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!

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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
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Please let me know if these are too difficult to read and I will rewrite in equation.
\[y^{\frac{4}{9}} + \frac{1}{2}+ \frac{9}{16}y^4 = (y^\frac{2}{3} + \frac{3}{4}y^2)^2\] like that?
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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because i am not sure that is it right
Shoot, no, let me re-write; \[(y^4)/9 + (1/2) + 9/(16y^4)\]
= \[(y^2/3 + (3/4y^2)) ^2\]
ooooooooooh ok
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.
\[\frac{y^4}{121}+\frac{y^2}{2}+\frac{137}{16}\]?
Yup, I've gotten that far.
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.
problem is 137/16 is not a perfect square
let me think. if you have any hope of doing this it has to look like \[(\frac{x^2}{11}+b)^2\]
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?
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 \[b=\frac{11}{4}\]
then you will have \[(\frac{y^2}{11}+\frac{11}{4})^2+\frac{16}{16}\]
Wait a second - I just expanded \[(y^2/11 + 11/(4y^2)^2 = y^4/121 + 121/(16y^4) + 1/2\]
aka \[(\frac{y^2}{11}+\frac{11}{4})^2+1\]
oh crap how did the y end up in the denominator??
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.
i thought you were doing this one y^4/121 + y^2/2 + 137/16
I am. But earlier I wrote 11/4y, which I meant as (11)/(4y)
So dy/dx = (y^2)/11 + 11/(4y^2)
Then Length = Sqrt(1 + (dy/dx)^2)
=Sqrt(1 + ((y^2/11) + (11)/(4y^2))^2)
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.
oh ok i see. don't expect to get a perfect square. you will probably have to adjust somehow
I guess that will do it, hehe. Let me see if I can manage to adjust this.
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
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.
Ok - let me do that (in a new thread or here?)
put up a new thread will be best
I tried once asking about Arc Length, and explaining where I got stuck, but got no answers so I repost this.
Ok, will do. I hope to see you there, and thank you for all the help you have been giving! i really appreciate it!
except site has slowed to a crawl

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