Problem 15
Question
Flying a kite \(A\) girl flies a kite at a height of 300 \(\mathrm{ft}\) , the wind carrying the kite horizontally away from her at a rate of 25 \(\mathrm{ft} /\) sec. How fast must she let out the string when the kite is 500 ft away from her?
Step-by-Step Solution
Verified Answer
The girl must let out the string at 20 ft/sec.
1Step 1: Understand the Problem
We need to determine the rate at which the girl must let out the string when the kite is 500 ft away from her. The kite is flying at a fixed height of 300 ft, and it is moving horizontally away from her at 25 ft/sec. The string forms a right triangle with the vertical height and horizontal distance.
2Step 2: Define Variables
Let the horizontal distance from the girl to the kite be \( x \) ft, the length of the string be \( s \) ft, and the constant height of the kite be \( h = 300 \) ft. We are given \( \frac{dx}{dt} = 25 \) ft/sec and need to find \( \frac{ds}{dt} \) when \( s = 500 \) ft.
3Step 3: Use the Pythagorean Theorem
The relationship between the string length \( s \), the horizontal distance \( x \), and the height \( h \) is given by the Pythagorean theorem: \( s^2 = x^2 + h^2 \). Substitute \( h = 300 \) ft into the equation: \( s^2 = x^2 + 300^2 \).
4Step 4: Differentiate with Respect to Time
Differentiate both sides of the equation with respect to time \( t \) to obtain the relationship between the rates: \( 2s \frac{ds}{dt} = 2x \frac{dx}{dt} \). Simplifying gives \( s \frac{ds}{dt} = x \frac{dx}{dt}. \)
5Step 5: Solve for \( \frac{ds}{dt} \)
First, use the Pythagorean theorem to solve for \( x \) when \( s = 500 \) ft: \( 500^2 = x^2 + 300^2 \). This simplifies to \( x^2 = 500^2 - 300^2 \), or \( x = \sqrt{500^2 - 300^2} = 400 \) ft. Substitute \( x = 400 \) ft and \( \frac{dx}{dt} = 25 \) ft/sec into the differentiated equation:\[ 500 \frac{ds}{dt} = 400 \times 25 \]Solving for \( \frac{ds}{dt} \) yields \( \frac{ds}{dt} = \frac{400 \times 25}{500} = 20 \) ft/sec.
Key Concepts
Pythagorean theoremdifferentiationapplied mathematics
Pythagorean theorem
When flying this kite, we can imagine the setup as a right-angled triangle.The kite string serves as the hypotenuse, which is the longest side of the triangle that stretches from the girl to the kite.The horizontal movement away from her forms one leg of the triangle, while the constant height above the ground forms the other leg.
If we consider the Pythagorean theorem, it states that in a right triangle, the square of the hypotenuse (\( s^2 \)) is equal to the sum of the squares of the other two sides (\( x^2 \) + \( h^2 \)).
This is why we use this equation: \[ s^2 = x^2 + h^2 \]
Substituting the given height, we have: \[ s^2 = x^2 + 300^2 \]
This relationship allows us to tie together the string length, the height, and the horizontal distance.
Understanding how these three dimensions connect is crucial as it sets the stage for determining related rates, which is important when the kite moves.
If we consider the Pythagorean theorem, it states that in a right triangle, the square of the hypotenuse (\( s^2 \)) is equal to the sum of the squares of the other two sides (\( x^2 \) + \( h^2 \)).
This is why we use this equation: \[ s^2 = x^2 + h^2 \]
Substituting the given height, we have: \[ s^2 = x^2 + 300^2 \]
This relationship allows us to tie together the string length, the height, and the horizontal distance.
Understanding how these three dimensions connect is crucial as it sets the stage for determining related rates, which is important when the kite moves.
differentiation
Differentiation helps us observe how changes in one variable affect another.In this case, we want to know how fast the length of the kite string is changing.
We can use calculus to analyze the relationship among the rates of change of the string, height, and horizontal distance.
By differentiating the Pythagorean equation, \( s^2 = x^2 + 300^2 \), with respect to time \( t \), we use the chain rule to find:
With the known horizontal speed and the computed horizontal distance, we can solve for the rate at which the girl must let out the string.
We can use calculus to analyze the relationship among the rates of change of the string, height, and horizontal distance.
By differentiating the Pythagorean equation, \( s^2 = x^2 + 300^2 \), with respect to time \( t \), we use the chain rule to find:
- \( 2s \frac{ds}{dt} = 2x \frac{dx}{dt} \)
- This simplifies to \( s \frac{ds}{dt} = x \frac{dx}{dt} \)
With the known horizontal speed and the computed horizontal distance, we can solve for the rate at which the girl must let out the string.
applied mathematics
In applied mathematics,
we use mathematical concepts to solve real-world problems.
This kite-flying problem is an excellent example of applying
related rates to measure dynamic situations.
You can see that by understanding geometry and calculus, we can determine how fast a change occurs, such as the string being let out.
These principles have everyday applications, from predicting the speed of an object from a moving car to analyzing population changes.
In our problem, we calculated the rate at which the string should be released. The girl's observation, mixed with our mathematical conclusions, allows us to understand these concepts in action.
By relating rates with practical situations, you build a vital skill set for both academic and everyday problem-solving.
You can see that by understanding geometry and calculus, we can determine how fast a change occurs, such as the string being let out.
These principles have everyday applications, from predicting the speed of an object from a moving car to analyzing population changes.
In our problem, we calculated the rate at which the string should be released. The girl's observation, mixed with our mathematical conclusions, allows us to understand these concepts in action.
By relating rates with practical situations, you build a vital skill set for both academic and everyday problem-solving.
Other exercises in this chapter
Problem 14
In Exercises \(13-16,\) differentiate the functions and find the slope of the tangent line at the given value of the independent variable. $$ k(x)=\frac{1}{2+x}
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Show that the linearization of \(f(x)=(1+x)^{k}\) at \(x=0\) is \(L(x)=1+k x .\)
View solution Problem 15
Find the first derivatives of the functions in Exercises \(11-18\) . $$ f(x)=\sqrt{1-\sqrt{x}} $$
View solution Problem 15
In Exercises \(9-18,\) write the function in the form \(y=f(u)\) and \(u=g(x) .\) Then find \(d y / d x\) as a function of \(x .\) $$ y=\sec (\tan x) $$
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