Solution to Problem 242


Correct solutions were submitted by Ron Welch, Lou Cairoli, Nancy Schwarzkopf, Bill Webb, Cee Ann Franklin, Brett Hendricks, Juan Carlos Marivela, Bryan Fluhrer, Will Shelton, Francisco de Leon-Sotelo y Esteban and Adolfo Alvarez Buelta.

To use calculus, we start by parametrizing the curve followed by the string.
In rectangular coordinates, this might be x=(4/2pi)cos t, y=(4/2pi)sin t, z=(3/2pi)t with t from 0 to 8pi.
In cylindrical coordinates, we could get r=4/2pi and theta=t with the same z and the same range on t.
We then set up an integral to calculate the arc length.
In rectangular coordinates, the differential of arc length is the square root of dx2+dy2+dz2.
In cylindrical coordiantes, it is the square root of dr2+r2dtheta2+dz2.
Either way, this simplifies to 5/2pi.
Integrating from 0 to 8pi, we get that the length is 20 (cm).

The simplicity of this result suggests that a more elementary method exists.
Consider unrolling the cylinder.
Then we can think of the string following the diagonal of a rectangle 16 cm wide (4 cm of circumference, traversed once for each of the four revolutions) and 12 cm high.
By the Pythagorean Theorem, this has a length of 20 cm.

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