# Math Help - Length of Curves

1. ## Length of Curves

Find the length of the curve

$x = \frac{y^3}{3} + \frac{1}{4y}$ from $y=1$ to $y=3$

$g'(y) = y^2 + \frac{y^{-2}}{4}$

$L = \int \sqrt{1 + \left(y^2 + \frac{y^{-2}}{4}\right)^2}dy$

$L = \int \sqrt{1 + y^4 + \frac{1}{2} + \frac{y^{-4}}{16}}dy$

$L = \int \sqrt{1 + y^4 + \frac{3}{2} + \frac{y^{-4}}{16}}dy$

It says it in the book that 1 + \left(\frac{dx}{dy}\right)^2 is a perfect square but I'm not seeing that. Any help?

Edit: I think I found my error. :|

2. The problem starts here

g'(y) = y^2 - 1/(4y^2)

g' (y) ^2 = y^4 -1/2 + 1/(16y^2)

1 + g' (y)^2 = y^4 +1/2 + 1/(16y^2) = [y^2 +1/(4y^2)]^2