Show that the curvature of the curve is given by $|{\bf H}|^{−3}|{\bf H}×({\bf H·∇}){\bf H}|$. 
The tangent vector at each point on a curve is parallel to a
  non-vanishing vector field ${\bf H}({\bf x})$. Show that the curvature
  of the curve is given by $|{\bf H}|^{−3}|{\bf H}×({\bf H·∇}){\bf H}|$.

I've never tackled a question like this and I'm struggling a bit, so I think it would help to see one worked example.
I have all the definitions in front of me, but they're not helping much. I can't see any way of doing it that doesn't require at some point differentiation with respect to $s$ (the arclength) and I don't know how we could get anything into that form.
Any help is appreciated.
Thanks
 A: Let $c(t)$ be the curve. For some non-vanishing scalar field $\lambda$, $c'(t) = \lambda(c(t))\mathbf{H}(c(t))$. By the product rule, $$c''(t) = \frac{d}{dt}(\lambda(c(t)))\, \mathbf{H}(c(t)) + \lambda(c(t))\frac{d}{dt}(\mathbf{H}(c(t)))$$ By the chain rule, $$\frac{d}{dt}(\lambda(c(t)) = \nabla \lambda(c(t))\cdot c'(t) = \nabla \lambda(c(t)) \cdot\lambda(c(t))\mathbf{H}(c(t)) = \lambda(c(t))[\nabla \lambda\cdot \mathbf{H}](c(t))$$
and
\begin{align}\frac{d}{dt}(\mathbf{H}(c(t))) &= \langle c'(t)\cdot \nabla H^1(c(t)), c'(t)\cdot \nabla H^2(c(t)), c'(t)\cdot \nabla H^3(c(t))\rangle\\
&=\langle \lambda\mathbf{H}(c(t))\cdot \nabla H^1(c(t)), \lambda\mathbf{H}(c(t))\cdot \nabla H^2(c(t)), \lambda \mathbf{H}(c(t))\cdot \nabla H^3(c(t))\rangle\\
&=\lambda[(\mathbf{H}\cdot \nabla)\mathbf{H}](c(t))
\end{align}
Consequently, $$c'\times c'' = \lambda \mathbf{H}\times(\lambda \mathbf{H}[\nabla \lambda \cdot \mathbf{H}] + \lambda^2[(\mathbf{H}\cdot \nabla)\mathbf{H}]) = \lambda^3 [\mathbf{H}\times (\mathbf{H}\cdot \nabla)\mathbf{H}]$$ and the curvature of the curve at a point $\mathbf{x} = c(t_0)$ is 
$$\frac{\lvert c'(t_0)\times c''(t_0)\rvert}{\lvert c'(t_0)\rvert^3} = \frac{\lvert \lambda\rvert^3\lvert [\mathbf{H}\times (\mathbf{H}\cdot \nabla)\mathbf{H}](\mathbf{x})\rvert}{\lvert\lambda\rvert^3\lvert \mathbf{H}(\mathbf{x})\rvert^3} = \lvert \mathbf{H}\rvert^{-3}\lvert \mathbf{H}\times(\mathbf{H}\cdot \nabla)\mathbf{H}\rvert(\mathbf{x}) $$
