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# polar tangential angle

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

Given a polar curve

$\displaystyle r=r(\varphi)$ | (1) |

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

$\displaystyle\frac{P^{{\prime}}Q}{PQ}\;=\;\frac{(r\!+\!dr)d\varphi}{dr}\;=\;% \frac{r\!+\!dr}{\frac{dr}{d\varphi}}.$ | (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

$\tan\psi\;=\;\frac{r}{\frac{dr}{d\varphi}}$ |

of the polar tangential angle, i.e.

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

## Mathematics Subject Classification

51-01*no label found*53A04

*no label found*

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