pi-related neatness

[javasaurus]

Suppose you have a string that's about 31.4 meters long (actually 10 pi meters for those who care). Then it can be wrapped tightly around a cylinder with diameter 10 meters. If you add 6.3 meters to the string, you can form the lengthened string into a circle, with the cylinder at the center, and the space between string and cylinder is 1 meter all the way around.

The Earth is about 12,756,000 meters in diameter, or about 40,000,000 meters in circumference. Wrap a string (it'll be approximately 40Mm long) tightly around the Earth. Now make the string 6.3 meters (yes, meters!) longer, as in the previous example. If you form the string into a circle with the earth at its center (like the cylinder before), how much gap do you get between the Earth and the string?

The answer may surprise you!
You get the same gap as with the cylinder, 1 meter, all the way around!

Comments

[javasaurus.livejournal.com]

Okay, got an answer for the "what does it mean."

First, I'll point out that the same derivative thing works with squares. If r is half the length of a square's side, then the area is 4*r^2, and the border is length 8*r. I'm going to use squares to explain, but the same idea works for circles.

Recall that derivative is rate of change. In this case, how fast the area changes as r increases. Suppose r increases by a very small amount, h. Then the area increases by 8*r*h + 4*h^2. But h is very small, so the second term is negligible, and the increase is just the 8rh, or 8r per unit of h. This is like adding a thin layer to the outside of the square.

That's kind of loosely stated, but it gives the idea.