How Do You Solve High Degree Polynomials in Calculus Problems?

[b2386]

Hi everyone,

I am having trouble finding an answer for this problem.

The problem requires me to find the maximum value of r for this equation:

$$r^{2}e^{\frac{-r}{a}}(1-\frac{r}{a}+\frac{r^2}{4a^{2}})$$

I took the derivative and set it equal to zero but keep getting a high degree polynomial that I am not sure how to solve for r.

Any help would be appreciated.

 

[FrogPad]

Hi everyone,

I am having trouble finding an answer for this problem.

The problem requires me to find the maximum value of r for this equation:

$$r^{2}e^{\frac{-r}{a}}(1-\frac{r}{a}+\frac{r^2}{4a^{2}})$$

I took the derivative and set it equal to zero but keep getting a high degree polynomial that I am not sure how to solve for r.

Any help would be appreciated.

You will get a polynomial of degree 4. You can factor it and solve for each of the factors individually. This is not necessarily enjoyable, but you can do it. You could use Newton's method, or you could graph it and solve numerically.

What did you get when you differentiated it?

 

[b2386]

Once I took out all the extra fluff, I am left with this:

$$2-\frac{r}{a}-3r+\frac{r^2}{a^2}+4r^2-\frac{r^3}{4a^3}=0$$

Could you explain how to factor this polynomial using Newtons method? I am not familiar with it.

Also, I know r is equal or close to 5a

 

[FrogPad]

Once I took out all the extra fluff, I am left with this:

$$2-\frac{r}{a}-3r+\frac{r^2}{a^2}+4r^2-\frac{r^3}{4a^3}=0$$

Could you explain how to factor this polynomial using Newtons method? I am not familiar with it.

Also, I know r is equal or close to 5a

what happened to your exponential?

Also, if you think r=5a is one solution, then plug that in into the derivative and you will get zero if it is a solution.

Newton's method is a root finding algorithm. You can look it up and try it if you would like, but it would probably be better to wait on that. Are you in precalc? Cause you are doing calculus right now... so I'm a little unsure about where you are mathematically.

 

[HallsofIvy]

Once I took out all the extra fluff, I am left with this:

$$2-\frac{r}{a}-3r+\frac{r^2}{a^2}+4r^2-\frac{r^3}{4a^3}=0$$

Could you explain how to factor this polynomial using Newtons method? I am not familiar with it.

Also, I know r is equal or close to 5a

"Newton's method" is not a method for factoring- it's a method for finding numerical approximations to solutions to equations.

In any case, your polynomial is incorrect.
Multiplying that "r²" into the polynomial part, you have
$$y= e^{-\frac{r}{a}}\left(\frac{r^4}{4a^2}-\frac{r^3}{a}+ r^2\right)$$
$$y'= -\frac{1}{a}e^{-\frac{r}{a}}\left(\frac{r^4}{4a^2}-\frac{r^3}{a}+ r^2\right)+ e^{-\frac{r}{a}}\left(\frac{r^3}{a^2}-\frac{3r^2}{a}+ 2r\right)$$
Setting that equal to 0, you can, of course, divide through by the exponential to get rid of it and then you have
$$-\frac{r^4}{a^3}+ \frac{r^3}{a^3}- \frac{r}{a}+ \frac{r^3}{a^3}-\frac{3}{a}r^2+ r= 0$$
(Notice that equal powers of r have equal powers of a in the denominator!)
You immediately factor out an "r" (remembering that one solution is r= 0) and multiply through by [itex]-a^3[/tex] to get
$$r^3- 2ar^2+ 4a^2r- 2a^3= 0$$

You can get rid of the a by letting x= r/a so r= ax and the equation becomes
$$a^3x^3= -2a^3x^2+ 4a^3x-2a^3= 0$$
or
$$x^3- 2x^2+ 4x- 2= 0$$
If that has any rational roots, they can only be 1, -1, 2, or -2, and it's easy to check that none of those work. This clearly has an irrational root between 0 and 1 and, graphing, it appears that that is the only real root.