# Math

Here is the question :

A Ferris wheel is 40 meters in diameter and boarded from a platform that is 5 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 4 minutes. How many minutes of the ride are spent higher than 37 meters above the ground?

I'm thinking about the unit circle with this, but I know we also probably need to make a graph because they give us a period other than 2pi. The period would be 4 right? Because it fully rotates in 4 minutes. I don't know what the amplitude would be. Maybe 35 because the maximum is 40 and it is 5 meters above ground.

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height of center = 25
height = 25 + 20 sin [2 pi t/4 - pi/2}

I use 2 pi t/4 because that gives 0 at t = 0 and a full revolution, 2 pi, at t = 4.
I put in the phase of -pi/2 so that at t = 0 we have 25 +20 sin -pi/2
= 25 - 20 = 5 meters, the starting point at t = 0
=================================
Now, all of that said, it is not necessary
we need to know what fraction of the circumference is above 37 meters or how much of the circle is more than 37-5 = 32 meters above bottom point of circle
which is 12 meters above center.
T is angle from horizontal through center to 12 feet above center
sin T = 12/20
T = 37 degrees approximately
so out of the 180 degree upward swing, the rider spends 180 - (37+90) = 53 degrees above 37 meters on the rise
or a total of 106 degrees including the start of the fall
the wheel spends 4 minutes going from bottom to top so
time above = (106/360)4 = 1.18 minutes

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posted by Damon

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