Question

One airplane is approaching an airport from the north at 181 kn/hr. A second airplane approaches from the east at 278 km/hr. Find the rate at which the distance between the planes changes when the southbound plane is 30 km away from the airport and the westbound plane is 15 km from airport.

Answer 1

Answer:

The value  is  $\frac{dR}{dt} = -286.2 \ km/hr$

Explanation:

From the question we are told that  

   The speed of the airplane from the north is $\frac{dN}{dt} = -181 \ km /hr$

The negative sign is because the direction is towards the south

  The speed of the airplane from the east is  $\frac{dE}{dt} = -278 \ km/hr$

The negative sign is because the direction is towards the west

   The distance of the southbound plane from the airport is  $N = 30 \ km$

   The distance of the westbound plane is  $E = 15 \ km$

Generally the distance between the plane is mathematically represented using Pythagoras theorem  as

    $R^2 = N^2 + E^2$

Next differentiate implicitly this equation to obtain the rate at which the distance between the planes changes

So

     $2R\frac{dR}{dt} = 2N \frac{dN}{dt} + 2E\frac{dE}{dt}$

Here

     $R = \sqrt{N^2 + E^2}$

=>    $R = \sqrt{30^2 + 15^2}$

=>    $R = \sqrt{30^2 + 15^2}$

=>    $R =33.54 \ m$

    $2(33.54) * \frac{dR}{dt} = 2( 30)*(-181) + 2*15*(-278)$

=>   $67.08 * \frac{dR}{dt} = -19200$

=>   $\frac{dR}{dt} = -286.2 \ km/hr$