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worm problem, PEDAGOGY



On March 1, 1997 JACK L. URETSKY wrote:

Hi all-
I've barely been following the worm problem, but it doesn't look
ill-formed to me.

The length after n steps is L_n=L_0 + n (in km). with L_0=1.
The worm position after n+1 steps is x_(n+1) = x_n + 1 +x_n(1km/L_n).
But L_n = (n+1)km
so x_(n+1) = x_n + 1 + x_n(1/(n+1))
or x_(n+1) - x_n((n+2)/(n+1)) =1
the solution, satisfying the initial conditions is
x_n = (n+1)Sum(k=2 to n+1)(1/k) and
x_n = L_n when Sum(k=2 to n+1)(1/k) = 10^5.
Since the sum diverges, there is clearly a solution.
........................................................................
On February 24 Leigh Palmer wrote:
"At the end of one second the worm has marched one centimeter toward the
other end of the rope. At that instant he is instantaneously transported
to the 2 centimeter mark on a rope which is now two kilometers in length.
By calculation you will see that at the end of the second second he will
be three centimeters along his way on the two kilometer rope, which will
then lengthen to three kilometers, taking the worm to the 4.5 cm point,
etc."
***********************************************************

I also now agree that the above wording is not ambigous. But this was not
clear to me when I read the problem the first three times. Translating words
into mathematical equations is not always easy, even to us, teachers. I have
a little problem with your last sentence, Jack; I do not know how to check
your solution against that of John or Uri. (See my "post scriptum" below)

Playing a student again, I would like to say that teachers often overestimate
our capability of seeing clearly through their ways of thinking. Trying to be
"as short as possible" may be intellectually elegant (their inner circle) but
it is not always pedagogically effective. For example, I would include, in
the formulation of the problem, the numerical outcome of step #3.
Redundancy may often help some students. More specifically, I would end the
problem by saying:

"This can be generalized; the bug would move 1 cm to the right in every
step if the rope were stationary but it covers more distance because the
rope is being stretched. The constant speed of the rope is 100000 cm/s
at the tractor's end, 50000 cm/s in the middle, etc. Thus the additional
distance coverd by the bug, during the third step, is not 1 cm but 2 and
5/6 cm"

The whole point of this exercise, as clearly stated in the first message,
was to deal with large numbers. A teacher should try to bring students,
as quickly as possible, to this issue. I know what Donald will say. "Real
scientists, real engineers ...". But we are teachers and teaching is based
on artificial examples. I agree that hard problems should also be given,
sometimes. In another context I would use this example to emphasize that
the average bug's speed is 2 cm/s during the 1st second, 2.5 during the
2nd second, 2.833 during the third, etc. The easiest way to convince
students about this would be to ask them to draw the rope, with the
labeled makings (initially 1 cm apart), at t=0, 1, 2, 3 seconds; one
picture above the other.
Ludwik Kowalski

Would somebody like to discuss F=m*a when a tractor is accelerating the
bug (of finite mass) by stretching a massless rope? That is not QCD.
.......................................................................
POST SCRIPTUM:
I use your formula: x_(n+1) = x_n + 1 + x_n(1/(n+1))

We know that n=1 --> x(1)=2
thus for n=2 --> x(2)=x(1)+1+x(1)/3=2+1+1/3=4+1/3
But the wording of the problem states that n=2 ---> x(2)=4+1/2
Likewise n=3 --> x(3)=x(2)+1+x(2)/4 = 4.33333+1+4.3333/4=6.4166
This disagrees with my interpretation of the problem which leads
to x(3)=4.5+2.8333=7.3333
Am I misinterpreting something again?
...................................**************************************
There are so many ways of being wrong and only one or two of being right.
The only way to stay out of trouble is to keep quiet.