Question

Find the expressions for the unit vectors in cylindrical coordinate system, p, φ,2. in terms of x, ý, 2. Find the time-derivative of each. Hints: Unit vector p is defined in (x, y) plane. Remember that α -φ is perpendicular to a The easiest way to find φ is to express rho through φ and add 90 degrees

Answer 1

Answer:

Provided that the cylindrical coordinate system is given by the coordinates

$\rho,\,\phi, \,z$

The unit vectors are:

$\hat{\rho}=\hat{x}\cos(\phi)+\hat{y}\sin(\phi)$

$\hat{\phi}=-\hat{x}\sin(\phi)+\hat{y}\cos(\phi)$

$\hat{z}=\hat{z}$

With their time derivatives being:

$\frac{d \hat{\rho}}{dt}=\hat{\phi}\frac{d \phi}{dt}$

$\frac{d \hat{\phi}}{dt}=-\hat{\rho}\frac{d \phi}{dt}$

$\frac{d \hat{z}}{dt}=0$

Explanation:

Let's start by writing the coordinate transformations:

$\rho=\sqrt{x^2+y^2}$, $x=\rho\cos(\phi)$

$\phi=\atan(y/x)$, $y=\rho\sin(\phi)$

$z=z$.

For the unit vector $\hat{\rho}$ we have:

$\hat{\rho}=\frac{\vec{\rho}}{\rho}=\frac{x\hat{x}+y\hat{y}}{\rho}=\hat{x}\cos(\phi)+\hat{y}\sin(\phi)$

For the unit vector $\hat{\phi}$ we have:

$\hat{\phi}=\hat{x}\cos(\phi+90)+\hat{y}\sin(\phi+90)$

By adding 90 degrees to $\hat{\rho}$, we have (see the 2nd attachment),

$\hat{\phi}=-\hat{x}\sin(\phi)+\hat{y}\cos(\phi)$

It is not hard to see that $\hat{z}=\hat{z}$.

From this we can write the following useful expressions that we'll use later on to determine the time derivatives:

$\begin{array}{ccc}\frac{\partial \hat{\rho}}{\partial \rho}=0&\frac{\partial \hat{\phi}}{\partial \rho}=0&\frac{\partial \hat{\rho}}{\partial z}=0\\\frac{\partial \hat{\rho}}{\partial \phi}=-\hat{x}\sin(\phi)+\hat{y}\cos(\phi)=\hat{\phi}\\&\frac{\partial \hat{\phi}}{\partial \phi}=-\hat{x}\cos(\phi)-\hat{y}\sin(\phi)=-\hat{\rho}&\frac{\partial \hat{z}}{\partial \phi}=0\\\frac{\partial \hat{\rho}}{\partial z}=0&\frac{\partial \hat{\phi}}{\partial z}=0&\frac{\partial \hat{z}}{\partial z}=0\end{array}$

Now, knowing all of the above the time derivatives come in a straightforward way:

$\frac{d \hat{\rho}}{d t}=\frac{\partial \hat{\rho}}{\partial \rho}.\frac{d \rho}{dt}+\frac{\partial \hat{\rho}}{\partial \phi}.\frac{d \phi}{d t}+\frac{\partial \hat{\rho}}{\partial z}.\frac{d z}{d t}=\hat{\phi}.\frac{d \phi}{d t}$

$\frac{d \hat{\phi}}{d t}=\frac{\partial \hat{\phi}}{\partial \rho}.\frac{d \rho}{dt}+\frac{\partial \hat{\phi}}{\partial \phi}.\frac{d \phi}{d t}+\frac{\partial \hat{\phi}}{\partial z}.\frac{d z}{d t}=-\hat{\rho}.\frac{d \phi}{d t}$

$\frac{d \hat{\rho}}{d t}=\frac{\partial \hat{z}}{\partial \rho}.\frac{d \rho}{dt}+\frac{\partial \hat{z}}{\partial \phi}.\frac{d \phi}{d t}+\frac{\partial \hat{z}}{\partial z}.\frac{d z}{d t}=0$