Olympiad problem - struggling with polar coordinates This is a Physics Olympiad problem; and I am still struggling with it. I will quote it here:

A particle moves along a horizontal track following the trajectory $r=r_{0}e^{-k\theta}$, where $\theta$ is the angle made by the position vector with the horizontal. Recall that the velocity in polar coordinates is $\frac{d\vec r}{dt}=\dot r \hat r+r \dot \theta \hat \theta$. If at $t=0, \theta=0$ the velocity is $v_{0}$, find $\theta$ dependence of velocity and the angle the velocity vector makes with the radial vector.

The answer is that $v(\theta)=v_{0}$ and is independent of $\theta$; and $\alpha(\theta)=\tan^{-1}(\pm\frac{1}{k})$ and independent of $\theta$. 
It makes sense, as the equation resembles a logarithmic spiral, and the results hold for it. But how do I prove it?
I tried differentiating it all I could, but I am always getting an extra $e^{-k\theta}$ in the term for velocity. 
$\vec v=\frac{d}{dt}(r_{0}e^{-k\theta}) \hat r + r_{0}e^{-k\theta}\frac{d\theta}{dt}$=$-kr_{0}e^{-k\theta}\dot \theta\hat r + r_{0}e^{-k\theta}\dot \theta \hat \theta$ =$r_{0}e^{-k\theta}\dot \theta(-k\hat r+\hat \theta)$. ------(1)
At $\theta=0$,
$\vec v= \vec v_{0}=r_{0}\dot \theta(-k\hat r+\hat \theta)$ ------------------(2)
But then how do I prove that (1) and (2) are the same? i.e. how do I get rid of $e^{-k\theta}$ in the first equation? Or do I have to approach it differently? But I want to stick to polar coordinates.
Please help. I am posting this after struggling for 2 straight days, and now I am completely frustrated. What's even more annoying that it is only a 4 mark (2 marks for each part) problem.
Thanks in advance!
 A: Since there is no time dependence in the given equation (in other words, you have only the trajectory in terms of some parameter -- $\theta$ -- with no given connection to the time), you cannot arrive at the velocity by differentiating. 
Of course $k = \cot\alpha$ for a well-chosen $\alpha$ ... ($\alpha = \arctan(k^{-1})$) :)
But why does this prove anything, did you learn some advanced features of the logarithmic spiral? I would say, to prove the constancy of the velocity you need to show that the force (i.e. the derivative of the velocity, which is given in the problem) is always perpendicular to it...

Well, since I do not understand what you are doing, unfortunately, and my hints don't seem to help (well, they went in another direction than the problem wants to guide you, therefore the confusion, sorry...), I will just post a direct solution...:
take the given formula $$ \vec v = \frac{\mathrm d}{\mathrm dt}\vec r = \dot r\hat r + r\dot\theta\hat\theta$$
now use the given trajectory to reduce the number of variables: $$\dot r = \frac{\mathrm dr}{\mathrm d\theta}\dot\theta \mathrm{\qquad with\qquad} \frac{\mathrm dr}{\mathrm d\theta} = -kr$$
Now combine and extract the factor $r\dot\theta$: $$\vec v = r\dot\theta(
-k\hat r + \hat\theta)$$
Obviously, the direction has always the same angle (given by the vector in parentheses). The magnitude $v$ is $r\dot\theta$. There is probably a more elegant way to proceed now, but I would argue like this:
Assume $v$ is constant and show that this is a valid solution, since the solution is unique, you are done now. So, let's go:
A constant $v$ means, the force is always perpendicular to $\vec v$, and has therefore a constant angle to $\hat r$. Thus the tangential force is constant and the torque therefore is proportional to $r$.
The decrease of angular momentum (which is $mr^2\dot\theta$) has to be proportional to the torque, that is, to $r$. This is consistent, since $d/dt(r^2\dot\theta) = \dot r$ for constant $r\dot\theta$. And $r$ is proportional to $\dot r$.
