Problem 4 - 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 \(\delta(x)\) . $$1 \leq x \leq 2, \quad \delta(x)=\frac{8}{x^{3}}$$

Step-by-Step Solution

Answer

The mass is 3, and the center of mass is \( \frac{4}{3} \).

Step 1: Understand the Formula for Mass

The mass of the wire, denoted as \( M \), is calculated by integrating the density function \( \delta(x) \) over the given interval \([a, b]\). The formula for the mass is given by \[ M = \int_{a}^{b} \delta(x) \, dx. \]

Step 2: Set Up the Mass Integral

Here, the interval is from \( x = 1 \) to \( x = 2 \), and the density function is \( \delta(x) = \frac{8}{x^3} \). Thus, the integral to find the mass is: \[ M = \int_{1}^{2} \frac{8}{x^3} \, dx. \]

Step 3: Compute the Mass Integral

To integrate \( \frac{8}{x^3} \), rewrite the integrand as \( 8x^{-3} \). The antiderivative is \( -4x^{-2} \), thus: \[ M = \left[-4x^{-2} \right]_{1}^{2} = \left(-4 \cdot 2^{-2}\right) - \left(-4 \cdot 1^{-2}\right) = -1 + 4 = 3. \] Therefore, the mass \( M = 3 \).

Step 4: Understand the Formula for Center of Mass

The center of mass \( \overline{x} \) of the wire is calculated using the formula \[ \overline{x} = \frac{1}{M} \int_{a}^{b} x \cdot \delta(x) \, dx. \] This involves an integral of the product of \( x \) and the density function.

Step 5: Set Up the Integral for Center of Mass

Given \( x \) and \( \delta(x) = \frac{8}{x^3} \), the integral becomes: \[ \int_{1}^{2} x \cdot \frac{8}{x^3} \, dx = \int_{1}^{2} \frac{8}{x^2} \, dx. \]

Step 6: Compute the Center of Mass Integral

To integrate \( \frac{8}{x^2} \), rewrite the integrand as \( 8x^{-2} \). The antiderivative is \( -8x^{-1} \), thus: \[ \overline{x} = \frac{1}{3} \left[-8x^{-1}\right]_{1}^{2} = \frac{1}{3} \left(-8 \cdot 2^{-1} + 8 \cdot 1^{-1}\right) = \frac{1}{3} \left(-4 + 8\right) = \frac{4}{3}. \] Hence, the center of mass \( \overline{x} = \frac{4}{3} \).

Key Concepts

Mass Calculation

In calculus, mass calculation for objects like a wire on a specific interval involves integrating the density function over the specified range. Here, we are tasked with finding the mass of a wire spanning from 1 to 2 units with a density function \( \delta(x) = \frac{8}{x^3} \). To find its mass \( M \), we use the integral:

• Formula: \( M = \int_{a}^{b} \delta(x) \, dx \)
• Given Interval: 1 to 2
• Density Function: \( \delta(x) = \frac{8}{x^3} \)

Rewriting the integrand as \( 8x^{-3} \), the antiderivative is \( -4x^{-2} \). By evaluating this from 1 to 2, we find:\[ M = \left[ -4x^{-2} \right]_{1}^{2} = -1 + 4 = 3. \]Thus, the mass \( M \) of the wire is 3.

Center of Mass

The center of mass represents the balancing point of the wire if it were to be hung horizontally. Calculating the center of mass \( \overline{x} \) involves finding the weighted average position, using both the density of the wire and its position along the x-axis.

• Formula: \( \overline{x} = \frac{1}{M} \int_{a}^{b} x \cdot \delta(x) \, dx \)
• For this wire: Use \( \delta(x) = \frac{8}{x^3} \)

First, we set up the integral by multiplying \( x \) with the density function: \( x \cdot \frac{8}{x^3} = \frac{8}{x^2} \).We rewrite it again as \( 8x^{-2} \), whose antiderivative is \( -8x^{-1} \). Calculating this from 1 to 2 gives:\[ \overline{x} = \frac{1}{3} \left[ -8x^{-1} \right]_{1}^{2} = \frac{1}{3}(-4 + 8) = \frac{4}{3}. \]Therefore, the center of mass of the wire is \( \frac{4}{3} \).

Density Function

A density function defines how mass is distributed along an object, such as our wire here. It varies with position, which in our given problem is along the x-axis.

• Example Density Function Used: \( \delta(x) = \frac{8}{x^3} \)
• This implies that the density decreases significantly as x increases.

Such functions are crucial because they offer insights into how an object's mass isn't uniformly distributed. The density function also impacts both mass calculation and the determination of the center of mass by providing the necessary component for integration within the given interval. Knowing the density function allows us to assess exactly where mass is concentrated or sparse, ultimately affecting physical behaviors and properties like balance and stability.