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Circumference of ellipse

  1. May 10, 2012 #1
    Here is the simple question.

    When I differentiate the area of a circle, I got the parameter. As the Area is the sum of Circumference.

    But why cannot I get the circumference of ellipse by differentiate the area with respect to the radius (I set it as the longer side). The answer I got is 2πa√1-e^2, where e is eccentricity and a is half the longer side.


    THank you!
     
  2. jcsd
  3. May 10, 2012 #2

    tiny-tim

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    hi tomz! :smile:
    no, you get the perimeter (from the greek "peri-", meaning "about") :wink:
    because the gap between two ellipses of the same shape isn't of equal thickness all the way round :smile:
     
  4. May 10, 2012 #3
    Ah, thanks. I think i got it... Thanks.
     
  5. May 12, 2012 #4
    Could you clarify what you mean by this? What is the visual - where is the gap / thickness?
     
  6. May 12, 2012 #5

    HallsofIvy

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    Look at the graphs of
    [tex]\frac{x^2}{4}+ \frac{y^2}{9}= 1[/tex]
    and
    [tex]\frac{x^2}{4.01}+ \frac{y^2}{9.01}= 1[/tex]

    The distance between the two ellipses, as measured along a line through the origin (I started to say "on a line perpendicular to both ellipes" but there is not such a line through any given point on one ellipse), is not the same for all such lines.
     
  7. May 12, 2012 #6
    alright - I see what you mean, now - how does this come to bare on the area to circumference issue? Perhaps the more relevant question is why the derivative of the area of the circle is equal to it's circumference ... thanks in advance!
     
  8. May 12, 2012 #7

    HallsofIvy

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    A circle happens to have the property that if have two circles with the same center but different radii then the distance between the two circles, as measured along a radius, is always the same. That means that if you set up the "area integral" by taking a ring of inner radius r and thickness "[itex]\Delta x[/itex]" it has area approximately [itex]2\pi r[/itex], its circumference, times the thickness [itex]\Delta r[/itex]. Other figures, not having that property, cannot be set up as the integral of the circumference.
     
    Last edited: May 13, 2012
  9. May 12, 2012 #8
    i.e. [itex]dA = 2\pi x~dx[/itex]? So if it was possible to express the circumference of an ellipse strictly in terms of x, then we could do the same? (not sure 'bout the [itex]\Delta r[/itex] above ... did you means [itex]\Delta x[/itex]?)
     
  10. May 13, 2012 #9

    tiny-tim

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    but then it wouldn't be ∫ (circumference) dr,

    the dr would be a d(f(r)) … totally different formula! :redface:
     
  11. May 13, 2012 #10
    Suppose you took an arbitrary ellipse, and then for each point P on the ellipse, you drew the tangent line at that point, and then the normal line (perpendicular to the tangent line) at that point. And you specified the point Q on the normal line which is a distance x from P. What would the locus of all such points Q look like? Would it be an ellipse, or something weirder? I'm pretty sure that it would be continuous, because I think I can prove that doing this procedure for any continuous curve will yield a continuous curve.
     
  12. May 14, 2012 #11
    Imagine what happens to the locus of points Q as the radii are reduced equally. The original ellipse appears to shrink in greater proportion along the minor axis than the major axis. This will eventually cause the minor axis to be reduced to zero, while the radii increase as you move from the minor to the major axis. This figure resembles the infinity symbol, definitely not an ellipse.
     
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