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Shortest Distance Between Two Points

April 9th, 2009 |

Author: Merrin

We can use calculus of variations to solve the problem of the shortest function connecting two points. We already know how to write down the arclength for an arbitrary function between two points.

$J=\int _{x_{1} }^{x_{2} }\sqrt{1+\dot{y}^{2} } dx$

The end points of this problem are specified as

$(x_{1} ,y(x_{1} )) \qquad (x_{2} ,y(x_{2} ))$

We can calculate the Euler-Lagrange equation for this problem and solve it to find the curve that minimizes the arclength.

$\cfrac{d}{dx} \cfrac{\partial f}{\partial \dot{y}} -\cfrac{\partial f}{\partial y} =0$

${\cfrac{d}{dx} \cfrac{\dot{y}}{\sqrt{1+\dot{y}^{2} } } =0} \\\\ {\cfrac{\dot{y}}{\sqrt{1+\dot{y}^{2} } } =c} \\\\ {\dot{y}^{2} =c(1+\dot{y}^{2} )} \\ {\dot{y}^{2} =\cfrac{c}{1-c} } \\\\ {\dot{y}=m} \\\\ {y=mx+b}$

Notice that even though the differential equation turns out to be first order there are two integration constants. One comes from the first step of the first integral of the Euler-Lagrange equation. It is no effort to solve for the constants of a linear function in terms of the endpoints.

$y=\cfrac{y(x_{2} )-y(x_{1} )}{x_{2} -x_{1} } (x-x_{1} )+y_{1}$