Question

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!

Answer 1

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

Answer 2

\[y^{\frac{4}{9}} + \frac{1}{2}+ \frac{9}{16}y^4 = (y^\frac{2}{3} + \frac{3}{4}y^2)^2\]
like that?

Answer 3

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.

Answer 4

because i am not sure that is it right

Answer 5

Shoot, no, let me re-write;

\[(y^4)/9 + (1/2) + 9/(16y^4)\]

Answer 6

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

Answer 7

ooooooooooh ok

Answer 8

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.

Answer 9

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

Answer 10

Yup, I've gotten that far.

Answer 11

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.

Answer 12

problem is 137/16 is not a perfect square

Answer 13

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

Answer 14

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?

Answer 15

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
\[b=\frac{11}{4}\]

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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.

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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.

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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.

Answer 31

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

Answer 32

put up a new thread will be best

Answer 33

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

Answer 34

Ok, will do. I hope to see you there, and thank you for all the help you have been giving! i really appreciate it!

Answer 35

except site has slowed to a crawl