Standing 500 feet away, you watch a hot-air balloon lift off and then rise at a constant rate of 4 ft/s. You continue to watch it as it rises, slowly leaning your head back to keep your gaze fixed on it as it dwindles into a small dot against the blue sky. How fast is your angle of elevation to the balloon changing when the balloon is 50 feet above the ground? How fast is your angle of elevation to the balloon changing when the balloon is 225 feet above the ground?

To solve this problem, we will use related rates in the context of right triangle trigonometry. Let's define the variables and the relationships between them.

1. Let \( \theta \) be the angle of elevation from you to the balloon.
2. Let \( y \) be the height of the balloon above the ground.
3. Let \( x \) be the horizontal distance from you to the point directly below the balloon (which remains constant at 500 feet).

The relationship between these variables can be described using the tangent function:
\[
\tan(\theta) = \frac{y}{x}
\]
Since \( x = 500 \) feet is constant, we can rewrite this as:
\[
\tan(\theta) = \frac{y}{500}
\]

To find the rate at which \( \theta \) is changing, we will differentiate both sides of this equation with respect to time \( t \).

First, differentiate the left side using the chain rule:
\[
\sec^2(\theta) \frac{d\theta}{dt}
\]

Next, differentiate the right side:
\[
\frac{1}{500} \frac{dy}{dt}
\]

Equating both sides, we get:
\[
\sec^2(\theta) \frac{d\theta}{dt} = \frac{1}{500} \frac{dy}{dt}
\]

Since \( \sec(\theta) = \frac{1}{\cos(\theta)} \), we can express \( \sec^2(\theta) \) as:
\[
\sec^2(\theta) = 1 + \tan^2(\theta)
\]

Substitute \( \tan(\theta) = \frac{y}{500} \) into the equation:
\[
1 + \left( \frac{y}{500} \right)^2
\]

Therefore, the equation becomes:
\[
\left(1 + \left( \frac{y}{500} \right)^2\right) \frac{d\theta}{dt} = \frac{1}{500} \frac{dy}{dt}
\]

Given that \( \frac{dy}{dt} = 4 \) ft/s (the rate at which the balloon is rising), we can solve for \( \frac{d\theta}{dt} \):
\[
\frac{d\theta}{dt} = \frac{4}{500 \left(1 + \left( \frac{y}{500} \right)^2\right)}
\]

Now, we can find \( \frac{d\theta}{dt} \) for the two given heights.

### When the balloon is 50 feet above the ground:
\( y = 50 \)

\[
\frac{d\theta}{dt} = \frac{4}{500 \left(1 + \left( \frac{50}{500} \right)^2\right)} = \frac{4}{500 \left(1 + \left( 0.1 \right)^2\right)} = \frac{4}{500 \times 1.01} = \frac{4}{505} \approx 0.00792 \text{ radians per second}
\]

### When the balloon is 225 feet above the ground:
\( y = 225 \)

\[
\frac{d\theta}{dt} = \frac{4}{500 \left(1 + \left( \frac{225}{500} \right)^2\right)} = \frac{4}{500 \left(1 + \left( 0.45 \right)^2\right)} = \frac{4}{500 \times 1.2025} = \frac{4}{601.25} \approx 0.00665 \text{ radians per second}
\]

So, the rates at which your angle of elevation to the balloon is changing are approximately:
1. \( 0.00792 \) radians per second when the balloon is 50 feet above the ground.
2. \( 0.00665 \) radians per second when the balloon is 225 feet above the ground.

Would you like more details or have any questions?

Here are 5 questions you might consider asking next:
1. How do you find the rate of change of distance between two moving objects?
2. Can you explain how to use implicit differentiation in related rates problems?
3. How do you solve a related rates problem involving a ladder sliding down a wall?
4. What are the common applications of related rates in physics?
5. How do you set up a related rates problem involving volume changes?

**Tip:** When solving related rates problems, carefully identify and label all variables and constants, and determine how they are related before differentiating with respect to time.

Learn how to calculate the rate of change of angle of elevation in a related rates problem involving a hot-air balloon. Detailed step-by-step solution provided for heights of 50 feet and 225 feet above the ground.

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