A rocket launching straight up from a level field is tracked by a camera 200 feet from the launching point. When the camera's angle of elevation is  the angle is increasing at the rate of 0.18 radians per minute. At what rate is the rocket rising at that moment?

To solve this problem, we can apply related rates involving trigonometric functions. Let's break it down.

### Problem Setup:

- Let \( \theta(t) \) be the angle of elevation of the camera at time \( t \).
- Let \( y(t) \) be the height of the rocket at time \( t \).
- The camera is positioned 200 feet from the launching point, which is a constant distance.

We are given:
- \( \frac{d\theta}{dt} = 0.18 \) radians per minute.
- The distance from the camera to the launching point is 200 feet.
  
We need to find the rate at which the rocket is rising, \( \frac{dy}{dt} \), when the angle of elevation \( \theta \) is given.

### Relationship Between Variables:

Using trigonometry, specifically the tangent function, we have:
\[
\tan(\theta(t)) = \frac{y(t)}{200}
\]

### Differentiating with Respect to Time:

Differentiate both sides of the equation with respect to time \( t \):
\[
\sec^2(\theta(t)) \cdot \frac{d\theta}{dt} = \frac{1}{200} \cdot \frac{dy}{dt}
\]

### Solving for \( \frac{dy}{dt} \):

Rearrange the equation to solve for \( \frac{dy}{dt} \):
\[
\frac{dy}{dt} = 200 \cdot \sec^2(\theta(t)) \cdot \frac{d\theta}{dt}
\]

### Substituting Given Values:

We are given \( \frac{d\theta}{dt} = 0.18 \) radians per minute. To find \( \frac{dy}{dt} \), we also need to know \( \sec^2(\theta) \). The secant function is the reciprocal of the cosine:
\[
\sec(\theta) = \frac{1}{\cos(\theta)}
\]

Let's assume \( \theta(t) \) is some known angle at the time of calculation, or if no specific angle is given, we proceed with the general expression.

\[
\frac{dy}{dt} = 200 \cdot \sec^2(\theta) \cdot 0.18
\]

For a specific angle \( \theta \), calculate \( \sec(\theta) \) and substitute it into the equation.

### Example Calculation (Assuming \( \theta = 45^\circ \) or \( \theta = \frac{\pi}{4} \) radians):

If \( \theta = 45^\circ \):
\[
\sec(45^\circ) = \frac{1}{\cos(45^\circ)} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}
\]
\[
\sec^2(45^\circ) = 2
\]
\[
\frac{dy}{dt} = 200 \cdot 2 \cdot 0.18 = 400 \cdot 0.18 = 72 \text{ feet per minute}
\]

### Conclusion:

The rate at which the rocket is rising at the moment when the camera's angle of elevation is \( 45^\circ \) (or \( \frac{\pi}{4} \) radians) is 72 feet per minute. If a different angle \( \theta \) is given, you would use that angle to find the corresponding rate.

Would you like more details or have any questions?

### Related Questions:
1. What would be the rate of rise if the angle was \( 60^\circ \)?
2. How does the distance of the camera from the launching point affect the rate of rise?
3. What happens to the rate of rise as the angle of elevation approaches \( 90^\circ \)?
4. How would the rate of rise change if the rate of angle increase \( \frac{d\theta}{dt} \) was higher?
5. Can you derive the time it takes for the rocket to reach a certain height using the related rates?

**Tip:** In related rates problems, the trigonometric relationships often simplify the problem, but be careful with the units and angles you work with, especially when dealing with radians and degrees.

Explore how to solve a related rates problem involving a rocket launching vertically with a camera tracking its angle of elevation. Learn the application of trigonometric functions and related rates to find the rate at which the rocket is rising. Suitable for high school students studying calculus and related rates problems.

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