Understand if a curve is parametrized by arc length or not

Question

Show that the curve
$$\alpha(t)=(t,1+\frac{1}{t},\frac{1}{t}-t), \quad t\in(0,\infty)$$
is a plane curve.

I know $\tau$ must be zero for curve being plane. However, I want to determine the given curve is parametrized by arc length or not before starting to solve problem. I thought the magnitude of its first derivative must be 1 to be arc length parametrization. However I couldn't apply my idea since the magnitude is t-dependent.

Answer 1

Directly:

$$\begin{align*}&\alpha'=\left(1\,,\,\,-\frac1{t^2}\,,\,\,-\frac1{t^2}-1\,\right)\\{}\\ &\alpha''=\left(0\,,\,\,\frac2{t^3}\,,\,\frac2{t^3}\right)\\{}\\ &\alpha'''=\left(0\,,\,-\frac6{t^4}\,,\,\,-\frac6{t^4}\right)\end{align*}$$

Also

$$\alpha'\times\alpha''\times\alpha'''=\alpha'\bullet(\alpha''\times\alpha''')=$$

$$=\left(1\,,\,\,-\frac1{t^2}\,,\,\,-\frac1{t^2}-1\,\right)\bullet\left(0,0,0\right)=0\implies$$

$$\tau=\frac{\alpha'\times\alpha''\times\alpha'''}{||\alpha'\times \alpha''||^2}=0$$

Answer 2

Irrespective of w.r.t. $s$ or $t$, $\tau =0 $ always, it cannot be determined in this vanishing case. (Not repeating derivation made by Joanpemo already). You can detect by finding non-vanishing curvature instead.