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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 Finally | Comment 23 of 29 |

OK I think I have it now.
New concept from my previous one.
Dog anchored at east end of building.  Dog tries to go as far as possible so that rope hugs building then at point T, rope diverges from the building at a tangent.  Point D is the dog's extreme position.  Theta is angle from center of building to the tangent point T.  R is radius of building.  r is distance from center of building.

Plan:  For a given theta, find point T, the length of rope on and off the building, the slope of the rope between T and D, and the coordinates of point D.  Find parametric equations for x(theta) and y(theta) representing the extreme positions of the dog, ie point D.  Then find sqrt(x^2 + y^2) which is r(theta).  Then find double integral of r dr dtheta  as theta goes from 0 to pi, and as r goes from 0 to r(theta).  That should be the area of the north half of the bldg plus the north half of roaming area.  Double it.  Subtract off the area of the building.  Add in the big half circle of dog country (where the rope never touches the building).  And that should be it.
(I'm leaving out the word "theta", every sin or cos is of theta unless otherwise noted)
Point T is (r cos, r sin)
Slope of rope is  theta + pi/2
Length of rope touching bldg:  R*theta
Total rope length:  R*pi
Length of rope not on bldg:  R*(pi - theta)
Point D:
   x(theta)/R= cos + R*(pi - theta)*cos(theta+pi/2)
   y(theta)/R= sin + R*(pi - theta)*sin(theta+pi/2)

   x(theta)/R= cos - pi*sin + theta*sin
   y(theta)/R= sin + pi*cos - theta*cos

   (x^2 + y^2)/R^2= 1 + (pi - theta)^2
   so r(dog)= R*sqrt(1 + (pi - theta)^2)

Double Integral:
inner part is Int(r=0 to r(dog)) r dr   =  r^2/2
which is   R^2*(1 + (pi - theta)^2)/2
which is R^2/2 *(theta^2 -2pi*theta + theta^2 + 1)
Outer integral is from theta =0 to pi of the above line
which is R^2/2 *(pi^3/3 + pi)
double it:  R^2 *(pi^3/3 + pi)
subtract pi*R^2 (the bldg):  R^2 * pi^3 / 3
Add the big half circle whose radius is the total rope length (rope is pi*R, big half circle is R^2 * pi^3 / 2
Total is (1/2 + 1/3) pi^3 R^2
  =(5/6) pi^3 100
  =2583.856        which agrees with Richard's answer

  Posted by Larry on 2004-10-30 13:27:45
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