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This problem involves finding the rate of change of the angle of elevation using related rates. Let's break it down step-by-step.

### Given Information:
- The balloon rises at a constant rate of \( \frac{dy}{dt} = 4 \) meters per second.
- The horizontal distance between the observer and the point on the ground directly beneath the balloon is \( x = 50 \) meters.
- We are asked to find the rate of change of the angle of elevation \( \theta \) when the balloon is \( y = 50 \) meters above the ground.

### Step 1: Set up a relationship between \( y \) and \( \theta \)
We can use trigonometry to relate the height of the balloon \( y \) and the angle of elevation \( \theta \). Specifically, we use the tangent function because the tangent of the angle of elevation is the ratio of the opposite side (the balloon’s height) to the adjacent side (the distance between the observer and the point directly under the balloon):

\[
\tan(\theta) = \frac{y}{x}
\]

### Step 2: Differentiate with respect to time \( t \)
We differentiate both sides of the equation with respect to \( t \), using implicit differentiation. Remember that \( x \) is constant, so its derivative is zero.

\[
\frac{d}{dt} \left( \tan(\theta) \right) = \frac{d}{dt} \left( \frac{y}{x} \right)
\]

Using the chain rule on the left-hand side:

\[
\sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{x} \cdot \frac{dy}{dt}
\]

### Step 3: Solve for \( \frac{d\theta}{dt} \)
Now, solve for \( \frac{d\theta}{dt} \):

\[
\frac{d\theta}{dt} = \frac{1}{\sec^2(\theta)} \cdot \frac{1}{x} \cdot \frac{dy}{dt}
\]

Since \( \sec^2(\theta) = 1 + \tan^2(\theta) \), we can rewrite this as:

\[
\frac{d\theta}{dt} = \frac{1}{1 + \tan^2(\theta)} \cdot \frac{1}{x} \cdot \frac{dy}{dt}
\]

### Step 4: Evaluate at \( y = 50 \)
When the balloon is 50 meters above the ground, \( y = 50 \) meters and \( x = 50 \) meters. Therefore:

\[
\tan(\theta) = \frac{y}{x} = \frac{50}{50} = 1
\]

Thus,

\[
\sec^2(\theta) = 1 + \tan^2(\theta) = 1 + 1^2 = 2
\]

Now, substitute the known values into the equation:

\[
\frac{d\theta}{dt} = \frac{1}{2} \cdot \frac{1}{50} \cdot 4 = \frac{4}{2 \cdot 50} = \frac{4}{100} = 0.04 \text{ radians per second}
\]

### Final Answer:
The rate of change of the angle of elevation when the balloon is 50 meters above the ground is \( 0.04 \) radians per second.

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Would you like further explanation on any part of this process?

Here are 5 related questions to deepen your understanding:
1. How would the rate of change of the angle of elevation change if the balloon rose at a different speed?
2. What happens to the rate of change of the angle of elevation as the balloon rises higher and higher?
3. Can you derive a similar equation for a different trigonometric function, such as sine or cosine?
4. How does the distance between the observer and the balloon’s starting point affect the rate of change of the angle of elevation?
5. What would the rate of change of the angle of elevation be if the observer were standing directly under the balloon?

**Tip:** Whenever you use trigonometric functions in related rates problems, make sure you’re familiar with their derivatives, especially the chain rule!

A balloon rises at a rate of 4 meters per second from a point on the ground 50 meters from an observer. Find the rate of change of the angle of elevation of the balloon from the observer when the balloon is 50 meters above the ground.

Learn how to calculate the rate of change of the angle of elevation as a balloon rises at a rate of 4 meters per second, given a 50-meter horizontal distance from the observer. Detailed explanation of related rates and trigonometric concepts.

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