I was the one who asked this question, all the way back in 2012. I figure I should answer it, now that I've finally figured it out after all these years.
Going by Rahul's comment, you have the spiral distance function given by: $ D^2_{\theta,T}=R_\theta^2+\left|T\right|^2-2\left|T\right|R_\theta \cos\left(\theta_T - \theta\right) $ which is the cosine rule. This function is periodic, such that $D_{\theta,T}$ is at a minimum when $\theta_T-\theta=2\pi n \rightarrow \theta = \theta_T - 2\pi n$. Substituting this into the function above removes the cosine term, and the derivative then becomes invertible. $ D^2\left(\theta_T - 2\pi n\right)=R^2\left(\theta_T - 2\pi n\right)+\left|T\right|^2-2\left|T\right|R\left(\theta_T - 2\pi n\right) $
$ -2\pi \frac{d}{dn} D^2\left(\theta_T - 2\pi n\right)=-4\pi R\left(\theta_T - 2\pi n\right) R^\prime\left(\theta_T - 2\pi n\right) + 4\pi\left|T\right|R^\prime\left(\theta_T - 2\pi n\right) = 0 $
$ 4\pi R\left(\theta_T - 2\pi n\right) R^\prime\left(\theta_T - 2\pi n\right) = 4\pi\left|T\right|R^\prime\left(\theta_T - 2\pi n\right) $
$ R\left(\theta_T - 2\pi n\right) = \left|T\right| \rightarrow \theta = R^-\left(\left|T\right|\right) $
Substituting this back into the spiral distance function, we have:
$ D^2_T = R^2\left(R^-\left(\left|T\right|\right)\right)+\left|T\right|^2-2\left|T\right|R\left(R^-\left(\left|T\right|\right)\right) \cos\left(\theta_T - R^-\left(\left|T\right|\right)\right) $ $ \rightarrow D^2_T = 2\left|T\right|^2-2\left|T\right|^2 \cos\left(\theta_T - R^-\left(\left|T\right|\right)\right) = 2\left|T\right|^2 \text{ver}\left(\theta_T - R^-\left(\left|T\right|\right)\right) $
Therefore, the distance of any given point $T$ to a spiral curve defined by $R_\theta$ is $ D_{R,T} = \left|T\right|\sqrt{2\text{ver}\left(\theta_T - R^-\left(\left|T\right|\right)\right)} $
Happy times!