## Polares

Tiene por origen O y vectores unitaros, $r ^$ y $θ ^$

Relación entre los vectores unitarios

$\begin{array}{l}\stackrel{^}{r}=\mathrm{cos}\theta ·\stackrel{^}{i}+\mathrm{sin}\theta ·\stackrel{^}{j}\hfill \\ \stackrel{^}{\theta }=-\mathrm{sin}\theta ·\stackrel{^}{i}+\mathrm{cos}\theta ·\stackrel{^}{j}\hfill \end{array}$

$\begin{array}{l}\frac{d\stackrel{^}{r}}{dt}=\left(-\mathrm{sin}\theta ·\stackrel{^}{i}+\mathrm{cos}\theta ·\stackrel{^}{j}\right)\frac{d\theta }{dt}=\frac{d\theta }{dt}\stackrel{^}{\theta }\\ \frac{d\stackrel{^}{\theta }}{dt}=-\left(\mathrm{cos}\theta ·\stackrel{^}{i}+\mathrm{sin}\theta ·\stackrel{^}{j}\right)\frac{d\theta }{dt}=-\stackrel{^}{r}\frac{d\theta }{dt}\end{array}$

• Vector posición

• $\stackrel{\to }{r}=r\stackrel{^}{r}$

• $\stackrel{\to }{v}=\frac{d\stackrel{\to }{r}}{dt}=\frac{d\left(r\stackrel{^}{r}\right)}{dt}=\frac{dr}{dt}\stackrel{^}{r}+r\frac{d\stackrel{^}{r}}{dt}=\frac{dr}{dt}\stackrel{^}{r}+r\frac{d\theta }{dt}\stackrel{^}{\theta }$

• Vector aceleración

• $\begin{array}{l}\frac{{d}^{2}\stackrel{\to }{r}}{d{t}^{2}}=\frac{d\stackrel{\to }{v}}{dt}=\frac{{d}^{2}r}{d{t}^{2}}\stackrel{^}{r}+\frac{dr}{dt}\frac{d\stackrel{^}{r}}{dt}+\left(r\frac{{d}^{2}\theta }{d{t}^{2}}+\frac{dr}{dt}\frac{d\theta }{dt}\right)\stackrel{^}{\theta }+r\frac{d\theta }{dt}\frac{d\stackrel{^}{\theta }}{dt}=\hfill \\ \frac{{d}^{2}r}{d{t}^{2}}\stackrel{^}{r}+\frac{dr}{dt}\frac{d\theta }{dt}\stackrel{^}{\theta }+\left(r\frac{{d}^{2}\theta }{d{t}^{2}}+\frac{dr}{dt}\frac{d\theta }{dt}\right)\stackrel{^}{\theta }-r{\left(\frac{d\theta }{dt}\right)}^{2}\stackrel{^}{r}=\hfill \\ \left(\frac{{d}^{2}r}{d{t}^{2}}-r{\left(\frac{d\theta }{dt}\right)}^{2}\right)\stackrel{^}{r}+\left(r\frac{{d}^{2}\theta }{d{t}^{2}}+2\frac{dr}{dt}\frac{d\theta }{dt}\right)\stackrel{^}{\theta }\hfill \end{array}$

## Cilíndricas

Tiene por origen O y vectores unitaros, $ρ ^$, $φ ^$ y $k ^$

Relación entre los vectores unitarios

$\begin{array}{l}\stackrel{^}{\rho }=\mathrm{cos}\phi \stackrel{^}{i}+\mathrm{sin}\phi \stackrel{^}{j}\\ \stackrel{^}{\phi }=-\mathrm{sin}\phi \stackrel{^}{i}+\mathrm{cos}\phi \stackrel{^}{j}\end{array}$

$\begin{array}{l}\frac{d\stackrel{^}{\rho }}{dt}=\left(-\mathrm{sin}\phi \stackrel{^}{i}+\mathrm{cos}\phi \stackrel{^}{j}\right)\frac{d\phi }{dt}=\frac{d\phi }{dt}\stackrel{^}{\phi }\\ \frac{d\stackrel{^}{\phi }}{dt}=\left(-\mathrm{cos}\phi \stackrel{^}{i}-\mathrm{sin}\phi \stackrel{^}{j}\right)\frac{d\phi }{dt}=-\frac{d\phi }{dt}\stackrel{^}{\rho }\end{array}$

