When trying to calculate arc length, what is the easiest way to approach the $(dy/dx)^2$ portion?
If I have: $x = \frac{1}{3}\sqrt{y}(y-3),\qquad 1\leq y\leq 9;$
I take the derivative of the function and get $\frac{1}{2}y^{1/2} - \frac{1}{2}y^{-1/2}$.
Next I have to square the derivative and I got $\frac{1}{4}y^{1/4} + \frac{1}{2} + \frac{1}{4}y^{-1}$ after adding the 1 from the formula (for arc length) to it.
Now to condense everything into the formula up to that point I would have:
$L = \int_1^9 \sqrt{ \frac{1}{4}y^{1/4} + \frac{1}{2} + \frac{1}{4}y^{-1}}$
Now in order to get rid of that radical I would have to get some sort of perfect square but the trouble is sometimes it's difficult to see it right away, and I don't really see it in this one. Is there a better way to go about these problems other than just "looking" at it and trying to figure it out?a