To compare and contrast: the rectangular, cylindrical and spherical coordinates systems.

To compare and contrast: the rectangular, cylindrical and spherical coordinates systems.

Question
Alternate coordinate systems
asked 2020-12-28
To compare and contrast: the rectangular, cylindrical and spherical coordinates systems.

Answers (1)

2020-12-29
Given:
Point is represented as \(\displaystyle{\left({r},\theta,{z}\right)}.\)
Proof:
A point in cylindrical coordinate is represented by \(\displaystyle{\left({r},\theta.{z}\right)}.\)
\(\displaystyle{\left({r},\theta\right)}\) are polar coordinates
z is the usual - coordinates
The rectangular (x, y, z) and the spherical coordinates \(\displaystyle{\left({r},\theta,{z}\right)}\) are related as
\(\displaystyle{x}={r} \cos{\theta}\)
\(\displaystyle{y}={r} \sin{\theta}\)
\(\displaystyle{z}={z}\)
These are used to convert from cylindrical to rectangular coordinates.
Rectangular to cylindrical coordinates:
\(\displaystyle{r}^{2}={x}^{2}+{y}^{2}\)
\(\displaystyle \tan{\theta}=\frac{y}{{x}}\)
\(\displaystyle{z}={z}\)
A point in spherical coordinate is represented by \(\displaystyle{\left(\rho,\theta,\varphi\right)}.\)
The rectangular (x, y, z) and the spherical coordinates \(\displaystyle{\left(\rho,\theta,\varphi\right)}\) are related as
\(\displaystyle{x}=\rho \sin{\varphi} \cos{\theta}\)
\(\displaystyle{y}=\rho \sin{\varphi} \sin{\theta}\)
\(\displaystyle{z}=\rho \cos{\varphi}\)
These are used to convert from spherical to rectangular coordinates.
Rectangular to spherical coordinates:
\(\displaystyle\rho^{2}={x}^{2}+{y}^{2}+{z}^{2}\)
\(\displaystyle \tan{\theta}=\frac{y}{{x}}\)
\(\displaystyle\varphi={{\cos}^{ -{{1}}}{\left(\frac{z}{\sqrt{{{x}^{2}+{y}^{2}+{z}^{2}}}}\right)}}\)
Spherical to rectangular coordinates:
\(\displaystyle{r}=\rho \sin{\varphi}\)
\(\displaystyle\theta=\varphi\)
\(\displaystyle{z}=\rho \cos{\varphi}\)
Cylindricalto spherical coordinates:
\(\displaystyle\rho^{2}=\sqrt{{{r}^{2}+{z}^{2}}}\)
\(\displaystyle\theta=\theta\)
\(\displaystyle\varphi={{\cos}^{ -{{1}}}{\left(\frac{z}{\sqrt{{{r}^{2}+{z}^{2}}}}\right)}}\)
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