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?

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

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

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.

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)

To solve this problem, we can use trigonometry and related rates to find the speed of the plane. Let's start by labeling the given information:

1. The altitude of the plane = 6 km
2. The angle of elevation = π/3
3. The rate at which the angle of elevation is decreasing = π/3 rad/min

Now, let's set up a right triangle to represent the situation. The tracking telescope represents the position of the observer on the ground, the base of the triangle, and the plane's altitude represents the height of the triangle. The hypotenuse represents the line of sight between the observer and the plane.

Since the plane is flying horizontally, the hypotenuse remains constant. We need to find the rate at which the base is shrinking, which will help us determine the speed of the plane.

Let:
- b = base of the triangle (distance between observer and plane)
- h = height of the triangle (altitude of the plane)
- θ = angle of elevation

Using trigonometry, we have:
tan(θ) = h / b

Differentiating both sides of the equation with respect to time (t), we get:
sec^2(θ) * dθ/dt = (dh/dt) / b

We are given:
θ = π/3
dθ/dt = -π/3 rad/min (since the angle is decreasing)
dh/dt = 0 (since the altitude remains constant at 6 km)

Substituting these values, we can solve for db/dt, the rate at which the base is shrinking.

sec^2(π/3) * (-π/3) = 0 / b

Simplifying the equation:
(2/√3) * (-π/3) = 0 / b
-2π / (3√3) = 0 / b

Since the right-hand side of the equation is 0, the left-hand side equals 0 as well:

-2π / (3√3) = 0

This implies that the rate at which the base is shrinking is 0. Therefore, the plane is not moving horizontally at that moment.

Hence, the speed of the plane at that time is 0 km/hr or 0 m/s.