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09.04 Applications of Circles: London Eye
Transcript of 09.04 Applications of Circles: London Eye
1. Name of the Ferris wheel - The London Eye
2. Diameter of the wheel - 120m
3. Number of cars / compartments - 32
4. Circumference of the wheel - 2πr = 2π(60) = 376.99 ≈ 377m
5. Area of the wheel - πr^2 = π(60)^2 = 11,309.73 ≈ 11,310m^2
6. Measure of a central angle in degrees - 360 / 32 = 11.25 degrees
7. Measure of a central angle in radians - 2π(x) / 360
2π(11.25) /360 = .1963495408 ≈ .20 radians
8. Arc length between two cars or compartments - 2πr(x) / 360
2π(60)(11.25) /360 = 11.78
9. Area of a sector between two cars or compartments - πr^2(x) / 360
π(60^2)(11.25) / 360 = 353.43
d = 120m
10. If a smaller replica of the Ferris wheel was constructed, what conclusions could you draw about the central angle of the original wheel and replica? What conclusions could you draw about the arc length of the original Ferris wheel and replica?
If a small replica of the Ferris wheel is constructed, it would still have a wheel that is a circle, and would have 32 compartments. And to find the central angle, I did 360/32 for the original wheel. I will have to do the same thing for the replica too. So the central angle of the original wheel and the replica, would be the same. But the arc length will be different. To find the arc length of a circle, you will do 2πr(x) / 360. The arc length of the original wheel will be greater than the arc length of the replica. This is because the radius of the 2 wheels will be different. Therefore, the arc length of the replica will not be the same as the replica.
11. Imagine the center of the Ferris wheel is located at (0, 0) on a coordinate grid and the radius lies on the x-axis. Write an equation of a circle for your Ferris wheel, and sketch an image of what your Ferris wheel would look like on the grid.
Formula for a circle: (x-h)^2 + (y-k)^2 = r^2
(x-0)^2 + (y-0)^2 = 3600
x^2 + y^2 = 3600