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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?

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

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

Yup, I've gotten that far.

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
\[(\frac{x^2}{11}+b)^2\]

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 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)

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

put up a new thread will be best

except site has slowed to a crawl