A plane flying horizontally at an altitude of 2 mi and a speed of 420 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 5 mi away from the station.

well, after t hours, the plane has gone 420t miles.

So, the distance z is found using

z^2 = 2^2 + (420t)^2
so, that means that z is changing at

2z dz/dt = 2*420^2 t

now just plug in your numbers.

To find the rate at which the distance from the plane to the station is increasing, we need to use the concept of related rates. We can start by drawing a diagram to visualize the problem.

Let's label the following information:
- The altitude of the plane is 2 mi.
- The speed of the plane is 420 mi/h.
- The distance from the plane to the radar station is represented by the line segment "x."

We want to find dX/dt, which represents the rate at which the distance from the plane to the station is increasing. We are given that the plane is 5 mi away from the station (x = 5). We need to find dX/dt when x = 5.

To relate the variables, we can use the Pythagorean theorem: x^2 = y^2 + z^2, where y is the altitude of the plane and z is the distance the plane has traveled horizontally.

First, let's find z, the distance the plane has traveled horizontally:
Speed = Distance/Time.
Therefore, Time = Distance / Speed.
The plane is flying horizontally, so the time it takes to travel z miles is given by z / 420.

Now, let's differentiate the equation x^2 = y^2 + z^2 with respect to time (t):
2x(dx/dt) = 2y(dy/dt) + 2z(dz/dt).

Since we want to find dX/dt and we know dx/dt = 0 (the distance x is not changing), the equation becomes:
2x(0) = 2y(dy/dt) + 2z(dz/dt).

Considering the given values, we substitute:
2(5)(0) = 2(2)(dy/dt) + 2(√(5^2 - 2^2))(dz/dt).

Simplifying further, we have:
0 = 4(dy/dt) + 2√(21)(dz/dt).

Rearrange the equation to isolate dy/dt, which represents the rate at which the altitude of the plane is changing:
4(dy/dt) = -2√(21)(dz/dt).
(dy/dt) = (-1/2) * √(21)(dz/dt).

Since we want to find dz/dt when x = 5, we can calculate it using the equation for time we derived earlier:
z = 420 * (z / 420) (simplifying time = distance/speed)
1 = dz/dt.

Now, substitute dz/dt = 1 into the equation for dy/dt:
(dy/dt) = (-1/2) * √(21)(1).

Simplifying further, we have:
(dy/dt) = -√21/2.

Therefore, the rate at which the distance from the plane to the station is increasing when it is 5 mi away from the station is -√21/2 mi/h.