Answer
$r(t)= 4 \cos t \space i+4 \sin t \space j$ and $0 \leq t \leq 2 \pi$
Work Step by Step
Re-arrange the given equation as follows: $\dfrac{x^2}{16}+\dfrac{y^2}{16}=1$ ....(1)
Use Trigonometric identity such as: $\cos^2 t+\sin^2 t=1$ ...(2)
On comparing the both above equations (1) and (2), we have: $\cos^2 t=\dfrac{x^2}{16} \implies x=4 \cos t$
and $\sin^2 t=\dfrac{y^2}{16} \implies y =4 \sin t$
Therefore, $r(t)= 4 \cos t \space i+4 \sin t \space j$ and $0 \leq t \leq 2 \pi$
