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Let $c: I rightarrow mathbb{R}^3$ be a regular curve, $V: I rightarrow mathbb{S}^2$ a vector field and $a < b$. Then we call
$$
f: (a,b)times I rightarrow mathbb{R}^3, quad f(s,t):= c(t) +sV(t)
$$
a ruled surface.

Show that $f$ has gaussian curvature $K(s, t) leq 0$.

For the first fundamental form, I obtained
$$
G = begin{pmatrix}
1 & langle V, c'rangle \
langle V, c'rangle & lvert c' + sV' rvert^2
end{pmatrix}
$$

and for the second fundamental form
$$
B = frac{1}{lvert V times ( c' + sV') rvert} begin{pmatrix}
0 & langle V', V times (c' + sV') rangle \
langle V', V times (c'+sV') rangle & *
end{pmatrix}
$$
and thus (with $V perp V'$) that
$$
K = frac{det B}{det G} = frac{-2 langle V', Vtimes c' rangle}{lvert c' + sV' rvert^2 -2langle V, c' rangle} cdot frac{1}{lvert V times (c' + sV') rvert}.
$$

I know that this question was already answered here: Gaussian and Mean Curvatures for a Ruled Surface.

However, there are additional assumptions made such as $c' perp V'$ and $lvert V' rvert = 1$. I don't know how to apply that as my case is a bit more general.

Two comments:

First, you can see on geometric grounds that any ruled surface has $Kle 0$, as the rulings are asymptotic curves, and one cannot have asymptotic curves when $K>0$.

Second, you know that $det G>0$ (always) and it's clear (from the $0$ in the $11$-entry) that $det Ble 0$. So $det B/det G le 0$.