If X surface and $ E=1+v^2 $, $ F=0 $, $ G=1 $, $ e=0 $, show that $ a(t)=X(uo,vo+t) $ is a straight curve Let $X : U \to \mathbb{R}^3$ be a regular surface with $E =1 + v^2$ , $F = 0$ , $G=1$ , $e=0$ Show that the curve $ a(t)=X(uo,vo+t)$ (for constant $ (uo,vo) $ at $ U $ and $ t $ belong at $ (-\epsilon,\epsilon) $) is straight. where $ E=\mathopen{<}Xu,Xu\mathclose{>} $, $ G=\mathopen{<}Xv,Xv\mathclose{>} $, $ F=\mathopen{<}Xu,Xv\mathclose{>} $ and $ e=\mathopen{<}N,Xuu\mathclose{>} $, $ g=\mathopen{<} N,Xvv \mathclose{>} $, $f=\mathopen{<}N,Xuu\mathclose{>}$, $ N $ is the Gauss map and $ Xu $, $ Xv $, $ Xuu $, $ Xvv$ are the first and second partial derivatives of $ X $. To show that i think we have to show that $ a''(t)=0 $ for every $ t \in (-\epsilon, \epsilon) $. but $ a''(t)=Xvv(uo,vo+t) $. From Christoffel's symbols we can show that $ Xvv=gN $ So that is not zero... Am i thinking something wrong? Thanks in advance.

Only because $a'$ has constant length $1$ can you say you want to show $a''=0$. In general, the $v$-curves will be lines if and only if $X_{vv}$ is a multiple of $X_v$ at every point.

OK, so here's what you need to do. Calculate the Gaussian curvature, using $E$ and $G$, and then compare this with the extrinsic formula for the Gaussian curvature (in terms of $E,F,G$ and $e,f,g$). You should get a formula for $f$ and see that it depends only on $v$. Then use one of the Codazzi equations to deduce that $g=0$ (because $e_v-f_u=0=-g\Gamma^2_{11}$ and $\Gamma^2_{11}\
e 0$).

By the way, a helicoid admits such a parametrization with the $u$-curves helices and the $v$-curves lines.