so I have this Polar Coordinates Homework Problem, and I don't know really know how to start/prove it. The problem goes as follows:
Let $f$ be a differentiable function, and let $P$ be the point $(\theta,f(\theta))$ on the polar graph of $r = f(\theta).$ Let $\alpha$ be the angle between the tangent line to the graph at $P,$ and the line $OP,$ where $O$ is the origin. Assuming that $f'(\theta) \neq 0,$ show that
$\tan\alpha=\frac{f(\theta)}{f'(\theta)}$
From what I've understood from this problem, if I extended a tangent line to P all the way to the x-axis, and called that point Q, I make $\angle PQO=\pi-\theta-\alpha,$ beyond that, I am lost.
Best Answer
Hint: Firstly, we have: $$x=r\cos \theta$$ and $$y=r\sin \theta $$ Then, we have, $$\frac {dy}{dx}=\frac{\frac {dy}{d\theta}}{\frac {dx}{d\theta}}$$ Be careful to note that $r=f(\theta)$. Now, slope of tangent is known. Also, slope of line $OP$ is simply $\tan \theta$. Can you find out the angle between two lines, if the slopes of both lines are known?
Note: With respect to a coordinate system, if slope of two lines is given by $m_1$ and $m_2$, then the angle between the two lines, $\alpha$, is given by- $$\tan \alpha=\frac {|m_2-m_1|}{|1+m_1m_2|}$$