One of my favorite facts about circles is the relationship between their area and their circumference. For a circle with radius r, its area is 𝜋r2 and its circumference is 2𝜋r. What’s neat here (or rather, one thing that’s neat here) is that if you take the derivative of the area formula with respect to r, you get the circumference formula! In other words, d(𝜋r2)/dr = 2𝜋r. Amazing, right?
(For those of you who are accustomed to using tau rather than pi, this still works. The area of a circle is 𝜏r2/2 and its circumference is 𝜏r. Once again, d(𝜏r2/2)/dr = 𝜏r.)
Inspired by this fact, let’s define the term “differential radius.” The differential radius r of a shape with area A and perimeter P (both functions of r) has the property that dA/dr = P. (Note that A always scales with r2 and P always scales with r.)
For example, consider a square with side length s. Its differential radius is r = s/2. The square’s area is s2, or 4r2, and its perimeter is 4s, or 8r. Sure enough, dA/dr = d(4r2)/dr = 8r = P.
What is the differential radius of an equilateral triangle with side length s?
The area of an equilateral triangle is $A = \dfrac{s^2\sqrt{3}}{4}$ and the perimeter is $P = 3s$. Let $s = k \cdot r$.
Starting from :
$$ \dfrac{\partial A}{\partial r} = P$$
$$ \left[ \dfrac{k^2 \cdot r^2 \sqrt{3}}{4} \right]' = 3 \cdot k \cdot r$$
$$ \dfrac{k^2r \sqrt{3}}{2} = 3 \cdot k \cdot r$$
$$k = 2\sqrt{3}$$
What is the differential radius of a rectangle with sides of length a and b? Your answer should be in terms of both a and b. Oh, and kudos if you can illustrate your solution geometrically!
The area of a rectangle is $A = ab$ and the perimeter is $P = 2a + 2b$.
Let $a = b \cdot m$. The area is $A = b^2m$ and the perimeter is $P = 2b(m+1)$.
Let $b = n \cdot r$. Starting from :
$$ \dfrac{\partial A}{\partial r} = P$$
$$ \left[ m \cdot n^2 \cdot r^2 \right]' = 2 \cdot (m+1) \cdot n \cdot r$$
$$ 2mn^2r = 2(m+1)nr$$
$$mn = m + 1$$
$$n = \dfrac{1}{m} + 1$$
Substituting,
$$b = r \cdot \left(\dfrac{1}{m} + 1\right)$$
$$r = \dfrac{b}{\dfrac{1}{m}+1}$$
$$r = \dfrac{1}{\dfrac{1}{bm} + \dfrac{1}{b}}$$
Or in other words, $r$ is the harmonic mean of the two sides of the rectangle. Let's see what that looks like...