Show That a Curve Is Regular
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Consider the curve given by

$\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. $

Problem: Prove that $\alpha$ is regular for all $t$.

Since a curve is regular at $t_0$ iff $\alpha'(t_0) \neq 0$ the curve seems to violate this at $t=0$. Perhaps this has something to do with the curve not being continuously differentiable at that point, but I have a feeling I'm missing something big here.

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1 Answer

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Well, $\alpha$ is regular for $t \neq 0$ since the first component of $\alpha'(t)$ is $1\neq 0$. But this actually holds for $t=0$ too (use the definition of derivative in case you don't see it immediately).

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David Miller

David Miller

Executive Financial & Market Analyst

David Miller brings 15 years of experience in global economics, personal finance strategy, and market dynamics. He specializes in turning complex economic trends into actionable insights for everyday readers.

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