Given a curve $y$ having constant curvature with $y=f(x)$ with $f''(x) > 0 $. I have to prove that this curve is a part of a circle.
My Try
Assume curvature of the curve as $c$. Then, I can write $$c = \frac{f''(x)}{(1+f'(x)^2)^{\frac 32}}$$ After this, I am unable to solve the non-linear differential equation.
Is my approach wrong? Thanks in advance !
If you separate the variables in the equation for $g=f'$ you obtain $$ \frac{dg}{(1+g^2)^{3/2}}=cdx $$ With the substitution $g=\tan\theta$ the integral on the left is just $\sin\theta+const.$ so $$ \sin\theta=cx+D $$ If we choose as our initial direction $\theta=0$ (the curve starts horizontally at $x=0$) you get $D=0$. Since $\sin\theta=\tan\theta/\sqrt{1+\tan^2\theta}=g/\sqrt{1+g^2}$ we have $$ \frac{g}{\sqrt{1+g^2}}=cx. $$ Solving for $g$: $$ g(x)=\frac{cx}{\sqrt{1-c^2x^2}}, $$ for $|x|\in[0,1/c)$. Integrating once again $$ f(x)=-\frac{\sqrt{1-c^2x^2}}{c}+E $$ Choosing the initial point of the curve at the origin, we find $E=\frac 1{c}$. Finally, $$ f(x)=\frac 1{c}-\frac{\sqrt{1-c^2x^2}}{c} $$ which is the lower half of the circle centered at $(0,1/c)$ with radius $1/c$, as expected (the radius is the reciprocal of the curvature).