Question

farrell boards a ferris wheel at the 3 - oclock position and rides the ferris wheel for several rotations. the ferris wheel has a radius 13 meters long, and when farrell boards the ferris wheel he is 17 meters above the ground. imagine an angle with its vertex at the center of the ferris wheel that subtends the path farrell travels.
a. complete the definition for function ( f ) that expresses farrells horizontal distance to the right of the center of the ferris wheel (in meters) in terms of the number of radians, ( \theta ), the angle has swept out since the ride started. (remember you can type \theta\ for ( \theta ).)
( f(\theta)= ) 13cos(theta)
preview
b. complete the definition for function ( g ) that expresses farrells distance above the ground (in meters) in terms of the number of radians, ( \theta ), the angle has swept out since the ride started. (remember you can type \theta\ for ( \theta ).)
( g(\theta)= ) 13sin(theta)
preview

Explanation:

Step1: Recall the definitions of sine and cosine in a right - triangle

In a right - triangle with hypotenuse \(r\), adjacent side \(x\), opposite side \(y\) and angle \(\theta\), we have \(\cos\theta=\frac{x}{r}\) and \(\sin\theta=\frac{y}{r}\). For a Ferris wheel with radius \(r = 13\) meters.

Step2: Define the functions based on the problem description

a. For the horizontal distance \(f(\theta)\):
We know that if the radius of the Ferris wheel is \(r = 13\) meters, and using the cosine function (adjacent side in the right - triangle formed with the center of the Ferris wheel). If the center of the Ferris wheel is at a certain height (but for the horizontal distance from the right of the center), we have \(f(\theta)=13\cos\theta\) (since \(\cos\theta=\frac{f(\theta)}{13}\), then \(f(\theta)=13\cos\theta\)).
b. For the vertical distance (height above the ground) \(g(\theta)\):
The center of the Ferris wheel is at a height of \(17\) meters. Using the sine function (opposite side in the right - triangle formed with the center of the Ferris wheel). We know that \(\sin\theta=\frac{h}{13}\) (where \(h\) is the vertical displacement from the center). So the height above the ground \(g(\theta)=17 + 13\sin\theta\)

Answer:

a. \(f(\theta)=13\cos\theta\)
b. \(g(\theta)=13\sin\theta + 17\)