Natural parameterization of the following curve 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!

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.
