Question

A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 3 2 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 8 ft from the wall

Answer 1

dα/dt = - 0,25 rad/sec

Explanation:

The vertical wall, the ground from the vertical wall up to the ladder, and the ladder shape a right triangle. angle α will be the angle between the ladder and the ground, then:

cos α = distance between the wall and the down end of the ladder (x) / length of the ladder

cos α = x/ L

- sin α * dα/dt = dx/dt * L /L²

- sin α * dα/dt = dx/dt /L

When the down end of the ladder is 8 ft apart from the wall

cos α = 8 / 10 = 0,8 then cos² α + sin²α = 1

sin² α = 1 - cos²α sin²α = 1 - (0.8)² sin²α = 0.36

sin α = 0.6

dα/dt = ?

dx/ dt = 3/2

L = 10 ft

x = 8 ft

By substitution

- 0.6*dα / dt = (3/2) / 10

dα/dt = - 1.5 / 6

dα/dt = - 0,25 rad/sec

(-) means the angle is decreasing