# How the rotatable airplane-attached-camera exactly photographing the ground?

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.

• What is the rate of change of angle at the angle 30 degrees?
– Nij
Commented Apr 5, 2022 at 9:10

You have one simple relation

$$\tan(\theta) = \dfrac{10}{x}$$

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, $$x = 10 \tan(60^\circ)$$, $$\theta = 30^\circ$$

$$\theta' =\dfrac{10000}{100 (3) \left(\dfrac{4}{3}\right) } = 25 rad/hr = \dfrac{1}{144} rad/sec = 0.398^\circ / sec$$

• The book says that the correct answer is $~25~\mathrm{rad}/\mathrm{hr}~$ Commented Apr 5, 2022 at 11:01
• I've reviewed my calculations, the answer I got after revision is $75 \text{ rad/hr}$ Commented Apr 5, 2022 at 11:13
• Your diagram shows $30^\circ$ in the wrong place. Actually ,$\theta = 30^\circ$, and this case, the book is right that $\theta' = 25 rad/hour$. I'll modify the diagram and the solution. Commented Apr 5, 2022 at 11:20