Answer
(a) The distance from the television camera to the rocket is increasing at a rate of $~~360~ft/s$
(b) The angle of elevation is increasing at a rate of $~~0.096~rad/s$
Work Step by Step
(a) Let $y$ be the rocket's height.
Let $x$ be the distance from the camera to the launching pad.
Let $D$ be the distance between the camera and the rocket.
We can find $D$:
$D^2 = x^2+y^2$
$D = \sqrt{x^2+y^2}$
$D = \sqrt{4000^2+3000^2}$
$D = 5000~ft$
We can find $\frac{dD}{dt}$:
$D^2 = x^2+y^2$
$2D~\frac{dD}{dt} = 2x~\frac{dx}{dt}+ 2y~\frac{dy}{dt}$
$\frac{dD}{dt} = \frac{x~\frac{dx}{dt}+ y~\frac{dy}{dt}}{D}$
$\frac{dD}{dt} = \frac{(4000)(0)+ (3000)(600)}{5000}$
$\frac{dD}{dt} = 360~ft/s$
The distance from the television camera to the rocket is increasing at a rate of $~~360~ft/s$
(b) Let $\theta$ be the camera's angle of elevation.
We can find $\frac{d\theta}{dt}$:
$\frac{y}{D} = sin~\theta$
$\frac{1}{D}~\frac{dy}{dt}-\frac{y}{D^2}~\frac{dD}{dt} = cos~\theta~\frac{d\theta}{dt}$
$\frac{d\theta}{dt}=(\frac{1}{cos~\theta})(\frac{1}{D}~\frac{dy}{dt}-\frac{y}{D^2}~\frac{dD}{dt})$
$\frac{d\theta}{dt}=(\frac{1}{4/5})[\frac{1}{5000}~(600)-\frac{3000}{5000^2}~(360)]$
$\frac{d\theta}{dt}= 0.096~rad/s$
The angle of elevation is increasing at a rate of $~~0.096~rad/s$