Question

Harold is in an airplane that is flying at a constant height of 4505 feet away from a fixed observation point. Maude, whose eyes are 5 feet from the ground, is standing at this point and watching the plane; the angle between her line of sight (the line line between her eyes and the plane) and the horizontal is called the angle of elevation. If, at a particular moment in time, the angle of elevation is 60◦ and the plane is flying at a rate of 22,600 feet per minute, determine the rate at which the angle of elevation is changing. Give you answer in degrees per minute.

Answer 1

Always draw a diagram for these problems. You will see that if the plane is x feet away from Maude's position, then

tanθ = 4500/x
so,
sec^2θ dθ/dt = -4500/x^2 dx/dt
When θ=60°, that gives x = 2250, so
4 dθ/dt = -4500/2250^2 * 22600
dθ/dt = -226/45 ≈ -5.02

Of course, that is in radians/min, so just convert that to °/min