Characterization of reparametrizations of a straight line?

Question

Let $\alpha$ be a regular curve. How to show $\alpha$ is a reparametrization of a straight line $t\mapsto p+tq$ if and only if $\alpha^{''}(t)$ and $\alpha^{'}$ are collinear?

Answer 1

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$Since $\alpha''(t)$ is proportional to $\alpha'(t)$ for each $t$ and $\alpha'$ is non-vanishing, there exists a differentiable function $k$ such that $$ \alpha''(t) = k(t) \alpha'(t). \tag{1} $$

Let $q = \alpha'(0)$, and let $v$ be an arbitrary vector orthogonal to $q$.

By equation (1), the function $$ f(t) = \Brak{\alpha'(t), v} $$ satisfies $$ f'(t) = \Brak{\alpha''(t), v} = k(t)\, f(t). \tag{2} $$

Any non-vanishing function $g(t) = e^{\int k(t)\, dt}$ (well-defined up to a non-zero multiplicative constant) also satisfies $g'(t) = k(t)\, g(t)$, so the quotient rule and equation (2) imply $$ \left(\frac{f}{g}\right)'(t) = 0, $$ so that $f/g$ is constant. Since $f(0) = 0$, $f(t) = \Brak{\alpha'(t), v} = 0$ for all $t$.

Since $v$ was an arbitrary vector orthogonal to $q = \alpha'(0)$, $\alpha'(t)$ is proportional to $q = \alpha'(0)$ for all $t$: There exists a function $A$ such that $\alpha'(\xi) = A(\xi)q$ for all $\xi$. It follows that $$ \alpha(t) = \alpha(0) + \int_{0}^{t} \alpha'(\xi)\, d\xi = \alpha(0) + \alpha'(0) \int_{0}^{t} A(\xi)\, d\xi, $$ which is a reparametrization of the line through $p = \alpha(0)$ with direction $q = \alpha'(0)$.