I am looking at the following exercise:
Show that the following are equivalent conditions on a surface patch $\sigma (u, v)$ with first fundamental form $Edu^2 + 2F dudv + Gdv^2$ :
- $E_v = G_u = 0$.
- $σ_{uv}$ is parallel to the standard unit normal $N$.
- The opposite sides of any quadrilateral formed by parameter curves of $\sigma$ have the same length.
$$$$
I have done the folowing:
$(1)\Rightarrow (2)$:
We have that $E_v=0 \Rightarrow \sigma_{uv}\cdot \sigma_u=0$ and $G_u=0\Rightarrow \sigma_{uv}\cdot \sigma_v=0$.
So, $\sigma_{uv}$ is perpendicular to $\sigma_u$ and $\sigma_v$. Since the unit normal, $N$, is perpendicular to $\sigma_u$ and $\sigma_v$, we have that $\sigma_{uv}$ is paralle to $N$.
$$$$
$(2) \Rightarrow (1)$:
We have that $\sigma_{uv}$ is parallel ot the unit normal $N$. Since $\sigma_u$ and $\sigma_v$ are tangent to the surface, so perpendicular to $N$, we have that they are perpendicular also to $\sigma_{uv}$. That means that $\sigma_{uv}\cdot \sigma_u=0$ and $\sigma_{uv}\cdot \sigma_v=0$.
Since $E_v=2\sigma_{uv}\cdot \sigma_v$ we have that $E_v=0$ and since $G_u=2\sigma_{uv}\cdot \sigma_u$ we have that $G_u=0$.
$$$$
Is everything correct so far?
How can we show that $(1)-(3)$ are equivalent? Could you give me some hints?
$$$$
$$$$
$\quad$ If $\gamma$ is a curve lying in the image of a surface patch $\boldsymbol{\sigma}$, we have $$\gamma(t)=\boldsymbol{\sigma}(u(t),v(t))$$ for some smoth functions $u(t)$ and $v(t)$. Then, denoting $d/dt$ by a dot, we have $\dot\gamma=\dot{u}\boldsymbol{\sigma}_u+\dot{v}\boldsymbol{\sigma}_v$ by the chain rule, so $$\langle\dot\gamma,\dot\gamma\rangle=E\dot u^2+2F\dot u\dot v+G\dot v^2,$$ and the length of $\gamma$ is given by $$\int(E\dot u^2+2F\dot u\dot v+G\dot v^2)^{1/2}dt.\tag{6.1}$$
$$$$
$$$$
EDIT1:
$(3) \Rightarrow (1)$:
From the IVT we have $$\left (\sqrt{E(u,\tilde{v})}\right )_v=\frac{\sqrt{E(u,v^\ast )}-\sqrt{E(u,0)}}{v^\ast}, \ \ \tilde{v}\in (0,v^\ast)$$
So $$\frac{E_v(u,v)}{2\sqrt{E(u,v)}} \big |_{v=\tilde{v}}=\frac{\sqrt{E(u,v^\ast )}-\sqrt{E(u,0)}}{v^\ast}$$
Taking the derivative we have $$\int_0^{\epsilon}\frac{E_v(u,\tilde{v})}{2\sqrt{E(u,\tilde{v})}}du=\int_0^{\epsilon}\frac{\sqrt{E(u,v^\ast )}-\sqrt{E(u,0)}}{v^\ast}du=\frac{\int_0^{\epsilon}\sqrt{E(u,v^\ast )}du-\int_0^{\epsilon}\sqrt{E(u,0)}du}{v^\ast}=0$$
So $$\int_0^{\epsilon}\frac{E_v(u,\tilde{v})}{2\sqrt{E(u,\tilde{v})}} du=0 \Rightarrow E_v(u,\tilde{v})=0$$
Is this correct?
Or is the last implication wrong?
$$$$
EDIT2:
Could you give me also a hint how we could show the following?
$$$$
How can we find such a reparametrization?