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Date	November 2020	Marks available	3	Reference code	20N.2.SL.TZ0.S_8
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Find	Question number	S_8	Adapted from	N/A

Question

The following diagram shows a water wheel with centre $O$ and radius $10$ metres. Water flows into buckets, turning the wheel clockwise at a constant speed.

The height, $h$ metres, of the top of a bucket above the ground $t$ seconds after it passes through point $A$ is modelled by the function

$h (t) = 13 + 8 \cos (\frac{π}{18} t) - 6 \sin (\frac{π}{18} t)$ , for $t \geq 0$ .

A bucket moves around to point $B$ which is at a height of $4.06$ metres above the ground. It takes $k$ seconds for the top of this bucket to go from point $A$ to point $B$ .

The chord $[AB]$ is $17.0$ metres, correct to three significant figures.

Find the height of point $A$ above the ground.

[2]

a.i.

Calculate the number of seconds it takes for the water wheel to complete one rotation.

[2]

a.ii.

Hence find the number of rotations the water wheel makes in one hour.

[2]

a.iii.

Find $k$ .

[3]

Find $A \hat{O} B$ .

[3]

Determine the rate of change of $h$ when the top of the bucket is at $B$ .

[2]

Markscheme

valid approach (M1)

eg $h (0), 13 + 8 \cos (\frac{π}{18} \times 0) - 6 \sin (\frac{π}{18} \times 0), 13 + 8 \times 1 - 6 \times 0$

$21$ (metres) A1 N2

[2 marks]

a.i.

valid approach to find the period (seen anywhere) (M1)

eg $(36, 21)$ , attempt to find two consecutive max/min, $50.3130 - 14.3130$

$\frac{2 π}{\frac{π}{18}}, b = \frac{2 π}{period},$

$36$ (seconds) (exact) A1 N2

[2 marks]

a.ii.

correct approach (A1)

eg $\frac{60 \times 60}{36}, 1.6666$ rotations per minute

$100$ (rotations) A1 N2

[2 marks]

a.iii.

correct substitution into equation (accept the use of $t$ ) (A1)

eg $4.06 = 13 + 8 \cos (\frac{π}{18} \times k) - 6 \sin (\frac{π}{18} \times k)$

valid attempt to solve their equation (M1)

$11.6510$

$11.7$ A1 N3

[3 marks]

METHOD 1

evidence of choosing the cosine rule or sine rule (M1)

eg ${AB}^{2} = {OA}^{2} + {OB}^{2} - 2 \times OA \times OB \cos (A \hat{O} B), \frac{\sin (A \hat{O} B)}{AB} = \frac{\sin (O \hat{A} B)}{OB}$

correct working (A1)

eg $\cos (A \hat{O} B) = \frac{10^{2} + 10^{2} - 17 . 0^{2}}{2 \times 10 \times 10}, - 0.445, \frac{\sin (A \hat{O} B)}{17.0} = \frac{\sin (\frac{π}{2} - \frac{1}{2} A \hat{O} B)}{10},$

$\frac{\sin (O \hat{A} B)}{10} = \frac{\sin (π - 2 \times O \hat{A} B)}{17.0}$

$2.03197, 116.423 °$

$2.03 (116 °)$ A1 N3

METHOD 2

attempt to find the half central angle (M1)

eg $\sin (\frac{1}{2} A \hat{O} B) = \frac{\frac{1}{2} AB}{OA}$

correct working (A1)

eg $2 \times \sin^{- 1} (\frac{8.5}{10})$

$2.03197, 116.423 °$

$2.03 (116 °)$ A1 N3

METHOD 3

valid approach to find fraction of period (M1)

eg $\frac{k}{36}, \frac{11.6510}{36}$

correct approach to find angle (A1)

eg $\frac{k}{36} \times 2 π$

$2.03348, 116.510 °$ ( $2.04203$ using $11.7$ )

$2.03 (117 °)$ A1 N3

[3 marks]

recognizing rate of change is $h'$ (M1)

eg $h' (k), h' (11.6510), 0.782024$

$- 0.782024$ ( $- 0.768662$ from $3 sf$ )

rate of change is $- 0.782 ({ms}^{- 1})$ A1 N2

( $- 0.769 ({ms}^{- 1})$ from $3 sf$ )

[2 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

a.iii.

[N/A]

Syllabus sections

Topic 2—Functions » AHL 2.9—HL modelling functions

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