Orthogonal differentiable family of curves This problem is out of section 4-4 in M. do Carmos' Differential Geometry of Curves and Surfaces:

We say that a set of regular curves on a surface $S$ is a differentiable family of curves on $S$ if the tangent lines to the curves of the set make up a differentiable field of directions. Assume that a surface $S$ admits two differentiable orthogonal families of geodesics. Prove that the Gaussian curvature of $S$ is zero.

As a tip, he says:

Parametrize a neighborhood of $p\in S$ in such a way that the two families of
  geodesics are coordinate curves (Corollary 1, Sec. 3-4). Show that this implies
  that $F = 0, E_v = 0 = G_u$. Make a change of parameters to obtain that $\bar{F} = 0, \bar{E} = \bar{G} = 1$.

I know that for every $p\in S$ we can find a neighbourhood $U\subset S$ of $p$ and a parametrization $X:I\times J\to U$ such that the coordinate curves $\alpha_{v_0}(u)=X(u,v_0)$ belong to one family and $\beta_{u_0}(v)=X(u_0, v)$ belong to the other. That way,
$$F=<X_u,X_v>(u_0,v_0)=<\alpha'_{v_0}(u_0), \beta'_{u_0}(v_0)>=0$$
I also know that if I can prove that $E_v=G_u=0$, then I can use the formula for orthogonal parametrizations:
$$K=-\frac{1}{2\sqrt{EG}}\left\{\left(\frac{G_u}{\sqrt{EG}}\right)_u+\left(\frac{E_v}{\sqrt{EG}}\right)_v\right\}$$
to prove that $K=0$, but I don't know how to prove that. Besides, I don't see the need of changing parameters to get $\bar{E}=\bar{G}=1$. Any hints?
 A: It is possible to show that a surface which admits two orthogonal families of geodesic is isometric with a plane (i.e. the Gaussian curvature is zero) in three steps:


*

*we show the the coordinate lines are orthogonal

*we show that once parametrized the two families of geodesics by coordinate curves $E_v=0$ and $G_u=0$

*we use the fact that when $F=0$, $E_v=0$ and $G_u=0$ the Gaussian curvature is zero


First step
Consider two plane curves $\alpha(t)$ and  $\beta(t)$ in the domain of $X$, and let $\gamma(t)=X \circ \alpha$ and $\delta(t)=X \circ \beta$ be the corresponding curves on the regular surface S. Then at the point $p= X(\alpha(t_0))=X(\beta(t_0))$, the angle $\theta$ between the two curves is
\begin{equation}
\cos{\theta}= \frac{\gamma' \cdot \delta'}{|\gamma'||\delta'|}
\end{equation}
In the basis ${X_u,X_v}$ the above equation becomes
\begin{equation}
\cos{\theta}= \frac{X_u \cdot X_V}{|X_u||X_v|}= \frac{F}{E G}
\end{equation}
Thus, all the coordinate curves of a parametrization are orthogonal if and only if $F=0$ for all $(u,v)$ in the domain of $X$.
Second step
Given the metric tensor $f$ 
\begin{equation}
f=\sqrt{E(u,v)u'^{2}+2 u' v' F(u,v)+G(u,v) v'^{2}}
\end{equation}
we compute the following partial derivative:
\begin{equation}
\frac{\partial f}{\partial u}=\frac{1}{2 f} \big(u'^{2} E_u+2 u' v' F_u+v'^2 G_u \big)
\end{equation}
\begin{equation}
\frac{\partial f}{\partial u'}=\frac{1}{f} \big(u' E+v' F \big)
\end{equation}
\begin{equation}
\frac{\partial f}{\partial v}=\frac{1}{2 f} \big(u'^{2} E_v+2 u' v' F_v+v'^2 G_v \big)
\end{equation}
\begin{equation}
\frac{\partial f}{\partial v'}=\frac{1}{f} \big(u' F+ v' G \big)
\end{equation}
where $E_x$, $F_x$, and $G_x$ are the derivative of $E$, $F$, and $G$ with respect to the coordinate x ={u,v}, while $u'$ and $v'$ are the derivative withy respect to the parameter $t$. On the curve $v=constant$, $u$ can be taken as a parameter, then the curve is $u=t$, $v=c$. We also have that $u'=1$ and $v'=0$. By substituting these values in the previous equations we have:
\begin{equation}
\frac{\partial f}{\partial u}=\frac{1}{2 \sqrt{E}} E_u
\end{equation}
\begin{equation}
\frac{\partial f}{\partial u'}=\sqrt{E}
\end{equation}
\begin{equation}
\frac{\partial f}{\partial v}=\frac{1}{2 \sqrt{E}} E_v
\end{equation}
\begin{equation}
\frac{\partial f}{\partial v'}=\frac{F}{\sqrt{E}}
\end{equation}
According to the Euler-Lagrange theorem in the calculus of variations, the parametric equation $(u(t),v(t))$ that optimize the metric $f$ must satisfy the differential equations:
\begin{equation}
\frac{\partial f}{\partial u}-\frac{d}{dt} \big (\frac{\partial f}{\partial u'} \big)=0
\end{equation}
\begin{equation}
\frac{\partial f}{\partial v}-\frac{d}{dt} \big (\frac{\partial f}{\partial v'} \big)=0
\end{equation}
By substituting the above values, after some calculations we have that the curve $v= constant$ is a geodesic if
\begin{equation}
E (E_v- 2 F_u) + E_u F =0
\end{equation}
This condition is therefore necessary. Conversely if the above condition is satisfied, it is possible to prove that the curve $v= constant$ is a geodesic. We can now repeat the same computation for a curve $u=constant$. In this case, $u=constant$ is a geodesic if and only if
\begin{equation}
G (G_u- 2 F_v) + G_v F =0
\end{equation}  
Third step
We can then conclude that for orthogonal parametrizations two families of geodesics are coordinate curvesmusrt satisfy the conditions $F=0$, $E_v=0$, and $G_u=0$. Thus the metric tensor becomes:
\begin{equation}
f=\sqrt{E(u)u'^{2}+G(v) v'^{2}}
\end{equation}
In the situation where one considers an orthogonal parametrization of a neighborhood of a surface, the Gaussian curvature can be written as
\begin{equation}
K=-\frac{1}{\sqrt{E G}} \big[\frac{\partial }{\partial u} \big(\frac{1}{\sqrt{E}} \frac{\partial \sqrt{G}}{\partial u} \big)+\frac{\partial }{\partial v}\big(\frac{1}{\sqrt{G}} \frac{\partial \sqrt{E}}{\partial v} \big)\big]
\end{equation}
Since $E=E(u)$ and $G=G(v)$, we have that the Gaussian curvature is zero. The formula I used for the Gaussian curvature can be derived by the usual definition of the Gaussian curvature computed as the radio of the determinants of the second and first fundamental form, by writing the determinal of the second fundamental form by using the triple product. The computation is quite long.
