I'm interested in the general case when we have a curve $(x,f(x))$ parameterized by $x$ to find a parametrization $x=g(t)$ such that $ds/dt=1$ along the curve.
So far what I came up: $$\frac{ds}{dt}^2 = \frac{dx}{dt}^2 + \frac{dy}{dt}^2 = \frac{dx}{dt}^2 + \frac{dy}{dx}^2 \frac{dx}{dt}^2 = \frac{dx}{dt}^2 \left(\frac{dy}{dx}^2 + 1 \right) = 1$$ $$ \frac{dx}{dt} = \sqrt{\frac{dy}{dx}^2 + 1 }^{-1}$$ However this does not tell me anything about $g$. Is there anyway to derive this in the general case or I do need a concrete function $f$ to solve this.
The transformation that reparametrizes a (regular) curve to a unit speed is always of the form $g(t) = \pm s(t) + c$ where $c$ a constant and $s(t)$ the arc length.
Let $\gamma(t)$ denote the regular curve, $\gamma'(x) = (v_1(x), v_2(x))$ the reparametrized curve, where $x = g(t)$ such that $\gamma'(g(t)) = \gamma(t)$. Then
$$\frac{d\gamma'}{dx} = \frac{d\gamma}{dt}\frac{dt}{dx}$$
If $\gamma'$ is unit speed with x, then $\frac{ds}{dx} = \| \frac{d \gamma'}{dx} \| = 1$ so that $\| \frac{d\gamma}{dt}\| |\frac{dt}{dx}| = 1$. Therefore,
$$\frac{dx}{dt} = \pm \| \frac{d\gamma}{dt} \| = \frac{ds}{dt}$$ by definition, and thus $x = g(t) = \pm s(t) + c$.