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

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

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

of the polar tangential angle, i.e.

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