Point-Point Distance--2-D

Given two points in the Plane, find the curve which minimizes the distance between them. The Line Element is given by

$\begin{displaymath} ds=\sqrt{dx^2+dy^2}, \end{displaymath}$

(1)

so the Arc Length between the points

and

$\begin{displaymath} L = \int ds = \int_{x_1}^{x_2} \sqrt{1+{y'}^2}\,dx, \end{displaymath}$

(2)

where $y'\equiv dy/dx$ and the quantity we are minimizing is

$\begin{displaymath} f=\sqrt{1+{y'}^2}. \end{displaymath}$

(3)

Finding the derivatives gives

$\displaystyle {\partial f\over \partial y}$	$\textstyle =$	$\displaystyle 0$	(4)
$\displaystyle {d\over dx}{\partial f\over\partial y'}$	$\textstyle =$	$\displaystyle {d\over dx}[(1+{y'}^2)^{-1/2} y'],$	(5)

so the Euler-Lagrange Differential Equation becomes

$\begin{displaymath} {\partial f\over \partial y}-{d\over dx}{\partial f\over\partial y'} = {d\over dx} \left({y'\over\sqrt{1+{y'}^2}}\right)=0. \end{displaymath}$

(6)

Integrating and rearranging,

$\begin{displaymath} {y'\over\sqrt{1+{y'}^2}} = c \end{displaymath}$

(7)

$\begin{displaymath} y'^2=c^2(1+{y'}^2) \end{displaymath}$

(8)

$\begin{displaymath} y'^2(1-c^2)=c^2 \end{displaymath}$

(9)

$\begin{displaymath} y'={c\over\sqrt{1-c^2}} \equiv a. \end{displaymath}$

(10)

The solution is therefore

$\begin{displaymath} y=ax+b, \end{displaymath}$

(11)

which is a straight Line. Now verify that the Arc Length is indeed the straight-line distance between the points.

and

are determined from

$\displaystyle y_1$	$\textstyle =$	$\displaystyle ax_1+b$	(12)
$\displaystyle y_2$	$\textstyle =$	$\displaystyle ax_2+b.$	(13)

Writing (12) and (13) as a Matrix Equation gives

$\begin{displaymath} \left[{\matrix{y_1\cr y_2\cr}}\right]=\left[{\matrix{x_1 & 1\cr x_2 & 1\cr}}\right]\left[{\matrix{a\cr b\cr}}\right] \end{displaymath}$

(14)

$\displaystyle \left[\begin{array}{c}a\\ b\end{array}\right]$	$\textstyle =$	$\displaystyle \left[\begin{array}{cc}x_1 & 1\\ x_2 & 1\end{array}\right]^{-1}\left[\begin{array}{c}y_1\\ y_2\end{array}\right]$
	$\textstyle =$	$\displaystyle {1\over{x_1-x_2}}\left[\begin{array}{cc}1 & -1\\ -x_2 & x_1\end{array}\right]^{-1}\left[\begin{array}{c}y_1\\ y_2\end{array}\right],$	(15)

$\displaystyle a$	$\textstyle =$	$\displaystyle {y_1-y_2\over x_1-x_2} = {y_2-y_1\over x_2-x_1}$	(16)
$\displaystyle b$	$\textstyle =$	$\displaystyle {x_1y_2-x_2y_1\over x_1-x_2}$	(17)

$\displaystyle L$	$\textstyle =$	$\displaystyle \int_{x_1}^{x_2} \sqrt{1+y'^2}\,dy = (x_2-x_1)\sqrt{1+a^2}$
	$\textstyle =$	$\displaystyle (x_2-x_1)\sqrt{1+\left({y_2-y_1\over x_2-x_1}\right)^2}$
	$\textstyle =$	$\displaystyle \sqrt{(x_2-x_1)^2+(y_2-y_1)^2},$	(18)

as expected.

The shortest distance between two points on a Sphere is the so-called Great Circle distance.

References

Arfken, G. Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, pp. 930-931, 1985.