Consider a sphere for example.

It's volume is calculated by the formula: $\frac 4 3 \pi r^3$

The derivative of that is $4\pi r^2$ which represents the sphere's surface area.

The derivative of that is $8\pi r$.

Does that quantity represent something tangible about a sphere? Clearly, it is some sort of linear measurement, but of what, I'm not sure.

Just curious.

  • $\begingroup$ Consider the disc and the circle. $\endgroup$ – hamam_Abdallah Aug 3 '17 at 22:39
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    $\begingroup$ @Salahamam, yes I see that the derivative of the circle's area is it's circumference, but I am having trouble picturing what the Surface Circumference of a sphere is? Why 8πr? $\endgroup$ – Octopus Aug 3 '17 at 22:42
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    $\begingroup$ My other related question is: integrate the circle's area and you get 1/3⋅πr³, but what is that to the circle? Maybe nothing, but maybe there is a real way to visualize what these quantities represent. $\endgroup$ – Octopus Aug 3 '17 at 22:45
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    $\begingroup$ If you think of the derivative of the volume as $\lim\limits_{h\to0}\frac{V(r+h)-V(r)}h$, and interpret it geometrically, it becomes easier to see why it should equal $SA(r)$. However, $\lim\limits_{h\to0}\frac{SA(r+h)-SA(r)}h$ has no obvious geometrical significance. $\endgroup$ – Akiva Weinberger Aug 3 '17 at 22:50
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    $\begingroup$ Try thinking of the sphere as covered with lines of latitude and longitude, like a globe, but very finely, so the little areas are almost plane. Now inflate the sphere. The areas will all tear apart like postage stamps and these tears will expose a bit more of the area. It has to be the case that the new area exposed is $8\pi r\;dr$, and it would also be the case if the stamps were defined by arbitrary surface coordinates. Does that line of thinking lead anywhere? $\endgroup$ – Philip Roe Aug 3 '17 at 23:07

Maybe first review why the derivative of the volume is the surface area: It's because $4\pi r^2 \,dr$ is the incremental change in volume when you increase the radius by $dr.$ It makes sense that $4\pi r^2$ is the surface area since the then $4\pi r^2\,dr$ is the volume of a spherical shell, which represents the added volume.

So it's certainly true that $ 8\pi r \,dr $ is the incremental change in surface area when we increase the radius of the sphere by $dr.$ But is there a nice way to picture the difference in surface area in terms of a geometrical object? I can't think of any.


A general identity says \begin{align} & \Big(\text{size of boundary}\Big) \times \Big( \text{rate of motion of boundary} \Big) \\[5pt] = {} & \Big( \text{rate of change of size of bounded region} \Big) \tag 1 \end{align} (This equality has no standard name as far as I know; I have sometimes called it the boundary rule.)

It can be looked at this way: $$ \frac{d(\text{size of bounded region})}{d(\text{location of boundary})} = \text{size of boundary} $$ In particular $$ \frac{d(\text{volume of sphere})}{d(\text{radius of sphere})} = \text{surface area of sphere} $$ Can we view the surface of the sphere as growing by virtue of the motion of a boundary, when the radius grows? I don't know how to do that, nor how to find any sort of size of such a supposed boundary, nor what it would mean to speak of the amount by which the location of such a boundary changes (and hence to speak of a rate of motion of the boundary).

Therefore I don't see how such a geometric interpretation of the second derivative can be given. However, the fact that I don't know how to do it doesn't mean it can't be done. But it does make one wonder why I am typing this answer. Part of the reason is that I do know how to do such a thing with a cube.

Consider the cube $[0,s]^n$ in $n$-dimensional Euclidean space. As $s$ changes, there are $n$ moving boundaries, each of which is an $(n-1)$-dimensional cube $$ \underbrace{ \cdots \times [0,s]\times \cdots \times [0,s]} {} \times \{s\} \times {} \underbrace{[0,s]\times \cdots \times [0,s]\times\cdots} {}. $$ The volume of this boundary is $s^{n-1}$ and its rate of motion is the rate of change of $s,$ and thus from $(1)$ we conclude that $\dfrac{ds^n}{ds} = ns^{n-1}.$ The factor $n$ in front of $s^{n-1}$ comes from the fact that that is how many such boundaries there are.

So what about the second derivative? Some of those $n$ components of the whole boundary interface with others and some interface with the motionless $(n-1)$-dimensional coordinate hyperplanes. Those latter interfaces make up the "boundary of the boundary". Each of the $(n-1)$-dimensional components of the boundary of the whole cube has $n-1$ such $(n-2)$-dimensional components of its boundary. Hence $n(n-1)s^{n-2}$ is the size of the "boundary of the (moving part of the) boundary".

How can such an idea be applied to the sphere? At this point I don't know . . .


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