The LDraw parts library is one source for this sort of information. It doesn't come directly from TLG, but it is used by a number of modelling tools and is well supported by the community.
Here's the file representing 11477 (11477.dat):
0 Slope Brick Curved 2 x 1
0 Name: 11477.dat
0 Author: Owen Burgoyne [C3POwen]
0 !LDRAW_ORG Part UPDATE 2013-01
0 !LICENSE Redistributable under CCAL version 2.0 : see CAreadme.txt
0 BFC CERTIFY CCW
0 !HISTORY 2013-07-21 [PTadmin] Official Update 2013-01
1 16 0 0 0 1 0 0 0 1 0 0 0 1 s\11477s01.dat
1 16 10 24.972 20 0 -20 0 -28.9719 0 -28.9719 -40 0 40 48\1-8cyli.dat
The file format is line based. Lines beginning with 0 are comments, and lines beginning with 1 reference a subfile with an applied transformation matrix.
This is the line representing the curve on the top of the element:
1 16 10 24.972 20 0 -20 0 -28.9719 0 -28.9719 -40 0 40 48\1-8cyli.dat
This references the file 1-8cyli.dat. We can dig into that file to see exactly how it approximates the curve as a series of polygons, but the primitives reference is perhaps more useful. This tells us that the file referenced is 1/8 of cylinder.
We can experiment a little by creating a "part" that we can open in something like LeoCAD. Here's a file that will render the full (4-4) cylinder scaled 20x:
1 16 0 0 0 20 0 0 0 20 0 0 0 20 48\4-4cyli.dat
And here's a render of the same:
That shows a cylinder with a 20 LDU radius. It is 20 LDU because we multiplied the identity cylinder primitive by the following transformation matrix:
20 0 0
0 20 0
0 0 20
Here's the render showing the 1-8 cylinder:
Now we can inspect the transformation matrix from the line in 11477.dat:
1 16 10 24.972 20 0 -20 0 -28.9719 0 -28.9719 -40 0 40 48\1-8cyli.dat
Here's the 3x3 matrix from that line:
0 -20 0
-28.9719 0 -28.9719
-40 0 40
This matrix is harder to reason about, as it represents both scaling and rotation to orient the curve appropriately. We can determine scaling factors by computing the magnitude of the rows, so the scaling factors are:
20
40.9725
56.5485
20 LDU gives us the 1 module width of our part. The other factors give us the shape of our curve.
Without the rotation needed to orient the curve correctly, we could use the following simplified transformation matrix:
40.9725 0 0
0 20 0
0 0 56.5485
We can represent this in an LDraw part as:
1 16 0 0 0 40.9725 0 0 0 20 0 0 0 56.5485 48\1-8cyli.dat
Here's the render of that showing the curve we are expecting:
This shows us that the curve profile is 1/8 of a unit circle that has been scaled by 56.5485 LDU in the x dimension and 40.9725 LDU in the y dimension according to the LDraw catalog.
If it helps, here's the curve in Desmos:
Update:
As Michael pointed out, this LDraw part is not precisely accurate. The true curve of this part appears to simply be a circular arc with radius 80 LDU or 4 modules. I've confirmed this by comparing against a round part with a 4 module radius such as 74611, and the shapes appear to be identical.
Here's an updated graph showing the LDraw version in red and the true radius in blue: