# CoV1

Find the path that minimizes the arclength of the curve between $(x_{0},y_{0})=(0,0)\,$ and $(x_{1},y_{1})=(1,1)\,$.

The goal is to minimize arclength as $x\,$ goes from $0$ to $1$. This quantity is described in this integral:

$\int _{0}^{1}(1+y'^{2})^{{1/2}}dx\,$

The Euler-Lagrange equation is defined as:

$EL:{\frac {d}{dx}}{\frac {\partial f}{\partial y'}}-{\frac {\partial f}{\partial y}}=0\,$ where $f\,$ is the integrand in the integral above.

Compute the derivatives for the given problem.

${\frac {\partial f}{\partial y}}=0\,$ since $f\,$ is independent of $y\,$.

${\frac {\partial f}{\partial y'}}={\frac {1}{2}}(1+y'^{2})^{{-1/2}}2y'=(1+y'^{2})^{{-1/2}}y'\,$

${\frac {d}{dx}}{\frac {\partial f}{\partial y'}}=(1+y'^{2})^{{-1/2}}y''-(1+y'^{2})^{{-3/2}}y'^{2}y''\,$

Simplify the expression by factoring out $y''\,$ and putting the common denominator $(1+y'^{2})^{{3/2}}\,$ on the bottom.

$EL:{\frac {y''(1+y'^{2}-y'^{2})}{(1+y'^{2})^{{3/2}}}}={\frac {y''}{(1+y'^{2})^{{3/2}}}}=0\,$

This equation will hold only if $y'\,$ is constant so that $y''=0\,$.

In that case,

$y(x)=c_{1}x+c_{2}\,$ and since $y(0)=0,y(1)=1\,$,

$c_{2}=0\,$ and $1=c_{1}+c_{2}\,$ so that $c_{1}=1\,$

Therefore,

$y(x)=x\,$