Find the geodesic and normal curvatures of a surface 
For the surface $\sigma(u,v)= (\frac{\cos v}{\cosh u}, \frac{\sin v}{\cosh u}, \tanh u)$, compute the geodesic curvature and the normal curvature of:
(i) a meridian $v$ = constant,
(ii) a parallel $u$ = constant.
Which of these curves are geodesics?

I know that a meridian is a geodesic and its geodesic curvature $\kappa_g = 0$.
From the textbook, the normal curvature $\kappa_n = \gamma''·N$ and the geodesic curvature $\kappa_g = \gamma'' · (N \times \gamma')$, where $N$ is the unit normal of the surface, and $\gamma$ is a unit-speed curve in $\mathbb{R}^3$.
How can I find the $\gamma$ in the formula? Thanks!
 A: By keeping $u$ or $v$ constant in the parametrisation $\sigma$ of the surface and keeping the other parameter "variable", you obtain the parametrisation of the coordinate lines.
For instance, to parametrise a meridian, keeping $v$ constant gives
$$
   \gamma(t) = \sigma(t, v_0) = \left(\frac{\cos v_0}{\cosh t}, \frac{\sin v_0}{\cosh t}, \tanh t\right).
$$ 
In the same way, if you keep $u$ fixed and let the other parameter of $\sigma$ be the variable of the curve, you get parametrisations of the parallels.
A: Normal and geodesic curvature of curves on a surface can be computed using the following formulae:
\begin{equation}
k_n=\frac{q}{s'^2} \\
k_g=\frac{p}{s'^3}
\end{equation}
where
\begin{equation}
q=L_{11} u'^2+2 u' v' L_{12}+v'^2 L_{22} \\
p=\left[\Gamma_{11}^2 u'^3+(\Gamma_{22}^2-2 \Gamma_{12}^1) u' v'^2+(u' v''-u'' v')+(2 \Gamma_{12}^2-\Gamma_{11}^1)u'^2v'-\Gamma_{22}^1 v'^3\right] \sqrt{Det(g)}
\end{equation}
where $\Gamma_{jk}^i$ are the Christoffel symbols of the second kind, $L_{ij}$ the coefficient of the second fundamental form and g the first fundamental form. Independently of the case u=const or v=const, the product $u' v'$ are zero since or v is constant or u is constant. Thus, the above expressions reduce to:
\begin{equation}
q=L_{11} u'^2+v'^2 L_{22} \\
p=\left[\Gamma_{11}^2 u'^3-\Gamma_{22}^1 v'^3\right] \sqrt{Det(g)}
\end{equation}
In addition $s'=\sqrt{g_{11}u'^2+2 g_{12} u'v'+g_{22}v'^2}$ reduces to $s'=\sqrt{g_{11}u'^2+g_{22}v'^2}$. We can now consider the two cases.


*

*v=const


In such a case we have 
\begin{equation}
q=L_{11} u'^2 \\
p=\left[\Gamma_{11}^2 u'^3\right] \sqrt{Det(g)}\\
s'=\sqrt{g_{11}u'^2}
\end{equation}


*u =const


In such a case we have
\begin{equation}
q=v'^2 L_{22} \\
p=\left[-\Gamma_{22}^1 v'^3\right] \sqrt{Det(g)}
s'=\sqrt{g_{22}v'^2}
\end{equation}
When computing the non zero coefficient of the first fundamental form for $\sigma$, we have $g_{11}=g_{22}=\rm sech^2(u)$, while  non zero coefficient of the second fundamental form are $L_{11}=L_{22}=-\rm sech^2(u)$. When replacing these coefficient in the above expression we have
1. v=const
\begin{equation}
k_n=-1 \\
k_g=0
\end{equation}
because $\Gamma_{11}^2=0$. Since the geodesic curvature is zero, the curve is a geodesic.


*

*u=const
\begin{equation}
k_n=-1 \\
k_g=-\rm sech{(u_0)}
\end{equation}
where $u_0$ is the constant.
The definition of normal and geodesic curvature in your textbook can be written in term of p and q for practical computation.

