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Rearranging a formula (segment area)

  1. Nov 5, 2006 #1
    Hey
    I’ve been given a problem to solve involving the minor segment of a circle. I know that the formula to solve for a segment is:

    [tex]S = \frac{1}{2}r^2 (\theta - \sin \theta )[/tex]

    However in this problem I’ve been given the segment area and the radius of the circle and been ask to find the minor angle. So far the best I have managed in rearranging the formula is:

    [tex] \theta - \sin \theta = \frac{{2S}}{{r^2 }}[/tex]

    I am confused on what to do with the sin theta part, so if somebody out there is able to help it will be greatly appreciated. Or am I wasting my time and this is not possible?
    Many thanks in advance
    Pavadrin
     
  2. jcsd
  3. Nov 5, 2006 #2

    verty

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    The only way I can fathom to solve this is to plot the graphs of theta and sin(theta) over the interval {0; pi} and read the answer off the graph.
     
  4. Nov 5, 2006 #3
    okay thanks for that, didn't even consider graphing it :)
     
  5. Nov 5, 2006 #4
    You can use derivatives. If you have been given a function f(x). this function has a local minimum/maximum at those x values where [tex]\frac{df(x)}{dx} = 0[/tex]. To be sure you are looking at a local minimum, you need to make a sign chart of the first derivative you just calculated. At those x-points when the sign of the [tex]\frac{df(x)}{dx} = 0[/tex] changes from - to + , you are in a minimum.

    In your case you have a function S of both r and [tex]\theta[/tex]. The [tex]\theta[/tex] corresponding to the minimum segment can be found by taking the first derivative of S with respect to r ([tex]\theta[/tex] is considered to be a constant) and solve that equation for [tex]\theta[/tex].

    this should get you started

    regards
    marlon
     
  6. Nov 5, 2006 #5

    arildno

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    If you can assume that the angle is tiny, you may approximate the sine with a finite sum, say [tex]\sin(\theta)\approx\theta-\frac{\theta^{3}}{6}[/tex]
    This will yield:
    [tex]\frac{\theta^{3}}{6}\approx\frac{2S}{r^{2}}\to\theta\approx(\frac{12S}{r^{2}})^{\frac{1}{3}}[/tex]
     
  7. Nov 6, 2006 #6

    HallsofIvy

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    marlon, what does this problem have to do with finding a minimum?
     
  8. Nov 6, 2006 #7
    Well, isn't he supposed to find the angle corresponding to the minimum segment surface ?

    marlon
     
  9. Nov 6, 2006 #8

    Office_Shredder

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    No, he's trying to find what angle a segment intersects on a circle given the radius of the circle and the length of the segment
     
  10. Nov 6, 2006 #9
    Opps then i misread the question.

    marlon
     
  11. Nov 7, 2006 #10

    HallsofIvy

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    Perhaps you misread " the minor segment ". Two radii divide a circle into two arcs. Unless the radii are part of the same diameter, one arc is smaller than the other: the "minor arc" and, by extension, forms the "minor segment".
     
  12. Nov 7, 2006 #11
    KABOOMMMM... Indeed you are right : i completely misread that question. Thanks for the clarification.

    regards
    marlon
     
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