Chapter 9: Review

The chapter problem, solved

The landing drone

The drone starts $40$ metres above the pad, and its height is

h(t) = 40e^{-t/4} - 2t

metres, $t$ seconds after the descent begins. It lands when $h(t) = 0$ .

Step 1. Bracket the landing time. Substitute $t = 5$ and $t = 6$ :

h(5) = 40e^{-1.25} - 10 = 1.4602 > 0, \qquad h(6) = 40e^{-1.5} - 12 = -3.0748 < 0.

The sign changes and $h$ is continuous, so the landing time lies between $5$ and $6$ seconds. The first-contact prediction is settled: the drone lands before $6$ seconds.

Step 2. Differentiate. The derivative is

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

Step 3. Iterate with Newton–Raphson. Use $t_{n+1} = t_n - \dfrac{h(t_n)}{h'(t_n)}$ :

t_0 = 5,

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

t_2 = 5.3001 - \frac{0.0315}{-4.6579} = 5.3069, \qquad t_3 = 5.3069 - \frac{0.00002}{-4.6535} = 5.3069.

The landing time is $5.307$ seconds to three decimal places.

Step 4. Interpret the answer. The rotors stop after about $5.3$ seconds. The third decimal place comes from the model, not from the real landing. Air movement, the shape of the drone and the exact start of the timing all change the landing by more than $0.001$ seconds, so the extra digits do not describe the real event.

The river survey

The engineer records the depth every two metres across a twelve metre channel.

Distance across (m)	$0$	$2$	$4$	$6$	$8$	$10$	$12$
Depth (m)	$0$	$1.2$	$2.5$	$3.1$	$2.4$	$0.9$	$0$

Step 1. Estimate the area. The readings are equally spaced with $h = 2$ and the interior depths add to $10.1$ . The trapezium rule gives

A \approx \frac{2}{2}\Bigl[0 + 0 + 2(10.1)\Bigr] = 20.2

square metres. This is smaller than the rectangle $12 \times 2 = 24$ , so the first-contact prediction is settled the other way.

Step 2. Give the limits. On each strip, take the smaller and the larger of its two readings:

A_{\text{lower}} = 2(0 + 1.2 + 2.5 + 2.4 + 0.9 + 0) = 14.0,

A_{\text{upper}} = 2(1.2 + 2.5 + 3.1 + 3.1 + 2.4 + 0.9) = 26.4.

The true area lies between $14.0$ and $26.4$ square metres. The estimate $20.2$ sits inside the limits.

Step 3. Find the flow rate. The water flows at about $0.8$ metres per second. The flow rate is the area multiplied by the speed:

Q \approx 20.2 \times 0.8 = 16.16.

The flow rate is about $16$ cubic metres per second, or $16$ m³/s.

Step 4. Say what the answer means. The area and the flow rate are estimates. The survey measured the depth at six points, and the depth between them is unknown. A second survey with more readings would change the number. The model also assumes that the speed is the same everywhere across the channel, which a real river does not guarantee.

Mixed practice — choose the method first

For each question, say which method you will use before you calculate. The items mix this chapter with earlier chapters.

Show that $e^{-x} = x$ has a root between $0$ and $1$ .
For $f(x) = x^3 - 4x + 1$ , write down $f'(x)$ .
Use Newton–Raphson on $x^3 - 4x + 1 = 0$ with $x_0 = 0.3$ . Find $x_2$ to four decimal places.
Use $x_{n+1} = \dfrac{1}{2}\left(x_n + \dfrac{3}{x_n}\right)$ with $x_0 = 2$ to find $x_2$ .
Use the trapezium rule with four strips to estimate $\displaystyle\int_0^1 (x^3 + 1)\,dx$ .
Find the exact value of $\displaystyle\int_0^1 (x^3 + 1)\,dx$ .
State whether the four-strip estimate $1.265625$ for $\displaystyle\int_0^1 (x^3 + 1)\,dx$ is too large or too small.
Using four strips and rectangles, give a lower limit and an upper limit for $\displaystyle\int_0^1 (x^3 + 1)\,dx$ .
Differentiate $y = x^2 \ln x$ . This uses Chapter 7.
Evaluate $\displaystyle\int_0^1 \left(3x^2 + 2x\right)dx$ . This uses Chapter 8.
Solve $2^x = 10$ , giving an exact answer for $x$ . This uses Chapter 6.
A van's speed is recorded every two seconds: $0, 6, 10, 12, 12$ metres per second at $t = 0, 2, 4, 6, 8$ seconds. Estimate the distance travelled.
Show that $x + \ln x = 3$ has a root between $2$ and $3$ .
For $f(x) = x + \ln x - 3$ , state the root to three decimal places.

