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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

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    @Ryan: I was about to leave when I typed the comment, so I was wrong about (i). I had misinterpreted your original function, and realized the mistake when I edited the question; only to then be too rushed to fix the comment. Sorry if I misdirected you on that score.2011-07-17

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Updated. I assume there is a typo in the title and in the first sentence of the question, as I commented, and that you want to evaluate $\displaystyle\int_{1}^{9}\sqrt{1+\left( \dfrac{dx}{dy}\right) ^{2}}\mathrm{d}y$, where $x=\dfrac{1}{3}\sqrt{y}\left( y-3\right) $. Its derivative is

$\dfrac{\mathrm{d}x}{\mathrm{d}y}=\dfrac{1}{2}y^{1/2}-\dfrac{1}{2}y^{-1/2}.$

So

$\left( \frac{1}{2}y^{1/2}-\frac{1}{2}y^{-1/2}\right) ^{2}=\frac{1}{4}y-% \frac{1}{2}+\frac{1}{4}y^{-1},$

and

$1+\left( \frac{1}{2}y^{1/2}-\frac{1}{2}y^{-1/2}\right) ^{2}=\frac{1}{4}y+% \frac{1}{2}+\frac{1}{4}y^{-1}.$

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?

Hint:

$\frac{1}{4}y+\frac{1}{2}+\frac{1}{4}y^{-1}=\frac{1}{4}\frac{\left( y+1\right) ^{2}}{y},$

or use the completing the square technique.

Note: most of the times inside the radical you have a function $f(y)$ which is not a perfect square nor anything similar. What you get as integrand is a radical $R(y)=\sqrt{f(y)}$. And you have to integrate it using the normal integration techniques: substitution or by parts. But it is not guaranteed that the integral has a closed form. However, in the present case you do obtain a closed form.

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    As I comented above this answer assumes there is a typo in the title and in the first sentence of the question. Instead of $dy/dx$ should be $dx/dy$, and OP wants to evaluate $\int_{1}^{9}\sqrt{1+\left( \frac{dx}{dy}\right) ^{2}}dy,$ because the derivative $dy/dx$ and the corresponding integral would become too complicated.2011-07-16