I'm supposed to prove that the only plane curve (up to a direct isometry) whose signed curvature is $\kappa_{s}(s)=\frac{1}{\sqrt{s}}$ ($s$ is the unit-speed parametrization) is the circle involute.
I'm having trouble with this because my calculations just don't add up: let's say we have a unit circle $\alpha(s)=(\cos{s},\sin{s})$, so its involute would be $\gamma(s)=(\cos{s}+s\sin{s}, \sin{s}-s\cos{s})$, if I'm not mistaken. One of many ways to find the signed curvature is as $\kappa_{s}^{\gamma}(s)=\frac{x'(s)y{''}(s)-x{''}(s)y'(s)}{\mathcal{v}^3}$, where $x$ and $y$ are the coordinates of $\gamma$. This can't possibly result in $\frac{1}{\sqrt{s}}$, because of the sines and cosines. Am I going wrong somewhere, or is the formulation of the problem wrong?