Answer
$\lt 2t+1,2t-1,3t^2 \gt$, $\lt 2,2,6t \gt$ , $\sqrt {9t^4+8t^2+2}$
Work Step by Step
Given: $r(t)=\lt t^2+t,t^2-t,t^3 \gt$
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,
$v(t)=r'(t)=\lt 2t+1,2t-1,3t^2 \gt$
$a(t)=v'(t)=\lt 2,2,6t \gt$
$s(t)=|v(t)|=\sqrt {(2t+1)^2+(2t-1)^2+(3t^2)^2}=\sqrt {9t^4+8t^2+2}$
Hence, the required answers are:
$\lt 2t+1,2t-1,3t^2 \gt$, $\lt 2,2,6t \gt$ , $\sqrt {9t^4+8t^2+2}$
