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The Watchdog (Posted on 2004-10-28) Difficulty: 5 of 5
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?

See The Solution Submitted by Erik O.    
Rating: 3.4000 (5 votes)

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Solution Pi Dog Night | Comment 16 of 29 |

First, my answer:
(200/3)*Pi^3 + 200*Pi - 400  =  2295.4 sq ft

Semi disc area is 50 Pi^3
Then add the 2 'horns'

To figure one horn, I drew a circle with the dog tethered at the western most point, or point 'W'.  Angle 'a' measured from the west to east line.  Let 'a' vary from zero to pi/2.  When a is zero, dog is stuck on the east side of the pen.   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. 
Where is point L?  Well, the chord length from point W to point L is 20 cos(a).

So the area of one horn is the double integral of the area:
r dr da
as angle a goes from zero to pi/2, and as r goes from point L to the end of the rope.
Integ{'a' from 0 to pi/2}  Integ{'r' from 20 cos(a) to 20cos(a)+20a}  r dr da
The inner integral becomes r^2 / 2 :
[(20cos(a) + 20a)^2 - (20 cos(a))^2]/2
[200 a^2 + 200 a cos(a) ]
So, one horn is Integ{'a' from 0 to pi/2} [200 a^2 + 200 a cos(a) ]   da
= (25/3) pi^3 + 200 Integ{0 to pi/2} a cos(a) da
This last integral is:   cos(a) + a*sin(a) =  -1 + pi/2
= (25/3) pi^3 - 200 + 100 pi   (but this is only 1 of 2 horns)
Total area = 50 pi^3  + (50/3) pi^3  + 200 pi - 400

2295.4 sq ft

  Posted by Larry on 2004-10-29 14:07:03
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