Consider the curve $\gamma(t)=(t,{1\over t}),t\in \mathbb R-\{0\}$.Does,then does this curve have a singularity at some point?I suspect that it has none because only problem is at $0$ which is not in the domain of the variable $t$.So,I think the curve is regular as $|| \dot \gamma(t)||\neq 0$ for each $t\in \mathbb R-\{0\}$.
2026-03-26 09:39:52.1774517992
Singularity of the given curve.
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That is correct. It is a regular curve, since the derivative $\|(1,-1/t^2)\|\geq1$ is never zero.