## anonymous 4 years ago 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)

1. anonymous

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

2. anonymous

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

3. anonymous

you are off by a minus sign

4. anonymous

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

5. anonymous

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.

6. anonymous

oh wow isn't that annoying!

7. anonymous

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

8. anonymous

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

9. anonymous

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?

10. anonymous

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.