Related Rates Question

Question

A road running north to south crosses a road going east to west at the point P. Car A is driving north along the first road, and an airplane is flying east above the second road. At a particular time the car is 15 kilometers to the north of P and traveling at 55 km/hr, while the airplane is flying at speed 200 km/hr 10 kilometers east of P at an altitude of 2 km. How fast is the distance between the car and the airplane changing?

Answer 1

Let t = 0 represent the time spoken of where the car is 15 km north of point P and the plane is 10 km east. At that time the distance (using a 3-D version of the Pythagorean Theorem is sqrt(15² + 10² + 2²) After that, the distance of the car grows by 55t and the distance of the plane grows by 200t so the distance, plane to car, is defined by d(t) = sqrt[(15+55t)² + (10+200t)² + 2²). Multiplying this out, I get:

d(t) = (43025t² + 5650t + 329)^1/2

To find the rate of change, take the derivative:

d'(t) = 1/2(43025t² + 5650t + 329) ^-1/2 (2(43025)t + 5650)

d'(t) = (86050t + 5650)/[2(43025t² + 5650t + 329)^1/2]

To find the rate of change at that moment (t = 0), just evaluate d'(0) which gives the rate of change of 155.75 km/hr