Rotating Frames of Reference: Doubt while deriving the velocity I am following an online Chapter on Coriolis force, where the author develops the equations for a rotating frame of reference. The Figure and most of the notation used in the derivation can be understood from the diagram shown below:
 
The $x$ and $y$ coordinates (in the rotating frame) are related to the $X$ and $Y$ (in the fixed frame) by the following equations:

Taking the time derivative we get: 

Now, I have the following questions based on $dx\over{dt}$ and $dy\over{dt}$
1) What do these quantities represent? Are they the components of velocity of a particle observed from the rotating frame of reference?
2) If we further simply the expressions for $dx\over{dt}$ and $dy\over{dt}$, we can write:
$$\frac{dx}{dt}=\frac{dX}{dt}\cos(\Omega t)+\frac{dY}{dt}\sin(\Omega t)+\Omega y$$
$$\frac{dy}{dt}=-\frac{dX}{dt}\sin(\Omega t)+\frac{dY}{dt}\sin(\Omega t)-\Omega x$$
Is there any significance of the last terms? 
 A: 
What do these quantities represent? Are they the components of velocity of a particle observed from the rotating frame of reference?

$\dot x$ and $\dot y$ are indeed the components of the velocity of the particle observed from the rotating frame.

Is there any significance of the last terms?

Yes, indeed it has.

The vectorial relation 
$$\mathrm d\mathbf A_\mathrm S~=~ \mathrm d\mathbf A_\mathrm{S'} + (\omega \times \mathbf A)~\mathrm dt \tag 1$$
shows how the differential of $\mathbf A$ as observed from an inertial frame $\rm S$ is related to the differential of the same observed from a rotating-frame $\rm S'$; the rightmost term connects the two or more accurately transforms the change as viewed from $\rm S'$ to the change as viewed from $\rm S\,.$
$(\omega\times \mathbf A)\,\mathrm dt$ is the rotating effect due to the change of the unit vectors of $\rm S'$ over the course of time due to its rotation and it must be added to the change exhibited in the non-inertial frame to get the actual change.
$(1)$ is a very powerful relation as $\bf A$ can be 'any vector as you please'.
At the present context, let we choose $\bf A$ to be the position vector $\mathbf r\;;$ then $(1)$ infers
$$\mathbf v~=~ \mathbf v' +  \underbrace{\mathbf \omega\times \mathbf r}_{\textrm{velocity of the coordinate system}~~ \mathrm S'} \;.$$
It can now be easily decomposed to the components that OP wrote. 
A: The last term seems to be the effect of the addition of tangential velocity component $\vec\omega\times \vec r$. 
$\vec r = x \hat x + y \hat y$
$\vec\omega = \omega \hat z$
$\vec\omega\times \vec r = \omega x \hat z \times \hat x +\omega y\hat z\times \hat y = \omega x\hat y-\omega y\hat x$
But since it's the rotating frame that is rotating with rate $\omega$ and not the object in the rotating frame, subtracting $\vec\omega\times \vec r$ will make up for the effect of the rotation.
Hence the velocity correction with respect to x-coordinate is $\omega y\hat x$ and the velocity correction for the y-coordinate is $-\omega x\hat y$
