Answer
$r(t)=ti+(\dfrac{t^2-1}{2})j+(\dfrac{t^2+1}{2})k$
Work Step by Step
The parametric equations of a circle having radius $r$ are:
$x=r \cos t ; y =r \sin t$
Here, we have $z=\sqrt {x^2+y^2}$ and $z=1+y$
This gives: $x^2+y^2=1+2y+y^2$
or, $y=\dfrac{x^2-1}{2}$
Let us set $x=t$, then we have $y=\dfrac{t^2-1}{2}$
Thus, $z=\dfrac{t^2-1}{2}+1=\dfrac{t^2+1}{2}$
Hence, the parametric equation in the vector form is:
$r(t)=ti+(\dfrac{t^2-1}{2})j+(\dfrac{t^2+1}{2})k$
