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Math Help - Wheel Trig Problem

  1. #1
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    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

    please use diagrams
    Last edited by stones44; November 19th 2007 at 06:02 PM.
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  2. #2
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    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)

    Last edited by Soroban; November 20th 2007 at 10:24 AM.
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  3. #3
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    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...
    Attached Thumbnails Attached Thumbnails Wheel Trig Problem-untitled.jpg  
    Last edited by stones44; November 19th 2007 at 04:47 PM.
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  4. #4
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    wait..what if the pebble is in the first quadrant..than it would be r+rsintheta
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  5. #5
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    Since . we have: . .(a)
    why is it 2x/5pi

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