Chapter 6, Problem 81RE

Solution

To find: Parametric equations for the given situation that model Henry’s position as a function of time

  $x=40\sin \frac{2\pi }{15}t$ and $y=50-40\cos \frac{2\pi }{15}t$

Given information:

Lowest point of wheel is 10 ft above the ground and its radius is 40 ft. Wheel completes one revolution is 15 seconds. The starting point of revolution is given at $\left(0,10\right)$ .

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 Henry’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 $\cos \theta =\frac{\text{base}}{\text{hypotenuse}}$ .

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

The wheel complete one revolution or $2\pi$ distance in every 15 seconds. Represent $\theta$ as a function of time (in seconds).

  $\theta =\frac{2\pi }{15}t$

Substitute $\theta =\frac{2\pi }{15}t$ into $x=40\sin \theta$ and $y=50-40\cos \theta$ to get the parametric equations.

  $x=40\sin \frac{2\pi }{15}t$

  $y=50-40\cos \frac{2\pi }{15}t$

Therefore, the parametric equations are $x=40\sin \frac{2\pi }{15}t$ and $y=50-40\cos \frac{2\pi }{15}t$ .