A plane flies horizontally at an altitude of 6 km and passes directly over a tracking telescope on the ground. When the angle of elevation is π/3, this angle is decreasing at a rate of π/3 rad/min. How fast is the plane traveling at that time?

Answer 1

At t=0, let x=0, so that at time t,

x/6 = cotθ
x = 6cotθ

dx/dt = -6sec^2θ dθ/dt
when θ=π/3, we have x = 6/√3 and secθ = 2
dx/dt = -6(2)(-π/3) = 4π km/min

Answer 2

rats - that's

dx/dt = -6csc^2θ dθ/dt
when θ=π/3, we have x = 6/√3 and cscθ = 2/√
dx/dt = -6(4/3)(-π/3) = 8π/3 km/min

Answer 3

At a time of t min after the plane passed over the tracking telescope, let the horizontal distance be x km
Let the angle of elevation be Ø

Make a sketch, I get
cot Ø = x/6
x = 6cotØ
dx/dt = -6 csc^2 Ø dØ/dt

when Ø = π/3 and dØ/dt = -π/3

dx/dt = -6( csc^2 (π/3)) (-π/3)
= -6((4/3)(-π/3)
= 24π km/min

check my arithmetic, should have written it out first.

Answer 4

I can add more rats!!

last step:
dx/dt = -6( csc^2 (π/3)) (-π/3)
= -6((4/3)(-π/3)
= 24π/9 km/min
= 8π/3 km/min
(which is Steve's answer)