On page 252 of do Carmo's book "Differential Geometry of Curves and Surfaces" there is a proof given of the following proposition:

Let $x(u,v)$ be an orthogonal parametrization (that is, $F=0$) of a neighborhood of an oriented surface $S$, and $w(t)$ be a differentiable field of unit vectors along the curve $x(u(t), v(t))$. Then $$\left[\frac{Dw}{dt}\right]=\frac{1}{2\sqrt{EG}}\left\{G_u\frac{dv}{dt}-E_v\frac{du}{dt}\right\}+\frac{d\varphi}{dt}$$ where $\varphi(t)$ is the angle from $x_u$ to $w(t)$ in the given orientation.

Here is up to the relevant part the proof. I don't understand how the last equality in the below proof comes about. That equality is $\left\langle \left(\frac{x_u}{\sqrt{E}}\right)_u,\frac{x_v}{\sqrt{G}} \right\rangle=-\frac{1}{2}\frac{E_v}{\sqrt{EG}}$ and I don't see why the partial by $u$ on the left hand side does not effect the denominator $\sqrt{E}$ give that $E$ should be defined as $x_u \cdot x_u$

Proof. Let $e_1 = x_u/\sqrt{E}$, $e_2=x_v/\sqrt{G}$ be the unit vectors tangent to the coordinate curves. Observe that $e_1\wedge e_2=N$, where $N$ is the given orientation of $S$. By using Lemma 2, we may write $$\left[\frac{Dw}{dt}\right]=\left[\frac{De_1}{dt}\right]+\frac{d\varphi}{dt}$$ where $e_1(t)=e_1(u(t),v(t))$ is the field $e_1$ restricted to the curve $x(u(t),v(t))$. Now $$\left[\frac{De_1}{dt}\right]=\left\langle\frac{de_1}{dt},N\wedge e_1\right\rangle=\left\langle\frac{de_1}{dt},e_2\right\rangle=\left\langle(e_1)_u,e_2\right\rangle\frac{du}{dt}=\left\langle(e_1)_v,e_2\right\rangle\frac{dv}{dt}$$. On the other hand, since $F=0$, we have $$\langle x_{uu},x_v \rangle=-\frac{1}{2}E_v$$, and therefore $$\left\langle (e_1)_u,e_2 \right\rangle=\left\langle \left(\frac{x_u}{\sqrt{E}}\right)_u,\frac{x_v}{\sqrt{G}} \right\rangle=-\frac{1}{2}\frac{E_v}{\sqrt{EG}}$$


Well, because $x_u \cdot x_v = 0$, so

\begin{align} \left\langle \left(\frac{x_u}{\sqrt{E}}\right)_u,\frac{x_v}{\sqrt{G}} \right\rangle&= \left\langle \frac{x_{uu}}{\sqrt E} + \left(\frac{1}{\sqrt E}\right)_u x_u, \frac{x_v}{\sqrt G}\right\rangle\\ & = \left\langle \frac{x_{uu}}{\sqrt E}, \frac{x_v}{\sqrt G}\right\rangle \end{align}

  • $\begingroup$ Oh I totally forgot about the orthogonal parametrization condition while reading the proof. $\endgroup$ – user782220 Sep 14 '18 at 17:45

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