Am i correct in this derivation of equation of motion? The problem at hand was deriving the vector (as a function of time) that traces out the trajectory of a bug moving with velocity v that always holds an angle $\alpha$ relative to a point (a lamp or something, the solution is apparently a logarithmic spiral, $0<\alpha <\dfrac{\pi}{2}$).
I used polar coordinates, so the vector r has the form $ \textbf{r} =r(t)\cdot \bf\hat{r}(t) $ where $$ r(t) =r_0-v\cdot \cos(\alpha) t $$ is the magnitude of the distance between the bug and the point, and $$ \hat {\textbf{r}}(t) =\cos(\theta)\hat{i}+\sin(\theta)\hat{j} $$ is the unit vector along the radius of the bug-point in terms of the unit vectors of the X,Y axis.
The problem arises when we realize that $\theta$ itself is a function of time, and not in a trivial way. When deriving the relationship between them, my reasoning was the following: 
We know that $\theta = \omega \cdot t$ where omega is some angular velocity.
The arc-length of a curve is $l_{arc} = r\cdot\theta$ so differentiating this with respect to time we get: $$\dfrac{dl_{arc}}{dt} =v_{tangential} =\dot{r}\theta+r\omega$$ since "r" is a function of time, $\dot{r} =-v\cos(\alpha) $. Factoring this for $\theta$, it becomes $ \theta =\dfrac{v_{tan}-r(t) \omega}{-v\cos(\alpha)}$ . Substituting in r(t), $v_{tan} =v\sin(\alpha) $ and doing a bit of algebra, we arrive at the following differential equation: $$ \theta(t)=\dot{\theta}\big(\dfrac{r_0}{\cos(\alpha)}-vt\big)-\tan(\alpha)$$ to which the solution i think is $\theta =-\exp\Bigg(\dfrac{t}{\dfrac{r_0}{\cos(\alpha)}-vt} \Bigg)-\tan(\alpha) $
Putting this all together, we get: $$  \textbf{r}(t) =[r_0-v\cos(\alpha)t] \Bigg(\cos\Big(\exp\Bigg(\dfrac{t}{\dfrac{r_0}{\cos(\alpha)}-vt}\Bigg)+\tan(\alpha)\Big)\hat{\textbf{i}}-\sin\Big(\exp\Bigg(\dfrac{t}{\dfrac{r_0}{\cos(\alpha)}-vt}\Bigg)+\tan(\alpha)\Big)\hat{\textbf{j}}$$
Is this- or more like, was my- derivation reasonable?
EDIT: The solution to the differential equation is actually: $$ \theta =\exp\Bigg(\dfrac{-\ln\Big(\dfrac{r_0}{\cos(\alpha)}-vt\Big)}{v}\Bigg)-\tan(\alpha)$$
 A: I think everything you presented was right (except for the solution of the differential equation which I don't know how to solve... so I'm trusting you on this point).
If I can advise you something, I think that using $\dot{\theta}$ AND $\omega$ (which are two notations of the same think in your case) can lead you to some problems. So, may be avoid using both of them in the same equation (use $\dot{\theta}$ everywhere and replace it by $\omega$ only at the end).
But I'm not an expert at all, so maybe I just don't see an horrible mistake!
Edit
In polar coordinates you can define the position of any object with two components: the angle et the "distance" from the origin to your point. So you get, if $M$ is your object and $R$ is your distance, $\vec{OM}=R\hat{R}$
So if you derive this (warning: $\frac{d}{dt}(\hat{R})\neq 0$), you have the following velocity as the one you found:
$$\vec{v}=\dot{R}\hat{R}+R\dot{\theta}\hat{\theta}$$
And so, because you said that your radius was constant, $R=C^{ste}$, so :
$$\vec{v}=\vec{0}+R\dot{\theta}\hat{\theta}=R\dot{\theta}\hat{\theta}$$
And this makes sense, your speed is turned in the direction of $\hat{\theta}$. The speed direction is constantly being updated, and is always tangent to the trajectory: that's what creates circular motions.
How circular motion works.
Hoverwise, if $R\neq C^{ste}$, you can imagine this as if there were a "normal" circular motion, with the $\dot{R}\hat{R}$ parameter "pulling" the circular orbit through space.
It's like a rotation movement also "moving" into space.
