Date: Jan 25, 2002 6:10 PM Author: Kirby Urner Subject: Movie metaphor: does it fly? 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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