 1 + h$	$y = x^2 + 1$	gradient of the chord
$1$	$2$	$5$	$\dfrac{5 - 2}{1} = 3$
$0.5$	$1.5$	$3.25$	$\dfrac{3.25 - 2}{0.5} = 2.5$
$0.1$	$1.1$	$2.21$	$\dfrac{2.21 - 2}{0.1} = 2.1$

The values close in on $2$ , so the gradient at $P$ is about $2$ .

2. A curve has gradient function $f'(x) = 2x - 4$ . Find the value of $x$ where the gradient is zero. Then say whether the curve is rising or falling at $x = 0$ and at $x = 3$ .

Hint

The gradient is zero when the expression for $f'(x)$ equals zero. To say whether the curve rises or falls, work out the sign of $f'(x)$ at each value of $x$ .

Next step

Solve $2x - 4 = 0$ for the zero gradient. Then substitute $x = 0$ and $x = 3$ into $2x - 4$ and record the sign of each answer.

Full working

Set $2x - 4 = 0$ , so $2x = 4$ and $x = 2$ . The gradient is zero at $x = 2$ .

At $x = 0$ , $f'(0) = -4$ , which is negative, so the curve is falling. At $x = 3$ , $f'(3) = 2$ , which is positive, so the curve is rising.

3. A stone falls. Its distance from the top of a cliff after $t$ seconds is $s = 3t^2 + 2t$ metres. Find the average speed between $t = 1$ and $t = 3$ seconds.

Hint

Average speed is the change in distance divided by the change in time. Work out $s$ at the two times first.

Next step

$s(1) = 3 + 2 = 5$ m and $s(3) = 27 + 6 = 33$ m. Divide the change in distance by the $2$ seconds between the two times.

Full working

\text{average speed} = \frac{s(3) - s(1)}{3 - 1} = \frac{33 - 5}{2} = 14 \text{ m/s}.

This is the average rate over two seconds, not the speed at one instant.

4. Use chords to estimate the gradient of $y = x^3$ at $x = 2$ . Take $h = 1$ , $h = 0.5$ and $h = 0.1$ , and give each chord gradient to two decimal places.

Hint

The point is $(2, 8)$ . The chord runs to $(2 + h, (2 + h)^3)$ . Work out the cube of each new $x$ , then apply $\dfrac{f(2 + h) - 8}{h}$ .

Next step

For $h = 0.5$ the chord runs to $x = 2.5$ , where $y = 15.625$ . The chord gradient is $\dfrac{15.625 - 8}{0.5}$ . Repeat for the other two values of $h$ .

Full working

$h$	$x = 2 + h$	$y = x^3$	gradient of the chord
$1$	$3$	$27$	$\dfrac{27 - 8}{1} = 19.00$
$0.5$	$2.5$	$15.625$	$\dfrac{15.625 - 8}{0.5} = 15.25$
$0.1$	$2.1$	$9.261$	$\dfrac{9.261 - 8}{0.1} = 12.61$

As $h$ shrinks the gradient approaches $12$ , so the gradient at $x = 2$ is about $12$ .

5. A curve rises from $(-2, -1)$ to a peak at $(0, 3)$ . It then falls to $(2, -1)$ and rises after that. State the sign of $f'(x)$ on the rising part, at the peak, and on the falling part. Then describe the shape of the graph of $f'(x)$ .

Hint

A rising curve has a positive gradient, a falling curve has a negative gradient, and the gradient is zero at a peak.

Next step

Mark the rising part from $x = -2$ to $x = 0$ , the peak at $x = 0$ , and the falling part from $x = 0$ to $x = 2$ . Write the sign of the gradient on each part, then transfer those signs to a graph.

Full working

On the rising part, $f'(x) > 0$ . At the peak, $f'(0) = 0$ . On the falling part, $f'(x) < 0$ . After $x = 2$ the curve rises again, so $f'(x) > 0$ once more.

The graph of $f'(x)$ starts above the $x$ -axis, meets it at $x = 0$ and again at $x = 2$ , and sits below the axis between those two points. Outside the interval it sits above the axis.

6. A cup of tea cools. Its temperature after $t$ minutes is $\theta$ degrees Celsius. The temperature is $90^\circ$ C at $t = 0$ and $65^\circ$ C at $t = 5$ . Find the average rate of change of the temperature. Give the units, and explain the sign.

Hint

The average rate is the change in temperature divided by the time taken. The temperature falls, so expect a negative answer.

Next step

The change is $65 - 90 = -25$ degrees over $5$ minutes. Divide the change by the time.

Full working

\frac{\theta(5) - \theta(0)}{5 - 0} = \frac{65 - 90}{5} = -5.

The average rate of change is $-5$ degrees Celsius per minute. The sign is negative because the drink is cooling, so its temperature is falling.