Chapter 15: Problem 2
Determine whether \(\mathbf{F}\) is a conservative vector field. If so, find a potential function for it. $$ \mathbf{F}(x, y)=3 y^{2} \mathbf{i}+6 x y \mathbf{j} $$
Short Answer
Expert verified
Yes, \( \mathbf{F} \) is a conservative vector field. The potential function is \( f(x, y) = 3xy^2 \).
Step by step solution
01
Understand the Conditions for a Conservative Vector Field
A vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} \) is conservative if the curl of \( \mathbf{F} \) is zero. In a two-dimensional field, this translates to checking if \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \).
02
Identify Components of the Vector Field
Given \( \mathbf{F}(x, y) = 3y^2 \mathbf{i} + 6xy \mathbf{j} \), identify \( P(x, y) = 3y^2 \) and \( Q(x, y) = 6xy \).
03
Compute the Partial Derivatives
Calculate \( \frac{\partial P}{\partial y} \) and \( \frac{\partial Q}{\partial x} \): - \( \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(3y^2) = 6y \) - \( \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(6xy) = 6y \)
04
Verify the Conservative Condition
Since \( \frac{\partial P}{\partial y} = 6y \) and \( \frac{\partial Q}{\partial x} = 6y \), they are equal, confirming that \( \mathbf{F} \) is conservative.
05
Find the Potential Function
To find the potential function \( f(x, y) \), determine \( \frac{\partial f}{\partial x} = P(x, y) \) and \( \frac{\partial f}{\partial y} = Q(x, y) \). Integrate each one:1. Integrate \( P(x, y) = 3y^2 \) with respect to \( x \): \[ f(x, y) = 3y^2 x + g(y) \] 2. Integrate \( Q(x, y) = 6xy \) with respect to \( y \): \[ f(x, y) = 3xy^2 + h(x) \]Combine the two results to find \( f(x, y) = 3xy^2 + C \), where \( C \) is a constant.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Potential Function
When dealing with conservative vector fields, a potential function is a key concept. A vector field is said to be conservative if there exists a scalar function, often called the potential function, from which the vector field can be derived as a gradient. This means that the vector field represents the rate of change of the potential function in space.
In the given exercise, the vector field \( \mathbf{F}(x, y) = 3y^2 \mathbf{i} + 6xy \mathbf{j} \) has been verified to be conservative. This verification means there is a potential function \( f(x, y) \) such that the vector field \( \mathbf{F} \) is the gradient of \( f(x, y) \).
In the given exercise, the vector field \( \mathbf{F}(x, y) = 3y^2 \mathbf{i} + 6xy \mathbf{j} \) has been verified to be conservative. This verification means there is a potential function \( f(x, y) \) such that the vector field \( \mathbf{F} \) is the gradient of \( f(x, y) \).
- The potential function was found by integrating the components of the vector field.
- Integration requires us to handle each component separately.
- The resulting expression \( f(x, y) = 3xy^2 + C \) combines these integrations where \( C \) is an arbitrary constant.
Partial Derivatives
Partial derivatives play a crucial role in understanding vector fields, particularly when examining whether a vector field is conservative. A partial derivative measures how a function changes as one of its variables changes, while the other variables are held constant.
In our example, to determine whether \( \mathbf{F}(x, y) \) is a conservative vector field, we need to find the partial derivatives of its components \( P(x, y) \) and \( Q(x, y) \):
In our example, to determine whether \( \mathbf{F}(x, y) \) is a conservative vector field, we need to find the partial derivatives of its components \( P(x, y) \) and \( Q(x, y) \):
- \( \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(3y^2) = 6y \)
- \( \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(6xy) = 6y \)
Curl of a Vector Field
The curl of a vector field is a vector operator that describes the rotation or the "twisting" force of a field around a point. In two dimensions, the concept simplifies, focusing primarily on the differences in partial derivatives of the vector field's components.
This exercise leverages the fact that in a 2D vector field, for it to be conservative, the curl must be zero. In mathematical terms, this means:
In \( \mathbf{F}(x, y) = 3y^2 \mathbf{i} + 6xy \mathbf{j} \), both partial derivatives of the components matched (\( 6y \)), confirming that the curl is indeed zero. This zero curl is the hallmark of a conservative field, suggesting that it is possible to find a potential function that exactly describes \( \mathbf{F} \). This function dictates the path-independent nature of the field, which is a fundamental property of conservative vector fields.
This exercise leverages the fact that in a 2D vector field, for it to be conservative, the curl must be zero. In mathematical terms, this means:
- \( \frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y} \)
In \( \mathbf{F}(x, y) = 3y^2 \mathbf{i} + 6xy \mathbf{j} \), both partial derivatives of the components matched (\( 6y \)), confirming that the curl is indeed zero. This zero curl is the hallmark of a conservative field, suggesting that it is possible to find a potential function that exactly describes \( \mathbf{F} \). This function dictates the path-independent nature of the field, which is a fundamental property of conservative vector fields.