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Date	May 2021	Marks available	1	Reference code	21M.2.AHL.TZ1.9
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 1
Command term	Show that	Question number	9	Adapted from	N/A

Question

Two boats $A$ and $B$ travel due north.

Initially, boat $B$ is positioned $50$ metres due east of boat $A$ .

The distances travelled by boat $A$ and boat $B$ , after $t$ seconds, are $x$ metres and $y$ metres respectively. The angle $θ$ is the radian measure of the bearing of boat $B$ from boat $A$ . This information is shown on the following diagram.

Show that $y = x + 50 cot θ$ .

[1]

a.

At time $T$ , the following conditions are true.

Boat $B$ has travelled $10$ metres further than boat $A$ .
Boat $B$ is travelling at double the speed of boat $A$ .
The rate of change of the angle $θ$ is $- 0.1$ radians per second.

Find the speed of boat $A$ at time $T$ .

[6]

b.

Markscheme

$\tan θ = \frac{50}{y - x}$ OR $cot θ = \frac{y - x}{50}$ A1

$y = x + 50 cot θ$ AG

Note: $y - x$ may be identified as a length on a diagram, and not written explicitly.

[1 mark]

a.

attempt to differentiate with respect to $t$ (M1)

$\frac{d y}{d t} = \frac{d x}{d t} - 50 {(cosec θ)}^{2} \frac{d θ}{d t}$ A1

attempt to set speed of $B$ equal to double the speed of $A$ (M1)

$2 \frac{d x}{d t} = \frac{d x}{d t} - 50 {(cosec θ)}^{2} \frac{d θ}{d t}$

$\frac{d x}{d t} = - 50 {(cosec θ)}^{2} \frac{d θ}{d t}$ A1

$θ = arctan 5 (= 1.373 \dots = 78.69 \dots °)$ OR ${cosec}^{2} θ = 1 + {cot}^{2} θ = 1 + {(\frac{1}{5})}^{2} = \frac{26}{25}$ (A1)

Note: This A1 can be awarded independently of previous marks.

$\frac{d x}{d t} = - 50 (\frac{26}{25}) \times - 0.1$

So the speed of boat $A$ is $5.2 ({ms}^{- 1})$ A1

Note: Accept $5.20$ from the use of inexact values.

[6 marks]

b.

Examiners report

[N/A]

a.

[N/A]

b.

Syllabus sections

Topic 3— Geometry and trigonometry » AHL 3.9—Reciprocal trig ratios and their pythagorean identities. Inverse circular functions

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Topic 3— Geometry and trigonometry