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Why is (a) and (c) not true? Show me the right way to write them

Let's go through each part of the problem.

### Problem Summary
A kite is moving horizontally at an altitude of \( 30 \, \text{m} \) with a string that is being unspooled. The kite's horizontal velocity is \( 3 \, \text{m/s} \) at a certain moment. We need to find the rate at which the angle \( \theta \) between the string and the horizontal is changing when the length of the string unspooled is \( 60 \, \text{m} \).

### Part (a): Expressing \( \theta \) as a Function of \( x \)
To express \( \theta \) as a function of \( x \), we can use trigonometry. We know that:
- The height of the kite is \( 30 \, \text{m} \), which is constant.
- The length of the string (hypotenuse) is \( 60 \, \text{m} \).
  
Using the right triangle formed by the string, height, and horizontal distance \( x \), we can write:
\[
\cos(\theta) = \frac{x}{60}
\]
Therefore, solving for \( \theta \):
\[
\theta = \cos^{-1}\left(\frac{x}{60}\right)
\]

This is what was entered in the answer box for part (a), which seems correct. However, it was marked incorrect, which might be due to formatting or needing a more precise explanation.

### Part (b): Value of \( x \)
Given that \( z = 60 \) m and \( h = 30 \) m, we can find \( x \) using the Pythagorean theorem:
\[
x = \sqrt{z^2 - h^2} = \sqrt{60^2 - 30^2} = \sqrt{3600 - 900} = \sqrt{2700} = 30\sqrt{3} \, \text{m}
\]
The answer provided for \( x \) is correct here.

### Part (c): Calculating \( \frac{d\theta}{dt} \)
To find \( \frac{d\theta}{dt} \), we need to differentiate \( \theta = \cos^{-1} \left( \frac{x}{60} \right) \) with respect to \( t \).

1. **Differentiate \( \theta = \cos^{-1} \left( \frac{x}{60} \right) \) with respect to \( t \):**

\[
\frac{d\theta}{dt} = \frac{-1}{\sqrt{1 - \left(\frac{x}{60}\right)^2}} \cdot \frac{1}{60} \cdot \frac{dx}{dt}
\]

2. **Substitute the values of \( x \), \( \frac{dx}{dt} \), and simplify:**
   - \( x = 30\sqrt{3} \) (from part (b))
   - \( \frac{dx}{dt} = 3 \, \text{m/s} \)
   
   Substituting \( x = 30\sqrt{3} \):
   \[
   \cos(\theta) = \frac{x}{60} = \frac{30\sqrt{3}}{60} = \frac{\sqrt{3}}{2}
   \]
   So \( \sin(\theta) = \sqrt{1 - \cos^2(\theta)} = \sqrt{1 - \left(\frac{\sqrt{3}}{2}\right)^2} = \frac{1}{2} \).

   Plugging this in:
   \[
   \frac{d\theta}{dt} = \frac{-1}{\sqrt{1 - \left(\frac{\sqrt{3}}{2}\right)^2}} \cdot \frac{1}{60} \cdot 3 = \frac{-1}{\frac{1}{2}} \cdot \frac{1}{60} \cdot 3 = -\frac{1}{10} \, \text{rad/s}
   \]

So, the correct answer for \( \frac{d\theta}{dt} \) should be:
\[
\frac{d\theta}{dt} = -\frac{1}{10} \, \text{rad/s}
\]
The negative sign indicates that \( \theta \) is decreasing. 

### Corrected Answers
1. **(a)**: \( \theta = \cos^{-1} \left( \frac{x}{60} \right) \)
2. **(c)**: \( \frac{d\theta}{dt} = -\frac{1}{10} \, \text{rad/s} \)

Let me know if you need further clarification or have more questions!

### Follow-up Questions
1. How does \( \frac{d\theta}{dt} \) change if the kite’s horizontal velocity changes?
2. What would happen to \( \theta \) over time if the kite's velocity increased?
3. How would the length of the string affect \( \frac{d\theta}{dt} \) if it were longer?
4. Could we use a different trigonometric function to express \( \theta \) in terms of \( x \)?
5. How would \( \frac{d\theta}{dt} \) be affected if the kite's altitude was changing?

### Tip
When working with trigonometric relationships in related rates problems, make sure to correctly differentiate implicit functions with respect to time to capture all changing variables.

Explore a calculus problem involving a kite gliding horizontally with a changing angle between its string and the horizontal line. Learn how to find the rate at which this angle changes using trigonometric functions and related rates. Suitable for college-level calculus students.

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