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 whose radius is $r$ are given as follows: $x=r \cos t ; y =r \sin t$
From the given problem, we have
$z=\sqrt {x^2+y^2}$ and $z=1+y$
Now, $z=\sqrt {x^2+y^2} \implies x^2+y^2=1+2y+y^2$
This yields: $y=\dfrac{x^2-1}{2}$
Plug $x=t$, Now, we get $y=\dfrac{t^2-1}{2}$
This implies that $z=\dfrac{t^2-1}{2}+1=\dfrac{t^2+1}{2}$
Hence, we have the parametric equation in the vector form as follows;
$r(t)=ti+(\dfrac{t^2-1}{2})j+(\dfrac{t^2+1}{2})k$
