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 $x^2+y^2=1$
Write down the parametric equations of a circle of radius $r$.
$x=r \cos t ; y =r \sin t$
We need to write the parametric equations of a circle with radius $1$.
$x=(1) \cos t ; y =(1) \sin t$
Further, $z=x^2-y^2$
$\implies z=\cos^2 t -\sin^2 t=\cos (2t)$
Thus, $r(t)=\cos ti+\sin t j+\cos 2t k$ ; $0 \leq t \leq 2 \pi$
