The rate at which the distance from the plane to the station increases is 537 miles/hour. Computed using Pythagoras Theorem and Differentiation.
We assume the station to be at point A.
The plane when directly over the station, takes point C.
As it is flying horizontally continuously at an altitude of 1 mile, AC = b = 1 mile.
The plane flies continuously, and when it is at a distance of 3 miles from the station, it is at point B.
The distance between the plane and the station of 3 miles can be shown as AB = c = 3 miles.
The movement of the plane from C to B can be shown as BC = a.
Now, from the given case, we get a right triangle ABC.
By Pythagoras' Theorem,
a² + b² = c² ... (i).
or, a² + 1² = 3²,
or, a² = 9 - 1 = 8,
or, a = 2√2 miles.
We are told that the plane is moving at a speed of 570 miles/hour.
This can be shown as, da/dt = 570 miles/hour.
As the plane is moving horizontally, the vertical distance between the plane and the station doesn't change, that is, db/dt = 0.
In the question, we are asked to find the rate at which the distance from the plane to the station is increasing, that is, we are asked to find dc/dt.
Differentiating (i), with respect to time t, we get:
2a(da/dt) + 2b(db/dt) = 2c(dc/dt).
Substituting values of a = 2√2 miles, b = 1 mile, c = 3 miles, da/dt = 570 miles/hour, and db/dt = 0, we get:
2(2√2)(570) + 2(1)(0) = 2(3)(dc/dt),
or, 2280√2 + 0 = 6(dc/dt),
or, dc/dt = 380√2 miles/hour = 537.40 miles/hour ≈ 537 miles/hour.
Thus, the rate at which the distance from the plane to the station increases is 537 miles/hour. Computed using Pythagoras Theorem and Differentiation.
Learn more about the Pythagoras Theorem and Differentiation at
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