Steel this integral!

Are you ready for some Monday Night Calculus? With the NFL season having started only a few days ago, it is time to perform some gridiron mathematics! Our calculations will focus on the logo for the Pittsburgh Steelers seen below.

The outer edge of the emblem is a circle, which has the well-known ratio of its circumference to its diameter equalling π. Let’s now find the ratio of the perimeter of the red emblem to its diameter, defined to equal the distance from one point to the diametrically opposite point. This shape is often called an astroid or if you want to sound quite mathematical, a 4-cusped hypocycloid. The curve can be formed parametrically by graphing
$x = frac{1}{4}(3cos(t) + cos(3t))$ and $y = frac{1}{4}(3sin(t) - sin(3t))$ for t = 1 to t = 2π. This produces the curve:

which is the shape we’ll tackle with some Calculus computations! The diameter of this asteroid is 2 since two diametrically opposite points are (1,0), and (-1,0).

The upper left quarter of the asteroid is represented by the function
$f(x) = left(1 - x^{2/3}right)^{3/2}$ for x from 0 to 1. We need only to find the length of this curve and multiply the result by 4. From Calculus, we know that the formula for the length of the curve $y = f(x)$ from 0 to 1 is
$L = int_0^1 sqrt{1 + [f'(x)]^2}dx.$ For $f(x) = left(1 - x^{2/3}right)^{3/2}$, $f'(x) = -sqrt{1-x^{2/3}}/x^{1/3}$. We seem to have a computational pile up when we consider finding $L = int_0^1sqrt{1 + [f'(x)]^2}$. We find an opening with a push of algebra and find a clear path to the equivalent expression
$L = int_0^1 1/x^{1/3} dx$. Need a replay of that computation? Try WolframAlpha by clicking here.

The curve of $x^{-1/3}$ is plotted below

We are finding the area under a curve that increases without bound as x approaches 0 from the right. How improper! Indeed, this requires an improper integral! Getting down in the dirt of integration, we find
$L = int_0^1 1/x^{1/3} dx = 3/2.$ You can do this by hand or pass the computation to WolframAlpha by clicking here. The length of a quarter of the astroid equals 3/2; so, the perimeter of the complete asteroid would be 6! We have our desired ratio of the perimeter to the diameter of the asteroid, which equals is 6/2 or 3. Things went from being quite irrational with the circle to rational with the asteroids. Nice exercising of our mathematical minds…now let’s go out and toss a ball around trying create perfect spirals, of course!

These ideas opened a lecture to my Calculus classes by my colleague Mike Mossinghoff who graciously filled in for me as I travelled. The students were treated to this example as a way to motivate improper integrals. Thanks, Mike!