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I'd for a long time thought of parabolas as semi-circles.

However, if you take half of a circle the ends will - look - parallel, where as parabolas continue to extend horizontally and infinitely.

Are the only similarities then the fact that they're curved?

Are sinusoidal graphs connected semi-circles or parabolas?

If they were connected semi-circles, wouldn't the gradient at $x=0$ be $\infty$ or $\frac10$

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    $\begingroup$ Projectively, circles and parabolas are the basically same (and they are also basically the same as hyperbolas). I'm currently writing an answer. In particular, the parabola $y=x^{2}$ does not extend horizontally. $\endgroup$ – Will R Dec 23 '16 at 20:27
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    $\begingroup$ @WillR Can't wait :D $\endgroup$ – Simply Beautiful Art Dec 23 '16 at 20:31
  • $\begingroup$ Actually, I don't quite have time right this second; I'll come back in a couple of hours and finish what I've started, provided no one else provides a good answer along the same lines. $\endgroup$ – Will R Dec 23 '16 at 20:31
  • $\begingroup$ @WillR :P Guess I'll be waiting then. $\endgroup$ – Simply Beautiful Art Dec 23 '16 at 20:34
  • $\begingroup$ @WillR As will I! $\endgroup$ – Tobi Dec 23 '16 at 20:34
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Notice: I hope the pictures explain themselves and that you enjoy pondering over this post :-)

Circles and parabolas are both second degree equations, meaning that they can be created by cutting open a cone (conic sections): (You could also make these by shining a flashlight)

enter image description here

This is related to eccentricity.

enter image description here

You could imagine eccentricity to be how much these graphs are stretched: ($e=$ eccentricity)

circle: $e=0$

ellipse: $0<e<1$

parabola: $e=1$

hyperbola: $e>1$


Clearly sinusoidal graphs are related to circles (see here for original post):

enter image description here

while probably not what you were thinking of, sinusoidal functions look parabolic locally:

enter image description here

which is due to Taylor's theorem.

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  • $\begingroup$ The image helped me differentiate between the two. $\endgroup$ – Tobi Dec 23 '16 at 20:15
  • $\begingroup$ Am I right in thinking hyperbolas and Ellipses are just transformed circles and parabolas $\endgroup$ – Tobi Dec 23 '16 at 20:16
  • $\begingroup$ @Tobi Sort of? I'd recommend you look into eccentricity. $\endgroup$ – Simply Beautiful Art Dec 23 '16 at 20:19
  • $\begingroup$ A circle with the appropriate center and radius, superimposed on the graph that shows the local similarity of the sine curve and parabola, will also match the sine and parabola closely for small values of $x,$ for the same reason that the other two match closely there. $\endgroup$ – David K Dec 23 '16 at 20:36
  • $\begingroup$ @DavidK Hm, guess I could overlap all of them. $\endgroup$ – Simply Beautiful Art Dec 23 '16 at 20:38
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I should perhaps begin this answer by saying that, in many (often important) ways, parabolas and circles are very different. For example, a particle on a parabolic path will shoot off to infinity, while a particle on a circular (or, generally, elliptical) path will not; if you're flying a spaceship around a planet and you're running out of fuel, you might care about this difference!

Also, sinusoidal graphs are very different to circles: at the roots of $\sin$ we have a gradient of $\pm1$, so the tangent line never goes vertical like it would on a circle.

Aside: This fact about the roots of $\sin$ is the reason why we can make the approximation $\sin{\theta}\approx\theta$ for $\theta\approx0.$ For exactly the same reason we see, for example, that $\cos{\theta}\approx\pi/2-\theta$ for $\theta\approx\pi/2.$

Sinusoidal graphs are also different to parabolas, but I haven't (yet) managed to come up with so easy a way to see the difference. At worst you can just try to figure out what the correct parabola would be and deduce that no such parabola exists.


I mentioned projective geometry in the comments and I'd like to stay true to my word. As I said, I don't want to get bogged down in technical details, but the answer is a bit long nonetheless.

