anonymous
  • anonymous
Finding the perfect square of a fraction. I am having some real problems finding the perfect square of a fraction. For example; y^4/9 + 1/2 + 9/16y^4 = (y^2/3 + 3/4y^2)^2 But no MATTER WHAT I do I can not get this answer to replicate. I always get different forms of what I assume are the same answer. But I am trying to do arc length, and I need the answer to be a perfect square to get rid of the square root. I need the perfect square of; y^4/121 + y^2/2 + 137/16 ANY help, please!
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
Please let me know if these are too difficult to read and I will rewrite in equation.
anonymous
  • anonymous
\[y^{\frac{4}{9}} + \frac{1}{2}+ \frac{9}{16}y^4 = (y^\frac{2}{3} + \frac{3}{4}y^2)^2\] like that?
anonymous
  • anonymous
The actual question I am trying to solve is; Find the length of the curve x = y^3/33 + 11/4y on 3<= y <= 5 I am fine until I get to Sqrt(1+ dy/dx), at which point I can never get a perfect square to eliminate the root.

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anonymous
  • anonymous
because i am not sure that is it right
anonymous
  • anonymous
Shoot, no, let me re-write; \[(y^4)/9 + (1/2) + 9/(16y^4)\]
anonymous
  • anonymous
= \[(y^2/3 + (3/4y^2)) ^2\]
anonymous
  • anonymous
ooooooooooh ok
anonymous
  • anonymous
I will give my first born to you if you help me with this problem. Been killing myself trying to get the perfect square (even on that one which I KNOW IS RIGHT!) but I cannot duplicate. It is driving me bonkers.
anonymous
  • anonymous
\[\frac{y^4}{121}+\frac{y^2}{2}+\frac{137}{16}\]?
anonymous
  • anonymous
Yup, I've gotten that far.
anonymous
  • anonymous
Now how do you get the perfect square of that? I tried factoring on my TI-89 and Wolframalpha, but was never able to get anything resembling a perfect square. Not entirely sure how to do it by hand. Calculators have crippled me.
anonymous
  • anonymous
problem is 137/16 is not a perfect square
anonymous
  • anonymous
let me think. if you have any hope of doing this it has to look like \[(\frac{x^2}{11}+b)^2\]
anonymous
  • anonymous
Oh wtf. I mean, this is only the second Arc Length problem I've ever done and they are getting into that kind of stuff?
anonymous
  • anonymous
well you are going to have to add something we see from what i wrote that if you multiply out you will get a middle term of \[\frac{2b}{11}y^2\]and if you want that to be \[\frac{y^2}{2}\] that means \[\frac{2b}{11}=\frac{1}{2}\] so \[b=\frac{11}{4}\]
anonymous
  • anonymous
then you will have \[(\frac{y^2}{11}+\frac{11}{4})^2+\frac{16}{16}\]
anonymous
  • anonymous
Wait a second - I just expanded \[(y^2/11 + 11/(4y^2)^2 = y^4/121 + 121/(16y^4) + 1/2\]
anonymous
  • anonymous
aka \[(\frac{y^2}{11}+\frac{11}{4})^2+1\]
anonymous
  • anonymous
oh crap how did the y end up in the denominator??
anonymous
  • anonymous
I am really not sure. Just tried looking up to find if I wrote it wrong, but I couldn't see it. I was looking at all my scratch work and noticed I wrote it differently then above.
anonymous
  • anonymous
i thought you were doing this one y^4/121 + y^2/2 + 137/16
anonymous
  • anonymous
I am. But earlier I wrote 11/4y, which I meant as (11)/(4y)
anonymous
  • anonymous
So dy/dx = (y^2)/11 + 11/(4y^2)
anonymous
  • anonymous
Then Length = Sqrt(1 + (dy/dx)^2)
anonymous
  • anonymous
=Sqrt(1 + ((y^2/11) + (11)/(4y^2))^2)
anonymous
  • anonymous
And thats where I get stuck on every problem. Getting rid of that square root by making it a perfect square. One of the things i posted was an example that gives the correct answer, but I was never able to duplicate it.
anonymous
  • anonymous
oh ok i see. don't expect to get a perfect square. you will probably have to adjust somehow
anonymous
  • anonymous
I guess that will do it, hehe. Let me see if I can manage to adjust this.
anonymous
  • anonymous
ok i just took the derivative of \[\frac{y^3}{33}+\frac{11}{4y}\] and got \[\frac{y^2}{11}-\frac{11}{4y^2}\] added 1, took the square root, and asked wolfram for the integral. it is a hot mess and so there must be some error somewhere
anonymous
  • anonymous
it would probably be best if you reposted the original problem, just as you got it. then i bet you would get lots of good answers.
anonymous
  • anonymous
Ok - let me do that (in a new thread or here?)
anonymous
  • anonymous
put up a new thread will be best
anonymous
  • anonymous
I tried once asking about Arc Length, and explaining where I got stuck, but got no answers so I repost this.
anonymous
  • anonymous
Ok, will do. I hope to see you there, and thank you for all the help you have been giving! i really appreciate it!
anonymous
  • anonymous
except site has slowed to a crawl

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