The line is the shortest path between two points

Let $p, q$ be two points of $\mathbb{R}^n$ , $u$ be a unit vector and let $\gamma$ be a curve passing through those points. (that is there exist $a, b$ such that $\gamma(a)=p , \; \gamma(b)=q$ .) Prove that the shortest path between these two points is the line.

Solution

The following facts hold:

(1) $\begin{equation*} \gamma'(t)\cdot u \leqslant \left \| \gamma'(t) \right \| \end{equation*}$

(2) $\begin{equation*} \overrightarrow{q}-\overrightarrow{p}=\overrightarrow{\gamma}(b)-\overrightarrow{\gamma}(a)=\int_{a}^{b}{\overrightarrow{\gamma}\,'(t)\,{\rm d} t}\end{equation*}$

Integrating $(1)$ we get that:

(3) $\begin{equation*} \int_{a}^{b}{\overrightarrow{\gamma}\,'(t)\cdot\overrightarrow{u}\,dt} \leqslant \int_{a}^{b} {\| \overrightarrow{\gamma}\,'(t) \|\,dt}=\mathcal{L}(\overrightarrow{\gamma})\end{equation*}$

where $\mathcal{L}(\overrightarrow{\gamma})$ is the length of the curve between the points $p, q$ . Taking $\overrightarrow{ u}=\frac{\overrightarrow{q}-\overrightarrow{p}}{ \|\overrightarrow{q}-\overrightarrow{p} \|}$ we have

$\begin{align*} \|{\overrightarrow{q}-\overrightarrow{p}}\|&= \frac{\|{\overrightarrow{q}-\overrightarrow{p}}\|^2}{\|{\overrightarrow{q}-\overrightarrow{p}}\|} \\ &=\bigl({\overrightarrow{q}-\overrightarrow{p}}\bigr)\cdot\frac{\overrightarrow{q}-\overrightarrow{p}}{\|{\overrightarrow{q}-\overrightarrow{p}}\|} \\ &= \bigl({\overrightarrow{\gamma}(b)-\overrightarrow{\gamma}(a)}\bigr)\cdot\overrightarrow{ u}\\ &\stackrel{(2)}{=}\int_{a}^{b}{\overrightarrow{\gamma}\,'(t)\,dt}\cdot\overrightarrow{u}\\ &=\int_{a}^{b}{\overrightarrow{\gamma}\,'(t)\cdot\overrightarrow{u}\,dt} \\ &\stackrel{(1)}{\leqslant} \int_{a}^{b} { \| \overrightarrow{\gamma}\,'(t) \|\,dt}\\ &\stackrel{(3)}{=}\mathcal{L}(\overrightarrow{\gamma})\,. \end{align*}$

The line is the shortest path between two points

They might interest you:

Convex function

A rational harmonic series

Double series

Seemous 2017/4

Vanishing double summation

Value of parameter

Limit with harmonics

A contour integral

Circumscribed and inscribed circle

Leave a Reply Cancel reply