He uses a nice trick to approximate the real thing, however this algorithm wouldn't be able to deal with curved objects.
This is a nice little post that reasonably describes a technique that has been used by a fair number of professional, commercial products. That anyone would try to downplay it to make themselves feel superior is silly.
To provide some actual value, most of the the images in the post are actually interactive. It's pretty cool. I somehow read the whole thing and didn't notice.
"Today, I will show you how to make something like this: (move your mouse around in the box below)"
"The demo below just draws a bunch of line segments and tracks your mouse position."
"Here's what all that math looks like: (move your mouse over the box)"
It's not a sin.
Why would you think that was my goal? Did i say the author was stupid or didn't pay attention? I simply pointed out a limitation that might not be obvious to others at first sight and which wasn't discussed in the article itself, thus also providing for others an easy starting point to improve on the algorithm as presented.
Heck, in the circles* i run the first part of my comment is even praise. Figuring out ways to approximate "real" results so they can be done in realtime is not easy.
*Have an example of EXTREMELY competent faking from those circles: https://www.youtube.com/watch?v=MJfceF0syK8
http://i.imgur.com/G5U43ZU.png
Calculate the edge points of the circle, draw a line between them, and then your shadow is simply a rectangle. Overlay the circle over the shadow to hide the line.
This will work with circles and ovals, but not more complex curves.
cx'(t) = a-x(t) cy'(x) = b-y(t)
You need to solve these two equations for values of c and t. The difficulty of this task depends mostly on the difficulty of the curve itself.
If the curve has equation y = f(x) (or in other words a simple representation x(t) = t and y(t) = f(x(t)) = f(t)), then you have
c = a-x cf'(x) = b-f(x)
Therefore (a-x)f'(x) = b-f(x). If f(x) is a polynomial of degree n, the problem is reduced to root-finding of a polynomial of degree n so you'll have at most n roots to deal with. So in general it won't be easy. This reduces to the code in the post if you use line segments.