Problem
All the followings happen on $\mathbb{R}^3$
$r_1(t)=p(u(t),v(t))$ is a 'line of curvature' on a surface $p(u,v)$. $q(t,s):=p(u(t),v(t))+sn(u(t),v(t)).$ ($n(u,v)$ is unit normal vector of $p(u,v)$)
Find the Gaussian curvature of $q(t,s)$.
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My trial
Gaussian curvature = $\frac{L_1N_1-M_1^2}{E_1G_1-F_1^2}$
and the fact that $r_1$ is a line of curvature means $\frac {L_2(u'(t))^2+2M_2u'(t)v'(t)+N_2(v'(t))^2}{E_2(u'(t))^2+2F_2u'(t)v'(t)+G_2(v'(t))^2} = K_1 (K_1$ is one of two principal curvature)
After some calculating, I got $G_1=|n|^2=1, F_1=sn\cdot n'=0, E_1=...$
But I don't know how to calculate $L_1N_1-M_1^2$. I know those definitions but it's very hard to calculate $L_1=-q_t\cdot \partial_t(\frac {q_t\times q_s}{|q_t\times q_s|}), N_1= ...$
Maybe I'm wrong anywhere above or I'm trying long way around.
I really appreciate your help.
Let $q(t,s)=p(t)+s n(t)$ a surface formed by the family of stright lines normal to the surface parametrized by $p(u,v)$ along the line of curvature parametrized by $p(u(t),v(t))$. A parametrization \begin{equation} q(t,s)= \alpha(t)+ s w(t) \end{equation} where $\alpha(t)$ and $w(t)$ are curves in $\mathbb{R}^3$, are ruled surfaces. The curve $\alpha(t)$ is called the directrix or base curve of the ruled surface, and $w(t)$ is called the director curve. The Gaussian curvature of a ruled surface can be written as: \begin{equation} K=-\frac{\lambda^2}{\lambda^2+s^2} \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 a special curve, called line of stricture, which satisfies the equation \begin{equation} \beta(t)=\alpha(t)-\frac{\alpha_t(t) \cdot w_t(t)}{|w_t(t)|^2} w(t) \end{equation} As you can see the Gaussian curvature of a ruled surface is always negative or zero. More exactly $K=0$ if the mixed product $w_t(t) \cdot(\beta_t(t) \times w(t))$ is zero. In our case the curve $\alpha(t)=p(t)$, while $w(t)=n(t)$. So we have \begin{equation} n_t(t) \cdot(\beta_t(t) \times n(t))=n_t(t) \cdot(\frac{d}{dt} (p(t)-\frac{p_t(t) \cdot n_t(t)}{|n_t(t)|^2} \times n(t))) = n_t(t) \cdot(p_t(t) \times n(t)) \end{equation}
The Gaussian curvature is zero if:
A smooth curve $\gamma$ on a regular surface S of class $C^2$ is called a line of curvature if the tangent vector of $\gamma$ is a principal vector of S at all points of $\gamma$. We can now use the Rodrigues’s theorem which says that the derivative of a normal $n(P)$ to a regular surface S of class $C^k$ with $k \geq 2$ along some direction, is parallel to it if and only if this direction is the principal vector of the surface at $P$ and the coefficient of proportionality is equal to $−k$, where $k$ is the principal curvature of S at P corresponding to this principal vector. So we can write: \begin{equation} n_t(t)=-k p_t(t) \end{equation} In a parametric surface a curve is a line of curvature if satisfied the Rodrigues’s theorem. Then by replacing $n_t(t)$ in the formula for the Gaussian curvature of $q(t,s)$ we have \begin{equation} n_t(t) \cdot(p_t(t) \times n(t)) = -k p_t(t) \cdot(p_t(t) \times n(t)) = 0 \end{equation} Then K = 0$.