Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Problem Statement:**
Find the center of mass of the thin rod lying along the \( x \)-axis from \( x = 0 \) to \( x = 3 \) with density \( \delta(x) = x + 2 \).
**Solution:**
1. **Determine the Mass of the Rod:**
To find the center of mass, we first need to calculate the total mass \( M \) of the rod. The mass can be found by integrating the density function \( \delta(x) \) over the length of the rod:
\[
M = \int_{0}^{3} \delta(x) \, dx = \int_{0}^{3} (x + 2) \, dx
\]
Perform the integration:
\[
M = \int_{0}^{3} x \, dx + \int_{0}^{3} 2 \, dx
\]
The integrals can be solved as follows:
\[
\int_{0}^{3} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{3} = \frac{3^2}{2} - \frac{0^2}{2} = \frac{9}{2}
\]
\[
\int_{0}^{3} 2 \, dx = 2x \Big|_{0}^{3} = 2(3) - 2(0) = 6
\]
Therefore, the total mass of the rod is:
\[
M = \frac{9}{2} + 6 = \frac{9}{2} + \frac{12}{2} = \frac{21}{2}
\]
2. **Determine the Center of Mass:**
The center of mass \( \bar{x} \) of the rod can be found using the formula:
\[
\bar{x} = \frac{1}{M} \int_{0}^{3} x \delta(x) \, dx
\]
Substitute \( \delta(x) = x + 2 \):
\[
\bar{x} = \frac{1}{M} \int_{0}^{3} x (x + 2) \, dx = \frac{1}{M} \int_{0}^{3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1e2c0fb4-fe7c-46b9-9b1b-ce7222ffa53b%2F1d599f4b-f09a-46c4-9665-5979d970116c%2Fjjohyt_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the center of mass of the thin rod lying along the \( x \)-axis from \( x = 0 \) to \( x = 3 \) with density \( \delta(x) = x + 2 \).
**Solution:**
1. **Determine the Mass of the Rod:**
To find the center of mass, we first need to calculate the total mass \( M \) of the rod. The mass can be found by integrating the density function \( \delta(x) \) over the length of the rod:
\[
M = \int_{0}^{3} \delta(x) \, dx = \int_{0}^{3} (x + 2) \, dx
\]
Perform the integration:
\[
M = \int_{0}^{3} x \, dx + \int_{0}^{3} 2 \, dx
\]
The integrals can be solved as follows:
\[
\int_{0}^{3} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{3} = \frac{3^2}{2} - \frac{0^2}{2} = \frac{9}{2}
\]
\[
\int_{0}^{3} 2 \, dx = 2x \Big|_{0}^{3} = 2(3) - 2(0) = 6
\]
Therefore, the total mass of the rod is:
\[
M = \frac{9}{2} + 6 = \frac{9}{2} + \frac{12}{2} = \frac{21}{2}
\]
2. **Determine the Center of Mass:**
The center of mass \( \bar{x} \) of the rod can be found using the formula:
\[
\bar{x} = \frac{1}{M} \int_{0}^{3} x \delta(x) \, dx
\]
Substitute \( \delta(x) = x + 2 \):
\[
\bar{x} = \frac{1}{M} \int_{0}^{3} x (x + 2) \, dx = \frac{1}{M} \int_{0}^{3
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