Show that , for all $(s_{0},t_{0})\in [0,1]\times [0,a]$, the curves $s\to f(s,t_{0})$, $t\to f(s_{0},t)$ are orthogonals.

Question

Let $f:[0,1]\times [0,a]\to M$ a parameterized surface such that for all $t_{0}\in[0,a]$, the curve $s\to f(s,t_{0})$, $s\in [0,1]$, is a parameterized geodesic by arc lenght , orthogonal to the curve $t\to f(0,t)$, $t\in[0,a]$, in the point $f(0,t_{0})$.

Show that , for all $(s_{0},t_{0})\in [0,1]\times [0,a]$, the curves $s\to f(s,t_{0})$, $t\to f(s_{0},t)$ are orthogonals.

Take the derivate of $\left\langle \dfrac{\partial f}{\partial s},\dfrac{\partial f}{\partial t}\right\rangle$ in relation to $s$, for the vector field along the curve we have that $$\dfrac{d}{ds}\left\langle \dfrac{\partial f}{\partial s},\dfrac{\partial f}{\partial t}\right\rangle=\left\langle\dfrac{D}{ds}\dfrac{\partial f}{\partial s},\dfrac{\partial f}{\partial t}\right\rangle+\left\langle \dfrac{\partial f}{\partial s},\dfrac{D}{ds}\dfrac{\partial f}{\partial t}\right\rangle$$ With $t\in I$. Morover is M is a differential manifold with a conection D simetric and a parameterized surface then, $$\dfrac{D}{\partial v}\dfrac{\partial s}{\partial u}=\dfrac{D}{\partial u}\dfrac{\partial s}{\partial v}$$ and by condition of geodesic $$\dfrac{D}{ds}\dfrac{\partial f}{\partial s}=0$$ Therefore $$\dfrac{d}{ds}\left\langle \dfrac{\partial f}{\partial s},\dfrac{\partial f}{\partial t}\right\rangle=\dfrac{1}{2}\dfrac{d}{dt}\left\langle \dfrac{\partial f}{\partial s},\dfrac{\partial f}{\partial s}\right\rangle=0$$ This is the idea of book, but I can not see how finishing the problem, thanks!

Answer 1

Actually it's a bit to easy for an answer so I do not know if I missed something.

From $\frac{d}{ds}\langle\frac{\partial f}{\partial s},\frac{\partial f}{\partial t}\rangle = 0$ you know that $\langle\frac{\partial f}{\partial s},\frac{\partial f}{\partial t}\rangle $ is constant in $s$. In addition you know $\langle\frac{\partial f}{\partial s},\frac{\partial f}{\partial t}\rangle (0,t_0) = 0 $ for all $t_0 \in [0,a]$ from the orthogonality premise. From this it follows that $\langle\frac{\partial f}{\partial s},\frac{\partial f}{\partial t}\rangle (s_0, t_0) = 0$ for all $(s_0, t_0) \in [0,1] \times [0,a]$ which is, what was asked.

Did you ask for explanations of the steps from the book?

Answer 2

You did everything correctly, but the last part you need is just an observation: we have the inner product $\left \langle \frac{\partial f}{\partial s}, \frac{\partial f}{\partial t} \right\rangle$ is independent of the parameter $s$, hence this inner product remains the same regardless of what value of $s$ we have inputted. This means that all points $(s,t)$ we have

$$ \left\langle \frac{\partial f}{\partial s},\frac{\partial f}{\partial t}\right\rangle_{f(s,t)} \;\; =\;\; \left\langle \frac{\partial f}{\partial s},\frac{\partial f}{\partial t}\right\rangle_{f(0,t)}\;\; =\;\;0 $$

where it was an assumption in the problem that these two vectors are taken to be orthogonal when $s=0$.