how can I parameterize on a 3 dimensional Cartesian plane a thick helix?

Question

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"Function on a 3 dimensional Cartesian plane that represents a helix?"

I was wondering though how to parameterize a helix with some thickness to it, like a torus but spiraling up and down like a helix

Answer 1

Let $$ H(t) = (\cos t, \sin t, ct) $$ That's your basic helix, with "slope" $c$. Its tangent vector is $$ v(t) = (-\sin t , \cos t, c) $$ which happens to be a constant length. That means that its derivative is perpendicular to it, so we can write $$ n(t) = (-\cos t, -\sin t, 0) $$ to get a vector perpendicular to $H(t)$. Geometrically, this just points from the helix towards its axis. Now by taking $v \times n$, we can get a vector perpendicular to both: $$ w(t) = (c\sin t, -c\cos t, 1). $$ (Small confession: I didn't do the cross product, so this might be some scalar multiple of the actual result.)

The length of $w$ is $$ \|w\|^2 = \sqrt{1 + c^2} $$ so we let $$ u(t) = \frac{1}{\sqrt{1 + c^2}} (c \sin t, -c \cos t, 1) $$ which is a nice unit vector like $n(t)$.

Now we can define $$ K(s, t) = H(t) + r\cos(s) n(t) + r\sin(s) u(t) $$ and the graph of $K$ (as $s$ ranges from $0$ to $2\pi$, and $t$ ranges over whatever part of the real line you like) will be a tubular helix; it'll look best for a small value of $r$ like $\frac{1}{10}$, because $r$ is the radius of the "tube", and our base helix has a radius of $1$.