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A developable surface is a ruled surface $x(s,t)=\alpha(s)+t\beta(s)$, where $\alpha(s)$ is a unit-speed curve and $|\beta(s)|=1$, such that the tangent planes along each ruling are parallel. Show that the Gaussian curvature of such surface is $K=0$.

I am trying to show this using the equations of $\kappa_1$ and $\kappa_2$ which are the eigenvalues of the matrix $L$, which is the Weingarten map matrix. In computing the coefficients of the second fundamental form $L_{ij}:=\langle x_{ij},n \rangle$ with $i,j\in \{s,t\}$, I only find that $L_{tt}=0$, and I don't see how the calculations will simplify.

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The condition that the tangent spaces are constant in $t$ implies that $x_s(s,t+h)$ is in the tangent space $\operatorname{span}(x_s(s,t),x_t(s,t))$for all $t$ and $h$; so differentiating we see that $x_{st}$ lies in this tangent space too. But then $L_{st} = \langle x_{st} , n \rangle = 0$, so we know the second fundamental form is $$\left(\begin{matrix} \langle x_{ss} , n \rangle & 0 \\ 0 & 0 \end{matrix}\right)$$ which obviously has a zero eigenvalue.

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    $\begingroup$ Why do we know $x_{st}$ lies in the tangent space? $\endgroup$ – Jeze Ken Nov 3 '15 at 3:37
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    $\begingroup$ @JezeKen: $x_{st}(s,t) = \lim_{h\to 0} (x_{s}(s,t+h) - x_{s}(s,t))/h$. Since both the $x_{s}$ appearing here lie in the tangent space, their difference does too, and thus the derivative does. $\endgroup$ – Anthony Carapetis Nov 3 '15 at 3:40
  • $\begingroup$ This is a very clear answer. Thank you. $\endgroup$ – Jeze Ken Nov 3 '15 at 3:43

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