Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Did you notice that you do not have $y$ as a function of $x$, but rather $x$ as a function of $y$, so that your formula should use $\frac{dx}{dy}$, not $\frac{dy}{dx}$? – Arturo Magidin May 24 '11 at 5:31
Note the artificiality of the problem, a feature it shares with the majority of arc length problems in calculus books. If you change the $1/8$ to $1/8.1$, you end up with something you cannot integrate in closed form. – André Nicolas May 24 '11 at 6:12
+1 for trying before asking – Ross Millikan May 24 '11 at 12:20
up vote 5 down vote accepted

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.

share|cite|improve this answer
Ah yes you're right about the dx/dy part, missed that. And wow I feel so stupid after realizing what I did on the squaring part. That's a mistake you should only make in middle school! Think I've been studying too much today. Thanks! – Ryan May 24 '11 at 5:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.