We are tasked with finding the rate of change of the angle between the string of a kite and the horizontal ground, as the kite moves horizontally. This is a related rates problem, and the angle is getting smaller as the kite moves.

Problem Setup:

• The kite is 100 feet above the ground, so the height $h = 100$ ft.
• The kite is moving horizontally at a speed of 3 ft/s.
• The length of the string is 200 feet.
• We need to find how fast the angle $\theta$ between the string and the horizontal is changing (in radians per second).

Step 1: Relating the Variables

Let:

• $\theta$ be the angle between the string and the horizontal.
• $x$ be the horizontal distance from the point directly below the kite to where the string meets the ground.

By the Pythagorean theorem, we know that the length of the string, the height, and the horizontal distance are related by: $x^2 + h^2 = s^2$ where:

• $s = 200$ ft (the length of the string),
• $h = 100$ ft (the height of the kite).

Substituting $h = 100$ into the equation: $x^2 + 100^2 = 200^2$ $x^2 + 10000 = 40000$ $x^2 = 30000 \quad \Rightarrow \quad x = \sqrt{30000} = 100\sqrt{3} \text{ ft}.$

Now, we use the trigonometric relationship between the angle $\theta$, the height $h$, and the horizontal distance $x$: $\tan \theta = \frac{h}{x}.$

Step 2: Differentiating with Respect to Time

We need to differentiate $\theta$ with respect to time $t$. Using implicit differentiation on both sides of $\tan \theta = \frac{h}{x}$, we get: $\sec^2 \theta \cdot \frac{d\theta}{dt} = -\frac{h}{x^2} \cdot \frac{dx}{dt}.$ We know that $\frac{dx}{dt} = 3$ ft/s (since the kite is moving horizontally), and $h = 100$ ft.

We also need to find $\sec^2 \theta$. Using the relationship: $\cos \theta = \frac{x}{s} = \frac{100\sqrt{3}}{200} = \frac{\sqrt{3}}{2},$ so $\sec \theta = \frac{1}{\cos \theta} = \frac{2}{\sqrt{3}} \quad \Rightarrow \quad \sec^2 \theta = \frac{4}{3}.$

Step 3: Plugging in the Values

Now we can plug everything into the differentiated equation: $\frac{4}{3} \cdot \frac{d\theta}{dt} = -\frac{100}{(100\sqrt{3})^2} \cdot 3.$ First, simplify $(100\sqrt{3})^2$: $(100\sqrt{3})^2 = 10000 \times 3 = 30000,$ so the equation becomes: $\frac{4}{3} \cdot \frac{d\theta}{dt} = -\frac{100}{30000} \cdot 3 = -\frac{100}{10000} = -\frac{1}{100}.$ Thus, $\frac{d\theta}{dt} = -\frac{3}{4} \cdot \frac{1}{100} = -\frac{3}{400} \text{ radians per second}.$ Since the angle is getting smaller, the rate of change is negative. However, we are asked for the positive value of the rate: $\boxed{\frac{3}{400}} \text{ radians per second}.$

Would you like more details or further clarification?

Here are 5 related questions for deeper understanding:

1. How would the problem change if the height of the kite varied over time?
2. What is the rate of change of the string length if both the height and horizontal velocity were changing?
3. How would you adjust the solution if the kite’s speed increased to 5 ft/s?
4. Can you derive a general formula for $\frac{d\theta}{dt}$ if the string length changes as well?
5. How does this problem illustrate the concept of implicit differentiation?

Tip: When solving related rates problems, always identify and relate the variables before differentiating implicitly.