A ladder 10 ft long rests against a wall. If bottom of ladder slides away from the wall at 1.5 ft/s, how fast is angle between the ladder and ground changing when bottom of ladder is 8 ft from wall?

Answer 1

The problem is asking for dθ/dt. The problem gives us dx/dt, so if we can get dθ/dx we have the solution (from dθ/dt = dθ/dx * dx/dt), so let's try writing θ in terms of x.

The distance between the wall and the bottom of the ladder is x = 10 (ft) cos(θ). If we rearrange this in terms of θ, we get θ = arccos(x/10).

Now, dθ/dx = -1 / (√100 - x²) after we evaluate the derivative.

We get dθ/dt = -1 / (√100 - x²) * 1.5 ft/s. Substitute x = 8ft in and you get dθ/dt = -0.25 rad/s when the ladder is 8ft from the wall.