Home / IB Mathematics SL 2.5 Modelling with the various functions AI HL Paper 2- Exam Style Questions

IB Mathematics SL 2.5 Modelling with the various functions AI HL Paper 2- Exam Style Questions- New Syllabus

IBDP Maths AI HL - Paper 2 - All Topics
Question

A wind turbine is built on horizontal ground. 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. The end of one of the blades is represented by point \( C \). 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 \( h \).

(b) The blades complete \( 12 \) rotations per minute under normal conditions, moving at a constant rate.
(i) Find the time, in seconds, it takes for the blade \( [BC] \) to make one complete rotation.
(ii) Calculate the angle, in degrees, that the blade \( [BC] \) turns through in one second.

(c) The height \( h \) is modelled by \( h(t) = 90 – 40 \cos(72t) \), \( t \geq 0 \), where \( t \) is time in seconds.
(i) Write down the amplitude of the function.
(ii) Find the period of the function.

(d) Sketch the function \( h(t) \) for \( 0 \leq t \leq 5 \), clearly labelling the coordinates of the maximum and minimum points.

(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.

The wind speed increases and the blades rotate faster, but still at a constant rate.

(f) Given that point \( C \) is now higher than \( 110 \) m for \( 1 \) second during each complete rotation, find the time for one complete rotation.

▶️ Answer/Explanation
Markscheme

(a)(i)
Maximum \( h = 90 + 40 = 130 \) m (A1)
(a)(ii)
Minimum \( h = 90 – 40 = 50 \) m (A1)
Explanation:
The height is the tower height plus/minus the blade length.
Result:
Max: \( 130 \) m, Min: \( 50 \) m

(b)(i)
Time = \( 60 \div 12 = 5 \) seconds (A1)
(b)(ii)
Angle = \( 360 \div 5 \) (M1)
= \( 72^\circ \) (A1)
Explanation:
One rotation is \( 360^\circ \). Divide by the time for one rotation.
Result:
Time: \( 5 \) s, Angle: \( 72^\circ \)

(c)(i)
Amplitude = \( 40 \) (A1)
(c)(ii)
Period = \( 360 \div 72 = 5 \) seconds (A1)
Explanation:
The amplitude is the coefficient of the cosine term. The period is \( \frac{360}{\text{coefficient of } t} \).
Result:
Amplitude: \( 40 \), Period: \( 5 \) s

(d)

Sketch: Cosine graph starting at minimum (\( 0, 50 \)), rising to maximum at (\( 2.5, 130 \)), falling to minimum at (\( 5, 50 \)). (A1)
Maximum point labelled (\( 2.5, 130 \)). (A1)
At least one minimum point (e.g., (\( 0, 50 \)) or (\( 5, 50 \))) labelled. (A1)
Explanation:
The graph is a cosine curve shifted vertically and reflected.
Result:
Correct sketch with labels

(e)(i)
\( h = 90 – 40 \cos(72 \times 2) \) (M1)
\( h = 122 \) m (122.3606…) (A1)
(e)(ii)
Set \( 100 = 90 – 40 \cos(72t) \) (M1)
\( \cos(72t) = -0.25 \), \( 72t = 104.477^\circ \) or \( 255.523^\circ \) (A1)
\( t = 1.45107… \) or \( 3.54892… \) seconds
Time above \( 100 \) m = \( 3.54892 – 1.45107 = 2.09784… \) seconds (A1)
Explanation:
Solve the equation for \( h = 100 \). The time above is the difference between the two times in one period.
Result:
(i) \( 122 \) m, (ii) \( 2.10 \) s

(f)
METHOD 1
Set \( 90 – 40 \cos(a t^*) = 110 \) (M1)
\( \cos(a t^*) = -0.5 \)
\( a t^* = 120^\circ, 240^\circ \) (A1)
\( 1 = \frac{240}{a} – \frac{120}{a} \) (M1)
\( a = 120 \) (A1)
Period = \( 360 \div 120 = 3 \) seconds (A1)
METHOD 2

Attempt at diagram
\( \cos \alpha = \frac{20}{40} \) (or recognizing special triangle)
Angle made by \( C \), \( 2\alpha = 120^\circ \)
One third of a revolution in \( 1 \) second
Hence one revolution = \( 3 \) seconds


Explanation:
Find the angular speed required for the blade to spend \( 1 \) second in the arc where \( h > 110 \) m, using the geometry of the triangle.
Result:
Period is \( 3 \) seconds

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