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IB Mathematics SL 4.5 Concepts of trial, outcome, equally likely outcomes AI SL Paper 2 - Exam Style Questions - New Syllabus

IB DP Maths AI SL Paper 2 - All Topics

Question

A wind turbine is designed so that the rotation of the blades generates electricity. The turbine is built on level ground and is made up of a vertical tower and three blades.
The point A is on the base of the tower directly below point B at the top of the tower. The height of the tower, AB, is \(90\) m. The blades of the turbine are centred at B and are each of length \(40\) m. This is shown in the following diagram.
The end of one of the blades of the turbine is represented by point C on the diagram. Let \(h\) be the height of \(C\) above the ground, measured in metres, where \(h\) varies as the blade rotates.

(a) Find the

(i) maximum value of \(h\).
(ii) minimum value of \(h\). [2]
The blades of the turbine complete \(12\) rotations per minute under normal conditions, moving at a constant rate.
(b) (i) Find the time, in seconds, it takes for the blade \([BC]\) to make one complete rotation under these conditions.
(ii) Calculate the angle, in degrees, that the blade \([BC]\) turns through in one second. [3]
The height, \(h\), of point \(C\) can be modelled by the following function. Time, \(t\), is measured from the instant when the blade \([BC]\) first passes \([AB]\) and is measured in seconds.
\(h(t)=90-40\cos(72t^{\circ}),\quad t\ge0\)
(c) (i) Write down the amplitude of the function.
(ii) Find the period of the function. [2]
(d) Sketch the function \(h(t)\) for \(0\le t\le 5\), clearly labelling the coordinates of the maximum and minimum points. [3]
(e) (i) Find the height of \(C\) above the ground when \(t=2\).
(ii) Find the time, in seconds, that point \(C\) is above a height of \(100\) m, during each complete rotation. [5]
Looking through her window, Riya has a partial view of the rotating wind turbine. The position of her window means that she cannot see any part of the wind turbine that is more than \(100\) m above the ground. This is illustrated in the following diagram.
(f) (i) At any given instant, find the probability that point \(C\) is visible from Riya’s window.
(ii) The wind speed increases. The blades rotate at twice the speed, but still at a constant rate. At any given instant, find the probability that Riya can see point \(C\) from her window. Justify your answer. [5]
▶️ Answer / Explanation
Solution

(a)

Tower top at \(90\) m; blade length \(40\) m.
(i) \(h_{\max}=90+40=\boxed{130\text{ m}}\). (ii) \(h_{\min}=90-40=\boxed{50\text{ m}}\).

[2]

(b)

(i) \(12\) rev/min \(\Rightarrow\) period \(=\dfrac{60}{12}=\boxed{5\text{ s}}\).
(ii) One full rotation \(360^\circ\) in \(5\) s \(\Rightarrow\) angle per second \(=\dfrac{360^\circ}{5}=\boxed{72^\circ\text{/s}}\).

[3]

(c)

From \(h(t)=90-40\cos(72t^\circ)\): amplitude \(=\boxed{40}\). Period \(=\dfrac{360^\circ}{72^\circ\!/\text{s}}=\boxed{5\text{ s}}\).

[2]

(d)

Over \(0\le t\le 5\): minima at \(t=0,5\) with \(h=50\); maximum at \(t=2.5\) with \(h=130\). Sketch cosine shape; label \((0,50)\), \((2.5,130)\), \((5,50)\); correct concavity near minima.

[3]

(e)

(i) \(h(2)=90-40\cos(144^\circ)\). Since \(\cos(144^\circ)=\cos(180^\circ-36^\circ)=-\cos36^\circ\approx-0.809016\),
\(h(2)\approx 90-40(-0.809016)=90+32.3606=\boxed{122\text{ m (to 3 s.f.)}}\).

(ii) \(h(t)>100 \Rightarrow 90-40\cos(72t^\circ)>100 \Rightarrow \cos(72t^\circ)<-\tfrac14\).
Let \(\theta=72t^\circ\). Threshold angles: \(\theta_1=\arccos(-\tfrac14)\approx 104.4775^\circ\), \(\theta_2=360^\circ-104.4775^\circ=255.5225^\circ\).
Time above \(100\) m each rotation: \[ \Delta t=\frac{\theta_2-\theta_1}{360^\circ}\times 5 =\frac{151.045^\circ}{360^\circ}\times 5 \approx \boxed{2.10\text{ s}}. \] (Crossing times: \(t_1=\tfrac{104.4775}{72}\approx 1.4511\) s, \(t_2=\tfrac{255.5225}{72}\approx 3.5489\) s.)
[5]

(f)

(i) In one period \(T=5\) s, visible when \(h\le 100\). From (e)(ii), time above \(100\) m is \(2.09784\) s, so time visible \(=5-2.09784=2.90216\) s.
Probability (uniform in time) \(=\dfrac{2.90216}{5}=\boxed{0.580}\) (to 3 d.p.).

(ii) Doubling the speed halves both \(T\) and the above-threshold duration; the proportion of time visible is unchanged. Probability remains \(\boxed{0.580}\).

[5]
Total Marks: 20
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