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With the information given:

$$x=\frac{y^4}{8}+\frac{1}{4y^2}\,,\ \ 1 \le y \le 2$$

I must find the exact length of the curve.

I use this formula to find it: $$\sqrt{1+\left(\frac{dx}{dy}\right)^2}\ dy $$

So of course, I should find what 1 + (dx/dy)^2 is.

This is what I got:

$$\frac{y^6 -y^{-6}+2}{4} $$

So . . .

$$\int_1^2\sqrt{\frac{y^6-y^{-6}+2}{4}} \, dy$$

I don't know... It just looks really funky to me. Should I do a $u$ substitution? Like $u=y^6-y^{-6}+2$?

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There was a little mistake made when squaring, you should have $$\frac{y^6+2+y^{-6}}{4}\tag{1}$$ inside the square root. And the square root of (1) is very nice, in our interval it is $$\frac{y^3+y^{-3}}{2}.$$

Remark: Many arclength exercises, including this one, are rather contrived. The coefficients were carefully chosen to make the thing we are integrating "magically" simplify.

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  • $\begingroup$ so should I use u substitution to evaluate this? u = y^3 + y^-3 $\endgroup$ – user106039 Sep 28 '14 at 20:09
  • $\begingroup$ To integrate, no substitution. You are just integrating powers. For $y^3$ you get $\frac{1}{4}y^4$. For $y^{-3}$ you get $\frac{1}{-2}y^{-2}$. $\endgroup$ – André Nicolas Sep 28 '14 at 20:13
  • $\begingroup$ I'm hard-pressed to think of an arclength exercise that isn't either contrived (such as this one) or computationally trivial (such as a circle.) I suppose the case of the arc-length of an ellipse (and the connection to elliptic functions thereof) is a counter-example, but that's not really at an 'exercise' level. $\endgroup$ – Semiclassical Sep 29 '14 at 2:43
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Observe that $$ \frac{dx}{dy}=\frac{1}{2}\left(y^{3}-y^{-3}\right), $$ hence $$ 1+\left(\frac{dx}{dy}\right)^2=\frac{1}{2}\left(y^{3}+y^{-3}\right)^2 $$ and thus $$ \sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy=\int_{1}^{2} \frac{(y^3+y^{-3})\,dy}{2}. $$

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