Once​ Kate's kite reaches a height of 55 ft ​(above her​ hands), it rises no higher but drifts due east in a wind blowing 5 ft divided by s. How fast is the string running through​ Kate's hands at the moment that she has released 103 ft of​ string?
Question content area bottom
Part 1
At the moment that Kate has released 103 ft of​ string, it is running through her hands at a rate of about 
  
4.22 
ft divided by s.
ft cubed divided by s.
ft squared divided by s.
ft divided by s.
ft.
​(Round to two decimal places as​ needed.)

This problem involves related rates, where we need to determine how fast the string is moving through Kate's hands when a certain amount of string has been released.

Let's break this down into a right triangle scenario, where:

- The height of the kite, \( h \), remains constant at 55 feet.
- The length of the string released is \( s = 103 \) feet.
- The horizontal distance from Kate to the kite is \( x \), which increases due to the wind.

We can apply the Pythagorean theorem to relate these variables:

\[
s^2 = x^2 + h^2
\]

Substitute \( h = 55 \) ft into the equation:

\[
s^2 = x^2 + 55^2
\]
\[
s^2 = x^2 + 3025
\]

Differentiate both sides with respect to time \( t \), using implicit differentiation:

\[
2s \frac{ds}{dt} = 2x \frac{dx}{dt}
\]

We are given the rate at which the horizontal distance is changing, \( \frac{dx}{dt} = 5 \, \text{ft/s} \). We also know that at the moment when \( s = 103 \, \text{ft} \), we need to find \( \frac{ds}{dt} \), the rate at which the string is moving through Kate's hands.

First, we calculate \( x \) when \( s = 103 \) ft using the Pythagorean theorem:

\[
103^2 = x^2 + 3025
\]
\[
10609 = x^2 + 3025
\]
\[
x^2 = 7584
\]
\[
x = \sqrt{7584} \approx 87.1 \, \text{ft}
\]

Now, substitute the known values into the differentiated equation:

\[
2(103) \frac{ds}{dt} = 2(87.1)(5)
\]

Simplify and solve for \( \frac{ds}{dt} \):

\[
206 \frac{ds}{dt} = 871
\]
\[
\frac{ds}{dt} = \frac{871}{206} \approx 4.22 \, \text{ft/s}
\]

Therefore, the string is running through Kate's hands at a rate of approximately \( \boxed{4.22 \, \text{ft/s}} \).

---

Would you like more details or have any other questions?

Here are 5 related questions:

1. How would the rate of string release change if the wind speed increased?
2. What if the kite's height changed? How would this affect the string's speed?
3. How does the Pythagorean theorem apply to other real-life related rates problems?
4. Can we derive a general formula for any wind speed and kite height?
5. What assumptions are made in this problem about the shape of the triangle?

**Tip**: Always check if variables such as height remain constant in related rates problems; this simplifies the equation

Once Kate's kite reaches a height of 55 ft (above her hands), it rises no higher but drifts due east in a wind blowing 5 ft/s. How fast is the string running through Kate's hands at the moment that she has released 103 ft of string?

Solve a related rates problem involving a kite that reaches a height of 55 ft and drifts due east. Learn how to calculate the speed at which the string is running through Kate's hands when she has released 103 ft of string. Includes step-by-step application of the Pythagorean Theorem and implicit differentiation.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation