9.5 Numerical methods in context — Practise with help

Work each question closed book first. Decide which method you need before you calculate, and write the units with the answer.

Warm-up: remember

For $f(x) = x^3 - 2x - 5$ , work out $f(2)$ and $f(3)$ . What do the signs tell you?
Use $x_{n+1} = \dfrac{x_n + 5}{3}$ with $x_0 = 1$ to find $x_2$ .

Answers

$f(2) = -1$ and $f(3) = 16$ . The sign changes, so a root lies in $(2, 3)$ .
$x_1 = \dfrac{1 + 5}{3} = 2$ and $x_2 = \dfrac{2 + 5}{3} = \dfrac{7}{3} = 2.3333$ .

If the first part was hard, revise change of sign in Section 9.1. If the second was hard, revise recurrence relations in Section 9.2.

Practise with help

1. A drone's height is $h(t) = 40e^{-t/4} - 2t$ metres. Show that the drone lands between $5$ and $6$ seconds.

Hint

Substitute both times into the height model. The drone lands when the height is zero.

Next step

Use $e^{-1.25} = 0.2865$ and $e^{-1.5} = 0.2231$ .

Full working

h(5) = 40(0.2865) - 10 = 1.4602 > 0, \qquad h(6) = 40(0.2231) - 12 = -3.0748 < 0.

The height changes sign, and the model is continuous, so the drone lands between $5$ and $6$ seconds.

2. For the drone in question 1, find $h'(t)$ and use Newton–Raphson with $t_0 = 5$ to find $t_1$ .

Hint

Differentiate $40e^{-t/4}$ and $-2t$ separately.

Next step

$h'(t) = -10e^{-t/4} - 2$ , so $h'(5) = -4.8650$ .

Full working

h'(t) = -10e^{-t/4} - 2.

t_1 = 5 - \frac{1.4602}{-4.8650} = 5.3001.

3. Continue question 2 to find $t_2$ , and state the landing time to three decimal places.

Hint

Use $t_1$ in the same formula. Carry the full value, not the rounded one.

Next step

$h(5.3001) = 0.0315$ and $h'(5.3001) = -4.6579$ .

Full working

t_2 = 5.3001 - \frac{0.0315}{-4.6579} = 5.3069.

One more step gives $5.3069$ again, so the landing time is $5.307$ seconds to three decimal places.

4. A drug's concentration is $C(t) = 6te^{-0.4t}$ milligrams per litre, $t$ hours after a dose. Form an equation for the time when the concentration is $2$ milligrams per litre, and bracket the root.

Hint

Set $C(t) = 2$ and divide both sides by $2$ . Then multiply by $e^{0.4t}$ .

Next step

The equation becomes $3t = e^{0.4t}$ , so use $f(t) = 3t - e^{0.4t}$ .

Full working

6te^{-0.4t} = 2 \quad\Longrightarrow\quad 3te^{-0.4t} = 1 \quad\Longrightarrow\quad 3t = e^{0.4t}.

Let $f(t) = 3t - e^{0.4t}$ . Then

f(0.35) = 1.05 - e^{0.14} = -0.1003 < 0, \qquad f(0.4) = 1.2 - e^{0.16} = 0.0265 > 0.

The sign changes, so the root lies in $(0.35, 0.4)$ hours.

5. Use Newton–Raphson from $t_0 = 0.4$ to find the time in question 4 to two decimal places.

Hint

Differentiate $f(t) = 3t - e^{0.4t}$ . The derivative of $e^{0.4t}$ is $0.4e^{0.4t}$ .

Next step

$f'(0.4) = 3 - 0.4e^{0.16} = 2.5306$ , so one step gives $0.4 - \dfrac{0.0265}{2.5306}$ .

Full working

f'(t) = 3 - 0.4e^{0.4t}, \qquad t_1 = 0.4 - \frac{0.0265}{2.5306} = 0.3895.

The next step gives $0.3895$ again, so the time is $0.39$ hours to two decimal places. That is about $23.4$ minutes.

6. A car's speed is recorded every two seconds for ten seconds.

Time (s)	$0$	$2$	$4$	$6$	$8$	$10$
Speed (m/s)	$0$	$5$	$9$	$12$	$14$	$15$

Estimate the distance travelled, and give lower and upper limits.

Hint

Distance is the area under the speed–time graph. The readings are equally spaced with $h = 2$ .

Next step

The interior heights are $5, 9, 12, 14$ , so they add to $40$ and twice that is $80$ .

Full working

s \approx \frac{2}{2}\Bigl[0 + 15 + 2(5 + 9 + 12 + 14)\Bigr] = 15 + 80 = 95.

The car travels about $95$ metres.

The speed increases throughout, so the lower sum uses the left-hand heights and the upper sum uses the right-hand heights:

s_{\text{lower}} = 2(0 + 5 + 9 + 12 + 14) = 80, \qquad s_{\text{upper}} = 2(5 + 9 + 12 + 14 + 15) = 110.

The distance lies between $80$ and $110$ metres.

7. A tank empties through a valve. The depth $d$ centimetres after $t$ minutes is recorded in a table, and the depth does not follow a formula you can integrate. You need the area under the depth–time graph for the first ten minutes. Which method will you use, and why?

Hint

Ask whether the unknown is a root of an equation or an area under a graph.

Next step

The depth readings are heights, and the unknown is the area beneath them. There is no formula, so an exact integral is not available.

Full working

Use the trapezium rule. The readings are equally spaced, so $h = \dfrac{b - a}{n}$ is the gap between readings. Substitute the depths into

A \approx \frac{h}{2}\Bigl[y_0 + y_n + 2\left(y_1 + \cdots + y_{n-1}\right)\Bigr].

Give the answer in centimetre-minutes, then convert if the question needs square metres. State that the answer is an estimate, and give limits from the smaller and larger readings on each strip.