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I have a riemannian manifold $M$ and a smooth curve $\alpha$. I want to take a variation of $\alpha$ and apply the first variation formula of arc length but I want to know if it is possible to take the curves of the variation to be geodesics.

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It is not possible in general. For example, if you take a Hadamard manifold, wich is a complete riemannian manifold for what the exponential map is a diffeomorphism in some point $p$, there will exist just one geodesic joinning two given points, hence you will not be able to take variations (with fixed extremes) by geodesics. In some cases you will be able: on the sphere, a geodesic joinning two opposite points can be variated by geodesics.

More generally, it is not true that you can approach a given smooth curve by a geodesic, for example in the plane $\mathbb{R}^2$: if you take a general curve joinning two given points, its lenght can be much greater than the distance between these two points. If you have a curve near another (for example in the $C^1$ topology), the arc lenght must be near too. On another hand, you can approach a given smooth curve by broken geodesics, making an argument with totally convex neighborhoods.

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Are you, by case, trying to prove that a geodesic wich minimizes distance between a point and a compact submanifold is orthogonal to the submanifold? –  matgaio May 6 '12 at 4:42
    
I'm working with regular and critical point of the distance function. –  Chu May 7 '12 at 15:02

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