What's the equation of this parametric surface? disclaimer: my math is sketchy at best AND english is not my first language, so... i might have some issues naming things - but i'll try my best to be clear :)
given this parametric curve:

see it on wolfram alpha.
anyway, its equation is (sorry if wrong notation):
$$
\begin{bmatrix}
x \\ y \\ z
\end{bmatrix}
=
\begin{bmatrix}
3\cos(t) + \cos(t)\cos(6t) \\
3\sin(t) + \sin(t)\cos(6t) \\
\sin(6t)
\end{bmatrix}
,t=0...2\pi
$$
now, if a circle goes along that curve, and it stays perpendicular to the curve as it goes along, it makes... what? how would you call that thing? maybe a "twisted torus"?
and the question is: what is the parametric equation of the surface of that "twisted torus"?
i hope this question makes sense – it does, in my mind. anyway, thanks for your time!
 A: Let's call your things $x(t), y(t), z(t)$, OK? 
Let 
$$
v(t)  = \begin{bmatrix}
x'(t) \\
y'(t) \\
z'(t)
\end{bmatrix} \\
T(t) = v(t) / \| v(t) \|. 
$$
Then $T$ will be tangent to your curve at each time $t$. 
Do the same thing with $x'', y'', z''$ to get 
$$
w(t)  = \begin{bmatrix}
x''(t) \\
y''(t) \\
z''(t)
\end{bmatrix} \\
u(t) = w(t) - ( u(t) \cdot T(t) ) T(t) \\
N(T) = u(t) / \| u(t) \|. 
$$
Then $N(t)$ will be perpendicular to $T(t)$ at each point. Finally, let 
$$
B(t) = T(t) \times N(t).
$$
Now: 
$$
S(s, t) = T(t) + r \sin(s) N(t) + r \cos(s) B(t)
$$
will, as $s$ ranges from $0$ to $2\pi$, and $t$ ranges over its usual range, and for small enough values of $r$ (like $r = 0.1$) sweep out a tube around your curve. 
What I've done is construct for you the Frenet-Serret frame for the curve, which assumes that the curvature (the length of the vector $u(t)$) is never zero; if it is, you have to use the "Bishop frame", which ... takes more work to write out. I think that for your curve, the F-S frame will work fine. 
