I want to describe a parabolic trough of the form $z=x^2$ and give it a twist, like a torsion in $y$ direction. Does anybody know how I can do that?
Imagine this is the trough and the $z$ direction would be my $y$ direction. Thats the best image I could find to make it clear. It doesn't matter what direction the twist is in, really as long as it looks like that.
On
If the parabola is to be rotated about its focus, then we must have parabolic sections of the form $$z = ax^2 - \frac{1}{4a}, \quad a \ne 0.$$ Without loss of generality we may suppose $a > 0$, and suppose that for each unit increase in $y$, the parabola is rotated by some angle $\theta$. This yields the parametrization $$\begin{align*} x(u,v) &= u \cos (\theta v) + \left( au^2 - \frac{1}{4a} \right) \sin (\theta v) \\ y(u,v) &= v \\ z(u,v) &= \left( au^2 - \frac{1}{4a}\right) \sin (\theta v) - u \sin (\theta v). \end{align*}$$
The below animation corresponds to varying $\theta \in [0,2]$ for a fixed $a = 0.5$:
The below animation corresponds to varying $a \in [0.05, 2]$ for a fixed $\theta = 1$:
For some reason, now I have a craving for pasta.
If the parabola has equation $y = x^{2} - a$, and is rotated about the origin at angular speed $k$ as the "horizontal" section moves along the $z$-axis, the resulting surface may be given the parametric description \begin{align*} x(u, v) &= u\cos(kv) - (u^{2} - a)\sin(kv), \\ y(u, v) &= u\sin(kv) + (u^{2} - a)\cos(kv), \\ z(u, v) &= v. \end{align*} The plot below shows $a = 1$ and $k = \pi/4$, for $-1 \leq u, v \leq 1$.
(This gives a couple of parameters to play around with, and should suggest how to rotate at non-uniform speed, or have the shape of the section change with height, or....)