Aladder 10 ft long rests against a vertical wall. if the bottom of the ladder slides away from the wall at a rate of 1.1 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 6 ft from the wall? (that is, find the angle's rate of change when the bottom of the ladder is 6 ft from the wall.)

Explanation:

If the distance between the bottom of the ladder and the wall is x, then:

cos θ = x / 10

Taking derivative with respect to time:

-sin θ dθ/dt = 1/10 dx/dt

Substituting for θ:

-sin (acos(x / 10)) dθ/dt = 1/10 dx/dt

Given that x = 6 and dx/dt = 1.1:

-sin (acos(6/10)) dθ/dt = 1/10 (1.1)

-0.8 dθ/dt = 0.11

dθ/dt = -0.1375

The angle is decreasing at 0.1375 rad/s.

i am pretty sure it is c because when people are closer together, they cannot move as freely. picture being in a mostly full elevator, for instance. the people are right next to you, and you can barely move.