## Calculus (3rd Edition)

The space curve given by $$r(t) = \langle t, t^3, t^2+1 \rangle$$ intersects the xy-plane when $z=0$, that is, we have the equation $$t^2+1=0\Longrightarrow t^2=-1$$ which has no solution in $R$. Hence, the curve does not intersect the xy-plane.