The Watchdog - Solution
by
Erik Oosterwal
Thanks to my friend, David Byrden, we
now have a nice concise solution to this problem.
Radius of the dog house is 10.
L, the length of chain, is 10 PI
I put my dog on the east side of the building.
As long as he hunts to the east, he has the full length of chain free, and
he can cover a half a disc... with area PI L2 / 2 .....
50 PI3 square feet.
p.s. what is a "foot". You quaint people...
As he hunts to the west, his free chain length reduces linearly as his angle
increases...the chain turns through PI, from north to south, before it runs
out. For an angle increment "dA", and chain length X, the area patrolled
is X2 dA / 2.
So..integrate from 0 to PI, the function X2 / 2 times dA
with respect to A....and substitute for X the function 10 ( PI - A )
That makes it: integrate ( 10 ( PI - A ) )2 / 2 times
dA
which is
integrate 50 ( PI2 -2•PI•A +
A2 ) times dA
and the integral is
50 ( A•PI2 - PI•A2 +
A3 / 3 ) between limits 0 and PI....making it
50 ( PI•PI2 - PI•PI2 +
PI3 / 3 ) minus 50 ( 0•PI2 -
PI•02 + 03 / 3 )
which makes
50 ( PI3 - PI3 +
PI3 / 3 )
which makes
50•PI3 / 3
and we must double, since I only considered going one way around the building..
So, the total area, east and west, is:
50•PI3 + PI3•100 / 3
making
PI3•250 / 3
All original puzzles and solutions are © Erik Oosterwal 1993-2008
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