Question

1. A 12-foot ladder is leaning against a wall that makes a 90-degree angle with the ground. A troublemaker is slowly pulling the bottom of the ladder along the ground, away from the wall at a rate of 2 feet per minute when the top of the ladder is 6 feet above the ground.

b. At that moment, what is the rate of change of the angle between the ladder and the ground?

Answer 1

b) Is similar to part a but instead of using Pythagorean equations use
cos(θ)=b/12
h = 6\[b = \sqrt{144-6^2}= 6\sqrt{3}\] and
\[\theta = \frac{\pi}{6}\]
\[\frac{d}{dt}[cos(\theta )] = \frac{d}{dt}[\frac{b}{12}]\]
\[-sin(\theta ) \frac{d\theta}{dt}= \frac{1}{12} \frac{db}{dt}\]
since db/dt = 2.
\[\frac{d\theta }{dt} = \frac{-1}{2sin(\theta)}2\]

\[\frac{d\theta }{dt} = \frac{-1}{sin(\theta)}\]
Hense , at h = 6.
\[\frac{d\theta }{dt} = -2\]

Answer 2

This is in radians BTW.^^^^^