1. The problem statement, all variables and given/known data Mr. Wilson is standing near the top of a ladder 24 feet long which is leaning against a vertical wall of his house. Dennis the little boy next door, ties a rope from his tricycle to the bottom of the ladder and starts to pull the foot of the ladder away from the house wall. The bottom end of the ladder begins to slide away from the wall at the rate of 1 foot per second. How fast is the angle between the top of the ladder and the wall changing when the foot of the ladder is 8 feet from the wall? 2. Relevant equations Again, simple derivatives. 3. The attempt at a solution Well, here's my best guess: Listing the knowns: z = 24, x' = 1, x = 8. One of my relating equations is x^2 + y^2 = z^2; therefore, xx' + yy' = 0. While the second relating equation is tan(theta) = x/y, and consequently sec^2(theta)(theta)' = (x'y - y'x)/y^2. Or similarly: (theta)' = (x'y - y'x)/(sec^2(theta)y^2). A few intermediate calculations give y' = -8/sqrt(512), y = sqrt(512), and theta = arctan(8/sqrt(512)). Plugging everything in produces (theta)' = 0.0405.