With reference to the following image:

enter image description here

the blue curve has trivially a parametrization:

$$(x, y, z) = (\cos\theta, \, \sin\theta, \, 0) \; \; \; \text{with} \; \theta \in [0,\,2\pi)$$

I would like to determine the parametric equations of the red curve, very badly drawn in Paint, where I mean a sinusoidal curve along the blue circumference.

Although I thought about it a lot, I still couldn't figure out how to derive these parametric equation. Any ideas?

  • 3
    $\begingroup$ It looks like a curve of the form $$ r(\theta) = r_0 + A \cos{(\omega \theta )} $$ But it's not very clear what you mean with the hand-drawn curve. Is it supposed to come out of the $z=0$-plane? $\endgroup$
    – Matti P.
    Jan 2, 2019 at 12:13

1 Answer 1


You can try$$\theta\mapsto\left(\cos\theta,\sin\theta,\frac{\cos(8\theta)}8\right),$$for instance.

enter image description here

  • 3
    $\begingroup$ The $z$ coordinate had to be a waving line again, and so I thought about $\cos(8\theta)$, but then the waves would go too high and too low. That's why I divided by $8$. $\endgroup$ Jan 2, 2019 at 12:45
  • 3
    $\begingroup$ No need for that. Just consider:$$\theta\mapsto\left(\cos\theta+\frac{\cos(8\theta)}8,\sin\theta,0\right).$$ $\endgroup$ Jan 2, 2019 at 13:02
  • 1
    $\begingroup$ @TeM If you want the red curve in the Z=0 plane, then you probably want to oscillate in the direction of the unit normal vector rather than in the x direction. Here's a comparison: math3d.org/cCvATTaY $\endgroup$ Jan 2, 2019 at 19:00

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