Merry-go-round question

Question

Assume the camera is placed 6 feet from the center of the merry-go-round and the diameter of the merry-go-round is 6 feet. Build a trigonometric function that models the distance (feet) of the rider from the camera as a function of time (seconds). Be sure to use function notation to express your function model.

Answer 1

I suggest drawing a figure with the following quantities.

Let

C be the center of the merry-go-round,

A be the point where the camera is located,

B(t) be the point where the rider is located at any time t,

γ(t) be the angle ACB(t) at time t,

r = 3 ft be the radius of the merry-go-round,

d = 6 ft be the distance of the camera from the center, i.e., the length of AC,

ω (unknown) be angular velocity,

c(t) = length of AB(t) (to be found) be the distance of the rider from the camera.

Actually the radius of the merry-go-round is immaterial;

the important thing is the distance of the rider from the center.

As that is not given, I will assume that he is at the edge;

that is, the length of CB(t) = r.

In order to make the answer simple I will assume

1) ω is constant

2) γ(0) = 0

With these assumptions

γ(t) = ωt

Derive c(t) from the triangle AB(t)C by the cosine theorem

c(t) = √(r² + d² - 2rdcos(γ(t)) = √(45 - 36cos(ωt) = 3√(5 - 4cos(ωt))