Let $\alpha: I \subset \mathbb R \to \mathbb R^3$ be a regular curve in space (not necessarily Parameterized by arc length).. 
Let $\alpha: I \subset \mathbb R \to \mathbb R^3$ be a regular curve in space (not necessarily Parameterized by arc length). Show that:
$$k(r) = \frac{\vert \alpha’(r) \times \alpha’’(r) \vert}{\vert \alpha’(r)\vert^3}$$
$$\tau(r) = \frac{\langle\alpha'(r) \times \alpha'''(r), \alpha''(r) \rangle}{\vert \alpha'(r ) \times \alpha''(r)\vert^2}$$

Differentiating with respect to $t$ the expression $\beta(s(t))=\alpha(t)$.
$$\frac{d\beta}{ds}\frac{ds}{dt} = \alpha'(t)\tag1$$
$$\frac{d^2\beta}{ds^2}\left(\frac{ds}{dt}\right)^2 + \frac{d\beta}{ds}\frac{d^2s}{dt^2 } = \alpha''(t)\tag2$$
$$\frac{ds}{dt} = \vert \alpha'(t)\vert\tag3$$
$$\frac{d^2s}{dt^2} = \frac{\langle\alpha''(t),\alpha'(t)\rangle}{\vert \alpha'(t)\vert}\tag4$$
of (1) and (2) that
$$\alpha'(t) \times \alpha''(t) = \left(\frac{ds}{dt}\right)^3\frac{d\beta}{ds}\times\frac{d^2\beta }{ds^2}$$
Therefore,
$$\vert \alpha'(t) \times \alpha''(t) \vert = \left\vert \frac{ds}{dt}\right\vert^3 \left\vert \frac{d^2\beta}{ds^ 2}\right\vert$$
Where we use the fact that $\beta$ is parameterized by the arc length and therefore $\frac{d\beta}{ds}$ is orthogonal to $\frac{d^2\beta}{ds^2} $.We conclude, using (3) that
$$k(s(t)) = \left\vert\frac{d^2\beta}{ds^2}\right\vert = \frac{\vert \alpha'(t) \times \alpha''(t)\vert}{\vert\alpha'(t)\vert^3}$$
I'm missing some details that I can't fill in.
Thanks for any help.
 A: HINT: The idea is to show that the value of the expression " does not change under a parametrization" ( we have to express that precisely) and moreover, it's OK for the parametrization by arc length.
$\bf{Added:}$ Let us show that the expressions are "invariant under change of parametrization"
Say we have $r \mapsto \alpha(r)$ a curve, $\alpha\colon I \to \mathbb{R}^3$, and a change of parameter
$r\colon J \to I$, $r = r(t)$.  Now let's see how the expressions change.
We got $\beta(t) = \alpha(r(t))$, so
$$\beta'(t) = \alpha'(r(t)) \cdot r'(t) \\
\beta''(t) = \alpha''(r(t)) \cdot r'(t)^2 + \alpha'(r(t))\cdot r''(t)$$ and so
$$\beta'(t) \times \beta''(t) = r'(t)^3 \cdot (\alpha(r(t)) \times \alpha'(r(t))$$
and we conclude
$$\frac{|\beta'(t) \times \beta''(t)|}{|\beta'(t)|^3} =  \frac{|\alpha'(r(t)) \times \alpha''(r(t))|}{|\alpha'(r(t))|^3}$$
This is what we mean by invariance.
(Notice that in the plane, with an orientation, the curvature can be negative, that is captured in the formula).
The other formula works in a similar way.
Now, when $\alpha$ IS parametrized by arc length it is easy to check that we get the curvature. Indeed, in that case we have $|\alpha'(s)|equiv 1$, and, as a consequence $\alpha'(s) \perp\alpha''(s)$, and so $k(s) = |\alpha''(s)|$.
A: Follows my derivation of the curvature formula
$\kappa(t) = \dfrac{\vert \alpha''(t) \times \alpha'(t) \vert}{\vert \alpha'(t) \vert^3},  \tag 0$
which I believe to be reasonably canonical.  I have not yet been able to derive the expression for torsion, though I have put considerable effort into my attempts.  I will keep working on it and edit my solution, if I find one, into this answer.
