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.
no subject
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.