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I'm trying to find a unit speed parametrisation for the curve $\alpha: (0, \infty) \to \mathbb{R}^3$ s.t $$\alpha(t)= 0.5 (t, 1/t, \sqrt{2} \log(t)).$$

However, $$s(t) = 0.5 (t - \frac{ 1}{ t} ),$$ and as $t \to 0$, $s \to \infty $, so I don't know how to interpret this result, or how to find $t(s)$.

I also tried reparametrization of the given curve with $x \to e^x$, but that gave me $$s = \int_{-\infty}^\infty (e^x)*(3 + e^{-4x})^{0.5}dx,$$ but this integral also does not converge.

Question:

How to find a unit speed parametrisation for such a curve ?

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Solve for $t$:

$$t-\frac1t=2s$$ or $$t^2-2st-1=0,$$ which gives $$t=s\pm\sqrt{s^2+1}.$$

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  • $\begingroup$ But, zero is not in the domain of $\alpha$, so you cannot take $t_0 = 0$ in the integral. $\endgroup$ – onurcanbektas Mar 16 at 13:24
  • $\begingroup$ so $t - \frac{ 1}{t } \not = s $ in that sense. $\endgroup$ – onurcanbektas Mar 16 at 13:24
  • $\begingroup$ Plus, if we even try to make that the limit of the lower bound of the integral goes to zero $0$, the integral for $s$ diverges. $\endgroup$ – onurcanbektas Mar 16 at 13:25
  • $\begingroup$ @onurcanbektas: what integral ? Are you trying to find the total curve length ?? You've been asking for a unit speed parameterization, I am giving you one. $\endgroup$ – Yves Daoust Mar 16 at 13:51
  • $\begingroup$ Then, the question is how did you find it ? $\endgroup$ – onurcanbektas Mar 16 at 13:56

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