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So let's suppose we have a surface $M$ that is embedded in $\mathbb{R}^3$ with an orthogonal parametrization. Further, assume that the parameter curves (i.e., $X(u$0$, v)$ and $X(u, v$0$)$ ) are geodesics that are unparametrized (i.e., not necessarily unit speed).

What can we say about the Gauss curvature of $M$?

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    I do believe (entirely from intuition) that Gauss curvature at a specific point $X(u_0, v_0)$ has the same sign as the scalar product of the second derivatives of the two curves at that point. Also, in an informal way, proportional to that scalar product, and inverse proportional to the size of the first derivative of the curves.2011-12-10
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    By "unparametrized" do you mean "affinely parametrized?" If you actually mean unparametrized then the parameter curves in an orthogonal parametrization are automatically geodesics, and so this is basically no restriction on the curvature. If you do mean "affinely parametrized," I would try writing down the geodesic equations for the geodesics pointing in the $u$ and $v$ directions to get a system of equations that the metric coefficients must satisfy and then go from there.2012-03-05

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