Two airplanes are flying in the air at the same height: airplane A is flying east at 340 mi/h and airplane B is flying north at 420mi/h.

If they are both heading to the same airport, located 60 miles east of airplane A and 80 miles north of airplane B, at what rate is the distance between the airplanes changing?

Round to the nearest tenth of a mile per hour.

Note: If you do not use derivatives and calculus, you will receive a 0.

dL/dt = 1707, 2 m/h

Step-by-step explanation:

Airplanes A and B flying east and north and the airport all, form a right triangle; the hypotenuse is L distance between the airplanes and position of airplanes A and B the legs. Therefore we can write:

L² = x² + y² (x and y position of airplanes)

Differentiation in relation to time on both sides of the equation

2*L*dL/dt = 2*x*dx/dt + 2*y * dy/dt (1)

In this expression we know:

x = 60 miles dx/dt = 340 miles / hour

y = 80 miles dy/dt = 420 miles / hour

We have to calculate L for the particular moment and then we can solve for DL/dt

L² = x² + y² ⇒ L² = (60) ² + (80) ² ⇒ L² = 360 + 640

L² = 1000 ⇒ L = 31.63 m

Plugging all the values in equation (1)

2*31.63 * dL/dt = 2*60*340 + 2*80*420

dL/dt = (20400 + 33600) / 31.63

dL/dt = 1707, 24 m/h