My problem is somewhat long but I hope someone can shed some light on this.
Full disclosure - I have posted in different forums (NOT StackExchange) to seek help but no fruitful answers so far.
Here it is:
Take the following equation:
$y(x) = \frac{b(1-x)}{b(1-x)+(1-a)x}$
The curvature K of the above polynomial is a function of the first and second order derivative of y(x) as follows:
$=\frac {|y''|}{\left(1+{y'}^{2}\right)^{\frac{3}{2}}}$
Now, I am interested in the maximum curvature of this polynomial, so we need to differentiate k with respect to x and find its roots:
$k'=\frac{d}{dx}k$
Hence, I am interested in finding the roots of k' as a function of a and b, particularly for values of x between 0 and 1. I know a solution exists because graphically it is evident as can be observed below:
That said, obtaining an general algebraic solution as a function of a and b (parameters that can be adjusted but that will always be between 0 and 1) has been a challenge!
Any input you may have would be GREATLY appreciated!
Many thanks in advance!
J
By introducing $c=\frac{1-a}{b}-1$ you can simplify your function to
$$y(x)=\frac{1-x}{1+cx}.$$
We find the maximum of $k$ by finding the maximum of $k^2$. We arrive to the function
$$\left(2c\left(1+c\right)\right)^{2}\frac{\left(1+cx\right)^{6}}{\left(\left(1+cx\right)^{4}+\left(1+c\right)^{2}\right)^{3}}.$$
We can forget the constant factor, take the cube root (it's always positive so maxed when cube root is maxed) and make the change of variable $t=(1+cx)^2$. This leads to maximizing the function
$$g(t)= \frac{t}{t^2+(1+c)^2}$$
and we find its maximum at $t=\pm (1+c) = \left|\frac{1-a}{b}\right|$. Winding everything back, we have that the curvature is maximized at
$$x_{\max} = \frac{\sqrt{\pm(1+c)}-1}{c} = \frac{\sqrt{\left|\frac{1-a}{b}\right|}-1}{\frac{1-a}{b}-1}.$$
Tested here.