1. ## Wheel Trig Problem

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 you have traveled. Diameter is 5pi inches

L=distance of the pebble from the left-most edge of the tire after you have traveled x inches. Write en equation relating L and X….work in degrees and radians? Degrees first I guess than convert…I dno.

Then…..determine the first positive x value that makes L = 1

2. 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 $5\pi$ inches.

$L$ = distance of the pebble from the leftmost edge of the tire.
$x$ = horizontal displacement of the wheel.

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

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

* E       O         *
D *-+-------*         *
* :     / |         *
:   / θ |
*: /     |        *
C* - - - +A      *
*     |     *
- - - - - * * * - - - - - -
B
Let the radius be $r = OB = OC = OD$
Let $L = DE$.

When the wheel has turned $\theta$ radians,
. . it has moved a distance $x$ equal to arc(BC): . $x \:=\:r\theta\quad\Rightarrow\quad\theta \:=\:\frac{x}{r}$ .[1]

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

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

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

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

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

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

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

3. yeah the distance L.....its DC not DE....the pebble is on the circumference somewhere....

oh wait..but it said horizontal distance...OK!

heres what i have so far...

4. wait..what if the pebble is in the first quadrant..than it would be r+rsintheta

5. Since . we have: . .(a)
why is it 2x/5pi

shouldnt it be x/2.5pi...are they the same thing?