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Let's break down the problem in the image step by step:

### Problem Summary:
We have a drawbridge of length \( L = 40 \) m that is raised by a cable attached to a rotating wheel of radius \( r = 2 \) m. The wheel rotates at a constant rate of 3 revolutions per minute, and at time \( t = 0 \), the bridge is in a horizontal position. We are asked to:

1. **(a)** Find the rate at which \( \theta \), the angle between the bridge and its vertical position, is changing when \( \theta = \frac{\pi}{6} \).

### Information and Variables:
- Bridge length: \( L = 40 \) m.
- Wheel radius: \( r = 2 \) m.
- Wheel rotates at 3 revolutions per minute.
- We are looking for \( \frac{d\theta}{dt} \) when \( \theta = \frac{\pi}{6} \).

### Step 1: Understanding the Geometry
The length of the bridge remains constant at \( L = 40 \) m. The angle \( \theta \) is measured between the bridge and the vertical position.

The cable's length, based on the rotational motion of the wheel, will be decreasing as the wheel pulls the cable. The distance \( s \) between the pivot point at the wheel and the point \( P \) (the end of the bridge) is related to \( \theta \) by:
\[
s = L \cos(\theta)
\]
This gives the length of the bridge projected vertically.

### Step 2: Rate of Change of the Cable
The wheel rotates at 3 revolutions per minute. One revolution corresponds to a change of \( 2\pi r \) in the length of the cable. Hence, the rate at which the cable is being pulled (or shortened) is:
\[
\frac{ds}{dt} = -\left(3 \, \text{rev/min} \times 2\pi \times 2 \, \text{m/rev}\right) = -12\pi \, \text{m/min}
\]
The negative sign indicates the cable is getting shorter as the bridge is being raised.

### Step 3: Relating \( s \) and \( \theta \)
Since \( s = L \cos(\theta) \), differentiate both sides with respect to \( t \):
\[
\frac{ds}{dt} = -L \sin(\theta) \frac{d\theta}{dt}
\]
Substituting the known values:
\[
-12\pi = -40 \sin(\theta) \frac{d\theta}{dt}
\]

### Step 4: Solve for \( \frac{d\theta}{dt} \)
At \( \theta = \frac{\pi}{6} \), we know that \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \). Substituting this into the equation:
\[
-12\pi = -40 \times \frac{1}{2} \times \frac{d\theta}{dt}
\]
Simplifying:
\[
-12\pi = -20 \frac{d\theta}{dt}
\]
Solving for \( \frac{d\theta}{dt} \):
\[
\frac{d\theta}{dt} = \frac{12\pi}{20} = \frac{3\pi}{5} \, \text{radians per minute}
\]

### Final Answer:
The rate at which \( \theta \) is changing when \( \theta = \frac{\pi}{6} \) is \( \frac{3\pi}{5} \) radians per minute.

---

Would you like a deeper explanation of any step, or do you have any additional questions?

Here are 5 relative questions for further exploration:
1. How does the rate of change of the angle depend on the rotational speed of the wheel?
2. What would happen to \( \frac{d\theta}{dt} \) if the radius of the wheel were doubled?
3. Can we find \( \theta \) as a function of time given the rotational speed of the wheel?
4. How would the problem change if the bridge length were not constant?
5. What would the equation look like if the bridge started at an initial angle \( \theta_0 \) instead of a horizontal position?

**Tip**: Always keep track of units, especially when dealing with angular and linear velocities!

Find the rate at which θ, the angle between the bridge's position at time t and its vertical position, is changing when θ = π/6. The drawbridge has length L = 40 m, and a rotating wheel with radius r = 2 m is pulling the bridge with a constant rotation rate of 3 revolutions per minute.

Learn how to calculate the rate of change of the angle θ for a drawbridge with a rotating wheel. This problem involves trigonometry and related rates, with a bridge of length 40 meters and a wheel rotating at 3 revolutions per minute.

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