#### To determine

**To find:** The rate of change of the angle between the ladder and the ground.

#### Answer

The rate of change of the angle between the ladder and the ground is dθdt=−18 rad/s.

#### Explanation

**Given:**

A ladder of 10 ft long is rests against a vertical wall and the on the horizontal ground.

Suddenly the bottom of the ladder slides away from the ground at a rate of 1 ft/s.

**Formula used:**

Chain rule: dydx=dydu⋅dudx

**Calculation:**

y be the vertical length of the wall at which the upper end of ladder is rests, and x be the distance from the wall to lower end of the ladder and θ be the angle between the horizontal ground and lower end of the ladder.

When the lower end of the ladder start moving away from wall with speed 1 ft/s. Due to this the horizontal distance increase and the vertical distance decreases and the angle between the ladder and ground become also decreases as shown in the figure-2.

Since the distance x and the angle θ changes with the time t.

Therefore, x and θ are function of the time t.

Form Figure 1.

cos θ=basehypotaneous=x10x=10cos θ

Differentiate x with respect to the time t .

ddt[x]=ddt[10cos θ]dxdt=10ddθ[cos θ]dθdt [Qdydx=dydu⋅dudx]=−(10sin θ)dθdt

On further simplification the rate of change of angle is as follows.

dθdt=−(110sinθ)dxdt

When x=6,

y=102−62 [by Pythagoras Theorem]=100−36=64=8

Thus, the value of sin *θ* is,

sinθ=810=45

Substitute, sinθ=810, dxdt=1 in dθdt

dθdt=−110×45(1)=−540=−18 rad/s

Therefore, the rate of change of the angle between the ladder and the ground is dθdt=−18 rad/s.