Gradient of a parametric form I want to find the gradient of a parametric form. So say you have the form
$$(x,y,z) = (f(u,v),g(u,v),h(u,v))$$ and now I want to find and the gradient in parametric form. How do I do that?
The background is that I want to plot certain geometric objects given in parametric form but then I also want to plot the gradient field which I need in a parametric description aswell.
 A: $\newcommand{\dd}{\partial}\newcommand{\vec}{\mathbf{x}}$Not sure I understand exactly, but here are some observations that may help. The two vector-valued partial derivatives
$$
\vec_{u} = \left(\frac{\dd f}{\dd u}, \frac{\dd g}{\dd u}, \frac{\dd h}{\dd u}\right),\quad
\vec_{v} = \left(\frac{\dd f}{\dd v}, \frac{\dd g}{\dd v}, \frac{\dd h}{\dd v}\right)
$$
are tangent to the surface, along the curves $v = \text{const.}$ and $u = \text{const.}$, respectively. Their cross product $\vec_{u} \times \vec_{v}$ is orthogonal to the surface.
For your "twisted-parabola" surfaces,
$$
(x, y, z) = \bigl(u\cos(kv)−(u^{2}−a)\sin(kv), u\sin(kv)+(u^{2}−a)\cos(kv), v\bigr),
$$
this cross product is everywhere non-zero. In fact, $\vec_{u}$ is "horizontal" at each point (i.e., parallel to the $(x, y)$-plane), and so is tangent to the parabola that sweep out the surface, while $\vec_{v}$ is nowhere horizontal.
From your first comment, perhaps you want the plane through the origin spanned by $\vec_{u}$ and $\mathbf{e}_{3} = (0, 0, 1)$.
