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| DK |
 Posted: Thu Apr 24, 2008 8:27 pm |
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In article <3s41149872tqp1jo8nnvqo2chbphvst8a4@4ax.com>, quasi@null.set wrote:
Quote: On Thu, 24 Apr 2008 06:45:02 -0700 (PDT), "dwwoelfel@gmail.com"
dwwoelfel@gmail.com> wrote:
What's wrong with using implicit differentiation? You'd get dy/dx = -x/
y.
There's nothing wrong with it.
But even before differentiating, it's obvious from the graph that the
slopes are not constant (of course if they were, it would be a line,
not a circle).
The OP was confused about something -- I'm not exactly sure what.
The OP is a spammer who is confused just about *everything*! Either
a dedicated troll or a full-blown insanity. |
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| foolsrushout |
| Posted: Thu Apr 24, 2008 8:37 pm |
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DK wrote:
Quote: In article <3s41149872tqp1jo8nnvqo2chbphvst8a4@4ax.com>, quasi@null.set wrote:
On Thu, 24 Apr 2008 06:45:02 -0700 (PDT), "dwwoelfel@gmail.com"
dwwoelfel@gmail.com> wrote:
What's wrong with using implicit differentiation? You'd get dy/dx = -x/
y.
There's nothing wrong with it.
But even before differentiating, it's obvious from the graph that the
slopes are not constant (of course if they were, it would be a line,
not a circle).
The OP was confused about something -- I'm not exactly sure what.
The OP is a spammer who is confused just about *everything*! Either
a dedicated troll or a full-blown insanity.
Another of those chicken/egg conundrums. |
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| foolsrushout |
| Posted: Thu Apr 24, 2008 11:14 pm |
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David R Tribble wrote:
Quote: quasi wrote:
Circles, do have constant curvature.
foolsrushout wrote:
What happens to a circle in an expanding universe?
Which universe did you have in mind?
Any of them that's expanding. Why did you ask? Do
you have a preference for some particular choice? |
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| foolsrushout |
| Posted: Thu Apr 24, 2008 11:19 pm |
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mitch.nicolas.raemsch@gmail.com wrote:
Quote: On Apr 24, 6:10 pm, David R Tribble <da...@tribble.com> wrote:
quasi wrote:
Circles, do have constant curvature.
foolsrushout wrote:
What happens to a circle in an expanding universe?
Which universe did you have in mind?...
He is talking about only one.
As the circle gets bigger if it were to get infinite it would be an
infinite straight line. The arc of an infinite circle is just a
straight line.
You cannot obtain a tangent for a circle because it cannot be touched
by two straight lines.
Mitch Raemsch; Twice Nobel Laureate 2008
It is a lot more fun that that. Consider the two
dimensional object expanding in three dimensions. |
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| Uncle Al |
| Posted: Fri Apr 25, 2008 4:15 pm |
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DK wrote:
Quote:
In article <3s41149872tqp1jo8nnvqo2chbphvst8a4@4ax.com>, quasi@null.set wrote:
On Thu, 24 Apr 2008 06:45:02 -0700 (PDT), "dwwoelfel@gmail.com"
dwwoelfel@gmail.com> wrote:
What's wrong with using implicit differentiation? You'd get dy/dx = -x/
y.
There's nothing wrong with it.
But even before differentiating, it's obvious from the graph that the
slopes are not constant (of course if they were, it would be a line,
not a circle).
The OP was confused about something -- I'm not exactly sure what.
The OP is a spammer who is confused just about *everything*! Either
a dedicated troll or a full-blown insanity.
The OP is dogshit attracting flies.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/lajos.htm#a2 |
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| David R Tribble |
| Posted: Sat Apr 26, 2008 5:30 am |
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quasi wrote:
Quote: Circles, do have constant curvature.
foolsrushout wrote:
Quote: What happens to a circle in an expanding universe?
David R Tribble wrote:
Quote: Which universe did you have in mind?
foolsrushout wrote:
Quote: Any of them that's expanding. Why did you ask? Do
you have a preference for some particular choice?
I can think of three geometric/mathematical "universes" in which
circles are defined, and that's just in three dimensions.
I can think of many more metric spaces where "circles" don't
act like the normal 2-D Euclidean object.
In our own physical universe, it's quite likely that there is
no such thing as a continuous circle because of the discrete
nature of the underlying quantum foam. In which case, the
question is without meaning.
In other words, circles are only abstract mathematical constructs
and as such have no "real" connection to our physical universe.
So I'm asking what you mean by "universe". |
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| foolsrushout |
| Posted: Sat Apr 26, 2008 1:25 pm |
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David R Tribble wrote:
Quote: quasi wrote:
Circles, do have constant curvature.
foolsrushout wrote:
What happens to a circle in an expanding universe?
David R Tribble wrote:
Which universe did you have in mind?
foolsrushout wrote:
Any of them that's expanding. Why did you ask? Do
you have a preference for some particular choice?
I can think of three geometric/mathematical "universes" in which
circles are defined, and that's just in three dimensions.
I can think of many more metric spaces where "circles" don't
act like the normal 2-D Euclidean object.
In our own physical universe, it's quite likely that there is
no such thing as a continuous circle because of the discrete
nature of the underlying quantum foam. In which case, the
question is without meaning.
