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Why can't you parametrise a nonsingular curve by its arclength?

Is it simply because the following arclength doesn't exist where $\alpha '(t) =0$?

$$s(t) = \int_{t_0}^{t} |\alpha '(t)| dt$$ Thanks.

the_fox
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T J. Kim
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  • Whats the definition of non regular ? – Rene Schipperus Apr 19 '18 at 02:28
  • More specifically I'm referring to parametrised differentiable curves that are not regular at all points, hence the first derivative of the curve, $\alpha ' (t) $ is zero at some points of t, where $\alpha : I \rightarrow R^n$,$ t \in I$ (interval). – T J. Kim Apr 19 '18 at 02:30
  • I think the parameterization, is $\alpha\circ s^{-1}$ and this wont be differentiable cause you have to divide by zero. – Rene Schipperus Apr 19 '18 at 02:44

1 Answers1

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If you take a look at this post, you can see that in order to get $\lvert \beta'(t) \rvert =1$ it is necessary to divide by $\alpha'(t)$ for each $t\in I$...

Alternatively, $\frac {dt}{ds}$ needs to exist for each $t$ in the domain of the curve.