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Why is the derivative of a circle's area its perimeter (and similarly for spheres)?

We all know that the volume of a sphere is:

$V = \frac{4}{3}\pi r^{3}$

and its surface area is

$S = 4 \pi r^2$

Now we see that

$\frac{dV}{dr} = S$

As well, the area of a circle is

$A = \pi r^2$

The circumference is

$C = 2 \pi r$

Now we see again that

$\frac{dA}{dr}=C$

There may be more that I have not noticed. Why does this relationship occur?

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marked as duplicate by The Chaz 2.0, t.b., Kannappan Sampath, Jim Belk, Asaf Karagila Apr 8 '12 at 21:16

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Imagine the area between circles of radius $r$ and $r+\epsilon$. –  anon Apr 8 '12 at 16:33
    
It is worth to look at math.stackexchange.com/questions/67039/… –  Norbert Apr 8 '12 at 16:43
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1 Answer 1

Let $K_1$ be a pleasant region, fixed from now on, like a sphere of radius $1$, or a half-sphere of radius $17$, or a regular tetrahedron of side $9.5 \pi$.

Now let $K_r$ be the region obtained by scaling all linear dimensions of $K_1$ by the scaling factor $r$.

Let $V(r)$ be the volume of $K_r$, and let $S(r)$ be the surface area of $K_r$. Note that because of the way volumes and surface areas scale, $$V(r)=r^3 V(1)\quad\text{and}\quad S(r)=r^2S(1).$$ Thus $$V'(r)=3r^2 V(1)=r^2 S(1)\tfrac{3V(1)}{S(1)}=\tfrac{3V(1)}{S(1)}S(r).$$ Thus for any starting body $K_1$, the derivative $V'(r)$ is a constant times the surface area $S(r)$.

In the case where $K_1$ is the sphere of unit radius, the constant happens to be $1$. For other starting shapes $K_1$, or other measures of linear dimension, the constant will be different. But the important point is that there always is a constant $C=C_{K_1}$ such that $V(r)=C S(r)$.

Similar considerations apply in two dimensions to $S'(r)$ and a linear feature such as the perimeter $P(r)$ of $K_r$. The idea can be extended to higher dimensions.

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