How can I find the curves whose curvature is $k(s)=\frac{1}{as+b}$? I'm trying to find the curves whose curvature is $k(s)=\frac{1}{as+b}$. I guess that they are Logarithmic Spirals.
Parametric form of Logarithmic spirals:
$x(t)=ae^{(bt)}\cos(t)$  and $y(t)=ae^{(bt)}\sin(t).$
I'm having problems, I have tried to find de angle or to find $T(s)$ and $N(s)$. We know for Frenet that $T'(s)=k(s)N(s)=\alpha''(s)$ 
$N'(s)=-k(s)T(s)$.
I don't know how to solve it.
 A: Let $\alpha$ be a curve with curvature $\kappa (s)=\dfrac{1}{as+b}$. From the definition of $\kappa (s)$, we have that $\kappa (s)=\theta '(s)$. Therefore$$\theta (s)=\int \kappa (s)\ ds=\int \frac{1}{as+b}\ ds=\frac{\ln (|as+b|)}{a}+c_0$$with $c_0\in \mathbb{R}$.
Then an arc length parametrization of the curve is$$\alpha (s)=\left (x_0+\int \cos (\theta (s))\ ds,y_0+\int \sin (\theta (s))\ ds\right )$$with $x_0,y_0\in \mathbb{R}$.
Substituting the value of $\theta (s)$, we get$$\alpha (s)=\left (x_0+\int \cos \left (\frac{\ln (|as+b|)}{a}+c_0\right )\ ds,y_0+\int \sin \left (\frac{\ln (|as+b|)}{a}+c_0\right )\right ).$$Solving this gives $\alpha (s)=(x(s),y(s))$, where\begin{align*}x(s) & =x+(as+b)\frac{\sin \left (\frac{\ln (|as+b|)}{a}+c\right )+a\cos \left (\frac{\ln (|as+b|)}{a}+c\right )}{a^2+1} \\
y(s) & =y+(as+b)\frac{a\sin \left (\frac{\ln (|as+b|)}{a}+c\right )-\cos \left (\frac{\ln (|as+b|)}{a}+c\right )}{a^2+1}
\end{align*}with $x,y,c\in \mathbb{R}$.
A: Another way of looking at this is as follows: A natural, or intrinsic, equation is one that can characterize a curve by a relation independent of coordinates; see, for example D.J. Druik (Lectures on Classical Differential Geometry: Second Edition, Dover Books on Mathematics, 1988). Consider an equation for the curvature as a function of length, say, $\kappa=\kappa(s)$ and Euler’s integral solution for the tangent angle of plane curves
$$\theta =\int{\kappa \left( s \right)\,ds;\,\,\,\,\,\,\,\,{d\theta }/{ds=\kappa \left( s \right)}\;}$$
The curve is solved with the parametric equations
$$x=\int{\frac{\cos \theta }{\kappa }\,d\theta };\,\,\,\,\,\,\,\,y=\int{\frac{\sin \theta }{\kappa }\,d\theta }$$
We can therefore express the equation for plane curves in the complex plane as
$$z\left( s \right)=\int{{{e}^{i\theta \left( s \right)}}\,ds}=\int{{{e}^{i\int{\kappa \left( s \right)ds}}}\,ds}$$
