wrote:
On May 1, 1:25 pm, tadchem wrote:
On May 1, 4:09 pm, wrote:

In order to derive a tangent line on a changing curve you must do
calculus. You must pick two points to get the the tangent line and get
the same slope. Calculus is just an estimation. You cannot become
infinitely accurate because you cannot make infinite calculations. You
could make a million!
Mitch Raemsch

You are pushing my patience to the infinitesimal limit...

In order to be precise calculus must make infinite calculations.

Right over his head.


--
--Bryan

"Androcles" wrote in message ...

[snip]

Yes, I concur. The tangent to a circle is constructed with straight edge and
compass.

I doubt Raemsch understands Xeno's paradox even though he appears to be
recreating it. Take his statement:
"In order to be precise calculus must make infinite calculations."

Let's do it. We make infinite calculations of [(x+h)^2 - x^2]/h for all h
not equal zero.

We have not calculated h at the very value it is needed. We must have made
oo-1
calculations.

Note that we don't take a limit here, we leave out one element of the set in
the domain
and find the only value in the codomain that remains.

Androcles takes a limit:
http://users.telenet.be/vdmoortel/di...les/Limit.html
Androcles, inventor of the differentation constant:
http://users.telenet.be/vdmoortel/di...DiffConst.html

Dirk Vdm

Mitch:


wrote:

[...]
Calculus needs to hone in an infinite amount of times to come to a
precise answer.



You are bluffing, aren't you?

Calculus is never done "an infinite amount of times".

It is done once in principle to set up the formula for the
demonstration, and the particular values are found at only at discrete
selections of instances.

It is clear that you don't employ or understand a rational definition of
the concept of infinity, especially regarding the particular and
universal aspects of the concept and its potential versus its impossible
particular mathematical demonstrations.






A tangent line is really the shortest cord.




That is not true. Are you bluffing and providing misinformation or what?

You spelled chord wrong.


Two definitions:

A chord is a straight line that cuts a circle at two points.

A tangent line is a straight line that touches a circle at a single point.

The definitions of both concepts may be extended under certain specified
contexts to include the types of curves, for example, circular arcs,
B-spline curves or conic sections. A circle is sufficient for the
current example, however.

The short length of a selected chord is merely a non-essential
characteristic of a chord. The essential defining characteristic, or
differentia, of a chord is that it cuts a circle at two points. Even if
the selected straight line, and you must provide that the line is
straight in your definition, is very short, it must have the two
specified points either on it or at its endpoints and that also lie on
the circle. The short length specification is irrelevant to the definition.

The two definitions clearly differentiate the two types of straight
line, the differentia, from each other and from all other straight lines
in the wider class. Under no conditions is a chord the same as a
tangent. Two points are not one point.

You obviously do not understand the principles of the proper
construction of a definition.


Ralph Hertle

On May 2, 5:52 am, Ralph Hertle wrote:
Mitch:

wrote:

[...]

Calculus needs to hone in an infinite amount of times to come to a
precise answer.

You are bluffing, aren't you?

Calculus is never done "an infinite amount of times".

It is done once in principle to set up the formula for the
demonstration, and the particular values are found at only at discrete
selections of instances.

It is clear that you don't employ or understand a rational definition of
the concept of infinity, especially regarding the particular and
universal aspects of the concept and its potential versus its impossible
particular mathematical demonstrations.



A tangent line is really the shortest cord.

That is not true. Are you bluffing and providing misinformation or what?

You spelled chord wrong.

Two definitions:

A chord is a straight line that cuts a circle at two points.

A tangent line is a straight line that touches a circle at a single point.

The definitions of both concepts may be extended under certain specified
contexts to include the types of curves, for example, circular arcs,
B-spline curves or conic sections. A circle is sufficient for the
current example, however.

The short length of a selected chord is merely a non-essential
characteristic of a chord. The essential defining characteristic, or
differentia, of a chord is that it cuts a circle at two points. Even if
the selected straight line, and you must provide that the line is
straight in your definition, is very short, it must have the two
specified points either on it or at its endpoints and that also lie on
the circle. The short length specification is irrelevant to the definition.

The two definitions clearly differentiate the two types of straight
line, the differentia, from each other and from all other straight lines
in the wider class. Under no conditions is a chord the same as a
tangent. Two points are not one point.

You obviously do not understand the principles of the proper
construction of a definition.

Ralph Hertle



You are trying to talk sense to a professional troll.

wrote:
Calculus needs to hone in an infinite amount of times to come to a
precise answer.

