Converting Cylindrical Coordinates to Orthogonal and Spherical Coordinates?

[mathmari]

Hey!

We are given the following point in cylindrical coordinates. We have to write in orthogonal and spherical coordinates.
The point is $\left (2, \frac{\pi}{2}, -4\right )$.

First of all, do orthogonal coordinates mean cartesian coordinates?? (Wondering)

The cylindrical coordinates are of the form $(r, \theta , z)$, that are defined by $x=r \cos \theta , y=r \sin \theta , z=z$.

The orthogonal coordinates are of the form $(x, y, z)$.

$x=r \cos \theta=2 \cos \frac{\pi}{2}=0 , y=r \sin \theta =2 \sin \frac{\pi}{2}=2 , z=z=-4$

So, the orthogonal coordinates of the point are $(0, 2, -4)$.
The spherical coordinates are of the form $(\rho, \theta , \phi)$, where $\rho =\sqrt{x^2+y^2+z^2}, \theta=\arctan \left (\frac{y}{x}\right ), \phi=\arccos \left (\frac{z}{\rho}\right )$.

$\rho=\sqrt{x^2+y^2+z^2}=\sqrt{0^2+2^2+(-4)^2}=2 \sqrt{5}, \theta=\arctan \left (\frac{y}{x}\right )=\arctan \left ( \frac{2}{0}\right ) \Rightarrow \theta=\frac{\pi}{2}, \phi=\arccos \left (\frac{z}{\rho}\right )=\arccos \left (\frac{-4}{2\sqrt{5}}\right )=\arccos \left ( \frac{-2}{\sqrt{5}}\right )$

So, the spherical coordinates of the point are $\left (2 \sqrt{5}, \frac{\pi}{2}, \arccos \left ( \frac{-2}{\sqrt{5}}\right )\right )$.
Is the formulation correct?? (Wondering)

Could I improve something?? (Wondering)

 

[I like Serena]

Hey!

We are given the following point in cylindrical coordinates. We have to write in orthogonal and spherical coordinates.
The point is $\left (2, \frac{\pi}{2}, -4\right )$.

First of all, do orthogonal coordinates mean cartesian coordinates?? (Wondering)

The cylindrical coordinates are of the form $(r, \theta , z)$, that are defined by $x=r \cos \theta , y=r \sin \theta , z=z$.

The orthogonal coordinates are of the form $(x, y, z)$.

$x=r \cos \theta=2 \cos \frac{\pi}{2}=0 , y=r \sin \theta =2 \sin \frac{\pi}{2}=2 , z=z=-4$

So, the orthogonal coordinates of the point are $(0, 2, -4)$.
The spherical coordinates are of the form $(\rho, \theta , \phi)$, where $\rho =\sqrt{x^2+y^2+z^2}, \theta=\arctan \left (\frac{y}{x}\right ), \phi=\arccos \left (\frac{z}{\rho}\right )$.

$\rho=\sqrt{x^2+y^2+z^2}=\sqrt{0^2+2^2+(-4)^2}=2 \sqrt{5}, \theta=\arctan \left (\frac{y}{x}\right )=\arctan \left ( \frac{2}{0}\right ) \Rightarrow \theta=\frac{\pi}{2}, \phi=\arccos \left (\frac{z}{\rho}\right )=\arccos \left (\frac{-4}{2\sqrt{5}}\right )=\arccos \left ( \frac{-2}{\sqrt{5}}\right )$

So, the spherical coordinates of the point are $\left (2 \sqrt{5}, \frac{\pi}{2}, \arccos \left ( \frac{-2}{\sqrt{5}}\right )\right )$.
Is the formulation correct?? (Wondering)

Could I improve something?? (Wondering)

Hi!

It's perfect. (Nod)

If anything, be careful with $\theta=\arctan \left (\frac{y}{x}\right )$, because it is not generally true.
Properly, it should be something like $\theta=\operatorname{atan2}(y, x)$.
That's because the range of $\arctan$ is limited to $(-\pi/2,\pi/2]$, while $\theta$ really has the full range.

 

[mathmari]

Nice! (Happy)

I applied the formula $\rho=\sqrt{z^2+r^2},\ \ \theta=\theta , \ \ \phi=\arctan \left (\frac{r}{z}\right )$ for the point in cylindrical coordinates $\left (1, -\frac{\pi}{6}, 0\right )$.

To calculate $\phi$ do we write it as followed?? $$\phi=\arctan \left (\frac{1}{0}\right )$$
Or is there an other way to write it?? (Wondering)

Do we say: "Since it $\frac{1}{0}$ is not defined and we know that $\tan $ is not defined at $\frac{\pi}{2}$, we conclude that $\phi=\frac{\pi}{2}$." ?? (Wondering)

Is the formulation correct?? Could I improve something?? (Wondering)

 

[I like Serena]

Nice! (Happy)

I applied the formula $\rho=\sqrt{z^2+r^2},\ \ \theta=\theta , \ \ \phi=\arctan \left (\frac{r}{z}\right )$ for the point in cylindrical coordinates $\left (1, -\frac{\pi}{6}, 0\right )$.

To calculate $\phi$ do we write it as followed?? $$\phi=\arctan \left (\frac{1}{0}\right )$$
Or is there an other way to write it?? (Wondering)

Do we say: "Since it $\frac{1}{0}$ is not defined and we know that $\tan $ is not defined at $\frac{\pi}{2}$, we conclude that $\phi=\frac{\pi}{2}$." ?? (Wondering)

Is the formulation correct?? Could I improve something?? (Wondering)

You should add that the range of $\phi$ in spherical coordinates is [$0,\pi$], which is why $\frac{\pi}{2}$ is the correct answer as opposed to $-\frac{\pi}{2}$. (Nerd)

 

[mathmari]

Ok... Thank you very much! (Smile)