I have the following parametrization in cylindrical coordinates:
$$R = 2, \quad \theta = \sin(t^2), \quad z = 2 \cos(t^2)$$
I know that it is the intersection of two perpendicular cylinders $x^2 + z^2 = 4$ and $x^2 + y^2 = 4$. But that is because I've plotted it. What I wish to know is how to identify the parametrization, basically demonstrate that the parametrization is the intersection. I've tried going to standard basis but it gets very complicated.
On the left is the curve here. On the right is half the intersection of the two cylinders $x^2 + y^2 = 4$ and $x^2 + z^2 = 4$ (the other half is another circle of the same radius, but slanted 90 degrees to this one - I don't know how to convince Wolfram Alpha to plot both). As you can see, they are not the same curve.
Of course, since $R = 2$ always, this curve does lie on the cylinder $x^2 + y^2 = 4$. But if you project the other curve onto the $yz$ plane, you get this:
Note that while the vertical direction extends to $\pm 2$, the horizontal direction only extends to $\pm 2\sin 1 \approx 1.68$
If we convert the parametric equations from cylindrical to cartesian coordinates, we get $$(x,y,z) = (2\cos(\sin t^2), 2\sin (\sin t^2), 2\cos t^2)$$
If we project onto the $xy$ plane, we get $$(x,y) = (2\cos(\sin t^2), 2\sin (\sin t^2)) = 2(\cos \theta, \sin \theta)$$where $\theta = \sin t^2$. Note that the points $2(\cos \theta, \sin \theta)$ lie on the circle $x^2 + y^2 = 4$. But the angle $\theta$ is restricted to the range $-1$ to $1$ radians, since it is equal to the sine of another value. So this projection is one one side of the cylinder.
The projection onto the $xz$ plane is $(x,z) = (2\cos(\sin t^2),2\cos t^2)$ This shape looks similar to a parabola, but is not quite one. In addition to being somewhat mishappen, the values of $x$ are limited to $0$ to $2\sin 1 \approx = 1.68$.
So while your curve looks kind of like a circle drawn on a cylinder, it is not one.