Volume = $\frac{4}{3}\pi r^{3}$
First Derivative is the surface area of the sphere = $4\pi r^{2}$
Second derivative is = $8\pi r$ ...
But what does this physically represent, or does it have any physical meaning?
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Related: https://math.stackexchange.com/questions/2216646/physical-significance-of-8-pi-r-the-second-derivative-of-the-volume-of-a-sphe?rq=1 – CBBAM Mar 07 '23 at 03:03
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2The volume can be obtained by "summing" a bunch of surfaces of volume $4\pi r^2 dr$. Essentially you're building up the sphere by spherical shells. The fundamental theorem says then that $\frac{dV}{dr} = 4\pi r^2$. The surface cannot be similarly obtained since the surface exists at fixed radius, i.e. $dr = 0$. As to whether $8\pi r$ has any meaning, I can't say. – Charles Hudgins Mar 07 '23 at 03:03
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I presume it means the rate of lengthening for the circumferences of the 4 circles that compose the sphere's surface area (the black lines between different colors on the basketball below), since each circumference is 2(pi)r (each color is equivalent to the area of one circle with the same radius as the sphere).
Nate
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