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 re: WD-40 please! - spoilers by Charlie)

Charlie, that makes so much sense. I was wondering why my integration was 1 dimension (and not 2 as an area should be), and it’s because the base of the triangles you talked about brings in the other dimension.

Ok, with Charlie’s help, this is what I get for the area of both horns:

S = pi*D/2 – pi*D*theta/(2pi) = pi*D/2 – D/2*theta = D/2(pi – theta)

2B = 2* (integral from 0 to pi) S^2/2 dtheta
= (Int. 0-pi) S^2 dtheta
= (Int. 0-pi) (D/2(pi – theta))^2 dtheta
= (Int. 0-pi) (D/2)^2(pi – theta)^2 dtheta
= D^2/4 (Int. 0-pi) (pi^2 – 2pi*theta + theta^2) dtheta
= D^2/4 [pi^2*theta – pi*theta^2 + 1/3*theta^3] (0-pi)
= D^2/4 [pi^3– pi^3 + 1/3*pi^3]
= D^2/4 [1/3*pi^3]
= 1/12*D^2*pi^3

So the total patrolling area is

A + 2B = 1/8*pi^3*D^2 + 1/12*D^2*pi^3
= 5/24*pi^3*D^2
= 2583.85639 sq. ft.

I think that’s better. Thanks Charlie! =)

Posted by nikki on 2004-10-28 17:08:31