Continuous polar equation of Sqrt spiral Please help find polar coordinate equation  $ r =f(\theta) $ or $ \theta =g(r) $ 
where $r$ and $\theta$ are continuous functions of  $n$ as given: 
SqrtSpiral
EDIT1:
n is a discrete (discontinuous) variable ( n = 1,2,3..) with
$$ r_n = \sqrt n ;$$
$$ r_{n+1}^2 -r_n^2 = 1 ;  \\ r_n = \cot\Delta\theta =\cot (\theta_{n+1} - \theta_n );$$
All known spirals in the plane are either 
$ r= f(\theta) $ or parametrically $ r(t),\theta(t), $ 
into which form Sqrt spiral is not castable.
I tried to adjust Pythogorean relation
$$ m^2+n^2 , m^2-n^2 , 2 m n $$
with no luck, so I posted here for help.
 A: This is the spiral of Theodorus. See here
http://en.wikipedia.org/wiki/Spiral_of_Theodorus
for references. I particularly recommend the preprint to be found here:
http://www.sam.math.ethz.ch/~joergw/Papers/theopaper.html
A: I have tried a different approach to the continuous sqrt spiral that I was hoping would include all the points on the spiral of Theodorus. It does not quite do that for the initial points but it agrees very well as it approaches the spiral of Archimedes asymptotically for large $n$. I'll present the analysis here. Let me start with the discrete spiral of Theodorus, which I develop as follows:
$$r(n)=\sqrt{n}\\\Delta\theta(n)=\text{tan}^{-1}\left(\frac{1}{\sqrt{n}}\right)\\\theta(n)=\sum_{k=1}^n \Delta\theta(k)\\z(n)=r(n)e^{i\theta(n)}$$
Now, by a simple extension of the idea, let $t$ be a continuous variable and let
$$r(t)=\sqrt{t}\\d\theta(t)=\text{tan}^{-1}\left(\frac{1}{\sqrt{t}}\right) dt\\\theta(t)=\int_0^t \text{tan}^{-1}\left(\frac{1}{\sqrt{t}}\right) dt=\sqrt{t}+t\ \text{tan}^{-1}\left(\frac{1}{\sqrt{t}}\right) -\text{tan}^{-1}\left(\sqrt{t}\right) \\z(t)=r(t)e^{i\theta(t)} $$
Let me digress for a moment to point out that as $t\to \infty$, $\theta(t)\to 2\sqrt{t}-\pi/2$, and $z(t)\to -i\sqrt{t} e^{-i\sqrt{t}}$, hence the spiral of Archimedes.
The figure below shows a comparison of the discrete spiral of Theodorus (circles) and the present sqrt spiral (line). Clearly, this continuous spiral does not include the spiral of Theodorus, but it is a definitely a sqrt spiral. (NOTE: the continuous spiral had to be rotated to align properly with the discrete one. I'm not exactly sure why that is.)

A: $r_{n+1}^2 -r_n^2 = 1 ;  \\ r_n = \cot\Delta\theta =\cot (\theta_{n+1} - \theta_n ) ;$
I'll write $t$ for $\theta$ since I'm lazy.
From your equations,
$\cot^2 (t_{n+2} - t_{n+1} ) 
=\cot^2 (t_{n+1} - t_n ) +1
$,
so
$t_{n+2}$
is a function of
$t_{n+1}$
and
$t_{n}
$.
Since
$\cot(a-b)
=\dfrac{1+\tan a \tan b}{\tan a -tan b}
$,
letting
$u_n
=\tan(t_n)
$,
$\dfrac{(1+u_{n+2} u_{n+1})^2}{(u_{n+2}-u_{n+1})^2}
=\dfrac{(1+u_{n+1} u_{n})^2}{(u_{n+1}-u_{n})^2}+1
$.
This is the relation among
the $u_i$.
Letting 
$g
=\dfrac{(1+u_{n+1} u_{n})^2}{(u_{n+1}-u_{n})^2}+1
$,
$(1+u_{n+2} u_{n+1})^2
=g(u_{n+2}-u_{n+1})^2
$
or,
letting
$p = u_{n+2}$
and
$q = u_{n+1}$,
$(1+pq)^2 = g(p-q)^2$
or
$1+2pq+p^2q^2
=g(p^2-2pq+q^2)
$
or
$p^2(q^2-g)
+p(2-2gq)
+gq^2-1
=0
$,
or
$p^2(q^2-g)
+2p(1-gq)
+gq^2-1
=0
$.
The discriminant of this
divided by $4$
 is
$\begin{array}\\
((1-gq)^2-(q^2-g)(gq^2-1)
&=(1-2gq+g^2q^2)-(gq^4-q^2(g^2+1)+g)\\
&=(1-2gq+g^2q^2)-(gq^4-q^2(g^2+1)+g)\\
&=1-2gq+g^2q^2-gq^4+q^2g^2+q^2-g\\
&=1-2gq+2g^2q^2-gq^4+q^2-g\\
\end{array}
$
I don't know where to go from here,
so I'll stop.
