This is Chapter 1.1, Problem 7 of Modern Differential Geometry of Curves and Surfaces with Mathematica, by Alfred Gray, et al.
Find a unit speed parametrization of the semicubical parabola $t\to\left(t^{2},t^{3}\right)$, valid for $t>0.$
They give this expression for the unit tangent vector.
In seeking a unit speed parametrization of $sc\left(t\right)=\left(t^{2},t^{3}\right),$ a computation shows that
$$\frac{sc^{\prime}\left(t\right)}{\left\Vert sc^{\prime}\left(t\right)\right\Vert }=\frac{\left(t^{2},t^{3}\right)}{\sqrt{t^{2}\left(4+9t^{2}\right)}}.$$
This fairly trivial computation is the square root of
$$ds^{2}=dx^{2}+dy^{2}=(2tdt)^{2}+\left(3t^{2}dt\right)^{2}=t^{2}\left(4+9t^{2}\right)dt^{2}.$$
I can also find this expression for arc length as a function of $t$
$$s\left(t\right)=\frac{1}{27}\left(9t^{2}+4\right)^{3/2}-\frac{8}{27},$$
which is given in the book, obtainable using Mathematica, and derivable using hyperbolic trigonometric substitution.
But, Mathematica tells me there is no closed form inverse, giving $t$ as a function of $s,$ which is what I would expect a unit speed parametrization would be. Is there such a closed form solution?
I'm wondering if all they are really asking for is the computation I provided above.
$s$ looks completely invertible – peel one function off at a time: $$t=\sqrt{(((27(s+8/27))^{2/3}-4)/9}$$ $$t=\frac{\sqrt{(27s+8)^{2/3}-4}}3$$