Let us consider a point P in cylindrical co-ordinate system, having cylindrical co-ordinates $(\rho,\ \phi,\ z)$. The cartesian co-ordinates of the point P is $(x,\ y,\ z)$. Let’s draw a vertical line PQ on the X-Y plane, which is parallel to the z-axis, as shown in Fig.1.

OQ is represented by $\rho$, which makes an angle $\phi$ with the X-axis.

From Fig. 1 we can write,

$x=\rho\ \cos\phi$

$y=\rho\ \sin\phi$

and, $z=z$

If $\vec{r}$ is the position vector with respect to the origin O, then we can write

$\vec{r}=x\ \hat{i}+y\ \hat{j}+z\ \hat{k}\tag{1}$

where, $\hat{i}$, $\hat{j}$ and $\hat{k}$ is the unit vector along the X-axis, Y-axis and Z-axis respectively.

Now, equation (1) can be written as,

$\vec{r}=\rho\ \cos\phi\ \hat{i}+\rho\ \sin\phi\ \hat{j}+z\ \hat{k}\tag{2}$

The unit vectors $\hat{\rho}$, $\hat{\phi}$ and $\hat{z}$ in the cylindrical pola co-ordinate system in terms of the unit vecctors in certesian co-ordinate can be written as,

$\hat{\rho}=\cos\phi\ \hat{i}+ \sin\phi\ \hat{j}\tag{i}$

$\hat{\phi}=-\sin\phi\ \hat{i}+\cos\phi\ \hat{j}\tag{ii}$

and, $\hat{z}=\hat{k}\tag{iii}$

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Again,

Differentiating $\hat{\rho}$ with respect to time $t$ we get,

$\frac{d\hat{\rho}}{dt}=\frac{d}{dt}(\cos\phi\ \hat{i}+ \sin\phi\ \hat{j})$

$or,\ \frac{d\hat{\rho}}{dt}=\frac{d}{d\phi}(\cos\phi\ \hat{i}+ \sin\phi\ \hat{j})\frac{d\phi}{dt}$

$or,\ \frac{d\hat{\rho}}{dt}=(-\sin\phi\ \hat{i}+\cos\phi\ \hat{j})\dot{\phi}$

$or,\ \displaystyle{\frac{d\hat{\rho}}{dt}=\dot{\phi}\ \hat{\phi}}\tag{iv}$ using equation (ii)

Differentiating $\hat{\phi}$ with respect to time $t$, we get

$\frac{d\hat{\phi}}{dt}=\frac{d}{dt}(-\sin\phi\ \hat{i}+\cos\phi\ \hat{j})$

$or,\ \frac{d\hat{\phi}}{dt}=\frac{d}{d\phi}(-\sin\phi\ \hat{i}+\cos\phi\ \hat{j})\frac{d\phi}{dt}$

$or,\ \frac{d\hat{\phi}}{dt}=-(\cos\phi\ \hat{i}+\sin\phi\ \hat{j})\dot{\phi}$

$or,\ \displaystyle{\frac{d\hat{\phi}}{dt}=-\dot{\phi}\hat{\rho}}\tag{v}$ using equation (i).

Differentiating $\hat{z}$ with respect to time $t$ we get,

$\displaystyle{\frac{d\hat{z}}{dt}=0}\tag{vi}$

Multiplying both side of equation (i) by $\cos\phi$ and equation (ii) by $\sin\phi$ and then subtracting, we get

$\hat{i}=\cos\phi\ \hat{\rho}-\sin\phi\ \hat{\phi}\tag{vii}$

Multiplying both side of equation (i) by $\sin\phi$ and equation (ii) by $\cos\phi$ and then adding, we get

$\hat{j}=\sin\phi\ \hat{\rho}+\cos\phi\ \hat{\phi}\tag{viii}$

Now putting value of $\hat{i}$, $\hat{j}$ and $\hat{k}$ in the equation (2), by using equaions (vii), (viii) & (iii), we get

$\vec{r}=\rho\ \cos\phi\ (\cos\phi\ \hat{\rho}-\sin\phi\ \hat{\phi})+\rho\ \sin\phi\ (\sin\phi\ \hat{\rho}+\cos\phi\ \hat{\phi})+z\ \hat{z}$

$or,\ \vec{r}=(\rho\ \sin^2\phi+\rho\ \cos^2\phi)\hat{\rho}+(\rho\ \cos\phi\ \sin\phi-\rho\ \cos\phi\ \sin\phi)\hat{\phi}+z\ \hat{z}$

$or,\ \vec{r}=\rho\ \hat{\rho}+z\ \hat{z}\tag{3}$

Now eqation (3) can be written as,

$\vec{r}=r_{\rho}\ \hat{\rho}+r_{\phi}\ \hat{\phi}+r_{z}\ \hat{z}\tag{4}$

where, $r_{\rho}=\rho$, $r_{\phi}=0$ and $r_{z}=z$ are the components of the position vectors $\vec{r}$ along the direction of the unit vectors $\hat{\rho}$, $\hat{\phi}$ and $\hat{z}$ respectively.

Expression for velocity in cylindrical polar co-ordinate system:

Since velocity is the time derivative of the displacement vector,

So, velocity $\displaystyle{\vec{v}=\frac{d\vec{r}}{dt}}$

Now differentiating equation (4) with respect to time t, we get

$\displaystyle{\frac{d\vec{r}}{dt}=\frac{dr_{\rho}}{dt}\ \hat{\rho}+r_{\rho}\frac{d\hat{\rho}}{dt}+\frac{dr_{\phi}}{dt}\hat{\phi}+r_{\phi}\frac{d\hat{\phi}}{dt}+\frac{dr_{z}}{dt}\hat{z}+r_{z}\frac{d\hat{z}}{dt}}$

$or,\ \displaystyle{\frac{d\vec{r}}{dt}=\frac{dr_{\rho}}{dt}\ \hat{\rho}+r_{\rho}\ \dot{\phi}\ \hat{\phi}+\frac{dr_{\phi}}{dt}\hat{\phi}+r_{\phi}(-\dot{\phi}\hat{\rho})+\frac{dr_{z}}{dt}\hat{z}}$
[ using equations (iv), (v) & (vi) ]

$or,\ \displaystyle{\frac{d\vec{r}}{dt}=\frac{d\rho}{dt}\ \hat{\rho}+\rho\ \dot{\phi}\ \hat{\phi}+\dot{z}\ \hat{z}}$

$or,\ \displaystyle{\frac{d\vec{r}}{dt}=\dot{\rho}\ \hat{\rho}+\rho\ \dot{\phi}\ \hat{\phi}+\dot{z}\ \hat{z}}\tag{5}$

So equation (5) is the expression for the velocity of a particle in a cylindrical polar co-ordinate system.