Problem. A ladder of $6$ meters is inclined against a wall, making up an angle $t$ with the wall. As the ladder slides down, the distance between the wall and the ladder's base increases up to its length. At what speed does the ladder slide when the angle is $t = \pi/6$? What about $t = \pi/3$?
Edit. The problem (as written by me) has an ambiguity which explains why my intuition failed in the first place. The ambiguity is that I don't specify which point of the ladder is of concern here. Different points of the ladder move in different speeds. See the accepted answer. Thanks to @DavidQuinn for first spotting the ambiguity.
A solution. The sine of $t$ equals the distance of interest, $x$, over the length of the ladder. In symbols, $\sin(t) = x(t)/6 \implies x(t) = 6 \sin(t)$. So we're interested in
$$x'\left(\frac{\pi}{6}\right) = 6 \cos\left(\frac{\pi}{6}\right) = 6 \times \frac{\sqrt{3}}{2} = 3 \sqrt{3},$$
meaning $3 \sqrt{3}$ meters per radian.
We're also interested in $$x'\left(\frac{\pi}{3}\right) = 6 \cos\left(\frac{\pi}{3}\right) = 6 \times \frac{1}{2} = 3,$$
meaning $3$ meters per radian. We're done.
Intuition and question. I've seen a ladder fall like that and it seems to me that it hits the ground with greatest speed possible. But, in this problem, the ladder seems to be losing speed as the angle opens up from $\pi/6$ to $\pi/3$. The cosine function is greater at angles near zero than at angles near $\pi/2$, which is consistent with the results. So I'm puzzled. Is the model wrong?
It's the $y$-coordinate of the point of contact with the wall that speeds up over time. The $x$-coordinate of the point of contact with the ground speeds up at first, but the slows down to $0$. You can see this if you imagine that the ladder somehow falls through the ground while still attached at its endpoints (you could implement this with rails and sliders) $-$ this $x$-coordinate will start to decrease.