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=

(b) What is the value of  x   at the moment in question? Give the exact value.

x=

(c) What is the value of  dθdt   at the same moment? Give the exact value, paying attention to the sign.

dθ/dt=

To approach this problem, we will break it down into parts by solving each section systematically.

### Part (a): Expressing \( \theta \) as a Function of \( x \)

Given:
- The kite flies horizontally at an altitude of \( y = 30 \) m.
- The string length, \( s = 60 \) m, forms an angle \( \theta \) with the horizontal.
- The kite’s horizontal position is denoted by \( x \).

From the given information, we have a right triangle where:
- The hypotenuse is the string length, \( s \).
- The opposite side to the angle \( \theta \) is the altitude, \( y = 30 \) m.
- The adjacent side is \( x \).

Using trigonometry, specifically the cosine function for this configuration:
\[
\cos(\theta) = \frac{x}{s}
\]
Rearrange to express \( \theta \) as a function of \( x \):
\[
\theta = \cos^{-1} \left( \frac{x}{s} \right)
\]
Since \( s = 60 \), we get:
\[
\theta = \cos^{-1} \left( \frac{x}{60} \right)
\]

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

At the given moment, we know the hypotenuse \( s = 60 \) m, and the opposite side \( y = 30 \) m. To find \( x \), we use the Pythagorean theorem:
\[
x = \sqrt{s^2 - y^2}
\]
Substituting \( s = 60 \) and \( y = 30 \):
\[
x = \sqrt{60^2 - 30^2} = \sqrt{3600 - 900} = \sqrt{2700} = 30\sqrt{3}
\]

So, the exact value of \( x \) is:
\[
x = 30\sqrt{3}
\]

### Part (c): Finding \( \frac{d\theta}{dt} \) at the Same Moment

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

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 values** at the moment in question:
   - \( x = 30\sqrt{3} \)
   - \( \frac{dx}{dt} = 3 \, \text{m/s} \)

3. Simplify the expression:
   \[
   \frac{d\theta}{dt} = -\frac{1}{\sqrt{1 - \left( \frac{30\sqrt{3}}{60} \right)^2}} \cdot \frac{1}{60} \cdot 3
   \]
   Simplify inside the square root:
   \[
   \frac{d\theta}{dt} = -\frac{1}{\sqrt{1 - \frac{3}{4}}} \cdot \frac{3}{60} = -\frac{1}{\sqrt{\frac{1}{4}}} \cdot \frac{3}{60} = -2 \cdot \frac{3}{60} = -\frac{1}{10}
   \]

Thus, the exact value of \( \frac{d\theta}{dt} \) at the moment in question is:
\[
\frac{d\theta}{dt} = -\frac{1}{10} \, \text{radians per second}
\]

### Summary of Answers
(a) \( \theta = \cos^{-1} \left( \frac{x}{60} \right) \)  
(b) \( x = 30\sqrt{3} \)  
(c) \( \frac{d\theta}{dt} = -\frac{1}{10} \) radians per second  

Would you like further clarification on any part of this solution?

---

1. How would the answer change if the kite’s horizontal velocity were different?
2. What if the altitude of the kite varied over time?
3. How can we interpret the negative sign in \( \frac{d\theta}{dt} \)?
4. How would we approach this problem if the kite followed a non-horizontal path?
5. What trigonometric identities could further simplify \( \theta \)?

**Tip:** Always check your trigonometric relationships against your triangle diagram for accuracy.

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 calculate the rate at which the angle between a kite's string and the horizontal decreases as it glides horizontally at 30 m altitude. With 60 m of string unspooled and a horizontal velocity of 3 m/s, this calculus problem applies related rates, inverse trigonometry, and the Pythagorean theorem. Suitable for undergraduate students in calculus.

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