Answer
$r(t)=ti+t^2j+4t^2+t^4k$ or, $r(t) =\lt t, t^2, 4t^2+t^4 \gt$
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 $z=4x^2+y^2$ and $y=x^2$
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 $y=x^2=t^2$
Thus, $z=4x^2+y^2=4t^2+(t^2)^2=4t^2+t^4$
Hence, the parametric equation in vector form is:
$r(t)=ti+t^2j+4t^2+t^4k$ or, $r(t) =\lt t, t^2, 4t^2+t^4 \gt$
