A watchdog is tied to the outside wall of a round building 20 feet in diameter. If the dog's chain is long enough to wind half way around the building, how large an area can the watchdog patrol?

(In reply to Pi Dog Night by Larry)

"When a is pi/2, dog is straight north of point W.   When a is between those extremes, consider the rope wrapped tightly on the circular fence until angle a is reached, then rope goes radially away from point W.  Call the point where the rope leaves the fence point 'L'. How much rope length is not touching the fence?  The same as the length of fence remaining between point L and the eastern point.   That length is 20*a. "

That is a linear relationship between  angle a and the free amount of rope, and is zero when a is zero and 20*pi/2 when a is pi/2.  You'd expect the free amount of rope to be 20*pi/4 when a is pi/4.  However, when the free amount of rope is 20*pi/4 (i.e., half the rope), the free half is tangent to the north point on the building (it is a building, not a fence). In cartesian coordinates it is 10+20*pi/4 feet east of the fixed tether and only 10 feet north of the tether, meaning the angle a from the tether is not pi/4, so the relationship between the angle a and the free amount of rope is not linear.

Posted by Charlie on 2004-10-29 17:31:38