+115 A carnival Ferris wheel with a radius 8 m makes one complete revolution every 20
seconds. The bottom of the wheel is 1 m off the ground. The ride lasts 40 seconds and
starts at the bottom of the wheel.
a) Find an equation of a function that models the height of a person riding the Ferris
wheel throughout their 40 second ride and give the corresponding graph.
b) Little Bobby is a bit afraid of heights and gets really nervous once above 15 meters.
During what time intervals on his ride should we expect little Bobby to be really
nervous?
Answer:
h(t) = -8 cos (π/10 t) + 9, between 7.7 and 12.3 seconds and between 27.7 and 32.3
seconds.
Note: I have the answer, I am looking comletely for the process! Thanks you!
Rewriteable as: -8 cos (π t /10) + 9 [if it looked confusing]
+9481 a)
Maybe this drawing will help...
The thing at the bottom right corner of the triangle is supposed to be somebody riding the Ferris wheel x)
All the lengths are in meters, and t is the number of seconds after the ride starts.
height of person = b + 1
height of person = 8 - 8 cos θ + 1
height of person = 9 - 8 cos θ
height of person = 9 - 8 cos( \(\frac{\pi t}{10}\) )
h(t) = -8 cos( \(\frac{\pi t}{10}\) ) + 9
If you have a question about where any part of this came from, please ask 
Here is a cool graph: https://www.desmos.com/calculator/csy92axqeq
You can choose the value of t using the slider and see where the point representing the person is at.
Also, you can turn on the folder called "graph of height" to see a graph of y = -8 cos( \(\frac{\pi x}{10}\) ) + 9
+9481 b)
You can try looking at that graph and moving the point until the y-value goes just above 15 .
But to find the answer without using that....
Let's find values of t which make h(t) = 15
\(15\ =\ -8\cos(\frac{\pi t}{10})+9\\~\\ 6\ =\ -8\cos(\frac{\pi t}{10})\\~\\ -\frac34\ =\ \cos(\frac{\pi t}{10})\)
There are two values of \(\frac{\pi t}{10}\) in the interval [0, 2π) and thus two values of t which make \(\cos(\frac{\pi t}{10})=-\frac34\)
They are:
\(\begin{array}{} \frac{\pi t}{10}\ =\ \arccos(-\frac34)&\quad&\frac{\pi t}{10}\ =\ 2\pi-\arccos(-\frac34)\\~\\ t\ =\ \frac{10}{\pi}\arccos(-\frac34)&&t\ =\ \frac{10}{\pi}(2\pi-\arccos(-\frac34))\\~\\ t\ \approx\ 7.7&&t\ =\ 20-\frac{10}{\pi}\arccos(-\frac34)\\~\\ &&t\ \approx\ 20-7.7\\~\\ &&t\ \approx\ 12.3 \end{array}\)
The height is 15 meters 7.7 seconds after the starting position and again 7.7 seconds before the starting position.
So we can see that Bobby will be really nervous when t is in the interval (7.7, 12.3)
Bobby will be really nervous again 20 seconds later when the Ferris wheel is back at the same spot, so
Bobby will be really nervous when t is in the interval (27.7, 32.3)
+9481 Here is another diagram:
The person is sitting at point B .
height of person = CD + DE
And DE = 1 because the bottom of the wheel is 1 m off the ground.
height of person = CD + 1
Now we just have to find the length of CD .
CD = AD - AC
And we know AD = 8 because the radius of the Ferris wheel is 8 m.
CD = 8 - AC
And we can find the length of AC by looking at △ABC.
cos θ = adjacent / hypotenuse
cos θ = AC / AB
And we know AB = 8 beause the radius of the Ferris wheel is 8 m.
cos θ = AC / 8
Multiply both sides of the equation by 8 .
8 cos θ = AC
So........
height of person = CD + DE = (AD - AC) + DE = ( 8 - 8 cos θ ) + 1 = -8 cos θ + 9
And....if you don't understand how to relate θ and t .....this might help: try to imagine what would that angle be after 10 seconds? (10 seconds is half of 20 seconds!) What would that angle be after 5 seconds? (5 seconds is a fourth of 20 seconds.)
