For a certain curve, the magnitude of the radius of curvature any point  {jatex options:inline}P(x_P,y_P){/jatex}  is equal to the length of the normal drawn from the point to the  {jatex options:inline}x{/jatex}  axis.
In the diagram,  {jatex options:inline}| PQ | = | r | =\frac{(1+ (\frac{dy}{dx})^2)^{3/2}}{ \frac{d^2y}{dx^2}}{/jatex}.

The normal at  {jatex options:inline}(x_P,y_P){/jatex}  is  {jatex options:inline}y-y_P = -1/ \frac{dy}{dx}_{(x_P,y_P)} (x-x_P){/jatex}. When  {jatex options:inline}y=0{/jatex},  {jatex options:inline}y_P \frac{dy}{dx}_{(x_P,y_P)}=x_Q-x_P{/jatex}.
Hence  {jatex options:inline}| PQ | = \sqrt{y_P^2 +(x_Q-x_P)^2}=y_P \sqrt{1+ (\frac{dy}{dx}_{(x_P,y_P)})^2}{/jatex}.
{jatex options:inline}| PQ | = | r | \rightarrow y_P \sqrt{1+ (\frac{dy}{dx}_{(x_P,y_P)})^2} =\frac{(1+ (\frac{dy}{dx})^2)^{3/2}}{ \frac{d^2y}{dx^2}} \rightarrow y_P | \frac{d^2y}{dx^2}_{(x_P,y_P)} | = 1+(\frac{dy}{dx})^2) {/jatex}. Since  {jatex options:inline}P{/jatex}  is a general point we can write  {jatex options:inline}y | \frac{d^2y}{dx^2}| = 1+(\frac{dy}{dx})^2) {/jatex}.
Suppose  {jatex options:inline}\frac{d^2y}{dx^2} \gt 0{/jatex}. Let  {jatex options:inline}\frac{dy}{dx}=p{/jatex}  then can write the last equation as  {jatex options:inline}1+p^2=yp \frac{dp}{dy} \rightarrow \frac{1}{y}dy= \frac{p}{1+p^2} dp \rightarrow \int \frac{1}{y}dy= \int \frac{p}{1+p^2} dp \rightarrow ln(y)+c = \frac{1}{2} ln(1+p^2){/jatex}.
This can be written as  {jatex options:inline}Ay^2=1+p^2= 1+ (\frac{dy}{dx})2 \frac{dy}{dx}=\sqrt{Ay^2-1}{/jatex}.
Separating variables and integrating again gives  {jatex options:inline}\frac{1}{\sqrt{A}} cos^{-1} y=x+B{/jatex}.
Similarly, letting  {jatex options:inline}\frac{d^2y}{dx^2} \lt 0{/jatex}  and solving gives the equation of a family of circles.