I was trying to write the kinetic energy by following, step by step, the construction of tangent vectors and then inducing the scalar product from $\mathbb{R}^3$. Theorically, all the construction is quite clear to me, but I find myself in trouble when trying to apply it (I guess it is not that clear). Let's take, for example, a Paraboloid. I can define a chart (which also is an atlas):
$\phi(\rho,\theta)=(\rho\cos\theta,\rho\sin\theta, \rho^2+1)$
Now, I could also think that $\rho=\rho(t)$ and $\theta=\theta(t)$. In this case, i have a curve on the manifold. I can now define a derivation:
$X_{p,\phi}f=(f\circ \phi(t))'(0)$
with $\phi(0)=p$. Now, it is easy to express this tangent vector thinking of it as immersed in $\mathbb{R}^3$, but I was trying to espress it in the basis $\bigg\{\frac{\partial}{\partial\rho},\frac{\partial}{\partial\theta}\bigg\}$. Frankly, and this makes me feel really stupid, i can't. The thing is that taking the vector I found and feeding it (two times, since it is a quadratic form) to the induced scalar product:
$S=(S_{11},S_{12},S_{21},S_{22})=(1,0,0,\rho^2)$
does not reproduce the kinetic energy I wanted, which is:
$T=\dot\rho^2(1+4\rho^2)+\rho^2\dot\theta^2$
I'm grateful for any help you can give me, because I really need to deeply understand these basic concepts!
Your value for $S_{11}$ is incorrect. Note that $$\frac{\partial}{\partial\rho} = (\cos\theta,\sin\theta,2r),$$ so its square length is $4\rho^2+1$. This will give you the kinetic energy you seek :)