Chapter 6.3, Problem 51E

Solution

To find:the parametric equations for the position of Ron as a function of time

The parametric equation for the position of the Ron is

  $x=35\cos \left(\frac{\pi }{6}t\right),$

  $y=15+35\sin \left(\frac{\pi }{6}t\right)$ .

Given information:

Given that

• a Ferris wheel with a radius of 35 ft
• the wheel turns counterclockwise at the rate of 12 sec/revolution
• the lowest point of the wheel is 15 ft above the ground level at the point $\left(0,15\right)$
• Ferris wheel starts $t=0$ with Ron at the point $\left(35,50\right)$

Formula used:

Any point on the circle ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ is expresses as $x=h+r\cos \theta ,$

  $y=k+r\sin \theta$ , where $\theta$ is an angle made with the positive x-axis with the line joining the center $\left(h,k\right)$ and the point $\left(x,y\right)$ .

Calculation:

With the given information, conclude that the Ferris wheel centered at $\left(0,50\right)$ .The equation of the circle is shown below.

  

Figure (1)

To get the parametric equations of the circle in terms of the parameter $\theta$ , substitute 35for $r$ , 0 for $h$ , and 50 for $k$ in the equations $x=h+r\cos \theta ,$

  $y=k+r\sin \theta$ :

  $x=35\cos \theta ,$

  $y=50+35\sin \theta$ .

Given that the wheel is turning at the rate 12 sec/revolution. That is, the wheel is turning $2\pi$ radians for 12 seconds or $\frac{2\pi }{12}=\frac{\pi }{6}$ radians/sec. So $\theta$ can be expressed as $\frac{\pi }{6}t$ radians. Thus, the parametric equation for the position of the Ron is

  $x=35\cos \left(\frac{\pi }{6}t\right),$

  $y=15+35\sin \left(\frac{\pi }{6}t\right)$ .