Not true. I can compute some limits using calculus in a finite time.

You are restating (very badly, at that) Zeno's so called-paradox.

Bob Kolker

On May 1, 3:09*pm, wrote:
In order to derive a tangent line on a changing curve you must do
calculus. You must pick two points to get the the tangent line and get
the same slope. Calculus is just an estimation. You cannot become
infinitely accurate because you cannot make infinite calculations. You
could make a million!

Mitch Raemsch

You are being an idiot. The value of calculus is to get the value of
the slope of the tangent withOUT two points, and in one step. I
realize this must seem like voodoo to some people.

On May 1, 8:24*pm, "Robert J. Kolker" wrote:
wrote:
In order to derive a tangent line on a changing curve you must do
calculus. You must pick two points to get the the tangent line and get
the same slope. Calculus is just an estimation. You cannot become
infinitely accurate because you cannot make infinite calculations. You
could make a million!

Review the limit concept and rethink the matter.

Bob Kolker

Consider motion at constant acceleration, like gravitational free-fall
near earth surface:

a = g --- v =v0 + gt --- s = s0 +v0t +gt^2/2, the Galilleo
equation.

That's fine, even Kolker, the dogmatic, can do this

Now, consider t + delta(t), which I denote as t+dt to be brief, not to
be confused with differential dt. It is an infinitesimal change from
t. For simplicity assume v0 and s0 equal to 0. Plug into equation for
s to get s+ds:

s+ds = g(t+dt)^2/2 = g[t^2+2tdt+(dt)^2]/2 or

s+ds = gt^2/2 + gtdt+g(dt)^2/2 or since, s = gt^2/2

ds = gtdt + g(dt)^2/2 or ds/dt = gt + gdt/2

Now, remember dt stands for delta(t) and must take the limit as
delta(t) --0

Sure, you can say that as dt -- 0, then v = ds/dt


But wait a second, delta(t) --- 0 should not mean delta(t) = 0,
otherwise, you cannot take the derivative in the first place since t +
dt = 0. But if delta(t) never gets to zero then:

v = ds/dt + gdt/2, a small error remains

Thus,

dt --- 0 must mean that it never reaches exactly zero so it is
correct to say that exact values of derivatives are indeterminate to
small second order errors and Mitch is correct, although he does not
know why and how and does not ake him less of an idiot

Mike

Mike wrote in message

On May 1, 8:24 pm, "Robert J. Kolker" wrote:
wrote:
In order to derive a tangent line on a changing curve you must do
calculus. You must pick two points to get the the tangent line and get
the same slope. Calculus is just an estimation. You cannot become
infinitely accurate because you cannot make infinite calculations. You
could make a million!

Review the limit concept and rethink the matter.

Bob Kolker

Consider motion at constant acceleration, like gravitational free-fall
near earth surface:

a = g --- v =v0 + gt --- s = s0 +v0t +gt^2/2, the Galilleo
equation.

That's fine, even Kolker, the dogmatic, can do this

Now, consider t + delta(t), which I denote as t+dt to be brief, not to
be confused with differential dt. It is an infinitesimal change from
t. For simplicity assume v0 and s0 equal to 0. Plug into equation for
s to get s+ds:

s+ds = g(t+dt)^2/2 = g[t^2+2tdt+(dt)^2]/2 or

s+ds = gt^2/2 + gtdt+g(dt)^2/2 or since, s = gt^2/2

ds = gtdt + g(dt)^2/2 or ds/dt = gt + gdt/2

Now, remember dt stands for delta(t) and must take the limit as
delta(t) --0

Sure, you can say that as dt -- 0, then v = ds/dt


But wait a second, delta(t) --- 0 should not mean delta(t) = 0,

uh-oh...

otherwise, you cannot take the derivative in the first place since t +
dt = 0.

ouch...

But if delta(t) never gets to zero then:

v = ds/dt + gdt/2, a small error remains

Good grief.


Thus,

dt --- 0 must mean that it never reaches exactly zero so it is
correct to say that exact values of derivatives are indeterminate to
small second order errors

GASP.

and Mitch is correct, although he does not
know why and how and does not ake him less of an idiot

Do you even *remotely* realize that you are an even bigger
idiot than he is?

