The problem reads: " You are given a string of fixed length $\ell$ with one end fastened at the origin $O$, and you are to place the string in the
$\left(x, y\right)$ plane with its other end on the x axis in such a way as to maximise the area between the string and the x axis. Show that the required shape is a semicircle ".
Hint: Show that the area can be written as
$$J(y) = \frac{1}{c}\int_{0}^{\ell} y(s)\sqrt{1 + (y'(s))^2}ds$$
where $s$ is the arc-length along the string, and then use the following first integral of the $\mbox{Euler-Lagrange}$ equation
$$
F – y'\,\frac{\partial F}{\partial y'} = C\quad\mbox{where}\quad
y' = dy/ds\quad\mbox{and}\quad F = y\sqrt{1 + (y')^2}
$$
My attempt at a solution:
I know that the rules of this site state that saying I really don't have a clue what to do is not allowed but I genuinely have been stuck on this for all of today and for hours last night as well. I know that if y were a function of x, then the following would work
$$\frac{\triangle{s^2}}{\triangle{x^2}} = 1 + \frac{\triangle{y^2}}{\triangle{x^2}}$$
and then taking the limit as x tends to 0 would give
$$ds = \sqrt{1 + (y')^2}dx$$
But due to the question stating that $y$ is a function of $s$, the arc-length. I have absolutely no clue where to go, I simply can not see how it is shown. Any help would be much appreciated, thank you !.
Best Answer
Your first equation seems to be wrong
The Area is given by $$J[y]=\int_0^\ell y dx$$
The length of the differential element: $$(ds)^2=(dx)^2+(dy)^2$$ $$dx=\sqrt{1-(y'(s))^2}ds$$ You get $$J[y]=\int_0^\ell y(s) \sqrt{1-(y'(s))^2}ds$$
Also see Isoperimetric problem in the calculus of variations