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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}}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\,


Calculus of Variations

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