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Science Forum Index » Physics Forum » Derivative of a circle?
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| Guest |
 Posted: Wed Apr 23, 2008 6:46 pm |
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I believe a circle is an unchanging curve and has no derivative.
Anyone ran into this before?
Mitch Raemsch |
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| Guest |
| Posted: Wed Apr 23, 2008 7:10 pm |
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On Apr 23, 9:04 pm, quasi <qu...@null.set> wrote:
Quote: On Wed, 23 Apr 2008 21:46:48 -0700 (PDT),
mitch.nicolas.raem...@gmail.com wrote:
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?
Quote:
But a circle certainly has tangent lines, and they do change.
Now if you represent the circle as the image of a vector function
r(t) = (cos(t),sin(t))
then, as a function from R to R^2, r is differentiable, with
derivative
r'(t) = (-sin(t),cos(t))
Note that the derivative is _not_ constant, as a vector, however it
does have constant _magnitude_, namely 1.
But circles aren't special in that regard.
All sufficiently smooth curves in R^n can be represented as the image
of a vector function whose derivative has constant magnitude 1.
Perhaps rather than "derivative", you actually meant "curvature".
Circles, do have constant curvature.
quasi |
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| quasi |
| Posted: Thu Apr 24, 2008 12:04 am |
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Guest
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On Wed, 23 Apr 2008 21:46:48 -0700 (PDT),
mitch.nicolas.raemsch@gmail.com wrote:
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).
But a circle certainly has tangent lines, and they do change.
Now if you represent the circle as the image of a vector function
r(t) = (cos(t),sin(t))
then, as a function from R to R^2, r is differentiable, with
derivative
r'(t) = (-sin(t),cos(t))
Note that the derivative is _not_ constant, as a vector, however it
does have constant _magnitude_, namely 1.
But circles aren't special in that regard.
All sufficiently smooth curves in R^n can be represented as the image
of a vector function whose derivative has constant magnitude 1.
Perhaps rather than "derivative", you actually meant "curvature".
Circles, do have constant curvature.
quasi |
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| foolsrushout |
| Posted: Thu Apr 24, 2008 12:25 am |
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quasi wrote:
[...]
Quote: Circles, do have constant curvature.
What happens to a circle in an expanding universe?
Does it remain a circle? |
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| Bill |
| Posted: Thu Apr 24, 2008 12:28 am |
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Guest
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"foolsrushout" <666@hotmail.com> wrote in message
news:fup5jl$3pi$1@aioe.org...
Quote: quasi wrote:
[...]
Circles, do have constant curvature.
Quote: What happens to a circle in an expanding universe?
Does it remain a circle?
Get some "Silly Putty" and find out... |
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| quasi |
| Posted: Thu Apr 24, 2008 12:34 am |
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Guest
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On Wed, 23 Apr 2008 22:10:08 -0700 (PDT),
mitch.nicolas.raemsch@gmail.com wrote:
Quote: On Apr 23, 9:04Â pm, quasi <qu...@null.set> wrote:
On Wed, 23 Apr 2008 21:46:48 -0700 (PDT),
mitch.nicolas.raem...@gmail.com wrote:
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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| quasi |
| Posted: Thu Apr 24, 2008 12:58 am |
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Guest
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On Thu, 24 Apr 2008 00:25:09 -0500, foolsrushout <666@hotmail.com>
wrote:
Quote: quasi wrote:
[...]
Circles, do have constant curvature.
What happens to a circle in an expanding universe?
Does it remain a circle?
Yes, sort of.
That is, yes, subject to some drastically simplifying assumptions ...
Assume the universe is a subset of R^3, varying with time, expanding
uniformly in all directions, with respect to some fixed center.
Thus, assume that a given point (x,y,z) has the trajectory
x(t) = x(0)*(1 + t)
y(t) = y(0)*(1 + t)
z(t) = z(0)*(1 + t)
Then if C is a circle in the universe at time t = 0, the points of C
remain in the shape of a circle as time changes.
Of course, for this analysis, I am ignoring relativistic effects, and
similarly, the question as to whether a curve is a circle is being
answered by an oracle, not by an observer.
