A ladder 10 meter long is leaning against a wall. If the top of the ladder is sliding down the wall at 3m/s, how fast is the angle between the top of the ladder and the wall changing when the foot of the ladder is 6 meter away from the wall?

Answer 1

let the angle you mentioned be Ø
Let the foot of the ladder be x ft from the wall, and the height of the ladder be y ft
cosØ = y/10
y = 10cosØ
dy/dt = -10sinØ dØ/dt

when x = 6
x^2+y^2=100
y = 8 , and sinØ = 8/10
Also we know dy/dt = -3 m/s

-3 = -10(8/10)(dØ/dt)
dØ/dt = 3/8 radians/s