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I have come across a question for the length of the loop of a given parametric curve defined by the equations

$x(t)= 4t-t^3$ ; $y(t)=-2+2t^2$

The loop is defined about the interval from t=-2 to t=2 or notice the symmetry by taking twice the length from 0 to 2. The problem lies in the calculation of the antiderivative with the traditional formula of

$\int_{-2}^2 \sqrt{(dx/dt)^2+(dy/dt)^2} dt$

This simplifies out to;

$\int_{-2}^2 \sqrt{9t^4-8t^2+16} dt$

As far as I know and have tried, this integral has no solvable real antiderivative as the root is complex. My best guess would be either to eliminate the parameter of the curve and potentially solve in Cartesian or rather find a polar interpretation and solve in polar (both of which I have tried but failed). If someone can help point me in the right direction it would be greatly appreciated!

aort01
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    I would have to say that this integral does not have an elementary antiderivative. This is an elliptic integral so it may be possible to define it in terms of one of the common elliptic integral functions but not in a way that I am aware of. If it helps the integral evaluates to $21.72473488516433$ with an estimated error of $2.41193640335144\times 10^{-13}$ – Henry Lee Oct 19 '22 at 22:22
  • Very well I will accept my defeat. I thought somehow there would be some mathematical manipulation that makes it work out nicely. Thank you anyways!! – aort01 Oct 19 '22 at 22:28
  • It's quite rare that arc lengths have closed forms. The $\sqrt{1+\ldots}$ in the integrand generally obstructs that. See MSE Q3321398 and Q20578. – Jam Oct 19 '22 at 22:34
  • If the integral was between $0$ and $2/3$, I have an almost exact solution. But, alas ... – Claude Leibovici Oct 20 '22 at 05:04
  • @ClaudeLeibovici How so, I'm curious? – aort01 Oct 20 '22 at 15:46

1 Answers1

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This is just an exercise with almost no interest.

As I wrote in comments, if the problem was $$\int_{0}^{\frac 23} \sqrt{9t^4-8t^2+16}\, dt$$ it could be possible to obtain almost the exact solution writing $$ \sqrt{9t^4-8t^2+16}=\sum_{n=0}^\infty a_n\,t^{2n}$$ where the coefficients, given by $$a_n=\frac{4(2n-3)\,a_{n-1}-9(n-3)\,a_{n-2}}{16\,n}\quad \quad\text{with} \quad a_0=4 \quad\text{and} \quad a_1=-1$$ decrease very fast to $0$.

Comuting the partial sums $$S_k=\sum_{n=0}^k \frac {a_n}{2n+1}\left(\frac 23\right)^{2n+1}$$ $$\left( \begin{array}{cc} k & S_k \\ 0 & 2.666666667 \\ 5 & 2.596066206 \\ 10 & 2.596065324 \\ 15 & 2.596065321 \\ \end{array} \right)$$

Just for your curiosity $$I=\int_0^x \sqrt{9t^4-8t^2+16}\, dt$$ is $$I=\frac{\sqrt{9 x^4-8 x^2+16}}{27}\,\,\, \Bigg[\cdots\Bigg]$$ $$\Bigg[\cdots\Bigg]=9 x+32 \left(\sqrt{2}-2 i\right) F\left(\sin ^{-1}\left(\frac{x}{2} \sqrt{1+2 i \sqrt{2}} \right)|-\frac{7+4 i \sqrt{2}}{9}\right)+16 \left(\sqrt{2}+i\right) E\left(\sin ^{-1}\left(\frac{x}{2} \sqrt{1+2 i \sqrt{2}} \right)|-\frac{7+4 i \sqrt{2}}{9}\right)$$

Make $x=\frac 23$ or whatever and ... enjoy !

Claude Leibovici
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