I have just learned what parametrization by arc length is and trying to gain some intuition on it I have tried to parametrize some curves. Lines, circles and helixes are easy using the following theorem (from my notes):
Let $\alpha:I \rightarrow \mathbb{R}^3$ be a differentiable (in this context $\alpha \in C^\infty(I)$) and regular curve. Then for each $a \in I$, $s:I \rightarrow J=s(I)$, $s(t)=\int_a^t||\alpha'(u)||du$ is a diffeomorphism. Furthermore, the curve $\alpha \circ s^{-1}:J \rightarrow \mathbb{R}^3$ is parametrized by arc length.
I found Descartes folium as an interesting example to apply this new theorem. My attempt is as follows:
$\alpha:(-1, +\infty) \rightarrow \mathbb{R}^3$ given by $$\alpha(t)=\left(\frac{3t}{1+t^3}, \frac{3t^2}{1+t^3}\right)$$ is the folium. From here, $$\alpha'(t)=\left(\frac{3-6t^3}{(1+t^3)^2}, \frac{6t-3t^4}{(1+t^3)^2}\right)$$
Clearly, $\alpha$ is differentiable, and equating both numerators to $0$ we see that there is no $t \in (-1, +\infty)$ such that $\alpha'(t)=0$. Using the theorem for $a=0$ we get that $$s(t)=\int_0^t 3\frac{\sqrt{x^8+4x^6-4x^5-4x^3+4x^2+1}}{(1+x^3)^2}dx$$ is a diffeomorphism, and we are looking for its inverse. However, the integral looks complicated and finding its inverse probably too. Is there any other way to do this?
Best Answer
There is no other way. There is simply no closed form for $s(t)$ and its inverse.
At least you could use the Joukowsky transform $u(t)=t+\frac1t$ to get $$s_j(u)=3\int_{u(t)}^\infty\frac{\sqrt{u^4-4u-6}}{(u^3-3u+2)\sqrt{u^2-4}}\,du$$ but ultimately this doesn't help anything since it's a hyperelliptic (degree-$6$) integral.
Your best option is just to numerically integrate and invert $s$ to get the arc length parametrisation.