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!

- anonymous

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- anonymous

Please let me know if these are too difficult to read and I will rewrite in equation.

- 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

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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## More answers

- anonymous

because i am not sure that is it right

- anonymous

Shoot, no, let me re-write;
\[(y^4)/9 + (1/2) + 9/(16y^4)\]

- anonymous

= \[(y^2/3 + (3/4y^2)) ^2\]

- anonymous

ooooooooooh ok

- 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

\[\frac{y^4}{121}+\frac{y^2}{2}+\frac{137}{16}\]?

- anonymous

Yup, I've gotten that far.

- 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

problem is 137/16 is not a perfect square

- anonymous

let me think. if you have any hope of doing this it has to look like
\[(\frac{x^2}{11}+b)^2\]

- 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

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

then you will have
\[(\frac{y^2}{11}+\frac{11}{4})^2+\frac{16}{16}\]

- anonymous

Wait a second - I just expanded \[(y^2/11 + 11/(4y^2)^2 = y^4/121 + 121/(16y^4) + 1/2\]

- anonymous

aka
\[(\frac{y^2}{11}+\frac{11}{4})^2+1\]

- anonymous

oh crap how did the y end up in the denominator??

- 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

i thought you were doing this one
y^4/121 + y^2/2 + 137/16

- anonymous

I am. But earlier I wrote 11/4y, which I meant as (11)/(4y)

- anonymous

So dy/dx = (y^2)/11 + 11/(4y^2)

- anonymous

Then Length = Sqrt(1 + (dy/dx)^2)

- anonymous

=Sqrt(1 + ((y^2/11) + (11)/(4y^2))^2)

- 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

oh ok i see. don't expect to get a perfect square. you will probably have to adjust somehow

- anonymous

I guess that will do it, hehe. Let me see if I can manage to adjust this.

- 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

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

Ok - let me do that (in a new thread or here?)

- anonymous

put up a new thread will be best

- anonymous

I tried once asking about Arc Length, and explaining where I got stuck, but got no answers so I repost this.

- 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

except site has slowed to a crawl

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