Roughly, I would agree with you. The differential geometry of curves and surfaces course I took wasn't the absolutely most rigorous, but this is how we did it. The only small remark I would make, is that it would make things more clear to note that since $\displaystyle \pm 1=\dot{x} \frac{dt}{du}$ on a connected set, $\displaystyle \dot{x}\frac{dt}{du}$ is continuous, we know that $\displaystyle \dot{x}\frac{dt}{du}=1$ simultaneously for all values or $\displaystyle \dot{x}\frac{dt}{du}=-1$ for all values. This was probably what you meant, but I think it's important to note--it at least puts things on a more rigorous standing.
NB: I obviously assumed that our curve was at least $C^1$.
I'll do the curvature, you try the torsion.
Let $\beta:[s_0,s_1] \to \mathbb R^3$ be the reparametrization of $\gamma:[t_0,t_1] \to \mathbb R^3$ by arc length i.e: $\gamma=\beta \circ s$ where $s:[t_0,t_1]\rightarrow[s_0,s_1]$ ; $s(t)=\int_{t_o}^t||\gamma'(u)||du$.
We have:
$$\gamma'
=\frac{\mathrm d\gamma}{\mathrm dt}
=\frac{\mathrm d\beta}{\mathrm ds} \cdot \frac{\mathrm ds}{\mathrm dt}
=T(s) \cdot||\gamma'(t)||\tag1$$
where $T(s)$ is the tangent vector of $\gamma$. From the equation we see that $T(s)=\gamma'(t)/||\gamma'(t)||$. Then differentiating $\gamma'$ and applying the chain rule, we get:
$$\gamma''
= \frac{\mathrm d^2\gamma}{\mathrm dt^2}
=\left( \frac{\mathrm dT}{\mathrm ds} \cdot \frac{\mathrm ds}{\mathrm dt}\right) \cdot ||\gamma'(t)||+ T(s) \cdot \left( \frac{\langle\gamma'(t),\gamma''(t)\rangle}{||\gamma'(t)||}\right)\tag2$$
And since $T'=k \cdot N$, where $N$ is the normal vector of $\gamma$, we get:
$$ \gamma''
=k \cdot N \cdot ||\gamma'||^2+\left( \frac{\langle\gamma',\gamma''\rangle}{||\gamma'||}\right)\cdot T(s)\tag3$$
Computing the binormal $B$:
$$B
= \frac{\gamma'(t)\times \gamma''(t)}{||\gamma'(t)\times \gamma''(t)||}
=T\times N\tag4$$
We get the following relation:
$$ \gamma'(t)\times \gamma''(t)=k \cdot ||\gamma'(t)||^3 \cdot T\times N\tag5$$
Finally:
$$N
= B\times T
= \left( \frac{\gamma'(t)\times \gamma''(t)}{||\gamma'(t)\times \gamma''(t)||}\right)\times \frac{\gamma'(t)}{||\gamma'(t)||}=\frac{k \cdot ||\gamma'(t)||^3 \cdot T\times N}{||\gamma'(t)\times \gamma''(t)||}\times \frac{\gamma'(t)}{||\gamma'(t)||}\tag6$$
Since $||N||=||T\times N||=1$ and $k\geq 0$, we get: $$k=\frac {||\gamma'(t)\times \gamma''(t)||}{||\gamma'(t)||^3}.$$
Best Answer
The parametrization of a curve does not affect the shape of the curve, only the speed which the curve is traversed. That is, $\gamma(f(t))$ is the same curve for any continuous $f$, only the direction and speed at which the curve is being traced is changed. We normally choose the arc length parametrization, since it’s arguably the “simplest” speed, 1 unit per second.