What is the velocity of an orthogonal matrix? Consider an orthogonal $n\times n$ matrix $\boldsymbol{A}$ so that$\boldsymbol{A}^T \boldsymbol{A} = \boldsymbol{A} \boldsymbol{A}^T = \boldsymbol{I}$. Suppose this matrix can vary with time, hence denoting it as $\boldsymbol{A}(t)$. I would like to obtain the velocity matrix $\frac{d}{d t} \boldsymbol{A}(t)$, which I denote as $\dot{\boldsymbol{A}}$.
The conditions that $\dot{\boldsymbol{A}}$ needs to satisfy can be easily obtained by differentiating all sides of the identity $\boldsymbol{A}^T \boldsymbol{A} = \boldsymbol{A} \boldsymbol{A}^T = \boldsymbol{I}$ with respect to $t$:
\begin{equation}
\boldsymbol{A}^T \dot{\boldsymbol{A}} + \dot{\boldsymbol{A}}^T \boldsymbol{A} = \boldsymbol{A} \dot{\boldsymbol{A}}^T + \dot{\boldsymbol{A}} \boldsymbol{A}^T = \boldsymbol{O}
\end{equation}
This essentially gives a linear system that $\dot{\boldsymbol{A}}$ needs to satisfy. My question is, whether the solution $\dot{\boldsymbol{A}}$ has a nice closed form expression in terms of $\boldsymbol{A}$ or its elements.
Thank you!
Golabi

Having reflected more on my comments, I think I can formulate this as an answer: No.
Let $\mathcal{A}$ be the set of smooth curves through the manifold of $O(n)$ of of orthogonal $n\times n$ matrices (i.e. time dependent orthogonal matrices, as in your question) such that for all $A \in \mathcal{A},$ $A(0) = I$. It can be shown that the tangent space to $I \in O(n)$ is the space of skew-symmetric $n\times n$ matrices. 
It is known from the theory of smooth manifolds (see Tu, An Introduction to Manifolds, Prop. 8.16) that for every tangent vector $X$ at a point $p$ in a smooth manifold, there exists (locally) a smooth curve through $p$ whose velocity at $p$ is $X$. This means that for every skew-symmetric $n \times n$ matrix $B$, we can find $A(t) \in \mathcal{A}$ such that  $\dot A(0) = B$. However, as defined above, $A(0) = I$ for all $A \in \mathcal{A}$. Therefore, there can be no prescriptive formula which determines $\dot A(0)$ in terms of $A(0) = I$.
Now, if we allow ourselves to consider the functions of $t$ for each matrix element of $A(t)$, since we cannot just use the values of these functions at a time $t$ to find $\dot A$ at $t$ as I have just shown, the only other logical option here is use a derivative of these functions. However, then we have a fairly easy and completely tautological answer:
$$
\dot A(t) = (\dot a_{ij}(t))_{ij} = \dot A(t).
$$
