How is $\tan\phi = (dy/d\theta) / (dx/d\theta)$?

Question

Consider the angle $\psi$ between the radius vector and the tangent line to a curve, $r = f(\theta)$, given in polar coordinates, as shown in Fig. 1. Show that $\psi = \tan^{−1} (r/(dr/d\theta))$.

http://www.math.usm.edu/lambers/cos702/cos702_files/docs/annote_figs/midsolve11.pdf

In step (4) of the pdf link above, how is $\tan\phi = \displaystyle\frac{\left(\frac{dy}{d\theta}\right)}{\left (\frac{dx}{d\theta}\right)}$?

Answer 1

Well I hope you can appreciate that the expression you wrote on the right is simply the derivative of $y$ with respect to $x$. Now consider consider the following triangle I will draw for you. By definition of the $\tan$ function you get the following. Obviously this isn't a rigorous proof but hopefully this will give you some intuition. Hope it helps.

EDIT: It won't let me upload the picture for some reason so here is a link. Hopefully someone can fix this for me.

https://i.stack.imgur.com/LVPdd.jpg

Obviously I have neglected to draw the function and only shown its tangent line but you get the point hopefully.

Answer 2

You can easily see that the tangent of a angle $\phi$ of a parametrized curve $(y(t),x(t))$, at $t=\theta$, with respect to the $x$-axis is \begin{equation}\tan{\phi} = \lim_{t\rightarrow \theta} \frac{y(t)-y(\theta)}{x(t)-x(\theta)}.\end{equation} Using L'Hospital Rule, and denoting $\frac{d}{dt}f$ by $f'$, we have \begin{equation}\lim_{t\rightarrow \theta} \frac{y(t)-y(\theta)}{x(t)-x(\theta)}= \left.\frac{\frac{d}{dt}(y(t)-y(\theta))}{\frac{d}{dt}(x(t)-x(\theta))}\right\vert_{t=\theta}=\frac{y'(\theta)}{x'(\theta)}\end{equation} and, since $\theta$ is arbitrary, we have that \begin{equation}\tan{\phi(t)}=\frac{y'(t)}{x'(t)}.\end{equation}