Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 30

A ladder 10ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1$\displaystyle \frac{\text{ft}}{s}$, At what rate is the angle between the ladder and the ground changing when the bottom of the ladder is 6ft from the wall?
Given: The length of the ladder = 10 ft.
The rate of change of distance of the bottom of the ladder to the wall = $\displaystyle\frac{1 \text{ft}}{s}$
Required: The rate of change of the angle between the ladder and the ground when the bottom of the ladder is 6 ft from the wall.




We will use the cosine function to equate all the given information
$\displaystyle \cos \theta = \frac{x}{10}; \text{ for } x = 6 \text{ ; } \theta = \cos^{-1} \left( \frac{6}{10}\right) = 53.13^{\circ}$

We take the derivative with respect to time, we get
$\displaystyle - \sin \theta \frac{d\theta}{dt} = \frac{1}{10} \left( \frac{dx}{dt} \right)$

$
\begin{equation}
\begin{aligned}
\frac{d \theta}{dt} &= \frac{\frac{\frac{dx}{dt}}{10} }{- \sin \theta}\\
\\
\frac{d \theta}{dt} &= \frac{\frac{-1 \text{ft}/s}{10}}{-\sin(53.13)}
\end{aligned}
\end{equation}\\
\boxed{\displaystyle \frac{d\theta}{dt} = - 0.125 \frac{^{\circ}}{s}} \qquad \text{, negative value because the rate is decreasing}
$