An air traffic controller spots 2 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 450 miles per hour. The other plane is 200 miles from the point moving at 600 miles per hour. At what rate is the distance between the planes decreasing? How much time does the air traffic controller have to get one of the planes on a different flight path?

At the moment in question, the distance d = 250

x^2 + y^2 = d^2
2xx' + 2yy' = 2dd'
2(150)(-450) + 2(200)(-600) = 2(250)d'

Now, find d'

750

To find the rate at which the distance between the planes is decreasing, we can use the concept of related rates.

Let's denote the distance between the planes as "D." We want to find dD/dt, the rate at which D is changing over time.

We can use the Pythagorean theorem to relate the distance between the planes, D, to the distances of each plane from the point of convergence. Let's call the distance of the first plane from the point of convergence "x" and the distance of the second plane from the point of convergence "y."

According to the Pythagorean theorem, we have:

x^2 + y^2 = D^2. ---(1)

Now, since the planes are moving, the distances x and y are changing with respect to time. Let's denote dx/dt as the rate of change of x and dy/dt as the rate of change of y.

Differentiating equation (1) implicitly with respect to time, we get:

2x(dx/dt) + 2y(dy/dt) = 2D(dD/dt).

Since the planes are moving at right angles to each other, x and y are changing independently. Therefore, dx/dt = 450 mph and dy/dt = 600 mph.

Substituting these values into the equation, we have:

2(150)(450) + 2(200)(600) = 2D(dD/dt).

Simplifying the equation, we get:

135,000 + 240,000 = 2D(dD/dt).

375,000 = 2D(dD/dt).

Now, let's solve for dD/dt:

dD/dt = 375,000 / (2D).

To find the value of D, we can use the Pythagorean theorem again:

150^2 + 200^2 = D^2.

Simplifying, we get:

22,500 + 40,000 = D^2.

62,500 = D^2.

D ≈ 250.

Now, substituting this value into the equation, we have:

dD/dt = 375,000 / (2 * 250).

dD/dt = 750 mph.

So, the distance between the planes is decreasing at a rate of 750 mph.

To find the time it takes for the air traffic controller to get one of the planes on a different flight path, we need to find the time it takes for the distance between the planes to become zero.

We can set D = 0 in the equation D ≈ 250, and solve for t:

150^2 + 200^2 = (450t)^2 + (600t)^2.

22,500 + 40,000 = (450t)^2 + (600t)^2.

62,500 = 202,500t^2.

t^2 = 62,500 / 202,500.

t^2 ≈ 0.3086.

t ≈ 0.5554 hours.

Therefore, the air traffic controller has approximately 0.5554 hours, or about 33.3 minutes, to get one of the planes on a different flight path.

To determine the rate at which the distance between the planes is decreasing, we can use the concept of related rates.

Let's consider the two planes as being at points A and B, with the intersection point being denoted as O. We are interested in finding the rate at which the distance AB (which is the distance between the two planes) is changing.

We are given the following information:
- Plane A is 150 miles from O and moving at a speed of 450 miles per hour.
- Plane B is 200 miles from O and moving at a speed of 600 miles per hour.

To find the rate at which AB is changing, we need to find expressions for AB and its derivatives with respect to time.

From the Pythagorean theorem, we know that the distance AB can be determined by:

AB = √(OA^2 + OB^2)

Differentiating both sides of the equation with respect to time (t), we get:

d(AB)/dt = (d(OA)/dt * OA + d(OB)/dt * OB) / AB

Now, let's calculate the derivatives d(OA)/dt and d(OB)/dt:

- d(OA)/dt is the rate at which the distance of plane A from O is changing. Since plane A is moving at a constant speed of 450 miles per hour, d(OA)/dt = 450 mph.

- d(OB)/dt is the rate at which the distance of plane B from O is changing. Since plane B is moving at a constant speed of 600 miles per hour, d(OB)/dt = 600 mph.

Substituting these values into the equation, we have:

d(AB)/dt = (450 * OA + 600 * OB) / AB

To find the distance between the two planes, we need to calculate OA and OB using the given information:

OA = 150 miles
OB = 200 miles

Substituting these values into the equation, we have:

d(AB)/dt = (450 * 150 + 600 * 200) / AB

Now, let's calculate AB:

AB = √(OA^2 + OB^2)
= √(150^2 + 200^2)
= √(22500 + 40000)
= √62500
= 250 miles

Substituting AB = 250 miles into the equation, we have:

d(AB)/dt = (450 * 150 + 600 * 200) / 250

Now, let's calculate the rate at which the distance between the two planes is decreasing:

d(AB)/dt = (67500 + 120000) / 250
= 187500 / 250
= 750 mph

Therefore, the distance between the two planes is decreasing at a rate of 750 miles per hour.

To determine the time the air traffic controller has to get one of the planes on a different flight path, we can use the time it would take for the distance AB to become zero. Since the rate at which AB is decreasing is 750 mph, we can calculate the time as:

Time = distance / rate
= 250 miles / 750 mph
= 1/3 hours
≈ 20 minutes

Hence, the air traffic controller has approximately 20 minutes to get one of the planes on a different flight path.