I have the following equation
$$y=\pi\ln(2x)$$
And when I ask Mathematica/WolframAlpha for the curvature I get
$$K=\frac{\pi x}{(x^2+\pi^2)^{3/2}}$$
However the formula for curvature of a plane curve is:
$$K = \frac{|y''|}{(1+y'^2)^{3/2}}$$
Which when using
$$y'=\frac{\pi}{x}$$ $$y''=-\frac{\pi}{x^2}$$
Which are both taken from WolframAlpha\Mathematica we find
$$K = \frac{\pi}{(1+ \frac{\pi^2}{x^2})^{3/2}x^2}$$
I've tested for equivalence between the two functions in Mathematica and they aren't the same. Can someone please explain how the Mathematica answer for curviture is reached or verify that my calculated curvature is correct?
They are both the same. Note $$\left(1 + \frac{\pi^2}{x^2}\right)^{3/2} = \left(\frac{x^2 + \pi^2}{x^2}\right)^{3/2} = \frac{(x^2 + \pi^2)^{3/2}}{x^3}$$ So $$\frac{\pi}{\left(1 + \frac{\pi^2}{x^2}\right)^{3/2}x^2} = \frac{\pi}{\frac{(x^2 + \pi^2)^{3/2}}{x}} = \frac{\pi x}{(x^2 + \pi^2)^{3/2}}$$