Answers and concise solutions

1. The two values

Answer: Write $f(x) = e^{-x} - x$ . Then $f(0) = 1 > 0$ and $f(1) = e^{-1} - 1 = -0.6321 < 0$ . The sign changes, so the root lies in $(0, 1)$ .

2. The derivative

Answer: Differentiate term by term: $f'(x) = 3x^2 - 4$ .

3. The value of x2

Answer: $f'(0.3) = -3.73$ gives $x_1 = 0.2536$ , and the next step gives $x_2 = 0.2541$ to four decimal places. The root is $0.2541$ .

4. The value of x2

Answer: $x_1 = \dfrac{1}{2}\left(2 + \dfrac{3}{2}\right) = \dfrac{7}{4}$ and $x_2 = \dfrac{97}{56} = 1.7321$ . The recurrence is Newton–Raphson for $x^2 - 3 = 0$ , so it finds $\sqrt{3} = 1.7321$ .

5. The estimate

Answer: Here $h = 0.25$ and the heights are $1, 1.015625, 1.125, 1.421875, 2$ . The estimate is $\dfrac{0.25}{2}\bigl[3 + 2(3.5625)\bigr] = 1.265625 = \dfrac{81}{64}$ .

6. The exact value

Answer: $\left[\dfrac{x^4}{4} + x\right]_0^1 = \dfrac{1}{4} + 1 = \dfrac{5}{4} = 1.25$ .

7. The comparison

Answer: The estimate is larger than $1.25$ , so it is an overestimate. The second derivative is $6x$ , which is positive on $[0, 1]$ , so the curve bends upwards and each chord lies above it.

8. The limits

Answer: The function increases, so the lower sum uses the left-hand heights: $\dfrac{1}{4}\left(1 + \dfrac{65}{64} + \dfrac{9}{8} + \dfrac{91}{64}\right) = \dfrac{73}{64} = 1.140625$ . The upper sum uses the right-hand heights: $\dfrac{1}{4}\left(\dfrac{65}{64} + \dfrac{9}{8} + \dfrac{91}{64} + 2\right) = \dfrac{89}{64} = 1.390625$ . So $1.140625 < \dfrac{5}{4} < 1.390625$ .

9. The derivative

Answer: A product, so use the product rule: $2x\ln x + x^2 \times \dfrac{1}{x} = 2x\ln x + x$ . A root-finding method on this derivative would need this step.

10. The value of the integral

Answer: $\Bigl[x^3 + x^2\Bigr]_0^1 = (1 + 1) - 0 = 2$ . The trapezium rule would estimate this value; on this interval it would overestimate, because the curve bends upwards.

11. The value of x

Answer: Take logarithms of both sides: $x\ln 2 = \ln 10$ , so $x = \dfrac{\ln 10}{\ln 2} = 3.32$ to three significant figures.

12. The distance

Answer: The distance is the area under the speed–time graph. With $h = 2$ , $s \approx \dfrac{2}{2}\bigl[0 + 12 + 2(6 + 10 + 12)\bigr] = 68$ metres.

13. The two values

Answer: Write $f(x) = x + \ln x - 3$ . Then $f(2) = -0.3069 < 0$ and $f(3) = 1.0986 > 0$ . The sign changes, so a root lies in $(2, 3)$ .

14. The root

Answer: Newton–Raphson from $x_0 = 2$ gives $2.2046$ , then $2.2079$ , so the root is $2.208$ to three decimal places. The bracket $(2, 3)$ confirms which root this is.

What you can now do

bracket a root by change of sign, and narrow the bracket toward an answer;
explain how the change-of-sign test can fail;
use a recurrence relation $x_{n+1} = g(x_n)$ to find a root approximately;
draw and interpret cobweb and staircase diagrams;
use Newton–Raphson, and explain where the formula comes from;
show that a recurrence is Newton–Raphson for a chosen function;
explain how a zero derivative, a cycle or a distant start makes a method fail;
estimate an area with the trapezium rule, from a formula or from a table;
use the second derivative to decide the direction of the error;
give lower and upper limits for an area using rectangles;
choose a numerical method for a problem in context, and give the answer with its units and its accuracy.

Your relevance reflection

Return to the prompt in the chapter opener: a quantity that had no simple formula. Write the quantity again. Now add three things you could not have said before this chapter: how to turn the quantity into an equation, how to find a value by repeating a calculation, and how to estimate an area from readings. Finish with one sentence about a measurement you would now handle differently.