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Length of the ellipse

  1. Mar 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the length of the ellipse [tex]9x^2 + 10y^2 = 90[/tex] correct to six decimal places.


    2. Relevant equations

    [tex]4L[/tex]arc in the first quadrant = [tex]L[/tex]ellipse

    3. The attempt at a solution

    Just checking to see if I did this right:

    [tex]9x^2 + 10y^2 = 90[/tex]

    [tex]x^2/10 + y^2/9 = 1[/tex]

    Therefore a = [tex]\sqrt{10}[/tex] and b = 3.

    This makes [tex]x = \sqrt{10}sin t[/tex] and [tex]y = 3 cos t[/tex]

    Since [tex]\int \sqrt{(dx)^2 + (dy)^2}[/tex] is the formula for arc length, do I just get: [tex]4\int \sqrt{10(cos t)^2 - 9(sin t)^2}[/tex]?

    And are the bounds just from 0 to [tex]\pi/2[/tex]?

    Thanks!
     
  2. jcsd
  3. Mar 5, 2009 #2

    Mark44

    Staff: Mentor

    This is the formula: [tex]\int \sqrt{(dx/dt)^2 + (dy/dt)^2}dt[/tex]
    There's a mistake in the line above. What do you get when you square dy/dt? Also, don't forget your dt in the integrand.
    Yes.
     
  4. Mar 5, 2009 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You should be aware that the calculation of the circumference of an ellipse results in what is known as an "elliptic integral" which cannot be done in terms of elementary functions.
     
  5. Mar 5, 2009 #4

    Mark44

    Staff: Mentor

    The instructions provided by the OP imply that numerical integration is to be performed.
     
  6. Mar 5, 2009 #5
    Oh, is it just that it should be positive instead of negative? That's the only thing I'm seeing, and yeah, I need to put the dt in there.

    So, other than that, is it correct?
     
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