Tracking a Car with a Drone Camera

A drone hovers above a road and keeps its camera locked on a passing car. As the car moves, the camera must turn, fastest when the car is directly below. Related rates turn the car's speed into the camera's turning speed.

How it works

Related rates

In one sentence

In a related rates problem, quantities linked by an equation are differentiated with respect to time (using the chain rule), giving an equation between their rates that is solved for the unknown rate.

General formula

\tan{\htmlData{sym=0}{\theta}} = \frac{\htmlData{sym=1}{{x}}}{\htmlData{sym=2}{{h}}} \quad\Rightarrow\quad \sec^2{\htmlData{sym=0}{\theta}}\,{\htmlData{sym=3}{\frac{d\theta}{dt}}} = \frac{1}{\htmlData{sym=2}{{h}}}{\htmlData{sym=4}{\frac{dx}{dt}}} \quad\Rightarrow\quad {\htmlData{sym=3}{\frac{d\theta}{dt}}} = \frac{\htmlData{sym=2}{{h}}}{{\htmlData{sym=2}{h}}^2 + {\htmlData{sym=1}{x}}^2}\,{\htmlData{sym=4}{\frac{dx}{dt}}}

The equation ties the angle to the position at every instant, so their rates are tied too; the chain rule converts one rate into the other.

What the symbols mean

$\theta$
Camera angle
Angle between straight down and the line to the car.
$x$
Car position
Distance along the road from the point below the drone.
$h$
Height
The drone's altitude.
$\frac{d\theta}{dt}$
Turning rate
How fast the camera must rotate.
$\frac{dx}{dt}$
Car speed
How fast the car moves along the road.

The geometry links the camera angle $\theta$ and the car's position $x$ : $\tan\theta = x/h$ .
Differentiate both sides with respect to time, using the chain rule on $\tan\theta$ .
Since $\sec^2\theta = 1 + x^2/h^2$ , solve for $\frac{d\theta}{dt} = \frac{h}{h^2 + x^2}\frac{dx}{dt}$ .
The turning rate is largest, $\frac{1}{h}\frac{dx}{dt}$ , when the car is right below ( $x = 0$ ).