Problem 6 - Thomas Calculus

Question

In Exercises \(1-6,\) find the mass \(M\) and center of mass \(\overline{x}\) of the linear wire covering the given interval and having the given density \(\delta(x)\) . $$\delta(x)=\left\\{\begin{array}{cc}{2-x,} & {0 \leq x<1} \\ {x,} & {1 \leq x \leq 2}\end{array}\right.$$

Step-by-Step Solution

Answer

Mass \(M = 3\), Center of Mass \(\overline{x} = 1\).

Step 1: Understand the Problem

We need to find the mass \(M\) and the center of mass \(\overline{x}\) of a linear wire. The density is piecewise-defined, with \(\delta(x) = 2-x\) for \(0 \leq x < 1\) and \(\delta(x) = x\) for \(1 \leq x \leq 2\).

Step 2: Find the Mass of the Wire

To find the mass \(M\), we integrate the density function over its entire range. This involves computing two integrals: one for \(0 \leq x < 1\) and another for \(1 \leq x \leq 2\).For \(0 \leq x < 1\): \[ M_1 = \int_{0}^{1} (2-x) \, dx \]For \(1 \leq x \leq 2\): \[ M_2 = \int_{1}^{2} x \, dx \]The total mass \(M = M_1 + M_2\).

Step 3: Calculate Each Integral for Mass

Let's compute each integral separately.For \(M_1\):\[M_1 = \int_{0}^{1} (2-x) \, dx = \left[ 2x - \frac{x^2}{2} \right]_{0}^{1} = (2(1) - \frac{1}{2}) - (2(0) - \frac{0}{2}) = \frac{3}{2}.\]For \(M_2\):\[M_2 = \int_{1}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{1}^{2} = \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{3}{2}.\]Thus, \(M = \frac{3}{2} + \frac{3}{2} = 3.\)

Step 4: Find the Center of Mass \(\overline{x}\)

The center of mass \(\overline{x}\) is given by the formula:\[\overline{x} = \frac{1}{M} \int_{a}^{b} x \cdot \delta(x) \, dx\]which breaks into two parts:\[\overline{x}_1 = \frac{1}{M} \int_{0}^{1} x(2-x) \, dx\]\[\overline{x}_2 = \frac{1}{M} \int_{1}^{2} x^2 \, dx\]\(\overline{x} = \overline{x}_1 + \overline{x}_2\).

Step 5: Calculate the Integrals for \(\overline{x}\)

Compute each integral:For \(\overline{x}_1\):\[\overline{x}_1 = \frac{1}{3} \int_{0}^{1} (2x - x^2) \, dx = \frac{1}{3} \left[ x^2 - \frac{x^3}{3} \right]_{0}^{1}\]\[= \frac{1}{3} \left( 1 - \frac{1}{3} \right) = \frac{1}{3} \times \frac{2}{3} = \frac{2}{9}.\]For \(\overline{x}_2\):\[\overline{x}_2 = \frac{1}{3} \int_{1}^{2} x^2 \, dx = \frac{1}{3} \left[ \frac{x^3}{3} \right]_{1}^{2}\]\[= \frac{1}{3} \left( \frac{8}{3} - \frac{1}{3} \right) = \frac{1}{3} \times \frac{7}{3} = \frac{7}{9}.\]So, \(\overline{x} = \frac{2}{9} + \frac{7}{9} = 1.\)

Key Concepts

Piecewise Functions

Piecewise functions are a type of mathematical function defined by different expressions depending on the region of the domain. This means that the function takes different forms, allowing us to represent complex situations in real-world problems. In the exercise, the density function \(\delta(x)\) is a piecewise function. It is defined by two different expressions based on the interval \(x\) belongs to.
This particular function can be expressed as:

• \(2-x\) for \(0 \leq x < 1\)
• \(x\) for \(1 \leq x \leq 2\)

Why Use Piecewise Functions?
Piecewise functions are used because they can model behaviors or systems that cannot be captured by a single mathematical expression. In our case, the density changes depending on the position along the wire, which is best described using two separate equations.

When dealing with piecewise functions, it is crucial to:

• Identify each segment of the domain.
• Understand how the expressions differ over the segments.
• Ensure continuity or account for transitions between pieces.

Density Functions

The concept of a density function plays a fundamental role in this type of problem. It describes how the mass is distributed along a length, area, or volume. For a one-dimensional object, such as a wire, the density function \(\delta(x)\) gives the mass per unit length at each point \(x\).
The purpose here is to determine how much mass is located at different points on the wire. Here, the exercise uses a piecewise density function that changes values over its domain.
Important Considerations:
For computation, it is necessary to:

• Express the density in terms of \(x\).
• Calculate the mass by integrating the density over the given interval.

The integration of these density functions over specific intervals enables us to determine key physical properties like the total mass and the center of mass of an object. Real World Application
Density functions are used in fields like material science and engineering to understand how different materials react under various conditions. For example, knowing how density varies across a material helps engineers design more efficient structures.

Integral Calculus

Integral calculus is essential for solving many real-world problems involving accumulated quantities over a continuous range. It allows us to sum up small changes, giving us a total value, whether it be area, mass, or another quantity. In our exercise, integral calculus is used to find both the mass and center of mass of the wire.

Steps in Integral Calculus Applied Here:

• Set up the integral of the density function over the specified intervals.
• Compute the integral to find the mass \(M\), first for each interval and then sum them.
• Use these integrals for determining the center of mass, which involves calculating weighted averages using the density function.

With piecewise functions, you need to solve separate integrals for each defined segment. Once each integral is calculated, summing the results or applying weighted formulas helps achieve the desired physical property, like the full mass or center of mass.

The Bottom Line
Understanding how integral calculus connects various concepts like mass and density allows us not only to solve mathematical problems but also to connect them to physical theories and applications. By breaking down complicated functions into simpler pieces, we can model the real world more accurately.