Chapter H.1, Problem 69E

To determine

To show: $\tan \psi =\frac{r}{\raisebox{1ex}{$dr$}\!\left/ \!\raisebox{-1ex}{$d\theta $}\right.}$

Explanation of Solution

Given: Let point P on the curve $r=f\left(\theta \right)$ , If $\psi$ is the angle between the tangent line at P and the radial line OP .

  

Slope of tangent, $m=\tan \varphi$

Slope of tangent at point P on polar curve, $\frac{dy}{dx}=\frac{\frac{dr}{d\theta }\sin \theta +r\cos \theta }{\frac{dr}{d\theta }\cos \theta -r\sin \theta }$

  $\because \tan \varphi =\frac{\frac{dr}{d\theta }\sin \theta +r\cos \theta }{\frac{dr}{d\theta }\cos \theta -r\sin \theta }$

But $\varphi =\theta +\psi$

  $\begin{array}{l}\tan \left(\theta +\psi \right)=\frac{\frac{dr}{d\theta }\sin \theta +r\cos \theta }{\frac{dr}{d\theta }\cos \theta -r\sin \theta }\\ \frac{\tan \theta +\tan \psi }{1-\tan \theta \tan \psi }=\frac{\frac{dr}{d\theta }\sin \theta +r\cos \theta }{\frac{dr}{d\theta }\cos \theta -r\sin \theta }\left[\therefore \tan \left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B}\right]\\ \frac{\tan \theta +\tan \psi }{1-\tan \theta \tan \psi }=\frac{\tan \theta +\frac{r}{dr/d\theta }}{1-\tan \theta \cdot \frac{r}{dr/d\theta }}\left[\text{Divide by }\frac{dr}{d\theta }\cos \theta \text{ numerator and denominator}\right]\\ \text{Compare both sides}\\ \tan \psi =\frac{r}{dr/d\theta }\end{array}$

Hence proved