An air traffic controller spots two planes at the same altitude converging on a point as they fly at right angles to each other. One plane is 150 miles from the point moving at 300 miles per hour. The other plane is 200 miles from the point moving at 400 miles per hour.(a)At what rate (in mi/hr) is the distance s between the planes changing?
Question
An air traffic controller spots two planes at the same altitude converging on a point as they fly at right angles to each other. One plane is 150 miles from the point moving at 300 miles per hour. The other plane is 200 miles from the point moving at 400 miles per hour.(a)At what rate (in mi/hr) is the distance s between the planes changing?
Solution
To solve this problem, we can use the Pythagorean theorem and the concept of related rates.
Step 1: Set up the Pythagorean theorem.
Let's denote the distance of the first plane from the point as x, the distance of the second plane from the point as y, and the distance between the planes as s. According to the Pythagorean theorem, we have:
s² = x² + y²
Step 2: Differentiate both sides with respect to time t.
2s(ds/dt) = 2x(dx/dt) + 2y(dy/dt)
We can simplify this to:
s(ds/dt) = x(dx/dt) + y(dy/dt)
Step 3: Substitute the given values.
We know that x = 150 miles, dx/dt = -300 miles per hour (the negative sign indicates that the distance is decreasing), y = 200 miles, and dy/dt = -400 miles per hour. We want to find ds/dt when s = sqrt(x² + y²) = sqrt((150 miles)² + (200 miles)²) = 250 miles.
Substituting these values into the equation gives:
250(ds/dt) = 150(-300) + 200(-400)
Step 4: Solve for ds/dt.
Solving the above equation for ds/dt gives:
ds/dt = (150(-300) + 200(-400)) / 250 = -340 miles per hour
The negative sign indicates that the distance between the planes is decreasing. So, the distance between the planes is decreasing at a rate of 340 miles per hour.
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