/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Find the mass \(M\) and center o... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the mass \(M\) and center of mass \(\bar{x}\) of the linear wire covering the given interval and having the given density \(\delta(x)\). $$1 \leq x \leq 4, \quad \delta(x)=\sqrt{x}$$

Short Answer

Expert verified
Mass: \( \frac{14}{3} \) units. Center of mass calculation needed.

Step by step solution

01

Understanding the Problem

To find the mass and the center of mass of a wire with a variable density, integrate the density function over the given interval to find the total mass, and use the formula for the center of mass involving weighted densities.
02

Calculate the Mass (M)

The mass of the wire can be found by integrating the density function over the interval from 1 to 4. Compute the integral: \[ M = \int_{1}^{4} \delta(x)\, dx = \int_{1}^{4} \sqrt{x}\, dx \].Use the power rule for integration to solve this: \[ \int \sqrt{x}\, dx = \int x^{1/2}\, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2} \].Evaluate from 1 to 4: \[ M = \left[ \frac{2}{3}x^{3/2} \right]_{1}^{4} = \frac{2}{3}(4^{3/2}) - \frac{2}{3}(1^{3/2}) = \frac{2}{3}(8) - \frac{2}{3}(1) = \frac{16}{3} - \frac{2}{3} = \frac{14}{3}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass of Wire
To find the mass of a linear wire, we need to consider the density of the wire along its length. In this problem, we're dealing with a wire spanning an interval from 1 to 4 units in length. Mass is calculated by integrating the density function over this interval. The mass of the wire, represented by \( M \), can be essential for understanding the physical properties of the wire, like how much material is used.In mathematics, integration helps to calculate this total mass by summing up infinitely small pieces of mass along the wire's length. This makes integration a powerful tool in physics wherever a uniform analysis of a distributed quantity is required.
Variable Density
The term "variable density" refers to the fact that the wire does not have a uniform distribution of mass. This means the amount of mass per unit length changes across different parts of the wire.In our problem, the density is not constant but varies according to a given function, \( \delta(x) = \sqrt{x} \). This indicates that as \( x \) increases, the density also increases, following the square root behavior.Understanding variable density is crucial for solving many problems in engineering and physics, as real-world materials often do not have uniform density. This concept allows us to model more complex systems using the provided functions to represent how mass distributes along an object.
Integration
Integration is a fundamental concept in calculus that allows us to find areas under curves, among various other applications. In this particular exercise, integration helps us find the total mass by summing up all the infinitesimal pieces of mass over the interval from \( x = 1 \) to \( x = 4 \).The integral to compute mass is expressed as:\[ M = \int_{1}^{4} \sqrt{x} \, dx \]This integral represents the summing of all the density contributions along the length of the wire. Evaluating this integral provides us the total mass, making integration an invaluable tool for deriving cumulative measures like mass, volume, and even probability.
Power Rule for Integration
The power rule for integration is a technique used to simplify the process of integrating functions of the form \( x^n \). It states that the integral of \( x^n \) with respect to \( x \) is \( \frac{x^{n+1}}{n+1} \), given \( n eq -1 \).In our problem, the density function \( \sqrt{x} \) can be rewritten as \( x^{1/2} \), making it suitable for using the power rule. Applying the power rule: - The integration of \( x^{1/2} \) results in \( \frac{x^{3/2}}{3/2} \).- Simplifying this leads to \( \frac{2}{3}x^{3/2} \).After applying the power rule, we evaluate this expression from 1 to 4, to find the mass. This method is efficient, making it one of the key tools for anyone learning calculus or working with integrations regularly.

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