Help with Setting Up/Simplifying Euler-Lagrangian

[stephenklein]

Homework Statement

Imagine we have a string of length $l$ in the $x y$ plane with endpoints $x=-a$ and $x=a$. The endpoints can move to maximize this area, but the total length of the string is fixed.

a) Show that $dx = \sqrt {1-y'^2}$ where $ds$ is a tiny length of the curve such that $ds = \sqrt {dx^2+dy^2}$ as in class.

b) The area of the shaded rectangle above is $ydx$, so the sum of all of those areas between $x=-a$ and $x=a$ gives us the total area. However, it’s easier to incorporate the fact that the length is fixed as $l$ by integrating with respect to s, the path length of moving along the string instead. Using the expression in part (a) to convert the $ydx$ integral to an integral with respect to $ds$, express the area under the string.

c) Using the Euler-Lagrange equations prove that the optimal shape of the string is a semicircle.

A couple hints:

1) The Euler-Lagrange equation should yield $\frac {dy} {ds} = \sqrt {1- \frac {y^2} {C^2}}$, which you can solve using separation of variables (and probably looking up the integral).

2) A semicircle of radius R has the equation $x^2+y^2=R$

Relevant Equations

Euler-Lagrange equation: $$\frac {\partial f} {\partial y} - \frac {d} {ds} \frac {\partial f} {\partial y'} = 0$$

I was able to work through parts (a) and (b). For part (c), I got $$\frac {\partial f} {\partial y} = \sqrt {1-y'^2}$$ and $$\frac {\partial f} {\partial y'} = \frac {-y y'} {\sqrt {1-y'^2}}$$ Taking $\frac {d} {ds}$ of the latter, I used the product rule for all three terms $y, y', (\sqrt{1-y'^2})^{-1/2}$ and my result was $$\sqrt {1-y'^2} + \frac {y y''+y'^2} {\sqrt {1-y'^2}} + \frac {y y'^2 y''} {(1-y'^2)^{3/2}} = 0$$ I'm unsure (even more, skeptical) that this result simplifies to the one given in the question. I'm confident in everything up until the last derivative, which has a lot of moving parts. Any thoughts?

EDIT: I simplified the above result to $y'=\sqrt {1+y y''}$, which agrees with a Wolfram widget I found that simplifies Lagrangian equations. I feel like I'm very close, but my result has a sum instead of a difference, and that pesky $y''$ term isn't in the given answer.

 

[Orodruin]

Homework Statement: Imagine we have a string of length $l$ in the $x y$ plane with endpoints $x=-a$ and $x=a$. The endpoints can move to maximize this area

What area? You have only told us about a string of length $l$. Which area is to be maximised?