Given that the Ferris diameter is 20 meters, with a rate of rotation of 1 turn in 8 minutes, we have;
a. Amplitude: A = 10 meters
Midline: h = 11 meters
Period, P = 8 minutes
b. h(t) = 10•cos((π/4)•t + π) + 11
c. 11 meters
The amplitude is the same as the radius of the Ferris wheel,
The radius of the Ferris wheel = 20 ÷ 2 = 10
Therefore;
[tex]the \: midline \: = \frac{max \: height \: + min \: height}{2} [/tex]
Therefore;
[tex]midline \: = \frac{10 + 10 + 1 + 1}{2} = 11[/tex]
The period is the time to complete one rotation, therefore;
b. h(t) = A•cos(B•t + C) + h
Where;
B = 2•π/P
When t = 0, h(t) = 1
Which gives;
h(0) = 1 = 10 × cos(B×0 + C) + 11
-10/10 = -1 = cos(C)
C = arcos(-1) = π
Therefore;
h(0) = 1
h(0) is the minimum value of h(t)
c. After 30 minutes, we have;
h(30) = 10•cos((π/4)×30 + π) + 11 = 11
Learn more about the Ferris wheel here:
https://brainly.com/question/86214