$$begin{align}

\y=x^2,y’=2x=tanpsi

\s=int_0^xsqrt{2t+1},dt

\s=frac{(2x+1)^{(3/2)}-1}{3}

\s=frac{(tanpsi+1)^{(3/2)}-1}{3}

end{align}$$

Is this correct?

And the curvature would be:

$$frac{dpsi}{ds}=frac{1}{frac{ds}{dpsi}}=frac{2}{sec^2psi(tanpsi+1)^{(1/2)}}$$

… right?