Question
The following diagram represents a large Ferris wheel at an amusement park.
The points P, Q and R represent different positions of a seat on the wheel.
The wheel has a radius of 50 metres and rotates clockwise at a rate of one revolution every 30 minutes.
A seat starts at the lowest point P, when its height is one metre above the ground.
Find the height of a seat above the ground after 15 minutes.
After six minutes, the seat is at point Q. Find its height above the ground at Q.
The height of the seat above ground after t minutes can be modelled by the function \(h(t) = 50\sin (b(t – c)) + 51\).
Find the value of b and of c .
The height of the seat above ground after t minutes can be modelled by the function \(h(t) = 50\sin (b(t – c)) + 51\).
Hence find the value of t the first time the seat is \(96{\text{ m}}\) above the ground.
Answer/Explanation
Markscheme
valid approach (M1)
e.g. 15 mins is half way, top of the wheel, \(d + 1\)
height \( = 101\) (metres) A1 N2
[2 marks]
evidence of identifying rotation angle after 6 minutes A1
e.g. \(\frac{{2\pi }}{5}\) , \(\frac{1}{5}\) of a rotation, \({72^ \circ }\)
evidence of appropriate approach (M1)
e.g. drawing a right triangle and using cosine ratio
correct working (seen anywhere) A1
e.g. \(\cos \frac{{2\pi }}{5} = \frac{x}{{50}}\) , \(15.4(508 \ldots )\)
evidence of appropriate method M1
e.g. height \(= {\rm{radius}} + 1 – 15.45 \ldots \)
height \(= 35.5\) (metres) (accept 35.6) A1 N2
[5 marks]
METHOD 1
evidence of substituting into \(b = \frac{{2\pi }}{{{\rm{period}}}}\) (M1)
correct substitution
e.g. period = 30 minutes, \(b = \frac{{2\pi }}{{30}}\) A1
\(b = 0.209\) \(\left( {\frac{\pi }{{15}}} \right)\) A1 N2
substituting into \(h(t)\) (M1)
e.g. \(h(0) = 1\) , \(h(15) = 101\)
correct substitution A1
\(1 = 50\sin \left( { – \frac{\pi }{{15}}c} \right) + 51\)
\(c = 7.5\) A1 N2
METHOD 2
evidence of setting up a system of equations (M1)
two correct equations
e.g. \(1 = 50\sin b(0 – c) + 51\) , \(101 = 50\sin b(15 – c) + 51\) A1A1
attempt to solve simultaneously (M1)
e.g. evidence of combining two equations
\(b = 0.209\) \(\left( {\frac{\pi }{{15}}} \right)\) , \(c = 7.5\) A1A1 N2N2
[6 marks]
evidence of solving \(h(t) = 96\) (M1)
e.g. equation, graph
\(t = 12.8\) (minutes) A2 N3
[3 marks]
Question
The following diagram shows a waterwheel with a bucket. The wheel rotates at a constant rate in an anticlockwise (counter-clockwise) direction.
The diameter of the wheel is 8 metres. The centre of the wheel, A, is 2 metres above the water level. After t seconds, the height of the bucket above the water level is given by \(h = a\sin bt + 2\) .
Show that \(a = 4\) .
The wheel turns at a rate of one rotation every 30 seconds.
Show that \(b = \frac{\pi }{{15}}\) .
In the first rotation, there are two values of t when the bucket is descending at a rate of \(0.5{\text{ m}}{{\text{s}}^{ – 1}}\) .
Find these values of t .
In the first rotation, there are two values of t when the bucket is descending at a rate of \(0.5{\text{ m}}{{\text{s}}^{ – 1}}\) .
Determine whether the bucket is underwater at the second value of t .
