Question

(1 point) an airplane, flying at 500 km/hr at a constant altitude of 5 km is approaching a camera mounted on the ground. let θθ be the angle of elevation above the ground at which the camera is pointed. when this angle is equal to π/4π/4, how many radians per hour is the rate of change for the camera's angle? give your answer in radians/hr.

Answer 1

The rate of change for the camera's angle is 20 rad/hr

Step-by-step explanation:

Refer the given figure.

We have

            $tan\theta =\frac{h}{b}$

Differentiating with respect to time

             $\frac{d}{d\theta}\left (tan\theta \right ) =\frac{d}{d\theta}\left (\frac{h}{b} \right )\\\\sec^2\theta\frac{d\theta}{dt}=\frac{b\frac{dh}{dt}-h\frac{db}{dt}}{b^2}$

We have

              h = 5 km

               $\frac{dh}{dt}=0\\\\\frac{db}{dt}=-500km/hr\\\\\theta =\frac{\pi }{4}\\\\tan\frac{\pi }{4}=\frac{h}{b}\Rightarrow 1=\frac{5}{b}\Rightarrow b=5km$

Substituting

                $sec^2\theta\frac{d\theta}{dt}=\frac{b\frac{dh}{dt}-h\frac{db}{dt}}{b^2}\\\\sec^2\frac{\pi }{4}\frac{d\theta}{dt}=\frac{5\times 0-5\times (-500)}{5^2}\\\\2\frac{d\theta}{dt}=\frac{1000}{25}\\\\2\frac{d\theta}{dt}=40\\\\\frac{d\theta}{dt}=20rad/hr$

The rate of change for the camera's angle = 20 rad/hr

Answer 2

PLZ GIVE MORE INFORMATION!!!