Consider F and C below. F(x, y) = 6xy? i + 6x²y j (re), e + cos{ nt), Gnt)), osts1 C: r(t) = (t + sin (a) Find a function f such that F = Vf. f(x, у) %3D (b) Use part (a) to evaluate Vf• dr along the given curve C.
Consider F and C below. F(x, y) = 6xy? i + 6x²y j (re), e + cos{ nt), Gnt)), osts1 C: r(t) = (t + sin (a) Find a function f such that F = Vf. f(x, у) %3D (b) Use part (a) to evaluate Vf• dr along the given curve C.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Vector Fields and Line Integrals**
### Consider \( \mathbf{F} \) and \( C \) below.
\[ \mathbf{F}(x, y) = 6xy^2 \ \mathbf{i} + 6x^2y \ \mathbf{j} \]
\[ C : \mathbf{r}(t) = \langle t + \sin\left(\frac{1}{2}\pi t\right), t + \cos\left(\frac{1}{2}\pi t\right) \rangle, \quad 0 \leq t \leq 1 \]
### Exercises:
**(a) Find a function \( f \) such that \( \mathbf{F} = \nabla f \).**
\[ f(x, y) = \ \_\_\_\_\_\_\_\_\_\_ \]
**(b) Use part (a) to evaluate \( \int_C \nabla f \cdot d\mathbf{r} \) along the given curve \( C \).**
\[ \boxed{\ \_\_\_\_\_\_\_\_\_\_ \} \]
### Explanation of the Diagram:
- **Vector Field \( \mathbf{F} \)**: This represents a vector field with \( x \) and \( y \) coordinates, and the vector components changing based on a function of both \( x \) and \( y \).
- **Curve \( C \)**: This curve is parameterized by \( t \) from 0 to 1 and includes trigonometric functions of \( t \) to describe its path in the \( xy \)-plane.
### Goals:
1. **To find the potential function \( f \)**:
- Identify a scalar function \( f(x, y) \) whose gradient \( \nabla f \) matches the given vector field \( \mathbf{F}(x, y) \).
2. **To evaluate the line integral**:
- Use the found potential function \( f \) to calculate the line integral over the path \( C \).
This exercise is part of a study in vector fields, potential functions, and line integrals, which are key concepts in multivariable calculus.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc6615292-9c23-445f-b631-819ab1904cf2%2Ff64dfd64-8f8d-4725-a8f5-aeb61858c805%2Fx5xgf9.png&w=3840&q=75)
Transcribed Image Text:**Vector Fields and Line Integrals**
### Consider \( \mathbf{F} \) and \( C \) below.
\[ \mathbf{F}(x, y) = 6xy^2 \ \mathbf{i} + 6x^2y \ \mathbf{j} \]
\[ C : \mathbf{r}(t) = \langle t + \sin\left(\frac{1}{2}\pi t\right), t + \cos\left(\frac{1}{2}\pi t\right) \rangle, \quad 0 \leq t \leq 1 \]
### Exercises:
**(a) Find a function \( f \) such that \( \mathbf{F} = \nabla f \).**
\[ f(x, y) = \ \_\_\_\_\_\_\_\_\_\_ \]
**(b) Use part (a) to evaluate \( \int_C \nabla f \cdot d\mathbf{r} \) along the given curve \( C \).**
\[ \boxed{\ \_\_\_\_\_\_\_\_\_\_ \} \]
### Explanation of the Diagram:
- **Vector Field \( \mathbf{F} \)**: This represents a vector field with \( x \) and \( y \) coordinates, and the vector components changing based on a function of both \( x \) and \( y \).
- **Curve \( C \)**: This curve is parameterized by \( t \) from 0 to 1 and includes trigonometric functions of \( t \) to describe its path in the \( xy \)-plane.
### Goals:
1. **To find the potential function \( f \)**:
- Identify a scalar function \( f(x, y) \) whose gradient \( \nabla f \) matches the given vector field \( \mathbf{F}(x, y) \).
2. **To evaluate the line integral**:
- Use the found potential function \( f \) to calculate the line integral over the path \( C \).
This exercise is part of a study in vector fields, potential functions, and line integrals, which are key concepts in multivariable calculus.
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