The projective geometry is the proper setting for talking about "points at infinity." You may or may not have heard that "parallel lines meet at infinity" (if you haven't, don't worry). This statement makes no sense in ordinary geometry, but (plane) projective geometry is specially designed and set up so that we can make such statements precise. In the projective plane $\mathbb{RP}^{2}$, there is a line of points "at infinity," and, just like there is always a unique line going through two given points, there is also always a unique point lying on two given lines: lines in general position meet at one point, just like in the affine plane $\mathbb{R}^{2}$, while parallel lines meet at infinity.

So far, what I've said about the projective plane is just hand-wavy algebra. But projective geometry really stems from trying to draw the world around us: we see the world in perspective, resulting in certain projective features in our vision.

For example, one can imagine that if we lived on a perfectly flat plane that extended infinitely in all directions, then the railroad tracks in this image would appear to meet at the horizon (they don't in the image due to the curvature of the Earth, but a plane has curvature zero, thereby removing that problem). But they're railroad tracks, so they must be parallel - parallel lines that "meet somewhere". The reason they would appear to meet at the horizon is that we would be seeing the projective plane around us, wherein the horizon line would be the line at infinity. This at least makes some intuitive sense: if you're standing on an infinite plane, and you still see a horizon (which you must do if, say, the ground is green and the sky is blue), then the horizon must be infinitely far away: the line at infinity. Moreover, the railroad tracks are parallel lines, so they must meet at a unique point on the line at infinity, i.e., at some point on the horizon.

Now, we can put coordinates on our projective plane: put coordinates on the usual Cartesian (or affine) plane and then imagine that you're standing at $(0,-1),$ looking straight up the $y$-axis. We get something like this:

enter image description here

Hopefully you can fill in the rest of the coordinate system with your imagination. Note that the "vertical" lines $x=7,6,5,\ldots,-5,-6,-7$ are parallel, and hence all meet at the same point on the horizon.

We can sketch the parabola $y=x^{2}$ on such a coordinate system. Here's what we get:

enter image description here

At this point (if my freehand drawing skills are good enough), you might be thinking "Will, that's not a parabola, that's an ellipse" (and if you aren't thinking that, then I need to improve). Indeed, I have more-or-less drawn an ellipse, but you even know in advance that it can't be an ellipse in the usual sense, because it meets the line at infinity.

So you see that, projectively, parabolas and ellipses are at least related, even if you aren't convinced they're the same. With an additional helping of mathematical formalism (which would take a good few hours of lectures to properly explain) one can make everything I've said so far completely rigorous, work in "homogeneous coordinates" and actually write down "projective changes of coordinates", that keep the underlying geometric objects the same (at least, projectively the same) but transform the corresponding polynomial equations into each other, and it turns out that there is indeed a projective change of coordinates that transforms $x^{2}+y^{2}=1$ into $y=x^{2}.$

So, from a projective standpoint, circular and parabolic paths aren't all that different. If you're travelling along a parabola in the projective plane, and if you experience time projectively as well, i.e., that your path is parametrized by $t\in\mathbb{RP}^{1}$ (the projective line with its own "point at infinity"), then you will "eventually" get back to where you started, just as if you were travelling along a circle the whole time (because, projectively, you were).


As a final note, I think it should be mentioned that, of course, projective hyperbolae are also just circles; the difference is that hyperbolae (like $xy=1$) meet the line at infinity in two places, whereas parabolae meet the line at infinity at just one place.

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  • $\begingroup$ @SimpleArt: In case you're still interested in this answer. $\endgroup$ – Will R Dec 29 '16 at 0:20
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The Circles and Parabolas are related as conic sections. If you take a normalized Semicircle and graph it together with a normalized Parabola shifted down by -1, then the Parabola will fit INSIDE the Semicircle within the -1≥x≥1 domain, and have 3 common points with the Semicircle at (0,-1) and (-1,0) and (+1,0). The same can be done with a normalized negative Cosine curve - it will fit inside both the Parabola and Semicircle, within the -1≥x≥1 domain and have the same 3 common points with the other two curves.

The graph of first derivatives (x/dx) of the Semicircle, Parabola and the Cosine curve, depicts the slopes, that these curves have at a given X coordinate with respect to the horizontal X axis. You correctly noticed, that the slope of the Semicircle becomes vertical (90º) at its ends (the x/dx derivative of the Semicircle becomes infinite when Y = 0), however the arms of the Parabola assume the slope angle of ±63º (rounded down) at the same coordinates. The Cosine curve's slope angle is ±45º at these coordinates.

