CoV1

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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 $y\,$ .

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

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

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,

$y(x) = x\,$

Calculus of Variations

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