Let $x(t)$ be a path of class $C^1$ that does not pass through the origin in $R^3$. If $x(t_0)$ is the point on the image of $x$ closest to the origin and $x'(t_0)\neq 0$, show that the position vector $x(t_0)$ is orthogonal to the velocity vector $x'(t_0)$.
Define the function $f(t)=x(t)\cdot x(t)$, then if the function attains its minimum at an interior point of the domain, $f'(t_0)=2x(t_0)\cdot x'(t_0)=0$, so we can get the desired result. However, my question is what if the domain of the path is a closed interval of the form $[a,b]$. Then the minimum may occur at the endpoints, in which case we can't use the interior extremum theorem. How can the result still be guaranteed in such general cases?
Consider: $$x: [0,1] \rightarrow \mathbb{R}^3:t \rightarrow (1+t,0,0) $$ The point closest to the origin is $x(0)$ but the velocity (wich we now see as the right derivative of $x$) is not $0$.