I have the following vector function $\vec{g}:\mathbb{R} \rightarrow \mathbb{R}^3/\vec{g} (t) = (cos(t),sin(t),3t)$ and I have to find its image and express it in cartesian form. What I did is: $$x=cos(t)$$ $$y=sin(t)$$ $$z=3t$$ Then $$x^2 + y^2=1$$ and $$z=3arcsin(y)$$ So my answer is $Im(\vec{g})=\{(x,y,z)\in\mathbb{R}^3/x^2+y^2=1, z=3arcsin(y) \}$. I think so far I'm correct but I have no idea how to graph such a set, so I'm guessing my solution for $z$ isn't a good one? Intuitively this seems to be some sort of spiral.
Basically my question is whether I should be able to graph this set as presented or is there a better solution that makes the graphing easier?