Question

An aeroplane passes directly overhead then flies horizontally away from an observer at an altitude of 5000 m and air speed of 200 m/s. At what rate is the angle of elevation to the observer changing at the instant when the angle of elevation is 60 degrees?

Answer 1

\[\tan \theta = \frac{5000}{200t} = \frac{25}{t}\]
take derivative
\[\sec^2 \theta \frac{d \theta}{dt} = -\frac{25}{t^2}\]
\[\frac{d \theta}{dt} = \frac{-25 \cos^2 \theta}{t^2}\]
Notice from above we can replace "t" with theta since
\[t = \frac{25}{\tan \theta}\]
substituting and simplifying you get
\[\frac{d \theta}{dt} = \frac{-\sin^2 \theta}{25}\]
plug in theta = 60