Question

A plane flies horizontally at an altitude of 7 km and passes directly over a tracking telescope on the ground. When the angle of elevation is π/4, this angle is decreasing at a rate of π/4 rad/min. How fast is the plane traveling at that time?

Answer 1

Answer:

The plane moving with velocity =$\frac{dx}{dt}=\frac{7\pi}{2} Km/min$

Explanation:

Let $\theta$  be angle of elevation and x(t)be the horizontal displacement of the plane  from the tracking slope

We are given that a plane flies horizontally at an altitude of 7 Km and passes directly over a tracking telescope on the ground .

The angle of elevation is $\frac{\pi}{4}$

The rate at which an angle of elevation decreasing =-$\frac{\pi}{4}$rad/min

$tan\theta=\frac{perpendicular}{base}$

$tan\theta=\frac{7}{x(t)}$

$x(t)=\frac{7}{tan\theta}=7 cot\theta$

$x(t)=7 cot\theta$

Differentiate w.r.t time

Then we get $\frac{dx}{dt}=- 7 cosec^2\theta \cdot \frac{d\theta}{dt}$

$\frac{dx}{dt}=-7 cosec^2\frac{\pi}{4}\cdot (-\frac{\pi}{4})$

$\frac{dx}{dt}=7 (\sqrt2)^2 \cdot\frac{\pi}{4}=\frac{7\pi}{2}$

$\frac{dx}{dt}=\frac{7\pi}{2} Km/min$

The plane moving with velocity =$\frac{dx}{dt}=\frac{7\pi}{2} Km/min$