Chapter 6, Problem 82RE

Solution

To find: Parametric equations for the given situation that model Sarah’s position as a function of time (in seconds) if she started from the point $\left(0,90\right)$ . At starting point $t=0$ .

  $x=-40\sin \frac{\pi }{9}t$ and $y=50+40\cos \frac{\pi }{9}t$

Given information:

Lowest point of wheel is 10 ft above the ground and its radius is 40 ft. Wheel completes one revolution is 18 seconds. Sarah is start moving from point $\left(0,90\right)$ at $t=0$ seconds.

Formula used:

  $\sin \theta =\frac{\text{perpendicular}}{\text{hypotenuse}}$ and $\cos \theta =\frac{\text{base}}{\text{hypotenuse}}$

Calculation:

Assume an arbitrary point $P\left(x,y\right)$ , which is Sarah’s location at an angle $\theta$ . Suppose the wheel is rotating in the anticlockwise direction.

  

Determine the $x$ -coordinate of point $P$ using the sine function. Apply the trigonometric ratio $\sin \theta =\frac{\text{perpendicular}}{\text{hypotenuse}}$ .

  $\begin{array}{c}\sin \theta =\frac{-x}{40}\\ x=-40\sin \theta \end{array}$

Determine the $y$ -coordinate of point $P$ using the cosine function. Apply the trigonometric ratio $\cos \theta =\frac{\text{base}}{\text{hypotenuse}}$ .

  $\begin{array}{c}\cos \theta =\frac{y-50}{40}\\ 40\cos \theta =y-50\\ y=50+40\cos \theta \end{array}$

The wheel complete one revolution or $2\pi$ distance in every 18 seconds. Represent $\theta$ as a function of time (in seconds).

  $\begin{array}{c}\theta =\frac{2\pi }{18}t\\ =\frac{\pi }{9}t\end{array}$

Substitute $\theta =\frac{\pi }{9}t$ into $x=-40\sin \theta$ and $y=50+40\cos \theta$ to get the parametric equations.

  $x=-40\sin \frac{\pi }{9}t$

  $y=50+40\cos \frac{\pi }{9}t$

Therefore, the parametric equations are $x=-40\sin \frac{\pi }{9}t$ and $y=50+40\cos \frac{\pi }{9}t$ .