IB DP Maths Topic 3.4 Applications SL Paper 2

 

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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.

[2]
a.

After six minutes, the seat is at point Q. Find its height above the ground at Q.

[5]
b.

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 .

[6]
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.

[3]
d.
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]

a.

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]

b.

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]

c.

evidence of solving \(h(t) = 96\)     (M1)

e.g. equation, graph

\(t = 12.8\) (minutes)     A2     N3

[3 marks]

d.

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 \(a = 4\) .

[2]
a.

The wheel turns at a rate of one rotation every 30 seconds.

Show that \(b = \frac{\pi }{{15}}\) .

[2]
b.

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 .

[6]
c.

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 .

[4]
d.
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]

a.

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]

b.

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]

c.

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]

d.

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.

[2]
a.

(i)     Show that the period of \(h\) is \(25\) minutes.

(ii)     Write down the exact value of \(b\) .

[2]
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 \(0 \le t \le 50\) .

[9]
bcd.

Find the value of \(a\) .

[3]
c.

Sketch the graph of \(h\) , for \(0 \le t \le 50\) .

[4]
d.

In one rotation of the wheel, find the probability that a randomly selected seat is at least \(105\) metres above the ground.

[5]
e.
Answer/Explanation

Markscheme

valid approach     (M1)

eg   \(13 + {\rm{diameter}}\) , \(13 + 122\)

maximum height \( = 135\) (m)     A1 N2

[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.

(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]

bcd.

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]

 

d.

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]

e.

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.

[3]
a.

Find the rate of change of the deer population on 1 May 2014.

[2]
b(i).

Interpret the answer to part (i) with reference to the deer population size on 1 May 2014.

[1]
b(ii).
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]

a.

evidence of considering derivative     (M1)

eg     \(P’\)

\(104.475\)

\(104\) (deer per month)     A1     N2

[2 marks]

b(i).

(the deer population size is) increasing     A1     N1

[1 mark]

b(ii).

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 \(t = 0\).

[2]
a.

The seat first reaches a height of 20 m after \(k\) minutes. Find \(k\).

[3]
b.

Calculate the time needed for the seat to complete a full rotation, giving your answer correct to one decimal place.

[3]
c.
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]

a.

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\(\,\,\,\,\,\)M16/5/MATME/SP2/ENG/TZ2/04.b/M, \(\cos 1.2k =  – \frac{3}{{15}}\)

1.47679

\(k = 1.48\)     A1     N2

[3 marks]

b.

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]

c.

Question

The following diagram shows the graph of \(f(x) = a\sin bx + c\), for \(0 \leqslant x \leqslant 12\).

N16/5/MATME/SP2/ENG/TZ0/10

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 \(a\).

[6]
a.

(i)     Write down the value of \(k\).

(ii)     Find \(g(x)\).

[3]
b.

(i)     Find \(w\).

(ii)     Hence or otherwise, find the maximum positive rate of change of \(g\).

[6]
c.
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]

a.

(i)     \(k = 2.5\)     A1     N1

(ii)     \(g(x) =  – 6\sin \left( {\frac{\pi }{6}(x – 2.5)} \right) + 11\)     A2     N2

[3 marks]

b.

(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]

c.

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 \(0 \leqslant t \leqslant 72\).

M17/5/MATME/SP2/ENG/TZ1/08

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?

[2]
a.i.

Find the difference in height between low tide and high tide.

[2]
a.ii.

Find the value of \(p\);

[2]
b.i.

Find the value of \(q\);

[3]
b.ii.

Find the value of \(r\).

[2]
b.iii.

There are two high tides on 12 December 2017. At what time does the second high tide occur?

[3]
c.
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]

a.i.

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]

a.ii.

valid approach     (M1)

eg\(\,\,\,\,\,\)\(\frac{{{\text{max}} – {\text{min}}}}{2},{\text{ }}0.9 \div 2\)

\(p = 0.45\)     A1     N2

[2 marks]

b.i.

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]

b.ii.

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]

b.iii.

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]

c.

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 \(p\).

[2]
a.

Find the value of \(q\).

[2]
b.

Use the model to find the depth of the water 10 hours after high tide.

[2]
c.
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]

a.

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]

b.

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]

c.

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.

[2]
a.

Find the value of a.

[3]
b.

Find when the seat is 30 m above the ground for the third time.

[3]
c.
Answer/Explanation

Markscheme

valid approach to find k      (M1)

eg   8 minutes is half a turn, + diameter, + 111 = 117

k = 6      A1 N2

[2 marks]

a.

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]

b.

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]

c.
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