Apostol, chapter 14.21, problem 1 (a review problem)
Here is the question:
Let r denote the vector from the origin to an arbitrary point on the parabola $y^2 = x$, let $\alpha$ be the angle that r makes with the tangent line, $0 \le \alpha \le \pi$, and let $\theta$ be the angle that r makes with the positive x-axis, $0\le \theta\le \pi$. Express $\alpha$ in terms of $\theta$.
How do I think: Represent r in polar coordinates in function of $\theta$, the tangent vector will be r'. The inner product of r and r' will give a number, function of $\theta$, which is function of the $\alpha$. But, since $\theta$ is function of t, a parameter inside r(t) and r'(t), I can't find an expression of r' or r only in function of $\theta$, or to find how to eliminate this parameter. And more, this way to look at the problem doesn't seem good enough.
Can anyone help?
The very first thing to do is to find a parametrization of the parabola in terms of the angle $\theta$. This is quite simple. If $$(x,y) = (||\boldsymbol r|| \cos \theta, ||\boldsymbol r|| \sin \theta)$$ is the standard polar coordinate transformation, then $y^2 = x$ gives $$||\boldsymbol r||^2 \sin^2 \theta = ||\boldsymbol r|| \cos \theta,$$ where upon solving for $||\boldsymbol r||$ yields $$||\boldsymbol r|| = \csc \theta \cot \theta.$$ This tells us that the desired parametrization is $$\boldsymbol r(\theta) = (x(\theta), y(\theta)) = (\cot^2 \theta, \cot \theta), \quad \theta \in (0,\pi),$$ which in retrospect should be quite obvious.
Once equipped with such a parametrization, it is straightforward to compute the tangent vector $\boldsymbol r'(\theta)$, the dot product $\boldsymbol r \cdot \boldsymbol r'$, and then the angle $\alpha$, which I leave as an exercise.