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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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As far as I can tell, you've (i) computed the derivative incorrectly; and you've (ii) squared incorrectly. –  Arturo Magidin Jul 16 '11 at 3:43
    
@Arturo - I checked both parts against the solutions manual and they are correct. Actually I lied. Only the derivative is correct. Edit: Ok yeah I'm dumb, I squared incorrectly and I think that threw the problem off for me the rest of the way. I'll try it again. –  Ryan Jul 16 '11 at 3:48
    
@Ryan: Yes, what is the square of $y^{1/2}$? And after you get it right, note that adding $1$ just changes the sign of the middle term. Does this suggest what your new expression might be the square of? –  André Nicolas Jul 16 '11 at 3:53
    
@Ryan: I posted an answer assuming there is a typo in the title and in the first sentence. Instead of $dy/dx$ should be $dx/dy$, and that you want 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. –  Américo Tavares Jul 16 '11 at 14:34
    
@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. –  Arturo Magidin Jul 17 '11 at 3:56

1 Answer 1

up vote 1 down vote accepted

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. –  Américo Tavares Jul 16 '11 at 14:36

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