Answer
$r(t)=\cos ti+\sin t j+\cos 2t k$ ; $0 \leq t \leq 2 \pi$
Work Step by Step
The parametric equations of a circle whose radius is $r$ are given as follows: $x=r \cos t ; y =r \sin t$
From the given problem, we have
$z=x^2-y^2$ and $x^2+y^2=1$
Now, we have the parametric equations of a circle of radius $1$ are as follows:
$x=\cos t ; y = \sin t$
and $z=x^2-y^2=\cos^2 t -\sin^2 t=\cos (2t)$
Hence, we have the parametric equation in the vector form as follows:
$r(t)=\cos ti+\sin t j+\cos 2t k$ ; $0 \leq t \leq 2 \pi$
