Home / IB Mathematics SL 3.7 Composite functions of the form AA SL Paper 1- Exam Style Questions

IB Mathematics SL 3.7 Composite functions of the form AA SL Paper 1- Exam Style Questions- New Syllabus

Question

A Ferris wheel with a diameter of 72 metres turns at a constant rate. Its lowest point is situated 8 metres above the ground. A passenger boards the wheel at this minimum height at time \( t = 0 \).
 
The wheel takes 12 minutes to complete one full rotation. The height, \( h \) metres, of the passenger above the ground after \( t \) minutes is modelled by the function \[ h(t) = a \cos(bt) + 44, \] where \( a \) and \( b \) are real constants.
(a) Determine the value of:
 (i) \( a \);
 (ii) \( b \).
(b) Draw a sketch of the function \( h \) for the interval \( 0 \leq t \leq 12 \), marking the coordinates of the peak and the start/end points.
Consider the first revolution of the wheel.
(c) Calculate the total time \( T \), in minutes, for which the passenger is at least 26 m above the ground.
(d) (i) Find the derivative \( h'(t) \).
 (ii) Hence, find the specific time at which the passenger’s height is decreasing at the maximum possible rate.

Most-appropriate topic codes (Mathematics: analysis and approaches guide):

SL 3.7: Circular functions \( \sin x, \cos x \); amplitude, periodic nature, and their graphs; real-life contexts — Part a, b, c
SL 5.6: Derivative of \( \sin x, \cos x \); the chain rule for composite functions — Part d(i)
SL 5.8: Optimization; local maximum and minimum points — Part d(ii) 
▶️ Answer/Explanation

(a)(i) Determine \( a \):
The radius of the wheel is \( 72 / 2 = 36 \) m, which represents the amplitude. Since the seat starts at the lowest point (a minimum), the cosine function is reflected.
\( \boxed{a = -36} \)

(a)(ii) Determine \( b \):
The period is 12 minutes. Using the relationship \( \text{Period} = \frac{2\pi}{b} \):
\( 12 = \frac{2\pi}{b} \implies b = \frac{2\pi}{12} = \frac{\pi}{6} \)
\( \boxed{b = \frac{\pi}{6}} \)

(b) Sketch the graph:
Graph of height vs time
• Minimum points: \( (0, 8) \) and \( (12, 8) \)
• Maximum point: \( (6, 80) \)
• The midline is at \( h = 44 \).

(c) Calculate \( T \):
Solve the inequality \( h(t) \geq 26 \):
\( -36\cos\left(\frac{\pi}{6}t\right) + 44 \geq 26 \)
\( -36\cos\left(\frac{\pi}{6}t\right) \geq -18 \)
\( \cos\left(\frac{\pi}{6}t\right) \leq \frac{1}{2} \)
On the interval \( [0, 12] \), the boundaries are \( \frac{\pi}{6}t = \frac{\pi}{3} \) and \( \frac{\pi}{6}t = \frac{5\pi}{3} \).
\( t_1 = 2 \) and \( t_2 = 10 \).
The duration is \( T = 10 – 2 = 8 \) minutes.
\( \boxed{T = 8} \)

(d)(i) Find \( h'(t) \):
Applying the chain rule:
\( h'(t) = -36 \left(-\sin\left(\frac{\pi}{6}t\right)\right) \cdot \frac{\pi}{6} \)
\( \boxed{h'(t) = 6\pi \sin\left(\frac{\pi}{6}t\right)} \)

(d)(ii) Maximum rate of decrease:
The rate is decreasing when \( h'(t) \) is negative. We find the minimum value of the sine component.
The fastest decrease occurs when \( \sin\left(\frac{\pi}{6}t\right) = -1 \).
\( \frac{\pi}{6}t = \frac{3\pi}{2} \implies t = 9 \).
\( \boxed{t = 9 \text{ minutes}} \)

Graph of the derivative function

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