Question

a ladder of 8 m os leaning against a wall begins to slide. If its upper end slides down the wall at a rate of 0.25 m/sec, at what rate is the angle between the ladder and the ground changing when the foot of the ladder is 5 m from the wall?

Answer 1

Let θ be the angle formed by the ladder and the ground
let the top of the ladder by y m above the ground

sinθ = y/8
8sinθ = y

given: dy/dt = -.25 m/sec, <----- negative, since it is sliding down
find dθ/dt when x = 5
then 25 + y^2 = 64
y = √39 and cosθ = √39/8

8cosθ dθ/dt = dy/dt
dθ/dt = -.25/(8√39/8) = -1/4√39 m/s