Chapter 10: Review

The chapter problem, solved

The rescue base is the origin $O$ of a map grid, with one unit to a kilometre. East is $\mathbf{i}$ and north is $\mathbf{j}$ .

Step 1. Write the data as vectors. The pad and the walker are at

\mathbf{p} = 2\mathbf{i} + \mathbf{j}, \qquad \mathbf{w} = 13\mathbf{i} + 7\mathbf{j}.

The two legs and the wind drift are

\mathbf{u} = 3\mathbf{i} + 4\mathbf{j}, \qquad \mathbf{v} = 4\mathbf{i} - 3\mathbf{j}, \qquad \mathbf{d} = \mathbf{i} + \mathbf{j}.

Step 2. Add the movements. Add the two legs first:

\mathbf{u} + \mathbf{v} = (3 + 4)\mathbf{i} + (4 - 3)\mathbf{j} = 7\mathbf{i} + \mathbf{j}.

Now add the wind drift:

\mathbf{u} + \mathbf{v} + \mathbf{d} = 8\mathbf{i} + 2\mathbf{j}.

Step 3. Find the final position. Add this total displacement to the pad:

\mathbf{f} = \mathbf{p} + 8\mathbf{i} + 2\mathbf{j} = (2 + 8)\mathbf{i} + (1 + 2)\mathbf{j} = 10\mathbf{i} + 3\mathbf{j}.

The drone finishes at $(10, 3)$ km. That is not the walker at $(13, 7)$ km.

Step 4. Find how far short the drone lands. The shortfall is the vector from the drone to the walker:

\vec{FW} = \mathbf{w} - \mathbf{f} = (13 - 10)\mathbf{i} + (7 - 3)\mathbf{j} = 3\mathbf{i} + 4\mathbf{j}.

Its magnitude is

|\vec{FW}| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5 \text{ km}.

The drone lands $5$ km from the walker.

Step 5. Find the direction of the walker from the drone. Both components of the shortfall are positive, so the walker is north-east of the drone. The angle from the positive $x$ -axis (east) is

\tan^{-1}\left(\frac{4}{3}\right) = 53.1^\circ,

so the walker lies $53.1^\circ$ north of east. As a bearing, measured clockwise from north, that is $90^\circ - 53.1^\circ = 36.9^\circ$ , a bearing of $037^\circ$ to the nearest degree.

Step 6. The final leg. The drone flies with velocity $\mathbf{v} = 6\mathbf{i} + 8\mathbf{j}$ km/h. Its speed is $\sqrt{6^2 + 8^2} = 10$ km/h. In half an hour it travels $0.5(6\mathbf{i} + 8\mathbf{j}) = 3\mathbf{i} + 4\mathbf{j}$ km, which is exactly the shortfall. So the drone reaches the walker after half an hour on this heading.

Step 7. The thrust against the wind. The wind force is $\mathbf{F} = -3\mathbf{i} - 4\mathbf{j}$ N. To hover, the thrust $\mathbf{T}$ must balance it:

\mathbf{T} = -\mathbf{F} = 3\mathbf{i} + 4\mathbf{j} \text{ N}.

The thrust has magnitude $5$ N and points $53.1^\circ$ north of east.

The chapter's answer. The drone finishes at $(10, 3)$ km, which is $5$ km south-west of the walker at $(13, 7)$ km. The walker lies $53.1^\circ$ north of east from the drone. A final leg of $3\mathbf{i} + 4\mathbf{j}$ km, flown at $10$ km/h, reaches the walker in half an hour. The thrust needed to hover against the wind is $5$ N at $53.1^\circ$ north of east.

Mixed practice — choose the method first

For each question, say which method you will use and why before you calculate. The items mix this chapter with earlier chapters, so the choice is not always obvious.

Add $\mathbf{a} = 3\mathbf{i} - 2\mathbf{j}$ and $\mathbf{b} = -2\mathbf{i} + 5\mathbf{j}$ .
Find the magnitude and the direction of $\mathbf{c} = 8\mathbf{i} - 6\mathbf{j}$ , giving the direction from the positive $x$ -axis.
Find the distance between $A(1, 2, 3)$ and $B(4, 6, 15)$ .
For $A(2, 3)$ and $B(7, 9)$ , find $\vec{AB}$ and the distance $AB$ .
Show that $A(2, 1)$ , $B(5, 7)$ and $C(8, 13)$ are collinear.
The points are $A(1, 2)$ and $B(11, 7)$ . Find the point $P$ on $AB$ with $AP : PB = 2 : 3$ .
Forces $\mathbf{F}_1 = 4\mathbf{i} - 3\mathbf{j}$ N and $\mathbf{F}_2 = -\mathbf{i} + 5\mathbf{j}$ N act at a point. Find the third force that keeps the point still, and its magnitude.
A particle starts at $\mathbf{r}_0 = \mathbf{i} - 2\mathbf{j}$ m and moves with velocity $\mathbf{v} = 4\mathbf{i} + 3\mathbf{j}$ m/s. Find its position after $2$ s and its distance from the origin then.
Solve $2x^2 - 7x + 3 = 0$ .
Solve $3^x = 20$ , giving an exact answer for $x$ .
Differentiate $y = 3x^2 - 4x + 1$ .
Find $\displaystyle\int \left(6x^2 + 2x\right)dx$ .
Solve $\sin x = \dfrac{1}{2}$ for $0 \le x \le 2\pi$ .
Prove that the sum of two consecutive odd numbers is a multiple of $4$ .
Use the trapezium rule with two strips to estimate $\displaystyle\int_0^2 x^2\,dx$ .
A drone moves with velocity $6\mathbf{i} + 8\mathbf{j}$ km/h for $1.5$ hours. Find its speed and the distance travelled.

