How to prove that a surface obtained by linking the points of two regular and simple curves is $C^{2}$? Let $\gamma_0(t): [a,b] \to \mathbb{R}^3$ and $\gamma(t): [a,b] \to \mathbb{R}^3$ be two regular (and so $C^{1}([a,b])$) and simple (injective on for $t \in [a,b)$) closed curves. 
Consider the surface $r: D=[a,b] \times [0,1] \to \mathbb{R}^3$
$$r(t,s)= s \gamma_0 (t)+(1-s) \gamma(t) $$ 
How can I prove that $r$ belongs to $C^{2}(D)$? 
 A: I first define the curve:
\begin{equation}
w(t) = \gamma(t)-\gamma_0(t)
\end{equation}
Then I can write the surface:
\begin{equation}
r(t,s) = \gamma(t)+s w(t)
\end{equation}
The surface $r(t,s)$ is a ruled surface. A ruled surface has a Gaussian curvature K which is always less or equal to zero. The Gaussian curvature can be found in sevealal book. The point is that if the Gaussian curvature exist, then r(t,s) must be twice continuously differentiable.
Now few details. Without loss of generality we can assume $|w(t)|²=1$. When doing so, it is possible to compute the line of stricture for the ruled surface as:
 \begin{equation}
\beta(t) = \gamma(t)-\frac{\gamma_t(t) \cdot w_t(t)}{|w_t(t)|²} w(t)
\end{equation}
The line of stricture is the special directrix $\beta(t)$ which passes through the central point of each generator. The Gaussian curvature can be written as:
 \begin{equation}
K = -\frac{\lambda²}{\lambda²+s²} \leq 0
\end{equation}
Where
\begin{equation}
\lambda = \frac{w_t(t) \cdot(\beta_t(t) \times w(t))}{|w_t(t)|²}
\end{equation}
$\beta(t)$ is not constant, i.e. $\beta_t(t) \neq 0$, the mixed product is non zero, the Gaussian curvature is defined and $r(t,s)$ is twice continuously differentiable.
