Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

A child is flying a kite. If the kite is 90 feet above the child's hand level and the wind is blowing it on a horizontal course at 5 feet per second, how fast is the child paying out cord when 150 feet of cord is out? (Assume that the cord remains straight from hand to kite, actually an unrealistic assumption.)

Step-by-Step Solution

Answer

The child is paying out cord at 4 feet per second.

Step 1: Understand the Problem

We are given a right triangle formed by the height of the kite, the horizontal distance from the child to a point directly below the kite, and the length of the cord. We need to find the rate at which the cord is being let out when a certain amount of cord is out.

Step 2: Identify the Known Values

The height (vertical side) of the triangle is 90 feet, the length of the hypotenuse (cord) at that instant is 150 feet, and the horizontal rate of change is 5 feet per second.

Step 3: Relate the Quantities Using Pythagoras' Theorem

We use Pythagoras' Theorem: \[ x^2 + y^2 = z^2 \]where \(x\) is the horizontal distance, \(y\) is the height (90 feet), and \(z\) is the length of the cord (150 feet).

Step 4: Differentiate the Pythagorean Theorem

Differentiate both sides of \[ x^2 + y^2 = z^2 \]with respect to time \(t\). This gives us: \[ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt} \]Since the height \(y\) is constant, \(\frac{dy}{dt} = 0\).

Step 5: Solve for the unknown rate \(\frac{dz}{dt}\)

Plug in the known values from the setup:\[ x = \sqrt{150^2 - 90^2} = 120 \text{ feet} \]\( \frac{dx}{dt} = 5 \text{ feet per second} \)Substitute into the differentiated equation:\[ 2(120)(5) = 2(150)\frac{dz}{dt} \]\[ 1200 = 300\frac{dz}{dt} \]Solve for \(\frac{dz}{dt}\):\[ \frac{dz}{dt} = \frac{1200}{300} = 4 \text{ feet per second} \]

Step 6: Conclusion

The rate at which the child is paying out cord is 4 feet per second when the cord is 150 feet long.

Key Concepts

Right Triangles

A right triangle is a critical part of understanding this problem. In a right triangle, one of the angles is exactly 90 degrees. This type of triangle forms when you have one vertical and one horizontal line meeting at a point and connected by a diagonal line called the hypotenuse. This exercise involves a right triangle created by:

• The height of the kite (vertical side).
• The horizontal distance (from the child to a point directly beneath the kite).
• The hypotenuse (the kite's cord).

When visualizing or dealing with a problem involving right triangles, always identify these three sides, as they will guide you in setting up equations or applying mathematical concepts. Understanding this setup is fundamental in tackling problems related to the Pythagorean theorem.

Pythagorean Theorem

The Pythagorean theorem is an essential mathematical principle used to solve problems involving right triangles. It states that in a right triangle, the square of the hypotenuse (\( z^2 \)) is equal to the sum of the squares of the other two sides (\( x^2 + y^2 \)).This theorem can be expressed as:\[ x^2 + y^2 = z^2 \]In the problem, we apply it to find missing lengths. You are given the height of the kite and the length of the cord, so with the Pythagorean theorem, you can solve for the horizontal distance, which helps in analyzing the rate change.Knowing how to apply this theorem gives you the power to dissect many geometric problems, especially those involving distances and movement like in the kite problem.

Differentiation

Differentiation is a core aspect of calculus that helps us understand how things change. When applied to the Pythagorean theorem in this problem, differentiation assists in determining the rate at which the cord is being let out.By differentiating the Pythagorean equation:\[ x^2 + y^2 = z^2 \] with respect to time, you can track how each component of the triangle changes over time. This leads to the equation:\[ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt} \]Since the height \( y \) does not change (the kite remains at a constant height), \( \frac{dy}{dt} = 0 \), simplifying the equation to focus on the hypotenuse and horizontal distance. Differentiation in calculus opens up a world of possibilities for understanding dynamics like speed and rate of change in various contexts.

Applied Calculus

Applied calculus is the use of calculus methods to solve practical problems in everyday life, such as the kite problem. It often entails understanding how different quantities are interrelated and how they change over time.In the exercise, calculus helps us solve for the rate at which the string is being let out (\( \frac{dz}{dt} \)). Using the differentiated form of the Pythagorean theorem along with the known rates of change (like the horizontal movement of the kite), you can determine the rate of change for the hypotenuse.The beauty of applied calculus lies in its ability to provide solutions to real-world problems, whether it's calculating how fast a kite is moving or determining other rates of change you may encounter in engineering, physics, and beyond. Embracing these concepts might seem challenging at first, but they can demystify many phenomena occurring around us.