Differential geometry question.

Question

Please explain how to solve this question. Thank you:)

And sorry for hand-writing.

Answer 1

Proof. If $E$ is constant, then $\sigma_u$ is constant, analogue $G$ constant then $\sigma_v$ is constant.

$F$ constant then the angle between $\sigma_u$ and $\sigma_v$ is contant.

Now

$L=-\sigma_u\cdot{\textbf{N}_u}$, $M=-\sigma_u\cdot{\textbf{N}_v}=-\sigma_v\textbf{N}_u$ and $N=-\sigma_v\cdot{\textbf{N}_v}$ are also constant, ie. $$\textbf{N}_u=cte_1$$ $$\textbf{N}_v=cte_2$$.

Therefore

$$\frac{d\textbf{N}}{ds}=\textbf{N}_u\frac{du}{ds}+\textbf{N}_v\frac{dv}{ds}v=cte_1\frac{du}{ds}+cte_2\frac{dv}{ds}.$$

Since $d\textbf{N}$ is ortogonal a $\textbf{N}$, the vector $$\textbf{W}=cte_1\frac{du}{ds}+cte_2\frac{dv}{ds}$$ lies the tangent plane in a point $p$ in the surface and the constant direction fixed $u, v.$.

Then define the surface (Cylinder)

$$\sigma(u, v)=\sigma(u)+v\textbf{W}.$$