I tried to answer the following question ( I believe there is a typo, It should say 'Show that the curve determined...') but had to resort to the solution for guidance. I don't understand where these three equations to begin with come from. What I do know:
$\gamma(s)$ is a plane curve with parameter arc length, so we can write $\gamma(s)=(x(s),y(s))$ and using polar co-ordinates we have $x(s)=r(\theta{(s)})cos(\theta(s))$ and $y(s)=r(\theta{(s)})sin(\theta(s))$ (obviously in calculations I would omit the extra (s) etc, but I'm just making my understanding clear).
I also know that signed curvature satisfies:
$\kappa=|\kappa(s)|$ and $t'=k_sn_s$. Am I missing something, or have I misunderstood something?
There's no reason to use polar coordinates, actually. But, yes, $\theta(s)$ is the angle the unit tangent at arclength $s$ makes with the positive $x$-axis. It is standard that the signed curvature is $\theta'(s)$ (the rate of turning of the unit tangent vector). If you don't know that, here's the easy argument: Write $\vec T(s) = (\cos\theta(s),\sin\theta(s))$. Then $$\vec T'(s) = (-\sin\theta(s),\cos\theta(s))\theta'(s) = \kappa_s(s)\vec N(s).$$ (Remember that in the case of plane curves, we always take $\vec T,\vec N$ to be a right-handed basis.)