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Date May 2022 Marks available 2 Reference code 22M.2.SL.TZ1.4
Level Standard Level Paper Paper 2 Time zone Time zone 1
Command term Find Question number 4 Adapted from N/A

Question

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

A square field with side 8m 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 4.5m 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 dVdt=0.3te-t.

Find the angle AÔB.

[3]
a.i.

Find the area of the shaded segment.

[5]
a.ii.

Find the area of a circle with radius 4.5m.

[2]
b.i.

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

[3]
b.ii.

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

[2]
c.

Markscheme

12AÔB=arccos44.5=27.266        (M1)(A1)

AÔB=54.53254.5°  (0.9517640.952 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 =12×4.52×sin54.532        (M1)


Note: Award M1 for correct substitution into formula.


=8.246218.25 m2        (A1)

OR

AB=2×4.52-42=4.1231

area triangle =4.1231×42        (M1)

=8.246218.25 m2        (A1)

 

finding area of sector

EITHER

area of sector =54.532360×π×4.52        (M1)

=9.636619.64 m2        (A1)

OR

area of sector =12×0.9517641×4.52        (M1)

=9.636619.64 m2        (A1)

 

THEN

area of segment =9.63661-8.24621

=1.39 m2  1.39040        A1 

 

[5 marks]

a.ii.

π×4.52         (M1)

63.6m2   63.6172m2        A1 

 

[2 marks]

b.i.

METHOD 1

4×1.39040...   (5.56160)         (A1)

subtraction of four segments from area of circle         (M1)

=58.1m2   58.055       A1 

 

METHOD 2

40.5×4.52×sin54.532+435.4679360×π×4.52         (M1)

=  32.9845+25.0707         (A1)

=58.1m2   58.055       A1 

 

[3 marks]

b.ii.

sketch of dVdt   OR   dVdt=0.110363   OR   attempt to find where d2Vdt2=0         (M1)

t=1 hour        A1 

 

[2 marks]

c.

Examiners report

Part (a)(i) proved to be difficult for many candidates. About half of the candidates managed to correctly find the angle AO^B. A variety of methods were used: cosine to find half of AO^B then double it; sine to find angle AO^B , then find half of AAO^B and double it; Pythagoras to find half of AB and then sine rule to find half of angle AO^B then double it; Pythagoras to find half of AB, then double it and use cosine rule to find angle AO^B . Many candidates lost a mark here due to premature rounding of an intermediate value and hence the final answer was not correct (to three significant figures).

In part (a)(ii) very few candidates managed to find the correct area of the shaded segment and include the correct units. Some only found the area of the triangle or the area of the sector and then stopped.

In part (b)(i), nearly all candidates managed to find the area of a circle.

In part (b)(ii), finding the area of the field reached by the goat proved troublesome for most of the candidates. It appeared as if the candidates did not fully understand the problem. Very few candidates realized the connection to part (a)(ii).

Part (c) was accessed by only a handful of candidates. The candidates could simply have graphed the function on their GDC to find the greatest value, but most did not realize this.

a.i.
[N/A]
a.ii.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.

Syllabus sections

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