Show That the Curve is Straight

[Euge]

Fix points $p,q\in \mathbb{R}^n$, and let $\gamma : [a,b] \to \mathbb{R}^n$ be a continuously differentiable curve from $p$ to $q$ whose arclength equals the Euclidean distance between the points, $|q - p|$. Prove that $\gamma$ lies on the straight line passing through $p$ and $q$.

 

[anuttarasammyak]

It’s just a question on reading the problem, what do a and b mean?

 

[pasmith]

So the starting point is $$\|\gamma(b) - \gamma(a)\| = \int_a^b \|\gamma'(t)\|\,dt$$ if $\gamma(a) = p$ and $\gamma(b) = q$, and we want to conclude that $$\gamma(t) = p + \gamma'(a)G(t)$$ for some $G: [a,b] \to \mathbb{R}$ with $\gamma'(a) = K(q - p)$ for some $K \in \mathbb{R}$, $G(a) = 0$, and $G(b) = K^{-1}$.

I think that without loss of generality we can take $a = 0$ and $b = 1$.

 

[pasmith]

Spoiler: Proof by contradiction

Suppose there exists $t_0 \in (0,1)$ such that $r= \gamma(t_0)$ is not on the straight line from $p$ to $q$, Then $$\begin{split}<br /> \int_a^b \|\gamma'(t)\|\,dt &= \int_a^{t_0} \|\gamma'(t)\|\,dt + \int_{t_0}^b \|\gamma'(t)\|\,dt \\<br /> &\geq \|r - p\| + \|q - r\| \\<br /> &> \|q - p\| \end{split}$$ since the straight line path gives the shortest possible path between $p$ and $q$. But this a contradiction, so each point of $\gamma$ lies on the straight line.

 

[Office_Shredder]

since the straight line path gives the shortest possible path between p and q

I believe the actual point of the challenge is to prove this is true.