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?

Creeeeek creeeeek

Sorry, that was my rusty calc causing that commotion.

Alrighty, I thought I had a solution, but I ended up getting 0 for something that should not have been zero. I’m still going to post it, though, because I think I am on the right track, but I just make a mistake somewhere. Hopefully one of you guys can point it out.

Ok, I am going to first define my setup. I draw a circle representing the building, with Point O in the center. I have the dog chained to the building at the bottom, labeled as point C. I kind of sketch out the basic patrolling area. I draw a line tangent to the building at Point C. Since I put point C at the bottom, this tangent line happens to be horizontal to me.

Now I have 3 chunks of area. A large semi-circle on the bottom, a "devil horn" shaped area on the left of the building, and another devil horn shape on the right of the building. I will call the area of the semi-circle A, and the area of the left horn B. The area of the right horn is just B as well. So the total patrolling area of the dog will be A+2B.

Well, A is easy enough to figure out because it is just a semicircle with radius equal to the dogs chain. If the diameter of the building is D, then the length of the dog’s chain is just L = (pi*D)/2. So A = (pi*L^2)/2 = pi/2*(pi*D/2)^2 = 1/8*pi^3*D^2.

Now for the tough part that requires WD-40. I’m just going to find the area of the left horn, which is B. I’ll ignore the right side because I can just double B to account for that. Ok, imagine the dog walking into area B, reaching for his limits, pulling the chain taut. As he walks around, part of his chain will be wrapped around the building, and part of it will be tangent to the building (since he is pulling it tight). I’m not sure if this is right, but the way I want to approach it is to calculate that length of change that is tangent to the building, "add them all up," and get my area.

Ok, let’s look at the dog somewhere in area B, still pulling the chain tight. The end of the wrapped length, and the beginning of the tangent length will be at point T. Pay attention to angle TOC. As the dog walks closer to area A, T moves closer to C, and angle TOC approaches 0. As the dog walks closer to the "top" of the building, T moves away from C, and angle TOC approaches 180.

So I can define my tangent length of rope in terms of theta = angle TOC. So at any given point, the wrapped part of the rope, W, is just theta/360 of the building’s circumference. So W = pi*D*theta/360. The straight and tangent part of the rope, S, is just L-W. So S = pi*D/2 – pi*D*theta/360 = pi*D/2(1-theta/90)

So now don’t I just integrate S(theta) dtheta from theta = 0 to theta = 180? Is this idea wrong, or is my following integration wrong?

(integral from 0 to 180) S(theta) dtheta
= (Int. 0-180) pi*D/2(1 – theta/90) dtheta
= pi*D/2 (Int. 0-180) (1 – theta/90) dtheta
= pi*D/2 [theta – (1/2)theta^2/90] (0-180)
= pi*D/2 [theta – theta^2/180] (0-180)
= pi*D/2 * [(180-180^2/180) – (0 – 0^2/180)]
= pi*D/2 * [180-180]
=0

bah!

Posted by nikki on 2004-10-28 15:12:59