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Let's solve this problem step-by-step.

### Step 1: Identify Given Values and Information
- The rocket travels vertically at a speed of \( 1400 \, \text{km/hr} \).
- The observer is located \( 18 \, \text{km} \) horizontally from the launch point.
- We are asked to find the rate at which the angle \( \theta \) (between the telescope's line of sight to the rocket and the ground) is changing, specifically \( \frac{d\theta}{dt} \), three minutes after liftoff.

### Step 2: Set Up the Related Rates Equation
Let:
- \( h \) represent the height of the rocket above the ground at time \( t \).
- \( \theta \) represent the angle between the telescope and the ground.

From trigonometry, we can write:
\[
\tan(\theta) = \frac{h}{18}
\]

### Step 3: Differentiate with Respect to Time
To find \( \frac{d\theta}{dt} \), we differentiate both sides of \( \tan(\theta) = \frac{h}{18} \) with respect to \( t \):
\[
\sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{18} \cdot \frac{dh}{dt}
\]

Solving for \( \frac{d\theta}{dt} \):
\[
\frac{d\theta}{dt} = \frac{\frac{1}{18} \cdot \frac{dh}{dt}}{\sec^2(\theta)}
\]

Since \( \sec^2(\theta) = 1 + \tan^2(\theta) \), we can rewrite this as:
\[
\frac{d\theta}{dt} = \frac{\frac{1}{18} \cdot \frac{dh}{dt}}{1 + \left(\frac{h}{18}\right)^2}
\]

### Step 4: Substitute Known Values
1. **Rocket's speed**: \( \frac{dh}{dt} = 1400 \, \text{km/hr} \).
2. **Height after 3 minutes**: Since \( 3 \) minutes is \( \frac{3}{60} = 0.05 \) hours, the height \( h \) at \( t = 3 \) minutes is:
   \[
   h = 1400 \times 0.05 = 70 \, \text{km}
   \]

3. **Substitute \( h = 70 \) km, \( \frac{dh}{dt} = 1400 \, \text{km/hr} \), and \( 18 \) km into the equation**:
   \[
   \frac{d\theta}{dt} = \frac{\frac{1}{18} \cdot 1400}{1 + \left(\frac{70}{18}\right)^2}
   \]

4. **Simplify the expression**:
   - Calculate \( \frac{70}{18} \approx 3.89 \), and then \( \left(3.89\right)^2 \approx 15.15 \).
   - So the denominator is \( 1 + 15.15 = 16.15 \).

   Now,
   \[
   \frac{d\theta}{dt} \approx \frac{\frac{1400}{18}}{16.15}
   \]

5. **Final Calculation**:
   - \( \frac{1400}{18} \approx 77.78 \).
   - So,
     \[
     \frac{d\theta}{dt} \approx \frac{77.78}{16.15} \approx 4.82 \, \text{rad/hr}
     \]

### Answer
\[
\frac{d\theta}{dt} \approx 4.82 \, \text{rad/hr}
\]

Would you like further details on any part of the solution?

### Related Questions:
1. How does the horizontal distance between the observer and the rocket affect \( \frac{d\theta}{dt} \)?
2. What would happen to \( \frac{d\theta}{dt} \) if the rocket's speed increased?
3. How does the time after launch affect the angle change rate \( \frac{d\theta}{dt} \)?
4. How can you derive the height \( h \) if only the angle \( \theta \) and horizontal distance are known?
5. What are some other applications of related rates in physics and engineering?

### Tip:
When dealing with related rates, always make sure to define each variable and check units carefully before proceeding with differentiation.

A rocket travels vertically at a speed of 1400 km/hr. The rocket is tracked through a telescope by an observer located 18 km from the launching pad. Find the rate at which the angle between the telescope and the ground is increasing 3 min after the lift-off.

Learn how to find the rate of angle change for a rocket traveling vertically at 1400 km/hr, observed from 18 km away. This related rates problem involves trigonometric differentiation and is suitable for calculus students.

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