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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}

The Principle of Least Action (PLA) states that the motion of a
particle of mass $m$ between points $P$ and $Q$ is a potential
field $V(x,y)$ is such that, for $h=const$, and $ds$ being the
distance element along the particle's path, the action

$$A=\ds\int_P^Q \,ds\sqrt{2m(h-V)}$$

is minimised. Show that if $V=mgy(x)$ (a vertical gravitational
potential field) and the particle is under moves horizontally
between $x=a$ and $x=b$ then the PLA requires minimisation of the
following functional integral:

$$A=\ds\int_a^b \,dx\sqrt{2m(h-mgy)(1+{y'}^2)}.$$

Determine the particle path, given that $m=1,\ g=10,\ h=1,\ a=0,\
b=1$ (in appropriate units). Sketch the path.



\medskip

{\bf Answer}
\begin{eqnarray*} A & = & \ds\int_P^Q ds \sqrt{2m(h-V)}\\ & = & \ds\int_{x=a}^{x=b}dx
\sqrt{1+{y'}^2}\sqrt{\underbrace{2m(h-V)}}\\ & = & \ds\int_a^b dx
\sqrt{2m(h-mgy)(1+{y'}^2)} \end{eqnarray*}

Set $m=1,\ g=10,\ h=1,\ a=0,\ b=1$ and we have

$$A=\ds\int_0^1 dx \sqrt{2(1-10y)(1+{y'}^2)}$$

So $F=\sqrt{2(1-10y)(1+{y'}^2)}=F(y,y')$ \un{only}

\newpage
So E-L becomes $y'\ds\frac{\pl F}{\pl y'}-F=const$

$\Rightarrow
y'\ds\frac{\sqrt{2(1-10y)}}{\sqrt{1+{y'}^2}}y'-\sqrt{2(1-10y)(1+{y'}^2)}=const$

$\Rightarrow
\ds\frac{\sqrt{2(1-10y)}}{\sqrt{1+{y'}^2}}=const=\sqrt{2}c$ say

Therefore $\ds\frac{1-10y}{1+{y'}^2}=c^2 \Rightarrow
{y'}^2=\left(\ds\frac{1}{c^2}-1\right)-\ds\frac{10y}{c^2}$

which integrates to give

$$\ds\int\ds\frac{dy}{\sqrt{\left(\frac{1}{c^2}-1\right)-\ds\frac{10y}{c^2}}}=\ds\int
 \,dx$$

$\Rightarrow
-2\ds\frac{c^2}{10}\sqrt{\left(\frac{1}{c^2}-1\right)-\frac{10y}{c^2}}=x+d$
$d=const$

$\Rightarrow
\un{y=\ds\frac{(1-c^2)}{10}-\ds\frac{10}{4c^2}(x+d)^2}$

A parabola. Would need boundary conditions to evaluate $c$ and
$d$.

\end{document}
