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Date	May 2022	Marks available	5	Reference code	22M.2.AHL.TZ1.2
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 1
Command term	Find	Question number	2	Adapted from	N/A

Question

A sector of a circle, centre $O$ and radius $4.5 m$ , is shown in the following diagram.

A square field with side $8 m$ has a goat tied to a post in the centre by a rope such that the goat can reach all parts of the field up to $4.5 m$ from the post.

^{[Source: mynamepong, n.d. Goat [image online] Available at: https://thenounproject.com/term/goat/1761571/}
^{This file is licensed under the Creative Commons Attribution-ShareAlike 3.0 Unported (CC BY-SA 3.0)}
^{https://creativecommons.org/licenses/by-sa/3.0/deed.en [Accessed 22 April 2010] Source adapted.]}

Let $V$ be the volume of grass eaten by the goat, in cubic metres, and $t$ be the length of time, in hours, that the goat has been in the field.

The goat eats grass at the rate of $\frac{d V}{d t} = 0.3 t e^{- t}$ .

Find the angle $AÔB$ .

[3]

a.i.

Find the area of the shaded segment.

[5]

a.ii.

Find the area of the field that can be reached by the goat.

[4]

Find the value of $t$ at which the goat is eating grass at the greatest rate.

[2]

The goat is tied in the field for $8$ hours.

Find the total volume of grass eaten by the goat during this time.

[3]

Markscheme

$(\frac{1}{2} AÔB =) arccos (\frac{4}{4.5}) = 27.266 \dots$ (M1)(A1)

$AÔB = 54.532 \dots \approx 54.5 °$ ( $0.951764 \dots \approx 0.952$ radians) A1

Note: Other methods may be seen; award (M1)(A1) for use of a correct trigonometric method to find an appropriate angle and then A1 for the correct answer.

[3 marks]

a.i.

finding area of triangle

EITHER

area of triangle $= \frac{1}{2} \times 4 . 5^{2} \times \sin (54.532 \dots)$ (M1)

Note: Award M1 for correct substitution into formula.

$= 8.24621 \dots \approx 8.25 m^{2}$ (A1)

$AB = 2 \times \sqrt{4 . 5^{2} - 4^{2}} = 4.1231 \dots$ (M1)

area triangle $= \frac{4.1231 \dots \times 4}{2}$

$= 8.24621 \dots \approx 8.25 m^{2}$ (A1)

finding area of sector

EITHER

area of sector $= \frac{54.532 \dots}{360} \times π \times 4 . 5^{2}$ (M1)

$= 9.63661 \dots \approx 9.64 m^{2}$ (A1)

area of sector $= \frac{1}{2} \times 0.9517641 \dots \times 4 . 5^{2}$ (M1)

$= 9.63661 \dots \approx 9.64 m^{2}$ (A1)

THEN

area of segment $= 9.63661 \dots - 8.24621 \dots$

$= 1.39 m^{2} (1.39040 \dots)$ A1

[5 marks]

a.ii.

METHOD 1

$π \times 4 . 5^{2} (63.6172 \dots)$ (A1)

$4 \times 1.39040 . . . (5.56160)$ (A1)

subtraction of four segments from area of circle (M1)

$= 58.1 m^{2} (58.055 \dots)$ A1

METHOD 2

angle of sector $= 90 - 54.532 \dots (\frac{π}{2} - 0.951764 \dots)$ (A1)

area of sector $= \frac{90 - 54.532 \dots}{360} \times π \times 4 . 5^{2} (= 6.26771 \dots)$ (A1)

area is made up of four triangles and four sectors (M1)

total area $= (4 \times 8.2462 \dots) + (4 \times 6.26771 \dots)$

$= 58.1 m^{2} (58.055 \dots)$ A1

[4 marks]

sketch of $\frac{d V}{d t}$ OR $\frac{d V}{d t} = 0.110363 \dots$ OR attempt to find where $\frac{d^{2} V}{d t^{2}} = 0$ (M1)

$t = 1$ hour A1

[2 marks]

recognizing $V = \int \frac{d V}{d t} d t$ (M1)

$\int_{0}^{8} 0.3 t e^{- 1} d t$ (A1)

volume eaten is $0.299 \dots m^{3} (0.299094 \dots)$ A1

[3 marks]

Examiners report

Generally, this question was answered well but provided a good example of final marks being lost due to premature rounding. Some candidates gave a correct three significant figure intermediate answer of 27.3˚ for the angle in the right-angles triangle and then doubled it to get 54.6˚ as a final answer. This did not receive the final answer mark as the correct answer is 54.5˚ to three significant figures. Premature rounding needs to be avoided in all questions.

a.i.

[N/A]

a.ii.

Unfortunately, many candidates failed to see the connection to part (a). Indeed, the most common answer was to assume the goat could eat all the grass in a circle of radius 4.5m.

Most candidates completed this question successfully by graphing the function. A few tried to differentiate the function again and, in some cases, also managed to obtain the correct answer.

This was a question that was pleasingly answered correctly by many candidates who recognized that integration was needed to find the answer. As in part (c) a few tried to do the integration ‘by hand’, and were largely unsuccessful.

Syllabus sections

Topic 3—Geometry and trigonometry » SL 3.2—2d and 3d trig

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Hide related questions

Topic 3—Geometry and trigonometry » SL 3.4—The circle, arc and area of sector, degrees only

Topic 3—Geometry and trigonometry

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Markscheme

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