# Shortest path between two points in polar coordinates

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Show that the shortest path between two given points in a plane is a straight line, using plane polar coordinates.

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##### Solution Summary

We show that the shortest path between two given points in a plane is a straight line,

using calculus of variations in polar coordinates. We give two derivations: using the Lagrangian, and the Hamiltonian. In the first case we show that the solution follows from conservation of the conjugate momentum, while in the latter case it follows from conservation of the Hamiltonian (analogous to conservation of energy).

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documentclass[a4paper]{article}

usepackage{amsmath,amssymb}

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rightranglerightrangle}

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begin{document}

title{Shortest path between two points}

date{}

author{}

maketitle

section{Polar coordinates}

Before we start, let's first derive the formula for the path length of an arbitrary path in polar coordinates. Consider two nearby points on a path at coordinates $(r,theta)$ and $(r + dr, theta + dtheta)$. The distance between these two points is $sqrt{dr^{2} + r^{2}dtheta^{2}}$. If we parametrize the path by specifying $theta$ as a function of $r$, the path length $S$, can then be written as:

begin{equation}label{thetapar}

S = int_{theta_{1}}^{theta_{2}}sqrt{r^{2}+haak{frac{dr}{dtheta}}^{2}}dtheta

end{equation}

Note that when we ask for the minimum path length we need to specify $r$ at $theta_{1}$ and $theta_{2}$ and ask which path which has these fixed end points is the shortest.

If we instead parametrize the path by specifying $r$ as a function of $theta$, then we can write:

begin{equation}label{rpar}

S = int_{r_{1}}^{r_{2}}sqrt{1+r^{2}haak{frac{dtheta}{dr}}^{2}}dr

end{equation}

And when we minimize this, we need to specify $theta$ at the two end points at $r_{1}$ and $r_{2}$.

Since we already know that the solution is a straight line, let's find out how a straight line looks like in polar coordinates. For a given line, there will be a point on that line that is closest to the origin. Let's call this point $P$. So, at the point $P$, $r$ is minimal, say $r=r_{0}$ and $theta$ will have some value $theta_{0}$. Because $r$ is minimal on the line at $P$, the line must be at right angles to the line connecting $P$ to the Origin at $r=0$. For some arbitrary point $Q$ on the line, consider the right triangle formed by the Origin, the point $P$ and the point $Q$. If the point has coordinates $(r,theta)$, then:

begin{equation}label{line}

coshaak{theta-theta_{0}}=frac{r_{0}}{r}

end{equation}

From the equations eqref{thetapar} ...

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