Ferris Wheel Height and a False Advertisement
An amusement park Ferris wheel has a diameter of 40 m. Its centre is 25 m above the ground, and it rotates once every 3 minutes. Riders board at the lowest point, which is 5 m above the ground, at time t = 0. The height h(t) metres of a rider above the ground is modelled by h(t) = 25 − 20cos(2πt/3). The operator advertises that 'for half of every ride, passengers are more than 35 m above the ground.' A school group plans to take photographs from the wheel and wants to know when, during the first revolution, the wheel is above 35 m and for how long. A safety inspector also asks whether the rider is ever above 40 m and, if so, for what fraction of the revolution. The operator's claim is being checked before the advertisement is printed.
1. Solve h(t) = 35 for the first revolution. State the times and the duration for which the rider is above 35 m, and compare with the operator's claim.
Read the worked answer
Set 25 − 20cos(2πt/3) = 35. Then −20cos(2πt/3) = 10, so cos(2πt/3) = −1/2.
During one revolution, 2πt/3 = 2π/3 or 4π/3, giving t = 1 or t = 2 minutes. The cosine is less than −1/2 between these angles, so h(t) > 35 for 1 < t < 2. The duration is 1 minute out of a 3-minute ride, which is one-third, not one-half. The advertisement is false.
2. Determine whether the rider is ever above 40 m and find the fraction of the revolution spent above 40 m.
Read the worked answer
Set 25 − 20cos(2πt/3) = 40. Then cos(2πt/3) = −3/4. Let α = arccos(−3/4) ≈ 2.419 rad. Then 2πt/3 = α or 2π − α, so t ≈ 1.155 min or t ≈ 1.845 min.
The duration above 40 m is about 1.845 − 1.155 = 0.690 min. As a fraction of the 3-minute revolution this is 0.690/3 ≈ 0.230, or about 23%. The maximum height is 45 m at t = 1.5 min, so the rider does go above 40 m, but not for half the ride.