Final answer:
The distance between two planes approaching an airport along perpendicular paths is decreasing at a rate of 615.38 mph. Using the principle of related rates and the Pythagorean theorem, the rate is calculated by differentiating distance with respect to time and plugging in the given rates and distances.
Step-by-step explanation:
The question concerns the rate at which the distance between two airplanes is changing as they fly towards the same airport along perpendicular paths. To solve this problem, we apply the Pythagorean theorem to find the distance between the planes at any given time. Since we are dealing with rates of change, we will differentiate with respect to time using related rates.
Let d represent the distance between the two planes, x the distance of Plane A from the airport, and y the distance of Plane B from the airport. At the given moment, x = 100 miles and y = 240 miles. According to the Pythagorean theorem, d2 = x2 + y2. Differentiating with respect to time t, we get 2ddd/dt = 2xdx/dt + 2ydy/dt.
Plane A is flying south towards the airport, so the rate of change of x is -520 mph (since it is decreasing). Plane B is flying west towards the airport, so the rate of change of y is -450 mph. Inserting these values and the initial distances into the differentiated equation, we can solve for dd/dt, which is the rate at which the distance between the planes is changing.
Doing the calculations, we get:
2ddd/dt = -2(100)(520) + -2(240)(450)
To solve for dd/dt, we need the current distance d, which we find using the initial positions of the planes:
d = √(1002 + 2402)
d = √(10000 + 57600)
d = √67600
d = 260 miles
Substituting d back into the rate equation:
2(260)dd/dt = -2(100)(520) + -2(240)(450)
520dd/dt = -104000 - 216000
dd/dt = -320000 / 520
dd/dt = -615.38 mph
Therefore, the distance between the two planes is decreasing at a rate of 615.38 mph.