In the above diagram, the curve depicted in black is given by $$r = f(\theta)$$ and PT is the tangent and NP is the normal to the curve at P.
Suppose if $$P = (r,\theta)$$
How to prove below equation? $$tan(\psi) = r / (dr/d\theta)$$
Ref:
$\endgroup$2 Answers
$\begingroup$Slope of a tangent in polar coordinate is given by $$\tan\gamma =m_{\text{tangent}}=\frac{r'\sin(\theta)+r\cos(\theta)}{r'\cos(\theta)-r\sin(\theta)}$$ Where $\gamma$ is the angle tangent makes with horizontal axis. $\gamma=\theta+\psi$ $$\tan\psi=\tan(\gamma-\theta)= \frac{\frac{r'\sin(\theta)+r\cos(\theta)}{r'\cos(\theta)-r\sin(\theta)}-\tan\theta}{1+\frac{r'\sin(\theta)+r\cos(\theta)}{r'\cos(\theta)-r\sin(\theta)}\tan\theta}=\frac{r}{r'}=\frac{r}{\frac{dr}{d\theta}}$$
$\endgroup$ $\begingroup$Consider the blue differential ( representing infinitesmal) right angled triangle $PQR$ in a direct geometrical approach. Right angle is at $Q$.
Now $P$ has polar coordinates $ (r, \theta),$ and
$$ OP \approx OQ= r$$ Normal $PN$ can be considered parallel to $QR$ in the limit as $ dr\rightarrow 0.$ Then,
$$\tan \psi= \dfrac{QR}{QP}=\dfrac{r \cdot d \theta}{dr} = \dfrac{r }{r^{'}} ;$$
Subtangent $OT= r \tan \psi = \dfrac{r^2}{r^{'}} $ and subnormal $ ON= \dfrac{r}{ \tan \psi }= {r^{'}};$
Also if $QR$ makes $\angle \phi$ to x-axis, then external angle is sum of two internal angles:
$$ \phi= \psi + \theta $$
