Minimum radius of curvature of a sinusoidal curve

Question

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.

Answer 1

If you minimize the numerator while maximizing the denominator, you win. Note that at the points in question, you have $dy/dx = 0$ and $|d^2y/dx^2|$ is as large as it can be.

Answer 2

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.

Answer 3

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 $\cot ( \omega x)=0$.