Answer
The plane is flying at a speed of $~~3.49~km/min$
Work Step by Step
Let $y$ be the plane's altitude.
Let $x$ be the horizontal distance between the telescope and the plane.
Let $\theta$ be the telescope's angle of elevation.
We can find $x$ when $\theta = \frac{\pi}{3}$:
$\frac{y}{x} = tan~\theta$
$x = \frac{y}{tan~\theta}$
$x = \frac{5}{tan~\frac{\pi}{3}}$
$x = 2.887$
We can find $\frac{dx}{dt}$:
$\frac{y}{x} = tan~\theta$
$\frac{1}{x}~\frac{dy}{dt}-\frac{y}{x^2}~\frac{dx}{dt} = sec^2~\theta~\frac{d\theta}{dt}$
$\frac{y}{x^2}~\frac{dx}{dt} = \frac{1}{x}~\frac{dy}{dt}-sec^2~\theta~\frac{d\theta}{dt}$
$\frac{dx}{dt} = (\frac{x^2}{y})[\frac{1}{x}~(0)-sec^2~\frac{\pi}{3}~\frac{d\theta}{dt}]$
$\frac{dx}{dt} = (\frac{2.887^2}{5})[-(2)^2~(-\frac{\pi}{6})]$
$\frac{dx}{dt} = 3.49~km/min$
The plane is flying at a speed of $~~3.49~km/min$
