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 0.8 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?

-0.133 radians/s is the angle between the ladder and the ground changing when the bottom of the ladder is 8 ft from the wall.

Step-by-step explanation:

Let

x = distance between the wall and ladder

Ф = angle between the ladder and ground

The rate of change is: dx/dt = 0.8 ft/s

We need to find dФ/dt when x = 8 ft/s

From the triangle in the figure we see that:

cos Ф = x/10

=> x = 10 cosФ

Now, the rate of change will be

d (x) / dt = d/dt (10cosФ)

dx/dt = - 10 sin Ф dФ/dt

dx/dt = 0.8 (given)

to find dФ/dt

we need to find sin Ф when x = 8

By Pythagoras theorem

(hypotenuse) ^2 = (Perpendicular) ^2 + (base) ^2

(10) ^2 = (8) ^2 + (Perpendicular) ^2

100 = 64 + (Perpendicular) ^2

=> (Perpendicular) ^2 = 100-64

(Perpendicular) ^2 = 36

Perpendicular = 6

sin Ф = Perpendicular/hypotenuse

sin Ф = 6/10 = 3/5

Putting values in:

dx/dt = - 10 sin Ф dФ/dt

0.8 = - 10 (3/5) dФ/dt

0.8 = - 6 dФ/dt

=> dФ/dt = 0.8 / - 6

dФ/dt = - 0.133 radians/s

So, - 0.133 radians/s is the angle between the ladder and the ground changing when the bottom of the ladder is 8 ft from the wall.