Question

Two commercial airplanes are flying at an altitude of​ 40,000 feet along with​ straight-line courses that intersect at right angles. Plane A is approaching the intersection point at a speed of 442 knots​ (nautical miles per​ hour; a nautical mile is 2000​ yards). Plane B is approaching the intersection at 481 knots. At what rate is the distance between the planes changing when A is 5 nautical miles from the intersection point and B is 12 nautical miles from the intersection​ point?

Answer 1

Answer:

The distance between the planes is decreasing at a rate of 614 knots.

Step-by-step explanation:

Let x be the distance A is from the intersection point, and let y be the distance B is from the intersection point. Let s be the distance between A and B, so s² = x² + y²

Note that s, x and y are functions of time, t, so to emphasize this we should write  

s(t)² = x(t)² + y(t)²

Differentiate both sides of the above equation, with respect to t, to get  

2s*ds/dt = 2x*dx/dt + 2y*dy/dt .  

Dividing by 2 gives  

s*ds/dt = x*dx/dt + y*dy/dt ,  

and dividing by s gives  

ds/dt = (1/s)*(x*dx/dt + y*dy/dt)  

Since

dx/dt = −442,

dy/dt = −481,

x = 5 nautical miles, and

y = 12 nautical miles, we have  

ds/dt = (1/√(5² + 12²))*(5*(−442) + 12*(−481)) = −7982/13 = − 614.  

Thus the distance between the planes is decreasing at a rate of 614 knots.