Answer/Explanation
Markscheme
METHOD 1
evidence of recognizing the amplitude is the radius (M1)
e.g. amplitude is half the diameter
\(a = \frac{8}{2}\) A1
\(a = 4\) AG N0
METHOD 2
evidence of recognizing the maximum height (M1)
e.g. \(h = 6\) , \(a\sin bt + 2 = 6\)
correct reasoning
e.g. \(a\sin bt = 4\) and \(\sin bt\) has amplitude of 1 A1
\(a = 4\) AG N0
[2 marks]
METHOD 1
period = 30 (A1)
\(b = \frac{{2\pi }}{{30}}\) A1
\(b = \frac{\pi }{{15}}\) AG N0
METHOD 2
correct equation (A1)
e.g. \(2 = 4\sin 30b + 2\) , \(\sin 30b = 0\)
\(30b = 2\pi \) A1
\(b = \frac{\pi }{{15}}\) AG N0
[2 marks]
recognizing \(h'(t) = – 0.5\) (seen anywhere) R1
attempting to solve (M1)
e.g. sketch of \(h’\) , finding \(h’\)
correct work involving \(h’\) A2
e.g. sketch of \(h’\) showing intersection, \( – 0.5 = \frac{{4\pi }}{{15}}\cos \left( {\frac{\pi }{{15}}t} \right)\)
\(t = 10.6\) , \(t = 19.4\) A1A1 N3
[6 marks]
METHOD 1
valid reasoning for their conclusion (seen anywhere) R1
e.g. \(h(t) < 0\) so underwater; \(h(t) > 0\) so not underwater
evidence of substituting into h (M1)
e.g. \(h(19.4)\) , \(4\sin \frac{{19.4\pi }}{{15}} + 2\)
correct calculation A1
e.g. \(h(19.4) = – 1.19\)
correct statement A1 N0
e.g. the bucket is underwater, yes
METHOD 2
valid reasoning for their conclusion (seen anywhere) R1
e.g. \(h(t) < 0\) so underwater; \(h(t) > 0\) so not underwater
evidence of valid approach (M1)
e.g. solving \(h(t) = 0\) , graph showing region below x-axis
correct roots A1
e.g. \(17.5\), \(27.5\)
correct statement A1 N0
e.g. the bucket is underwater, yes
[4 marks]
Question
A Ferris wheel with diameter \(122\) metres rotates clockwise at a constant speed. The wheel completes \(2.4\) rotations every hour. The bottom of the wheel is \(13\) metres above the ground.
A seat starts at the bottom of the wheel.
After t minutes, the height \(h\) metres above the ground of the seat is given by\[h = 74 + a\cos bt {\rm{ .}}\]
Find the maximum height above the ground of the seat.
(i) Show that the period of \(h\) is \(25\) minutes.
(ii) Write down the exact value of \(b\) .
(b) (i) Show that the period of \(h\) is \(25\) minutes.
(ii) Write down the exact value of \(b\) .
(c) Find the value of \(a\) .
(d) Sketch the graph of \(h\) , for \(0 \le t \le 50\) .
Find the value of \(a\) .
Sketch the graph of \(h\) , for \(0 \le t \le 50\) .
In one rotation of the wheel, find the probability that a randomly selected seat is at least \(105\) metres above the ground.
Answer/Explanation
Markscheme
valid approach (M1)
eg \(13 + {\rm{diameter}}\) , \(13 + 122\)
maximum height \( = 135\) (m) A1 N2
[2 marks]
(i) period \( = \frac{{60}}{{2.4}}\) A1
period \( = 25\) minutes AG N0
(ii) \(b = \frac{{2\pi }}{{25}}\) \(( = 0.08\pi )\) A1 N1
[2 marks]
(a) (i) period \( = \frac{{60}}{{2.4}}\) A1
period \( = 25\) minutes AG N0
(ii) \(b = \frac{{2\pi }}{{25}}\) \(( = 0.08\pi )\) A1 N1
[2 marks]
(b) METHOD 1
valid approach (M1)
eg \({\rm{max}} – 74\) , \(\left| a \right| = \frac{{135 – 13}}{2}\) , \(74 – 13\)
\(\left| a \right| = 61\) (accept \(a = 61\) ) (A1)
\(a = – 61\) A1 N2
METHOD 2
attempt to substitute valid point into equation for h (M1)
eg \(135 = 74 + a\cos \left( {\frac{{2\pi \times 12.5}}{{25}}} \right)\)
correct equation (A1)
eg \(135 = 74 + a\cos (\pi )\) , \(13 = 74 + a\)
\(a = – 61\) A1 N2
[3 marks]
(c)
A1A1A1A1 N4
Note: Award A1 for approximately correct domain, A1 for approximately correct range,
A1 for approximately correct sinusoidal shape with \(2\) cycles.
