Find the accelerations of a moving particle (or point) referred to rectangular axes which are not fixed in space 
Find the accelerations of a moving particle (or point) referred to rectangular axes $OX$ and $OY$ which are not fixed in space but which revolve in any manner about the origin $O$ in their own plane.


Let $OA$ be a fixed line in space, and let $\theta$ be the inclination of $OX$ at any time $t$. Let $P$ be the moving point. Draw $PM$ and $PN$ perpendiculars to $OX$ and $OY$ respectively.
The book says, The velocity of $P$ parallel to $OX$ is,
the velocity of $N$ parallel to $OX+$ the velocity of $M$ along $OM$
$$-y\frac{d\theta}{dt}+\frac{dx}{dt}$$
Similarly, the velocity of $P$ parallel to $OY$ is,
the velocity of $M$ parallel to $OY+$ the velocity of $N$ along $ON$
$$x\frac{d\theta}{dt}+\frac{dy}{dt}$$
The acceleration of $P$ parallel to $OX$,
acceleration of $N$ parallel to $OX+$ acceleration of $P$ relative to $N$
$$-\frac1y\frac{dy}{dt}\left(y^2\frac{d\theta}{dt}\right)+\frac{d^2x}{dt^2}-x\left(\frac{d\theta}{dt}\right)^2$$
Similarly, the acceleration of $P$ parallel to $OY$ is,
the acceleration of $M$ parallel to $OY+$ the acceleration of $N$ along $ON$
$$\frac1x\frac{d}{dt}\left(x^2\frac{d\theta}{dt}\right)+\frac{d^2y}{dt^2}-y\left(\frac{d\theta}{dt}\right)^2$$

I couldn't understand how they come up with those equations. In fact, They didn't give any intuition. It will be a great help if anyone briefly explain those.
 A: Suppose that we constructed a set of unit vectors $\hat{i}$ and $\hat{j}$ that are always pointing in the directions of $X$ and $Y$ respectively from $O$. As a result, we could represent the position vector of $P$ relative to $O$ as:
$$\vec{r}_{P/O} = x\hat{i} + y\hat{j}$$ where $x$ and $y$ are defined per your diagram.
In addition, we might consider constructing some unit vectors $\hat{a}$ and $\hat{b}$, which do not change over time, that point in the direction of $\hat{i}$ and $\hat{j}$ when $\theta = 0$. As a result, we can write the unit vectors $\hat{i}$ and $\hat{j}$ as:
$$\hat{i} = \cos{\theta}\hat{a} + \sin{\theta}\hat{b}$$
$$\hat{j} = \cos{\theta}\hat{b} - \sin{\theta}\hat{a}$$
Taking the time derivative of these expressions, while keeping in mind that $\hat{a}$ and $\hat{b}$ don't change over time, results in:
$$\dot{\hat{i}} = -\sin{\theta}\dot{\theta}\hat{a} + \cos{\theta}\dot{\theta}\hat{b} = \dot{\theta}\hat{j}$$
$$\dot{\hat{j}} = -\sin{\theta}\dot{\theta}\hat{b} - \cos{\theta}\dot{\theta}\hat{a} = -\dot{\theta}\hat{i}$$
Taking the time derivative of the position vector above, we arrive at the velocity of $P$ relative to $O$:
$$\vec{v}_{P/O} = \dot{x}\hat{i} + x\dot{\hat{i}} + \dot{y}\hat{j} + y\dot{\hat{j}}$$
Inserting the expressions for the time derivatives of $\hat{i}$ and $\hat{j}$ we found before, the velocity of the particle can be equivalently written as:
$$\vec{v}_{P/O} = \dot{x}\hat{i} + x\dot{\theta}\hat{j} + \dot{y}\hat{j} - y\dot{\theta}\hat{i}$$
Taking the dot product of this expression with either the $\hat{i}$ or $\hat{j}$ vectors results in the speeds your text had calculated:
$$\vec{v}_{P/O} \cdot \hat{i} = \dot{x} - y\dot{\theta}$$
$$\vec{v}_{P/O} \cdot \hat{j} = x\dot{\theta} + \dot{y}$$
For the acceleration, we can calculate the time derivative of the velocity and replace the same kinematic expressions for the time derivatives of the unit vectors:
$$\vec{a}_{P/O} = \ddot{x}\hat{i} + 2 \dot{x}\dot{\theta}\hat{j} + x\ddot{\theta}\hat{j} - x\dot{\theta}^2\hat{i} + \ddot{y}\hat{j} - 2 \dot{y}\dot{\theta}\hat{i} - y\ddot{\theta}\hat{i} + y\dot{\theta}^2\hat{j}$$
By taking the appropriate dot products with this expression, and some manipulation of the expressions for the scalars in front of each unit vector, you obtain the expressions for the components of the acceleration that your text calculated.
