Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

A straight wire 7 units long has density \(\delta(x)=\sqrt{x}\) at a point \(x\) units from one end. Find the distance from this end to the center of mass.

Step-by-Step Solution

Answer

The center of mass is 4.2 units from one end of the wire.

Step 1: Understanding the Problem

We need to find the center of mass of a wire that stretches from 0 to 7 units in length. The density of the wire is given as a function of position, \( \delta(x) = \sqrt{x} \), where \( x \) is the distance from one end of the wire.

Step 2: Define the Center of Mass Formula

The center of mass \( \bar{x} \) for a wire with variable density from \( x = a \) to \( x = b \) is given by \( \bar{x} = \frac{\int_a^b x \delta(x)\, dx}{\int_a^b \delta(x)\, dx} \). Here \( a = 0 \) and \( b = 7 \).

Step 3: Calculate the Denominator

The denominator is the total mass of the wire, calculated by \( \int_0^7 \sqrt{x} \, dx \). This integral evaluates as follows: \[\frac{2}{3}x^{3/2}\bigg|_0^7 = \frac{2}{3}(7^{3/2} - 0^{3/2}) = \frac{2}{3} \times 7^{3/2}.\]

Step 4: Calculate the Numerator

The numerator involves the integral \( \int_0^7 x \sqrt{x} \, dx = \int_0^7 x^{3/2} \, dx \). The antiderivative is \( \frac{2}{5}x^{5/2} \), which gives: \[\frac{2}{5}x^{5/2} \bigg|_0^7 = \frac{2}{5}(7^{5/2} - 0^{5/2}) = \frac{2}{5} \times 7^{5/2}.\]

Step 5: Solve for Center of Mass

Substitute the results of the integrals back into the center of mass formula: \( \bar{x} = \frac{\frac{2}{5} \times 7^{5/2}}{\frac{2}{3} \times 7^{3/2}} = \frac{3}{5} \times 7 = \frac{21}{5} = 4.2.\) The distance from one end to the center of mass is 4.2 units.

Key Concepts

Center of Mass

The center of mass is a crucial concept in physics and engineering. Imagine a long thin wire with variable thickness. We want to find the point where this wire balances perfectly if supported just at that point. This balancing point is known as the center of mass. For objects with uniform density, it aligns with the geometric center. However, with variable density, like our wire, it shifts.
To find the center of mass for a linear object like a wire, we use the center of mass formula:

• Calculate the entire mass of the object with an integral.
• Then find the weighted position of the mass by multiplying the position by its density at each point and then integrating.

This approach gives us the exact location of the center of mass by balancing the distribution of mass along the object's length.

Variable Density

When dealing with objects in calculus, density often varies across the object. Instead of being uniform, density might change at different points. In our problem, the density of the wire changes with respect to its length, described by the function \( \delta(x) = \sqrt{x} \).
Variable density affects how mass is distributed along the wire. This influences the center of mass's position. Calculating properties like total mass and center of mass with variable density requires integrating over the length of the object.
Given the function, integrate the density over the region to find the total mass. Then use this to evaluate where the mass is centered, adjusting for density changes.

Definite Integrals

Definite integrals are fundamental in calculus for calculating precise quantities like area, volume, and mass. With our wire, we use definite integrals to determine the mass along with the position where it is centered.
A definite integral over a specific range of a function gives us the accumulated value of that function between two points. For example, \(\int_0^7 \sqrt{x} \, dx \)computes the total mass of the wire. Additionally, \( \int_0^7 x \cdot \sqrt{x} \, dx \)evaluates the contribution of each piece of the wire's length towards the center of mass.
Through these integrals, we can precisely calculate significant characteristics of objects, particularly when dealing with variable density.

Mathematical Modeling

Mathematical modeling is the process of describing real-world phenomena with mathematical concepts to solve problems more effectively. In our exercise, we model the wire with math to predict and determine its center of mass.
We utilize a function to represent the wire's variable density. This function \( \delta(x) = \sqrt{x} \)serves as a model explaining how density varies along the wire's length. By integrating this function, we model the entire system mathematically, finding crucial features like total mass and the center of mass.

• Translate physical characteristics into mathematical equations.
• Solve equations using calculus to derive meaningful answers.

This approach leverages mathematics to simplify complex physical situations, advancing both our understanding and operational capabilities.