How to explain why a curve is on a cylindrical surface?

Question

The question may be a bit general but I'm unsure about how to define it.

I have a curve: $\vec r(t) = (2\cos(t),2\sin(t), 2t)$, for $0\le t \le 2\pi$,

The problem I'm trying to pose :

"Show that the given space curve lies on a cylindrical surface"

What I've tried to say is something like (due to $\cos(t)$ and $\sin(t)$): We rotate the curve about $z$ axis and move it up (proportionate to rotation t ) along $z$ axis to create this cylindrical surface.

Answer 1

Comment converted to answer, as requested.

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If you wanted to show that the curve lay in a(n origin-centered) sphere, you'd just have to show that the parameterizations of the $x$, $y$, $z$ coordinates satisfied the sphere equation $x^2+y^2+z^2=k^2$ for some $k$.

For your problem, just realize that the equation for a $z$-axis-aligned cylinder is $x^2+y^2=k^2$ for some $k$, and show that your parameterizations satisfy that relation.

Answer 2

The surface lies on a cylinder because x^2+y^2 =4, irrespective of z(t)!

Please note, z can be ANY function of t, meaning.. you can write any curved line on your cylinder.