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From: Rob Johnson on 18 Jan 2010 08:11
In article <2c9deced-a6f0-4f1b-930b-f33a9a8a2af4(a)m25g2000yqc.googlegroups.com>,
"Ken S. Tucker" <dynamics(a)vianet.on.ca> wrote:
>Hello Fredrick, nice to meet you.
>
>On Jan 15, 5:55 pm, Frederick Williams <frederick.willia...(a)tesco.net>
>wrote:
>> "Ken S. Tucker" wrote:
>>
>> > [...]
>>
>> Work an example for us. What's
>> the integral from 0 to 1 of x sqrt(dx)
>
>Very reasonable question.
>
>Let x=U^2 then d(x=U^2) => (dx = dU^2), then sub to Fredricks
>example,
>
>1
>$ U^2 dU = 1/3.
>0

What you have computed above is

|\1
| x d sqrt(x)
\| 0

not

|\1
| x sqrt(dx)
\| 0

No one is debating that the former makes sense. Frederick is
asking what the latter is.

Rob Johnson <rob(a)trash.whim.org>
take out the trash before replying
to view any ASCII art, display article in a monospaced font
From: Ostap S. B. M. Bender Jr. on 19 Jan 2010 03:05
On Jan 18, 1:05 am, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote:
> On Jan 17, 7:57 pm, "Ostap S. B. M. Bender Jr."
>
>
>
> <ostap_bender_1...(a)hotmail.com> wrote:
> > On Jan 17, 2:16 pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote:
>
> > > On Jan 16, 11:20 pm, "Ostap S. B. M. Bender Jr."
>
> > > <ostap_bender_1...(a)hotmail.com> wrote:
> > > > On Jan 13, 4:06 pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote:
>
> > > > > Hi Mr. Eastham.
>
> > > > > On Jan 13, 3:21 pm, Chip Eastham <hardm...(a)gmail.com> wrote:
>
> > > > > > On Jan 13, 12:45 pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote:
>
> > > > > > > To Dr. Lewis et al,
>
> > > > > > > On Jan 13, 8:12 am, "Robert H. Lewis" <rle...(a)fordham.edu> wrote:
>
> > > > > > > > > > >${-oo,+oo} F(x) (dx)^.5   (1)
>
> > > > > > > > > > Where did you run across this? The notation makes very little
> > > > > > > > > > sense.
> > > > > > > > > Jays doing just fine, David I think you need to take a basic
> > > > > > > > > calculus course, let me explain (using Jay's example), that I
> > > > > > > > > frequently encounter.
>
> > > > > > > > > V = dX/dT = Velocity.
> > > > > > > > > sqrt(V) = sqrt (dX/dT) , sqrt(dX) = sqrt(V)*sqrt(dT).
> > > > > > > > > That's SOP in calculus, from that a bit of algebraic massage
> > > > > > > > > produces *generally* a means to solve.
> > > > > > > > > Regards
> > > > > > > > > Ken S. Tucker
>
> > > > > > > >   SOP in Calculus?  Elementary course?   No way!
>
> > > > > > > >  I have been teaching calculus at all levels for close to 30 years, and I have never seen \sqrt(dx).  It is absurd on the face of it.  Perhaps some physicists have come up with this and found a coherent usage?
>
> > > > > > > Dr. Lewis, I too have taught and refined Calculus and very much
> > > > > > > respect the reasoning and notation conventions.
> > > > > > > I think it's reasonable to examine the 'differential coefficents',
> > > > > > > apart from the ratio, let's start here,
>
> > > > > > > V=dX/dT, sqrt(V)= sqrt(dX/dT)
>
> > > > > > > that's an expression of V by differential coefficients,
> > > > > > > The apparent confusion results from writing the above as,
>
> > > > > > > sqrt(V) = sqrt(dX)/sqrt(dT)   , THE STEP.
>
> > > > > > > Here we ask, does THE STEP represent a clear definition
> > > > > > > of symbolic logic?
>
> > > > > Above, I'm working on getting a handle on 'what is the
> > > > > problem'.
>
> > > > > > No, it is not clear by any definition of
> > > > > > derivative.
>
> > > > > I'll need to clarify, that we're a bit deeper than
> > > > > derivatives, specifically 'differential co-efficents'
> > > > > then a bit beyond that, a respected mathematician
> > > > > Herman Weyl was fond of.
>
> > > > > > The casual treatment of the
> > > > > > derivative dX/dT as a ratio of two values
> > > > > > mixed with taking a square root in the
> > > > > > numerator and denominator is lacking in
> > > > > > logical justification.
>
> > > > > Good enough!
> > > > > Allow me to move to integration ($) and then pose,
>
> > > > > dU = dV*dV , dV = sqrt(dU)
>
> > > > What is 'dV*dV'? A function? A number? A bird? A mammal? A mineral?
>
> > > The easiest I know is to begin with an Area=Length^2, I'll
> > > write as A=L^2, with A=L=0 to start, then
>
> > > d(Area) = d(Length)^2   ==    dA = dL^2.
>
> > You mean the area of a square? OK. Indeed, you can write:
>
> >  dA(L) = d(L^2)
>
> > But how do you get form this to (dL)^2 ?
>
> Well in a science experiment shining light through an aperature,
> we may wish to calculate the area of projection as we move the
> screen the image is projected upon, thinking in terms of a movie
> projector.
>
> We can use, (A=Area, r=distance from bulb, t=time)
>
> dA/dt = (dr/dt)^2  = dr^2 / dt^2,
>

