Prove: $\kappa^2v^4=|\alpha^{''}|^2-(\frac{dv}{dt})^2.$ 
Given a regular curve $\alpha:\mathbb R\to {\mathbb R}^3$,
  Prove: $$\kappa^2v^4=|\alpha^{''}|^2-\left(\frac{dv}{dt}\right)^2.$$
  ,where $\kappa$ is the curvature, $v$ is the rate of change of curve length.a

I know that $\kappa^2v^4=\dfrac{|\alpha'\times\alpha''|^2}{|\alpha'|^6}\cdot v^4$, but these expression don't resemble each other well. I also don't know how one can isolate $\alpha^{''}$ out of the cross product.
 A: $v\frac{dv}{dt}={\alpha'}\cdot{\alpha''}$
$\frac{dv}{dt}=\frac{\alpha'\cdot\alpha"}{|\alpha'|}$
$\kappa^2v^4=\frac{|\alpha'\times\alpha''|^2}{|\alpha'|^6}\cdot v^4=\frac{|\alpha'|^2|\alpha''|^2\sin^2\theta}{|\alpha'|^2}=|\alpha''|^2(1-\cos^2\theta)=|\alpha''|^2-|\alpha''|^2\frac{(\alpha'\cot \alpha'')^2}{\mid\alpha'\mid^2|\alpha''|^2}=|\alpha''|^2-(\frac{dv}{dt})^2$
A: HINT: $v^2=\alpha'\cdot\alpha’$, so what is $dv/dt$? 
Next, what is the geometric formula for $\|A\times B\|$?
A: Here'a a nice, quick way to get at it which doesn't us the vector cross product, but instead relies on the Frenet-Serret formulas:
In what follows, we make repeated us of the fact that
$v = \dfrac{ds}{dt}, \tag{1}$
this means that by the chain rule we may write, for any differentiable quantity $\lambda$, be it vector or scalar,
$\dfrac{d\lambda}{dt} = \dfrac{ds}{dt} \dfrac{d\lambda}{ds} = v\dfrac{d\lambda}{ds}; \tag{2}$
where $s$ is of course the arc-length along $\alpha(t)$.  I will also adopt the notation
$\lambda' = \dfrac{d \lambda}{dt} \tag{3}$
and
$\dot \lambda = \dfrac{d\lambda}{ds}, \tag{4}$
so that (2) may be written
$\lambda' = v \dot \lambda; \tag{5}$ 
bearing these observations in mind, we recall that the unit tangent vector $T$ to $\alpha(t)$ is given by
$T = v^{-1} \alpha' \tag{6}$
since $v = \Vert \alpha' \Vert$.  In then follows by the Frenet-Serret equation for the unit normal vector $N$ that
$\kappa N = \dot T = v^{-1}T' = v^{-1}(v^{-1} \alpha')' = v^{-1}(-v^{-2} v' \alpha' + v^{-1} \alpha'') = -v^{-3}v' \alpha' + v^{-2} \alpha''; \tag{7}$
since $N \cdot N$ = 1, (7) yields
$\kappa^2 = (\kappa N) \cdot (\kappa N) = v^{-6} (v')^2 \Vert \alpha' \Vert^2 + v^{-4} \Vert \alpha'' \Vert^2 - 2v^{-5} v' \alpha' \cdot \alpha'', \tag{8}$
whence, multiplying through by $v^4$:
$\kappa^2 v^4 = v^{-2} (v')^2 \Vert \alpha' \Vert^2 + \Vert \alpha'' \Vert^2 - 2v^{-1} v' \alpha' \cdot \alpha''. \tag{9}$
Inspecting (9), we see that by (6)
$v^{-2} (v')^2 \Vert \alpha' \Vert^2 = (v')^2 \Vert v^{-1} \alpha' \Vert^2 = (v')^2 \Vert T \Vert^2 = (v')^2; \tag{10}$
furthermore,
$v^{-1} v' \alpha' \cdot \alpha'' = v' (v^{-1} \alpha') \cdot \alpha'' = v' T \cdot \alpha''. \tag{11}$
Finally, since again by (6)
$\alpha' = vT, \tag{12}$
we have from (2) and (5)
$\alpha'' = v'T + vT' = v'T + v^2 \dot T, \tag{13}$
whence
$T \cdot \alpha'' = v' T \cdot T = v', \tag{14}$
using $\dot T = \kappa N$ and $T \cdot N = 0$.  When (14) is inserted into (11) we obtain
$v^{-1} v' \alpha' \cdot \alpha'' = (v')^2, \tag{15}$
and using (10) and (15) in (9) yields
$\kappa^2 v^4 = \Vert \alpha'' \Vert - (v')^2 = \Vert \alpha'' \Vert - (\dfrac{dv}{dt})^2, \tag{16}$
the desired objective.  QED.
Hope this helps.  Cheers,
and as ever,
Fiat Lux!!!
