Related Rates A plane flies at a speed 500 km/hr at a constant height of 12 km.

Question

A plane flies at a speed  km/hr at a constant height of  km. How rapidly is the angle of elevation to the plane changing when the plane is directly above a point  km away from the observer?

The angle of elevation is changing at ____  radians/hr (enter a positive value). Round your answer to 3 decimal places.

Answer 1

Let θ be the angle of elevation. Let s be the distance from the observer to the plane which is the

hypotenuse of a right triangle.

tan(θ) = 12/x

Taking the implicit derivative of both sides

sec²(θ) dθ/dt = -12/x² dx/dt Note: sec(θ) = 1/cos(θ)

Solving for dθ/dt

dθ/dt = -12 cos²(θ)/x² dx/dt cos(θ) = x/s

dθ/dt = -12 (x/s)² (1/x²) dx/dt

= (-12/s²) dx/dt

When x = 110, s²=110²+12²= 12244

dθ/dt = (-12/12244) (500) = -0.490 radians/hr

Doesn't say if the plane is flying toward or away from the observer. This assumes the plane is flying away from the observer, so the angle is decreasing at that rate.

Answer 2

The given rate is -500 km/hr, which is the rate at which the length of the horizontal leg of the right triangle decreases. (The right triangle is formed by 3 vertices: the observer, the plane, and the point on the ground directly below the plane.) The length of the vertical leg of this triangle remains constant: 12 km.

Since we want to know the rate of change of the angle opposite from the vertical leg, we write an equation relating that angle to the variable whose rate of change we know:

x: length of the horizontal leg of the right triangle ; dx/dt = - 500 ; x = 110 ; dΘ/dt = ?

cotΘ = x/12 (we can use cotangent here to simplify the derivative on the right hand side)

-csc²Θ·dΘ/dt = 1/12 dx/dt

dΘ/dt = -sin²Θ·(-500) / 12

dΘ/dt = [12/√(110²+12²)]²·500/12 = 6,000/12,244 rads/hr