Problem 26
Question
Determine whether or not the vector field is conservative. If it is, find a potential function. $$\langle y, 1\rangle$$
Step-by-Step Solution
Verified Answer
The vector field \(\langle y, 1\rangle\) is not conservative, thus there is no potential function.
1Step 1: Calculate the Curl of the Vector Field
The curl of a vector field \( F = \langle M, N\rangle \) in two dimensions is given by \( \partial N/\partial x - \partial M/\partial y \). Taking the partial derivative of N with respect to x gives 0 as there's no x-term in N. Meanwhile, the partial derivative of M with respect to y gives 1.
2Step 2: Check if the Vector Field is Conservative
A vector field is conservative if and only if the curl of the vector, defined as \( \partial N/\partial x - \partial M/\partial y \), equals zero. In this case, the curl is \( 0 - 1 = -1 \). Therefore, the vector field is not conservative.
3Step 3: Find a Potential Function
Since the vector field is not conservative, a potential function does not exist.
Key Concepts
Potential FunctionCurl of a Vector FieldPartial Derivatives
Potential Function
In vector calculus, a potential function is a scalar function whose gradient gives back the original vector field. If a vector field is conservative, there exists a potential function such that its gradient is equal to the vector field. This concept is particularly useful because it allows us to express a vector field in the simpler form of a gradient of a single function.
When we have a conservative vector field, it implies that it has no rotation or swirl within the field. This is mathematically related to the fact that the field's line integral around any closed path is zero. In terms of finding a potential function, once we confirm that a vector field is conservative, we look for a function \( f(x, y) \) such that the gradient \( abla f(x, y) \) equals the original vector field \( \langle M, N \rangle \).
In cases where the vector field is not conservative, as is the case in this exercise, a potential function does not exist. Thus, if we are required to find one, confirming the field's conservativeness is the first necessary step.
When we have a conservative vector field, it implies that it has no rotation or swirl within the field. This is mathematically related to the fact that the field's line integral around any closed path is zero. In terms of finding a potential function, once we confirm that a vector field is conservative, we look for a function \( f(x, y) \) such that the gradient \( abla f(x, y) \) equals the original vector field \( \langle M, N \rangle \).
In cases where the vector field is not conservative, as is the case in this exercise, a potential function does not exist. Thus, if we are required to find one, confirming the field's conservativeness is the first necessary step.
Curl of a Vector Field
The curl of a vector field is a measure of the field's tendency to rotate around a point. For a two-dimensional vector field \( F = \langle M, N \rangle \), the curl is calculated with the formula \((\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y})\). This formula helps determine whether a vector field possesses any rotational aspects.
In the original exercise, the calculation of the curl was crucial in determining the nature of the vector field. By evaluating \( \frac{\partial N}{\partial x} \), the lack of an \( x \)-term in \( N \), meant this derivative was zero. Evaluating \( \frac{\partial M}{\partial y} \) gave \( 1 \). Therefore, the curl of the vector field was computed to be \( 0 - 1 = -1 \).
Since the curl was not zero, this indicated that the vector field is not conservative. Understanding and calculating curl provides insight into whether a field is suitable for certain operations, like finding a potential function.
In the original exercise, the calculation of the curl was crucial in determining the nature of the vector field. By evaluating \( \frac{\partial N}{\partial x} \), the lack of an \( x \)-term in \( N \), meant this derivative was zero. Evaluating \( \frac{\partial M}{\partial y} \) gave \( 1 \). Therefore, the curl of the vector field was computed to be \( 0 - 1 = -1 \).
Since the curl was not zero, this indicated that the vector field is not conservative. Understanding and calculating curl provides insight into whether a field is suitable for certain operations, like finding a potential function.
Partial Derivatives
Partial derivatives are derivatives of functions with multiple variables, where all but one variable are held constant during differentiation. They provide important insights into how a function changes with respect to its variables.
In our exercise involving the vector field \( \langle y, 1 \rangle \), calculating partial derivatives was essential for understanding the components of the vector field. Specifically, to find the curl, we computed \( \frac{\partial N}{\partial x} \) and \( \frac{\partial M}{\partial y} \):
These derivatives informed us about the behavior of the vector field and confirmed the lack of conservativeness of the field. Thus, partial derivatives serve as vital tools in analyzing and solving problems related to vector fields, such as determining their curl or finding potential functions.
In our exercise involving the vector field \( \langle y, 1 \rangle \), calculating partial derivatives was essential for understanding the components of the vector field. Specifically, to find the curl, we computed \( \frac{\partial N}{\partial x} \) and \( \frac{\partial M}{\partial y} \):
- The partial derivative of \( N \) with respect to \( x \) was \( 0 \) because \( N \) does not contain \( x \).
- The partial derivative of \( M \) with respect to \( y \) was \( 1 \) as \( M = y \).
These derivatives informed us about the behavior of the vector field and confirmed the lack of conservativeness of the field. Thus, partial derivatives serve as vital tools in analyzing and solving problems related to vector fields, such as determining their curl or finding potential functions.
Other exercises in this chapter
Problem 26
Determine whether the given vector field is conservative and/or incompressible. $$\left\langle e^{y}, x e^{y}+z^{2}, 2 y z-1\right\rangle$$
View solution Problem 26
Find the flux of \(\mathbf{F}\) over \(\partial Q\). \(Q\) is bounded by \(x+2 y+3 z=12\) and the coordinate planes, \(\mathbf{F}=\left\langle x^{2} y, 3 x, 4 y
View solution Problem 26
Evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}.\) \(\mathbf{F}(x, y)=\left(4 x y-2 x, 2 x^{2}-x\right), C\) is the portion of the parabola \(y=x^{2}\) from \
View solution Problem 26
Use a line integral to compute the area of the given region. The region bounded by \(y=x^{2}\) and \(y=2 x\)
View solution