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Homework Help: Maximising the Area of a Chord.

  1. Aug 22, 2010 #1
    1. The problem statement, all variables and given/known data
    This is a problem within a problem. I need to differentiate the area of a chord of find the maximum area (and hopefully, in the process, radius).

    2. Relevant equations
    I found this equation on another site:
    A=R^2[(Pi/180*c - sin c)]/2
    Where:
    • C is the central angle in degrees.
    • R is the radius of the circle of which the segment is a part.
    • A is the area of the chord.
    (I think I typed it out correctly. Here's the website: http://www.mathopenref.com/segmentarea.html)

    3. The attempt at a solution
    First problem: How can I differentiate the equation when the is denominator is 2? My central angle is 180 Degrees, so it worked out to be:
    A=R^2[(Pi - sin 180)]/2
    Something is terribly wrong, I'm sure. Is it even the right formula? There is also this formula:
    http://mathworld.wolfram.com/Chord.html

    Sorry it's a little messy, but my thought pattern is crazy at the moment. Thanks for any help.
     
  2. jcsd
  3. Aug 22, 2010 #2

    HallsofIvy

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    Science Advisor

    First, a chord is a line and so has no "area". I think you mean the area of the portion of a segment of a circle lieing between the chord and the circle.

    But I don't see what your difficulty is. Having a "2" in the denominator is just another constant- it multiplies the final derivative.

    If [itex]A= R^2(c- sin(c))/2[/itex] (note that I have removed the "pi/180" from your formula. That would be assuming that c is in degrees and the derivative of sin(x) is cos(x) only if x is in radians so I am treating c as being in radians to begin with.)

    [itex]A= (R^2/2)(c- sin(c))[/itex] so that [itex]A'= (R^2/2)(1- cos(c))[/itex].

    And what do you mean your "central angle is 180 degrees"? I thought you were trying to find the central angle that will give the largest area. Are you just checking to be sure the area formula works?

    If c= 180 degrees, which is [itex]\pi/2[/itex] radians, then sin(c)= -1 so [itex]A= R^2\pi/2[/itex], the area of half of a circle as you would expect.
     
  4. Aug 22, 2010 #3
    Ah, well there's a problem with my assignment right off the bat. I think I may be approaching it wrong.

    The question asks us to minimise the materials needed for a gutter, in this case a circle. We then have to compare it to other shapes. My logic is to first find the maximum radius of the chord (effectively, where the gutter lets water in) and then set it as the radius for the circumference of the circle. Is that the right logic?
     
    Last edited: Aug 22, 2010
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