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 need to find out how fast the child is letting out the cord (rate of change of the length of the cord) when 150 feet of cord is out, given that the kite is moving horizontally at 5 feet per second.

Step 2: Visualize the scenario with a right triangle

We can visualize the problem with a right triangle where the horizontal distance from the child to the kite is one leg of the triangle, the height of the kite above the child's hand is the other leg, and the length of the cord is the hypotenuse.

Step 3: Assign variables

Let \( x \) represent the horizontal distance from the child to the kite, \( y \) represent the height of the kite (which is 90 feet), and \( z \) represent the length of the cord. The rate at which the horizontal distance is changing is \( \frac{dx}{dt} = 5 \) feet per second.

Step 4: Apply Pythagorean theorem

According to the Pythagorean theorem, we have the equation \( x^2 + y^2 = z^2 \). Since \( y \) is constant at 90 feet, the equation becomes \( x^2 + 90^2 = z^2 \).

Step 5: Differentiate with respect to time

Differentiate both sides of \( x^2 + 90^2 = z^2 \) with respect to time \( t \):\( 2x \frac{dx}{dt} + 0 = 2z \frac{dz}{dt} \).This simplifies to \( x \frac{dx}{dt} = z \frac{dz}{dt} \).

Step 6: Solve for \( \frac{dz}{dt} \)

We want \( \frac{dz}{dt} \). Thus, solving for it, we get:\[ \frac{dz}{dt} = \frac{x \frac{dx}{dt}}{z} \].

Step 7: Find \( x \) for \( z = 150 \)

Using the Pythagorean equation with \( z = 150 \), we solve for \( x \):\[ x^2 + 90^2 = 150^2 \]\[ x^2 + 8100 = 22500 \]\[ x^2 = 14400 \]\[ x = 120 \]Thus, the horizontal distance \( x \) is 120 feet.

Step 8: Calculate \( \frac{dz}{dt} \)

Substitute \( x = 120 \), \( z = 150 \), and \( \frac{dx}{dt} = 5 \) into the equation \( \frac{dz}{dt} = \frac{x \frac{dx}{dt}}{z} \):\[ \frac{dz}{dt} = \frac{120 \cdot 5}{150} \]\[ \frac{dz}{dt} = 4 \] feet per second.

Key Concepts

Pythagorean theorem

The Pythagorean theorem is a fundamental principle in geometry, especially useful in problems involving right triangles. This theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. It can be expressed as:\[x^2 + y^2 = z^2\]In our kite problem, the child, kite, and cord form a right triangle:

• One leg represents the horizontal distance from the child to the kite. Let's call this distance \(x\).
• The other leg is the height of the kite above the child's hand, which remains constant at 90 feet, denoted as \(y\).
• The hypotenuse is the length of the cord, denoted as \(z\).

This equation helps us understand the spatial relationship between the kite, its height, and the cord length. By using specific values for \(y\) and \(z\), we can calculate \(x\), thus facilitating further calculations when the kite moves.

Differentiation

Differentiation is a key concept in calculus that deals with how functions change with respect to variables, often involving rates of change. In problems involving related rates, like our kite example, differentiation helps us decipher how one variable impacts another over time.
When applying this concept to the kite problem, we first set up the Pythagorean theorem:\[x^2 + 90^2 = z^2\]Since the variables are functions of time, we differentiate both sides with respect to time \(t\):\[2x \frac{dx}{dt} + 0 = 2z \frac{dz}{dt}\]This step requires applying the chain rule, a vital differentiation technique. Simplifying this gives:\[x \frac{dx}{dt} = z \frac{dz}{dt}\]With known values for \(x\), \(z\), and \(\frac{dx}{dt}\), we solve for \(\frac{dz}{dt}\). Here, differentiation enables us to determine the rate at which the child pays out the cord based on the kite's horizontal speed.

Kinematics

Kinematics is a branch of physics that studies motion without considering the forces that cause it. This concept is applied in related rates problems where motion and change over time are evaluated, much like in our kite-flying exercise.
In the problem, the kite's motion is horizontal, and its speed is provided as 5 feet per second. This rate affects the change in length of the cord, making it a classic kinematics challenge.

• We know \(\frac{dx}{dt} = 5\) ft/s, representing the horizontal speed of the kite.
• We need to find \(\frac{dz}{dt}\), the rate at which the length of the cord is increasing.

Kinematics helps determine how changes in one aspect of the motion, such as horizontal distance, influence another, such as the cord length. By examining these variable changes over time, and using the right equations, we can accurately map the motion effects in a given scenario like the child's kite-flying activity.