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