# why $x^2 = y^3$ is not smooth?

I read a definition of a smooth curve on the plane: A smooth curve is a map from $[a,b] \to \mathbb R^2: t\mapsto ( f(t),g(t) )$, where $f$ and $g$ are infinitely differentiable functions.

According to this definition, $x^2 = y^3$ can be parametrized by $t \mapsto (t^2, t^3)$, so it should be smooth? But it has a special point at $(0,0)$? Can anyone help me please?

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Usually you require that the derivative is never zero, which in this case happens at $t = 0$. –  Javier Badia Aug 24 '13 at 19:11
smooth with nonvanishing velocity vector. When $t=0$ the derivative of $(t^2, t^3)$ vanishes –  Will Jagy Aug 24 '13 at 19:11
Also, if you draw an actual picture, it's kind of pointy. –  Will Jagy Aug 24 '13 at 19:14
It is smooth if a tiny bug with myopia witting on it thinks it is sitting on a straight line. –  André Nicolas Aug 24 '13 at 19:18
This couldn't matter less, but the parameterization should be $t\mapsto(t^3,t^2)$. –  Jonathan Y. Aug 24 '13 at 19:41

Indirect answer: Let me invert the role of $x$ and $y$: if $y^2=x^3$ is smooth then it is an elliptic curve in short Weierstrass form: $$y^2=x^3+ax+b,$$ with $a=b=0.$ Its discriminant is $\Delta=-16(4a^3+27b^2)=0$. As a consequence, it is a singular curve, i.e. it isn't smooth.