Only if this last A1 awarded, award A1 for max/min in approximately correct positions.
[4 marks]
Total [9 marks]
METHOD 1
valid approach (M1)
eg \({\rm{max}} – 74\) , \(\left| a \right| = \frac{{135 – 13}}{2}\) , \(74 – 13\)
\(\left| a \right| = 61\) (accept \(a = 61\) ) (A1)
\(a = – 61\) A1 N2
METHOD 2
attempt to substitute valid point into equation for h (M1)
eg \(135 = 74 + a\cos \left( {\frac{{2\pi \times 12.5}}{{25}}} \right)\)
correct equation (A1)
eg \(135 = 74 + a\cos (\pi )\) , \(13 = 74 + a\)
\(a = – 61\) A1 N2
[3 marks]
A1A1A1A1 N4
Note: Award A1 for approximately correct domain, A1 for approximately correct range,
A1 for approximately correct sinusoidal shape with \(2\) cycles.
Only if this last A1 awarded, award A1 for max/min in approximately correct positions.
[4 marks]
setting up inequality (accept equation) (M1)
eg \(h > 105\) , \(105 = 74 + a\cos bt\) , sketch of graph with line \(y = 105\)
any two correct values for t (seen anywhere) A1A1
eg \(t = 8.371 \ldots \) , \(t = 16.628 \ldots \) , \(t = 33.371 \ldots \) , \(t = 41.628 \ldots \)
valid approach M1
eg \(\frac{{16.628 – 8.371}}{{25}}\) , \(\frac{{{t_1} – {t_2}}}{{25}}\) , \(\frac{{2 \times 8.257}}{{50}}\) , \(\frac{{2(12.5 – 8.371)}}{{25}}\)
\(p = 0.330\) A1 N2
[5 marks]
Question
The population of deer in an enclosed game reserve is modelled by the function \(P(t) = 210\sin (0.5t – 2.6) + 990\), where \(t\) is in months, and \(t = 1\) corresponds to 1 January 2014.
Find the number of deer in the reserve on 1 May 2014.
Find the rate of change of the deer population on 1 May 2014.
Interpret the answer to part (i) with reference to the deer population size on 1 May 2014.
Answer/Explanation
Markscheme
\(t = 5\) (A1)
correct substitution into formula (A1)
eg \(210\sin (0.5 \times 5 – 2.6) + 990,{\text{ }}P(5)\)
\(969.034982 \ldots \)
969 (deer) (must be an integer) A1 N3
[3 marks]
evidence of considering derivative (M1)
eg \(P’\)
\(104.475\)
\(104\) (deer per month) A1 N2
[2 marks]
(the deer population size is) increasing A1 N1
[1 mark]
Question
The height, \(h\) metros, of a seat on a Ferris wheel after \(t\) minutes is given by
\[h(t) = – 15\cos 1.2t + 17,{\text{ for }}t \geqslant 0{\text{.}}\]
Find the height of the seat when \(t = 0\).
The seat first reaches a height of 20 m after \(k\) minutes. Find \(k\).
Calculate the time needed for the seat to complete a full rotation, giving your answer correct to one decimal place.
