# Show the curve $\alpha$ is differentiable and regular

Consider the map: $$\alpha(t) = \left\{ \begin{array}{ll} (t,0,e^{-1/t^2}) & t>0 \\ (t,e^{-1/t^2},0) & t<0 \\ (0,0,0) & t=0 \end{array} \right.$$

a. Prove that $\alpha$ is a differentiable curve

b. Prove that $\alpha$ is regular for all $t$.

This question is out of an old textbook, Riemannian Geometry of Curves and Surfaces, 1976. I believe the question is incorrect based on the definitions. As written, that curve is neither differentiable nor regular.

Could I get some justification of this and the statements of definitions from other sources? Thanks.

• Why do you think it is neither differentiable nor regular? – Lays Nov 15 '13 at 22:49
• What does "regular at $t$" mean? Maybe that the tangent vector is not the zero vector? – coffeemath Nov 15 '13 at 23:29
• As expected you know that $e^{\frac{-1}{t^2}}$ is $C^\infty$ at zero. Perhaps it explain somethings. – Hoseyn Heydari Nov 16 '13 at 19:42

Denoting $\alpha=(\alpha_1,\alpha_2,\alpha_3)$, you have:
• $\forall t\in\mathbb{R}, \alpha_1(t)=t$;
• $\alpha_2(t)=e^{-t^2/2} \ \ \textrm{if} \ \ t<0$ else $\alpha_2(t)=0$;
• $\alpha_3(t)=e^{-t^2/2} \ \ \textrm{if} \ \ t>0$ else $\alpha_3(t)=0$.
It is a well-known fact that $\alpha_2$ and $\alpha_3$ are smooth.
The first component $\alpha_1$ is clearly smooth and its derivative is $\alpha_1'\equiv 1$. It follows that $\alpha$ is a smooth curve and that its derivative $\alpha'$ does not vanish. The second statement means that $\alpha$ is regular.