You're too much an aetherist.
Quote: In other words, circles are only abstract mathematical constructs
and as such have no "real" connection to our physical universe.
"A circle is the set of all points an equal distance
(r) from a fixed point, or center (h,k). The set can
be expressed as the solutions of the equation..."
www.math.utah.edu/~forde/1270/glossary/c.html
For the discussion you've entered into in responding
to my question a circle can most certainly have a
"real" connection to our physical universe. There's
nothing wrong with defining a subset of all available
points in some region of space to use for an example.
But please don't confuse the model with the reality.
Because I write circle doesn't mean I can't have a
real circle simply by defining points in space,
unless for some reason you think space doesn't exist.
Since we've been describing a circle as "the locus
of all points" <attach restrictive parameters here>
there is a question whether a circle is a discontinuous
structure of adjoining points although we draw it with
a continuous (if there is such a thing) line with a
compass, or whether the "locus" can be considered
continuous. As you bang your fist on the desk while
reading this decide once again whether or not the
surface is discrete or continuous, and how that
affects the pain you feel with your (technically
described) discontinuous body when you hit too hard.
It seems we experience the universe rather differently
from the way we might expect to given the models we
generally accept the "really" universe to be.
As amusing as this petty distraction is to the main
question I raised, repeated here:
Quote: What happens to a circle in an expanding universe?
I think the question I raised can lead one down a
number of more interesting paths to investigate.
As I mentioned in another posting, this question
brings into play the effects of a universe of at
least three dimensions expanding in all directions
available on a two dimensional structure (series of
points).
Next consider the circle to be the rim of a two
dimensional disk. Expand your mind to consider
the entire disk living in space that's expanding
in three dimensions!
Then carry the problem further to include one of
my favorite constructs, two skewed line segments.
Those present a two dimensional construct requiring
no less than three dimensions to describe them.
Quote: So I'm asking what you mean by "universe".
<http://www.theage.com.au/ffximage/2007/05/29/universe3_narrowweb__300x345,0.jpg>
"How many dimensions do you see?"  |
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| DAVID GREENE |
| Posted: Sun Apr 27, 2008 8:38 pm |
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"quasi" <quasi@null.set> wrote in message:
You know what? Just put the damn circle in polar coordnates.
Then take the derivative with respect to angle. QED.
Quote: I believe a circle is an unchanging curve and has no derivative.
Anyone ran into this before?
Curves don't have derivatives -- functions do (some of them).
All changing curves unlike the circle have a derivative at every
point. Is this not applied calculus?
No, it's not precise terminology.
Find a calculus book, and look up "derivative".
You'll see that they define the derivative of a function, not the
derivative of a curve.
However, if f is a differentiable function from, say, R to R, then the
_slope_ of the tangent line to the graph of the curve y = f(x) at a
given point (x0,f(x0)) on the curve does equal f'(x0).
Now a circle is not the image of a function of x -- it fails the
vertical line test, but that's easily fixed by separately considering
the upper and lower semicircles.
But take a look at the slopes. Do they appear constant to you?
quasi |
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| bill |
| Posted: Mon Apr 28, 2008 6:16 pm |
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On Apr 23, 9:46 pm, mitch.nicolas.raem...@gmail.com wrote:
Quote: I believe a circle is an unchanging curve and has no derivative.
Anyone ran into this before?
Mitch Raemsch
The equation of the function that plots as a circle
with center at the origin is
y = (k^2 - x^2)^(0.5)
This function may have a derivative.
Bill J |
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| Uncle Al |
| Posted: Tue Apr 29, 2008 3:34 pm |
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bill wrote:
Quote:
On Apr 23, 9:46 pm, mitch.nicolas.raem...@gmail.com wrote:
I believe a circle is an unchanging curve and has no derivative.
Anyone ran into this before?
Mitch Raemsch
The equation of the function that plots as a circle
with center at the origin is
y = (k^2 - x^2)^(0.5)
This function may have a derivative.
(x - h)^2 + (y - k)^2 = r^2
It's not a function. Each abscissa does not have a unique ordinate.
The derivative is OBVIOUSLY the tangent to the circle at any given
circumferential point.
Hind gut-fermenting fundie Christians are invited to debate the
thickness of the circle's circumference (analytic method being given
twice in the bible).
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/lajos.htm#a2 |
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| foolsrushout |
| Posted: Tue Apr 29, 2008 8:05 pm |
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Uncle Al wrote:
Quote: bill wrote:
On Apr 23, 9:46 pm, mitch.nicolas.raem...@gmail.com wrote:
I believe a circle is an unchanging curve and has no derivative.
Anyone ran into this before?
Mitch Raemsch
The equation of the function that plots as a circle
with center at the origin is
y = (k^2 - x^2)^(0.5)
This function may have a derivative.
(x - h)^2 + (y - k)^2 = r^2
It's not a function. Each abscissa does not have a unique ordinate.
The derivative is OBVIOUSLY the tangent to the circle at any given
circumferential point.
Hind gut-fermenting fundie Christians are invited to debate the
thickness of the circle's circumference (analytic method being given
twice in the bible).
I would have thought you'd be more interested in the
circles in your personal future. :-)
http://hhhknights.com/curr/human/2/hellinferno.html |
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