Re: displacement and graphs

["John S. Denker" <jsd@MONMOUTH.COM>]

At 08:07 AM 10/10/01 -0500, Rick Tarara wrote:
> A little sidetrack here:

A valuable track IMHO.

> I've always had troubles at the beginning of intro courses trying to keep my
> nomenclature consistent.  The problem as I see it is as follows:
>
> 1)  We usually start out defining velocity as the (change in
> position)/(change in time).

Actually, it's the LIMIT of the difference quotient.  This sounds like a
technicality, but I suspect it is close to the heart of the matter.

> 2)  The change in position can be considered a vector--the displacement--so
> we can also say that the velocity is the displacement/(change in time).

No.  The AVERAGE velocity might be given by such a formula, but not the
velocity per se.

> 3)  However, we can also express any position as a displacement vector
> (relative to some origin) and therefore the change in position is equivalent
> to a change in displacement.  Hence we can say that velocity is the (change
> in displacement)/(change in time).  In a calc course we will end up with v =
> ds/dt.

Again, providing the appropriate limit is taken.

> 4)  Of course (3) is consistent with (2) since the change in displacement is
> indeed another displacement vector.  This is really no different then our
> definition of acceleratiion as (change in velocity)/(change in time) since
> the change in velocity is indeed another velocity--but we don't tend to talk
> that way.

We don't?  Every discussion of longitudinal wave propagation I've seen
talks about changes in the displacement.

> 5)  So--(change in position)/(change in time); displacement/(change in
> time), (change in diplacement)/(change in time) are all equivalent---but I
> suspect confusing to students.  It is difficult to stay with just one of
> these however when working through various examples.

Confusing?  Maybe they need a homework assignment:  Prove that
        (d/dt)[y + const] = (d/dt)[y]