SOLUTION: A Ferris wheel is 45 meters in diameter and boarded from a platform that is 2 meters above the ground. The six o'clock position on the Ferris wheel is level with the loading platfo

Question 1149303: A Ferris wheel is 45 meters in diameter and boarded from a platform that is 2 meters above the ground. The six o'clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 10 minutes. How many minutes of the ride are spent higher than 28 meters above the ground?

Answer by greenestamps(13203) (Show Source): You can put this solution on YOUR website!

The height above the ground is a sinusoidal function.

The minimum height is 2; the maximum height is 2+45 = 47; the midline is (2+47)/2 = 24.5.

We can model a revolution of the Ferris wheel as starting at the loading platform, which is the minimum height of the ride; so we can model the height with a negative cosine function:

For this problem, we don't need to use the 10-minute rotation time of the Ferris wheel in our function; we can use a "plain" cosine function and simply determine what fraction of one revolution is spent above 28 feet.

Here is a graph of a bit more than one period of the function, along with the constant function 28, showing the ride at its minimum height at 0 degrees of revolution and again at 360 degrees:

To find the fraction of a period during which the height of the ride is above 28 feet, you can use a graphing calculator to find the points of intersection of the two graphs.

Algebraically, you can find the two angles when the height of the ride is 28 feet by solving the equation

You can find those angles by either of those methods; I leave that to you.

The final answer is that the ride is at or above 28 feet for approximately 162.1 degrees of every 360-degree revolution.

to 3 decimal places.

Then, since the period of revolution is 10 minutes, the ride is above 28 feet for 4.5 minutes each revolution.

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