Calculus II - Arc Length Problem

Hello, I am having some serious problems calculating the Arc Length of a curve. Here is the problem;

(Putting in a new post to use equation calculator)

Question

Calculus II - Arc Length Problem

Hello, I am having some serious problems calculating the Arc Length of a curve. Here is the problem;

(Putting in a new post to use equation calculator)

anonymous · Answer

$$x = (y^3/33) + 11/(4y) $$
on 3<= y <= 5

anonymous · Answer

So, I take the derivative. No problem;

$$dy/dx = (y^2/11) + (11/(4y^2) $$

Then Length =  $$\sqrt(1 + (dy/dx)^2) $$

So;

$$Length = \sqrt(1 + (y^2/11 + 11/(4y))^2)$$

anonymous · Answer

you are off by a minus sign

anonymous · Answer

$$\int_3^5\sqrt{1+(\frac{y^2}{11}-\frac{11}{4y^2})^2}dy$$

anonymous · Answer

Holy god. If that's the problem I have been having this whole time I'm not sure whether I'm going to laugh or cry. Probably a bit of both.

anonymous · Answer

oh wow isn't that annoying!

anonymous · Answer

happens to me all the damn time, that is why it is good to get another pair of eyes

anonymous · Answer

$$\int_3^5\sqrt{\frac{y^4}{121}+\frac{121}{16 y^4}+\frac{1}{2}}$$

anonymous · Answer

Ok, maybe you can tell me this - when I type 

((y^2/11) - (11/(4y^2))^2 on my calculator it gives me;

((4y^4 - 121)^2)/1936y^4

Any idea why?

anonymous · Answer

Same thing on Wolfram - if I type your above equation (everything under the root) and tell it to factor it gives me the same as above.