Answer/Explanation
Markscheme
valid approach (M1)
eg\(\,\,\,\,\,\)\(h(0),{\text{ }} – 15\cos (1.2 \times 0) + 17,{\text{ }} – 15(1) + 17\)
\(h(0) = 2{\text{ (m)}}\) A1 N2
[2 marks]
correct substitution into equation (A1)
eg\(\,\,\,\,\,\)\(20 = – 15\cos 1.2t + 17,{\text{ }} – 15\cos 1.2k = 3\)
valid attempt to solve for \(k\) (M1)
eg\(\,\,\,\,\,\), \(\cos 1.2k = – \frac{3}{{15}}\)
1.47679
\(k = 1.48\) A1 N2
[3 marks]
recognize the need to find the period (seen anywhere) (M1)
eg\(\,\,\,\,\,\)next \(t\) value when \(h = 20\)
correct value for period (A1)
eg\(\,\,\,\,\,\)\({\text{period}} = \frac{{2\pi }}{{1.2}},{\text{ }}5.23598,{\text{ }}6.7–1.48\)
5.2 (min) (must be 1 dp) A1 N2
[3 marks]
Question
The following diagram shows the graph of \(f(x) = a\sin bx + c\), for \(0 \leqslant x \leqslant 12\).
The graph of \(f\) has a minimum point at \((3,{\text{ }}5)\) and a maximum point at \((9,{\text{ }}17)\).
The graph of \(g\) is obtained from the graph of \(f\) by a translation of \(\left( {\begin{array}{*{20}{c}} k \\ 0 \end{array}} \right)\). The maximum point on the graph of \(g\) has coordinates \((11.5,{\text{ }}17)\).
The graph of \(g\) changes from concave-up to concave-down when \(x = w\).
(i) Find the value of \(c\).
(ii) Show that \(b = \frac{\pi }{6}\).
(iii) Find the value of \(a\).
(i) Write down the value of \(k\).
(ii) Find \(g(x)\).
(i) Find \(w\).
(ii) Hence or otherwise, find the maximum positive rate of change of \(g\).
Answer/Explanation
Markscheme
(i) valid approach (M1)
eg\(\,\,\,\,\,\)\(\frac{{5 + 17}}{2}\)
\(c = 11\) A1 N2
(ii) valid approach (M1)
eg\(\,\,\,\,\,\)period is 12, per \( = \frac{{2\pi }}{b},{\text{ }}9 – 3\)
\(b = \frac{{2\pi }}{{12}}\) A1
\(b = \frac{\pi }{6}\) AG N0
(iii) METHOD 1
valid approach (M1)
eg\(\,\,\,\,\,\)\(5 = a\sin \left( {\frac{\pi }{6} \times 3} \right) + 11\), substitution of points
\(a = – 6\) A1 N2
METHOD 2
valid approach (M1)
eg\(\,\,\,\,\,\)\(\frac{{17 – 5}}{2}\), amplitude is 6
\(a = – 6\) A1 N2
[6 marks]
(i) \(k = 2.5\) A1 N1
(ii) \(g(x) = – 6\sin \left( {\frac{\pi }{6}(x – 2.5)} \right) + 11\) A2 N2
[3 marks]
(i) METHOD 1 Using \(g\)
recognizing that a point of inflexion is required M1
eg\(\,\,\,\,\,\)sketch, recognizing change in concavity
evidence of valid approach (M1)
eg\(\,\,\,\,\,\)\(g”(x) = 0\), sketch, coordinates of max/min on \({g’}\)
\(w = 8.5\) (exact) A1 N2
METHOD 2 Using \(f\)
recognizing that a point of inflexion is required M1
eg\(\,\,\,\,\,\)sketch, recognizing change in concavity
evidence of valid approach involving translation (M1)
eg\(\,\,\,\,\,\)\(x = w – k\), sketch, \(6 + 2.5\)
\(w = 8.5\) (exact) A1 N2
(ii) valid approach involving the derivative of \(g\) or \(f\) (seen anywhere) (M1)
eg\(\,\,\,\,\,\)\(g'(w),{\text{ }} – \pi \cos \left( {\frac{\pi }{6}x} \right)\), max on derivative, sketch of derivative
attempt to find max value on derivative M1
eg\(\,\,\,\,\,\)\( – \pi \cos \left( {\frac{\pi }{6}(8.5 – 2.5)} \right),{\text{ }}f'(6)\), dot on max of sketch
3.14159
max rate of change \( = \pi \) (exact), 3.14 A1 N2
[6 marks]
Question
At Grande Anse Beach the height of the water in metres is modelled by the function \(h(t) = p\cos (q \times t) + r\), where \(t\) is the number of hours after 21:00 hours on 10 December 2017. The following diagram shows the graph of \(h\) , for \(0 \leqslant t \leqslant 72\).
