Chapter 4.6, Problem 45E

b.

To determine

A circle wheel of radius revolves around its centre eight times every second.

Answer to Problem 45E

Express $\theta$ as a function of time $t$ .

  $\theta (t)=16\pi t(t\text{ is in sec })$

Explanation of Solution

As a function of time $t$ , express $\theta$ .

Given: $8$ rotations occur every second.

  $\begin{array}{c}\frac{1}{8}\text{sec}\mapsto 2\pi (=\theta )\text{=5}.\text{8} \text{cm1sec}\\ \text{16}\pi (\text{=}\theta )\text{=5}.\text{8} \text{cmtsec}\\ \text{16}\pi \text{t}(\text{=}\theta )\end{array}$

Conclusion:

Express $\theta$ as a function of time $t$ .

  $\theta (t)=16\pi t(t\text{ is in sec })$

c.

To determine

A point on the edge of the wheel's rate of horizontal and vertical movement when it is in the position indicated by $\theta =\pi /4,\pi /2,\text{and }\pi .$

Answer to Problem 45E

  $\begin{array}{c}\theta =\frac{\pi }{4}\\ \frac{dx}{dt}=-16\pi \sqrt{2}\\ \frac{dy}{dt}=16\pi \\ \sqrt{2}\theta =\frac{\pi }{2}\\ \frac{dx}{dt}=-32\pi \\ \frac{dy}{dt}=0\\ \theta =\pi \\ \frac{dx}{dt}=0\\ \frac{dy}{dt}=-32\pi \end{array}$

Explanation of Solution

Given: A point on the edge's rate of vertical movement and its horizontal movement.

  $\begin{array}{c}\theta =16\pi t\\ \frac{d\theta }{dt}=16\pi \end{array}$

  $\theta =16\pi t$ The derivative should be found in terms of $t.$

  $\begin{array}{c}dx=2\cos \theta \\ \frac{dx}{dt}=-2\sin \theta \\ \frac{d\theta }{dt}=-2\sin \theta \\ (16\pi )=-32\pi \sin \theta \end{array}$

According to a), $x=2\cos \theta$ . After calculating the derivative in terms of $t.$ , replace it with $\frac{d\theta }{dt}=16\pi$ .

  $\begin{array}{c}\theta =\frac{\pi }{4}:\\ \frac{dx}{dt}=-32\pi \\ \sin \frac{\pi }{4}=-32\pi \\ \left(\frac{\sqrt{2}}{2}\right)=-16\pi \\ \sqrt{2}\theta =\frac{\pi }{2}:\\ \frac{dx}{dt}=-32\pi \\ \sin \frac{\pi }{2}=-32\\ \pi (1)=-32\pi \\ \theta =\pi :\frac{dx}{dt}=-32\pi \\ \sin \pi =-32\\ \pi (0)=0\end{array}$

At the provided angles of $\theta =\frac{\pi }{4},\frac{\pi }{2},$ and $\pi$ calculate the value of $\frac{dx}{dt}$ .

  $\begin{array}{c}y=2\sin \theta \\ \frac{dy}{dt}=2\cos \theta \\ \frac{d\theta }{dt}=2\cos \theta \\ (16\pi )=32\pi \cos \theta \end{array}$

  $y=2\sin \theta$ After calculating the derivative in terms of $t,$ substitute in $\frac{d\theta }{dt}=16\pi .$

  $\begin{array}{c}\theta =\frac{\pi }{4}\\ \frac{dy}{dt}=32\pi \\ \cos \frac{\pi }{4}=32\pi \\ \left(\frac{\sqrt{2}}{2}\right)=16\pi \\ \sqrt{2}\theta =\frac{\pi }{2}\\ \frac{dy}{dt}=32\pi \\ \cos \frac{\pi }{2}=32\pi \\ (0)=0\theta =\pi \\ \frac{dy}{dt}=32\pi \\ \cos \pi =32\pi \\ (-1)=-32\pi \end{array}$

Determine $\frac{dy}{dt}$ at the specified angles. $\theta =\frac{\pi }{4},\frac{\pi }{2},$ and $\pi .$

Conclusion:

  $\begin{array}{c}\theta =\frac{\pi }{4}\\ \frac{dx}{dt}=-16\pi \sqrt{2}\\ \frac{dy}{dt}=16\pi \\ \sqrt{2}\theta =\frac{\pi }{2}\\ \frac{dx}{dt}=-32\pi \\ \frac{dy}{dt}=0\\ \theta =\pi \\ \frac{dx}{dt}=0\\ \frac{dy}{dt}=-32\pi \end{array}$