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Movie metaphor: does it fly?
Posted:
Jan 25, 2002 6:10 PM
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This might be more physics than calculus, but I'll run it through
here for peer review/feedback just in case.
Seems that the "movie metaphor" (hence w/o quotes) is an obvious
one to deploy when introducing calculus. t is the x axis, and
we want students to really experience their own lives, each
waking instant, as a function of time i.e. MyLife(t) is what
you experience right now (including memories of MyLife(t-m)).
From this "whole life" image space (range of MyLife(t)), we go
to film as a first approximation. Each frame represents the
action that goes on in delta t.
I've gotten flak for this: people say the change occurs *between*
the frames, whereas a frame is a frozen instant, not action per
any delta. I disagree. The camera shutter is open for a discrete
time frame, and what we see is a superposition of images over that
time, which may end up being blurred if the shutter speed was
inappropriately short. So the frame of film itself represents
a time interval, and intervals *between* frames we'll safely set
to zero.
Next comes dimensional analysis. What we see in a frame is a massive
body (a Volkswagon) traveling left to right, i.e. in the next frame
it'll pick up further to the right, so per the time interval shown,
we impute a change in position, even if the shutter speed is such
that the image remains sharp. So we have a massive body, with a
velocity (mv), undergoing a change in location (mvd). And it does all
this per time interval (mvd/t). Dimensional analysis: mvd = action
= pd (p = mv). mvd/t = mvv = mv^2 = energy. frequency = 1/t (e.g.
hertz) i.e. mvd/t = action * hertz = hf (Planck's constant *
frequency)
= E = energy (E = hf is an important physics formula).
In other words, each action-packed frame of film represents and
energy amount, is a momentum (Volkswagon with velocity) translating
for a distance (d) per a time interval (1/delta t or, at the limit
1/dt).
I like this (if it works) because then each frame of file represents
and "energy bucket". As the film plays out, we're watching energy
being expended. If I tell you each frame is 1/125th of a second
(a standard camera shutter speed), then you have so-and-so amount
of energy being expending in SIGMA delta-t time (add all the frame
times for the total time). But if the shutter speed is 1/1250 (10
times faster), then the *power* of what you're seeing increases,
because the same amount of action (SIGMA mvd) occurs in 10 times
less time. And E/t = Power. When I show you a Saturn V booster
leaving the ground much more quickly than you're expecting, that's
a much higher power booster (probably breaks laws of physics, in terms
of structural stability, i.e. a *real* Saturn V booster couldn't
handle the stress of a 10-times-faster take off -- not to mention
the astronauts, subject to all that extra G-force).
What my aim is here, for those who have lost me, is to introduce the
movie metaphor in a way that commensurates with the standard
physics vocab, *and* is faithful to the calculus, in the sense that
we see the integral of dt frames as a sum of energy content, and
therefore a length of film (definite integral) as commensurate
(in dimensional analysis terms) with a quantity of energy per a
total quantity of time. The same energy in a shorter time would
represent more power. And the same energy in longer time, would
represent less power.
Is this a good synching of physics and calculus? I'd like to see it
used more, if so, as the segue from "movie metaphor" to "my life as
I experience it now" is easy, and therefore makes the calculus come
alive as a lived experience.
Kirby
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