Show that curvature and torsion are $κ = \frac{\left|γ'∧γ''\right|}{\left|γ'\right|^3}$ and $\tau = \frac{(γ'∧γ'')\cdot γ'''}{\left|γ'∧γ''\right|^2}$ 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 $\tau = \frac{(\gamma^{\prime} ∧\gamma^{\prime\prime})\cdot \gamma^{\prime\prime\prime}}{|\gamma^{\prime} ∧\gamma^{\prime\prime}|^2}$?
 A: 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: $$k=\frac {||\gamma'(t)\times \gamma''(t)||}{||\gamma'(t)||^3}.$$
A: Let $I=[a,b]$ and  $t(s):[0,L]\to I$ be a differentiable function whit $s: [0,L]\to [a,b]$ inverse differentiable and 
$$
L=\int_{a}^{b} \|\gamma^\prime(t)\| \; d\, t \qquad\mbox{ and }\qquad  s(t)=\int_{0}^{t}  \|\gamma^\prime(t)\| \; d\, t
$$
Fix the notation $\kappa(t)=\kappa(t(s))=\kappa(s)$  and $\gamma(t)=\gamma(t(s))=\gamma(s)$. We have by definition of curvature  $\kappa(s)=\|\gamma^{\prime\prime}(s)\|$.
Note that
\begin{align}
\frac{d^2}{d t^2}\gamma(s)=
&
\frac{d}{d t}\left[\frac{d}{dt}\gamma(s) \right]
\\
=
&
\frac{d}{d t}\left[\gamma^{\prime}(s)\cdot \left(\frac{d s}{dt}\right)\right]
\\
=
&
\left[ 
\gamma^{\prime\prime}(s)\cdot \left(\frac{d s}{dt}\right) +
\gamma^{\prime}(s)\cdot \frac{d s}{dt}\left(\frac{d s}{dt}\right)
\right].
\end{align}
Using Frenet-Serret equations 
$$
\left\{
\begin{array}{rl}
\gamma^\prime(s)=&T(s)\\
T^\prime(s)=& \kappa(s)\cdot N(s)\\
N^{\prime}(s)= &-\kappa(s)\cdot T(s) + \tau(s)\cdot B(s)\\
B^\prime(s)=& \tau(s)\cdot B(s)
\end{array}
\right.
$$
we have 
$$
\left\{
\begin{array}{rl}
\frac{d^2}{d\, t^2}\gamma(s)=& \frac{d\,s}{d\,t} T(s) \\
\frac{d^2}{d\,t^2}\gamma(s)=& \kappa(s)N(s)\left(\frac{ds}{dt}\right)+T(s)\left(\frac{d^2 s}{d\,t^2} \right)
\end{array}
\right.
$$
Calculate the cross product below
\begin{align}
\gamma^{\prime}(t)\times \gamma^{\prime\prime}(t)=
&
\frac{d }{dt}\gamma(s)\times \frac{d^2 }{dt^2}\gamma(s)
\end{align}
knowing that $\{T(s), N(s), B(s)\}$ is a basis of orthonormal vectors. After replacing the equations $\frac{d\,s}{d\,t}= \|\gamma^\prime(t)\|$ and $\frac{d^2\,s}{d\,t^2}=-\frac{\|\gamma^\prime(t)\|}{|\langle \gamma^\prime,\gamma^{\prime\prime}(t)\rangle|}$. Use too the equation
$$
\langle u\times v, w \rangle = det(u,v,w) \quad \forall w\in\mathbb{R}^3
$$
for simplifications. After doing some algebraic manipulations to obtain the curvature.The formula for calculating the torsion is analogous. Just obtain the torsion use the expression below.
$$
\left\langle 
\gamma^\prime(t)\times \gamma^{\prime\prime}(t),\gamma^{\prime\prime\prime}(t)
\right\rangle
=
\left\langle
\frac{d}{d\,t}\gamma(s)\times\frac{d^2}{d\,t^2}\gamma(s) \, ,\, \frac{d^3}{d\,t^3}\gamma(s)
\right\rangle
$$
