I am looking at the following exercise:
Suppose that the first fundamental form of a surface patch $\sigma (u, v)$ is of the form $E(du^2 + dv^2)$.
Prove that $\sigma_{uu} + \sigma_{vv}$ is perpendicular to $\sigma_u$ and $\sigma_v$.
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From the first fundamental form we have that $G=E$ and $F=0$.
We also have that $\sigma$ is conformal.
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Could you give me a hint how we could show that $\sigma_{uu} + \sigma_{vv}$ is perpendicular to $\sigma_u$ and $\sigma_v$ ?
You know $$\sigma_u\sigma_v = 0 $$ which implies $$\sigma_{uu}\sigma_{v}+\sigma_{u}\sigma_{uv}=0\\ \sigma_{vu}\sigma_{v}+\sigma_{u}\sigma_{vv}=0$$
You also know $$\sigma_u^2=\sigma_v^2$$ which implies $$2\sigma_{uv}\sigma_u=2\sigma_{vv}\sigma_{v}\\ 2\sigma_{uu}\sigma_u=2\sigma_{uv}\sigma_{v} $$
Insert the first two implications into the second to arrive at
$$\sigma_u(\sigma_{uu}+\sigma_{vv})=0\\ \sigma_v(\sigma_{uu}+\sigma_{vv})=0 $$