Maximum Curvature of sin(x)

Nov 2018
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4
USA
$$y = sin x, -\pi \leq 0 \leq \pi, y' = cos x, y'' = -sin x$$

$$\kappa = \frac{|-sin x|}{(1 + cos^2 x)^{\frac{3}{2}}}$$

Considering x only from 0 to pi lets me remove the absolute value bars.

$$\frac{d\kappa}{dx} = \frac{cos x}{(1 + cos^2 x)^{\frac{3}{2}}} + \frac{3cos x sin^2 x}{(1 + cos^2 x)^{\frac{5}{2}}}$$

$$\frac{d\kappa}{dx} = 0 \longrightarrow \frac{cos x}{(1 + cos^2 x)^{\frac{3}{2}}} = -\frac{3cos x sin^2 x}{(1 + cos^2 x)^{\frac{5}{2}}} \longrightarrow 1 + cos^2 x = -3sin^2 x = -2sin^2 x -sin^2 x \longrightarrow 2 = -2 sin^2 x \longrightarrow sin^2 x = -1$$

This gives me an imaginary root. Wolfram tells me this function has real roots: https://www.wolframalpha.com/input/?i=d/dx+(sinx/(1+(cosx)^2)^1.5)
 

romsek

MHF Helper
Nov 2013
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$\dfrac{d \kappa}{dx} = 0~~ \forall x = \pi\left(n - \dfrac 1 2 \right),~n\in \mathbb{Z}$

and these are maxima which can be shown by evaluating the 2nd derivative at these points.

$\left . \dfrac{d^2\kappa}{dx} \right |_{x=\pi\left(n - \dfrac 1 2 \right)} = -3$
 
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Nov 2018
32
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I was looking to determine the points of maximum curvature analytically, not confirm them.
 

romsek

MHF Helper
Nov 2013
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3,073
California
I was looking to determine the points of maximum curvature analytically, not confirm them.
The derivative has two terms with factors of $\cos(x)$ in the numerator.

It should be pretty clear that $x = \left(n-\dfrac 1 2\right)\pi$ is going to be a solution to the derivative being zero.

In your derivation you implicitly assume $\cos(x) \neq 0$ when you divide by it.

Thus you derived the roots of the equation when this isn't the case.

The real roots occur when it is the case.
 
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Nov 2018
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Thanks. I drove past the solution... again.