I have put in the formula for arc length and am fairly sure I have the derivatives figured correctly. The problem is, I'm unable to put the equation into a form that is integratable.
Thanks for pointing that out. Anyway, I seem to have made another mess of it. I've been doing this calculus for hours now and can't seem to see straight anymore so I'm going to take a break. This is what I have so far but it is not pretty. Maybe trig substitution isn't the way to go but I can't see how else to do it.
$\displaystyle dx=\frac{1}{\sqrt{1-t^2}}$
$\displaystyle dy=\frac{-t}{1-t^2}$
Arc length then is
$\displaystyle \int_0^{\frac{1}{2}}\sqrt{\frac{1}{1-t^2}+\frac{t^2}{(1-t^2)^2}}dt$
Seeing that our quantity inside our squareroot is equal to
$\displaystyle \frac{(1-t^2)}{(1-t^2)^2}+\frac{t^2}{(1-t^2)^2}=\frac{1}{(1-t^2)^2}$
So we have
$\displaystyle \int\frac{dt}{1-t^2}$
if you look all the values within our limits of integration give a positive result, so we can forgoe our absolute value. So we have
$\displaystyle \int_1^{\frac{1}{2}}\frac{dt}{1-t^2}={\rm{arctanh}}(t)\bigg|_{0}^{\frac{1}{2}}=\fr ac{\ln(3)}{2}$