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This problem involves related rates, specifically calculating the rate at which the angle of elevation is changing as the balloon rises.

Let’s define the variables:

- \( x \) = the horizontal distance from Madeleine to the balloon (this is constant at 400 meters).
- \( y \) = the vertical distance from the ground to the balloon (which is changing over time). 
- \( \theta \) = the angle of elevation from Madeleine's perspective (the angle we are asked about).

We are given the following:

- \( \frac{dy}{dt} = 8 \) meters/second (the balloon's vertical velocity).
- When \( y = 700 \) meters, we need to find \( \frac{d\theta}{dt} \) (the rate of change of the angle of elevation).

### Step 1: Set up the equation
Using trigonometry, we have:
\[
\tan(\theta) = \frac{y}{x}
\]
Thus:
\[
\theta = \tan^{-1}\left(\frac{y}{x}\right)
\]

### Step 2: Differentiate with respect to time
Now, differentiate both sides with respect to time \( t \):
\[
\frac{d\theta}{dt} = \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \frac{1}{x} \cdot \frac{dy}{dt}
\]
Substitute \( x = 400 \) meters, \( y = 700 \) meters, and \( \frac{dy}{dt} = 8 \) meters/second into the equation:
\[
\frac{d\theta}{dt} = \frac{1}{1 + \left(\frac{700}{400}\right)^2} \cdot \frac{1}{400} \cdot 8
\]

### Step 3: Simplify
First, calculate the term inside the parentheses:
\[
\left(\frac{700}{400}\right)^2 = \left(\frac{7}{4}\right)^2 = \frac{49}{16}
\]
Thus, the equation becomes:
\[
\frac{d\theta}{dt} = \frac{1}{1 + \frac{49}{16}} \cdot \frac{1}{400} \cdot 8
\]
\[
\frac{d\theta}{dt} = \frac{1}{\frac{65}{16}} \cdot \frac{1}{400} \cdot 8
\]
\[
\frac{d\theta}{dt} = \frac{16}{65} \cdot \frac{1}{400} \cdot 8
\]
Now simplify:
\[
\frac{d\theta}{dt} = \frac{128}{65} \cdot \frac{1}{400} = \frac{128}{26000}
\]
\[
\frac{d\theta}{dt} = \frac{32}{6500} = 0.004923 \text{ radians per second}
\]

So, the rate at which the angle of elevation is changing is approximately \( 0.00492 \) radians per second.

Would you like a more detailed breakdown of any of the steps, or do you have any further questions?

Here are 5 related questions to expand your understanding:
1. How does the rate of elevation change if the balloon's speed increases?
2. What happens to \( \frac{d\theta}{dt} \) as \( y \to \infty \)?
3. How would the equation change if Madeleine were moving horizontally toward or away from the balloon?
4. What would the rate of change be if Madeleine were positioned at a different horizontal distance?
5. How does the angle of elevation change over time if the balloon's ascent slows down?

**Tip**: When solving related rates problems, make sure to differentiate implicitly, keeping track of how each variable changes with respect to time!

While visiting Chateau d'Oex in the Swiss Alps, Madeleine watches a hot air balloon launch from 400 meters away. If the balloon is rising vertically at a rate of 8 meters/second, at what rate is the angle of elevation changing from Madeleine's perspective when the balloon is 700 meters in the air?

Learn how to solve a related rates problem involving a hot air balloon rising at 8 meters/second. This calculus-based problem shows how to find the rate at which the angle of elevation changes from a distance of 400 meters as the balloon ascends to 700 meters.

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