A plane flying with a constant speed of 19 km/min passes over a ground radar station at an altitude of 7 km and climbs at an angle of 25 degrees. At what rate is the distance from the plane to the radar station increasing 5 minutes later?
The distance is increasing at ___ km/min.

To find the rate at which the distance from the plane to the radar station is increasing, we need to use related rates in calculus. Here’s the detailed solution:

Given:
- Speed of the plane: \( 19 \, \text{km/min} \)
- Altitude of the plane when it passes over the radar station: \( 7 \, \text{km} \)
- Angle of climb: \( 25^\circ \)

First, define the variables:
- Let \( t \) be the time in minutes.
- Let \( x(t) \) be the horizontal distance traveled by the plane from the point directly above the radar station.
- Let \( y(t) \) be the altitude of the plane at time \( t \).
- Let \( s(t) \) be the distance from the plane to the radar station at time \( t \).

Given:
- The horizontal component of the velocity is \( 19 \cos(25^\circ) \, \text{km/min} \).
- The vertical component of the velocity is \( 19 \sin(25^\circ) \, \text{km/min} \).

The position of the plane after \( t \) minutes:
- \( x(t) = (19 \cos(25^\circ)) \cdot t \)
- \( y(t) = 7 + (19 \sin(25^\circ)) \cdot t \)

The distance from the plane to the radar station at any time \( t \) is given by:
\[ s(t) = \sqrt{x(t)^2 + y(t)^2} \]

To find the rate at which the distance \( s \) is increasing, we need to find \( \frac{ds}{dt} \) when \( t = 5 \).

First, compute \( x(5) \) and \( y(5) \):
\[ x(5) = 19 \cos(25^\circ) \cdot 5 \]
\[ y(5) = 7 + 19 \sin(25^\circ) \cdot 5 \]

Now compute:
\[ x(5) = 19 \cdot \cos(25^\circ) \cdot 5 \approx 19 \cdot 0.9063 \cdot 5 = 86.097 \, \text{km} \]
\[ y(5) = 7 + 19 \cdot \sin(25^\circ) \cdot 5 \approx 7 + 19 \cdot 0.4226 \cdot 5 = 46.175 \, \text{km} \]

The distance \( s \) at \( t = 5 \) is:
\[ s(5) = \sqrt{x(5)^2 + y(5)^2} = \sqrt{(86.097)^2 + (46.175)^2} \approx \sqrt{7412.698 + 2132.125} = \sqrt{9544.823} \approx 97.706 \, \text{km} \]

Now, we need to find \( \frac{ds}{dt} \) using the chain rule:
\[ s^2 = x^2 + y^2 \]
Differentiate both sides with respect to \( t \):
\[ 2s \frac{ds}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \]
\[ s \frac{ds}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt} \]

Substitute the known values:
- \( x = 86.097 \)
- \( y = 46.175 \)
- \( \frac{dx}{dt} = 19 \cos(25^\circ) \approx 17.219 \)
- \( \frac{dy}{dt} = 19 \sin(25^\circ) \approx 8.029 \)
- \( s = 97.706 \)

\[ 97.706 \frac{ds}{dt} = 86.097 \cdot 17.219 + 46.175 \cdot 8.029 \]
\[ 97.706 \frac{ds}{dt} = 1481.408 + 370.634 \]
\[ 97.706 \frac{ds}{dt} = 1852.042 \]
\[ \frac{ds}{dt} \approx \frac{1852.042}{97.706} \approx 18.956 \, \text{km/min} \]

Therefore, the rate at which the distance from the plane to the radar station is increasing 5 minutes later is approximately \( 18.956 \, \text{km/min} \).

Would you like more details on any step or have any other questions? Here are some additional questions you might find interesting:

1. How does the horizontal distance change over time for a different climbing angle?
2. What would be the altitude of the plane after 10 minutes at the same climbing angle?
3. How would the problem change if the plane were descending instead?
4. What is the horizontal distance from the radar station after 3 minutes?
5. How do you calculate the speed of the plane relative to the ground?

**Tip:** Always break down complex related rates problems into smaller steps, defining variables and their relationships clearly before differentiating.

Learn how to solve a related rates problem involving a plane flying at a constant speed passing over a radar station. Understand the application of trigonometry and calculus to find how fast the distance between the plane and the radar station is increasing. Suitable for advanced high school students.

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