The point \({\text{A}}(6.25,{\text{ }}0.6)\) represents the first low tide and \({\text{B}}(12.5,{\text{ }}1.5)\) represents the next high tide.
How much time is there between the first low tide and the next high tide?
Find the difference in height between low tide and high tide.
Find the value of \(p\);
Find the value of \(q\);
Find the value of \(r\).
There are two high tides on 12 December 2017. At what time does the second high tide occur?
Answer/Explanation
Markscheme
attempt to find the difference of \(x\)-values of A and B (M1)
eg\(\,\,\,\,\,\)\(6.25 – 12.5{\text{ }}\)
6.25 (hours), (6 hours 15 minutes) A1 N2
[2 marks]
attempt to find the difference of \(y\)-values of A and B (M1)
eg\(\,\,\,\,\,\)\(1.5 – 0.6\)
\(0.9{\text{ (m)}}\) A1 N2
[2 marks]
valid approach (M1)
eg\(\,\,\,\,\,\)\(\frac{{{\text{max}} – {\text{min}}}}{2},{\text{ }}0.9 \div 2\)
\(p = 0.45\) A1 N2
[2 marks]
METHOD 1
period \( = 12.5\) (seen anywhere) (A1)
valid approach (seen anywhere) (M1)
eg\(\,\,\,\,\,\)\({\text{period}} = \frac{{2\pi }}{b},{\text{ }}q = \frac{{2\pi }}{{{\text{period}}}},{\text{ }}\frac{{2\pi }}{{12.5}}\)
0.502654
\(q = \frac{{4\pi }}{{25}},{\text{ 0.503 }}\left( {{\text{or }} – \frac{{4\pi }}{{25}},{\text{ }} – 0.503} \right)\) A1 N2
METHOD 2
attempt to use a coordinate to make an equation (M1)
eg\(\,\,\,\,\,\)\(p\cos (6.25q) + r = 0.6,{\text{ }}p\cos (12.5q) + r = 1.5\)
correct substitution (A1)
eg\(\,\,\,\,\,\)\(0.45\cos (6.25q) + 1.05 = 0.6,{\text{ }}0.45\cos (12.5q) + 1.05 = 1.5\)
0.502654
\(q = \frac{{4\pi }}{{25}},{\text{ }}0.503{\text{ }}\left( {{\text{or }} – \frac{{4\pi }}{{25}},{\text{ }} – 0.503} \right)\) A1 N2
[3 marks]
valid method to find \(r\) (M1)
eg\(\,\,\,\,\,\)\(\frac{{{\text{max}} + {\text{min}}}}{2},{\text{ }}0.6 + 0.45\)
\(r = 1.05\) A1 N2
[2 marks]
METHOD 1
attempt to find start or end \(t\)-values for 12 December (M1)
eg\(\,\,\,\,\,\)\(3 + 24,{\text{ }}t = 27,{\text{ }}t = 51\)
finds \(t\)-value for second max (A1)
\(t = 50\)
23:00 (or 11 pm) A1 N3
METHOD 2
valid approach to list either the times of high tides after 21:00 or the \(t\)-values of high tides after 21:00, showing at least two times (M1)
eg\(\,\,\,\,\,\)\({\text{21:00}} + 12.5,{\text{ 21:00}} + 25,{\text{ }}12.5 + 12.5,{\text{ }}25 + 12.5\)
correct time of first high tide on 12 December (A1)
eg\(\,\,\,\,\,\)10:30 (or 10:30 am)
time of second high tide = 23:00 A1 N3
METHOD 3
attempt to set their \(h\) equal to 1.5 (M1)
eg\(\,\,\,\,\,\)\(h(t) = 1.5,{\text{ }}0.45\cos \left( {\frac{{4\pi }}{{25}}t} \right) + 1.05 = 1.5\)
correct working to find second max (A1)
eg\(\,\,\,\,\,\)\(0.503t = 8\pi ,{\text{ }}t = 50\)
23:00 (or 11 pm) A1 N3
[3 marks]
Question
The depth of water in a port is modelled by the function \(d(t) = p\cos qt + 7.5\), for \(0 \leqslant t \leqslant 12\), where \(t\) is the number of hours after high tide.
