To solve this problem, let's go through each part step by step.

Part (a): Diagram and Expressing $\theta$ as a Function of $x$

1. Diagram: Draw a right triangle where:

   • The vertical leg is the altitude of the kite, $h = 30 \, \text{m}$.
   • The hypotenuse is the length of the string, $s = 60 \, \text{m}$.
   • The horizontal leg is the distance of the kite from the origin on the ground, $x$, where the string is attached.
   • The angle $\theta$ is between the hypotenuse (string) and the horizontal leg (ground).

2. Expressing $\theta$ as a function of $x$: Using trigonometry, we have: $\sin(\theta) = \frac{h}{s} = \frac{30}{s}$ Since $s = 60 \, \text{m}$, this simplifies to: $\sin(\theta) = \frac{30}{60} = \frac{1}{2}$

   Then, using the right triangle relationships, we can also express: $\cos(\theta) = \frac{x}{s} \quad \Rightarrow \quad \cos(\theta) = \frac{x}{60}$

Therefore, solving for $\theta$ in terms of $x$: $\theta = \cos^{-1} \left( \frac{x}{60} \right)$

Part (b): Finding the Value of $x$ at the Moment in Question

Since $\sin(\theta) = \frac{1}{2}$, $\theta = \frac{\pi}{6}$ radians. Using the cosine relationship:

$\cos\left(\frac{\pi}{6}\right) = \frac{x}{60}$

Since $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$: $\frac{\sqrt{3}}{2} = \frac{x}{60}$

Solving for $x$: $x = 60 \cdot \frac{\sqrt{3}}{2} = 30 \sqrt{3} \approx 51.96 \, \text{m}$

Thus, the exact value of $x$ is: $x = 30 \sqrt{3} \, \text{m}$

Part (c): Finding the Value of $\frac{d\theta}{dt}$

To find $\frac{d\theta}{dt}$, we differentiate the expression $\cos(\theta) = \frac{x}{60}$ with respect to time $t$: $-\sin(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{60} \cdot \frac{dx}{dt}$

We know:

• $\sin(\theta) = \frac{1}{2}$
• $\frac{dx}{dt} = 3 \, \text{m/s}$

Substituting these values: $-\frac{1}{2} \cdot \frac{d\theta}{dt} = \frac{1}{60} \cdot 3$

Solving for $\frac{d\theta}{dt}$: $\frac{d\theta}{dt} = -\frac{3}{60 \cdot \frac{1}{2}} = -\frac{3}{30} = -\frac{1}{10} \, \text{rad/s}$

Thus, the rate at which the angle $\theta$ decreases is: $\frac{d\theta}{dt} = -\frac{1}{10} \, \text{rad/s}$

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Summary of Answers:

• (a) $\theta = \cos^{-1} \left( \frac{x}{60} \right)$
• (b) $x = 30 \sqrt{3} \, \text{m}$
• (c) $\frac{d\theta}{dt} = -\frac{1}{10} \, \text{rad/s}$

Would you like further clarification on any of these steps?

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Follow-Up Questions:

1. How would the result change if the kite's altitude was different?
2. What would happen to $\frac{d\theta}{dt}$ if the horizontal velocity of the kite changed?
3. How can we interpret $\frac{d\theta}{dt}$ physically in this scenario?
4. What is the significance of the negative sign in $\frac{d\theta}{dt}$?
5. How would you find the rate of change of the string length with respect to time?

Tip:

When analyzing trigonometric relationships in moving objects, look for constants and use them to simplify differentiation tasks.