• Vector posición

• $\stackrel{\to }{r}=\rho \stackrel{^}{\rho }+z\stackrel{^}{k}$

• $\begin{array}{l}\stackrel{\to }{v}=\frac{d\stackrel{\to }{r}}{dt}=\frac{d\rho }{dt}\stackrel{^}{\rho }+\rho \frac{d\stackrel{^}{\rho }}{dt}+\frac{dz}{dt}\stackrel{^}{k}\\ \stackrel{\to }{v}=\frac{d\rho }{dt}\stackrel{^}{\rho }+\rho \frac{d\phi }{dt}\stackrel{^}{\phi }+\frac{dz}{dt}\stackrel{^}{k}\end{array}$

• Vector aceleración

• $\begin{array}{l}\frac{{d}^{2}\stackrel{\to }{r}}{d{t}^{2}}=\frac{d\stackrel{\to }{v}}{dt}=\frac{{d}^{2}\rho }{d{t}^{2}}\stackrel{^}{\rho }+\frac{d\rho }{dt}\frac{d\stackrel{^}{\rho }}{dt}+\frac{d\rho }{dt}\frac{d\phi }{dt}\stackrel{^}{\phi }+\rho \frac{{d}^{2}\phi }{d{t}^{2}}\stackrel{^}{\phi }+\rho \frac{d\phi }{dt}\frac{d\stackrel{^}{\phi }}{dt}+\frac{{d}^{2}z}{d{t}^{2}}\stackrel{^}{k}=\\ \frac{{d}^{2}\rho }{d{t}^{2}}\stackrel{^}{\rho }+\frac{d\rho }{dt}\frac{d\phi }{dt}\stackrel{^}{\phi }+\frac{d\rho }{dt}\frac{d\phi }{dt}\stackrel{^}{\phi }+\rho \frac{{d}^{2}\phi }{d{t}^{2}}\stackrel{^}{\phi }-\rho {\left(\frac{d\phi }{dt}\right)}^{2}\stackrel{^}{\rho }+\frac{{d}^{2}z}{d{t}^{2}}\stackrel{^}{k}\\ \frac{{d}^{2}\stackrel{\to }{r}}{d{t}^{2}}=\left(\frac{{d}^{2}\rho }{d{t}^{2}}-\rho {\left(\frac{d\phi }{dt}\right)}^{2}\right)\stackrel{^}{\rho }+\left(\rho \frac{{d}^{2}\phi }{d{t}^{2}}+2\frac{d\rho }{dt}\frac{d\phi }{dt}\right)\stackrel{^}{\phi }+\frac{{d}^{2}z}{d{t}^{2}}\stackrel{^}{k}\end{array}$

## Esféricas

Tiene por origen O y vectores unitaros, $r ^$, $φ ^$ y $θ ^$

Relación entre los vectores unitarios

El vector unitario $r ^$ tiene la misma dirección que el vector $r →$ pero su módulo es la unidad

$\stackrel{^}{r}=\mathrm{sin}\theta ·\mathrm{cos}\phi \stackrel{^}{i}+\mathrm{sin}\theta ·\mathrm{sin}\phi \stackrel{^}{j}+\mathrm{cos}\theta \stackrel{^}{k}$

Como apreciamos en la figura

$\stackrel{^}{\phi }=-\mathrm{sin}\phi \stackrel{^}{i}+\mathrm{cos}\phi \stackrel{^}{j}$

Utilizamos el producto vectorial para obtener la expresión de $θ ^$

$\stackrel{^}{\theta }=\stackrel{^}{\phi }×\stackrel{^}{r}=|\begin{array}{ccc}\stackrel{^}{i}& \stackrel{^}{j}& \stackrel{^}{k}\\ -\mathrm{sin}\phi & \mathrm{cos}\phi & 0\\ \mathrm{sin}\theta ·\mathrm{cos}\phi & \mathrm{sin}\theta ·\mathrm{sin}\phi & \mathrm{cos}\theta \end{array}|=\mathrm{cos}\theta ·\mathrm{cos}\phi \stackrel{^}{i}+\mathrm{cos}\theta \mathrm{sin}\phi \stackrel{^}{j}-\mathrm{sin}\theta \stackrel{^}{k}$

