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 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?

Answer 1

-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.

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.