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Thepotemich [5.8K]

4 years ago

6

One airplane is approaching an airport from the north at 181 kn/hr. A second airplane approaches from the east at 278 km/hr. Fin

d 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.

1 answer:

umka2103 [35]4 years ago

4 0

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

$R = \sqrt{N^2 + E^2}$

=> $R = \sqrt{30^2 + 15^2}$

=> $R = \sqrt{30^2 + 15^2}$

=> $R =33.54 \ m$

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

=> $67.08 * \frac{dR}{dt} = -19200$

=> $\frac{dR}{dt} = -286.2 \ km/hr$

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