A watchdog is tied to the outside wall of a round building 20 feet in diameter If the dog s

Radius of the dog house is 10.

L, the length of chain, is 10 PI

Suppose dog is 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 L^2 / 2 ..... 50 PI ^3 square feet.

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 X^2 dA / 2.

So..integrate from 0 to PI, the function X^2 / 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 ( PI^2 -2•PI•A + A^2 ) times dA

and the integral is

50 ( A•PI^2 - PI•A^2 + A^3 / 3 ) between limits 0 and PI...
making it

[50 ( PI•PI^2 - PI•PI^2 + PI^3 / 3 )] -[ 50 ( 0•PI^2 - PI•0^2 + 0^3 / 3 ) ]

which makes

50 ( PI3 - PI3 + PI3 / 3 )

which makes

50•PI^3 / 3

and we must double it as we have only considered going one way around the building..
So, the total area, east and west = 50•PI^3 + PI3•100 / 3

making

(PI^3)*250 / 3
13 years agoHelpfull: Yes(1) No(0)
50 Pi^3

the length of chain is 10 PI.
13 years agoHelpfull: Yes(0) No(0)
10 pi
the length of the patrol should be equal to the circumference of round building
13 years agoHelpfull: Yes(0) No(1)

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?