A kite glides horizontally at an altitude of  30  m while we unspool the string. Consequently, the angle made between the string and the horizon diminishes. We would like to determine the rate at which this angle decreases once  60  m of string has been unspooled, given that, at that instant, the kite‘s horizontal velocity is  3  m/s.
 
To solve this problem, let  θ   be the angle in radians made between the string and the horizontal,  x   the kite’s horizontal position in meters since being attached to the ground, and  t   the time in seconds.
 
We further suppose that the string is straight and taut.
 
(a) Sketch a diagram of this question and use it to express  θ   as a function of  x  .
 
θ=      Preview Change entry mode rad


(b) What is the value of  x   at the moment in question? Give the exact value.
 
x=      Preview Change entry mode m


(c) What is the value of  dθdt   at the same moment? Give the exact value, paying attention to the sign.
 
dθdt=      Preview Change entry mode rad/s

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:
\[
\frac{d\theta}{dt} = -\frac{1}{10} \, \text{rad/s}
\]

---

### 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?

---

### 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.

A kite glides horizontally at an altitude of 30 m while we unspool the string. Consequently, the angle made between the string and the horizon diminishes. We would like to determine the rate at which this angle decreases once 60 m of string has been unspooled, given that, at that instant, the kite‘s horizontal velocity is 3 m/s.

Explore how to determine the rate of decrease in the angle between a kite string and the horizontal ground when the kite flies at a 30 m altitude with 60 m of string unspooled and a 3 m/s horizontal speed. This detailed problem solution covers trigonometric relationships, differentiation, and related rates. Ideal for advanced high school students.

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