Question
Asked Nov 5, 2019
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One airplane is approaching an airport from the north at 203 km/hr. a second airplane approaches from the east at 300 km/hr. find the rate at which the distance between the planes changes when the southbound plane is 32 km away from the airport and the westbound plane is 16 km from the airport.

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Expert Answer

Step 1

Let in the below figure P represents the airport, Q represents the southbound plane and R represents the westbound plane.

Then x = 16 km, y = 32 km

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у P х

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Step 2

Using Pythagoras theorem in triangle PQR

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x2y2 Differentiate both sides with respect to t dy = Z dt dx dz --(ii) dt dt Hereis the velocity of west bound plane dt andis the velocity of east bound plane. dx = -203 km/hr dt - 300 km/hr and dt Substitute x 16, y = 32 in (i) and find z z2=16 +32 = 1280 x 35.78

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Step 3

Now find the change of distance between two planes b...

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dz dx = x- dt dy dt dt dz (35.78)(16)(-300)(32)(-203) dt dz 4800 6496 dt 35.78 dz-11296 315 km/hr dt 35.78

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Tagged in

Math

Calculus

Derivative