A plane flying horizontally at an altitude of 3 mi and a speed of 520 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 has a total distance of 4 mi away from the station. (Round your answer to the nearest whole number.)


This is a related rates problem where we need to determine how fast the distance between the plane and the radar station is increasing at a certain moment.

### Step 1: Set up the problem

We know:
- The plane is flying horizontally at an altitude of 3 miles.
- The horizontal speed of the plane is 520 miles per hour.
- We want to find the rate at which the distance between the plane and the radar station is increasing when the total distance is 4 miles.

We can model this situation using the Pythagorean Theorem. Let:
- \( x(t) \) represent the horizontal distance between the plane and the radar station.
- \( z(t) \) represent the total distance between the plane and the radar station.
- The altitude \( y \) of the plane is constant at 3 miles.

The relationship between these quantities is given by the Pythagorean theorem:
\[
z^2 = x^2 + y^2
\]
Since the altitude is constant at 3 miles:
\[
z^2 = x^2 + 3^2 \quad \text{or} \quad z^2 = x^2 + 9
\]

### Step 2: Differentiate with respect to time

We need to find how fast \( z \) is changing, i.e., \( \frac{dz}{dt} \), when \( z = 4 \) miles. Using implicit differentiation with respect to time \( t \):
\[
2z \frac{dz}{dt} = 2x \frac{dx}{dt}
\]
Simplifying:
\[
z \frac{dz}{dt} = x \frac{dx}{dt}
\]

### Step 3: Solve for \( \frac{dz}{dt} \)

Now, solve for \( \frac{dz}{dt} \):
\[
\frac{dz}{dt} = \frac{x}{z} \frac{dx}{dt}
\]

We know:
- \( z = 4 \) miles,
- \( y = 3 \) miles (constant),
- \( \frac{dx}{dt} = 520 \) miles per hour (the horizontal speed of the plane).

To find \( x \), use the Pythagorean theorem when \( z = 4 \):
\[
z^2 = x^2 + 3^2
\]
\[
4^2 = x^2 + 9
\]
\[
16 = x^2 + 9 \quad \Rightarrow \quad x^2 = 7 \quad \Rightarrow \quad x = \sqrt{7}
\]

Now substitute \( x = \sqrt{7} \), \( z = 4 \), and \( \frac{dx}{dt} = 520 \) into the equation:
\[
\frac{dz}{dt} = \frac{\sqrt{7}}{4} \times 520
\]
\[
\frac{dz}{dt} \approx 0.6614 \times 520 \approx 343.92
\]

### Step 4: Final answer

Rounding to the nearest whole number:
\[
\frac{dz}{dt} \approx 344 \text{ miles per hour}
\]

Thus, the rate at which the distance from the plane to the radar station is increasing is approximately **344 miles per hour**.

Do you have any questions, or would you like more details on any step?

Here are 5 related questions you might explore:
1. What would happen if the altitude of the plane changes?
2. How does the rate of change vary as the plane gets farther from the radar station?
3. How would you solve this problem if the plane were descending at a constant rate?
4. Can you find the distance rate if the plane flies directly away from the radar station at a constant speed?
5. How would the answer change if the speed of the plane were doubled?

**Tip:** When solving related rates problems, always start by writing an equation that relates all variables before differentiating.

A plane flying horizontally at an altitude of 3 mi and a speed of 520 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 has a total distance of 4 mi away from the station. (Round your answer to the nearest whole number.)

Solve a related rates problem involving a plane flying at 520 mi/h at an altitude of 3 miles. Learn how to use the Pythagorean Theorem and implicit differentiation to find the rate at which the distance from the plane to a radar station is increasing when it is 4 miles away. Suitable for Grades 11-12.

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