Question

An airplane flies west at 150m/s at an altitude of 1000m. A searchlight on the ground in the same vertical plane must be kept on the airplane. what is the rate of revolution of the searchlight when the airplane is 500m due east of the searchlight?

Answer 1

arghh it erased my diagram, anyway
\[\tan \theta = \frac{1000}{x}\]

Answer 2

\[\sec^2 \theta \frac{d \theta}{dt} = -\frac{1000}{x^2} \frac{dx}{dt}\]
we are given:
\[\frac{dx}{dt} = -150\]
\[\frac{d \theta}{dt} = \cos^2 \theta \frac{1000(150)}{x^2}\]

Answer 3

Got it! Thx!

Answer 4

also keep in mind you may need to divide by 2pi to get "revolutions per sec"