An airplane is flying horizontally on a straight line at a speed of$~1000~\mathrm{km}/\mathrm{hr}~$, at an elevation of$~10\mathrm{km}~$. An automatic camera is photographing a point directly ahead on the ground. How fast must the camera be turning when the angle between the path of the plane and the line of sight to the point is$~30^{\circ}~$?
Currenlty I've been struggling to understand the following meaning.
How fast must the camera be turning when the angle between the path of the plane and the line of sight to the point is$~30^{\circ}~$?
I've depicted the following to try to get it.
Can anyone explain me what the problem statement exactly asking for a solver?
I even can't determine the location where angle$~\theta_{}~$should be taken in the first place.
$ \tan(\theta) = \dfrac{10}{x} $
Differentiating
$ \sec^2(\theta) \theta' = -10 \dfrac{x'}{x^2} $
From which,
$\theta' = - \dfrac{10 x'}{x^2 \sec^2(\theta) } $
Substituting, $x' =- 1000 $ km/h, $x = 10 \tan(60^\circ) $, $\theta = 30^\circ$
$\theta' =\dfrac{10000}{100 (3) \left(\dfrac{4}{3}\right) } = 25 rad/hr = \dfrac{1}{144} rad/sec = 0.398^\circ / sec $