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Isoperimetric problem

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data
    A plane curve with length [itex]l[/itex] has its end points at [itex](0, 0)[/itex] and [itex](a, 0)[/itex] on the positive [itex]x[/itex]-axis. Show that the area [itex]A[/itex] under this curve is given by [tex]A = \int_0^l y \sqrt{1 - y'^2}ds,[/tex] where [itex]y' = dy/dx[/itex], Find the function [itex]y(s)[/itex] and the value of [itex]a[/itex] which maximises [itex]A[/itex] and in turn determine that the curve in the [itex]x - y[/itex] plane is a semi-circle.


    2. Relevant equations
    The Euler-Lagrange equation


    3. The attempt at a solution
    I have managed to show that the area is given by that integral (by making the substitution [itex]ds^2 = dx^2 + dy^2[/itex]). I applied the Euler-Lagrange equation to the integrand and I got the solution [tex]y(s) = c\mathrm{sin}(\frac{s}{c} + k),[/tex] where [itex]c[/itex] and [itex]k[/itex] are arbitrary constants.

    I think this is correct so far, but I'm unsure how to find the "value of [itex]a[/itex] which maximises [itex]A[/itex]", and I don't know how to express the curve in terms of [itex]x[/itex]. Do we need to use some relationship between [itex]s[/itex] and [itex]x[/itex]?
     
  2. jcsd
  3. Oct 2, 2011 #2
    Well, you know that this curves goes between (0,0) and (a,0) right? So in particular, y(0)=0 and y(l) = 0. Those should allow you to get rid of some constants.
     
  4. Oct 2, 2011 #3
    Ok, so y(0) = 0 implies k = 0 and y(l) = 0 implies l/c = π, so c = l/π and y(s) = (l/π)sin(πs/l).

    I'm still not entirely sure how I'm supposed to find the value of a that maximises the area or the equation in terms of x and y.
     
  5. Oct 2, 2011 #4
    It seems to me that you should be able to calculate l using a.
     
  6. Oct 2, 2011 #5
    Well, I know that [itex]l = \int_0^a \sqrt{1 + y'^2}dx[/itex] but in order to evaluate this integral I would need to know y in terms of x, which is what I'm trying to find.
     
  7. Oct 2, 2011 #6
    Does anyone else have any ideas?
     
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