Dirk Vdm

On May 2, 1:50*pm, "Dirk Van de moortel" dirkvandemoor...@ThankS-NO-
SperM.hotmail.com wrote:
Mike wrote in message

*





On May 1, 8:24 pm, "Robert J. Kolker" wrote:
wrote:
In order to derive a tangent line on a changing curve you must do
calculus. You must pick two points to get the the tangent line and get
the same slope. Calculus is just an estimation. You cannot become
infinitely accurate because you cannot make infinite calculations. You
could make a million!

Review the limit concept and rethink the matter.

Bob Kolker

Consider motion at constant acceleration, like gravitational free-fall
near earth surface:

a = g --- v =v0 + gt --- *s *= s0 +v0t +gt^2/2, the Galilleo
equation.

That's fine, even Kolker, the dogmatic, can do this

Now, consider t + delta(t), which I denote as t+dt to be brief, not to
be confused with differential dt. It is an infinitesimal change from
t. For simplicity assume v0 and s0 equal to 0. Plug into equation for
s to get s+ds:

s+ds = g(t+dt)^2/2 = *g[t^2+2tdt+(dt)^2]/2 or

s+ds = gt^2/2 + gtdt+g(dt)^2/2 *or since, *s = gt^2/2

ds = gtdt + g(dt)^2/2 *or ds/dt = gt + gdt/2

Now, remember dt stands for delta(t) and must take the limit as
delta(t) --0

Sure, you can say that as dt -- 0, then v = ds/dt

But wait a second, delta(t) --- 0 should not mean delta(t) = 0,

uh-oh...

otherwise, you cannot take the derivative in the first place since t +
dt = 0.

ouch...

But if delta(t) never gets to zero then:

v = ds/dt + gdt/2, *a small error remains

Good grief.



Thus,

dt --- 0 must mean that it never reaches exactly zero so it is
correct to say that exact values of derivatives are indeterminate to
small second order errors

GASP.

and Mitch is correct, although he does not
know why and how and does not ake him less of an idiot

Do you even *remotely* realize that you are an even bigger
idiot than he is?

Dirk Vdm- Hide quoted text -

- Show quoted text -

I was waiting for you imbecile. This exact derivation I gave is from
the book Mathematical Experience and it was the argument of Berkeley
against Newtonian infinitesimal anf fluxions.

But you are not only an imbecile, you are a degenerate species found
in degraded human byproducts.

Mike

Mike wrote in message

On May 2, 1:50 pm, "Dirk Van de moortel" dirkvandemoor...@ThankS-NO-
SperM.hotmail.com wrote:
Mike wrote in message







On May 1, 8:24 pm, "Robert J. Kolker" wrote:
wrote:
In order to derive a tangent line on a changing curve you must do
calculus. You must pick two points to get the the tangent line and get
the same slope. Calculus is just an estimation. You cannot become
infinitely accurate because you cannot make infinite calculations. You
could make a million!

Review the limit concept and rethink the matter.

Bob Kolker

Consider motion at constant acceleration, like gravitational free-fall
near earth surface:

a = g --- v =v0 + gt --- s = s0 +v0t +gt^2/2, the Galilleo
equation.

That's fine, even Kolker, the dogmatic, can do this

Now, consider t + delta(t), which I denote as t+dt to be brief, not to
be confused with differential dt. It is an infinitesimal change from
t. For simplicity assume v0 and s0 equal to 0. Plug into equation for
s to get s+ds:

s+ds = g(t+dt)^2/2 = g[t^2+2tdt+(dt)^2]/2 or

s+ds = gt^2/2 + gtdt+g(dt)^2/2 or since, s = gt^2/2

ds = gtdt + g(dt)^2/2 or ds/dt = gt + gdt/2

Now, remember dt stands for delta(t) and must take the limit as
delta(t) --0

Sure, you can say that as dt -- 0, then v = ds/dt

But wait a second, delta(t) --- 0 should not mean delta(t) = 0,

uh-oh...

otherwise, you cannot take the derivative in the first place since t +
dt = 0.

ouch...

But if delta(t) never gets to zero then:

v = ds/dt + gdt/2, a small error remains

Good grief.



Thus,

dt --- 0 must mean that it never reaches exactly zero so it is
correct to say that exact values of derivatives are indeterminate to
small second order errors

GASP.

and Mitch is correct, although he does not
know why and how and does not ake him less of an idiot

Do you even *remotely* realize that you are an even bigger
idiot than he is?

Dirk Vdm- Hide quoted text -

- Show quoted text -

I was waiting for you imbecile.

Were you waiting for me imbecile?

Dirk Vdm