Minimum radius of curvature of a sinusoidal curve Intuitively, the minimum radius of curvature is obtained at the point of highest amplitude. How do I prove it mathematically?
I proceeded by assuming the curve, $y=A\sin\omega x$. We know that radius of curvature is given by, $$\rho= \frac{[1+(\frac{dy}{dx})^2]^{1.5}}{\frac{d^2y}{dx^2}}$$
So I proceeded by substituting the respective values and differentiating $\rho$ with respect to $x$ and putting it equal to zero but each time I got vague results instead of getting $T/4$ or $3T/4$, assuming $\omega= 2\pi/T$.
Any help is appreciated.

Using the definition, 
$$\rho=-\frac{\csc ( \omega x ) \left(1+A^2 \omega ^2 \cos ^2( \omega x )\right)^{3/2}}{A
   \omega ^2}$$ $$\frac{d\rho}{dx}=\frac{\cot ( \omega x ) \csc ( \omega x ) \sqrt{1+A^2 \omega ^2 \cos ^2( \omega x)}
   \left(2 A^2 \omega ^2+1-A^2 \omega ^2 \cos (2  \omega x )\right)}{A \omega }$$ The derivative can only cancel when $\cot ( \omega x)=0$.
