A plane flying horizontally at an altitude of 1 mi and a speed of 560 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station. (Round your answer to the nearest whole number.)

when the plane has traveled x miles, the distance d is found using

d^2 = 1 + x^2
2d dd/dt = 2x dx/dt

when d=2, we have x=√3

2(2) dd/dt = 2√3 (560)
dd/dt = 485 mph

To find the rate at which the distance from the plane to the station is increasing, we need to differentiate the given information.

Let's assume that the distance between the plane and the station is represented by the variable "x" (in miles), and the time is represented by the variable "t" (in hours).

Since the plane is flying horizontally, the distance between the plane and the radar station is equal to its altitude, which is 1 mile.

At a given time t, the distance x between the plane and the radar station can be represented by the Pythagorean theorem as follows:

x^2 = 1^2 + (560t)^2

To find the rate at which the distance is increasing, we need to differentiate both sides of the equation with respect to time t:

2x(dx/dt) = 0 + 2(560t)(560)

Simplifying the equation:

2x(dx/dt) = 1120t(560)

dx/dt = 1120t(560) / 2x

We need to find the rate of change when x = 2 miles away from the station, which means using x = 2 in the equation:

dx/dt = 1120t(560) / 2(2)

dx/dt = 1120t(280) / 4

dx/dt = 280 * 280t / 4

dx/dt = 19600t

Now, we can substitute t = 2 (as given in the problem):

dx/dt = 19600 * 2

dx/dt = 39200

Therefore, the rate at which the distance from the plane to the station is increasing when it is 2 miles away from the station is approximately 39,200 miles per hour.

To solve this problem, we can use the concept of related rates, which involves finding the rate of change of one quantity with respect to another related quantity.

Let's consider the distance between the plane and the radar station as our related quantities. Let's define the distance between the plane and the station as "x," and the rate at which the distance is changing as "dx/dt," where dt represents the change in time.

Given:
- The plane is flying horizontally at an altitude of 1 mi.
- The speed of the plane is 560 mi/h.
- We need to find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station.

We can start by drawing a diagram to visualize the problem. Since the plane is flying horizontally, the altitude does not affect the horizontal distance between the plane and the station. Keep in mind that the distance remains constant.

Let's consider a right triangle formed by the plane, the station, and the distance between them. The distance between the plane and the station is the hypotenuse of the triangle, and the horizontal distance is the adjacent side.

Using the Pythagorean theorem, we can express the distance between the plane and the station, x, in terms of the horizontal distance, y:

x^2 = y^2 + 1^2 (altitude of 1 mi)

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

(2x) * (dx/dt) = (2y) * (dy/dt) + 0 (since the altitude is constant and has no change in time)

We can simplify this equation further by dividing both sides by 2:

x * (dx/dt) = y * (dy/dt)

Now we have an equation relating the rates of change of x and y.

At the given moment, when the plane is 2 mi away from the station, we can substitute x = 2 in the equation above.

2 * (dx/dt) = y * (dy/dt)

We need to find dx/dt, the rate at which the distance from the plane to the station is changing. To do that, we need to find the values of y and (dy/dt).

Since the plane is flying horizontally, the rate of change of the horizontal distance, (dy/dt), is equal to the speed of the plane. Hence, (dy/dt) = 560 mi/h.

Now, we need to find y, the horizontal distance between the plane and the station when the plane is 2 mi away.

Using the Pythagorean theorem equation:

x^2 = y^2 + 1^2

Substituting x = 2:

2^2 = y^2 + 1

4 = y^2 + 1

y^2 = 4 - 1

y^2 = 3

y = √3

Thus, when the plane is 2 mi away from the station, the horizontal distance, y, between the plane and the station is √3 mi.

Now we can substitute these values in the equation:

2 * (dx/dt) = (√3) * (560)

To find dx/dt, we isolate it:

(dx/dt) = (√3 * 560) / 2

(dx/dt) = 980√3

Now, rounding to the nearest whole number, the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is approximately 1690 mi/h.