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An aircraft is 8 km above the ground and flying horizontally at a speed of 400 km/h. How far is the distance between the aircraft and the radio beacon increasing 3 minutes after the aircraft passes directly over the beacon? could someone please explain.

Answer 1

"How far is the distance between the aircraft and the radio beacon increasing 3 minutes after the aircraft passes directly over the beacon?" do you mean how fast instead of how far? If so than the solution is as follows:
Let the height the plane is flying at be h = 8 km. Let the distance it has traveled along the ground (that is, perfectly horizontally) be x. Then dx/dt = 400 km/h. Call the distance between the plane and the radio beacon L. Then h, x, and L form a right triangle. By the Pythagorean Theorem,
\[L ^{2}=x^{2}+h^{2}\]
So\[L=\sqrt{x^2+h^2}\] now we differentiate with respect to time,
\[dL/dt=(x/\sqrt{x^2+h^2})dx/dt\]
After 3 minutes the aircraft has gone (horizontally) (3 min)x(400 km/h) = (0.05 h)x(400 km/h) = 20 km. Plug in x = 20 km, h = 8 km and dx/dt = 400 km/h and you get dL/dt = 371.39 km/h, which is about 103.15 m/sec. Note: if you're having trouble following, draw a picture and you'll see what I'm talking about.