Is it possible to compute geodesic without induced metric Suppose a manifold embedding $i:M\to N$ into Riemannian manifold $(N,g)$ is given by $f(x)=0$, where $f:M\to R^m$ is a smooth vector-valued function. Now if it is very hard to parameterize the submanifold $M$ with $\text{dim}M$ parameters, can we still compute the geodesic on $M$?
For a very simple example, given an isometry embedding $i:S^n\to R^{n+1}$ using $f(x)=|x|^2-1$, without parameterize the sphere by spherical coordinate, do we know the geodesics of sphere are great circles?
 A: This is sort of a soft question, so some of this answer may ramble.
Computing the geodesics of a general Riemannian manifold is (as far as I can tell) not an easy problem. (I'm not even sure how to make sense of this problem in general without parameterizing $M$.) However, for specific Riemannian manifolds one can often use tricks to compute geodesics that don't require a parameterization. 
For instance, let's show that the geodesics on $S^2$ are great circles. We imagine $S^2$ as embedded in $\mathbb{R}^{3}$. Then take a point $p \in S^2$ and any tangent vector $v \in T_p S^2$. We can think of $v$ as living in $\mathbb{R}^{3}$ in the usual way. 
Consider the reflection $R$ of $\mathbb{R}^{3}$ through the plane spanned by $p$ and $v$. This is obviously an isometry of $S^2$ whose fixed-point-set is exactly the great circle through $v$. On the other hand, since $R$ leaves $v$ fixed, it must also fix the geodesic through $v$. Thus the great circle must exactly be this geodesic.
The exact same argument works for $S^n$, of course.
As another example, this method determines geodesics in the hyperbolic plane. We use the model where $\mathbb{H}^2$ is the complex upper half-plane with the metric $\tfrac{1}{y^2} g_{{}_\text{Euclid}}$. By symmetry of the metric, the reflection in the imaginary is an isometry; since this isometry fixes only the imaginary axis, we conclude that the imaginary axis is the geodesic through $i$ in the vertical direction. Now one shows (standard calculation) that $SL(2, \mathbb{R})$ acts by isometries on $\mathbb{H}^2$ (via Mobius transformations), and since this isometry group is transitive on tangent vectors (there is an isometry taking any tangent vector to any other), this allows us to easily compute all the geodesics.
This sort of idea is very useful whenever the isometry group of $M$ is known and relatively large. But even when the isometry group of $M$ is not so large, one might still be able to exploit symmetries to at least determine some of the geodesics, which can still be useful. For instance, consider the case of $M$ a surface obtained by revolving a smooth curve about the $x$-axis in $\mathbb{R}^3$. Again exploiting the fact that reflection through any plane containing the $x$-axis is an isometry, one sees that the curves running along the surface parallel to the $x$-axis are geodesics. This doesn't determine all the geodesics on $M$, but it is a start.
