Area and circumference of a square?

I've encountered this weird question: Show that the derivative of the area of a circle with respect to its radius is equal to its circumference. Show that this relationship holds between the area and circumference of a square, too. Explain what you mean by the radius of a square.

I don't have any trouble with the first part, $A=\pi r^2$, $A'= 2\pi r$ and $c = 2\pi r$. Now for the square, my speculation was that the "radius" is half of the diagonal ($\frac{r\sqrt 2}{2}$) but it's derivative is not equal to the perimeter (or "circumference"). Something that makes sense in terms of the numbers is $A$ equal the whole diagonal ($r \sqrt 2)^2$. But I don't know if it means anything on a geometry basis.

What are your thoughts on this?

Thanks!

• The question is asking about the relationship between the perimeter and the derivative of area, not the perimeter and the derivative of radius. – Travis Oct 7 '15 at 7:06

Let the "radius" of a square be the perpendicular distance from the centre to a side, or equivalently the radius of the inscribed circle, or equivalently half the side.

Then the area of a square of radius $r$ is $4r^2$, and the perimeter is $8r$. Note that the perimeter is the derivative of the area with respect to $r$.

Remark: Let the "radius" $r$ of a regular $n$-gon be the radius of the inscribed circle. Join the centre $O$ to two adjacent vertices $A$ and $B$. Let $2\theta=\angle AOB$. Then the side $AB$ is equal to $2r\cot\theta$, so the perimeter is $(2n\cot\theta)r$.

The area of the polygon turns out to be $(n\cot\theta) r^2$. Thus again, if we measure the "size" of the polygon using $r$, the perimeter is the derivative of area! If we think of the polygon as made up of "shells" then the fact this should happen becomes geometrically clear.

• +1. I believe this is true in higher dimensions too, e.g for a ball and sphere, for Platonic solids, and for n-spheres and regular polytopes – Henry Oct 7 '15 at 7:53

This is sort of a research problem: the goal is to formulate on your own a notion of radius that leads to nice properties. Even better if you can come up with some heuristic or intuitive rationale for the formulation beyond the mere fact it has the nice property you're looking for.

To just find a good definition of radius, we could "cheat". If $s$ is the side of a square, then $A = s^2$ and $P = 4s$. Whatever "radius" might mean, it should satisfy

$$16 A(r) = P(r)^2$$

and consequently

$$16 A'(r) = 2 P(r) P'(r)$$

If we insist that $A'(r) = P(r)$, then that means $P'(r) = 8$, from which we can quickly deduce that $r$ should be half the length of the side.

I say "cheat" above, but this sort of approach is a standard thing to do -- if you want something to be true, then you assume it is true, and then solve for what you need to have in order for it to be true.

As for a rationale why this is reasonable, consider a circle of radius $r$, and another circle of slightly larger radius $r'$, and convince yourself geometrically that the difference in the areas should be approximately the perimeter times the change in radius: that is,

$$A(r') - A(r) \approx P(r) (r' - r)$$

Once this idea becomes intuitive, now imagine the same thing with squares; the same intuition should tell you that a radius has to be perpendicular to the sides (so that the change in radius is perpendicular).