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Date May 2022 Marks available 3 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 4.5m, 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 the field that can be reached by the goat.

[4]
b.

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

[2]
c.

The goat is tied in the field for 8 hours.

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

[3]
d.

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        (M1)

area triangle =4.1231×42

=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.

METHOD 1

π×4.52  63.6172         (A1)

4×1.39040...   (5.56160)         (A1)

subtraction of four segments from area of circle         (M1)

=58.1m2   58.055       A1 

 

METHOD 2

angle of sector =90-54.532  π2-0.951764         (A1)

area of sector =90-54.532360×π×4.52  =6.26771         (A1)

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

total area =4×8.2462+4×6.26771

 

=58.1m2   58.055       A1 

 

[4 marks]

b.

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

t=1 hour        A1 

 

[2 marks]

c.

recognizing V=dVdtdt         (M1)

080.3te-1dt         (A1)

volume eaten is 0.299m3   0.299094        A1 

 

[3 marks]

d.

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.

b.

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.

c.

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.

d.

Syllabus sections

Topic 5—Calculus » SL 5.5—Integration introduction, areas between curve and x axis
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