Let's go through each part of this problem to identify where corrections are needed.

Problem Breakdown

A kite glides horizontally at an altitude of $30$ m as string is unwound. We need to determine the rate at which the angle between the string and the horizontal ($\theta$) decreases when $60$ m of string has been unspooled. Given:

• The kite's horizontal velocity ($\frac{dx}{dt}$) is $1 \, \text{m/s}$.
• $\theta$ is the angle between the string and the horizontal.
• $x$ is the kite's horizontal position (distance from where the string is anchored).

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

We are given a setup where:

• The altitude of the kite is $30$ m.
• The hypotenuse (length of the string unspooled) is $60$ m.

Using trigonometry: $\cos(\theta) = \frac{x}{60}$ Solving for $\theta$: $\theta = \arccos\left(\frac{x}{60}\right)$

Thus, the correct answer in terms of $\theta$ as a function of $x$ is: $\theta = \arccos\left(\frac{x}{60}\right)$

If this was not entered as shown, please ensure this exact expression is used for part (a).

Part (b): Determine the Value of $x$

We have a right triangle with the vertical leg $30$ m (altitude) and hypotenuse $60$ m (string length). To find $x$ (the horizontal distance):

Using the Pythagorean theorem: $x = \sqrt{60^2 - 30^2} = \sqrt{3600 - 900} = \sqrt{2700} = 30\sqrt{3} \, \text{m}$

Therefore, the exact value for $x$ is $30\sqrt{3}$ meters, which matches the answer you entered. This part is correct.

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

We are given $\frac{dx}{dt} = 1 \, \text{m/s}$ and need to find $\frac{d\theta}{dt}$ when $x = 30\sqrt{3}$.

From part (a), we have: $\theta = \arccos\left(\frac{x}{60}\right)$

Differentiating both sides with respect to time $t$: $\frac{d\theta}{dt} = -\frac{1}{\sqrt{1 - \left(\frac{x}{60}\right)^2}} \cdot \frac{1}{60} \cdot \frac{dx}{dt}$

Substitute $x = 30\sqrt{3}$ and $\frac{dx}{dt} = 1$:

1. First, calculate $\frac{x}{60} = \frac{30\sqrt{3}}{60} = \frac{\sqrt{3}}{2}$.
2. Substitute into the derivative: $\frac{d\theta}{dt} = -\frac{1}{\sqrt{1 - \left(\frac{\sqrt{3}}{2}\right)^2}} \cdot \frac{1}{60} \cdot 1$
3. Simplify the square root: $= -\frac{1}{\sqrt{1 - \frac{3}{4}}} \cdot \frac{1}{60} = -\frac{1}{\sqrt{\frac{1}{4}}} \cdot \frac{1}{60} = -\frac{1}{\frac{1}{2}} \cdot \frac{1}{60} = -\frac{2}{60} = -\frac{1}{30}$

Therefore, the correct value of $\frac{d\theta}{dt}$ is $-\frac{1}{30}$ rad/s.

Summary of Corrections

• (a) The correct expression is $\theta = \arccos\left(\frac{x}{60}\right)$.
• (c) The correct rate of change $\frac{d\theta}{dt}$ is $-\frac{1}{30}$ rad/s, not $+\frac{1}{2}$.

Would you like further clarification on any step, or do you have additional questions?

Additional Related Questions:

1. How would the answer change if the altitude of the kite were different?
2. How can we find the vertical component of the kite’s velocity in this situation?
3. What if the kite’s horizontal velocity was variable? How would this affect $\frac{d\theta}{dt}$?
4. How does the rate of change of the angle depend on the length of the string unspooled?
5. If we had the length of the string change over time, how would we approach this?

Tip:

In problems involving trigonometric rates of change, always start by defining relationships and differentiating implicitly with respect to time to set up the problem correctly.