Chapter 4.1, Problem 24E

To determine

To find: The rate at which the angle between the ladder and ground changes.

Answer to Problem 24E

The rate at which the angle between the ladder and ground changes is $\frac{1}{8}$ .

Explanation of Solution

Given information: The distance between the bottom of the ladder and wall is 6ft.

Calculation:

Consider the Figure.

  

Consider the equation.

  $\begin{array}{l}{x}^2+y^2=100\\ 6^2+y^2=100\\ y=8\end{array}$

The angle is given as,

  $\tan \theta =\frac{y}{x}$

Differentiate above equation.

  $\begin{array}{c}{\sec }^2\theta \frac{d\theta }{dt}=\frac{x\frac{dy}{dt}-y\frac{dx}{dt}}{x^2}\\ {\sec }^2\theta \frac{d\theta }{dt}=\frac{\left(6\right)\left(\frac{-3}{4}\right)-\left(-8\right)\left(1\right)}{6^2}\\ \frac{d\theta }{dt}=\frac{-25}{72{\sec }^2\theta }\\ =\frac{-25}{72\left(1+{\tan }^2\theta \right)}\\ =\frac{-25}{72\left(1+\frac{y^2}{x^2}\right)}\\ =\frac{-25}{72\left(1+\frac{8^2}{6^2}\right)}\\ =-\frac{1}{8}\end{array}$