If you take the derivative of the volume of a sphere, $$\frac{4}{3}\pi r^3$$ you get its surface area, $$4\pi r^2$$ If you differentiate again, you get $$8 \pi r$$ Does this have any physical (or other kind of) significance, besides being $4$ times the length of a great circle on the sphere?
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13It also represents how much the volume accelerates as you increase the radius. – J126 Apr 03 '17 at 21:14
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9Similar to the above comment, it is the rate of change of the surface area with respect to the radius. – Tyberius Apr 03 '17 at 21:16
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2If you generalize into $\boldsymbol{n}$-sphere, the second derivative equals to surface area of $(n-2)$-sphere times $\dfrac{2n\pi R^2}{n-1}$, that is $$(V_{n+1})''=(S_{n})'=\frac{2n\pi R^2}{n-1}S_{n-2}$$ – Ng Chung Tak Apr 03 '17 at 21:54
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The lenght of 4 maximum circles? – Dac0 Apr 04 '17 at 04:35
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Is it possible that it is the sum of all the circumferences of all parallel cross sections? – Digcoal Nov 29 '21 at 00:14
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$8\pi r = 4\pi \times \ell = 2\times H (4\pi r^2)$ where $\ell = 2r$ is the mean width for a ball with radius $r$ and $H = \frac1r$ is the mean curvature for corresponding sphere.
Given any convex body $K$ in $\mathbb{R}^3$ and unit vector $n \in S^2$, let $w(K,n)$ be the width of shadow when one project $K$ orthogonally to the line $\{ \lambda n : \lambda \in \mathbb{R} \}$, the mean width of $K$ is the angular average of $w(K,n)$ for $n$ varies over $S^2$:
$$\ell(K) \stackrel{def}{=} \frac{1}{4\pi} \int_{S^2} w(K,n)$$
For any $\epsilon > 0$, let $B_\epsilon = \{ x \in \mathbb{R}^3 : |x| \le \epsilon \}$ be the sphere centered at origin with radius $\epsilon$.
If one form the Minkowski sum of $K$ with $B_\epsilon$, i.e.
$$K_\epsilon = \{ x + y : x \in K, y \in B_\epsilon \}$$
The volume of $K_\epsilon$ depends on $\epsilon$ polynomially,
$$\verb/Vol/(K_\epsilon) = \verb/Vol/(K) + \verb/Area/(K)\epsilon + 2\pi\ell(K)\epsilon^2 + \frac{4\pi}{3} \epsilon^3$$
This means in general, we have
$$\left.\frac{d^2}{d\epsilon^2}\verb/Vol/(K_\epsilon) \right|_{\epsilon=0}= 4\pi \ell(K)$$
In particular, for a sphere with radius $r$, we have
$$(B_r)_\epsilon = B_{r+\epsilon} \quad\implies\quad \frac{d^2}{dr^2}\verb/Vol/(B_r) = \frac{d^2}{d\epsilon^2} \verb/Vol/((B_r)_\epsilon) = 4\pi \ell(B_r) = 4\pi(2r) = 8\pi r$$
When $K$ is smooth enough, the mean width of $K$ is related to the average of mean curvature $H$ over the surface bounding $K$. In fact
$$\ell(K) = \frac{1}{2\pi}\int_{\partial K}H \quad\implies\quad \left.\frac{d^2}{d\epsilon^2}\verb/Vol/(K_\epsilon)\right|_{\epsilon=0} = 2\int_{\partial K}H$$ For a sphere with radius $r$, the expression $8\pi r$ can be decomposed as
$$8 \pi r = 2 \times \frac1r(4\pi r^2)$$
and the RHS can be interpreted as twice the integral of mean curvature $H = \frac1r$ over a surface of area $4\pi r^2$.
achille hui
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