Problem 2 - 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 \(\delta(x)\) . $$-3 \leq x \leq 3, \quad \delta(x)=1+3 x^{2}$$

Step-by-Step Solution

Answer

Mass is 60 units; center of mass is at \( x = 0 \).

Step 1: Determine the Mass of the Wire

The mass of the wire can be found using the integral of the density function over the given interval. The mass \( M \) is given by:\[ M = \int_{-3}^{3} \delta(x) \, dx = \int_{-3}^{3} (1 + 3x^2) \, dx. \]Let's calculate this integral step-by-step.

Step 2: Evaluate the Integral for Mass

Break the integral into two parts and solve:1. \( \int_{-3}^{3} 1 \, dx = [x]_{-3}^{3} = 3 - (-3) = 6 \).2. \( 3 \int_{-3}^{3} x^2 \, dx = 3 \left[ \frac{x^3}{3} \right]_{-3}^{3} = [x^3]_{-3}^{3} = 3^3 - (-3)^3 = 27 + 27 = 54 \).Combine results: \[ M = 6 + 54 = 60. \]

Step 3: Determine the Center of Mass

The center of mass \( \overline{x} \) is given by:\[ \overline{x} = \frac{1}{M} \int_{-3}^{3} x \cdot \delta(x) \, dx = \frac{1}{60} \int_{-3}^{3} x(1 + 3x^2) \, dx. \]Let's calculate this integral to find \( \overline{x} \).

Step 4: Evaluate the Integral for Center of Mass

Expand and solve the integral:1. \( \int_{-3}^{3} x \, dx = \left[ \frac{x^2}{2} \right]_{-3}^{3} = 0 - 0 = 0 \).2. \( 3 \int_{-3}^{3} x^3 \, dx = 3 \left[ \frac{x^4}{4} \right]_{-3}^{3} = 3\left(\frac{3^4}{4} - \frac{(-3)^4}{4}\right) = 3(0) = 0 \).Thus, the integral evaluates to zero:\[ \int_{-3}^{3} x(1+3x^2) \, dx = 0. \]Hence, \[ \overline{x} = \frac{1}{60} \times 0 = 0. \]

Key Concepts

Mass

When we talk about mass, we're referring to the amount of matter contained in an object. For a wire or any linear object, this can be described by a mathematical quantity obtained through calculus. Here, the mass is calculated by integrating its density function \(\delta(x)\) over a specified interval.
For example, given the density function \(\delta(x) = 1 + 3x^2 \), over the interval from \(-3\) to \(3\), the mass (\(M\)) is determined as follows:

• First, integrate the constant in the density function, \(\int_{-3}^{3} 1 \, dx\), resulting in 6.
• Then, integrate the variable term, \(3\int_{-3}^{3} x^2 \, dx\), resulting in 54.
• Add these two results together to find the total mass: \ M=6+54=60 \.

The ability to calculate the mass through integration allows us to understand how mass is distributed across an object, especially when the density is not uniform.

Density Function

The density function \(\delta(x)\) characterizes how mass is distributed over a particular region. Specifically for linear objects like wires, the density function describes the mass per unit length along the wire.
In the exercise, you are given a density function \(\delta(x) = 1 + 3x^2\). This function indicates that the density increases with the square of the distance x from the center point.

• The term \(1\) can be thought of as a base density across the wire.
• The term \(3x^2\) means that as you move away from the center, the density increases quadratically.

Knowing the density function is crucial as it directly impacts calculations of mass and center of mass. If the density were constant, the calculations would be simpler. However, with a function, calculus techniques like integration become essential to figure out the actual distribution and quantities of mass.

Integral Calculus

Integral calculus is a vital tool in finding the mass and center of mass for objects with variable densities. It helps accumulate small quantities continuously, like tiny mass parts of a wire, to find the whole.
In the exercise, integral calculus was used to find both the mass and the center of mass:

• For mass, integrals sum up the tiny bits of mass, given by the density function, over the length of the wire.
• For the center of mass, the integral takes into account each segment's position and density, to find a balance point or average position. This is often signified by the integral \(\int x \delta(x) \, dx\).

Without integral calculus, accurately determining these properties for non-uniformly dense objects would be nearly impossible. This powerful mathematical tool turns complex physical descriptions into manageable computations, helping us understand and predict the behavior of physical systems.