Why is the by-product of a circle's area its border (and also in a similar way for rounds)?

When set apart relative to $r$, the by-product of $\pi r^2$ is $2 \pi r$, which is the area of a circle.

In a similar way, when the formula for a round's quantity $\frac{4}{3} \pi r^3$ is set apart relative to $r$, we get $4 \pi r^2$.

Is this simply a coincidence, or exists some deep description for why we should anticipate this?

2019-05-07 01:51:02
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Answers: 1

Consider raising the distance of a circle by an infinitesimally percentage, $dr$. This raises the area by an annulus (or ring) with internal distance $2 \pi r$ and also external distance $2\pi(r+dr)$. As this ring is exceptionally slim, we can visualize reducing the ring and afterwards squashing it bent on create a rectangular shape with size $2\pi r$ and also elevation $dr$ (the side of size $2\pi(r+dr)$ is close adequate to $2\pi r$ that we can overlook that). So the area gain is $2\pi r\cdot dr$ and also to establish the price of adjustment relative to $r$, we separate by $dr$ therefore we get $2\pi r$. Please keep in mind that this is simply an insightful, instinctive description in contrast to an official evidence. The very same thinking collaborates with a round, we simply squash it bent on a rectangle-shaped prism rather.

2019-05-08 22:50:29