Answer
$\begin{cases}
x=2\cos \dfrac{2\pi}{3}t\\
y=3\sin \dfrac{2\pi}{3}t
\end{cases}$
with $0\leq t\leq 3$
Work Step by Step
We are given the ellipse:
$\dfrac{x^2}{4}+\dfrac{y^2}{9}=1$
Graph the ellipse:
When $t=0$, the motion begins from $(2,0)$; therefore for $t=0$, we have:
$x=2$
$y=0$
Therefore a parametric set of equations for the ellipse is:
$\begin{cases}
x=2\cos (\omega t)\\
y=3\sin (\omega t)
\end{cases}$
When $t$ increases, $x$ decreases and $y$ increases; therefore $\omega>0$.
Determine $\omega$ using the revolution:
$\dfrac{2\pi}{|\omega|}=3$
$\omega=\dfrac{2\pi}{3}$
The parametric set of equations is:
$\begin{cases}
x=2\cos \dfrac{2\pi}{3}t\\
y=3\sin \dfrac{2\pi}{3}t
\end{cases}$
with $0\leq t\leq 3$
