Calculus
posted by George on .
A train is heading due west from St. Louis. At noon, a plane flying horizontally due north at a fixed altitude of 4 miles passes directly over the train. When the train has
traveled another mile, it is going 80 mph, and the plane has traveled another 5 miles and is going 500 mph. At that moment, how fast is the distance between the train and the plane increasing?
Is the answer 105 mph.

Measuring time (t) in hours from that moment, the vector distance R between train and plane at later times is
R = (1+80t) i + (5+ 500t) j + 4 k
where i, j and k are unit vectors west, north and vertical.
dR/dt = 80 i + 500 j
The magnitude of that rate of change is
sqrt[(80)^2 + (500)^2] = 506 mph.
This could also have been solved simply by taking the magnitude of the relative velocity vector, 80 i + 500 j mph