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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 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
up vote 3 down vote accepted

According to my Differential Geometry notes (had to look it up...), a curve is smooth if the partial derivatives (i.e. x' , y' , z' , ...) are not all zero for the same parameter t. In the given example, the curve is not smooth in the Origin, and such points are sometimes referred to as cusps. From a physical point of view: If a particle traces the curve, its speed should never become zero at a particular time.

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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.

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