To tackle this problem, let's break it down into parts based on the given information and questions:

Given Information

• The horizontal distances in the setup are $9$ meters to the left and $3$ meters to the right from the center where the rope is being pulled.
• $x$ represents the vertical distance from the ground level to the point where the rope is pulled up.
• The rate of change of $x$ with respect to time $\frac{dx}{dt} = 5$ meters per second.

Questions

Part (a)

Write $\theta$ as a function of $x$.

The angle $\theta$ is formed at the left attachment point of the rope. Using trigonometry:

• The distance opposite $\theta$ is $x$.
• The horizontal distance adjacent to $\theta$ is $9$.

Thus, we can express $\theta$ in terms of $x$ as: $\theta = \arctan\left(\frac{x}{9}\right)$

Part (b)

When $x = 10$, find the rate at which $\theta$ is changing, i.e., $\frac{d\theta}{dt}$.

To find $\frac{d\theta}{dt}$, we differentiate $\theta = \arctan\left(\frac{x}{9}\right)$ with respect to time $t$: $\frac{d\theta}{dt} = \frac{1}{1 + \left(\frac{x}{9}\right)^2} \cdot \frac{1}{9} \cdot \frac{dx}{dt}$

Plugging in $\frac{dx}{dt} = 5$ and $x = 10$:

1. Substitute $x = 10$ into the expression inside the square term: $\frac{d\theta}{dt} = \frac{1}{1 + \left(\frac{10}{9}\right)^2} \cdot \frac{1}{9} \cdot 5$
2. Simplify this expression to get the value of $\frac{d\theta}{dt}$.

Part (c)

Compare the rate of change of the left-hand piece of the rope (L) with the right-hand piece of the rope (R) when $x = 10$.

To determine if the rate of change of $L$ is greater than that of $R$, we can examine the distances:

1. The left side $L$ is longer than the right side $R$ (since the left side horizontal component is $9$ meters and the right side is $3$ meters).
2. Given that $x$ is increasing, the left side $L$ will experience a greater rate of change compared to $R$ because of the larger horizontal base.

Therefore, Yes, the rate of change of $L$ will be larger than that of $R$ when $x = 10$.

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

Here are 5 related questions for further exploration:

1. How would the result change if $x$ were increasing at a different rate?
2. What if the horizontal distances on either side of the pulling point were equal?
3. How does the angle $\theta$ behave as $x$ approaches zero?
4. How would we calculate the total length of rope needed for different values of $x$?
5. How would the rate of change of $\theta$ be affected if the distances on either side were reversed?

Tip: When working with related rates, carefully identify the relationships and differentiate with respect to time for accurate results.