I posted this question in the physics SE with no luck so I will try it here.
what I understand: Let $\mathbb{Q}$ be a smooth $n$-dim manifold and $q=(q^1,\dots, q^n):Q\to\mathbb{R}^n$ some local coordinates. Then the derivative of a curve, $r:t\mapsto r_t\in\mathbb{Q}$, is expressed in the local $q^i$ coordinate basis, $\boldsymbol{\partial}_i$ with dual 1-forms $\pmb{d}q^i$, as
$$ \pmb{v}_t = \tfrac{d}{dt} r_t = \tfrac{d}{dt}(q^i\circ r)_t \boldsymbol{\partial}_i =: \dot{q}^i(t)\boldsymbol{\partial}_i \; \in {T} _{r_t}\mathbb{Q} \quad,\qquad \dot{q}^i = \tfrac{d}{dt}(q^i\circ r) = \pmb{d}q^i\cdot\pmb{v} = v^i $$
(I should technically write $ {(\boldsymbol{\partial}_i)} _{r} $.) That is, the components of the velocity in the $q^i$-basis are equivalent to the derivatives of the coordinate functions along the curve: $v^i(t)=\dot{q}^i(t)$.
what I don't understand: what happens when the coordinates are also explicit functions of time? That is, local coordinates $\tilde{q}=(\tilde{q}^1,\dots,\tilde{q}^n):Q\times \mathbb{R} \to \mathbb{R}^n$. If the above is still correct for time-dependent case, we should get
$$ \pmb{v}_t=\tfrac{d}{dt} r_t = \tfrac{d}{dt}(q^i\circ r)_t \boldsymbol{\partial}_i = \tfrac{d}{dt}(\tilde{q}^i\circ r)_t \tilde{\boldsymbol{\partial}}_i \qquad\quad (*) $$
but it seems the last equality, for the the time-dependent coordinates, is wrong as illustrated by the following simple example.
Example: Let $\mathbb{Q}=\mathbb{E}^3$ be classic flat 3-space, let $q=(q^1,q^2, q^3)$ be some linear coordinates in a fixed/inertial basis, $\pmb{e}_i\in\mathbb{E}^3$, and let $\tilde{q}=(\tilde{q}^1,\tilde{q}^2,\tilde{q}^3)$ be linear coordinates in a rotating basis, $\tilde{\pmb{e}}_i(t)=\pmb{R}(t)\cdot\pmb{e}_i\in\mathbb{E}^3$. Since our manifold is a vector space, we have the privilege of regarding a point, $r\in\mathbb{E}^3$, as a displacement vector, $\pmb{r}=q^i\pmb{e}_i=\tilde{q}^i(t)\tilde{\pmb{e}}_i(t)$. It is well known the the velocity along some $r_t$ is expressed in the inertial and the rotating bases as:
$$ \pmb{v}_t = \tfrac{d}{dt} r_t = \dot{q}^i(t)\boldsymbol{\partial}_i = ( \dot{\tilde{q}}^i(t) + \tilde{\Omega}^i_j(t) \tilde{q}^j(t)) \tilde{\boldsymbol{\partial}}_i(t) = \tilde{v}^i(t)\tilde{\boldsymbol{\partial}}_i(t) \quad,\qquad \tilde{v}^i:= \pmb{v}\cdot\pmb{d}\tilde{q}^i = \dot{\tilde{q}}^i + \tilde{\Omega}^i_j \tilde{q}^j = C^i_j v^j = C^i_j v^j $$
where I've identified $\boldsymbol{\partial}_i\cong \pmb{e}_i$ and $\tilde{\boldsymbol{\partial}}_i\cong \tilde{\pmb{e}}_i$ (again, since $\mathbb{Q}=\mathbb{E}^3$ is a vector space) and where $\tilde{\Omega}^i_j(t)$ are the components of the rotating basis's angular velocity tensor (in the rotating basis). That is, $\tilde{\Omega}^i_j=C^i_k\dot{R}^k_j$ where $\tilde{\pmb{e}}_i(t) = R^j_i(t) \pmb{e}_j$ and $C^i_k R^k_j=\delta^i_j$ such that $\tilde{q}^i = C^i_j q^j$ and $q^i=R^i_j \tilde{q}^j$ and $\dot{R}=R\tilde{\Omega}$ and $\dot{C}=-\tilde{\Omega}C$. Note $v^i=\dot{q}^i$ but $\tilde{v}^i \neq \dot{\tilde{q}}^i$.
Now, $\tilde{q}^i(t):=\tilde{q}^i\circ r_t$ is expressed in terms of $\tilde{q}^i$ and $q^i$ as $\tilde{q}^i\circ r_t= \tilde{q}^i(t) = C^i_j(t) q^j(t)$ then
$$ \tfrac{d}{dt}(\tilde{q}^i\circ {r}) = \tfrac{d}{dt}(C^i_j q^j) = C^i_j\dot{q}^j + \dot{C}^i_j q^j = C^i_j\dot{q}^j -\tilde{\Omega}^i_k C^k_j q^j = C^i_j\dot{q}^j -\tilde{\Omega}^i_k \tilde{q}^k = \dot{\tilde{q}}^i \neq \tilde{v}^i $$
So, for the time-dependent rotating coordinates, $\tilde{q}^i$, the last equality in my equation (*) does NOT hold:
$$ \tilde{v}^i(t)\tilde{\boldsymbol{\partial}}_i(t) = \pmb{v}_t = \tfrac{d}{dt} r_t \quad \neq\quad \tfrac{d}{dt}(\tilde{q}^i\circ {r})_t\tilde{\boldsymbol{\partial}}_i(t) = \dot{\tilde{q}}(t)\tilde{\boldsymbol{\partial}}_i(t) $$ The same issue seems to arise whenever $\tilde{q}^i$ are coordinates relative to some moving frame or moving point. And it is very common to use such coordinates in things like celestial mechanics and robotics. I am confused how to deal with this in the framework of differential geometry.
question: For coordinate functions which are not explicit functions of time, $ \tfrac{d}{dt} r_t = \tfrac{d}{dt}(q^i\circ {r})_t\boldsymbol{\partial}_i$ is the proper formula for expressing velocity along a curve in a local coordinate basis. But this gives the wrong velocity if the coordinate functions are explicit functions of time. What is the roper expression for the latter case? I'm not asking how to express the velocity in a rotating/moving frame in flat space; I know how to do that using basic Newton/Euler mechanics. I want to do it using differential geometry. I want a general formula for the velocity along a curve on any smooth manifold expressed in an explicitly time-dependent coordinate basis.
Even if you don't wish to give the general answer, if you can simply resolve my example for rotating frames in flat space (in the language of differential geometry), that would also be appreciated.
If this is most easily addressed using the extended configuration space, $\mathbb{Q}\times\mathbb{R}$, feel free to do so.