A plane flying horizontally at an altitude of 2 miles 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 5 miles away from the station. (Round your answer to the nearest whole number.)


Let's define the variables and understand the given problem in mathematical terms.

1. **Altitude of the plane**: \( h = 2 \) miles.
2. **Speed of the plane**: \( v = 520 \) miles per hour.
3. **Distance from the plane to the radar station**: \( d = 5 \) miles.

The plane flies horizontally over the radar station, so we have a right triangle where:
- \( x \) is the horizontal distance from the plane to the point directly above the radar station.
- \( h \) is the vertical distance (altitude of the plane).
- \( d \) is the hypotenuse, which is the direct distance from the plane to the radar station.

The relationship between these distances can be described using the Pythagorean theorem:
\[ d^2 = x^2 + h^2 \]

We need to find the rate at which the distance \( d \) is increasing when \( d = 5 \) miles.

First, let's differentiate both sides of the Pythagorean theorem equation with respect to time \( t \):
\[ \frac{d}{dt}(d^2) = \frac{d}{dt}(x^2 + h^2) \]

Since \( h \) is constant (the altitude doesn't change), its derivative with respect to time is 0:
\[ \frac{d}{dt}(d^2) = \frac{d}{dt}(x^2) + \frac{d}{dt}(h^2) \]
\[ 2d \frac{dd}{dt} = 2x \frac{dx}{dt} \]

We can simplify this to:
\[ d \frac{dd}{dt} = x \frac{dx}{dt} \]

Now, we solve for \(\frac{dd}{dt}\):
\[ \frac{dd}{dt} = \frac{x}{d} \frac{dx}{dt} \]

We know that \(\frac{dx}{dt} = 520\) mi/h and \( d = 5 \) miles. We need to find \( x \) when \( d = 5 \).

Using the Pythagorean theorem again to find \( x \):
\[ 5^2 = x^2 + 2^2 \]
\[ 25 = x^2 + 4 \]
\[ x^2 = 21 \]
\[ x = \sqrt{21} \]

Now plug \( x \), \( d \), and \(\frac{dx}{dt}\) into the equation:
\[ \frac{dd}{dt} = \frac{\sqrt{21}}{5} \cdot 520 \]

Calculate the value:
\[ \frac{dd}{dt} = \frac{\sqrt{21}}{5} \cdot 520 \approx 95.999 \]

Rounding to the nearest whole number:
\[ \frac{dd}{dt} \approx 96 \]

So, the rate at which the distance from the plane to the station is increasing when the plane is 5 miles away from the station is approximately \( 96 \) miles per hour.

Do you have any questions or need further details?

Here are 8 related questions for further practice:
1. What is the rate of change of the horizontal distance when the plane is 7 miles away from the station?
2. How would the problem change if the plane's speed increased to 600 mi/h?
3. What if the altitude of the plane was 3 miles instead of 2 miles?
4. How do you calculate the vertical speed component if the plane is ascending or descending?
5. What is the horizontal distance when the plane is 10 miles away from the radar station?
6. How do you derive the equation for the rate of change of the distance in a non-right-angled triangle scenario?
7. What is the significance of the Pythagorean theorem in related rates problems?
8. How do related rates apply to real-life navigation and aviation?

**Tip**: Always check the units of measurement and convert them if necessary to ensure consistency when solving related rates problems.

Learn how to solve a related rates problem involving a plane flying horizontally at an altitude of 2 miles and a speed of 520 mi/h passing over a radar station. The problem requires using the Pythagorean theorem and differentiation to find the rate at which the distance from the plane to the station is increasing. Suitable for advanced high school students studying calculus and related rates problems.

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