Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cylindrical Coordinate System. Please check my answer

  1. Jan 31, 2012 #1
    1. The problem statement, all variables and given/known data

    (a) In cylindrical coordinates , show that [itex]\hat{r}[/itex] points along the x-axis is [itex]\phi[/itex] = 0 .

    (b) In what direction is [itex]\hat{\phi}[/itex] if [itex]\phi[/itex] = 90°

    2. Relevant equations

    3. The attempt at a solution

    here is my solution. for a.

    [itex] \vec{r} = \rho cos \phi \hat{i} + \rho sin \phi \hat{j} + z \hat{k} [/itex]

    [itex]\frac{\partial \vec{r}}{\partial \rho}[/itex] = [itex]\frac{\partial \rho}{\partial {\rho}}[/itex] cos [itex]\phi[/itex] [itex]\hat{i}[/itex] + [itex]\frac{\partial \rho}{\partial {\rho}}[/itex] sin [itex]\phi[/itex] [itex]\hat{j}[/itex]

    [itex]\frac{\partial \vec{r}}{\partial \rho}[/itex] = [itex] cos \phi \hat{i} + sin \phi \hat{j}[/itex]

    [itex]\left|\frac{\partial \vec{r}}{\partial \rho}\right|[/itex] = [itex]\sqrt{cos^2 \phi + sin^2 \phi}[/itex]

    [itex]\left|\frac{\partial \vec{r}}{\partial \rho}\right|[/itex] = 1

    [itex] \hat{r} = \frac{\frac{\partial \vec{r}}{\partial \rho}}{\left|\frac{\partial \vec{r}}{\partial \rho}\right|} [/itex]

    [itex] \hat{r} [/itex] = [itex] cos \phi \hat{i} + sin \phi \hat{j}[/itex]

    so if [itex] \phi [/itex] is 0.

    [itex] \hat{r} = \hat{i} [/itex]

    meaning [itex] \hat{r} [/itex] is pointing at the direction of the positive x-axis

    now for b.

    [itex] \vec{r} = \rho cos \phi \hat{i} + \rho sin \phi \hat{j} + z \hat{k} [/itex]

    [itex]\frac{\partial \vec{r}}{\partial \phi}[/itex] = [itex] \rho \frac{\partial cos \phi}{\partial {\phi}}[/itex] [itex]\hat{i}[/itex] + [itex] \rho \frac{\partial sin \phi}{\partial {\phi}}[/itex]

    [itex]\frac{\partial \vec{r}}{\partial \phi}[/itex] = [itex] - \rho sin \phi \hat{i} + \rho cos \phi \hat{j} [/itex]

    [itex]\left|\frac{\partial \vec{r}}{\partial \phi}\right|[/itex] = [itex]\sqrt{\rho^2 (sin^2 \phi + cos^2 \phi)}[/itex]

    [itex]\left|\frac{\partial \vec{r}}{\partial \phi}\right| = \rho [/itex]

    [itex] \hat{\phi} = \frac{\frac{\partial \vec{r}}{\partial \phi}}{\left|\frac{\partial \vec{r}}{\partial \phi}\right|} [/itex]

    [itex] \hat{\phi} = -sin \phi \hat{i} + cos \phi \hat{j} [/itex]

    so if phi is 90°

    [itex] \hat{\phi} = -\hat{i} [/itex]

    meaning [itex] \hat{\phi} [/itex] points along the negative x-axis

    I hope I'm correct. can someone please tell me if I did this right? Thanks
    Last edited: Jan 31, 2012
  2. jcsd
  3. Jan 31, 2012 #2
    You changed notation in b when the norm of the partial was taken (it should be the partial w.r.t [itex]\phi[/itex]), but other than that it is all correct.
  4. Jan 31, 2012 #3
    ow, I'm sorry about that. I just copy paste it from letter a's equation. wasn't able to change it. sorry. Thanks for the reminder though. :))
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook