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.

In full form:
${d \rho \over d x} = \left\{-\left(\cos ^2(x)+1\right)^{1.5} \csc (x)\right\}$
Set it to zero and find (for $x \in \mathbb{R}$) that $x = 1.5708 = \pi/2$, and obvious integral increments.
