Answer
$e^t[(\cos t -\sin t)i+(\sin t+\cos t) j+(t+1) k])$, $e^t[-2\sin t i+2\cos t j+(t+2) k])$, $e^t \sqrt {t^2+2t+3}$
Work Step by Step
As we are given that $r(t)=e^t(\cos t i+\sin t j+t k)$
Need to determine the velocity vector, acceleration vector and speed.
We have
$v(t)=r'(t)$ and $a(t)=v'(t)$ and speed is the magnitude of the velocity vector, that is $s(t)=|v(t)|$.
$v(t)=r'(t)=e^t[(\cos t -\sin t)i+(\sin t+\cos t) j+(t+1) k])$ [ given: $r(t)=e^t(\cos t i+\sin t j+t k)$]
Also, $a(t)=v'(t)=e^t[-2\sin t i+2\cos t j+(t+2) k])$
Thus, $s(t)=|v(t)|=e^t \sqrt {t^2+2t+3}$
