Chapter 5.6, Problem 45E

a.

To determine

To explain: how the parametric equation can be used to represent the motion of the wheel.

Answer to Problem 45E

The x and y values of the parametric equation would correspond to a point on the edge of the wheel as the wheel turns.

Explanation of Solution

Given information:

  $\text{Radius}=2\text{ ft}\text{.}$

  $x=2\cos \theta$

  $y=2\sin \theta$

  $f=8\text{ revolution/second}$

Calculation :

We have to explain how the parametric equation can be used to represent the motion of the wheel.

Let $\theta$ be minimum.

So,

  $\theta =0$

Therefore,

  $x=2\cos \theta$

Put $\theta =0$ ,

  $\begin{array}{l}x=2\cos \left(0\right)\\ x=2\left(1\right)\left[\text{Since }\cos \left(0\right)=1\right]\\ x=2\end{array}$

And

  $y=2\sin \theta$

Put $\theta =0$ ,

  $\begin{array}{l}y=2\sin \left(0\right)\\ y=2\left(0\right)\left[\text{Since }\sin \left(0\right)=0\right]\\ y=0\end{array}$

Again,

Let $\theta$ be minimum.

  $\theta ={180}^{\circ }$

Therefore,

  $x=2\cos \theta$

Put $\theta ={180}^{\circ }$ ,

  $\begin{array}{l}x=2\cos \left(180\right)\\ x=2\left(-1\right)\left[\text{Since }\cos \left(180\right)=-1\right]\\ x=-2\end{array}$

And

  $y=2\sin \theta$

Put $\theta ={180}^{\circ }$ ,

  $\begin{array}{l}y=2\sin \left(180\right)\\ y=2\left(0\right)\left[\text{Since }\sin \left(180\right)=0\right]\\ y=0\end{array}$

From $\theta =0$ , the x values start at 2 and from $\theta ={180}^{\circ }$ , $x=-2$ . Therefore, the x -values of the parametric equation will constantly swing from 0 to 2 to 0 to -2 to 0 and so on.

From $\theta =0$ , the x values start at 0, increases $y=2$ and decreases $y=-2$ . Therefore, the y- values of the parametric equation will constantly swing from 0 to 2 to 0 to -2 to 0 and so on.

Hence, the x and y values of the parametric equation would correspond to a point on the edge of the wheel as the wheel turns.

b.

To determine

To express: $\theta$ as a function of t .

Answer to Problem 45E

The value of $\theta$ is $16\pi t$ .

Explanation of Solution

Given information:

  $\text{Radius}=2\text{ ft}\text{.}$

  $x=2\cos \theta$

  $y=2\sin \theta$

  $f=8\text{ revolution/second}$

Calculation :

We have to express $\theta$ as a function of t when wheel is rotating 8 revolutions/second.

First, convert revolution per second to radians per second.

  $1\text{ revolution}=2\pi \text{ radians}$

Therefore,

  $\theta =2\pi \times 8\text{ radians/sec}\text{.}$

When $\theta$ will rotate t seconds, then

  $\begin{array}{l}\theta =2\pi \times 8\times t\\ \theta =16\pi t\end{array}$

Hence, the value of $\theta$ is $16\pi t$ .

c.

To determine

To find: the rate of horizontal and vertical movement when $\theta =\frac{\pi }{4}$ , $\frac{\pi }{2}$ and $\pi$

Answer to Problem 45E

The values of when-

  $\theta =\frac{\pi }{4}$ , $\frac{dx}{dt}=-16\pi \sqrt{2}$ and $\frac{dy}{dt}=16\pi \sqrt{2}$

  $\theta =\frac{\pi }{2}$ , $\frac{dx}{dt}=-32\pi$ and $\frac{dy}{dt}=0$

  $\theta =\pi$ , $\frac{dx}{dt}=0$ and $\frac{dy}{dt}=-32\pi$

Explanation of Solution

Given information:

  $\text{Radius}=2\text{ ft}\text{.}$

  $x=2\cos \theta$

  $y=2\sin \theta$

  $f=8\text{ revolution/second}$

Calculation :

We have to find the rate of horizontal and vertical movement when $\theta =\frac{\pi }{4}$ , $\frac{\pi }{2}$ and $\pi$

We have found $\theta$ in previous sums that is $\theta =16\pi t$

Differentiate with respect to t .

