## anonymous 4 years ago 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!

1. anonymous

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

2. 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?

3. 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.

4. anonymous

because i am not sure that is it right

5. anonymous

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

6. anonymous

= $(y^2/3 + (3/4y^2)) ^2$

7. anonymous

ooooooooooh ok

8. 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.

9. anonymous

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

10. anonymous

Yup, I've gotten that far.

11. 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.

12. anonymous

problem is 137/16 is not a perfect square

13. anonymous

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

14. 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?

15. 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 $b=\frac{11}{4}$

16. anonymous

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

17. anonymous

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

18. anonymous

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

19. anonymous

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

20. 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.

21. anonymous

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

22. anonymous

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

23. anonymous

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

24. anonymous

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

25. anonymous

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

26. 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.

27. anonymous

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

28. anonymous

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

29. 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

30. 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.

31. anonymous

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

32. anonymous

put up a new thread will be best

33. anonymous

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

34. 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!

35. anonymous

except site has slowed to a crawl