This is a related rate problem. The airplane is flying at a constant speed and altitude toward a point. The question is how fast is the view angle increasing as the plane flies closer. Note: the airplane may not appear in some browsers.[br][br]THE MATH[br][br]The math is simpler in Radians so find [math]\frac { d \theta}{dt}[/math] in radians per second, then convert to [math]^\circ[/math] per second.[br]The primary relation is [math]\tan \theta = \frac{alt}{dist}[/math] where [math]alt[/math] is the altitude and [math]dist[/math] is the distance in kft ( 1000 feet).[br]The speed is in Miles per Hour which is converted to Velocity by multiplying by [math]-\frac{ 66 (kft/sec)}{ 45000 MPH}[/math][br]Taking the derivative with respect to time of the tangent relation above gives[br] [math]\frac{ d \theta}{dt} \sec^2 \theta = \frac{-alt}{dist^2} \frac{d \; dist}{dt}[/math] [br]and [math] \frac{d \; dist}{dt} = Velocity = V[/math][br]Also from Pythagorean Theorem[br][math] \sec^2 \theta = \frac{dist^2 + alt^2}{dist^2}[/math][br]A little algebra then gives[br] [math]\frac{ d \theta}{dt} = - \frac{alt \; V}{ alt^2 + dist^2 }[/math][br]Then converting to degrees[br] [math]\frac{ d \alpha}{dt} = - \frac{180}{\pi} \frac{alt \; V}{ alt^2 + dist^2 }[/math]

As the plane flies closer, how does the rate of change of the angle change?[br]Is the rate the same at any altitude?[br]How does the rate of change of angle vary with the airplane's speed?