Problem 19 - Calculus: Early Transcendentals

Question

The linear density in a rod 8 m long is \( \frac{12}{\sqrt{x + 1}} \) kg/m, where \( x \) is measured in meters from one end of the rod. Find the average density of the rod.

Step-by-Step Solution

Answer

The average density of the rod is 6 kg/m.

Step 1: Understand the Problem

We need to find the average density of a rod with a variable linear density function given by \( f(x) = \frac{12}{\sqrt{x + 1}} \) kg/m, where \( x \) ranges from 0 to 8 meters.

Step 2: Recall the Formula for Average Density

The formula for the average density \( \bar{f} \) over an interval \([a, b]\) is given by: \[ \bar{f} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \] where \( f(x) \) is the density function.

Step 3: Set Up the Integral

Substitute \( a = 0 \), \( b = 8 \), and \( f(x) = \frac{12}{\sqrt{x + 1}} \) into the average density formula: \[ \bar{f} = \frac{1}{8-0} \int_{0}^{8} \frac{12}{\sqrt{x + 1}} \, dx \]

Step 4: Simplify and Evaluate the Integral

First, simplify the expression: \[ \bar{f} = \frac{1}{8} \int_{0}^{8} \frac{12}{\sqrt{x + 1}} \, dx \] Next, solve the integral using substitution. Let \( u = x + 1 \), then \( du = dx \) and change the limits from \( x = 0 \) to \( u = 1 \), and \( x = 8 \) to \( u = 9 \). The integral becomes: \[ \int_{1}^{9} \frac{12}{\sqrt{u}} \, du = 12 \int_{1}^{9} u^{-1/2} \, du \] Integrate: \[ = 12 \left[ 2u^{1/2} \right]_{1}^{9} = 24 [\sqrt{9} - \sqrt{1}] = 24 [3 - 1] = 48 \]

Step 5: Calculate the Average Density

Finally, compute the average density: \[ \bar{f} = \frac{1}{8} \times 48 = 6 \] kg/m.

Key Concepts

Linear Density Function

Linear density function is a way to describe how mass is distributed along a certain length of an object. In this case, the linear density function provided is \( f(x) = \frac{12}{\sqrt{x + 1}} \) kg/m. It tells us the density at any point \( x \) along the 8-meter rod. Linear density is crucial in understanding weight distribution, as not all materials are uniformly dense.
The important part is how this function helps us discover other properties of the rod, like its total mass or average density. By knowing the density at various points, we can integrate over the length to find these properties.

Definite Integral

The definite integral is a powerful mathematical tool used to calculate the accumulation of a quantity, such as area, volume, or, in this case, mass over a length. We use the definite integral to evaluate the total mass of the rod as the integration of its density function over the given interval.
Specifically, the formula for the average density over an interval \([a, b]\) is given by:

• \( \bar{f} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \)

This formula calculates the average density by summing the varying densities along the rod and then dividing by the rod's length (\({b-a}\)). It transforms our problem from one about densities to one about accumulations.

Substitution Method

The substitution method is a technique used to simplify an integral, making it easier to evaluate. It's especially helpful when dealing with composite functions, like \( \frac{12}{\sqrt{x + 1}} \). This method involves selecting a substitution that simplifies the integrand. In our exercise, we chose:

• Let \( u = x + 1 \)
• Then, \( du = dx \)
• Change limits: from \( x = 0 \) to \( u = 1 \), and \( x = 8 \) to \( u = 9 \)

Now, the integral becomes \( \int_{1}^{9} \frac{12}{\sqrt{u}} \, du \), which simplifies the problem significantly.
This method not only eases the calculation but also ensures accuracy in solving the integral.

Mathematical Problem Solving

Mathematical problem solving is akin to detective work, involving identifying the problem, gathering necessary formulas, and logically applying methods to reach a solution. The process in resolving our average density example entails:

• Understanding the problem: recognizing the role of each variable and function.
• Utilizing proper formulas: applying the average density and integral formulas effectively.
• Stepwise execution: breaking down the steps methodically to ensure each calculation is correct.

This problem-solving approach is crucial for handling complex problems, as it builds a bridge between understanding and practice. By following a structured method, we gain a deeper understanding and improve our ability to solve similar mathematical issues in the future.