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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Question
![**Problem Statement:**
Find \(\int_C x \, ds\), where \(C\) is the line segment in the plane joining \((0, 0)\) and \((3, 4)\).
---
**Solution Explanation:**
To solve this line integral, we'll follow these steps:
1. **Parametrize the Line Segment:**
- We need to find a parametrization for the line segment \(C\).
- The line segment can be parametrized by \(\mathbf{r}(t) = (1-t)(0, 0) + t(3, 4)\) for \(t \in [0, 1]\).
- This simplifies to \(\mathbf{r}(t) = (3t, 4t)\).
2. **Compute \(ds\):**
- The differential arc length \(ds\) is given by \(|\mathbf{r}'(t)| \, dt\).
- Calculate the derivative: \(\mathbf{r}'(t) = (3, 4)\).
- The magnitude is \(|\mathbf{r}'(t)| = \sqrt{3^2 + 4^2} = 5\).
- Thus, \(ds = 5 \, dt\).
3. **Express \(x\) in Terms of \(t\):**
- From \(\mathbf{r}(t) = (3t, 4t)\), we have \(x = 3t\).
4. **Evaluate the Integral:**
\[
\int_C x \, ds = \int_0^1 3t \cdot 5 \, dt = 15 \int_0^1 t \, dt
\]
- Compute \(\int_0^1 t \, dt\):
\[
\int_0^1 t \, dt = \left[ \frac{t^2}{2} \right]_0^1 = \frac{1}{2}
\]
- Substitute back:
\[
15 \int_0^1 t \, dt = 15 \cdot \frac{1}{2} = \frac{15}{2} = 7.5
\]
Thus, the value of the line integral is \(7.5\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F358deae0-b7d2-45f9-8056-32e7a5350251%2Fe8a412c5-ca16-414f-a262-1371d85a3818%2F1e8ptih_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find \(\int_C x \, ds\), where \(C\) is the line segment in the plane joining \((0, 0)\) and \((3, 4)\).
---
**Solution Explanation:**
To solve this line integral, we'll follow these steps:
1. **Parametrize the Line Segment:**
- We need to find a parametrization for the line segment \(C\).
- The line segment can be parametrized by \(\mathbf{r}(t) = (1-t)(0, 0) + t(3, 4)\) for \(t \in [0, 1]\).
- This simplifies to \(\mathbf{r}(t) = (3t, 4t)\).
2. **Compute \(ds\):**
- The differential arc length \(ds\) is given by \(|\mathbf{r}'(t)| \, dt\).
- Calculate the derivative: \(\mathbf{r}'(t) = (3, 4)\).
- The magnitude is \(|\mathbf{r}'(t)| = \sqrt{3^2 + 4^2} = 5\).
- Thus, \(ds = 5 \, dt\).
3. **Express \(x\) in Terms of \(t\):**
- From \(\mathbf{r}(t) = (3t, 4t)\), we have \(x = 3t\).
4. **Evaluate the Integral:**
\[
\int_C x \, ds = \int_0^1 3t \cdot 5 \, dt = 15 \int_0^1 t \, dt
\]
- Compute \(\int_0^1 t \, dt\):
\[
\int_0^1 t \, dt = \left[ \frac{t^2}{2} \right]_0^1 = \frac{1}{2}
\]
- Substitute back:
\[
15 \int_0^1 t \, dt = 15 \cdot \frac{1}{2} = \frac{15}{2} = 7.5
\]
Thus, the value of the line integral is \(7.5\).
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