# polar tangential angle

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)
Title polar tangential angle PolarTangentialAngle 2013-03-22 19:02:32 2013-03-22 19:02:32 pahio (2872) pahio (2872) 8 pahio (2872) Derivation msc 51-01 msc 53A04 LogarithmicSpiral