Cylindrical Coordinate System. Please check my answer

1. Jan 31, 2012

jhosamelly

1. The problem statement, all variables and given/known data

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

(b) In what direction is $\hat{\phi}$ if $\phi$ = 90°

2. Relevant equations

3. The attempt at a solution

here is my solution. for a.

$\vec{r} = \rho cos \phi \hat{i} + \rho sin \phi \hat{j} + z \hat{k}$

$\frac{\partial \vec{r}}{\partial \rho}$ = $\frac{\partial \rho}{\partial {\rho}}$ cos $\phi$ $\hat{i}$ + $\frac{\partial \rho}{\partial {\rho}}$ sin $\phi$ $\hat{j}$

$\frac{\partial \vec{r}}{\partial \rho}$ = $cos \phi \hat{i} + sin \phi \hat{j}$

$\left|\frac{\partial \vec{r}}{\partial \rho}\right|$ = $\sqrt{cos^2 \phi + sin^2 \phi}$

$\left|\frac{\partial \vec{r}}{\partial \rho}\right|$ = 1

$\hat{r} = \frac{\frac{\partial \vec{r}}{\partial \rho}}{\left|\frac{\partial \vec{r}}{\partial \rho}\right|}$

$\hat{r}$ = $cos \phi \hat{i} + sin \phi \hat{j}$

so if $\phi$ is 0.

$\hat{r} = \hat{i}$

meaning $\hat{r}$ is pointing at the direction of the positive x-axis

now for b.

$\vec{r} = \rho cos \phi \hat{i} + \rho sin \phi \hat{j} + z \hat{k}$

$\frac{\partial \vec{r}}{\partial \phi}$ = $\rho \frac{\partial cos \phi}{\partial {\phi}}$ $\hat{i}$ + $\rho \frac{\partial sin \phi}{\partial {\phi}}$

$\frac{\partial \vec{r}}{\partial \phi}$ = $- \rho sin \phi \hat{i} + \rho cos \phi \hat{j}$

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

$\left|\frac{\partial \vec{r}}{\partial \phi}\right| = \rho$

$\hat{\phi} = \frac{\frac{\partial \vec{r}}{\partial \phi}}{\left|\frac{\partial \vec{r}}{\partial \phi}\right|}$

$\hat{\phi} = -sin \phi \hat{i} + cos \phi \hat{j}$

so if phi is 90°

$\hat{\phi} = -\hat{i}$

meaning $\hat{\phi}$ 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. Jan 31, 2012

δοτ

You changed notation in b when the norm of the partial was taken (it should be the partial w.r.t $\phi$), but other than that it is all correct.

3. Jan 31, 2012

jhosamelly

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.

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