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As we know, the equation of elliptic is $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=r^2$$ But I want to use its parametric form,the $t$ is representative of the arc length.So how to parameterize the elliptic equation?

mayi
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    The usual analog for the $(r\cos t, r\sin(t)$ parameterizatoin of a circle is $(ar\cos t, br\sin(t)$ where $t$ again runs from $0$ to $2\pi$. But this is a parameterization in terms of angle from the origin, not arc length. I would not be shocked to learn that an ellipse can be parameterized in terms of its arc length using Jacobi elliptic functions sn and cn, with $k$ for those functions depending on the eccentricity of the ellipse. I would be shocked to learn that there is any more elementary parameterization in terms of arc length. – Mark Fischler Jun 12 '17 at 04:25
  • @MarkFischler I want to partition a elliptic into some isometric arc length.For the sake of it,I want to use a parameter of arc length to parameterize the elliptic equation. :) – mayi Jun 12 '17 at 05:01
  • Is there a way to prove that a function does not have an anti derivative or is it simply that no one has found one yet. For example $$ \int_{}^{} \sqrt{1 +a\cdot cos^{2}(x) }\hspace{3mm} dx $$

    Is there a way to prove that this integral cannot be expressed in terms of a finite number of elementary functions?

     – Jason Butler Jan 04 '20 at 18:58
  • See a newer post here. – Ng Chung Tak Oct 26 '21 at 06:36

1 Answers1

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As Mark Fischler commented, I do not see how this could be done.

Using $x=a\, r\cos(t)$, $y=a \,r\sin(t)$, the arc length is given by $$L=r\int_{0}^{u} \sqrt{a^2 \cos^2(t)+b^2 \sin^2(t)}\,dt=br\, E\left(u\left|1-\frac{a^2}{b^2}\right. \right)$$ You cannot extract explicitely $u$ from this equation except writing $$u=\text{InverseFunction}[\text{EllipticE},1,2]\left[\frac{L}{b r},1-\frac{a^2}{b^2}\right]$$ which is not the most convenient.

However, for cases, what you could do is to use parametric splines $x=f(L)$ and $y=g(L)$

Edit (six years later)

If you are optimistic and want to solve for $u$ the equation $$L=b\,r\, E\left(u\left|1-\frac{a^2}{b^2}\right. \right)$$ let $$A=\frac L{b\,r} \qquad \text{and}\qquad k=1-\frac{a^2}{b^2}$$ and use the series expansion $$E(u|k)=u-\frac{k}{6}u^3+\frac{(4-3 k) k }{120} u^5+O\left(u^7\right)$$ and power series reversion would give $$u=A+\frac{ k}{6}A^3+\frac{k (13 k-4)}{120} A^5 +O\left(A^7\right)$$

Claude Leibovici
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