Question

Two airplanes are flying in the air at the same height: airplane a is flying east at 250 mi/h and airplane b is flying north at 300mi/h.300mi/h. if they are both heading to the same airport, located 30 miles east of airplane a and 40 miles north of airplane b, at what rate is the distance between the airplanes changing?

Answer 1

If the airport is located at the origin, for units of miles and hours, we can write the equations of position for airplanes "a" and "b" in rectangular coordinates as
  a = (-30 +250t, 0)
  b = (0, -40 +300t)

The distance between these (moving) points can be computed in the usual way using the Pythagorean theorem.
  d = √((-30 +250t - 0)² +(0 - (-40 +300t))²)
  d = √(2500 -39000t +152500t²)

Then the rate of change of d is the derivative of this.
  d'(t) = (-19500 +152500t)/√(2500 -39000t +152500t²)

At the present time (t=0), the rate of change of distance between the planes is
  d'(0) = -19500/√2500 = -390

The distance between the planes is decreasing at 390 mi/h.