At high tide, the depth is 9.7 metres.
At low tide, which is 7 hours later, the depth is 5.3 metres.
Find the value of \(p\).
Find the value of \(q\).
Use the model to find the depth of the water 10 hours after high tide.
Answer/Explanation
Markscheme
valid approach (M1)
eg\(\,\,\,\,\,\)\(\frac{{{\text{max}} – {\text{min}}}}{2}\), sketch of graph, \(9.7 = p\cos (0) + 7.5\)
\(p = 2.2\) A1 N2
[2 marks]
valid approach (M1)
eg\(\,\,\,\,\,\)\(B = \frac{{2\pi }}{{{\text{period}}}}\), period is \(14,{\text{ }}\frac{{360}}{{14}},{\text{ }}5.3 = 2.2\cos 7q + 7.5\)
0.448798
\(q = \frac{{2\pi }}{{14}}{\text{ }}\left( {\frac{\pi }{7}} \right)\), (do not accept degrees) A1 N2
[2 marks]
valid approach (M1)
eg\(\,\,\,\,\,\)\(d(10),{\text{ }}2.2\cos \left( {\frac{{20\pi }}{{14}}} \right) + 7.5\)
7.01045
7.01 (m) A1 N2
[2 marks]
Question
At an amusement park, a Ferris wheel with diameter 111 metres rotates at a constant speed. The bottom of the wheel is k metres above the ground. A seat starts at the bottom of the wheel.
The wheel completes one revolution in 16 minutes.
After t minutes, the height of the seat above ground is given by \(h\left( t \right) = 61.5 + a\,{\text{cos}}\left( {\frac{\pi }{8}t} \right)\), for 0 ≤ t ≤ 32.
After 8 minutes, the seat is 117 m above the ground. Find k.
Find the value of a.
Find when the seat is 30 m above the ground for the third time.
Answer/Explanation
Markscheme
valid approach to find k (M1)
eg 8 minutes is half a turn, k + diameter, k + 111 = 117
k = 6 A1 N2
[2 marks]
METHOD 1
valid approach (M1)
eg \(\frac{{{\text{max}}\,\, – \,\,{\text{min}}}}{2}\) a = radius
\(\left| a \right| = \frac{{117 – 6}}{2},\,\,55.5\) (A1)
a = −55.5 A1 N2
METHOD 2
attempt to substitute valid point into equation for f (M1)
eg h(0) = 6, h(8) = 117
correct equation (A1)
eg \(6 = 61.5 + a\,{\text{cos}}\left( {\frac{\pi }{8} \times 0} \right),\,\,117 = 61.5 + a\,{\text{cos}}\left( {\frac{\pi }{8} \times 8} \right),\,\,6 = 61.5 + a\)
a = −55.5 A1 N2
[3 marks]
valid approach (M1)
eg sketch of h and \(y = 30,\,\,h = 30,\,\,61.5 – 55.5\,{\text{cos}}\left( {\frac{\pi }{8}t} \right) = 30,\,\,t = 2.46307,\,\,t = 13.5369\)
18.4630
t = 18.5 (minutes) A1 N3
[3 marks]