Question

A road running north to south crosses a road going east to west at the point P. Cyclist A is riding north along the first road, and cyclist B is riding east along the second road. At a particular time, cyclist A is 39 hundred meters to the north of P and traveling at 32 meters per second, while cyclist B is 52 hundred meters to the east of P and traveling at 24 meters per second. How fast is the distance between the two cyclists changing?

Answer 1

Answer:

  38.4 m/s (increasing)

Step-by-step explanation:

There are at least a couple of ways to work this. One is to write the equation for the distance between the cyclists and differentiate it.

  d^2 = (3900 +32t)^2 + (5200 +24t)^2

  (2d)dd/dt = 2(3900 +32t)(32) +2(5200 +24t)(24)

At t=0, this becomes ...

  dd/dt = (32·3900 +24·5200)/√(3900^2 +5200^2)

  = (124800 +124800)/√(15210000 [phone])

  = 249600/6500 = 38.4 . . . . meters per second