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Polar Coordinates

The polar coordinates $r$ and $\theta$ are defined by

$\displaystyle x$ $\textstyle =$ $\displaystyle r\cos\theta$ (1)
$\displaystyle y$ $\textstyle =$ $\displaystyle r \sin \theta.$ (2)

In terms of $x$ and $y$,
$\displaystyle r$ $\textstyle =$ $\displaystyle \sqrt{x^2 + y^2}$ (3)
$\displaystyle \theta$ $\textstyle =$ $\displaystyle \tan^{-1}\left({y\over x}\right).$ (4)

The Arc Length of a polar curve given by $r = r(\theta)$ is
\begin{displaymath}
s = \int^{\theta_2}_{\theta_1} \sqrt{r^2 +\left({dr\over d\theta}\right)^2}\, d\theta.
\end{displaymath} (5)

The Line Element is given by
\begin{displaymath}
ds^2 = r^2\,d\theta^2,
\end{displaymath} (6)

and the Area element by
\begin{displaymath}
dA = r\,dr\,d\theta.
\end{displaymath} (7)

The Area enclosed by a polar curve $r = r(\theta)$ is
\begin{displaymath}
A = {\textstyle{1\over 2}}\int^{\theta_2}_{\theta_1} r^2 \,d\theta.
\end{displaymath} (8)

The Slope of a polar function $r = r(\theta)$ at the point $(r,\theta)$ is given by
\begin{displaymath}
m = {r+\tan \theta {dr\over d\theta}\over -r \tan \theta +{dr\over d\theta}}.
\end{displaymath} (9)

The Angle between the tangent and radial line at the point $(r,\theta)$ is
\begin{displaymath}
\psi = \tan^{-1}\left({r\over {dr\over d\theta}}\right).
\end{displaymath} (10)

A polar curve is symmetric about the $x$-axis if replacing $\theta$ by $-\theta$ in its equation produces an equivalent equation, symmetric about the $y$-axis if replacing $\theta$ by $\pi -\theta$ in its equation produces an equivalent equation, and symmetric about the origin if replacing $r$ by $-r$ in its equation produces an equivalent equation.


In Cartesian coordinates, the Position Vector and its derivatives are


$\displaystyle {\bf r}$ $\textstyle =$ $\displaystyle \sqrt{x^2+y^2}\,\hat{\bf r}$ (11)
$\displaystyle \dot {\bf r}$ $\textstyle =$ $\displaystyle \dot{\hat {\bf r}}\sqrt{x^2+y^2}+\hat {\bf r}\,(x^2+y^2)^{-1/2}(x\dot x+y\dot y)$ (12)
$\displaystyle \hat {\bf r}$ $\textstyle =$ $\displaystyle {x\hat {\bf x}+y\hat {\bf y}\over \sqrt{x^2+y^2}}$ (13)
$\displaystyle \dot{\hat {\bf r}}$ $\textstyle =$ $\displaystyle {\dot x\hat {\bf x}+\dot y\hat {\bf y}\over\sqrt{x^2+y^2}}-{\textstyle{1\over 2}}(x^2+y^2)^{-3/2}(2)(x\dot x+y\dot y)(x\hat{\bf x}+y\hat{\bf y})$  
  $\textstyle =$ $\displaystyle {(x\dot y-y\dot x)(x\hat{\bf y}-y\hat{\bf x})\over(x^2+y^2)^{3/2}}.$ (14)


In polar coordinates, the Unit Vectors and their derivatives are

$\displaystyle {\bf r}$ $\textstyle \equiv$ $\displaystyle \left[\begin{array}{c}r\cos\theta\\  r\sin\theta\end{array}\right]$ (15)
$\displaystyle \hat {\bf r}$ $\textstyle \equiv$ $\displaystyle {{d{\bf r}\over dr}\over \left\vert{d{\bf r}\over dr}\right\vert} = \left[\begin{array}{c}\cos\theta\\  \sin\theta\end{array}\right]$ (16)
$\displaystyle \hat \boldsymbol{\theta}$ $\textstyle \equiv$ $\displaystyle {{d\boldsymbol{\theta}\over d\theta}\over \left\vert{d\boldsymbol...
...ght\vert} = \left[\begin{array}{c}-\sin \theta\\  \cos \theta\end{array}\right]$ (17)
$\displaystyle \dot{\hat {\bf r}}$ $\textstyle =$ $\displaystyle \left[\begin{array}{c}-\sin\theta\dot\theta\\  \cos\theta\dot\theta\end{array}\right] = \dot\theta\,\hat {\boldsymbol{\theta}}$ (18)
$\displaystyle \dot{\hat\boldsymbol{\theta}}$ $\textstyle =$ $\displaystyle \left[\begin{array}{c}-\cos\theta\dot\theta\\  -\sin\theta\dot\theta\end{array}\right] = -\dot\theta\hat {\bf r}$ (19)
$\displaystyle \dot {\bf r}$ $\textstyle =$ $\displaystyle \left[\begin{array}{c}-r\sin\theta\dot\theta+\cos\theta\dot r\\
...
...nd{array}\right] = r\dot\theta \,\hat\boldsymbol{\theta}+ \dot r \,\hat {\bf r}$  
$\displaystyle \ddot{\bf r}$ $\textstyle =$ $\displaystyle \dot r\dot\theta\,\hat\boldsymbol{\theta}+ r\ddot\theta\hat\bolds...
...a\dot{\hat\boldsymbol{\theta}} + \ddot r\hat {\bf r} + \dot r \dot{\hat{\bf r}}$  
  $\textstyle =$ $\displaystyle \dot r\dot\theta\hat\boldsymbol{\theta}+\dot r\ddot\theta\hat\bol...
...ta\hat{\bf r}) + \ddot r\hat {\bf r} + \dot r\dot\theta \hat\boldsymbol{\theta}$  
  $\textstyle =$ $\displaystyle (\ddot r-r{\dot\theta}^2)\hat {\bf r} + (2\dot r\dot\theta+r\ddot\theta)\hat{\boldsymbol{\theta}}$  
  $\textstyle =$ $\displaystyle (\ddot r-r{\dot\theta}^2)\hat {\bf r} + {1\over r} {d\over dt} (r^2\dot\theta) \hat{\boldsymbol{\theta}}.$ (20)

See also Cardioid, Circle, Cissoid, Conchoid, Curvilinear Coordinates, Cylindrical Coordinates, Equiangular Spiral, Lemniscate, Limaçon, Rose



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© 1996-9 Eric W. Weisstein
1999-05-25