finding gaussian curvature of a metric $ds^2=du^2+\lambda^2 dv^2$ For a surface with the metric $ds^2=du^2+\lambda^2 dv^2$, where $\lambda$ is a function of $u,v$; prove that the Gaussian curvature is given by $\displaystyle{K=-\frac{1}{\lambda}\frac{\partial^2 \lambda}{\partial u^2}}$.
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It is given that $E=1,F=0,G=\lambda^2,H=\sqrt{EG-F^2}=\lambda$, we take the parametrization $\textbf{r}\equiv \textbf{r}(u,v)$ throghout.
We know that $\displaystyle{K=\frac{LN-M^2}{EG-F^2}}$. So we need to calculate for $L,M$ and $N$; the second fundamental coefficients.
I started with $\displaystyle{L=-\textbf{r}_{1}.\bf{N}_1}$, $\bf{N}$ being the unit normal to the surface and $\displaystyle{\textbf{r}_1=\frac{\partial \textbf{r}}{\partial u}}$ and $\displaystyle{\textbf{N}_1=\frac{\partial \textbf{N}}{\partial u}}$. Then I used Weingarten equation $\displaystyle{\textbf{N}_1=\frac{1}{H^2}\bigg[(FM-GL)\textbf{r}_1+(LF-ME)\textbf{r}_2\bigg]}$, where $\displaystyle{\textbf{r}_2=\frac{\partial \textbf{r}}{\partial v}}$.
Afterwards, aiming to solve for $L$, I took negative dot product on both sides with $\textbf{r}_1$ and used the values $E=\textbf{r}_1.\textbf{r}_1=1$ and $F=\textbf{r}_1.\textbf{r}_2=0$. But I keep getting $L=L$ which is confusing as we should have a definite value of $L$ from here. I used the same technique for $\displaystyle{M=-\textbf{r}_{2}.\bf{N}_1}=-\textbf{r}_{1}.\bf{N}_2$ but keep getting same fallacy. What is possibly going wrong there? I also wonder if it is the correct approach to get the Gaussian curvature from the metric. Any help is appreciated.
 A: You don't even know that this metric arises from an embedding of the surface in $\Bbb R^3$. The point is that the Gaussian curvature can be computed just from the first fundamental form (metric), and there's a formula for that: If the metric is given by $ds^2 = E du^2 + G dv^2$, then
$$K = -\frac 1{2\sqrt{EG}}\left(\Big(\frac{E_v}{\sqrt{EG}}\Big)_v +
\Big(\frac{G_u}{\sqrt{EG}}\Big)_u\right).$$
A: (Not sure whether you'll like it, but at least it's short)
The curves $v=$const., parametrized by $u$, are geodesics, as they have zero geodesic curvature and are parametrized by length.
Hence, the vector field $X=\partial_v$ satisfies the Jacobi equation (geodesic deviation equation) $\ddot X + KX=0$ along these geodesics.
The vector field $Y=X/\lambda$ has unit length and is orthogonal to the geodesics, so it is invariant under the parallel transport along the geodesics: $\dot Y=0$.
We thus have $\ddot X=\ddot\lambda Y=(\ddot\lambda/\lambda) X$
As a result $K=-\ddot\lambda/\lambda$ where dot means the derivative wrt. the parameter of the geodesics, i.e. $K=-(\partial^2\lambda/\partial u^2)/\lambda$.
