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Let $\gamma : \left\{ \begin{array}{ccc} \mathbb{R} & \to & \mathbb{R} \\ t & \mapsto & (t^2,t^3) \end{array} \right.$ and $\Gamma= \gamma(\mathbb{R})$. Because of the singularity at $(0,0)$, there is no $\mathcal{C}^1$ regular parametrization of $\Gamma$; but there exists such parametrization for $\Gamma_+=\gamma([0,+ \infty[)$. However, $\Gamma_+$ seems not to admit $\mathcal{C}^2$ regular parametrization.

How do you show this result?

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up vote 1 down vote accepted

The short answer is: It is well known that Neil's parabola $\Gamma$ has infinite curvature at the origin. This already implies that $\Gamma_+$ cannot have a $C^2$ parametrization extending all the way to $0$ inclusive: The parametrization would also be $C^2$ with respect to arc length, but then $\kappa(s)=\dot x\ddot y-\dot y \ddot x$ would stay bounded.

In fact the parametrization with respect to arc length can be written down in elementary terms: From $$s(t):=\int_0^t \sqrt{4\tau^2+9\tau^4}\ d\tau ={1\over27}\ \bigl((4+9t^2)^{3/2}-8\bigr)\qquad(t\geq0) $$ you get $$t={1\over3}\ \sqrt{(8+27 s)^{2/3} -4}\qquad(s\geq0)\ ,$$ which you can plug into the original representation of $\Gamma_+$. Then check what happens for $s\to 0_+$.

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Why does the infinite curvature at the origin not depend on the parametrization? Why is it a property of $\Gamma$ itself and not of a given parametrization? – Seirios Sep 23 '12 at 17:16
I finally found a reference showing that curvature doesn't depend on the parametrization, so you answered my question. – Seirios Oct 11 '12 at 16:04

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