Answers and concise solutions

1. The sum

Answer: Add the matching components: $\mathbf{a} + \mathbf{b} = (3 - 2)\mathbf{i} + (-2 + 5)\mathbf{j} = \mathbf{i} + 3\mathbf{j}$ .

2. The magnitude and the direction

Answer: The magnitude is $\sqrt{8^2 + (-6)^2} = \sqrt{100} = 10$ . The vector is in the fourth quadrant, so the direction is $\tan^{-1}\left(\dfrac{6}{8}\right) = 36.9^\circ$ below the positive $x$ -axis, which is $-36.9^\circ$ .

3. The distance

Answer: $\vec{AB} = 3\mathbf{i} + 4\mathbf{j} + 12\mathbf{k}$ , so $AB = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$ .

4. The displacement and the distance

Answer: $\vec{AB} = \mathbf{b} - \mathbf{a} = 5\mathbf{i} + 6\mathbf{j}$ . The distance is $AB = \sqrt{5^2 + 6^2} = \sqrt{61} \approx 7.81$ .

5. The two displacements

Answer: $\vec{AB} = 3\mathbf{i} + 6\mathbf{j}$ and $\vec{BC} = 3\mathbf{i} + 6\mathbf{j}$ . The displacements are equal and share the point $B$ , so the three points are collinear.

6. The position vector of $P$

Answer: With $m = 2$ and $n = 3$ , $\vec{OP} = \dfrac{3\mathbf{a} + 2\mathbf{b}}{5} = \dfrac{(3, 6) + (22, 14)}{5} = \dfrac{(25, 20)}{5} = (5, 4)$ . Then $AP = (4, 2)$ and $PB = (6, 3)$ , the ratio $2 : 3$ .

7. The third force and its magnitude

Answer: The point is still, so the three forces sum to zero. Now $\mathbf{F}_1 + \mathbf{F}_2 = 3\mathbf{i} + 2\mathbf{j}$ N, so the third force is $-3\mathbf{i} - 2\mathbf{j}$ N. Its magnitude is $\sqrt{(-3)^2 + (-2)^2} = \sqrt{13} \approx 3.61$ N.

8. The position and the distance

Answer: After $2$ s, $\mathbf{r} = \mathbf{r}_0 + 2\mathbf{v} = (\mathbf{i} - 2\mathbf{j}) + (8\mathbf{i} + 6\mathbf{j}) = 9\mathbf{i} + 4\mathbf{j}$ m. The distance from the origin is $\sqrt{9^2 + 4^2} = \sqrt{97} \approx 9.85$ m.

9. The solutions

Answer: Factorise: $(2x - 1)(x - 3) = 0$ . So $2x - 1 = 0$ or $x - 3 = 0$ , giving $x = \dfrac{1}{2}$ or $x = 3$ . This uses the quadratics from Chapter 2.

10. The value of $x$

Answer: Take logarithms of both sides: $x\ln 3 = \ln 20$ , so $x = \dfrac{\ln 20}{\ln 3} \approx 2.73$ . This uses the logarithms from Chapter 6.

11. The derivative

Answer: Differentiate term by term: $3 \times 2x - 4 = 6x - 4$ . This uses the differentiation from Chapter 7.

12. The integral

Answer: Integrate term by term: $6 \times \dfrac{x^3}{3} + 2 \times \dfrac{x^2}{2} + C = 2x^3 + x^2 + C$ . This uses the integration from Chapter 8.

13. The solutions

Answer: The sine of $x$ is positive, so $x$ is in the first or second quadrant. The first solution is $x = \dfrac{\pi}{6}$ , and the second is $x = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}$ . This uses the trigonometric equations from Chapter 5.

14. The algebra of the proof

Answer: Let the two consecutive odd numbers be $2n + 1$ and $2n + 3$ for an integer $n$ . Their sum is $(2n + 1) + (2n + 3) = 4n + 4 = 4(n + 1)$ , which has a factor of $4$ . The sum is a multiple of $4$ . This is proof by deduction, from Chapter 1.

15. The estimate

Answer: With two strips, each of width $h = 1$ , the ordinates at $x = 0, 1, 2$ are $0, 1, 4$ . The estimate is $\dfrac{h}{2}\left(y_0 + 2y_1 + y_2\right) = \dfrac{1}{2}(0 + 2 + 4) = 3$ . This uses the trapezium rule from Chapter 9.

16. The speed and the distance

Answer: The speed is the magnitude of the velocity: $\sqrt{6^2 + 8^2} = 10$ km/h. The distance travelled is speed $\times$ time $= 10 \times 1.5 = 15$ km.

What you can now do

write a vector in column form and in $\mathbf{i}$ , $\mathbf{j}$ , $\mathbf{k}$ form, and recognise equal and parallel vectors;
find the magnitude and the direction of a vector, and keep the correct quadrant;
convert between component form and magnitude/direction form, including bearings;
add vectors, subtract them and multiply them by a scalar;
use position vectors to find a displacement, a distance and a midpoint;
prove that points are collinear, that a quadrilateral is a parallelogram and that two diagonals bisect each other;
divide a segment in a given ratio;
find a resultant force and an unknown force in equilibrium;
model motion with a starting position and a constant velocity.

Your relevance reflection

Return to the prompt in the chapter opener: a journey with more than one stage, and something that pushes you along the way. Write the stages and the push again. Now add three things you could not have said before this chapter: how to combine the stages into one displacement, how to find the distance and direction of the result, and how a vector keeps a size and a direction in the same calculation. Finish with one sentence about a journey where the push mattered more than the stage you chose.