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We are tasked with finding the rate of change of the angle of elevation (denoted as \( \theta \)) when the camera and rocket are 5516 feet apart.

### Step 1: Given Equation
The angle of elevation is given by:
\[
\theta = \tan^{-1}\left(\frac{x}{2000}\right)
\]
where \( x \) is the height of the rocket, and 2000 feet is the horizontal distance between the camera and the rocket.

### Step 2: Differentiate with Respect to Time (Implicit Differentiation)
We need to find \( \frac{d\theta}{dt} \), which is the rate of change of the angle with respect to time.

Using the chain rule to differentiate both sides of the equation with respect to \( t \):
\[
\frac{d\theta}{dt} = \frac{1}{1 + \left(\frac{x}{2000}\right)^2} \cdot \frac{1}{2000} \cdot \frac{dx}{dt}
\]
This formula can now be simplified to:
\[
\frac{d\theta}{dt} = \frac{2000}{2000^2 + x^2} \cdot \frac{dx}{dt}
\]

### Step 3: Use Pythagoras' Theorem to Find \( x \)
We know that the camera and the rocket are 5516 feet apart. The horizontal distance is fixed at 2000 feet. Using the Pythagorean theorem to solve for the height \( x \):
\[
5516^2 = 2000^2 + x^2
\]
\[
x^2 = 5516^2 - 2000^2
\]
\[
x^2 = 30426016 - 4000000
\]
\[
x^2 = 26426016
\]
\[
x = \sqrt{26426016} \approx 5140.8 \text{ feet}
\]

### Step 4: Substitute Values
Now, substitute the value of \( x \) and the horizontal distance into the expression for \( \frac{d\theta}{dt} \). We also need to know the rate of change of the height \( \frac{dx}{dt} \), but since it's not provided, we'll assume it's given or needs to be found separately. If we had \( \frac{dx}{dt} \), we would proceed by calculating:

\[
\frac{d\theta}{dt} = \frac{2000}{2000^2 + (5140.8)^2} \cdot \frac{dx}{dt}
\]

Now, do you have a value for \( \frac{dx}{dt} \) (the rate at which the height of the rocket is increasing)? If so, I can finalize the answer!

Let me know if you'd like more details or if you have any questions.

### Related Questions:
1. What is the formula for differentiating \( \tan^{-1} \)?
2. How does the chain rule apply to implicit differentiation in this case?
3. Can you calculate the rate of change if the horizontal distance is not fixed?
4. How can you find the height using other trigonometric functions in a related scenario?
5. How does the rate of change of the angle depend on the rocket's velocity?

**Tip:** In problems like these, keeping track of units (e.g., feet, seconds, radians) ensures accuracy when applying rates of change.

Find the rate of change of the angle of elevation after launch when the camera and the rocket are 5516 feet apart.

Learn how to calculate the rate of change of the angle of elevation of a camera during a rocket launch. This problem covers key trigonometric concepts like implicit differentiation and the Pythagorean theorem. Ideal for advanced high school or college students.

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