Since the plane is increasing in distance 3 minutes later with a constant speed of 24 km/min, we can compute the distance covered in 3 minutes first:
d = rt
d = (3) (24) = 72 km.
This is a diagonal distance at an angle of 35 degrees. Therefore it has horizontal and vertical components:
Horizontal: 72 cos 35°
Vertical: 72 sin 35°
Another vertical component is the altitude which is 6 km. We have to consider this if we are concern about the rate of the increase in distance from radar to plane:
Another Vertical Component: 6 km.
Let
x = horizontal component = 72 cos 35°
y + 6 = vertical component = 72 sin 35° + 6
R = distance from radar to plane
R2 = x2 + (y + 6)2
R2 = (72 cos 35°)2 + (72 sin 35° + 6)2
R ≈ 75.6 km
The vertical and horizontal components of the velocity of the plane at a rate of 24 km/min are
dx/dt = 24 cos 35°
dy/dt = 24 sin 35°
R, x and y are all functions of time (t). Therefore we can get the derivative of R2 = x2 + (y + 6)2 with respect to time, so we can get the rate of the distance from the plane to the radar station is increasing 3 minutes later (dR/dt):
2R (dR/dt) = 2x (dx/dt) + 2(y+6) (dy/dt)
2(75.6)(dR/dt) = 2(72•cos 35°)(24•cos 35°)+2(72•sin 35º + 6) (24•sin 35°)
dR/dt ≈ 1227 km/min.
