(adsbygoogle = window.adsbygoogle || []).push({}); 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

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# Cylindrical Coordinate System. Please check my answer

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