Calc
posted by Anonymous on .
Two commercial airplanes are flying at an altitude of 40,000 ft along straightline courses that intersect at right angles. Plane A is approaching the intersection point at a speed of 427 knots (nautical miles per hour; a nautical mile is 2000 yd or 6000 ft.) Plane B is approaching the intersection at 439 knots.
At what rate is the distance between the planes decreasing when Plane A is 5 nautical miles from the intersection point and Plane B is 5 nautical miles from the intersection point?

distanceapart^2=distaneAfromcollision^2+distanceBfromcollision^2
Take the derivative..
s ds/dt=2xdx/dt + 2y dy/dt where x, y are the distances from the intersection point, and s is sqrt(x^2+y^2)
you are given dx/dt, dy/dt, solve for ds/dt