Compute the velocity of an airplane given the rate of change of its distance to a point on the ground

An airplane is 8 miles above the ground flying at a constant velocity, at a constant altitude. (Assume the earth is flat.) There is a point $P$ on the ground directly under the airplane’s flight path. The distance between $P$ and the airplane is decreasing at a rate of 4 miles per minute when the distance is 10 miles. Find the velocity of the airplane.

Let $x$ be the distance from the point on the ground directly beneath the plane to the point $P$ , and let $y$ be the distance from the plane to $P$ . The following diagram illustrates the situation:

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We are trying to compute $\frac{dx}{dt}$ , the velocity of the airplane. We are given that the distance from the airplane to the point $P$ is changing at a rate of 4 miles per minute when $y = 10$ . Thus, adjusting units to mph, we have

$\frac{dy}{dt} = 240 \text{ mph} \qquad \text{when } y = 10 \text{ miles}.$

Furthermore, we can compute $x^2$ in terms of $y^2$ (by the Pythagorean identity) and then differentiate:

$x^2 = y^2 - 64 \quad \implies \quad \frac{dx}{dy} = \frac{y}{\sqrt{y^2 - 64}}.$

So, at $y = 10$ miles we have $\frac{dx}{dy} = \frac{5}{3}$ ; thus,

$\frac{dx}{dt} = \left( \frac{dy}{dt} \right) \left( \frac{dx}{dy} \right) = (240) \frac{5}{3} = 400 \text{ mph}.$

This was computed for $y = 10$ miles, but since we are given that the airplane is flying at a constant velocity, then this velocity is valid for all $y$ .

Stumbling Robot

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