Please Help! Finding roots of this particular polynomial

[JackDaniel87]

Hey guys,
Nice to be on here.
I have been banging my brain for the last two weeks trying to come up with an algebraic solution to the following question - to no avail.
Any input would be MUCH appreciated!
The problem is somewhat long but can be summarized as follows:

Begin with the following equation as a function of x. There are two parameters, a and b, that could take on arithmetic values but I am more interested in a general solution:

$$y(x)=\frac{b\left(1-x\right)}{b\left(1-x\right)+\left(1-a\right)x}$$
The curvature K of the above polynomial ought to be given by the the following differential equation which uses the first and second order derivatives of y(x), as follows:

$$?=\frac{|?″|}{{(1+?′^2)}^\frac{3}{2}}$$
Now, I am actually interested in the maximum curvature k - which is why we need to differentiate k with respect to x and find its roots:

$$?′=\frac{d}{dx}k$$
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 seen here, where the purple line (k') crosses the x-axis:






However, obtaining an algebraic solution as a function of a and b has been a challenge - hence my reaching out!

Any input you might have would be GREATLY appreciated!

Thank you in advance for any help you may offer!

-J

 

[blamocur]

You have a first order rational function which can be simplified as $y=\frac{px+q}{rx+s}$ -- do you know how to differentiate rational functions?

 

[JackDaniel87]

Yes. The issue isn’t the derivation. It’s finding the roots….any leads?

 

[Dr.Peterson]

Yes. The issue isn’t the derivation. It’s finding the roots….any leads?

Are you saying you want us to do all the work you did to get to an equation whose roots you want to find, and then tell you how to find its roots?

Please show us where you are stuck, so we don't have to do all that, and can focus on the part you need help with. (Showing your work might also reveal an error in your work before you get to the part you're asking about.)

Posting Guidelines (Summary)

Welcome to our tutoring boards! :) This page summarizes main points from our posting guidelines. As our name implies, we provide math help (primarily to students with homework). We do not generally post immediate answers or step-by-step solutions. We don't do your homework. We prefer to help...

www.freemathhelp.com

 

[JackDaniel87]

When I differentiate y(x) this is what the equation of the curvature becomes:

$$k =\frac{\left|-\frac{2b\left(a-1\right)\left(-b+1-a\right)}{\left(b\left(1-x\right)+x\left(1-a\right)\right)^{3}}\right|}{\left[1+\left(\frac{b\left(a-1\right)}{\left(b\left(1-x\right)+x\left(1-a\right)\right)^{2}}\right)^{2}\right]^{\frac{3}{2}}}$$
After that, I can't differentiate further...let alone find the roots of that derivative...

 

[blamocur]

You are right, it gets hairy in a hurry I looked at the general expression for $k^\prime$ and came to the conclusion that if $k^\prime=0$ then $\left(1+{y^\prime}^2\right)y^{\prime\prime\prime}-3{y^{\prime\prime}}^2=0$. I am not too confident in the last equation, so feel free to double check. Even if I haven't made any mistakes in my derivation I am not sure it is a helpful result: when you plugin your expression for $y$ you will most likely end up with high order polynomials where closed-form solutions might not exist.

 

[JackDaniel87]

Can I ask how did you arrive at such expression?

 

[BigBeachBanana]

Can you confirm if this is correct? If so, I will try to give a detailed solution, though it is tedious
$$x=-\dfrac{\sqrt{\left(a-1\right)b}-b}{b+a-1}$$

 

[blamocur]

Can I ask how did you arrive at such expression?

The expression for $k^\prime$ is a fraction, but for that fraction to be 0 it is enough to have its numerator to be equal to 0.

 

[JackDaniel87]

The numerator is a complex number...a and b are between 0 and 1, and there's clearly a real solution...so not sure your solution is correct
im afraid

 

[BigBeachBanana]

The numerator is a complex number...a and b are between 0 and 1, and there's clearly a real solution...so not sure your solution is correct
im afraid

Yes my apologies, there are actually 4 solutions (including complex roots):
$$x = \dfrac{\sqrt{\left(1-a\right)b}+b}{b+a-1}$$$$x = -\dfrac{\sqrt{\left(1-a\right)b}-b}{b+a-1}$$$$x = \dfrac{\sqrt{\left(a-1\right)b}+b}{b+a-1}$$$$x= -\dfrac{\sqrt{\left(a-1\right)b}-b}{b+a-1}$$

 

[JackDaniel87]

BigBeachBananas...my wife forgive me...but I COULD KISS YOU! would you be so kind as to show your work? PS: Number 2 is the winner!

 

[BigBeachBanana]

BigBeachBananas...my wife forgive me...but I COULD KISS YOU! would you be so kind as to show your work? PS: Number 2 is the winner!

Instead of me writing out my lengthy and strenuous solution, the online derivative calculator https://www.derivative-calculator.net/ can show you the work