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07312007  #1 
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calculus chicken problem?
Circumference of the earth is 25,000 miles. Imagine you have a wire that is 25,000 miles plus 3 feet, and you wrap the wire around the equator of the earth. You wrap the wire so that both ends meet and do not overlap. The wire is hovering above the equator, now is there enough room between the earth and the wire for a 5 inch tall chick to walk in between? Or will the chick simply step over the wire? Or will the wire be at a height where the chick cannot go under or over the wire? Keep in mind that the earth is not a perfect circle, it is more of an ellipsoid shape. Details please.

07312007  #2 
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The chicken will fit underneath.For each circle, the radius and the circumference are related bycirc = 2 * pi * radiusfor the earth itself: 25,000 mi = 2 * pi * radius_of_earthfor the wire:25,000 mi + 3 feet = 2 * pi * radius_of_wirerearrange and combine:radius_of_wire = (25000 mi + 3 feet) / (2 * pi) = 25000 mi / (2*pi) + 3 feet / (2*pi) = radius_of_earth + 3 feet / (2*pi) the height of the wire is the difference between the two radii,which is 3 feet / (2*pi), which works out to something like 56 inch.

07312007  #3 
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I don't really see any calculus in this problem, it's more just geometry.25000 miles + 3 feet = 132,000,003 feet25000 miles = 132,000,000 feetThe radius of the earth is 132,000,000 ft / (2pi)= 21,008,452.488 ftThe radius of the wire is...132,000,003 / (2pi)=21,008,452.966Subtract one from the other leaves0.477 ft. (about 6 inches)ANSWER:That's about 6 inches above the ground. I suppose a chicken could either go under or over the wire if it felt like it. I'm no chicken expert.If the wire were electricuted, then we could have BBQ chicken! (I guess I should go eat dinner, huh.)

07312007  #4 
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Assume the earth is a circle because the math is simpler.h = height of the wire above the groundC(Earth) = 25,000 = 2pi*rC(Wire) = 25,003 = 2pi*(r+h)2pi(r+h)  2pi*r = 2pi*h2pi*h = 3'h = 3/2pi = .477' = 5.73"The 5" chick can walk between

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