Let's analyze and solve the problem step by step based on the given diagram and information.

Given Information

1. An elastic rope is attached to the ground in a triangular setup, forming an angle $\theta$ with the horizontal.
2. The vertical distance $x$ is increasing at a rate of 5 meters/second.
3. The horizontal distances are 9 meters on the left side and 3 meters on the right side.

We are asked to find the following:

Part (a)

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

From the right triangle formed, we can express $\theta$ in terms of $x$ using trigonometry. The tangent of the angle $\theta$ can be written as: $\tan(\theta) = \frac{x}{6}$ since the total horizontal distance across the base is $9 + 3 = 12$ meters, and the midpoint of the base, from which the triangle's height extends, is 6 meters.

Therefore, $\theta = \tan^{-1} \left(\frac{x}{6}\right)$

So, the function for $\theta$ in terms of $x$ is: $\theta(x) = \tan^{-1} \left(\frac{x}{6}\right)$

Part (b)

When $x = 10$, determine the rate at which $\theta$ is changing:

To find $\frac{d\theta}{dt}$ when $x = 10$, we differentiate $\theta = \tan^{-1} \left(\frac{x}{6}\right)$ with respect to time $t$.

Using the chain rule: $\frac{d\theta}{dt} = \frac{d}{dx} \left(\tan^{-1} \left(\frac{x}{6}\right)\right) \cdot \frac{dx}{dt}$

First, find $\frac{d}{dx} \left(\tan^{-1} \left(\frac{x}{6}\right)\right)$: $\frac{d}{dx} \left(\tan^{-1} \left(\frac{x}{6}\right)\right) = \frac{1}{1 + \left(\frac{x}{6}\right)^2} \cdot \frac{1}{6} = \frac{1}{6 + \frac{x^2}{6}}$

Now substitute $\frac{dx}{dt} = 5$ m/s and $x = 10$ m: [ \frac{d\theta