   Chapter 3.9, Problem 32E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# How fast is the angle between the ladder and the ground changing in Example 2 when the bottom of the ladder is 6 ft from the wall?

To determine

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

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=dydududx

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.

cosθ=basehypotaneous=x10x=10cosθ

Differentiate x with respect to the time t .

ddt[x]=ddt[10cosθ]dxdt=10ddθ[cosθ]dθdt[Qdydx=dydududx]=(10sinθ)dθdt

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

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