Answer
$\begin{cases}
x=2\sin (2\pi t)\\
y=3\cos (2\pi t)
\end{cases}$
with $0\leq t\leq 1$
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 $(0,3)$; therefore, for $t=0$ we have:
$x=0$
$y=3$
Therefore a parametric set of equations for the ellipse is:
$\begin{cases}
x=2\sin (\omega t)\\
y=3\cos (\omega t)
\end{cases}$
When $t$ increases, $x$ increases and $y$ decreases, therefore $\omega>0$.
Determine $\omega$ using the revolution:
$\dfrac{2\pi}{|\omega|}=1$
$\omega=2\pi$
The parametric set of equations is:
$\begin{cases}
x=2\sin (2\pi t)\\
y=3\cos (2\pi t)
\end{cases}$
with $0\leq t\leq 1$
