# polar tangential angle

The angle, under which a polar curve is by a line through the origin, is called the *polar tangential angle* belonging to the intersection^{} point on the curve.

Given a polar curve

$r=r(\phi )$ | (1) |

in polar coordinates $r,\phi $, we derive an expression for the tangent^{} of the polar tangential angle $\psi $, using the classical differential^{} geometric method.

The point $P$ of the curve given by (1) corresponds to the polar angle^{} $\phi =\mathrm{\angle}POA$ and the polar radius $r=OP$. The “near” point ${P}^{\prime}$ corresponds to the polar angle $\phi +d\phi =\mathrm{\angle}{P}^{\prime}OA$ and the polar radius $r+dr=O{P}^{\prime}$. In the diagram, ${P}^{\prime}Q$ is the arc of the circle with $O$ as centre and $O{P}^{\prime}$ as radius. Thus, in the triangle-like figure $P{P}^{\prime}Q$ we have

$\frac{{P}^{\prime}Q}{PQ}}={\displaystyle \frac{(r+dr)d\phi}{dr}}={\displaystyle \frac{r+dr}{\frac{dr}{d\phi}}}.$ | (2) |

This figure can be regarded as an infinitesimal^{} right triangle^{} with the catheti ${P}^{\prime}Q$ and $PQ$. Accordingly, their ratio (2) is the tangent of the acute angle^{} $P$ of the triangle. Because the addend $dr$ in the last numerator in negligible compared with the addend $r$, it can be omitted. Hence we get the tangent

$$\mathrm{tan}\psi =\frac{r}{\frac{dr}{d\phi}}$$ |

of the polar tangential angle, i.e.

$\mathrm{tan}\psi ={\displaystyle \frac{r(\phi )}{{r}^{\prime}(\phi )}}.$ | (3) |

Title | polar tangential angle |
---|---|

Canonical name | PolarTangentialAngle |

Date of creation | 2013-03-22 19:02:32 |

Last modified on | 2013-03-22 19:02:32 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 8 |

Author | pahio (2872) |

Entry type | Derivation |

Classification | msc 51-01 |

Classification | msc 53A04 |

Related topic | LogarithmicSpiral |