Let $A$ and $B$ be positive constants. If $0 < a < b$, find a simple condition relating $A$ and $B$ that makes it possible to calculate the length of the arc of the curve

$$ y = Ax^4 + \frac{B}{x^2} $$

between $x=a$ and $x=b$ by means of an integral not involving a square root.

Here is what I have so far. Can't put into form that cancels the square root: https://i.stack.imgur.com/eVg7J.jpg

Edit (1) In essence, the problem is how to perform this integration:

$$ \int _a^b\:\sqrt{1+\left(\frac{d}{dx}\left(Ax^4+\frac{B}{x^2}\right)\right)^2}dx$$

  • 1
    $\begingroup$ What have you tried? I assume this doesn't mean you don't start with the usual square root expression for the arclength; only that the square root evaluates to something nice. $\endgroup$
    – mjqxxxx
    Apr 4, 2023 at 2:38
  • $\begingroup$ I took the derivative of the function. Squared it. Added 1. Could not find a squared factor that would cancel with the root in the arc length formula. $\endgroup$ Apr 4, 2023 at 3:16
  • $\begingroup$ lets set A =1 and B=1, then we have something not so messy. But it still can't be put into a square factor. Not as far as I can tell. Let alone with the Constants not being chosen yet. $\endgroup$ Apr 4, 2023 at 3:20
  • $\begingroup$ We can change A and B independent of eachother. But I don't see how they share some special relationship. This is what is confusing me. $\endgroup$ Apr 4, 2023 at 3:21
  • $\begingroup$ Not involving a square root.... in that the root and the square cancel? or no square root altogether? $\endgroup$ Apr 4, 2023 at 3:24

2 Answers 2


Your question winds down to simplifying $$\sqrt{1+\left(4Ax^3-2Bx^{-3}\right)^2}$$ by removing the square root. $$\sqrt{1+\left(4Ax^3-2Bx^{-3}\right)^2}=\sqrt{1+\left(4Ax^3-\dfrac{2B}{x^3}\right)^2}=\sqrt{1+\dfrac{\left(4Ax^6-2B\right)^2}{x^6}}\\=\sqrt{\dfrac{x^6+16A^2x^{12}+4B^2-16ABx^6}{x^6}}=\sqrt{\dfrac{16A^2x^{12}+4B^2+\left(1-16AB\right)x^6}{x^6}}\\=\dfrac{\sqrt{16A^2x^{12}+4B^2+\left(1-16AB\right)x^6}}{x^3}$$ Let $x^{6}=y$. $$\dfrac{\sqrt{16A^2y^2+4B^2+\left(1-16AB\right)y}}{x^3}$$ Notice $(4Ay+2B)^2=16A^2y^2+4B^2+16ABy$. We can write the numerator as $(4Ay+2B)^2$ if $$16AB=1-16AB\implies AB=\dfrac{1}{32}\implies A^2=\dfrac{1}{32^2B^2}$$ So,$$\dfrac{\sqrt{16A^2y^2+4B^2+\left(1-16AB\right)y}}{x^3}=\dfrac{\sqrt{\dfrac{16y^2}{32^2B^2}+4B^2+\dfrac{y}{2}}}{x^3}=\dfrac{\sqrt{\dfrac{y^2}{64B^2}+4B^2+\dfrac{y}{2}}}{x^3}\\=\dfrac{\sqrt{\left(\dfrac{y}{8B}+2B\right)^2}}{x^3}=\dfrac{\dfrac{y}{8B}+2B}{x^3}$$ Now, $\dfrac{y}{8B}=4Ay$.

Hence,$$\sqrt{1+\left(4Ax^3-2Bx^{-3}\right)^2}=\dfrac{4Ax^6+2B}{x^3}$$ if and only if $AB=\dfrac{1}{32}$.



$$y = Ax^4 + \frac{B}{x^2} \quad \implies \quad y'=4 A x^3-\frac{2 B}{x^3}$$ $$1+[y']^2=\frac { 16 A^2 x^{12}+ (1-16 A B)\,x^6+4 B^2}{x^6}$$ You wnat that the numerator be a square.

Consider the quadratic in $x^6$. What is the condition for a double roor ?

  • $\begingroup$ It is not easy to continue from this point (at least for me). $\endgroup$
    – NoChance
    Apr 4, 2023 at 22:16

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