Why?

A is "area"? Of what? How does it relate to r? How does it change with
time?

>
> Plug in a number like dr/dt=2 and the rate of change of the
> projected Area is dA/dt=4, so the projected Area of the image
> increased 4x as the distance from the bulb doubled.
>

I have no idea what you are saying, nor how this relates to '(dL)^2'.

>
>
> > > In the world of differentials, only change counts, absolute
> > > values are arbituary.
>
> > > In General Relativity oft used is
>
> > > ds^2 = g_uv dx^u dx^v = 0,
>
> > > (describes the light path)
> > > then some authors go to dA = ds^2.
>
> > > It looks ok to me, the integral
>
> > > $ dA = $ ds^2 = the same constant of integration,
>
> > Sure, an expression like  $ d(s^2) looks fine. But what does the
> > expression  $ (ds)^2 mean?
>
> Ostap, the subtle notation change is respected.
>

Respected by whom?

Are you trying to play games with formal notation? Wouldn't it be more
productive to work with substance instead?

>
> In GR ds=0 so we have   $ ds ds   and that gives,
>
> $ 0 ds = constant of integration.
>

I have no idea what you are saying.

>
> In physics I understand that constant to be number a "N"
> that is a scalar (invariant).
>

I have no idea what you are saying.

>
> > > PS:Just got dizzy, washed a naked girls back.
>
> > Are you a teenager?
>
> Betcha when I see 90 I'll still get dizzy bathing gals,
> wifey called me to scrub her back, it's tough work but
> gives me an excuse to stay alive ;-).
>

Your wife is very lucky.

From: Ken S. Tucker on 19 Jan 2010 03:18
On Jan 19, 12:05 am, "Ostap S. B. M. Bender Jr."
<ostap_bender_1...(a)hotmail.com> wrote:
> On Jan 18, 1:05 am, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote:
....

> > Well in a science experiment shining light through an aperature,
> > we may wish to calculate the area of projection as we move the
> > screen the image is projected upon, thinking in terms of a movie
> > projector.
>
> > We can use, (A=Area, r=distance from bulb, t=time)
>
> > dA/dt = (dr/dt)^2 = dr^2 / dt^2,
>
> Why?
>
> A is "area"? Of what? How does it relate to r? How does it change with
> time?

I'm afraid there's very little I can add.
Regards
Ken
....
From: Ostap S. B. M. Bender Jr. on 19 Jan 2010 03:25
On Jan 19, 12:18 am, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote:
> On Jan 19, 12:05 am, "Ostap S. B. M. Bender Jr."<ostap_bender_1...(a)hotmail.com> wrote:
> > On Jan 18, 1:05 am, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote:
>
> ...
>
> > > Well in a science experiment shining light through an aperature,
> > > we may wish to calculate the area of projection as we move the
> > > screen the image is projected upon, thinking in terms of a movie
> > > projector.
>
> > > We can use, (A=Area, r=distance from bulb, t=time)
>
> > > dA/dt = (dr/dt)^2  = dr^2 / dt^2,
>
> > Why?
>
> > A is "area"? Of what? How does it relate to r? How does it change with
> > time?
>
> I'm afraid there's very little I can add.
>

Good luck to you on your path to imminent scientific glory.
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