# Thread: Orthogonal families of curves

1. ## Orthogonal families of curves

Let E, F, G be the coefficients of the first fundamental form of a regular surface S in the parametrization \mathbf{x}:U \subset \mathbb{R}^2 \rightarrow S. Let \varphi (u,v) = c and \psi(u,v)=d be two families of regular curves on \mathbf{x}(U) \subset S. Prove that these two families are orthogonal if and only if

E\varphi_v \psi_v-F(\varphi_u\psi_v+\varphi_v\psi_u) +G\varphi_u\psi_u=0

All I can see is splitting up the terms were we would get (\mathbf{x}_u\varphi_v-\mathbf{x}_v\varphi_u)(\mathbf{x}_u\psi_v-\mathbf{x}_v\psi_u)=0 but that does not go anywhere.

(attached is the tex file and it's pdf build since the latex editor is not rendering)LaTeX1.texLaTeX1.pdf

2. Hint

For the first curve use:

Same considerations for the second one.

3. I don't see the relation. with the consideration of the second curve we would simply have (\varphi_v dv+\varphi_u du)(\psi_v dv+ \psi_u du) = (\varphi_v \psi_v) \langle dv,dv\rangle +\langle dv,du\rangle (\varphi_u\psi_v+\varphi_v\psi_u) +\langle du,du\rangle (\varphi_u\psi_u). But this does not give the desired result

4. We have

$d\vec{x}_{\varphi}=\vec{x}_udu+\vec{x}_vdv\;,\quad d\vec{x}_{\psi}=\vec{x}_u\delta u+\vec{x}_v\delta v$

then,

$d\vec{x}_{\varphi}\cdot d\vec{x}_{\psi}=Edu\delta u+F(dv\delta u+du \delta v)+Gdv \delta v=0\\ \Leftrightarrow E+F\left(\dfrac{dv}{du}+\dfrac{\delta v}{\delta u}\right)+G\dfrac{dv}{du}\dfrac{\delta v}{\delta u}=0$

Now, use

$0=\cos \alpha=\dfrac{d\vec{x}_{\varphi}\cdot d\vec{x}_{\psi}}{|d\vec{x}_{\varphi}||d\vec{x}_{\psi}|}\quad \textrm{\;and\;}\quad \dfrac{dv}{du}=-\dfrac{\varphi_u}{\varphi_v}\;,\;\dfrac{\delta v}{\delta u}=-\dfrac{\psi_u}{\psi_v}$