Understanding iso-parametric curves in the context of the First Fundamental Form I am reading the differential geometry notes here, and a statement about iso-parametric curves is confusing me.
The author writes, the angle between two curves on a parametric $r_1 = r(u_1(t), v_1(t))$, $r_2 = r(u_2(t), v_2(t))$ can be evaluated by considering
$$cos \omega = \frac{E du_1 du_2 + F(du_1dv_2 + dv_1du_2) + G dv_1 dv_2}{\sqrt{E du_1^2 + 2Fdu_1dv_1 + Gv_1^2}\sqrt{Edu_2^2 + 2Fdu_2dv_2 + Gdv_2^2}}$$
In particular, when two curves are the $u$ and $v$ iso-parametric curves, the above reduces to
$$cos \omega = \frac{r_u \cdot r_v}{||r_u|||r_v||}$$
where $ E = r_u \cdot r_u$, $F = r_u \cdot r_v$, $G = r_v \cdot r_v$,
Reading online, I see a definition of iso-parametric curves as : An iso-parametric curve of a surface is a curve obtained by varying one of the two surface parameters from $0$ to $1$, while leaving the value of the other parameter unchanged.
I'm not sure exactly what iso-parametric means in this context, or why this implies the last equation. Any insights appreciated.
 A: I haven't heard the term iso-parametric used in this sense, but apparently it refers to "being a coordinate curve":

*

*Holding $v$ constant and letting $u$ vary, obtaining $r_{1}(t) = r(t, v_{0})$ so that $r_{1}'(t) = r_{u}(r_{1}(t))$; or

*Holding $u$ constant and letting $v$ vary, obtaining $r_{2}(t) = r(u_{0}, t)$ so that $r_{2}'(t) = r_{v}(r_{2}(t))$.

If $p = r(u_{0}, v_{0})$, then by definition
\begin{align*}
  \|r_{1}'(u_{0})\|^{2} &= r_{u}(p) \cdot r_{u}(p) = E(p), \\
  r_{1}'(u_{0}) \cdot r_{2}'(v_{0}) &= r_{u}(p) \cdot r_{v}(p) = F(p), \\
  \|r_{2}'(v_{0})\|^{2} &= r_{v}(p) \cdot r_{v}(p) = G(p).
\end{align*}
Further, $du_{1} = dt$ and $dv_{1} = 0$ (because $(u_{1}, v_{1}) = (t, v_{0})$), and similarly $du_{2} = 0$ and $dv_{1} = dt$. Substituting all these into the first displayed equation (evaluated at $p$) causes most of the terms to drop out, leaving
\begin{align*}
  \cos \omega
  &= \frac{E\, du_{1} du_{2} + F\, (du_{1}\, dv_{2} + dv_{1}\, du_{2}) + G\, dv_{1} dv_{2}}{\sqrt{E\, du_{1}^{2} + 2F\, du_{1}\, dv_{1} + G\, v_{1}^{2}}\sqrt{E\, du_{2}^{2} + 2F\, du_{2}\, dv_{2} + G\, dv_{2}^{2}}} \\
  &= \frac{F\, dt^{2}}{\sqrt{E\, dt^{2}} \sqrt{G\, dt^{2}}}
  = \frac{F}{\sqrt{E} \sqrt{G}}
  = \frac{r_{u} \cdot r_{v}}{\|r_{u}\|\, \|r_{v}\|}.
\end{align*}
