I need to understand the velocity $\mathbf{r}^{\prime}(p)$ of a curve with parameter $p$, its acceleration $\mathbf{r}^{\prime\prime}(p)$ and the magnitude of a vector on a surface $S$ (or, 2-dimensional manifold).
The surface is defined as $S=S(u,v)$ via the 2 parameters (coordinates) $u$ and $v$
The curve is defined as $\mathbf{r}(p)=\left( \begin{array}{c} u(p)\\ v(p)\\ \end{array} \right)$ on the surface $S$.
Now, is the speed $\mathbf{r}^{\prime}(p)=\left( \begin{array}{c} u^{\prime}(p)\\ v^{\prime}(p)\\ \end{array} \right)$
and the acceleration
$\mathbf{r}^{\prime\prime}(p)=\left( \begin{array}{c} u^{\prime\prime}(p)\\ v^{\prime\prime}(p)\\ \end{array} \right)$ ?
How does one calculate the modulus, say, of the speed: is it
$|\mathbf{r}^{\prime}|=\sqrt{(u^{\prime}(p))^2+(v^{\prime}(p))^2}$?
This is how it would be done in Euclidean space, but am unsure if it is true on a manifold?
If $M$ is a smooth manifold and $r:\mathbb{R}\to M$ is a curve on $M$, then we have to look at the differential of $r$. At a point $y\in \mathbb{R}$, the differential of $r$ is a map $dr_y : T_y\mathbb{R} \to T_{r(y)}M$, where $T_aN$ denotes the tangent space to $N$ at a point $a\in N$. If our manifold $M$ is $n$-dimensional, then $T_{r(y)}M \cong \mathbb{R}^n$. In this way we can view the differential of $r$ as a mapping from $\mathbb{R}$ to $\mathbb{R}^n$, just like in the 'usual' Euclidean way.
In your case, since $r$ is a parametrized path, we are interested in the evaluation of $dr_y(\hat{x})$, where $\hat{x}$ is a unit vector in the "$x$" direction. This is called the pushforward of $\hat{x}$ by $r$, and from the definitions we see that $dr_{y}(\hat{x}) \in T_{r(y)}M\cong \mathbb{R}^n$. In other words, the pushforward of $\hat{x}$ by $r$ is an n-dimensional vector, which we can think of as the velocity of $r$ at the point $r(y)$. You can then compute the speed of $r$ by taking the magnitude of this vector, for example.