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To solve this problem, we need to find the rate of change of the angle of elevation \( \theta \) with respect to the height \( x \), when the distance between the camera and the rocket is 5516 feet.

### Step-by-step solution:

1. **Given relation**: 
   \[
   \theta = \tan^{-1} \left( \frac{x}{2000} \right)
   \]
   where \( \theta \) is the angle of elevation, and \( x \) is the height of the rocket.

2. **Differentiate with respect to \( x \)**:
   Using the chain rule:
   \[
   \frac{d\theta}{dx} = \frac{d}{dx} \left( \tan^{-1} \left( \frac{x}{2000} \right) \right)
   \]

   The derivative of \( \tan^{-1}(u) \) is \( \frac{1}{1 + u^2} \), so applying this:
   \[
   \frac{d\theta}{dx} = \frac{1}{1 + \left( \frac{x}{2000} \right)^2} \cdot \frac{1}{2000}
   \]

   Simplifying:
   \[
   \frac{d\theta}{dx} = \frac{2000}{2000^2 + x^2}
   \]

3. **Using the distance condition**:
   We know that the distance between the camera and the rocket is 5516 feet. This forms a right triangle, where the horizontal distance from the camera to the rocket is 2000 feet, and the height of the rocket \( x \) can be calculated using the Pythagorean theorem:
   \[
   2000^2 + x^2 = 5516^2
   \]

   Solving for \( x \):
   \[
   x^2 = 5516^2 - 2000^2
   \]
   \[
   x^2 = 30426176 - 4000000
   \]
   \[
   x^2 = 26426176
   \]
   \[
   x = \sqrt{26426176} \approx 5140.8 \text{ feet}
   \]

4. **Substitute \( x \) into the derivative**:
   Now, substitute \( x = 5140.8 \) into the expression for \( \frac{d\theta}{dx} \):
   \[
   \frac{d\theta}{dx} = \frac{2000}{2000^2 + 5140.8^2}
   \]
   \[
   \frac{d\theta}{dx} = \frac{2000}{4000000 + 26426176}
   \]
   \[
   \frac{d\theta}{dx} = \frac{2000}{30426176}
   \]
   \[
   \frac{d\theta}{dx} \approx 0.0000657 \text{ radians per foot}
   \]

### Final Answer:
\[
\frac{d\theta}{dx} \approx 0.00007 \text{ radians per foot (rounded to five decimal places)}
\]

Would you like more details or have any questions about this solution? Here are some related questions:

1. How is the derivative of \( \tan^{-1}(x) \) derived?
2. Can you explain the chain rule in this context in more detail?
3. How would the rate of change change if the camera were placed further away?
4. What other applications use the concept of angle of elevation changes?
5. How would you compute this if the rocket's motion wasn't vertical?

**Tip**: Always double-check the units in applied math problems to ensure the result makes sense. Here, radians per foot is appropriate for the rate of angle change per unit height.

A television camera at ground level is 2000 feet away from the launching pad of a space rocket that is set to take off vertically. The angle of elevation of the camera can be found by θ = tan^(-1)(x / 2000) where x is the height of the rocket. Find the rate of change of the angle of elevation after launch when the camera and the rocket are 5516 feet apart. Round to five decimal places.

This problem involves finding the rate of change of the angle of elevation for a rocket launch scenario using trigonometry and related rates. Given a camera at 2000 feet from the launch pad, learn how to calculate the rate when the distance between the camera and rocket is 5516 feet. The solution involves differentiation and the Pythagorean Theorem.

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