Let's solve each part of this problem step by step.

Given:

1. An elastic rope is attached to the ground, as shown in the diagram.
2. $x$ is the vertical distance, which is increasing at a rate of $\frac{dx}{dt} = 5$ meters/second.
3. We need to express $\theta$ as a function of $x$, calculate the rate of change of $\theta$ when $x = 10$, and compare the rates of change of the lengths $L$ and $R$ when $x = 10$.

Part (a): Express $\theta$ as a Function of $x$

1. Observe that the triangle in the diagram forms a right triangle.
2. We can use trigonometry to relate $\theta$ and $x$. Specifically, since $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$: $\tan(\theta) = \frac{x}{3}$
3. Solving for $\theta$: $\theta = \arctan\left(\frac{x}{3}\right)$

Thus, $\theta(x) = \arctan\left(\frac{x}{3}\right)$.

Part (b): Rate of Change of $\theta$ When $x = 10$

To find the rate of change of $\theta$ with respect to time $t$, we differentiate $\theta = \arctan\left(\frac{x}{3}\right)$ with respect to $t$:

1. Using the chain rule: $\frac{d\theta}{dt} = \frac{d}{dx} \left(\arctan\left(\frac{x}{3}\right)\right) \cdot \frac{dx}{dt}$
2. Differentiate $\arctan\left(\frac{x}{3}\right)$ with respect to $x$: $\frac{d}{dx} \left(\arctan\left(\frac{x}{3}\right)\right) = \frac{1}{1 + \left(\frac{x}{3}\right)^2} \cdot \frac{1}{3} = \frac{1}{3 + x^2}$
3. Substitute $\frac{dx}{dt} = 5$ and $x = 10$: $\frac{d\theta}{dt} = \frac{1}{3 + 10^2} \cdot 5 = \frac{1}{3 + 100} \cdot 5 = \frac{5}{103}$

So, the rate of change of $\theta$ when $x = 10$ is $\frac{5}{103}$ radians per second.

Part (c): Comparing the Rates of Change of $L$ and $R$

For this part, we need to determine if the rate of change of $L$ is larger than that of $R$ when $x = 10$.

1. The lengths $L$ and $R$ correspond to the hypotenuse of each triangle, calculated as: $L = \sqrt{x^2 + 9}$ and similarly for $R$.

2. Since the points where $L$ and $R$ attach are symmetric around the centerline, and both lengths depend on $x$ similarly, the rates of change of $L$ and $R$ will be equal in magnitude. Thus, neither is larger than the other.

Would you like further details on any specific part?

Here are 5 follow-up questions you might consider:

1. How would you calculate the rate of change of the hypotenuse $L$ in general terms?
2. What would happen to $\frac{d\theta}{dt}$ as $x$ increases significantly?
3. Can you derive $\frac{dL}{dt}$ using implicit differentiation for a given $x$?
4. How would you approach this problem if $x$ were decreasing instead of increasing?
5. Is there a way to determine the value of $\theta$ when the rate of $x$ is maximized?

Tip: For related rates problems, always identify each variable's relation before differentiating to simplify the process.