A hot air balloon rising straight up from a level field is tracked by a range finder 500 ft from the lift-off point. At the moment the range finder's elevation angle is `pi/4`, the angle is increasing at the rate of 0.14 rad/min. How fast is the balloon rising at that moment.

Answer

To solve the problem, we will use the relationship between the height of the balloon, the distance from the range finder to the lift-off point, and the angle of elevation. Here's a step-by-step solution:

### Step 1: Understand the Geometry
We have a right triangle formed by:
- The height of the balloon (let's denote it as \( y \)).
- The horizontal distance from the range finder to the lift-off point, which is constant at \( 500 \) ft.
- The angle of elevation \( \theta \) from the range finder to the balloon.

### Step 2: Set Up the Relationship
From the right triangle, we can express the height \( y \) in terms of the angle \( \theta \):
\[ \tan(\theta) = \frac{y}{500} \]
This implies:
\[ y = 500 \tan(\theta) \]

### Step 3: Differentiate with Respect to Time
To find how fast the balloon is rising, we need to differentiate \( y \) with respect to time \( t \):
\[ \frac{dy}{dt} = 500 \frac{d}{dt}(\tan(\theta)) \]
Using the chain rule, we have:
\[ \frac{dy}{dt} = 500 \sec^2(\theta) \frac{d\theta}{dt} \]

### Step 4: Substitute Known Values
We know:
- At the moment of interest, \( \theta = \frac{\pi}{4} \).
- The rate of change of the angle \( \frac{d\theta}{dt} = 0.14 \) rad/min.

First, we calculate \( \sec^2(\theta) \) at \( \theta = \frac{\pi}{4} \):
\[ \sec^2\left(\frac{\pi}{4}\right) = \left(\frac{1}{\cos\left(\frac{\pi}{4}\right)}\right)^2 = \left(\frac{1}{\frac{\sqrt{2}}{2}}\right)^2 = 2 \]

Now substituting the values into the differentiated equation:
\[ \frac{dy}{dt} = 500 \cdot 2 \cdot 0.14 \]

### Step 5: Calculate the Result
Calculating the above expression:
\[ \frac{dy}{dt} = 1000 \cdot 0.14 = 140 \text{ ft/min} \]

### Final Answer
The balloon is rising at a rate of \( 140 \) feet per minute at that moment.

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