Answer
$\lt -2 \sin t,3, 2 \cos t\gt$, $\lt -2 \cos t, 0, -2 \sin t\gt$ , $\sqrt {13}$
Work Step by Step
Given: $r(t)=\lt 2 \cos t,3t, 2 \sin t \gt$
Our aim is to calculate the velocity vector, acceleration vector and speed.
In order to calculate the all above terms we will use formulas, such as:
$v(t)=r'(t)$ and $a(t)=v'(t)$ and speed is the magnitude of the velocity vector, that is $s(t)=|v(t)|$.
Now,
$v(t)=r'(t)=\lt -2 \sin t,3, 2 \cos t\gt$
$a(t)=v'(t)=\lt -2 \cos t, 0, -2 \sin t\gt$
$s(t)=|v(t)|=\sqrt {(-2 \sin t)^2+(3)^2+(2 \cos t)^2}=\sqrt {13}$
Hence, the required answers are:
$\lt -2 \sin t,3, 2 \cos t\gt$, $\lt -2 \cos t, 0, -2 \sin t\gt$ , $\sqrt {13}$