But let me say -- I don't know much physics, so I don't really know
whether relativity, or effects such as delayed observations due to the
speed of light would change the answer or not.
quasi |
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| dwwoelfel@gmail.com |
| Posted: Thu Apr 24, 2008 3:45 am |
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What's wrong with using implicit differentiation? You'd get dy/dx = -x/
y. |
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| Dono |
| Posted: Thu Apr 24, 2008 4:21 am |
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Guest
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On Apr 23, 10:34 pm, quasi <qu...@null.set> wrote:
Quote: On Wed, 23 Apr 2008 22:10:08 -0700 (PDT),
mitch.nicolas.raem...@gmail.com wrote:
On Apr 23, 9:04 pm, quasi <qu...@null.set> wrote:
On Wed, 23 Apr 2008 21:46:48 -0700 (PDT),
mitch.nicolas.raem...@gmail.com wrote:
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
You are conversing with number 1 autistic imbecile in this forum,
"Nobel Prize Winner Mitch Raemsch". Don't expect anything remotely
coherent out of him, he's just trolling.  |
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| quasi |
| Posted: Thu Apr 24, 2008 9:09 am |
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Guest
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On Thu, 24 Apr 2008 06:45:02 -0700 (PDT), "dwwoelfel@gmail.com"
<dwwoelfel@gmail.com> wrote:
Quote: 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.
quasi |
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| Uncle Al |
| Posted: Thu Apr 24, 2008 10:40 am |
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mitch.nicolas.raemsch@gmail.com wrote:
Quote:
I believe a circle is an unchanging curve and has no derivative.
Anyone ran into this before?
Mitch Raemsch
The little shit can't do high school calculus.
(x - h)^2 + (y - k)^2 = r^2
Can anybody take a derivative? Math is soooo hard!
HEY FUCKFACE BABOON, the derivative at a given point is the tangent to
the circle at that point. Every such tangent has a twin across the
diameter.
If yer tooo stooopid to take a derivative, use limits
<http://www2.norwich.edu/rpodiac/cmjarticle2.pdf>
Use Greek construction. You should be good at that seeing as how you
have plucked more than a hundred things from your ass in recent days'
posts.
--
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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| Guest |
| Posted: Thu Apr 24, 2008 11:17 am |
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On Apr 23, 9:34 pm, quasi <qu...@null.set> wrote:
Quote: On Wed, 23 Apr 2008 22:10:08 -0700 (PDT),
mitch.nicolas.raem...@gmail.com wrote:
On Apr 23, 9:04 pm, quasi <qu...@null.set> wrote:
On Wed, 23 Apr 2008 21:46:48 -0700 (PDT),
mitch.nicolas.raem...@gmail.com wrote:
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
The derivative is the slope of the curve at any point. |
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| Simple Simon |
| Posted: Thu Apr 24, 2008 12:29 pm |
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Guest
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"Uncle Al" <UncleAl0@hate.spam.net> wrote in message
news:4810A9DB.51A0A620@hate.spam.net...
Quote: mitch.nicolas.raemsch@gmail.com wrote:
I believe a circle is an unchanging curve and has no derivative.
Anyone ran into this before?
Mitch Raemsch
The little shit can't do high school calculus.
(x - h)^2 + (y - k)^2 = r^2
Can anybody take a derivative? Math is soooo hard!
HEY FUCKFACE BABOON, the derivative at a given point is the tangent to
the circle at that point. Every such tangent has a twin across the
diameter.
If yer tooo stooopid to take a derivative, use limits
http://www2.norwich.edu/rpodiac/cmjarticle2.pdf
Please have them correct the statement:
"The ordinary derivative exists everywhere the symmetric derivative does,
but the reverse is not true."
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| David R Tribble |
| Posted: Thu Apr 24, 2008 4:10 pm |
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quasi wrote:
Quote: Circles, do have constant curvature.
foolsrushout wrote:
Quote: What happens to a circle in an expanding universe?
Which universe did you have in mind? |
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| Guest |
| Posted: Thu Apr 24, 2008 5:28 pm |
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On Apr 24, 6:10 pm, David R Tribble <da...@tribble.com> 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?...
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 |
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