Assume the parametrization is $C^2$ and regular, so its curvature is defined. Write $\mathbf v=\mathbf a'+y\mathbf b'$ and $W=|\mathbf v\times\mathbf b|>0$. Since $\sigma_x=\mathbf v$ and $\sigma_y=\mathbf b$, its <first fundamental form> is
$$
\boxed{I=|\mathbf a'+y\mathbf b'|^2\,dx^2+2(\mathbf a'+y\mathbf b')\cdot\mathbf b\,dx\,dy+|\mathbf b|^2\,dy^2.}
$$
For the unit normal $\mathbf n=(\mathbf v\times\mathbf b)/W$, the mixed coefficient of the <second fundamental form> is
$$
f=\mathbf n\cdot\mathbf b'=\frac{\mathbf a'\cdot(\mathbf b\times\mathbf b')}{W},
$$
where the term containing $\mathbf b'$ twice vanishes. Also $g=\mathbf n\cdot\sigma_{yy}=0$, while $EG-F^2=W^2$. Thus
$$
\boxed{K=-\frac{[\mathbf a'\cdot(\mathbf b\times\mathbf b')]^2}{W^4}.}
$$
The <Gaussian curvature> vanishes exactly when the displayed scalar triple product is zero. This is the <Gaussian curvature of a ruled surface> criterion; at a singular point with $W=0$, curvature and the claimed equivalence are not defined.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 3 / 14G / b / Solution

{c}
{title2=$K=-[\mathbf a'\cdot(\mathbf b\times\mathbf b')]^2/|\sigma_x\times\sigma_y|^4$}

For the regular $C^2$ <ruled surface> $\sigma(x,y)=\mathbf a(x)+y\mathbf b(x)$, put $W=|(\mathbf a'+y\mathbf b')\times\mathbf b|>0$. The second-form coefficient in the ruling direction is zero, and the mixed coefficient is $\mathbf a'\cdot(\mathbf b\times\mathbf b')/W$. The determinant formula for <Gaussian curvature> therefore gives the displayed nonpositive expression. Its vanishing is equivalent to the zero scalar triple product, only at regular points.

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 Incoming links: Gaussian curvature of a ruled surface