Curvature and torsion

Question

Show that the torsion for the space curve $x=2t+\frac{1}{t}-1 ,y= \frac{t^2}{t-1}, z=t+2$ is zero.

Actually when I apply the direct formula for curvature and torsion the problem becomes very complicated.

Answer 1

This question is wrong.

The torsion is NOT zero.

Remark: Assume $t\neq 0,1$.

The z-coordinate of $r^{'}(t)$ is 1.

So the z-coordinate of $r^{''}(t)$ and $r^{'''}(t)$ is 0.

Then to compute $(r^{'}\times r^{''})\cdot r^{'''}$,

$(r^{'}\times r^{''})\cdot r^{'''}=\frac{-12}{t^4 \cdot (t-1)^4} \neq 0$

Checking using this online calculator:

https://www.emathhelp.net/en/calculators/calculus-3/torsion-calculator/?rx=2t+%2B1%2Ft-1&ry=%28t%5E2%29%2F%28t-1%29&rz=t%2B2&t=