# Gaussian curvature of ruled surfaces

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 $$K = \frac{\det B}{\det G} = \frac{-2 \langle V', V\times c' \rangle}{\lvert c' + sV' \rvert^2 -2\langle V, c' \rangle} \cdot \frac{1}{\lvert V \times (c' + sV') \rvert}.$$

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.

First, you can see on geometric grounds that any ruled surface has $$K\le 0$$, 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 $$\det B\le 0$$. So $$\det B/\det G \le 0$$.