## Thomas' Calculus 13th Edition

$9x^2+y^2=9$ or, $y=3 \sqrt {1-x^2}$; $v(0)=6j$ and $a(0)=-4i$
We know that the velocity and acceleration is defined as: velocity is given as: $v(t)=r'(t)$ and acceleration is given as: $a(t)=v'(t)$ In the given problem, we have $x=\cos t; y=3 \sin 2t$ Re-write as follows: $9x^2+y^2=9 \\ \implies y=3 \sqrt {1-x^2}$ $v(t)=r'(t)=-2 \sin 2t+6 \cos 2t \\ \implies v(0)=6j$ and $a(t)=v'(t)=-4 \cos 2t-12 \sin 2t\\ \implies a(0)=-4i$