$\begin{array}{l}\frac{d\stackrel{^}{r}}{dt}=\left(\mathrm{cos}\theta ·\mathrm{cos}\phi \frac{d\theta }{dt}-\mathrm{sin}\theta \mathrm{sin}\phi \frac{d\phi }{dt}\right)\stackrel{^}{i}+\left(\mathrm{cos}\theta ·\mathrm{sin}\phi \frac{d\theta }{dt}+\mathrm{sin}\theta \mathrm{cos}\phi \frac{d\phi }{dt}\right)\stackrel{^}{j}-\mathrm{sin}\theta \frac{d\theta }{dt}\stackrel{^}{k}=\\ \left(\mathrm{cos}\theta ·\mathrm{cos}\phi \stackrel{^}{i}+\mathrm{cos}\theta ·\mathrm{sin}\phi \stackrel{^}{j}-\mathrm{sin}\theta \stackrel{^}{k}\right)\frac{d\theta }{dt}+\left(-\mathrm{sin}\phi \stackrel{^}{i}+\mathrm{cos}\phi \stackrel{^}{j}\right)\mathrm{sin}\theta ·\frac{d\phi }{dt}=\\ \frac{d\theta }{dt}\stackrel{^}{\theta }+\mathrm{sin}\theta ·\frac{d\phi }{dt}\stackrel{^}{\phi }\end{array}$

$\begin{array}{l}\frac{d\stackrel{^}{\theta }}{dt}=\left(-\mathrm{sin}\theta ·\mathrm{cos}\phi \frac{d\theta }{dt}-\mathrm{cos}\theta ·\mathrm{sin}\phi \frac{d\phi }{dt}\right)\stackrel{^}{i}+\left(-\mathrm{sin}\theta ·\mathrm{sin}\phi \frac{d\theta }{dt}+\mathrm{cos}\theta ·\mathrm{cos}\phi \frac{d\phi }{dt}\right)\stackrel{^}{j}-\mathrm{cos}\theta \frac{d\theta }{dt}\stackrel{^}{k}=\\ -\left(\mathrm{sin}\theta \mathrm{cos}\phi \stackrel{^}{i}+\mathrm{sin}\theta ·\mathrm{sin}\phi \stackrel{^}{j}+\mathrm{cos}\theta \stackrel{^}{k}\right)\frac{d\theta }{dt}+\left(-\mathrm{sin}\phi \stackrel{^}{i}+\mathrm{cos}\phi \stackrel{^}{j}\right)\mathrm{cos}\theta \frac{d\phi }{dt}=\\ -\frac{d\theta }{dt}\stackrel{^}{r}+\mathrm{cos}\theta \frac{d\phi }{dt}\stackrel{^}{\phi }\end{array}$

Utilizamos el producto vectorial para obtener la derivada de este vector unitario

$\begin{array}{l}\stackrel{^}{\phi }=\stackrel{^}{r}×\stackrel{^}{\theta }\\ \frac{d\stackrel{^}{\phi }}{dt}=\frac{d\stackrel{^}{r}}{dt}×\stackrel{^}{\theta }+\stackrel{^}{r}×\frac{d\stackrel{^}{\theta }}{dt}=\left(\frac{d\theta }{dt}\stackrel{^}{\theta }+\mathrm{sin}\theta ·\frac{d\phi }{dt}\stackrel{^}{\phi }\right)×\stackrel{^}{\theta }+\stackrel{^}{r}×\left(-\frac{d\theta }{dt}\stackrel{^}{r}+\mathrm{cos}\theta \frac{d\phi }{dt}\stackrel{^}{\phi }\right)=\\ \mathrm{sin}\theta ·\frac{d\phi }{dt}\stackrel{^}{\phi }×\stackrel{^}{\theta }+\mathrm{cos}\theta \frac{d\phi }{dt}\stackrel{^}{r}×\stackrel{^}{\phi }==\left(-\mathrm{sin}\theta ·\stackrel{^}{r}-\mathrm{cos}\theta ·\stackrel{^}{\theta }\right)\frac{d\phi }{dt}\end{array}$

