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

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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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