  $\begin{array}{l}\frac{d\theta }{dt}=16\pi \frac{dt}{dt}\\ \frac{d\theta }{dt}=16\pi \left[\text{Since, }\frac{dt}{dt}=1\right]\end{array}$

Therefore,

  $x=2\cos \theta$

Differentiate with respect to t .

  $\frac{dx}{dt}=2\sin \theta \frac{d\theta }{dt}$

Putting the value of $\frac{d\theta }{dt}=16\pi$ in above equation,

  $\begin{array}{l}\frac{dx}{dt}=-2\sin \theta \left[16\pi \right]\\ \frac{dx}{dt}=32\pi \sin \theta \end{array}$

When

  $\theta =\frac{\pi }{4}$ ,

  $\begin{array}{l}\frac{dx}{dt}=-32\pi \sin \left[\frac{\pi }{4}\right]\\ \frac{dx}{dt}=-32\pi \left(\frac{\sqrt{2}}{2}\right)\\ \frac{dx}{dt}=-16\pi \left(\sqrt{2}\right)\end{array}$

When

  $\theta =\frac{\pi }{2}$ ,

  $\begin{array}{l}\frac{dx}{dt}=-32\pi \sin \left[\frac{\pi }{2}\right]\\ \frac{dx}{dt}=-32\pi \left(1\right)\\ \frac{dx}{dt}=-32\pi \end{array}$

When

  $\theta =\pi$ ,

  $\begin{array}{l}\frac{dx}{dt}=-32\pi \sin \left[\pi \right]\\ \frac{dx}{dt}=-32\pi \left(0\right)\\ \frac{dx}{dt}=0\end{array}$

Again,

  $y=2\sin \theta$

Differentiate with respect to t .

  $\frac{dy}{dt}=2\cos \theta \frac{d\theta }{dt}$

Putting the value of $\frac{d\theta }{dt}=16\pi$ in above equation,

  $\begin{array}{l}\frac{dy}{dt}=2\cos \theta \left[16\pi \right]\\ \frac{dx}{dt}=32\pi \cos \theta \end{array}$

When

  $\theta =\frac{\pi }{4}$ ,

  $\begin{array}{l}\frac{dy}{dt}=32\pi \cos \left[\frac{\pi }{4}\right]\\ \frac{dy}{dt}=32\pi \left(\frac{\sqrt{2}}{2}\right)\\ \frac{dy}{dt}=16\pi \left(\sqrt{2}\right)\end{array}$

When

  $\theta =\frac{\pi }{2}$ ,

  $\begin{array}{l}\frac{dy}{dt}=32\pi \cos \left[\frac{\pi }{2}\right]\\ \frac{dy}{dt}=32\pi \left(0\right)\\ \frac{dy}{dt}=0\end{array}$

When

  $\theta =\pi$ ,

  $\begin{array}{l}\frac{dy}{dt}=32\pi \cos \pi \\ \frac{dx}{dt}=32\pi \left(-1\right)\\ \frac{dx}{dt}=-32\pi \end{array}$

Hence, the values of when

  $\theta =\frac{\pi }{4}$ , $\frac{dx}{dt}=-16\pi \sqrt{2}$ and $\frac{dy}{dt}=16\pi \sqrt{2}$

  $\theta =\frac{\pi }{2}$ , $\frac{dx}{dt}=-32\pi$ and $\frac{dy}{dt}=0$

  $\theta =\pi$ , $\frac{dx}{dt}=0$ and $\frac{dy}{dt}=-32\pi$