• Vector posición

• $\stackrel{\to }{r}=r\stackrel{^}{r}$

• $\frac{d\stackrel{\to }{r}}{dt}=\frac{dr}{dt}\stackrel{^}{r}+r\frac{d\stackrel{^}{r}}{dt}=\frac{dr}{dt}\stackrel{^}{r}+r\frac{d\theta }{dt}\stackrel{^}{\theta }+r\mathrm{sin}\theta ·\frac{d\phi }{dt}\stackrel{^}{\phi }$

• Vector aceleración

• $\begin{array}{l}\frac{{d}^{2}\stackrel{\to }{r}}{d{t}^{2}}=\frac{d}{dt}\left(\frac{d\stackrel{\to }{r}}{dt}\right)=\frac{{d}^{2}r}{d{t}^{2}}\stackrel{^}{r}+\frac{dr}{dt}\frac{d\stackrel{^}{r}}{dt}+\frac{dr}{dt}\frac{d\theta }{dt}\stackrel{^}{\theta }+r\frac{{d}^{2}\theta }{d{t}^{2}}\stackrel{^}{\theta }+r\frac{d\theta }{dt}\frac{d\stackrel{^}{\theta }}{dt}+\frac{dr}{dt}\mathrm{sin}\theta \frac{d\phi }{dt}\stackrel{^}{\phi }+r\mathrm{cos}\theta \frac{d\theta }{dt}\frac{d\phi }{dt}\stackrel{^}{\phi }+r\mathrm{sin}\theta \frac{{d}^{2}\phi }{d{t}^{2}}\stackrel{^}{\phi }+r\mathrm{sin}\theta \frac{d\phi }{dt}\frac{d\stackrel{^}{\phi }}{dt}=\\ \frac{{d}^{2}r}{d{t}^{2}}\stackrel{^}{r}+\frac{dr}{dt}\left(\frac{d\theta }{dt}\stackrel{^}{\theta }+\mathrm{sin}\theta ·\frac{d\phi }{dt}\stackrel{^}{\phi }\right)+\frac{dr}{dt}\frac{d\theta }{dt}\stackrel{^}{\theta }+r\frac{{d}^{2}\theta }{d{t}^{2}}\stackrel{^}{\theta }+r\frac{d\theta }{dt}\left(-\frac{d\theta }{dt}\stackrel{^}{r}+\mathrm{cos}\theta \frac{d\phi }{dt}\stackrel{^}{\phi }\right)+\frac{dr}{dt}\mathrm{sin}\theta \frac{d\phi }{dt}\stackrel{^}{\phi }+r\mathrm{cos}\theta \frac{d\theta }{dt}\frac{d\phi }{dt}\stackrel{^}{\phi }+r\mathrm{sin}\theta \frac{d\phi }{dt}\left(-\mathrm{sin}\theta ·\stackrel{^}{r}-\mathrm{cos}\theta ·\stackrel{^}{\theta }\right)\frac{d\phi }{dt}\end{array}$

$\begin{array}{l}\frac{{d}^{2}\stackrel{\to }{r}}{d{t}^{2}}=\left(\frac{{d}^{2}r}{d{t}^{2}}-r{\left(\frac{d\theta }{dt}\right)}^{2}-r{\mathrm{sin}}^{2}\theta {\left(\frac{d\phi }{dt}\right)}^{2}\right)\stackrel{^}{r}+\\ \left(r\frac{{d}^{2}\theta }{d{t}^{2}}+2\frac{dr}{dt}\frac{d\theta }{dt}-r\mathrm{sin}\theta ·\mathrm{cos}\theta {\left(\frac{d\phi }{dt}\right)}^{2}\right)\stackrel{^}{\theta }+\\ \left(r\mathrm{sin}\theta \frac{{d}^{2}\phi }{d{t}^{2}}+2\mathrm{sin}\theta \frac{dr}{dt}\frac{d\phi }{dt}+2r\mathrm{cos}\theta \frac{d\theta }{dt}\frac{d\phi }{dt}\right)\stackrel{^}{\phi }\end{array}$