<s>[INST] Calculating arc length of a curve, stuck on dy/dx part (algebra mostly) The equation is:
$$x=\frac{1}{8}y^4 + \frac{1}{4}y^{-2},\qquad 1\leq y\leq 2.$$
I have the formula. I'm not sure how to write it out but this is what it says:
Length is equal to the integral (with $b$ and $a$ for limits) of the square root of $1+(dy/dx)^2 dx$
So I first took the derivative of the equation and got $(1/2)y^3 - (1/2)y^{-3}$.
Now when I plug that back into the formula, I have to square it and I got $(1/4)y^6 - (1/4)y^{-6}$. I factored out a $1/4$ and turned the $y^{-6}$ into $1/y^6$. Just a mess at this point, that doesn't seem right. Can someone help me out with that? Haha I know it's ridiculous but my algebra skills are lacking. 
 [/INST] You squared incorrectly. The square of $a-b$ is not $a^2-b^2$. (Also, you should be using $\frac{dx}{dy}$, not $\frac{dy}{dx}$, because here your independent variable is $y$ and your dependent variable is $x$; your integral will be with respect to $y$, not with respect to $x$).
The square of $\displaystyle\frac{1}{2}y^3 - \frac{1}{2}y^{-3}$ is not just the difference of the squares (which is what you wrote). Rather, it is equal to:
$$\left(\frac{1}{2}y^3 - \frac{1}{2}y^{-3}\right)^2 = \frac{1}{4}y^6 - 2\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)y^3y^{-3} + \frac{1}{4}y^{-6} = \frac{1}{4}y^6+\frac{1}{4}y^{-6} - \frac{1}{2}.$$
When you add $1$, you get
$$\frac{1}{4}y^6 + \frac{1}{4}y^{-6} + \frac{1}{2} = \left(\frac{1}{2}y^3 + \frac{1}{2}y^{-3}\right)^2.$$
Remember: $(a+b)^2 = a^2 + 2ab+b^2$, and $(a-b)^2 = a^2-2ab+b^2$. The square of the sum is not the sum of the squares.
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