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Math Help - Related Rate of Change of an Angle

  1. #1
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    Related Rate of Change of an Angle

    Hello

    Out of pure maths interest, I was trying to solve the following problem:

    We have: point O, anywhere on the plan; line AB, anywhere on the plan; point M, moving at speed described by func s(t) on AB, from A.
    Find the rate of change of angle AOM (from time t), given that you can know everything about triangle ABO (lengths, angles...) but that this triangle is common (not iso, not rect, not equi....).

    I have created a graph of this using geometric methods... with GeoGebra. It looks a bit like a Gauss curve.

    When I now try to get the function for it, I tried several ways, and I couldn't. All I found was some strange graphs... (reversed U with vertical tangent on right side, graph in -1<y<1 -1<x<1; that was the latest... ?!)

    I tried searching on internet but to no avail.

    I know the basics of differentiation and am studying more.


    If anyone has an idea...?


    PS: I'll post a sketch when I've got time...
    Last edited by fxsapa; March 21st 2009 at 12:47 AM.
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  2. #2
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    Hello, fxsapa!

    We have: point O, anywhere on the plane; line AB, anywhere on the plane.
    Point M, moving on AB from A, at speed described by function s(t).

    Find the rate of change of \angle AOM (from time t),
    given that you can know everything about \Delta ABO (lengths, angles, ...)
    but that this triangle is scalene.
    Code:
                          O
                          *
                        */   *
                      * /       *
                 b  * θ/           *
                  *   /               *
                *    /                   *
              *     /                       *
            *    φ /                           *
        A *   *   *   *   *   *   *   *   *   *   * B
             x    M

    We have: . \Delta AOB,\:A = \angle OAB,\;\theta = \angle AOM,\;b = OA,\;x = AM
    . . . .and: . \phi = \angle AMO \:=\:180^o - A - \theta


    In \Delta OAM, apply the Law of Sines:

    . . \frac{\sin\theta}{x} \:=\:\frac{\sin(180-A-\theta)}{b} \quad\Rightarrow\quad b\sin\theta \;=\;x\sin([180 - A] - \theta)<br />

    . . b\sin\theta \;=\;x\bigg[\underbrace{\sin(180-A)}_{\text{This is }\sin A}\cos\theta - \underbrace{\cos(180-A)}_{\text{This is }-\cos A}\sin\theta\bigg]

    . . b\sin\theta \;=\;x\underbrace{\bigg[\sin A\cos\theta + \cos A\sin\theta\bigg]}_{\text{This is }\sin(\theta + A)}

    . . b\sin\theta \;=\;x\sin(\theta + A)


    Differentiate with respect to time:

    . . b\cos\theta\,\frac{d\theta}{dt} \;=\; x\cos(\theta + A)\,\frac{d\theta}{dt} + \sin(\theta + A)\,\frac{dx}{dt}

    . . b\cos\theta\,\frac{d\theta}{dt} - x\cos(\theta + A)\,\frac{d\theta}{dt} \;=\;\sin(\theta + A)\,\frac{dx}{dt}

    . . \bigg[b\cos\theta - x\cos(\theta+A)\bigg]\,\frac{d\theta}{dt} \;=\;\sin(\theta + A)\,\frac{dx}{dt}


    Therefore: . \frac{d\theta}{dt} \;=\;\frac{\sin(\theta+A)\,\frac{dx}{dt}}{b\cos\th  eta - x\cos(\theta +A)}


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  3. #3
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    Thx
    Last edited by fxsapa; January 13th 2010 at 01:22 PM. Reason: thx
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