Yet Again?
You need to be working in three dimensions, not two; and in that domain, there are asymptotes as well.
Cylindrical Asymptote - Wolfram Demonstrations Project
WHG
You need to be working in three dimensions, not two; and in that domain, there are asymptotes as well.
Cylindrical Asymptote - Wolfram Demonstrations Project
WHG
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Then please let someone who you trust more to explain it to you (if you are interested in how it works). It will require at least a working knowledge of a basic calculus. It doesn't approach a cylinder in the same way as a cone doesn't approach a cylinder. Or does it, in infinity?
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argumentum ad hominem
The issue you raise is:
"No. Asymptote [1] has its exact definiton and meaning and you can't use the term whatever way you like, if you want to be well understood. It doesn't approach a cylinder."
My response now is:
Your argument demonstrate the need for a calculus lesson, just not for me. A Cylindrical Asymptote exists and the surface of a Parabolic Horn (as shown by David) approaches one. Some variants also approach and asymptotic cone as well. Asymptotic surfaces exist. End of story. So move on. WHG
The issue you raise is:
"No. Asymptote [1] has its exact definiton and meaning and you can't use the term whatever way you like, if you want to be well understood. It doesn't approach a cylinder."
My response now is:
Your argument demonstrate the need for a calculus lesson, just not for me. A Cylindrical Asymptote exists and the surface of a Parabolic Horn (as shown by David) approaches one. Some variants also approach and asymptotic cone as well. Asymptotic surfaces exist. End of story. So move on. WHG
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No it doesn't. There would have to be a finite limit. There is not. Limit for sqrt(x) as x goes to infinity is also infinity. That does mean that the parabolic horn doesn't approach anything like cylinder. That's the end of story
It is a raw paper cone on my DIY test bench (i only trust in my own tests).
The indicated weight is only for the cone without the motor and suspension weight (around 70-80Gr)
The thruth lies less often than people, a 15 or 18 woofer with a less than 100Gr paper straight cone gives a THD shitstorm under 100hz.
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No it doesn't. There would have to be a finite limit. There is not. Limit for sqrt(x) as x goes to infinity is also infinity. That does mean that the parabolic horn doesn't approach anything like cylinder. That's the end of story
MaTHbat would suit you better
A Dinner of Crow for Me
The two legs of a parabola become increasingly more parallel as infinity is approached; i.e. slope and curvature are approaching zero (x-axis). In the limit, at infinity, they are parallel, and are parallel to the axis of the parabola as well. Technically this condition as you rightly point out, makes them not asymptotes. I took analytics class in the late 1950’s and have since forgotten that a parabola was unlike the hyperbola in this respect. The cylinder I had in mind, would have to have a radius of infinity. Thanks for the persistence in reminding me of the error I was making. WHG
No it doesn't. There would have to be a finite limit. There is not. Limit for sqrt(x) as x goes to infinity is also infinity. That does mean that the parabolic horn doesn't approach anything like cylinder. That's the end of story
The two legs of a parabola become increasingly more parallel as infinity is approached; i.e. slope and curvature are approaching zero (x-axis). In the limit, at infinity, they are parallel, and are parallel to the axis of the parabola as well. Technically this condition as you rightly point out, makes them not asymptotes. I took analytics class in the late 1950’s and have since forgotten that a parabola was unlike the hyperbola in this respect. The cylinder I had in mind, would have to have a radius of infinity. Thanks for the persistence in reminding me of the error I was making. WHG
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My pleasure. The point is that it's really not approaching anything. You even shouldn't think of an infinite radius at infinity, that leads you nowhere. It is not approaching infinity because infinity is not a number. If will still grow and grow, above any limit (no matter that the rate of grow is decreasing - it's not enough to make any difference). So there is no final radius to talk about. Nothing like that exists, there is no cylinder to approach.
Remember the infinities can be very tricky...
Remember the infinities can be very tricky...
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Limits
Sorry but it is still approaching a cylindrical shape, no other; even though at infinity, a bounding cylinder cannot exist. It is just simply not asymptotic. That is the only concession I am prepared to make here. Furthermore, for a hyperbolic horn at infinity, its asymptotic cone cannot exist as well. None the less, it is approaching a conical shape. WHG
Sorry but it is still approaching a cylindrical shape, no other; even though at infinity, a bounding cylinder cannot exist. It is just simply not asymptotic. That is the only concession I am prepared to make here. Furthermore, for a hyperbolic horn at infinity, its asymptotic cone cannot exist as well. None the less, it is approaching a conical shape. WHG
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N.B.
If you take a segment of parabola and incline it relative to the x-axis, and then rotate it about the x-axis, the resulting horn will approach a cone shape as well. In all cases, curvature is in decline, as a straight line is approached. Sometimes this line is not an asymptote.
WHG
If you take a segment of parabola and incline it relative to the x-axis, and then rotate it about the x-axis, the resulting horn will approach a cone shape as well. In all cases, curvature is in decline, as a straight line is approached. Sometimes this line is not an asymptote.
WHG
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y=x^2 is simplest form, or rotated y^2=x. Like mabat said, expands forever.
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Hi Earl,
y = ax^2 + bx + c
and the 1st derivitive is of course is
dy/dx = 2ax
Regards,
Bill
Well, it could expand forever and still approach a cylinder. But that is not the case of parabola. It's quite hard mathematical concept. For example the sum of infinite series of [1/n] (i.e. 1/1 + 1/2 + 1/3 + ...) grows with every added number and goes to infinity (i.e. there is no sum, it doesn't convegre). Now, the sum of [1/(n^2)] (i.e. 1/1 + 1/4 + 1/9 + ...) does grow as well as you add the numbers, BUT approaches the number (pi^2)/6 and never actually reaches this limit. Interesting, isn't it. It's because the second sum grows much, much slowly (and it's not enough even for infinity).
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