Natural parameterization of the following curve

Question

The given curve is: $ 27 x^{2} = 4 y^{3} $. I have to find the natural parameterization. First, I parameterized the curve: $$ 27 x^2 = 4 y^3 $$ $$ x = \pm \sqrt{\frac{4}{27} y^3} $$ What should I do when I have two possibilities? I took one possibility, for example: $$ x = \sqrt{\frac{4}{27} y^3} $$ Parameterization: $$ \alpha (t) = (\sqrt{\frac{4}{27} t^3},t) $$

The other steps: $$ \alpha ' (t) = (\frac{\sqrt{t}}{\sqrt{3}},1) $$ $$ || \alpha ' (t) || = \frac{1}{\sqrt{3}} \sqrt{t+3} $$ $$s(t) = \frac{1}{\sqrt{3}} \int_{0}^{t} \sqrt{u+3} du = \frac{1}{\sqrt{3}}(\frac{2}{3}(t+3)^{\frac{3}{2}}- 2\sqrt{3})$$ Did I go wrong at some point? It became too complicated, so I need help. Thanks!

Answer 1

What you did fine. Now, you invert $s$:$$s^{-1}(t)=\frac{3\sqrt[3]{(t+2)^2}}{\sqrt[3]4}-3.$$So, a natural parametrisation of your curve is$$t\mapsto\alpha\bigl(s^{-1}(t)\bigr)=\left(\frac{\left(\sqrt[3]{2(t+2)^2}-2\right)^{\frac32}}{\sqrt2},\left(\frac{3\sqrt[3]{(t+2)^2}}{\sqrt[3]4}-3\right)^2\right)$$

Answer 2

If we write $x=2t^3$ then we have $$\dfrac{y^3}{27}=\dfrac{x^2}{4}=t^6\implies y=3t^2.$$

Thus we get $$\alpha(t)=(2t^3,3t^2),$$ which is a nice parametrization to get the length of the curve.