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How would I parameterise this curve in 3D? I am confused since the diagrams deal with three variables in total – should I use complex numbers? I'm only used to two diagrams and haven't encountered a problem with three like this.

nomad609
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2 Answers2

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The projection in the $xy$-plane looks like an Archimedean spiral. Except that everything in the $y$-direction is doubled. The formulas $$ x=t \cos t,\qquad y=2t\sin t $$ match with the first figure perfectly. The given points correspond to $t=9\pi/2$, $t=5\pi$, $t=11\pi/2$ and $t=6\pi$ - all in the third revolution $t\in[4\pi,6\pi]$.

The two latter figures give the impression that the curve lies on the elliptical cone $$ (3x)^2+(3y/2)^2=(\pi z)^2. $$ Plugging in the first two equations gives $$ (3x)^2+(3y/2)^2=9t^2(\cos^2t+\sin^2t)=9t^2, $$ so we can solve that $z=3t/\pi$.

Here's a 3D view of that parametrized curve by Mathematica

enter image description here

together with a view from the side

enter image description here

How to? The Archimedean spirals (as well as the logarithmic spirals) occur in all books about polar coordinates. I had a bit of luck spotting that the four points you have fit on an Archimedean spiral, if you stretch it by a factor of two in the direction of $y$-axis. The latter two sketches give that we should always have $z\ge0, |x|\le \pi z/3, |y|\le2\pi z/3$, also implying that everything in the $y$-direction is stretched by a factor of two. Going from these data points to a curve on a cone is just 3D-imagination. Anyway, here is the curve $$ \left\{\begin{array}{cl}x&=t\cos t,\\ y&=2t\sin t,\\ z&=3t/\pi\end{array}\right. $$ one more time together with the surface of the above elliptical cone.

enter image description here

Jyrki Lahtonen
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  • Oops! I had accidentally halved some of the values of the parameter $t$ that match with the given points of the first diagram. Sorry about that. Should be fixed now. – Jyrki Lahtonen Oct 20 '16 at 16:14
  • Wow thank you, amazing work, Sorry about the error before, and thank you for providing such clear work. – nomad609 Oct 21 '16 at 07:52
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Hint:

in $xy$ plane, you use polar coordinates as follows.

$x=r\cos(t)$ and $y=r\sin(t)$ with

$r=ae^{-bt}$ and you choose the right parameters $a,b.$

to get $z$ , you replace $x$.

hamam_Abdallah
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  • A logarithmic spiral does come to mind from the first figure. But you need to take into account that it is bulging out in the direction of the $y$-axis, so this doesn't quite fit. – Jyrki Lahtonen Oct 20 '16 at 11:33
  • It remains the basis idea. – hamam_Abdallah Oct 20 '16 at 11:35
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    On second thought a logarithmic spiral does not fit at all. It never reaches the origin. Only approaches it. Use an Archimedean spiral instead. – Jyrki Lahtonen Oct 20 '16 at 11:36
  • try $ae^{-bt}+ce^{-dt}$ since you have $4$ conditions to satisfy. it goes to the origin. – hamam_Abdallah Oct 20 '16 at 11:39
  • @AbdallahHammam I have tried this, thank you for posting - although say I were to take the coordinates (6pi,0) then I have: 6pi = r cos(t) and 0 =r sin(t). Which naturally follows that tan(t)=0, hence arctan of both sides results in t=0? Which does not really folllow. Similarly if I let the coordinates be (0,9pi) then I have the equations: 9pi=rsin(t) [1] and 0=rcos(t) [2] and when I do [1] / [2] it isn't possible since 9pi/0 does not exist :l – nomad609 Oct 20 '16 at 11:54