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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.

Respuesta :

Answer:

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

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