TITLE: Show that curvature and torsion are $κ = \frac{\left|γ'∧γ''\right|}{\left|γ'\right|^3}$ and $\tau = \frac{(γ'∧γ'')\cdot γ'''}{\left|γ'∧γ''\right|^2}$
QUESTION [7 upvotes]: Suppose that $\gamma : I \to \mathbb{R}^3$ is a regular curve (not necessarily parameterised by arc length). How can we show that the curvature and torsion are given respectively by $\kappa = \frac{|\gamma^{\prime} ∧\gamma^{\prime\prime}|}{|\gamma^{\prime}|^3}$ and $\tau = \frac{(\gamma^{\prime} ∧\gamma^{\prime\prime})\cdot \gamma^{\prime\prime\prime}}{|\gamma^{\prime} ∧\gamma^{\prime\prime}|^2}$?

REPLY [5 votes]: 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: $$k=\frac {||\gamma'(t)\times \gamma''(t)||}{||\gamma'(t)||^3}.$$