Trivially, all of the slopes of these 3 curves are equal (to 0º) when the X coordinate is 0. Besides this, you can also see, that the slopes of the Semicircle and Parabola become equal (both = ±60º) at the ±(√3/2) X coordinate, which on the Semicircle is ±30º from the the horizontal X axis and the slope of the Cosine curve equals the slope of the Semicircle at the ±0.648138425908612 X coordinate, where both curves assume the slope of approximately ±40.4º (on the semicircle this slope appears approximately ±49.6º from the the horizontal X axis, because 90º-40.4º=49.6º). Calculation of this non-trivial slope equality between the Cosine curve and the Semicircle, can be done ONLY NUMERICALLY, which is an indicator that the Cosine curve is a different type of curve. Besides the trivial case at X=0, the slope of the normalized Cosine curve never equals the slope of the normalized Parabola for the same X coordinate within the -1≥x≥1 domain.

Finally, the curvatures of the Semicircle and Parabola are equal only at their apexes, that is at the (0,-1) point on the graph.

enter image description here

P.S.
It is possible to analyze other curving functions the same way. For example, the normalized Hyperbolic Cosine (shifted down by -2), will have the same 3 points common with the Semicircle, Parabola and the Cosine curve in the -1≥x≥1 domain, but it will fit between the Semicircle and the Parabola (the cosh() matches the Parabola better than -cos()) and the ends of its arms will have a ±60º slope when Y=0. The slope of the Hyperbolic Cosine is equal to the slopes of the Semicircle or the Cosine Curve at some X coordinate, but is never equal to the slope of the Parabola in the -1≥x≥1 domain, besides the trivial case at X=0. I am not including the cosh() on the graph, because the OP did not ask about it.

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Are sinusoidal graphs connected semi-circles or parabolas?

I hope you are not under the impression that these are the only possibilities. There are many different possible curves, infinitely many in fact, parabolas and semi-circles are only 2 examples. The sine curve is neither piecewise parabolical, nor piecewise semicircular. It is a different curve altogether.

The figure below contains a sine function, a parabola and a semicircle, all passing through $(0,0)$, $(\pi/2,1)$ and $(\pi,0)$. You can see that they are indeed distinct.

Half a period of the sine function, a parabola and a semicircle

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  • $\begingroup$ Do sinusoidal graphs intersect the x-axis at 45degrees? $\endgroup$ – Tobi Dec 23 '16 at 20:17
  • $\begingroup$ Yes, provided that the argument is in radians (not degrees). $\sin(x)=x+O(x^3)$ for small $x$. $\endgroup$ – Wouter Dec 23 '16 at 20:18
  • $\begingroup$ Yes, these are different curves, but you do nothing to explain how they are related, only how they are unrelated and seemingly insignificant. $\endgroup$ – Simply Beautiful Art Dec 23 '16 at 20:23
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Sinusoidal graphs can be approximated by parabolas in some regions ... depending on the type of sinusoid, the graph may be exactly replicated by linking parabolas together.

You are correct in saying that parabolas are linked to circles, but not in a simple way, as circles have tangents that rotate about completely as you rotate around the circle; parabolas do not.

Parabolas also are onto functions, whereas circles cannot be.

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    $\begingroup$ Not exactly replicated. One can't exactly replicate any non-trivial piece of a sinusoidal function with any mere polynomial. You're correct, though, that one function may approximate the other in a limited region. $\endgroup$ – David K Dec 23 '16 at 20:38
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    $\begingroup$ I should say that a sum of sinusoids (perhaps infinitely many) can be exactly replicated by linking parabolas together. But then anything can be, so it's not a very interesting answer, is it? $\endgroup$ – Anthony P Dec 23 '16 at 20:42
  • $\begingroup$ @AnthonyP Not anything. Please consider analytic functions and fourier series. $\endgroup$ – Simply Beautiful Art Dec 23 '16 at 20:50
  • $\begingroup$ @SimpleArt wow you guys really are precise! I'll need to be more careful :-) $\endgroup$ – Anthony P Dec 23 '16 at 21:02

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