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See Lowermost Image For A Higher Quality Rendering of The Wave Whose Function I'm trying to Find.

Important Note. I found the best fit thus far; but, it seems arbitrary or maybe I just don't understand why it works (see image above the lowest image!). (It's also not as good a fit as it could be.) It might be helpful for those answering to look at it and or try and explain / think about what's going on....

One thing that must be kept in mind is that a sine wave can be regarded as a helix viewed from the side: Viewed head on, my sine wave would be a special type of golden spiral.

If people wouldn't mind, I'd like to address this parametrically.....

(See images; the sine wave I'm trying to find a function for is in blue, and my attempts to find functions in desmos are in red.)

I've been working to find the function for creating this sine wave over a long length of time. Unfortunately, I can't seem to arrive at an accurate function. I've found functions that correspond to parts of it closely, after scaling (my image); but I just can't seem to make any progress with regard to the whole thing. I'd really be thankful for some help from someone more skilled / with access to technology capable of solving this problem. Thank you all for your time!

A few points:

  1. The wave uses the golden ratio, as you can see from my own work (but exactly how, in full, is part of the mystery).

  2. I suspect that the wave graphed in my first image (the uppermost one), is an accurate depiction of the sine wave's base. When I say 'base,' I mean where it cuts off if you plug in only ≤t or t≤ (exclusively appropriate plus or minus values), depending on how you write the equations. (Sine waves of this type always have such a cut off point.) This info could be used to understand scale, intercepts, and many other things.

I'm sure someone will have fun working on this; I certainly did for some time, but feel that I've reached my limit and would really like a true an correct answer!

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    $\begingroup$ seems like a $x=\sin(y)$ situation $\endgroup$ – Aops Vol. 2 Oct 6 '19 at 19:17
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    $\begingroup$ could it be a transformation of $x=\sin(y^2)$ $\endgroup$ – Aops Vol. 2 Oct 6 '19 at 19:19
  • $\begingroup$ @AopsVol.2, Thanks for the comment. The golden ratio is also meant to be involved. Do you know how that might fit in? $\endgroup$ – Jinny Ecckle Oct 6 '19 at 19:21
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    $\begingroup$ @JinnyEcckle No, what I'm getting at is that you seem to have a fundamental misunderstanding of how mathematics works. If you just draw a curve on a piece of paper, there is no "correct" formula for it. Of course you can guess at a formula and tease and coax until it fits the curve fairly well, but there is no one true formula - that isn't math, it's computer assisted drawing. (contd.) $\endgroup$ – Jack M Oct 7 '19 at 16:50
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    $\begingroup$ ...in your case, it seems likely that the hyperbola based answer you gave is the "correct" formula, in the sense that it might be what the original person who drew the curve used. If we're happy to make that assumption, then we can pose an actual math problem: which hyperbola provides the best fit to your drawing? $\endgroup$ – Jack M Oct 7 '19 at 16:51
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To me, it seems what you want is parametric equations for the projection onto the $y$$z$ plane of the curve on the surface of revolution$rz=1$ whose projection onto the $x$$y$ plane is some clockwise logarithmic spiral

$$\begin{align} \theta &=\tfrac{\pi}{2}t & r &= \phi^{-(at+b)}\text{.} \end{align}$$

If that's the case, then the equations must be of the form

$$\begin{align} y(t)&=\phi^{-(at+b)}\sin \tfrac{\pi t}{2}\text{,} & z(t)&=\phi^{at+b} \end{align}$$ up to reparametrization, e.g.,

$$\begin{align} y(t)&=\phi^{-t/2}\sin \tfrac{\pi t}{2}\text{,} & z(t)&=\phi^{t/2}\text{,} & t&\in [0,10]\text{:} \end{align}$$ Example parametrized curve

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  • $\begingroup$ Great work and answer. The issue is, I already know how to make spirals of this form (see my edited 'answer.') Although I used a different thought process to get there. My issue is in finding a spiral like this that fits my curve itself. Thank you! $\endgroup$ – Jinny Ecckle Oct 7 '19 at 17:09
  • $\begingroup$ I still think your answer will be very helpful, as it will give me the chance to think about the problem in a new way! $\endgroup$ – Jinny Ecckle Oct 7 '19 at 17:29
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This isn't an answer, per se, but something that may help: This sine wave is meant to fit perfectly on a correctly scaled hyperbola / moved. All sine waves of the form given will do this.

Note. In this case, however, I scaled / moved my image, not the hyperbola.

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    $\begingroup$ Im sorry, I can think of anything. Ill share this question so hopefully someone will. My initial idea was since the frequency and the amplitude both increase over time, to put the function as $$f(y)=g(y)\sin(\frac{2\pi}{h(y)})$$ and set $g(y)$ and $h(y)$ and increasing function (and thus increasing the period/amplitude). Sorry, I wasn't of any help. $\endgroup$ – Aops Vol. 2 Oct 6 '19 at 19:48
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    $\begingroup$ @AopsVol.2, No you're great! I'm just thankful that you took the time! I'm really excited to find the answer; I'm sure we'll get there! $\endgroup$ – Jinny Ecckle Oct 6 '19 at 19:50

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