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?

The semicircular area from the point of attachment to plus or minus pi/2 is 50*pi(cubed) - we agree on that.

Using the center of the building as the origin, I calculated the y value of the quasi-ellipse shape formed as the dog goes around the building and his travel distance from the building gets less and less as his leash wraps around the building:

y = SQRT(100-x(squared)) + 10*pi - 10*inverse cosine of x/10 (inverse cosine is in radians)

(10*pi - 10*inverse cosine of x/10 is the length of the leash minus the length of the arc of the building it is wrapped around.)

The integral of that, from -10 to 10, will give the area of the entire shape, but we need to subtract the area of the building, which is the integral of the first term.  So, just drop the first term and integrate, from x = -10 to 10,  10*pi - 10*inverse cosine of x/10, and multiply by two. The result is 200pi.

Total area is 200 pi + 50*pi(cubed).

 

 

Posted by lake on 2004-10-29 15:39:22