#### To determine

**To find:** The speed of the plane.

#### Answer

The rate of speed at which the plane is travelling horizontally is dxdt=109π km/min.

#### Explanation

**Given:**

Since the plane flies horizontally at an altitude of 5 km and passes directly over a tracking telescope over the ground.

The rate at which the angle of elevation is decreases π6 rad/min.

**Formula used:**

(1). Chain rule: dydx=dydu⋅dudx

**Calculation:**

Let at any time t, P be the position of the plane and A be the position of the telescope on the ground and B be the point on the ground just below the plane. And x be the distance between the telescope and point on the ground just below the plane and θ be the angle of elevation between the telescope and plane as shown in the Figure 1 given below.

Since the plane is flying horizontally.

Therefore, the horizontal distance x and the angle of elevation changes with the time t.

Since the angle of elevation decreasing at a rate of π6 rad/min.

That is dθdt=−π6 rad/min.

Obtain dxdt when θ=π3 rad.

In the ΔABP

cotθ=ABPB=x5x=5cotθ

Differentiate x with respect to the time t.

ddt(x)=ddt(5cotθ)dxdt=−5cosec2θdθdt [ ∵dydx=dydu⋅dudx]

Substitutes θ=π3 and dθdt=−π6 in dxdt.

dxdt=−5cosec2(π3)(−π6)=56π(23)2=56π(43)=106π rad/min

Therefore, the rate at which the plane is travelling horizontally dxdt=109π km/min.