A rocket rises vertically at 12 ft/s from a launch pad, passing a camera 12 ft away. When the rocket is 5 ft above the pad, how fast is the camera's angle of elevation increasing?

Core Concept

Here $\tan\theta=\dfrac{y}{12}$, so $\sec^2\theta\dfrac{d\theta}{dt}=\dfrac{1}{12}\dfrac{dy}{dt}$. When $y=5$, $\cos\theta=\dfrac{12}{13}$, so $\dfrac{d\theta}{dt}=\cos^2\theta\cdot\dfrac{12}{12}=\left(\dfrac{12}{13}\right)^2=\dfrac{144}{169}$ rad/s.