= Solution
Let $F=|\mathbf F|>0$ and choose an orientation of a <regular curve> following the <vector field>. Its <unit tangent vector> is $t=\sigma\mathbf F/F$, where $\sigma=\pm1$ is constant along a connected regular segment. <Arc length> differentiation along that <curve> is $d/ds=(\sigma\mathbf F/F)\cdot\nabla$. Therefore
$$
\frac{dt}{ds}
=\frac1F(\mathbf F\cdot\nabla)\left(\frac{\mathbf F}{F}\right)
=\frac{(\mathbf F\cdot\nabla)\mathbf F}{F^2}
-\frac{\mathbf F(\mathbf F\cdot\nabla)F}{F^3}.
$$
Taking the <cross product> with $t$ removes the term parallel to $\mathbf F$:
$$
t\times\frac{dt}{ds}
=\sigma\frac{\mathbf F\times(\mathbf F\cdot\nabla)\mathbf F}{F^3}.
$$
Where the <Frenet frame> exists, $t\times t'=\kappa(t\times n)=\kappa b$. Thus
$$
\boxed{\frac{\mathbf F\times(\mathbf F\cdot\nabla)\mathbf F}{F^3}
=\sigma\kappa b.}
$$
The printed sign corresponds to orientation along or against the field. Taking magnitudes gives the orientation-independent <curvature of an integral curve of a vector field>, $\kappa=|\mathbf F\times(\mathbf F\cdot\nabla)\mathbf F|/F^3$. This <scalar> formula also applies at zero curvature, where $b$ itself may be undefined.
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