/*! 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 14 Flying a Kite Inge flies a kite ... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 14

Flying a Kite Inge flies a kite at a height of 300 ft, the wind carrying the kite horizontally away at a rate of 25 \(\mathrm{ft} / \mathrm{sec} .\) How fast must she let out the string when the kite is 500 \(\mathrm{ft}\) away from her?

Short Answer

Expert verified
Inge must let out the string at a rate of 20 ft/sec when the kite is 500 ft away from her.

Step by step solution

01

Represent the problem as a triangle

The person, the kite, and the point directly above the person on the height of the kite can be thought of as forming a right triangle. The height from the ground to the kite is one leg of the triangle (300 ft), the horizontal distance of the person from the kite is the other leg of the triangle (let it be \(x\)), and the length of the string is the hypotenuse of the triangle (let it be \(y\)). According to Pythagorean Theorem, \(y^2 = x^2 + 300^2\).
02

Differentiate the equation

Differentiate both sides of the equation with respect to time \(t\), given \(dx/dt = 25 ft/sec\) (rate of change of horizontal distance). The derivative of \(y^2\) with respect to \(t\) is \(2y*(dy/dt)\). The derivative of \(x^2\) with respect to \(t\) is \(2x*(dx/dt)\}. The derivative of \(300^2\) with respect to \(t\) is 0 (since it's constant). Hence, the differentiated equation is \(2y*(dy/dt) = 2x*(dx/dt)\).
03

Solve for dy/dt

Solve the equation obtained in step 2 for \(dy/dt\), which represents how fast she must let out the string. Substituting \(x = 400 ft\) and \(y = 500 ft\) (from the Pythagorean theorem) and \(dx/dt = 25 ft/sec\), we get \(dy/dt = x*(dx/dt)/y = 400*25/500 = 20 ft/sec\).

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Key Concepts

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

Pythagorean Theorem
In our kite problem, the Pythagorean Theorem helps us understand the relationship between the different sides of our right triangle. The theorem tells us that in a right triangle, the sum of the squares of the two legs equals the square of the hypotenuse. So, the formula is:
  • For the legs: One leg is the vertical distance (300 ft), and the other is the horizontal distance (which we'll call \( x \)).
  • The hypotenuse is the string length (\( y \)).
Therefore, the equation \( y^2 = x^2 + 300^2 \) allows us to evaluate how these distances are interconnected as the kite moves. Understanding this basis is crucial when we start incorporating rates of change into the problem.
Differentiation
Differentiation is the process of finding how a function changes as its inputs change. In our problem, we differentiate with respect to time, which means we're examining how the triangle's dimensions change over time.
To differentiate the equation \( y^2 = x^2 + 300^2 \) implicitly, follow these steps:
  • Derive \( 2y \cdot \frac{dy}{dt} \) for \( y^2 \), because \( y \) changes over time.
  • Derive \( 2x \cdot \frac{dx}{dt} \) for \( x^2 \), since \( x \) also changes over time at a rate of 25 ft/sec.
  • Remember, \( 300^2 \) is constant, so its derivative is zero.
This gives us the equation \( 2y \cdot \frac{dy}{dt} = 2x \cdot \frac{dx}{dt} \), which helps us understand how changes in one part of the triangle impact the other parts.
Triangle Geometry
In triangle geometry, understanding the structure of the triangle is fundamental.
In this scenario, we consider how the kite, the person, and the vertical height together create a right triangle. The vertical line (300 ft) can be seen as constant, while the kite moving horizontally creates a dynamic horizontal leg.
  • The dynamic aspect here is represented by the horizontal component \( x \), which increases as the kite moves away.
  • The hypotenuse \( y \) of 500 ft represents the string's slanting distance and changes as the string is let out.
  • This understanding facilitates applying the Pythagorean Theorem, linking the geometry to the dynamic aspects described by rate of change.
Keeping this triangular perspective aids in setting up the problem correctly and understanding the ongoing movement.
Rate of Change
The rate of change is a measure of how fast a variable changes over time.
In the kite exercise, we have a few different rates:
  • Horizontal speed \( \frac{dx}{dt} = 25 \) ft/sec, describing how quickly the kite moves away horizontally.
  • The task is to find \( \frac{dy}{dt} \), which indicates how fast Inge needs to let out the kite string.
By rearranging the differentiated equation, \( \frac{dy}{dt} = \frac{x \cdot \frac{dx}{dt}}{y} \), we can calculate \( \frac{dy}{dt} \) using known values.When \( x = 400 \) ft and \( y = 500 \) ft, it follows that \( \frac{dy}{dt} = \frac{400 \cdot 25}{500} = 20 \) ft/sec.
This means Inge needs to let out the string at 20 ft per second to keep up with the kite's horizontal travel.

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Most popular questions from this chapter

Multiple Choice A cylindrical rubber cord is stretched at a constant rate of 2 \(\mathrm{cm}\) per second. Assuming its volume does no change, how fast is its radius shrinking when its length is 100 \(\mathrm{c}\) and its radius is 1 \(\mathrm{cm} ?\) $$\begin{array}{ll}{\text { (A) } 0 \mathrm{cm} / \mathrm{sec}} & {\text { (B) } 0.01 \mathrm{cm} / \mathrm{sec}} 67 {\text{ (C) } 0.02 \mathrm{cm} / \mathrm{sec}}$\\\ {\text { (D) } 2 \mathrm{cm} / \mathrm{sec}} & {\text { (E) } 3.979 \mathrm{cm} / \mathrm{sec}}\end{array}

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