one parameter family of diffeomorphism $\phi_{t}$ of $\mathbb{R}^2$ I have the following question: 
The one parameter family of diffeomorphism $\phi_{t}$ of $\mathbb{R}^2$ to itself for $t\in (\pi,\pi)$ is defined in polar coordinates $(r,\theta)$ by $$\phi_t(r,\theta)=(rcos(\theta+t),rsin(\theta+t))$$
and $\phi_{t}(0)=0$ for all $t$, i.e. origin is fixed.
Compute the derivative $X=\frac{d\phi_t}{dt}_{t=0}$.
Let $M_{s}:\mathbb{R}^2\rightarrow \mathbb{R}^2$ be the derivative of the map $\phi_s$ for a fixed $s$. What is the relation between $M_{s}(X)$ and $\frac{d\phi_t}{dt}_{t=s}$.

My efforts:
First I changed everything into Cartesian coordinates by putting $x=rcos(\theta)$ and $y=rsin(\theta)$. 
So $\phi_t(x,y)=(xcos(t)-ysin(t), xsin(t)+ycost(t))$. So $$\phi_t=\begin{pmatrix}
cost(t) & -sin(t) \\
sin(t) & cos(t)
\end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}$$.
Observe that this matrix is rotation matrix. 
Now $$\frac{d\phi_t}{dt}=\begin{pmatrix}
-sin(t) & -cos(t) \\
cos(t) & -sin(t)
\end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}$$
Putting $t=0$ we get $X(x,y)=(-y,x)$
Now I find $$M_s=\begin{pmatrix}
cos(s) & -sin(s) \\
sin(s) & cos(s)
\end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}$$
I hope my calculations are correct.

I am not able to solve the last part. Relation between $M_{s}(X)$ and $\frac{d\phi_t}{dt}_{t=s}$.
Second thing, I don't understand what is happening here geometrically. 
Is this question related to concept of flow, vector field, integral curves? (Just a wild guess!!)
I have done the calculations (as it was easy) but I don't understand the theory. What's going on here in background ? I guess it is related to vector field ...  
 A: $\bullet$ Just use the chain rule for smooth functions $f: \mathbb{R}^n \to \mathbb{R}^1$ i.e,
\begin{align*} \frac{d \phi_t}{d t}\Bigr|_{t = 0} &= \nabla \phi_0 \cdot \gamma'(0) \\ \\ &= \begin{pmatrix} \frac{\partial \phi_t}{\partial r}\Bigr|_{t=0} & \frac{\partial \phi_t}{\partial \theta}\Bigr|_{t = 0}  \end{pmatrix} \begin{pmatrix} \frac{\partial x}{\partial t}\Bigr|_{t=0} \\ \frac{\partial y}{\partial t}\Bigr|_{t=0} \end{pmatrix}\end{align*}
where,
\begin{cases}
 \gamma(t) = (r \cos ( \theta + t), r \sin (\theta + t))\\
 \\
\phi_t(r, \theta) = (x(r, \theta,t), y(r, \theta,t)) = (x,y)\end{cases}
$\bullet$ For the relationship between the derivatives , first recall that $\phi_{m+n} = \phi_{m} \circ \phi_n = \phi_n \circ \phi_m$.  Next we define the vector field,
$$\textbf{v}(x_0)=\frac{d}{dt}\Bigr|_{t = 0} \phi_t(x_0), x_0 \in \mathbb{R}^2$$
i.e  $\textbf{v}$ gives the initial rate of change of $x_0$ under the $1$-parameter subgroup. Now we just manipulate the leibniz notation,
$$ \frac{d \phi_t}{dt} \Bigr|_{t = s} = \frac{d (\phi_{\epsilon + s})}{d\epsilon} \Bigr|_{\epsilon = 0} = \frac{d}{d \epsilon} \Bigr|_{\epsilon = 0} (\phi_{\epsilon} \circ \phi_s) = \textbf{v}(\phi_s)$$
Hence, if you want to know how $\phi_s$ moves $p \in\mathbb{R}^2$, you just take the velocity field $\textbf{v}$, and translate it to the point $p$.
$\bullet$ Geometrically what is going on is that you are just taking a particle on a circle of radius $r$ centered at the origin and moving it along this circle by $t$ adding to the angle. 
