= Solution
Differentiate the <unit tangent vector> with $d/ds=v^{-1}d/du$ and $v'=(\mathbf x'\cdot\mathbf x'')/v$. This gives
$$
\boxed{\frac{d\mathbf t}{ds}=\frac{\mathbf x''}{v^2}-\frac{\mathbf x'(\mathbf x'\cdot\mathbf x'')}{v^4}.}
$$
The <curvature of a space curve> is $\kappa=|d\mathbf t/ds|$. Squaring the displayed expression and simplifying gives
$$
\kappa^2=\frac{v^2|\mathbf x''|^2-(\mathbf x'\cdot\mathbf x'')^2}{v^6}
=\frac{|\mathbf x'\times\mathbf x''|^2}{v^6},\qquad
\boxed{\kappa=\frac{|\mathbf x'\times\mathbf x''|}{|\mathbf x'|^3}.}
$$
The last equality is the <cross product> norm identity. Equivalently, the hinted <vector triple product> rewrites $d\mathbf t/ds=-\mathbf x'\times(\mathbf x'\times\mathbf x'')/v^4$.
Back to article page
