I have a distribution for which I have to find the point where the slope changes drastically. In visual terms, I have to find this point:
I though I could use derivatives, but for the following equation:
$$
y = -0.255ln(x) + 1.6889
$$
It seems I can't. How can I get what I need?
Best Answer
Let the curve be $y=-a \ln x +b.$ [The values of $a,b$ from the equation are then $a=0.255,\ b=1.6889,$ but note these values do not agree with the diagram in the question.]
Anyway it is not, for these curves, that one wants to maximize the second derivative, which here is $y''=a/x^2$ which goes to $+\infty$ as $x \to 0^+.$
The formula for curvature $\kappa(x)$ of the curve $y=y(x)$, when squared for ease of maximizing it, is $$\kappa(x)^2=\frac{(y'')^2}{(1+y'^2)^3},$$ and when we take the derivative of this and insert $y'=-a/x, \ y''=a/x^2$ and factor, we get for the derivative of squared curvature the expression $$\frac{-2a^2x(2x^2-a^2)}{(x^2+a^2)^4}.$$ So the maximal squared curvature (and thus the maximal curvature) occurs at $x=a/\sqrt{2}.$ For the values of $a,b$ noted at the top of this answer, taken from the displayed equation of the post, this gives the point of maximal curvature as about $(0.18031,2.12573).$ A more convincing thing would come if we knew the true values of $a,b$ for the equation which go with the diagram included in the question.