This is a homework problem I am having trouble with:
Show that if a planar unit speed curve $q(s)$ satisfies $$\kappa_s = \frac{1}{es+f}$$ for constants $e, f >0$, then the curve is a logarithmic Spiral.
Here is what I have so far.
$\kappa_s$ is the signed curvature of $q(t)$ and is defined by
$$\kappa_s = N_s \cdot \frac{dT}{ds}$$
so my starting point is
$$N_s \cdot \frac{dT}{ds} = \frac{1}{es+f}$$
Here's what i know about Logarithmic spirals:
Parametric form: $x(t)=ae^{bt}cos(t)$ and $y(t) = ae^{bt}sin(t)$.
Polar form: $r= ae^{b\theta}$ or $\theta = \frac{1}{b} \ln{(\frac{r}{a})}$
I honestly don't know how to get started on this problem. I have a suspicion I need to integrate both sides of the equation in the problem with respect to s, because this will give me a logarithm on the right hand side. But I wouldn't know how to tackle that integral on the left hand side, or relate that expression to r. Any general nudging in the right direction would be very helpful.