Suppose $r=\langle t^3+4t,t,2t^2\rangle$ is a vector function.
I want to find which curve has the following parametrization
$x=t^3+4t$, $y=t$, $z=2t^2$
$x=\frac{yz}{2}+4y$ (Is it Correct?)
even if my parametrization is not correct above equation will represent a surface but vector function $r$ is a curve traced out by the head of $r$ not a surface.
Where I'm wrong?