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Point-Point Distance--3-D

The Line Element is

\begin{displaymath}
ds=\sqrt{dx^2+dy^2+dz^2},
\end{displaymath} (1)

so the Arc Length between the points $x_1$ and $x_2$ is
\begin{displaymath}
L = \int ds = \int_{x_1}^{x_2} \sqrt{1+{y'}^2+z'^2}\,dx
\end{displaymath} (2)

and the quantity we are minimizing is
\begin{displaymath}
f=\sqrt{1+{y'}^2+z'^2}.
\end{displaymath} (3)

Finding the derivatives gives
$\displaystyle {\partial f\over \partial y}$ $\textstyle =$ $\displaystyle 0$ (4)
$\displaystyle {\partial f\over \partial z}$ $\textstyle =$ $\displaystyle 0$ (5)

and
$\displaystyle {\partial f\over\partial y'}$ $\textstyle =$ $\displaystyle {y'\over\sqrt{1+y'^2+z'^2}}$ (6)
$\displaystyle {\partial f\over\partial z'}$ $\textstyle =$ $\displaystyle {z'\over\sqrt{1+y'^2+z'^2}},$ (7)

so the Euler-Lagrange Differential Equations become
$\displaystyle {d\over dx} \left({y'\over\sqrt{1+{y'}^2+z'^2}}\right)$ $\textstyle =$ $\displaystyle 0$ (8)
$\displaystyle {d\over dx} \left({z'\over\sqrt{1+{y'}^2+z'^2}}\right)$ $\textstyle =$ $\displaystyle 0.$ (9)

These give
\begin{displaymath}
{y'\over\sqrt{1+{y'}^2+z'^2}}=c_1
\end{displaymath} (10)


\begin{displaymath}
{z'\over\sqrt{1+{y'}^2+z'^2}}=c_2.
\end{displaymath} (11)

Taking the ratio,
\begin{displaymath}
z'={c_2\over c_1}y'
\end{displaymath} (12)


\begin{displaymath}
{y'\over\sqrt{1+y'^2+\left({c_2\over c_1}\right)^2y'^2}} = c_1
\end{displaymath} (13)


\begin{displaymath}
y'^2={c_1}^2\left[{1+y'^2+\left({{c_2\over c_1}}\right)^2 y'^2}\right]= {c_1}^2+y'^2({c_1}^2+{c_2}^2),
\end{displaymath} (14)

which gives
\begin{displaymath}
y'^2 = {{c_1}^2\over 1-{c_1}^2-{c_2}^2} \equiv {a_1}^2
\end{displaymath} (15)


\begin{displaymath}
z'^2 = \left({c_2\over c_1}\right)^2 y'^2 = {{c_2}^2\over 1-{c_1}^2-{c_2}^2} \equiv {b_1}^2.
\end{displaymath} (16)

Therefore, $y'=a_1$ and $z'=b_1$, so the solution is
\begin{displaymath}
\left[{\matrix{x\cr y\cr z\cr}}\right]=\left[{\matrix{x\cr a_1x+a_0\cr b_1x+b_0\cr}}\right],
\end{displaymath} (17)

which is the parametric representation of a straight line with parameter $x \in [x_1,x_2]$. Verifying the Arc Length gives
\begin{displaymath}
L=\sqrt{1+{a_1}^2+{b_1}^2}\,(x_2-x_1)
\end{displaymath} (18)

where
\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_1\cr a_0\cr}}\right]
\end{displaymath} (19)


\begin{displaymath}
\left[{\matrix{z_1\cr z_2\cr}}\right] = \left[{\matrix{x_1 & 1\cr x_2 & 1\cr}}\right]\left[{\matrix{b_1\cr b_0\cr}}\right].
\end{displaymath} (20)

See also Point-Point Distance--1-D, Point-Point Distance--2-D, Point-Quadratic Distance



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© 1996-9 Eric W. Weisstein
1999-05-25