Question

An aeroplane is flying horizontally at a constant height of 1000m. At a certain instant the angle of elevation is 30 degrees and decreasing and the speed of the aeroplane is 480 km/h.
How fast is the distance between the aeroplane and the observation point changing at this instant?

Answer 1

The relationship between \(\theta\), the distance \(r\) to the plane (the diagonal in the triangle you drew) and the height \(h\) is that
\[r\sin\theta=h \Leftrightarrow r=\frac{h}{\sin\theta}\]
The speed of the airplane is given by \(dx/dt\), where \(x\) is the distance along the ground to the plane, so we have
\[x=r\cos\theta \Rightarrow \frac{dx}{dt}=\frac{dr}{dt}\cos\theta-r\sin\theta\]
But looking above, we see that \(r\sin\theta=h\) so
\[\frac{dx}{dt}=\frac{dr}{dt}\cos\theta-h\]
Now at the instant when \(\theta=\pi/6\) in radians, this speed is 480, this gives us
\[480=\frac{dr}{dt}\cos\frac{\pi}{6}-1000 \Leftrightarrow \frac{dr}{dt}=\frac{2}{\sqrt3}1480\]
This is what the question asks for, so we are done.

Answer 2

thankyou