Related Rates Question?

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 1.3 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:______ rad/s

Answer 1

The length of ladder which rests against a vertical wall h = 10 ft

The distance from the bottom of the ladder to the wall x = 8 ft

Let w = Height of the wall

The sliding rate of the bottom of the ladder from the wall dx/dt = 1.3 ft/s

The angle between the ladder and the ground = θ

 

When the ladder is 8 ft from the wall

Using Pythagoras's theorem, w² + x² = 10² ⇒  w² + 8² = 10² ⇒  w² = 100-64 ⇒ w = 6

Then sin θ = 6/10

 

When the ladder is sliding.

cos θ = x/10  ⇒  x = 10 cos θ.

Apply derivative each side with respect to time ⇒  dx/dt = -10 sin θ(dθ/dt).

⇒  1.3 = -10 (6/10)(dθ/dt)

⇒  1.3 = -6(dθ/dt).

⇒  dθ/dt = -1.3/6

⇒  dθ/dt = -0.2167 rad/s

(The negative sign indicates that as x increases, θ decreases)

Solution : The sliding rate of the angle between the ladder and the ground is - 0.2167 rad/s.