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
Given: $r(t)=e^t(\cos t i+\sin t j+t k)$
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,
$v(t)=r'(t)=e^t[(\cos t -\sin t)i+(\sin t+\cos t) j+(t+1) k])$
$a(t)=v'(t)=e^t[-2\sin t i+2\cos t j+(t+2) k])$
$s(t)=|v(t)|=e^t \sqrt {t^2+2t+3}$
Hence, the required answers are:
$e^t[(\cos t -\sin t)i+(\sin t+\cos t) j+(t+1) k])$, $e^t[-2\sin t i+2\cos t j+(t+2) k])$, $e^t \sqrt {t^2+2t+3}$
