An air traffic controller spots two airplanes at the same altitude converging to a point as they fly at right angles to each other. One airplane is 70 miles from the point and has a speed of 420 miles per hour. The other is 240 miles from the point and has a speed of 1440 miles per hour.

(a) At what rate is the distance between the planes changing? in mph?
(b) How much time does the controller have to get one of the airplanes on a different flight path?

(a) ðz/ðt = 1500miles/hr

(b) 10minutes

Step-by-step explanation:

Let x rep plane one

Let y rep plane two

ðx/ðt = 1440miles/hr = speed of plane one

ðy/ðt = 420miles/hr = speed of plane 2

ðz/ðt = ? = rate of change of distance btw two plane moving at right angle to meet at point o...

See attachment

Since x² + y² = z²... equation 1

So therefore (240² + 70²) ^½= z = 250 miles = distance between two planes

Differential rate of equation 1 gives

(2x* ðx/ðt ) + (2y* ðy/ðt) = 2z* ðz/ðt

Substituting to get ðz/ðt

(2*240*1440) + (2*70*420) = 2*250*ðz/ðt

750000 = 500 ðz/ðt

ðz/ðt = 1500miles/hr

B.

Since v = d/t where d = distance = 250miles

t = time = ? = time needed for controller to divert one of e planes , v = 1500miles/hr

t = d/v = 250/1500 = 10minutes

Sure! what is the code?