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Chapter 3: Problem 38

Once Kate's kite reaches a height of \(50 \mathrm{ft}\) (above her hands), it rises no higher but drifts due east in a wind blowing \(5 \mathrm{ft} / \mathrm{s} .\) How fast is the string running through Kate's hands at the moment that she has released \(120 \mathrm{ft}\) of string?

Short Answer

Expert verified
Answer: Approximately 1.95 ft/s.

Step by step solution

01

Understand the problem and draw a diagram

Let's visualize the problem. Make a sketch of the scenario - Draw a triangle with one vertex representing Kate's hands, the second vertex representing the position of the kite, and the third vertex being the point exactly 50 ft above Kate's hands. We notice that the triangle formed is a right-angled triangle. From the problem, we know the height (50 ft) and the length of the string (120 ft). Let the horizontal distance from Kate's hands to the point directly below the kite be x ft.
02

Apply the Pythagorean theorem

Now we use the Pythagorean theorem to find the relation between the height, the horizontal distance x, and the length of the string (120 ft). The Pythagorean theorem in this case can be written as: \(50^2 + x^2 = 120^2\)
03

Differentiate with respect to time

Differentiate both sides with respect to time (using \(t\) for time): \(\frac{d}{dt}(50^2 + x^2) = \frac{d}{dt}(120^2)\) We get: \(0 + 2x\frac{dx}{dt} = 0\) Now, from the problem, we know that the kite is moving horizontally at a constant rate of 5 ft/s, which is \(\frac{dx}{dt}\), so we can replace it: \(2x(5) = 0\)
04

Find x

To find the value of x, use the Pythagorean theorem again: \(x^2 = 120^2 - 50^2\) \(x = \sqrt{120^2 - 50^2}\) Evaluate the value of x: \(x \approx 93.49\) ft
05

Find the rate at which the string is running through Kate's hands

Now, to find the rate at which the string is running through Kate's hands, we can use the chain rule. Let's denote the length of the string as \(s\). Then we have: \(\frac{ds}{dt} = \frac{ds}{dx} \cdot \frac{dx}{dt}\) We know that \(\frac{dx}{dt} = 5\) ft/s. To find \(\frac{ds}{dx}\), use the Pythagorean theorem again: \(s^2 = 50^2 + x^2\) Differentiate both sides with respect to \(x\): \(2s\frac{ds}{dx} = 2x\) Now, substitute \(s = 120\) ft and \(x \approx 93.49\) ft, and solve for \(\frac{ds}{dx}\): \(240\frac{ds}{dx} = 2(93.49)\) \(\frac{ds}{dx} \approx 0.3896\)
06

Convert units and find the final answer

Finally, use the chain rule as written in Step 5. Substitute the values obtained: \(\frac{ds}{dt} = \left(0.3896\right) \cdot 5\) \(\frac{ds}{dt} \approx 1.95\) So, the string is running through Kate's hands at a rate of approximately \(1.95 \mathrm{ft/s}\).

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Key Concepts

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

Related Rates
Related rates problems are a type of problem in differential calculus where two or more quantities that are changing over time are related by an equation. These problems often involve determining how the rate of change of one quantity affects the rate of change of another.

In the case of Kate's kite, we seek to find how quickly the kite string is being released through her hands as the kite drifts eastward.
  • The problem helps us understand how simultaneous changes in different dimensions are connected.
  • The key is to set up a relationship between the quantities involved, like distance and speed, and then use differentiation to relate their rates of change over time.
  • Typically, this involves using derivatives to take into account how one variable changes as another changes.
Recognizing the connection between the rates of different variables is crucial to solve these types of problems.
Pythagorean Theorem
The Pythagorean Theorem is a fundamental principle in geometry, especially relating to right triangles. It postulates that in a right triangle, the square of the length of the hypotenuse is equal to the sum of squares of the other two sides. This theorem is instrumental in solving many geometry and calculus problems.

Applying the Pythagorean Theorem to the kite problem allows us to link the kite's height and horizontal distance to the hypotenuse, which is the length of the kite string. The equation used in this scenario is: \[ 50^2 + x^2 = 120^2 \]
  • The 50 ft represents the height of the kite.
  • The 120 ft is the length of the string, acting as the hypotenuse.
  • The horizontal distance "x" to be found is the kite's drift along the ground.
By utilizing this theorem, we can effectively solve for unknown parts of the triangle, which is essential in related rates problems when analyzing how changes in one part of the triangle affect another.
Right Triangle
A right triangle is a type of triangle that has one angle specifically set at 90 degrees. Knowing you are dealing with a right triangle not only helps in applying the Pythagorean Theorem but also in understanding the spatial relationships between different parts of the triangle.

In the context of the kite problem:
  • The right angle is crucial for defining the relationship between the ground (horizontal distance), the kite string, and the height.
  • The fixed height (50 ft) forms one of the triangle sides perpendicular to the ground.
  • The hypotenuse is the length of the string, which moves as the kite drifts horizontally.
Understanding the structure of a right triangle is essential because it simplifies the complexity of geometric relationships into predictable patterns, which are vital in related rate problems. By starting with this fundamental understanding, it becomes easier to visualize and solve the problem correctly.

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