A jet flies directly over a radar station at a constant height of 5km at a constant speed of 190km/hr. How fast is the angle of elevation of the jet changing when the horizontal distance from the radar station is 2km?

Let $LaTeX: \displaystyle x$ be the plane's horizontal distance from the station and let $LaTeX: \displaystyle h$ be the height of the plane. Drawing a diagram gives From the diagram $LaTeX: \displaystyle \theta = \tan^{-1}\left( \frac{h}{x} \right)$ . Taking the derivative gives: (don't forget that $LaTeX: \displaystyle h$ is constant) $LaTeX: \frac{ d\theta }{ dt } = \frac{1}{1+\left( \frac{h}{x}\right)^2}\cdot \left(-\frac{h}{x^2}\right)\left( \frac{dx}{dt} \right) = -\frac{h}{x^2+h^2}\frac{dx}{dt}$ Evaluating at the given values gives $LaTeX: \frac{d\theta}{dt} = -\frac{ 5 }{ 2^2+5^2}\cdot (190) = - \frac{950}{29} = -32.76$ Note: The solution is negative because counter-clockwise is a positive rotation for angles and the units are radians per hour.