Problem 224 - Calculus Volume 1

Question

For the following exercises, find the mass of the onedimensional object. A wire that is \(2 \mathrm{ft}\) long (starting at \(x=0\) ) and has a density function of \(\rho(x)=x^{2}+2 x \mathrm{lb} / \mathrm{ft}\).

Step-by-Step Solution

Answer

The mass of the wire is \(\frac{20}{3}\ lb\).

Step 1: Understand the Problem

We need to find the mass of a 1-dimensional object (a wire) of length 2 ft, given a density function. The density function is \(\rho(x) = x^2 + 2x\ lb/ft\), and the wire extends from \(x = 0\) to \(x = 2\).

Step 2: Set Up the Integral for Mass

The mass \(M\) of the wire can be found by integrating the density function over the length of the wire. The formula for mass is \(M = \int_{a}^{b} \rho(x) \, dx\), where \([a, b]\) is the interval over which the wire extends. For this problem, we need \(M = \int_{0}^{2} (x^2 + 2x) \, dx\).

Step 3: Compute the Integral

Calculate the integral \(\int_{0}^{2} (x^2 + 2x) \, dx\).Integrating the function term by term:\[\int (x^2 + 2x) \, dx = \int x^2 \, dx + \int 2x \, dx\]First term: \[\int x^2 \, dx = \frac{x^3}{3} + C\]Second term: \[\int 2x \, dx = x^2 + C\]

Step 4: Evaluate the Definite Integral

Substitute the limits of integration into the antiderivative and subtract:\[\left[\frac{x^3}{3} + x^2\right]_{0}^{2} = \left(\frac{2^3}{3} + 2^2\right) - \left(\frac{0^3}{3} + 0^2\right)\]Calculate:- \(\frac{8}{3} + 4 = \frac{8}{3} + \frac{12}{3} = \frac{20}{3}\)Thus, the mass is \(\frac{20}{3}\ lb\).

Key Concepts

Mass Calculation

In physical terms, the mass of an object refers to the amount of matter it contains. Calculating mass is essential when dealing with objects like wires or rods, as it helps in understanding their properties and behaviors. For one-dimensional objects, such as a wire, the mass is determined by its density and length.
The density function, \(\rho(x) = x^2 + 2x\ lb/ft\), represents how mass per unit length varies across the wire from one end to the other. Here, \(x\) is the position along the wire, starting from \(x=0\) to \(x=2\).
The formula for mass, \(M\), of a wire or rod is expressed as:

• \(M = \int_{a}^{b} \rho(x) \, dx\)

where \(a\) and \(b\) are the start and end points in the object's length. We use integration to sum up the contributions of the continuously varying density across the entire length of the wire. This approach provides an accurate representation of the total mass.

Definite Integral

The concept of a definite integral is central to solving problems involving continuous quantities such as area under a curve or total mass, as seen in this wire example. A definite integral is a tool that gives us the exact accumulation of quantities over a specified interval.
In our example, the wire stretches from \(x = 0\) to \(x = 2\), and we use a definite integral to compute the mass over this length. The notation often seen is:

• \(\int_{a}^{b} f(x) \, dx\)

This symbol conveys starting at point \(a\), integrating along to point \(b\), with \(f(x)\) representing the function being integrated.
Evaluating a definite integral means finding the antiderivative of the function, then substituting the upper and lower limits and finding the difference. This process gives a precise and exact total, like the mass of our wire being calculated as \(\frac{20}{3} \, lb\).

Integration Techniques

Integration is the mathematical process of finding an antiderivative, or the reversed process of differentiation. It's a critical component for solving problems involving curves, like determining the total mass of an object with a non-uniform density function.
One integration technique used in calculating the mass of the wire was the term-by-term method. We broke down the density function \(x^2 + 2x\) into two simpler parts: \(x^2\) and \(2x\). This allows us to integrate each part separately:

• The integral of \(x^2\) becomes \(\frac{x^3}{3}\)
• The integral of \(2x\) becomes \(x^2\)

After integrating, we evaluated the definite integral by substituting the limits from \(x = 0\) to \(x = 2\), finding the antiderivative values at these points, subtracting them, and obtaining the mass as \(\frac{20}{3} \, lb\).
Such integration techniques simplify and solve what initially appears as complex mathematical problems.