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We are dealing with a related rates problem. The balloon is rising vertically while a person on the ground is measuring the angle of inclination, \(\theta\), from a distance of 500 feet away from the point directly below the balloon.

### Given:
- Distance from the person to the point directly below the balloon = 500 ft.
- Angle of inclination \(\theta = \frac{\pi}{4}\) radians.
- Rate of change of the angle \( \frac{d\theta}{dt} = \frac{1}{10} \) radians per minute.

We need to find the rate at which the balloon is rising, i.e., the rate of change of the height \(h\) of the balloon with respect to time, \(\frac{dh}{dt}\), when \(\theta = \frac{\pi}{4}\).

### Step 1: Set up a relationship between height and angle
Using trigonometry, we can relate the height \(h\) of the balloon and the distance 500 ft (which remains constant) via the tangent function:
\[
\tan(\theta) = \frac{h}{500}
\]
Thus,
\[
h = 500 \cdot \tan(\theta)
\]

### Step 2: Differentiate with respect to time
Differentiate both sides of the equation \(h = 500 \cdot \tan(\theta)\) with respect to time \(t\):
\[
\frac{dh}{dt} = 500 \cdot \sec^2(\theta) \cdot \frac{d\theta}{dt}
\]

### Step 3: Plug in known values
At the moment when \(\theta = \frac{\pi}{4}\):
- \(\sec^2\left(\frac{\pi}{4}\right) = 2\) (since \(\sec\left(\frac{\pi}{4}\right) = \sqrt{2}\)).
- \(\frac{d\theta}{dt} = \frac{1}{10} \text{ radians per minute}\).

Now substitute these values into the equation:
\[
\frac{dh}{dt} = 500 \cdot 2 \cdot \frac{1}{10} = 100 \text{ feet per minute}.
\]

### Final Answer:
The balloon is rising at a rate of **100 feet per minute** when the angle is \(\frac{\pi}{4}\).

Would you like more details, or do you have any further questions? Here are some related questions you can explore:

1. How does the rate of change of the balloon's height depend on the angle \(\theta\)?
2. What happens to the rate of change of height as \(\theta\) approaches \(\pi/2\)?
3. How fast would the balloon rise if the angle \(\theta\) were \(\pi/6\) radians instead of \(\pi/4\)?
4. What would change if the person were standing 600 feet away instead of 500 feet?
5. Could you use a different trigonometric function to relate height and distance for this problem?

**Tip:** When solving related rates problems, always differentiate implicitly and remember to apply the chain rule to account for time-dependent variables.

A balloon is rising above a field, and a person 500 ft away from the spot on the ground underneath the balloon is watching it, measuring the angle of inclination, θ. When the angle is π/4 radians, the angle is increasing at 1/10 radians per minute. At that moment, how fast is the balloon rising?

Solve a related rates problem where a balloon rises above a field. Learn how to use trigonometric relationships and differentiation to find the rate at which the balloon rises. This problem involves finding the rate of change of the balloon's height when the angle of inclination is π/4 radians and increasing at 1/10 radians per minute.

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