Two airplanes are flying in the air at the same height. Airplane A is flying east at 200 mph and airplane B is flying north at 350 mph. If they are both heading to the same airport, located 50 miles east of airplane A and 120 miles north of airplane B, at what rate (in mph) is the distance between the airplanes changing?

We can solve this problem using the concept of related rates in calculus.

Let’s define the problem as follows:

- Airplane A is 50 miles west of the airport and is flying east at 200 mph.
- Airplane B is 120 miles south of the airport and is flying north at 350 mph.

We are interested in the rate at which the distance between the two airplanes is changing.

### Step 1: Define variables
Let:
- \( x(t) \) be the distance of airplane A from the airport (in miles) at time \( t \) (measured east-west).
- \( y(t) \) be the distance of airplane B from the airport (in miles) at time \( t \) (measured north-south).
- \( z(t) \) be the distance between the two airplanes (the quantity we want to find).

At time \( t = 0 \):
- Airplane A is 50 miles west of the airport, so \( x(0) = 50 \).
- Airplane B is 120 miles south of the airport, so \( y(0) = 120 \).

As both airplanes move:
- Airplane A’s distance from the airport is decreasing at a rate of 200 mph (eastwards). Therefore, \( \frac{dx}{dt} = -200 \).
- Airplane B’s distance from the airport is decreasing at a rate of 350 mph (northwards). Therefore, \( \frac{dy}{dt} = -350 \).

We need to find \( \frac{dz}{dt} \), the rate at which the distance between the airplanes is changing.

### Step 2: Apply the Pythagorean theorem
Since the airplanes are traveling at right angles to each other, the distance between them forms the hypotenuse of a right triangle with legs \( x(t) \) and \( y(t) \). So we can use the Pythagorean theorem:

\[
z^2 = x^2 + y^2
\]

### Step 3: Differentiate with respect to time \( t \)
Now, differentiate both sides of the equation with respect to \( t \):

\[
2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}
\]

Simplifying:

\[
z \frac{dz}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt}
\]

### Step 4: Substitute the known values
At the initial time \( t = 0 \), we know:
- \( x(0) = 50 \)
- \( y(0) = 120 \)
- \( \frac{dx}{dt} = -200 \)
- \( \frac{dy}{dt} = -350 \)

First, calculate \( z(0) \), the initial distance between the airplanes:

\[
z(0) = \sqrt{x(0)^2 + y(0)^2} = \sqrt{50^2 + 120^2} = \sqrt{2500 + 14400} = \sqrt{16900} = 130 \text{ miles}
\]

Now substitute the values into the differentiated equation:

\[
130 \frac{dz}{dt} = (50)(-200) + (120)(-350)
\]

\[
130 \frac{dz}{dt} = -10000 - 42000
\]

\[
130 \frac{dz}{dt} = -52000
\]

Now solve for \( \frac{dz}{dt} \):

\[
\frac{dz}{dt} = \frac{-52000}{130} = -400 \text{ mph}
\]

### Step 5: Interpret the result
The negative sign indicates that the distance between the airplanes is decreasing. Therefore, the rate at which the distance between the airplanes is changing is **400 mph** (decreasing).

### Summary:
The distance between the airplanes is decreasing at a rate of 400 mph.

Would you like a detailed explanation of any step?

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Here are 5 related questions to expand on this problem:
1. How would the problem change if airplane A was flying at a different speed?
2. What happens to the rate of change of distance if the airplanes are flying in the same direction?
3. How can we model this problem using vectors instead of the Pythagorean theorem?
4. What if one of the airplanes was flying at a changing speed? How would that affect the calculations?
5. Can we apply similar related rates problems to other types of vehicles, such as cars or ships?

**Tip:** When working with related rates problems, always differentiate the equation that relates the variables before plugging in the values.

Learn how to calculate the rate at which the distance between two airplanes flying towards an airport is changing using related rates. This problem applies the Pythagorean theorem and involves differentiation to find the solution. Suitable for students in Grades 11-12.

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