A Ferris wheel has a diameter of 120 m - Leaving Cert Mathematics - Question 8 - 2022

The height function h(t) is periodic, as it is based on a cosine function.

Period:

• The period of the cosine function is given by the formula: $Period = \frac{2\pi}{\text{frequency}}$ Here, frequency is 1 full rotation every 6 minutes (because it completes 10 rotations in 60 minutes). Therefore, the period is: $Period = 6 \text{ minutes}$

Range:

• The maximum height occurs when cos(t) = -1: $h_{max} = 72 + 60 = 132 \text{ m}$
• The minimum height occurs when cos(t) = 1: $h_{min} = 72 - 60 = 12 \text{ m}$
• Thus, the range is: $Range = [12, 132]$

To find when the height h(t) is greater than 42 m:

1. From the previously derived heights, we see:

   • h(0) = 12 m (not greater than 42)
   • h(1) = 42 m (equal to 42)
   • h(2) = 102 m (greater than 42)
   • h(3) = 132 m (greater than 42)
   • h(4) = 102 m (greater than 42)
   • h(5) = 42 m (equal to 42)
   • h(6) = 12 m (not greater than 42)
   • h(7) = 102 m (greater than 42)
   • h(8) = 42 m (equal to 42)

2. Thus, the intervals where h(t) > 42 m are:

   • From t = 2 to t = 5 (3 minutes above 42 m)
   • From t = 7 (above 42 m)

3. In one full period (6 minutes), point A is greater than 42 m for:

   • 3 minutes (during heights of 102 and 132 m both)
   • Repeating this for an additional period yields:
   • h(t) is above 42 m for a total of 8 minutes in 50 minutes.

Starting with the equation: $72 - 60 \, cos \left( \frac{\pi}{3} t \right) = 110$

1. Rearrange the equation: $-60 \, cos \left( \frac{\pi}{3} t \right) = 110 - 72$ $-60 \, cos \left( \frac{\pi}{3} t \right) = 38$ $cos \left( \frac{\pi}{3} t \right) = -\frac{38}{60} = -0.6333$

2. To find the values for t, we first determine the reference angle: $\theta = cos^{-1}(-0.6333)$

3. Calculating gives: $\theta \approx 0.874\text{ radians}$

4. The solutions for $t$ in the interval are: First Solution: Quadrant II gives: $t_1 = \frac{3}{\pi} \cdot \left(\pi - 0.874\right) \approx 1.069\text{ min}$

Second Solution: Quadrant III gives: $t_2 = \frac{3}{\pi} \cdot \left(\pi + 0.874\right) = \frac{3}{\pi} \cdot 4.016\approx 3.845\text{ min}$

5. Thus, the second time point A is at a height of 110 m is approximately 3.85 minutes, rounded to 2 decimal places.