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\begin{document}
In what follows you may assume that the following notation applies

$$y=y(x),\ y'=\ds\frac{dy}{dx}.$$

You may also assume that, unless otherwise stated, $y$ is a
sufficiently continuously differentiable function.

{\bf Question}

Find the continuous curve $y=y(x)$ between the points $(-1,0)$ and
$(1,0)$ for which

$$I=\ds\int_{-1}^1 \,dx \sqrt{(1-y)}\sqrt{1+{y'}^2}$$

is stationary.



\medskip

{\bf Answer}

Since $F=\sqrt{1-y}\sqrt{1+{y'}^2}$ is not a function of $x$, an
integral of the E-L equation is:

$y'\ds\frac{\pl F}{\pl y'}-F=const$.

$\Rightarrow \ds\frac{\pl F}{\pl
y'}=\ds\frac{\sqrt{1-y}}{\sqrt{1+{y'}^2}}\times \ds\frac{2y'}{2}$

Thus $\ds\frac{{y'}^2
\sqrt{1-y}}{\sqrt{1+{y'}^2}}-\sqrt{1-y}{\sqrt{1+{y'}^2}}=const$

$\Rightarrow \sqrt{\ds\frac{1-y}{1+{y'}^2}}=const=\alpha$ say

Therefore
$y'=\pm\left(\ds\frac{1}{\alpha^2}-1-\ds\frac{y}{\alpha^2}\right)^{\frac{1}{2}}$

Therefore $x+c=\mp 2\alpha ^2\left(\ds\frac{1}{\alpha
^2}-1-\ds\frac{y}{\alpha ^2}\right)^{\frac{1}{2}}$ (standard
integration)

or $(x+c)^2=4\alpha ^2\left(\ds\frac{1}\alpha
^2-1-\ds\frac{y}{\alpha ^2}\right)$

But $y=0$ at $x=\pm 1 \Rightarrow C=0,\ \alpha^2=\ds\frac{1}{2}$
and so \un{$y=\ds\frac{1}{2}(1-x^2)$}



\end{document}
