*(At last month’s big MathsJam conference, we asked a few people who gave particularly interesting talks if they’d like to write something for the site. A surprising number said yes. First to arrive in the submissions pile was this piece by Tom Button.)*

The formula for the surface area of a sphere, $A=4\pi r^{2}$, is the derivative of the formula for the volume of a sphere: $V=\frac{4}{3}\pi r^{3}$.

This result does not hold for a cube with side length $a$ if the surface area and volume are written in terms of $a$. However, if the surface area and volume are written in terms of half the side length, $r=\frac{a}{2}$, you get the surface area $A=24 r^{2}$, which is the derivative of the volume, $V=8 r^{3}$.

If you apply a similar technique to a tetrahedron with side length $a$, you need to write the surface area and volume formulae in terms of $r=\frac{\sqrt{6}}{12}a$. This gives $A=24\sqrt{3} r^{2}$ which is the derivative of the volume, $V=8\sqrt{3} r^{3}$.

$\frac{a}{2}$ is the inradius (radius of the insphere) of a cube with side length $a$, and $\frac{\sqrt{6}}{12}a$ is the inradius of a tetrahedron with side length $a$. This leads to a marvellous result:

*The formula for the surface area is the derivative of the formula for the volume if they are both written in terms of the inradius.*

The result holds for all the platonic solids. For example an octahedron with side length $a$ has inradius $r=\frac{\sqrt{6}}{6}a$. This gives surface area $A=12\sqrt{3} r^{2}$, which is the derivative of the volume, $V=4\sqrt{3} r^{3}$.

I think it also works for any polyhedron whose faces are all tangent to a single sphere.

The equivalent result holds in 2D for regular polygons: the formula for the perimeter is the derivative of the formula for the area if they are both written in terms of the radius of the incircle. I think this works for non-regular polygons too, provided that all the sides are tangents to the same circle.

*The original version of this was presented at the annual MathsJam conference, November 2012. David Fontaine’s table of properties of platonic solids was very useful.*

I liked this talk. I’ve seen the result that the surface area is the derivative of the volume before. It isn’t coincidence; the surface of a convex shape is the infinitesimal ‘extra bit’ you keep adding to make the shape bigger.

It makes sense that it should work for any polyhedron whose faces are tangent to a sphere. The sphere is a neat way of making sure all the faces move outward in a direction normal to themselves — if they didn’t then you’d have to have cosines in there weighting things, and it would stop being so neat.

I was going to ask if anyone could give a proof; that counts as a proof for me!

It does indeed generalise to all solids which have an insphere. Andrew’s assertions are perfectly correct, but we can prove this rigorously:

Let $V(r)$ be the volume as a function of the inradius $r$. Let $S$ and $T$ be geometrically similar polyhedra with concentric inspheres. The inradii of $S$ and $T$ are $r$ and $r+h$, respectively. Imagine extruding prisms of height $h$ on each of the faces of $S$, so that the tops of the prisms are incident with the faces of $T$. Note that the volume of the prisms is given by $Ah$, where $A$ is the surface area of the original polyhedron.

So, $V(r+h) = V(r) + Ah + \mathcal{O}(h^2)$. The $\mathcal{O}(h^2)$ term is the small excess volume near the edges of the polyhedron.

The definition of the derivative $\frac{dV}{dr}$ (evaluated at $r$) is given by the limit as $h$ tends to zero of $\frac{V(r+h)-V(r)}{h}$. This simplifies to $\frac{Ah + O(h^2)}{h} = A + \mathcal{O}(h)$. As $h$ tends to zero, this approaches $A$. Hence, the derivative $\frac{dV}{dr}$ is indeed the surface area of $S$.

Top stuff! I’ve LaTeXed your comment because I felt it would benefit from it.