A plane flies horizontally at an altitude of 3 km and passes directly over a tracking telescope on the ground. when the angle of elevation is π/6, this angle is decreasing at a rate of π/4 rad/min. how fast is the plane traveling at that time?

The solution is:tan (θ) = opp / adj tan (θ) = y/x xtan (θ) = y

Find x:

x = y/tan (θ)

So x = 3/tan (π/6)

Perform implicit differentiation to get the equation:

dx/dt * tan (θ) + x * sec² (θ) * dθ/dt = dy/dt

Since altitude remains the same, dy/dt = 0. Now ...

dx/dt * tan (π/6) + 3/tan (π/6) * sec² (π/4) * - π/4 = 0

changing the equation, will give us:

dx/dt = [3/tan (π/6) * sec² (π/6) * π/4} / tan (π/6) ≈ 12.83 km/min