One airplane is approaching an airport from the north at 203 km/hr. a second airplane approaches from the east at 300 km/hr

Question

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.

Answer 1

n=203t

e=300t

(203t-32)^2+(300t-16)^2=c^2

c is the diagonal distance between n and e.

Need more imfo (t).

Answer 2

This related rate problem is grounded in the Pythagorean theorem. The two planes approach the airport with right-angled paths. Since they are both approaching the airport, their distances from the airport are decreasing, as is their straight line distance from each other. We're concerned with the moment when the southbound plane is 32 km from the airport, and the westbound plane is 16 km away. At that moment, their straight line distance from each other is the hypotenuse of the right triangle that the paths form. This hypotenuse is √(32² + 16²) = 16√5. Now we're ready to set up the equation. We can let x, y, and z be the leg distances and the hypotenuse distance.

x² + y² = z²

These are all functions of time t, and we do implicit differentiation with respect to t.

2x(dx/ft) + 2y(dy/dt) = 2z(dz/dt)

This can be simplified a little by dividing by 2:

x(dx/ft) + y(dy/dt) = z(dz/dt)

Now we can substitute in all of our known values. Note that since distances are decreasing as the planes approach the airport, velocities are negative.

16(-300) + 32(-203) = 16√5(dz/dt)

-11296 = 16√5(dz/dt)

-315.73 = (dz/dt)

So the planes are approaching each other at a rate of -315.73 km/hr.