*doesn't*stay in the same plane. Obviously, polar curves are out of the question, since they are, well, in the plane. But what's one dimension up from polar coordinates?

*Spherical*coordinates $(r,\varphi,\theta)$, of course (or if you're a physicist, $(r,\theta,\phi)$).

^{1 }We fall back on a good ol' tool, parametrization.

To discover what its equation might be, we first consider an Archimedean spiral ($r = a\varphi$) in the $r\varphi$-plane (or $yz$-plane), or possibly a logarithmic one, like $e^{\varphi/3}$, which grows as it goes around (like some spirals last time) the "pole" which is now the $x$-axis. This ends up growing as you move clockwise, rather than counterclockwise, since $\varphi$ increases as it goes from north toward east.

This is most efficiently represented in a parametric manner: first, in the $yz$-plane:\[

x=0,

\] \[

y = at \sin t,

\] \[

z = at \cos t.

\]

gives the parametric equation of the Archimedean spiral in the $yz$-plane going clockwise. This gives $\varphi = t$ and $r = at$, which recovers the original polar equation $r = a\varphi$.

Archimedean Spiral in the $yz$-plane, with $a = 1/20$. |

How do we give it a twist? We reason that, after each revolution, we want it twisted out of the plane by some small amount (say, rotated about the $z$-axis by $b$ times a full revolution): we apply the rotation to the $x$ and $y$, giving:\[

x=at\sin t \cos (bt),

\] \[

y = at \sin t \sin(bt),

\] \[

z = at \cos t.

\]

Twisting out of the plane with parameter $b = 1/20$, also. |

Here's the original one I delivered to Yitz.. (it is actually exponential):

Pretty

*kewel*, right? (that was his exact word to describe it).^{1 }I always use the convention that if I indeed need to refer to a physics text, do calculations to be consistent... but I always write my $\phi$'s as non-curly in that context... I haven't yet figured how to draw my $\theta$s different though. Yes, I know there is $\vartheta$, but it hasn't caught on in my notation. It just doesn't feel enough like a $\theta$.

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