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 300 mi/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

Answer:

  decreasing at 390 miles per hour

Step-by-step explanation:

Airplane A's distance in miles to the airport can be written as ...

  a = 30 -250t . . . . . where t is in hours

Likewise, airplane B's distance to the airport can be written as ...

  b = 40 -300t

The distance (d) between the airplanes can be found using the Pythagorean theorem:

  d^2 = a^2 + b^2

Differentiating with respect to time, we have ...

  2d·d' = 2a·a' +2b·b'

  d' = (a·a' +b·b')/d

__

To find a numerical value of this, we need to find the values of its variables at t=0.

  a = 30 -250·0 = 30

  a' = -250

  b = 40 -300·0 = 40

  b' = -300

  d = √(a²+b²) = √(900+1600) = 50

Then ...

  d' = (30(-250) +40(-300))/50 = -19500/50 = -390

The distance between the airplanes is decreasing at 390 miles per hour.