Equivalent Equation for $\tau(s)=-\frac{\bigl(\alpha'(s)\wedge\alpha''(s)\bigr)\cdot\alpha'''(s)}{k^2(s)}$? Following do Carmo's Differential Geometry, these are chapter one exercises.
An exercise earlier in the chapter gives the following direction; Unless explicitly stated, $\alpha:I→\mathbb R^3$ is a curve parameterized by arc length $s$, with curvature $k(s)\neq$, for all $s\in I$. Then we are asked to derive equation $1$.
$$\tau(s)=-\frac{\bigl(\alpha'(s)\wedge\alpha''(s)\bigr)\cdot\alpha'''(s)}{k^2(s)}$$
I have already done this exercise and derived this torsion identity parameterized by $t$. Then, we come to another exercise that provides different parameters and gives the following definition: 
Let $α:I→\mathbb R^3$ be a regular parametrized curve (not necessarily by arc length) and let $β:J→\mathbb R^3$ be a reparametrization of $α(I)$ by the arc length $s=s(t)$, measured from $t_0∈I$. Let $t=t(s)$ be the inverse function of $s$ and set $dα/dt=α′$, $d^2α/dt^2=α′′$, etc. Prove
$$\tau(t)=-\frac{(\alpha'(t)\wedge\alpha''(t))\cdot\alpha'''(t)}{|\alpha'(t)\wedge\alpha''(t)|^2}$$.
I was wondering if there was a way to relate these two identities and avoid an entirely new derivation from the Frenet-Serret equations? If there isn't, would I use the chain rule and start with the normal tangent vector: 
$$\alpha'=\frac{d\alpha}{dt}=\frac{ds}{dt}\frac{d\alpha}{ds}?$$
 A: By specializing $t$ to $s$, one sees that the second formula implies the first one. So it is natural to prove that the second formula is invariant under reparametrization. For simplicity, assume $t,s$ are inverse to each other, $ds/dt=(dt /ds)^{-1}>0$, and $k>0$. Denote $d\alpha/ds$ by $\alpha'$, $d\alpha/dt$ by $\dot{\alpha}$, etc. Then one has $$\frac{(\alpha'\alpha''\alpha''')}{(\alpha'\times\alpha'')\cdot(\alpha'\times\alpha'')}=\frac{(\dot{\alpha}\ddot{\alpha}\dddot{\alpha})}{(\dot{\alpha}\times\ddot{\alpha})\cdot(\dot{\alpha}\times\ddot{\alpha})},\qquad (1)$$ where $(uvw):=(u\times v)\cdot w$.
To prove (1), one first apply chain rule $$\begin{array}{ccccccc}\dot{\alpha}&=&\alpha'(ds/dt)&&&&\\ \ddot{\alpha}&=&\alpha''(ds/dt)^2&+&A\alpha'&&\\ \dddot{\alpha}&=&\alpha'''(ds /d t)^3&+&B\alpha''&+&C
\alpha'\end{array},\qquad (2)$$ where $A,B,C$ are some scalars that play no roles in the final result. 
Using (2) and properties of cross product and dot product, one has $$(\dot{\alpha}\times\ddot{\alpha})\cdot\dddot{\alpha}=(\alpha'\times\alpha''(ds/dt )^3)\cdot\alpha'''(ds/dt)^3$$
$$=(\alpha'\times\alpha'')\cdot\alpha '''(ds/dt)^6$$ and $$\dot{\alpha}\times\ddot{\alpha}=\alpha'\times\alpha''(ds/dt )^3.\qquad (3)$$ From the two formulas in (3), the result of (1) follows.
As in the beginning remark, letting $s$ be the arc length parameter, one has $$|\alpha'\times\alpha''|^2=|\alpha''|^2=k^2,$$ which shows that OP's second formula implies the first. Working backward using (1), the first implies the second, hence the two formulas are equivalent.
