Chapter 6, Problem 80RE

Solution

To find: The parametric equations that model Lucinda’s position as a function of time $t$ .

The parametric equations as a function of $t$ are $x=35\cos \left(\frac{\pi t}{10}\right)$ and $y=50+35\sin \left(\frac{\pi t}{10}\right)$ .

Given data:

Radius of wheel $=35\text{ ft}$

Lowest point of Ferris wheel $=\left(0,15\right)$

Starting point $=\left(35,50\right)$

Starting time, $t=0$

Formula used:

A parametrization of a curve consists of the parametric equations and the interval of $t$ values.

The graph of the ordered pairs $\left(x,y\right)$ where $x=f(t)$ and $y=g(t)$ are functions defined on an interval $l$ of $t$ values is a parametric curve.

Calculation:

Figure 1 shows a circle with center $\left(0,50\right)$ and radius $35\text{ ft}$ models Ferris wheel.

  

Figure (1)

Compute relation between angle $\left(\theta \right)$ and time $\left(t\right)$ .

  $2\pi$ for every $20$ seconds.

  $\begin{array}{c}\theta =\frac{2\pi }{20}\times t\\ =\frac{\pi }{10}t\end{array}$

Compute parametrized equation.

  $\begin{array}{c}x=35\cos \left(\frac{\pi t}{10}\right)+0\\ =35\cos \left(\frac{\pi t}{10}\right)\\ y=50+35\sin \left(\frac{\pi t}{10}\right)\end{array}$

Therefore, the parametric equations as a function of $t$ are $x=35\cos \left(\frac{\pi t}{10}\right)$ and $y=50+35\sin \left(\frac{\pi t}{10}\right)$ .