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Let be $(M^n,g)$ a Riemannian manifold and $c: [0,l]\to M$ a geodesic with unit speed. Consider the parametric surface $f$ is $M$, given by

$$f(s,t)=\exp_{c(s)}(tn(s)),$$

where $(s,t)\in [0,l]\times (-\varepsilon,+\varepsilon)$, for $\varepsilon <<1$, and $n$ is a unit vector field along to $c$, that for each $s\in [0,l]$, we have $c'(s)\perp n(s)$.

We can say $f$ like a variation and we can explicite the variation field by

$$\frac{\partial f}{\partial t}=(d\exp_{c(s)})_{tn(s)}(n(s)).$$

But what about $\frac{\partial f}{\partial s}$? How can I to derive in $s$ the formula of $f$?

I don't know if I understood very well the variation fields things, but if I fix $t=0$, I have that $f(s,0)=c(s)$ is a curve in $M$, then by definition

$$\frac{\partial f}{\partial s}(s,t)=df_{(s,t)}(1,0)\text{ or }=df_{(s,t)}\left(\frac{\partial}{\partial s}\right), \text{if $(s,t)$ are coordinates in $\Bbb{R}^2$}.$$

So is correct to say that

$$\frac{\partial f}{\partial s}(s,0)=c'(s)?$$ Cause, it is very "plausible" when we think $\frac{\partial f}{\partial s}(s,t_0)$ like the variation field along to the curve $s\mapsto f(s,t_0)$, for $t_0$ fixed.

If that is fine, how can I show? This implies that

$$\frac{\partial f}{\partial s}\perp \frac{\partial f}{\partial t}?$$

Thanks so much.

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