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 whose radius is $r$ are given as follows: $x=r \cos t ; y =r \sin t$
From the given problem, we have
$z=4x^2+y^2$ and $y=x^2$
Also, $x^2+y^2=1+2y+y^2$
This implies that $y=\dfrac{x^2-1}{2}$
When we plug $x=t$, then we get $y=(x)^2=t^2$
This gives: , $z=4x^2+y^2$
or, $z=4t^2+(t^2)^2=4t^2+t^4$
Hence, we have the parametric equation in the vector form as follows:
$r(t)=ti+t^2j+4t^2+t^4k$ ;
or, $r(t) =\lt t, t^2, 4t^2+t^4 \gt$
