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\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\pl}{\partial}
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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 between the points $(0,0)$ and
$(\frac{\pi}{2},1)$ for which

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

is stationary.



\medskip

{\bf Answer}

$F=y\sqrt{1-{y'}^2}$ Again $F=F(y,y')$ only so

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

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

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

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

$\Rightarrow {y'}^12=1-\ds\frac{y^2}{\alpha^2}$

Hence $\ds\int \ds\frac{dy}{\sqrt{1-\frac{y^2}{\alpha^2}}}=\ds\int
dx \Rightarrow y=\alpha\sin\left(\ds\frac{x}{\alpha}+c\right)$

(standard integral) $c$=constant of integration

Now satisfy boundary conditions: $\left\{\begin{array} {rcl}
y(0)=0 &amp; \Rightarrow &amp; c=0\\ y\left(\ds\frac{\pi}{2}\right)=1 &amp;
\Rightarrow &amp; 1=\alpha \sin\left(\ds\frac{\pi}{2\alpha}\right)
\end{array} \right.$

Obvious solutions to the equation are $\alpha=\pm 1$. These are
the only two (sensible) ones. Why?

Plot $\ds\frac{1}{\alpha}$ and $\sin\ds\frac{\pi}{2\alpha}$
against $\ds\frac{1}{\alpha}$:

\newpage
PICTURE \vspace{1in}

$\ds\frac{1}{\alpha}=0$ is also a solution but is meaningless as
$\Rightarrow \alpha=\infty$, so a nonsensical solution.

Hence $y=\pm\sin(\pm x) \Rightarrow \un{y=\sin x \rm{uniquely}}$

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