# Finding a unit speed parametrisation for $\alpha(t)= 0.5 (t, 1/t, \sqrt{2} \log(t))$

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 ?

Solve for $$t$$:
$$t-\frac1t=2s$$ or $$t^2-2st-1=0,$$ which gives $$t=s\pm\sqrt{s^2+1}.$$
• But, zero is not in the domain of $\alpha$, so you cannot take $t_0 = 0$ in the integral. – onurcanbektas Mar 16 at 13:24
• so $t - \frac{ 1}{t } \not = s$ in that sense. – onurcanbektas Mar 16 at 13:24
• 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. – onurcanbektas Mar 16 at 13:25