Hello, stones44!

A pebble is wedged in the bottom of a tire.

The horizontal distance of the pebble from the left-most edge of the tire varies

sinusoidally with the distance the wheel has traveled. The diameter is $\displaystyle 5\pi$ inches.

$\displaystyle L$ = distance of the pebble from the leftmost edge of the tire.

$\displaystyle x$ = horizontal displacement of the wheel.

(a) Write an equation relating $\displaystyle L$ and $\displaystyle x.$

(b) Determine the first positive $\displaystyle x$ value that makes $\displaystyle L = 1$ Code:

* * *
* *
* *
* *
* E O *
D *-+-------* *
* : / | *
: / θ |
*: / | *
C* - - - +A *
* | *
- - - - - * * * - - - - - -
B

Let the radius be $\displaystyle r = OB = OC = OD$

Let $\displaystyle L = DE$.

When the wheel has turned $\displaystyle \theta$ radians,

. . it has moved a distance $\displaystyle x$ equal to arc(BC): .$\displaystyle x \:=\:r\theta\quad\Rightarrow\quad\theta \:=\:\frac{x}{r}$ .**[1]**

We see that: .$\displaystyle AC \:=\: OE\:=\:r\sin\theta$

. . Hence: .$\displaystyle L \:=\:r - r\sin\theta \:=\:r(1 - \sin\theta)$ .**[2]**

Substitute [1] into [2]: .$\displaystyle L \;=\;r\left(1 - \sin\frac{x}{r}\right)$

Since $\displaystyle r = \frac{5\pi}{2}$. we have: .$\displaystyle \boxed{L \;=\;\frac{5\pi}{2}\left(1 - \sin\frac{2x}{5\pi}\right)} $ .**(a)**

If $\displaystyle L = 1$, we have: .$\displaystyle 1 \;=\;\frac{5\pi}{2}\left(1 - \sin\frac{2x}{5\pi}\right) \quad\Rightarrow\quad \frac{2}{5\pi} \;=\;1 - \sin\frac{2x}{5\pi}$

. . $\displaystyle \sin\frac{2x}{5\pi} \;=\;1 - \frac{2}{5\pi}\;=\;\frac{5\pi-2}{5\pi} \quad\Rightarrow\quad \frac{2x}{5\pi} \;=\;\sin^{-1}\!\left(\frac{5\pi-2}{5\pi}\right) $

Therefore: .$\displaystyle x \;=\;\frac{5\pi}{2}\sin^{-1}\!\left(\frac{5\pi-2}{5\pi}\right) \;\approx\;\boxed{8.33\text{ inches}} $ .**(b)**