The hands of a clock in some tower are approximately

55

m and

1.51.5

m in length. How fast is the distance between the tips of the hands changing at​ 9:00? (Hint: Use the Law of​ Cosines.)

To solve this problem, we'll use the Law of Cosines to determine how fast the distance between the tips of the clock hands is changing at 9:00. The lengths of the clock hands are \( R_1 = 5.5 \) m (minute hand) and \( R_2 = 1.5 \) m (hour hand). Let's find the rate at which the distance between the tips of the hands changes.

### Step-by-step Solution:

1. **Setting Up the Problem:**

   Let \( \theta \) be the angle between the two hands of the clock, and \( D \) be the distance between the tips of the hands. At 9:00, the angle \( \theta \) between the minute and hour hands is \( 90^\circ \) (or \(\frac{\pi}{2}\) radians).

2. **Law of Cosines:**

   Using the Law of Cosines, we can express \( D \) as:
   \[
   D^2 = R_1^2 + R_2^2 - 2 R_1 R_2 \cos(\theta)
   \]

   Differentiating both sides with respect to time \( t \), we get:
   \[
   2D \frac{dD}{dt} = 2R_1 R_2 \sin(\theta) \frac{d\theta}{dt}
   \]

   Simplifying, we find:
   \[
   \frac{dD}{dt} = \frac{R_1 R_2 \sin(\theta) \frac{d\theta}{dt}}{D}
   \]

3. **Determine \(\theta\) and \( \frac{d\theta}{dt} \):**

   At 9:00, \( \theta = \frac{\pi}{2} \) radians. The hands of the clock move at different rates:
   - The minute hand completes a full revolution (360° or \( 2\pi \) radians) every 60 minutes, so its angular speed is \( \frac{2\pi}{60} = \frac{\pi}{30} \) radians per minute.
   - The hour hand completes a full revolution every 12 hours (720 minutes), so its angular speed is \( \frac{2\pi}{720} = \frac{\pi}{360} \) radians per minute.

   Thus, the rate of change of \( \theta \) is the difference between these two speeds:
   \[
   \frac{d\theta}{dt} = \frac{\pi}{30} - \frac{\pi}{360} = \frac{11\pi}{360} \, \text{radians per minute}
   \]

4. **Substitute the Values:**

   - \( R_1 = 5.5 \) m, \( R_2 = 1.5 \) m
   - \( \sin\left(\frac{\pi}{2}\right) = 1 \)
   - At \( \theta = \frac{\pi}{2} \), the distance \( D \) between the tips is:
     \[
     D = \sqrt{5.5^2 + 1.5^2} = \sqrt{30.25 + 2.25} = \sqrt{32.5} = 5.7 \, \text{m}
     \]

   Substituting into the derivative formula:
   \[
   \frac{dD}{dt} = \frac{5.5 \cdot 1.5 \cdot 1 \cdot \frac{11\pi}{360}}{5.7}
   \]
   \[
   \frac{dD}{dt} = \frac{8.25 \cdot \frac{11\pi}{360}}{5.7}
   \]
   \[
   \frac{dD}{dt} \approx \frac{90.75\pi}{360 \cdot 5.7} \approx \frac{90.75 \cdot 3.14159}{2052} \approx 0.139 \, \text{m/min}
   \]

### Final Answer:
The rate at which the distance between the tips of the hands is changing at 9:00 is approximately \( 0.139 \, \text{m/min} \).

Would you like to go into more detail on any step, or do you have any questions?

### Follow-up Questions:
1. How does the rate change if the angle between the hands is not \( 90^\circ \)?
2. What is the angular speed of each clock hand at other times of the day?
3. How would the calculation differ if the clock hands were of equal length?
4. How would you compute the distance between the tips of the hands at a different time, like 3:00?
5. Can you derive a general formula for the distance between the tips of the clock hands at any time?

### Tip:
Remember that at \( \theta = 90^\circ \), the trigonometric functions simplify calculations, making problems involving rates of change much easier to handle!

The hands of a clock in some tower are approximately 5.5 m and 1.5 m in length. How fast is the distance between the tips of the hands changing at 9:00? (Hint: Use the Law of Cosines.)

Explore how to calculate the rate of change in the distance between the tips of clock hands using the Law of Cosines at 9:00. This problem involves trigonometric concepts and differentiation techniques, with hands measuring 5.5 m and 1.5 m. Ideal for Grades 10-12 students.

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