Cylindrical Coordinate System. Please check my answer

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jhosamelly
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Homework Statement



(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°

Homework Equations


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-axisnow 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-axisI hope I'm correct. can someone please tell me if I did this right? Thanks
 
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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.
 
δοτ said:
You changed notation in b when the norm of the partial was taken (it should be ht epartial w.r.t φ), but other than that it is all correct.

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. :))