Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 39

At what rate is the plane's distance changing at the same moment.
Given: Altitude of the plane = 5km
Angle of elevator = $\displaystyle \frac{\pi}{3}$, when the rate of the angle is decreasing at $\displaystyle \frac{\pi}{5} \frac{\text{rad}}{\text{min}}$
Required: Speed of the airplane at that given moment solution.






Since the angle of elevation is decreasing, it is assumed that the plane is moving away to the telescope

We use $\tan \theta$ to solve for the unknown

$
\begin{equation}
\begin{aligned}
\sec^2 \theta \frac{d \theta}{dt} &= \frac{-5 \frac{dx}{dt}}{x^2}\\
\\
\frac{dx}{dt} &= \frac{x^2\sec^2 \theta \frac{d \theta}{dt}}{-5} = \frac{\left( \frac{5 \sqrt{3}}{3} \right)^2 \left( \frac{1}{\cos \left( \frac{\pi}{3}\right)^2 \left( - \frac{\pi}{6} \right)}\right)}{-5}
\end{aligned}
\end{equation}
$


$\boxed{\displaystyle \frac{dx}{dt} = \frac{10}{9} \pi \frac{\text{km}}{\text{min}}}$ make sure to be consistent with the mode of angle measurements whether radian or degree mode.