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To find \(\theta'(t)\) (the rate of change of the angle between the clock's minute and hour hands) at 2 o'clock in radians per minute, let's start by calculating the angle between the hands.

### Step 1: Calculating the angle \(\theta(t)\)

1. **Hour hand rotation:**
   - The hour hand completes 1 full rotation (or \(2\pi\) radians) in 12 hours.
   - Therefore, its angular speed is:  
     \[
     \omega_{\text{hour}} = \frac{2\pi}{12 \times 60} = \frac{\pi}{360} \text{ radians per minute.}
     \]
   - At 2 o'clock, the hour hand starts at:
     \[
     \theta_{\text{hour}} = \frac{2 \times 2\pi}{12} = \frac{\pi}{3} \text{ radians.}
     \]

2. **Minute hand rotation:**
   - The minute hand completes 1 full rotation (or \(2\pi\) radians) in 60 minutes.
   - Therefore, its angular speed is:  
     \[
     \omega_{\text{minute}} = \frac{2\pi}{60} = \frac{\pi}{30} \text{ radians per minute.}
     \]
   - At 2 o'clock (when \( t = 0 \) minutes), the minute hand starts at:
     \[
     \theta_{\text{minute}} = 0 \text{ radians.}
     \]

### Step 2: Finding the rate of change \(\theta'(t)\)

The angle between the minute and hour hands at any time \( t \) (in minutes) is given by the formula:
\[
\theta(t) = \left| \theta_{\text{minute}}(t) - \theta_{\text{hour}}(t) \right|,
\]
where:
- \(\theta_{\text{minute}}(t) = \omega_{\text{minute}} \times t = \frac{\pi}{30} t\).
- \(\theta_{\text{hour}}(t) = \frac{\pi}{3} + \omega_{\text{hour}} \times t = \frac{\pi}{3} + \frac{\pi}{360} t\).

Therefore:
\[
\theta(t) = \left| \frac{\pi}{30} t - \left( \frac{\pi}{3} + \frac{\pi}{360} t \right) \right|
\]
\[
\theta(t) = \left| \frac{\pi}{30} t - \frac{\pi}{3} - \frac{\pi}{360} t \right|
\]
\[
\theta(t) = \left| \frac{\pi}{30} t - \frac{\pi}{360} t - \frac{\pi}{3} \right|
\]
\[
\theta(t) = \left| \frac{12\pi}{360} t - \frac{\pi}{360} t - \frac{\pi}{3} \right|
\]
\[
\theta(t) = \left| \frac{11\pi}{360} t - \frac{\pi}{3} \right|
\]

### Step 3: Finding \(\theta'(t)\)

Now, take the derivative of \(\theta(t)\) with respect to \(t\):
\[
\theta'(t) = \frac{d}{dt} \left( \left| \frac{11\pi}{360} t - \frac{\pi}{3} \right| \right).
\]

Since we want to find the derivative at 2 o'clock (which corresponds to \( t = 0 \)), the absolute value function simplifies as:
\[
\theta'(0) = \frac{11\pi}{360} \text{ radians per minute.}
\]

### Final Answer
The rate of change of the angle between the clock's hands at 2 o'clock is:
\[
\theta'(0) = \frac{11\pi}{360} \text{ radians per minute.}
\]

Would you like more details or have any further questions?

Here are 5 related questions for practice:
1. What is the angle between the clock's hands at 3:00?
2. How does the rate of change of the angle vary over different hours?
3. What is the rate of change of the angle at 6:00?
4. Calculate the angle between the clock's hands at 2:30.
5. How many radians does the hour hand move in 5 minutes?

**Tip:** Remember that the minute hand moves 12 times faster than the hour hand!

Suppose θ(t) measures the minimum angle between a clock’s minute and hour hands in radians. What is θ'(t) at 2 o'clock? Give your answer in radians per minute.

Find the rate of change of the angle between the minute and hour hands of a clock at 2 o'clock in radians per minute. This problem involves trigonometry, angular velocity, and differentiation, suitable for high school students in Grades 11-12.

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