Let $P\subset M$ be a smooth regular submanifold of $M$, and $\alpha:(0,1)\to M$ be a smooth curve such that $\alpha(a,b)\subset P$. I want to see whether or not $\alpha'(t)\in T_{\alpha(t)}P$ for any $t\in(0,1)$ (as well as for $P$ a submanifold that doesn't have to be regular).
In my notes, I have that, taking $p\in P$, $T_pP$ is identified with the image in $T_pM$ through $(di)_p:T_pP\to T_pM$, where $i:P\hookrightarrow M$ is the inclusion. Proposition 3.9 from Lee's Smooth Manifolds tells us that $(di)_p$ is an isomorphism. But I don't really understand this characterization of $T_pP$ well, or if I can do anything at all in this problem.
I know that, taking $\left\{\left(\frac{\partial}{\partial x_i}\right)_p\right\}_{i=1}^n$ a base of $T_pM$ associated to the chart $(U,\varphi = (x_1,\ldots,x_n))$, one can represent the velocity of the curve as \begin{equation} \alpha'(t_0) = \sum_{i=1}^n(x_i\circ\alpha)'(t_0)\left(\frac{\partial}{\partial x_i}\right)_p. \end{equation} But I don't know how to represent $T_pP$ out of the information above. And I don't see if anything would change between $P$ being regular or not. Could anyone please help me out?
I've been gathering some things. If anything that I say is wrong, please let me know.
Since $\alpha:(0,1)\to M$ verifies that $\alpha(0,1)\subset P$, we can use a result which tells us that $\alpha_0:(0,1)\to P$ is continuous iff $\alpha$ is smooth. But $P$ is a regular submanifold, so it preserves the topology of $M$, which means that $\alpha_0$ is also smooth!
If we consider the inclusion $i:P\hookrightarrow M$, then $\alpha = i\circ\alpha_0$, which means that \begin{equation} \alpha'(t) = (i\circ\alpha)'(t) = (di)_{\alpha(t)}(\alpha_0'(t)). \end{equation} In my class notes, I have that $T_pP$ can be interpreted by the image of $(di)_p:T_pP\to T_pM$, so $(di)_{\alpha(t)}(\alpha_0'(t))\in T_{\alpha(t)}P$, which leads us to $\alpha'(t)\in T_{\alpha(t)}P$, q.e.d.
If we consider $P$ a non-regular submanifold, I've seen that the figure 8, defined with the structure given by the atlas $\mathcal A=\{(U,\varphi^{-1})\}$, where \begin{equation} \varphi:\left(\frac\pi2,\frac{5\pi}2\right)\to M,\;\;\varphi(t) = (\cos t,\sin 2t), \end{equation} could work as a counterexample. I still need to check if it works, though. I'll update when I have it.