/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 A girl flying a kite holds the s... [FREE SOLUTION] | 91Ó°ÊÓ

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A girl flying a kite holds the string 4 feet above ground level and lets out string at a rate of 2 feet per second as the kite moves horizontally at an altitude of 84 feet. a. Calculate the rate at which the kite is moving horizontally when 100 feet of string has been let out. b. How far is the kite (above the ground) from the girl at this time?

Short Answer

Expert verified
a) The kite is moving horizontally at approximately 3.33 feet/second. b) The kite is 84 feet above the ground.

Step by step solution

01

Understand the Geometry

Imagine a right triangle with one leg as the vertical distance from the girl's hand to the kite and the other leg as the horizontal distance the kite has moved. The hypotenuse of the triangle is the length of the string, 100 feet, including the 4 feet above the ground from the girl's hand.
02

Apply Pythagoras' Theorem

Use Pythagoras' theorem to relate the sides of the triangle. The vertical distance from the girl's hand to the kite is \( 84 - 4 = 80 \) feet. Let \( x \) be the horizontal distance. Then, the equation is \( x^2 + 80^2 = 100^2 \).
03

Solve for the Horizontal Distance

Substitute the values into the Pythagorean equation: \( x^2 + 6400 = 10000 \). Thus, \( x^2 = 3600 \), leading to \( x = 60 \) feet (since distance cannot be negative).
04

Differentiate with Respect to Time

Differentiate the equation of the triangle \( x^2 + 80^2 = s^2 \) with respect to time \( t \), where \( s = 100 \) feet is the string length. This gives \( 2x \frac{dx}{dt} = 2s \frac{ds}{dt} \).
05

Substitute Known Values and Solve for Rate

We know \( x = 60 \), \( s = 100 \) (constant here), and \( \frac{ds}{dt} = 2 \) feet per second. Substitute these into the equation: \( 2 \times 60 \times \frac{dx}{dt} = 2 \times 100 \times 2 \). Thus, \( 120 \frac{dx}{dt} = 400 \), which simplifies to \( \frac{dx}{dt} = \frac{400}{120} = \frac{10}{3} \approx 3.33 \) feet per second.
06

Calculate the Kite's Distance Above Ground

The vertical component of the kite is 80 feet (altitude from ground is 84 feet, subtract 4 feet at handle level). Therefore, the distance the kite is above the ground from the girl is 84 feet.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Right Triangle
A right triangle is a geometric shape that has one angle measuring precisely 90 degrees, also known as the right angle. The presence of a right angle simplifies calculations considerably, especially when solving problems related to kites, ladders, or any situation involving vertical and horizontal measurements.
In this context, the imaginary triangle involves the vertical line from the ground to the kite's altitude as one leg and the horizontal distance from the girl to the kite as the other. The kite string acts as the hypotenuse of this triangle.
This structure allows us to apply mathematical concepts like the Pythagorean theorem, helping determine unknown lengths or rates.
Pythagorean Theorem
The Pythagorean theorem is a fundamental principle in geometry, stating that in a right-angled triangle, the square of the hypotenuse length is equal to the sum of the squares of the other two sides. Formally expressed as \(c^2 = a^2 + b^2\), where \(c\) is the hypotenuse, and \(a\) and \(b\) are the triangle's legs.
For our kite scenario, we use this theorem to set up the equation \(x^2 + 80^2 = 100^2\). Here, \(x\) represents the horizontal distance, \(80\) is the vertical height from the girl's hand to the kite, and \(100\) is the hypotenuse, or the string's length.
This equation allows us to find the horizontal distance \(x\), knowing the string length and height, by solving for \(x\).
Differentiation
Differentiation is a process in calculus used to determine how a function changes as its input changes, often representing rates of change or slopes of functions. In related rates problems, differentiation helps us find how variables relate and change over time.
In the kite problem, we differentiate the Pythagorean theorem equation with respect to time: \(x^2 + 80^2 = s^2\). Applying differentiation, we get \(2x \frac{dx}{dt} = 2s \frac{ds}{dt}\).
Here, \(\frac{dx}{dt}\) is the rate at which the horizontal distance changes, and \(\frac{ds}{dt}\) is the rate at which the string length increases, helping solve for our desired rate of horizontal movement.
Horizontal Distance
Horizontal distance is the measurement along the flat, level ground between two points. In this problem, it's the distance the kite travels horizontally from the point where the girl stands.
This type of distance is crucial in many real-world applications including aviation, surveying, and navigation, where the horizontal component is often more relevant than height.
In the kite example, once we apply the Pythagorean theorem, the horizontal distance turns out to be 60 feet. With the aid of related rates, we later find the kite's horizontal movement speed to be approximately 3.33 feet per second. This calculation combines both geometry and calculus to provide an accurate measure of the kite's travel over time.

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