VCRTAPE
Tape thickness and speed.
BRIEF ABSTRACT
It's easy to estimate how thick a piece of paper is by measuring a ream of 500 sheets.  This (fairly easy) problem deals with estimating the tape thickness on a video cassette, and then asks for a function which represents the tape driver speed.  The problem is accessible to precalculus students, but accuracy and modeling issues also make it an appropriate application of polar integration!
GENERAL INFORMATION
FileName:  VCRTAPE
Full title:  Tape thickness and speed.
Last Update: 5/31/96
Developer: 
Aaron Klebanoff, Department of Mathematics,  Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA.
Contact:  
Aaron Klebanoff, Department of Mathematics, 
Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. 
Phone:  812-877-8151.  Email:  Klebanoff@rose-hulman.edu.  
FAX:  812-877-3198.
Support:  
The production of this material is supported by the National Science Foundation under Division of Undergraduate Education grant DUE-9352849:  Development Site for Complex, Technology-Based Problems in Calculus with Applications in Science and Engineering and the  Arvin Foundation of Columbus IN. 
STATEMENT OF PROBLEM
The following specifications were taken for a certain model of video cassette recorder.
	Tape Speed:			1.3125 inches per second
	Play Time:			2 hours
	Radius of tape spindle:	12 mm
	Outer radius of wound tape:	43 mm
1) How thick is the tape?
2) Derive a function that gives the tape spindle driver speed (in rotations per minute -- rpm) as a function of the radius.   Note that the tape speed is constant.
KEYWORDS
Angular speed, circumference, arc length in polar coordinates.
TEACHER NOTES
ISSUES RELATED TO THE PROBLEM 
Prerequisites
Algebra, circumference of a circle; (polar coordinates in calculus provides an optional approach.)
Time allotment - time management
Expectations 
Some students may need hints to Problem 2 such as: (1) what's the spindle speed when the radius is X? or Y?
Future payoffs
Extensions
A much harder problem is to determine the tape driver speed as a function of time.  Offer as a challenge problem.
References and Sources
POSSIBLE SOLUTION(S)
1) How thick is the tape?
We first convert all unit lengths to inches and unit times to seconds.
TapeSpeed = 1.3125
1.3125
PlayTime = 2 3600
7200
RadSpind = 12/25.4
0.472441
RadWound = 43/25.4
1.69291
TapeLength = TapeSpeed PlayTime
9450.
Approach 1:  Model with cylindrical shells.
The number n,
n = (RadWound - RadSpind)/Thick
1.22047
-------
 Thick
is the approximate number of shells required.
We add up the circumference of each shell to solve for the thickness.

Note that the first shell has circumference 2 p (RadSpind + Thick) and each shell around that has circumference given by 2 p (RadSpind + i*Thick) where i goes from 1 to n.  Using the fact that Sum[i, {i, 1, n}] = n(n+1)/2, we can simplify the sum of circumferences
	Sum[2 p (RadSpind + i*Thick), {i, 1, n}]
as
	2 p (RadSpind n + Thick n(n+1)/2) = 2 p n (RadSpind + (n + 1) Thick/2)
thickness = Flatten[
 Solve[2 Pi n(RadSpind + (n+1) Thick)/2 ==
         TapeLength, Thick]];
TapeThickness1 = Thick /. thickness
0.000687159
Approach 2:  Model as a spiral -- use polar coordinates.
As theta: 0 -> 2 p, we want the spiral to increase in radius by Thick.  We also require that when theta = 0, that the radius is RadSpind.  Finally, we want the length to be TapeLength when the radius is RadWound.
We know the total arc length.  So, our goal is to determine the arc length of a spiral as a function of the spacing until it becomes the correct length.
r[theta_] = RadSpind + theta Thick/(2 Pi)
           theta Thick
0.472441 + -----------
              2 Pi
rprime = r'[theta]
Thick
-----
2 Pi
PolarArcLength = Integrate[Sqrt[r[theta]^2 + rprime^2],
 {theta, 0, 2 Pi n}];
thickness = 
 FindRoot[PolarArcLength == TapeLength, {Thick, 0.0006}];
TapeThickness2 = Thick /. thickness
0.000878567
m = n /. Thick -> TapeThickness2;
PolarPlot[RadSpind + theta TapeThickness2/(2 Pi),
 {theta, 0, m 2 Pi},
 PlotPoints -> 400]

-Graphics-
2) Derive a function that gives the tape spindle driver speed (in rotations per minute -- rpm) as a function of the radius.
We require the tape speed to remain constant, so the spindle speed will have to vary with the radius of the tape:
SpindleSpeed[radius_] := TapeSpeed/(2 Pi radius)
Plot[SpindleSpeed[radius],
 {radius, RadSpind, RadWound},
 AxesLabel -> {"radius of wound tape [in]", "rpm"},
 PlotRange -> {{0, 2}, {0, 1}}]

-Graphics-
ISSUES IN SOLUTION
Neither of the tape thickness models offered necessarily yields the true tap thickness.  However, they both give good approximations.  Also, giving two solutions provides a nice way to check our work.