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

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}(d \phi)/(dt)`

`\frac{d \hat{\phi}}{dt}=-\hat{\rho}(d \phi)/(dt)`

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

Step-by-step explanation:

Let's start by writing the coordinate transformations:

`\rho=√(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}.(d \rho)/(dt)+\frac{\partial \hat{\rho}}{\partial \phi}.(d \phi)/(d t)+\frac{\partial \hat{\rho}}{\partial z}.(d z)/(d t)=\hat{\phi}.(d \phi)/(d t)`

`\frac{d \hat{\phi}}{d t}=\frac{\partial \hat{\phi}}{\partial \rho}.(d \rho)/(dt)+\frac{\partial \hat{\phi}}{\partial \phi}.(d \phi)/(d t)+\frac{\partial \hat{\phi}}{\partial z}.(d z)/(d t)=-\hat{\rho}.(d \phi)/(d t)`

`\frac{d \hat{\rho}}{d t}=\frac{\partial \hat{z}}{\partial \rho}.(d \rho)/(dt)+\frac{\partial \hat{z}}{\partial \phi}.(d \phi)/(d t)+\frac{\partial \hat{z}}{\partial z}.(d z)/(d t)=0`