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(π18t)-6 sin(π18t), for t≥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.
Calculate the number of seconds it takes for the water wheel to complete one rotation.
Hence find the number of rotations the water wheel makes in one hour.
Find k.
Find AˆOB.
Determine the rate of change of h when the top of the bucket is at B.
Markscheme
valid approach (M1)
eg h(0), 13+8 cos(π18×0)-6 sin(π18×0), 13+8×1-6×0
21 (metres) A1 N2
[2 marks]
valid approach to find the period (seen anywhere) (M1)
eg (36, 21), attempt to find two consecutive max/min, 50.3130-14.3130
2ππ18, b=2πperiod,
36 (seconds) (exact) A1 N2
[2 marks]
correct approach (A1)
eg 60×6036, 1.6666 rotations per minute
100 (rotations) A1 N2
[2 marks]
correct substitution into equation (accept the use of t) (A1)
eg 4.06=13+8 cos(π18×k)-6 sin(π18×k)
valid attempt to solve their equation (M1)
eg
11.6510
11.7 A1 N3
[3 marks]
METHOD 1
evidence of choosing the cosine rule or sine rule (M1)
eg AB2=OA2+OB2-2×OA×OB cos(AˆOB), sin(AˆOB)AB=sin(OˆAB)OB
correct working (A1)
eg cos(AˆOB)=102+102-17.022×10×10, -0.445, sin(AˆOB)17.0=sin(π2-12AˆOB)10,
sin(OˆAB)10=sin(π-2×OˆAB)17.0
2.03197 , 116.423°
2.03 (116°) A1 N3
METHOD 2
attempt to find the half central angle (M1)
eg sin(12AˆOB)=12ABOA
correct working (A1)
eg 2×sin-1(8.510)
2.03197 , 116.423°
2.03 (116°) A1 N3
METHOD 3
valid approach to find fraction of period (M1)
eg k36, 11.651036
correct approach to find angle (A1)
eg k36×2π
2.03348, 116.510° (2.04203 using 11.7)
2.03 (117°) A1 N3
[3 marks]
recognizing rate of change is h' (M1)
eg
( from )
rate of change is A1 N2
( from )
[2 marks]