The hands of a clock in some tower are approximately 4.5 m and 2.5 m in length. How fast is the distance between the tips of the hands changing at 9:00? (use law of cosines)

This problem can be solved using the Law of Cosines to relate the distance between the tips of the clock hands and the angle between the two hands as a function of time. 

### Step-by-step Solution:

1. **Label the Variables:**
   - Let \( L_1 = 4.5 \) m be the length of the hour hand.
   - Let \( L_2 = 2.5 \) m be the length of the minute hand.
   - Let \( \theta \) be the angle between the two hands of the clock.
   - Let \( d \) be the distance between the tips of the hands of the clock.
   
   According to the Law of Cosines:
   \[
   d^2 = L_1^2 + L_2^2 - 2L_1L_2 \cos(\theta)
   \]

2. **Determine the Angular Velocities:**
   - The minute hand completes one full revolution every 60 minutes, so its angular velocity is \( \omega_m = 2\pi/60 \) radians per minute, which simplifies to:
     \[
     \omega_m = \frac{\pi}{30} \text{ radians per minute.}
     \]
   - The hour hand completes one full revolution every 12 hours, so its angular velocity is \( \omega_h = 2\pi/(12 \times 60) \) radians per minute, which simplifies to:
     \[
     \omega_h = \frac{\pi}{360} \text{ radians per minute.}
     \]

3. **At 9:00, find the angle \(\theta\):**
   At 9:00, the minute hand points at 12 (i.e., 0 degrees or 0 radians), and the hour hand points at 9 (i.e., 270 degrees or \( \frac{3\pi}{2} \) radians). Therefore, the angle between the two hands is:
   \[
   \theta = \frac{\pi}{2} \text{ radians}.
   \]

4. **Differentiate the Law of Cosines:**
   Now we differentiate both sides of the equation \( d^2 = L_1^2 + L_2^2 - 2L_1L_2 \cos(\theta) \) with respect to time \( t \):
   \[
   2d \frac{dd}{dt} = 2L_1L_2 \sin(\theta) \frac{d\theta}{dt}
   \]
   Solving for \( \frac{dd}{dt} \) (the rate of change of the distance between the tips of the clock hands):
   \[
   \frac{dd}{dt} = \frac{L_1 L_2 \sin(\theta) \frac{d\theta}{dt}}{d}
   \]

5. **Rate of change of the angle \( \theta \):**
   The rate of change of the angle between the hands of the clock is the difference between the angular velocities of the minute and hour hands:
   \[
   \frac{d\theta}{dt} = \omega_m - \omega_h = \frac{\pi}{30} - \frac{\pi}{360} = \frac{11\pi}{360} \text{ radians per minute}.
   \]

6. **At \( \theta = \frac{\pi}{2} \), evaluate the expression:**
   Plug in the values:
   \[
   d^2 = L_1^2 + L_2^2 - 2L_1L_2 \cos\left(\frac{\pi}{2}\right)
   \]
   Since \( \cos\left(\frac{\pi}{2}\right) = 0 \), we have:
   \[
   d^2 = L_1^2 + L_2^2 = 4.5^2 + 2.5^2 = 20.25 + 6.25 = 26.5
   \]
   Therefore, \( d = \sqrt{26.5} \approx 5.15 \) m.

7. **Finally, calculate \( \frac{dd}{dt} \):**
   Using the expression for \( \frac{dd}{dt} \):
   \[
   \frac{dd}{dt} = \frac{L_1 L_2 \sin\left(\frac{\pi}{2}\right) \cdot \frac{11\pi}{360}}{d}
   = \frac{4.5 \cdot 2.5 \cdot \frac{11\pi}{360}}{5.15}
   \]
   Simplifying:
   \[
   \frac{dd}{dt} \approx \frac{11.25 \cdot \frac{11\pi}{360}}{5.15} \approx 0.268 \text{ m/min}.
   \]

### Final Answer:
The rate at which the distance between the tips of the hands is changing at 9:00 is approximately \( 0.268 \) meters per minute.

Would you like more details or further clarification?

Here are 5 related questions to deepen your understanding:
1. How would the answer change if the clock time were 3:00 instead of 9:00?
2. What happens to the rate of change of distance at 6:00?
3. How does the length of the clock hands affect the rate of change?
4. Can you apply a similar approach to find how fast the hands are moving relative to one another?
5. How would you compute the maximum rate of change of the distance?

**Tip:** In problems involving changing distances and angles, the Law of Cosines is a powerful tool, especially when you know the relationship between sides and angles!

Explore how to calculate the rate of change of distance between clock hands at 9:00 using the Law of Cosines. This problem involves trigonometry, angular velocities, and related rates, offering a detailed step-by-step solution. Suitable for advanced high school or introductory college-level students.

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