\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\pl}{\partial}
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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 curve joining the points $(0,\sqrt{2})$ and $(1,1)$ which
makes

$$I=\ds\int_0^1 \,dx y^{-1} (1+{y'}^2)^{\frac{1}{2}}$$

stationary, and show that for this curve
$I=arctanh\left(\ds\frac{1}{\sqrt{2}}\right)$.


\medskip

{\bf Answer}

Again $F=\ds\frac{\sqrt{1+{y'}^2}}{y}=F(y,y')$ only, so
$y'\ds\frac{\pl F}{\pl y'}-F=const$

Therefore $\ds\frac{\pl F}{\pl
y'}=\ds\frac{1}{2}\ds\frac{(1+{y'}^2)}{y}^{0-\frac{1}{2}}\times
2y'=\ds\frac{y'}{y\sqrt{1+{y'}^2}}$

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

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

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

$\Rightarrow \alpha x+c=\pm(1-\alpha^2y^2)^{\frac{1}{2}}$
(standard integration)

$\Rightarrow (\alpha x+c)^2=1-\alpha^2y^2$

$y(0)=\sqrt{2},\ y(1)=1 \Rightarrow c=0, \alpha^2=\ds\frac{1}{2}$
and so extremals is the circle \un{$x^2+y^2=2$}

In this case, $I=\ds\int_0^1
\ds\frac{1}{y}\left(x+\ds\frac{x^2}{y^2}\right)^{\frac{1}{2}}
\,dx$

$=\ds\int_0^1 \ds\frac{\sqrt{2}
\,dx}{2-x^2}=\un{arctanh\left(\ds\frac{1}{\sqrt{2}}\right)}$
standard integral

\end{document}