<s>[INST] $x(t)= a\cos(t)$ , $y(t)= b\sin(t)$ in terms of the arc length $S$ I'm trying to parameterize the ellipse $x(t)= a\cos(t)$ , $y(t)= b\sin(t)$ in terms of the arc length  $S$ but I don't know how to do it.
Supposing that $\gamma:[a,b]\to \mathbb{R}$ is a smooth curve  with $\gamma'(t)\neq 0$ for $t\in [a,b]$ , I know that $s(t)$= $\int_{a}^{t}\left\|  \gamma'(\psi)\right\|d\psi$ for $t\in [a,b]$ then I find the inverse funtion of $s$.
Can anybody help me find a way to express the parameterization for the ellipse? I’m looking for a solution in terms of sine amplitude and cosine amplitude.
 [/INST] There is no standard closed form but series solutions only

*

*Local canonical form starting from minor axis (clockwise convention):

\begin{align}
  k &= \sqrt{1-\frac{b^2}{a^2}} \\
  s(t) &= \int_0^t \sqrt{a^2\cos^2 \theta+b^2\sin^2 \theta} \, d\theta \\
  &= aE(t,k) \\
  t &= E^{-1} \left( \frac{s}{a},k \right) \\
  \begin{pmatrix}  x \\ y  \end{pmatrix} &=
  \begin{pmatrix}  a\sin t \\ b\cos t  \end{pmatrix} \\
  &=
  \begin{pmatrix}
    s-\frac{b^2 s^3}{6a^4}+\frac{b^2(13b^2-12a^2)s^5}{120a^8}+\ldots \\
    b-\frac{b s^2}{2a^2}+\frac{b(4b^2-3a^2)s^4}{24a^6}+\ldots
  \end{pmatrix} \\
  \kappa &= -\frac{b}{a^2}+\frac{3b(b^2-a^2)s^2}{2a^6}+
  \frac{b(b^2-a^2)(15a^2-19b^2)s^4}{8a^{10}}+\ldots \\
\end{align}

*

*Local canonical form starting from major axis (anti-clockwise convention):

\begin{align}
  k' &= \frac{b}{a} \\
  s(u) &= \int_0^u \sqrt{a^2\sin^2 \theta+b^2\cos^2 \theta} \, d\theta \\
  &= bE\left( u,\frac{ik}{k'} \right) \\
  u &= E^{-1}\left( \frac{s}{b},\frac{ik}{k'} \right) \\
  \begin{pmatrix}  x \\ y  \end{pmatrix} &=
  \begin{pmatrix}  a\cos u \\ b\sin u  \end{pmatrix} \\
  &=
  \begin{pmatrix}
    a-\frac{a s^2}{2b^2}+\frac{a(4a^2-3b^2)s^4}{24b^6}+\ldots\\
    s-\frac{a^2 s^3}{6b^4}+\frac{a^2(13a^2-12b^2)s^5}{120b^8}+\ldots
  \end{pmatrix} \\
  \kappa &= \frac{a}{b^2}+\frac{3a(b^2-a^2)s^2}{2b^6}+
  \frac{a(a^2-b^2)(19a^2-15b^2)s^4}{8b^{10}}+\ldots \\
\end{align}

*

*Plots of the partial sums for $s\in [0,3]$:



*

*Fourier series for eccentric angle:

\begin{align}
  ds &= a\sqrt{1-k^2\sin^2 t} \, dt \\
  t &= \frac{1}{2aE(k)} \sum_{n=1}^{\infty}
  \sin \frac{n\pi s}{2aE(k)} \int_{-2aE}^{2aE} E^{-1}
  \left( \frac{s'}{a}, k \right) \sin \frac{n\pi s'}{2aE(k)} ds' \\
  &= \sum_{n=1}^{\infty} \sin \frac{n\pi s}{2aE(k)} \int_{-\pi}^{\pi} 
  \underbrace{
    \frac{\tau \sqrt{1-k^2\sin^2 \tau}}{2E(k)}
    \sin \frac{n\pi E(\tau,k)}{2E(k)}
  }_{\text{expand as Maclaurin series in }k} \, d\tau \\
  &= \frac{\pi s}{2aE(k)}-
  \left( \frac{k^2}{8}+\frac{k^4}{16}+\ldots \right)
  \sin \frac{\pi s}{aE(k)}+\left( \frac{5k^4}{256}+\ldots \right)
  \sin \frac{2\pi s}{aE(k)}-\ldots \\
\end{align}

*

*The first term comes from sawtooth waveform:

$$\frac{\pi s}{2aE(k)}=\phi=2
\left(
  \sin \phi-\frac{\sin 2\phi}{2}+\frac{\sin 3\phi}{3}-\ldots
\right)$$

*

*Compare with the Jacobi amplitude here.


*See also the anti-clockwise version in my older post here.
</s>