Two planes leave the airport at the same time. One plane heads directly south at 550 mph. The other plane heads directly west at 300 mph. Assuming both fly at the same elevation, how quickly is the distance between them increasing 3 hours after takeoff (rounded to the nearest mile per hour)?
We proceed by applying the Pythagorean Theorem, which states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. In this case, the sides of the right triangle represent the distances traveled by the two planes.
Let $x$ represent the distance traveled by the southbound plane and $y$ represent the distance traveled by the westbound plane. We are interested in finding the rate of change of the distance between the planes, which can be represented by the length of the hypotenuse of the triangle.
We have a right triangle with the following sides:
Legs:
Southbound plane distance: $x = 550t$ miles
Westbound plane distance: $y = 300t$ miles
Hypotenuse:
Distance between the planes: $\sqrt{x^2 + y^2}$ miles
Taking the derivative with respect to time $t$, we find:
$$\frac{d}{dt}\left(\sqrt{x^2 + y^2}\right) = \frac{1}{2\sqrt{x^2 + y^2}} (2x\frac{dx}{dt} + 2y\frac{dy}{dt})$$
$$= \frac{1}{2\sqrt{x^2 + y^2}} (2x\cdot 550 + 2y\cdot(-300))$$
$$= \frac{x\cdot550 - y\cdot300}{\sqrt{x^2 + y^2}}$$
At $t=3$, we have $x = (550)(3) = 1650$ and $y = (300)(3) = 900$. Plugging these values in, we get:
$$\frac{(1650)(550) - (900)(300)}{\sqrt{(1650)^2 + (900)^2}}$$
$$= \frac{907,500 - 270,000}{\sqrt{3,025,000 + 810,000}}$$
$$= \frac{637,500}{\sqrt{3,835,000}}$$
$$\approx \frac{637,500}{1958.76}$$
$$\approx 326.26\text{ mph}$$
Thus, the distance between the planes is increasing at a rate of approximately $\boxed{326}$ mph three hours after takeoff.