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Movie metaphor: does it fly?
Posted:
Jan 25, 2002 6:10 PM


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(tm)).
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 actionpacked 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 soandso amount of energy being expending in SIGMA deltat 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 10timesfaster take off  not to mention the astronauts, subject to all that extra Gforce).
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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