Arc length of $y = \frac{x^3}{3} + \frac{1}{4x}$

Arc length of $y = \frac{x^3}{3} + \frac{1}{4x}$ over $1 \leq x \leq 2$

I know that the first thing I need to do is take the derivative.

$$y' = x^2 - 4x^{-2}$$

Then I take the integral on that range using the arc length formula.

$$\int_1^2 \sqrt{1 + (x^2-4x^{-2})^2}$$

$$(x^2-4x^{-2})^2 = -16x^{-4} - 8 + x^4$$

$$\int_1^2 \sqrt{-16x^{-4} - 7 + x^4 }$$

From here I have no idea how to factor this but I am pretty sure I must have messed up something before that.

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Look at the $4$ in your derivative. –  Argon Jun 6 '12 at 20:40
Note that the derivative is $x^2-\frac{x^{-2}}{4}$. With your version of it, you will never be able to integrate, and neither could I. –  André Nicolas Jun 6 '12 at 20:40
How do I write that differently? I guess I have to write $4^{-1} x^{-2}$ –  user138246 Jun 6 '12 at 20:46
That would be correct. I like to write it as a fraction, i.e. $$\frac{1}{4x^2}$$ –  Argon Jun 6 '12 at 20:47

1. Your derivative is wrong: $$\left(\frac{x^3}{3} +\frac{1}{4x}\right)' = \left(\frac{1}{3}x^3 + \frac{1}{4}x^{-1}\right)' = x^2 - \frac{1}{4}x^{-2} = x^2 - \frac{x^{-2}}{4}.$$

2. You squared incorrectly: if you square $x^2-4x^{-2}$, you get: $$(x^2-4x^{-2})^2 = x^4 - 8x^2x^{-2} + 16x^{-4} = x^4 - 8+16x^{-4}.$$ Note the plus sign on $16x^{-4}$; you have a minus sign.

If you square the correct function, you get $$\left( x^2 - \frac{1}{4}x^{-2}\right)^2 = x^4 - \frac{1}{2} + \frac{1}{16}x^{-4}.$$ So the integral would be $$\int_{1}^2 \sqrt{ x^4 + \frac{1}{2} + \frac{1}{16}x^{-4}}\,dx.$$ As for solving it, note that: $$x^4 + \frac{1}{2} + \frac {1}{16}x^{-4} = (x^2)^2 + 2x^2\left(\frac{1}{4}x^{-2}\right) + \left(\frac{1}{4}x^{-2}\right)^2 = \left( x^2 + \frac{1}{4}x^{-2}\right)^2.$$

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I do not follow the last line, specifically the $2x^2$ part where did that term come from? I mean how do you know to do this? –  user138246 Jun 6 '12 at 20:53
$$2x^2(\frac{1}{4}x^{-2})=\frac{1}{2}x^{-2+2}=\frac{1}{2}$$ –  Argon Jun 6 '12 at 20:56
@Jordan: I completed the square, and discovered that in fact it was a perfect square already. Since $(a+b)^2 = a^2+2ab+b^2$, I start with $x^4$ and notice that it is $(x^2)^2$. So I want $a=x^2$. The "middle term" has to be $2ab$; since $a=x^2$, then the middle term has to be $2x^2$ times something, and must total $\frac{1}{2}$. That is, I need $$\frac{1}{2}=2x^2(\text{something})$$and the something is $\frac{1}{4}x^{-2}$ (solve for something). This will be $b$. And then I notice that my final term is already $b^2$, so it's a perfect square. –  Arturo Magidin Jun 6 '12 at 20:58
I guess I never realized this before but if I have a series of things and each is a square their sqare root would just be that series without the square power? That is hard to think of considering that $(a + b)^2$ is not always equal to $(a^2 + b^2)$ –  user138246 Jun 6 '12 at 20:59
@Jordan: "Series" has a special meaning in calculus, which you do not mean here. And no, the square root of a sum of squares is not the sum of the square roots. That is not what we are doing here. We are taking the entire expression $x^4 + \frac{1}{2}+\frac{1}{16}x^{-4}$, and noting that this is in fact a single square: $(x^2 + \frac{1}{4}x^{-2})^2$. And the square root of a single square is its absolute value: $\sqrt{a^2} = |a|$. Not a "series of things". Just a single square. –  Arturo Magidin Jun 6 '12 at 21:02

The derivative is $x^2-\frac{x^{-2}}{4}$. Its square is $x^4 -\frac{1}{2}+x^{-4}$. Add $1$, we get $x^4+\frac{1}{2}+x^{-4}$. The square root of this is $x^2+\frac{x^{-2}}{4}$. That should not be hard to integrate.

Remark: Examples of arclength are often artificial. This because for most functions $f(x)$, $\sqrt{1+(f'(x))^2}$ is something horrible that cannot be integrated in terms of elementary functions. For example, if instead of our function we had $\frac{x^3}{3}+\frac{x^{-2}}{5}$, we would end up with something that cannot be integrated in terms of elementary functions. For the same reason, any little mistake in differentiating or squaring usually leads to something one cannot integrate. For that reason, I have mostly avoided putting such artificial examples on exams, since a minor slip can result in an excessive number of lost marks.

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