Question

A 20 foot ladder is sliding down a vertical wall at a constant rate of 2 feet per second.

a) How fast is the ladder moving away from the wall when the base of the ladder is 12 feet away from the wall?

b) Find the rate of change at which the angle between the ladder and the ground is changing when the base of the ladder is 16 feet from the wall.

Answer 1

when the ladder reaches height h, its base is x = √(400-h^2) from the wall

x^2 + h^2 = 400
2x dx/dt + 2h dh/dt = 0
when x=12, h=16, so
2(12) dx/dt + 2(16)(-2) = 0
dx/dt = 8/3 ft/s


when the ladder's base is x feet from the wall, the angle θ between the ladder and the ground thus satisfies

sinθ = h/20
cosθ dθ = dh/20
(16/20) dθ = -2/20
dθ = -1/8 radians/s

when x=12, sinθ = 4/5