The unit tangent vector $T(t)$ to the curve $\alpha(t)$ is clearly given by
$T(t) =\dfrac{\alpha'(t)}{\vert \alpha'(t) \vert} = \vert \alpha'(t) \vert^{-1}\alpha'(t), \tag 1$
where the magnitude of $\alpha'(t)$ is
$\vert \alpha'(t) \vert = \langle \alpha'(t), \alpha'(t) \rangle^{1/2}; \tag{2}$
we may find the rate of change of the magnitude of $\alpha'(t)$ along $\alpha(t)$:
$\vert \alpha'(t) \vert' = \dfrac{1}{2}\langle \alpha'(t), \alpha'(t) \rangle^{-1/2}(2\langle \alpha'(t), \alpha''(t) \rangle) = \langle \alpha'(t), \alpha'(t) \rangle^{-1/2}\langle \alpha'(t), \alpha''(t) \rangle$
$= \langle  \langle \alpha'(t), \alpha'(t) \rangle^{-1/2}\alpha'(t), \alpha''(t) \rangle = \langle  \vert \alpha'(t) \vert^{-1} \alpha'(t), \alpha''(t) \rangle = \langle T(t), \alpha''(t) \rangle. \tag{3}$
The speed or rate of change of arc-length $s$ along $\alpha(t)$ with respect to $t$ is
$\dfrac{ds}{dt} = \vert \alpha'(t) \vert, \tag 4$
from which
$\dfrac{dt}{ds} = \vert \alpha'(t) \vert^{-1} = \langle \alpha'(t), \alpha'(t) \rangle^{-1/2}. \tag 5$
We may also find the rate of change of the tangent vector $T(t)$ along $\alpha(t)$:
$\dfrac{dT(t)}{dt} = T'(t) = -\vert \alpha'(t) \vert^{-2} \vert \alpha'(t) \vert' \alpha'(t) +  \vert \alpha'(t) \vert^{-1}\alpha''(t), \tag 6$
and in light of (3) this may be written
$\dfrac{dT(t)}{dt} = T'(t) = -\vert \alpha'(t) \vert^{-2} \langle T(t), \alpha''(t) \rangle \alpha'(t) +  \vert \alpha'(t) \vert^{-1}\alpha''(t)$
$= -\vert \alpha'(t) \vert^{-1} \langle T(t), \alpha''(t) \rangle T(t) +  \vert \alpha'(t) \vert^{-1}\alpha''(t) = \vert \alpha'(t) \vert^{-1}(\alpha''(t) -  \langle T(t), \alpha''(t) \rangle T(t)). \tag 7$
We now introduce the first Frenet-Serret equation
$\dfrac{dT}{ds} = \kappa N; \tag{7.1}$
written in terms of the variable $t$ and the curve $\alpha(t)$ this becomes;
$\kappa(t) N(t) = \dfrac{dT(t)}{ds} = \dfrac{dT(t)}{dt}\dfrac{dt}{ds} = \vert \alpha'(t) \vert^{-1} \vert \alpha'(t) \vert^{-1}(\alpha''(t) -  \langle T(t), \alpha''(t) \rangle T(t))$
$= \vert \alpha'(t) \vert^{-2}(\alpha''(t) -  \langle T(t), \alpha''(t) \rangle T(t)); \tag 8$
since
$N(t) \cdot N(t) = T(t) \cdot T(t) = 1, \tag{8.1}$
we have
$\kappa^2(t) = \kappa(t) N(t) \cdot \kappa(t)N(t)$
$= \vert \alpha'(t) \vert^{-2}(\alpha''(t) -  \langle T(t), \alpha''(t) \rangle T(t)) \cdot \vert \alpha'(t) \vert^{-2}(\alpha''(t) -  \langle T(t), \alpha''(t) \rangle T(t))$
$= \vert \alpha'(t) \vert^{-4} (\alpha''(t) -  \langle T(t), \alpha''(t) \rangle T(t)) \cdot (\alpha''(t) -  \langle T(t), \alpha''(t) \rangle T(t))$
$= \vert \alpha'(t) \vert^{-4} (\langle \alpha''(t), \alpha''(t) \rangle - 2 \langle \alpha''(t), T(t) \rangle^2 + \langle \alpha''(t), T(t) \rangle^2)$
$= \vert \alpha'(t) \vert^{-4}(\langle \alpha''(t), \alpha''(t) \rangle - \langle \alpha''(t), T(t) \rangle^2).  \tag 9$
We apply the vector identity
$\vert \mathbf A \times \mathbf B \vert^2 = \vert \mathbf A \vert^2 \vert \mathbf B \vert^2 -
(\mathbf A \cdot \mathbf B)^2, \tag{10}$
where
$\mathbf A, \mathbf B \in \Bbb R^3, \tag{11}$
(cf. this wikepedia page) to the expression $\langle \alpha''(t), \alpha''(t) \rangle - \langle \alpha''(t), T(t) \rangle^2$,
taking
$\mathbf A = \alpha''(t), \tag{12}$
$\mathbf B = T(t); \tag{13}$
we immediately see that
$\langle \alpha''(t), \alpha''(t) \rangle - \langle \alpha''(t), T(t) \rangle^2 = \vert \alpha''(t) \times T(t) \vert^2; \tag{14}$
thus (9) becomes
$\kappa^2(t) = \vert \alpha'(t) \vert^{-4} \vert \alpha''(t) \times T(t) \vert^2, \tag{15}$
whence
$\kappa(t) = \vert \alpha'(t) \vert^{-2} \vert \alpha''(t) \times T(t) \vert.  \tag{16}$
We now recall (1) and substitute $\vert \alpha'(t) \vert^{-1} \alpha'(t)$ for $T(t)$ on the right-hand side of this equation, and obtain
$\kappa(t) = \vert \alpha'(t) \vert^{-2} \vert \alpha''(t) \times \vert \alpha'(t) \vert^{-1}\alpha'(t) \vert = \vert \alpha'(t) \vert^{-3} \vert \alpha''(t) \times \alpha'(t) \vert$
$= \dfrac{\vert \alpha''(t) \times \alpha'(t) \vert}{\vert \alpha'(t) \vert^3},  \tag{17}$
the desired result.
