Chapter 13: Problem 9
Determine whether or not \(\mathbf{F}\) is a conservative vector field. If it is, find a function \(f\) such that \(\mathbf{F}=\nabla f\). \(\mathbf{F}(x, y)=\left(\ln y+2 x y^{3}\right) \mathbf{i}+\left(3 x^{2} y^{2}+x / y\right) \mathbf{j}\)
Short Answer
Expert verified
\( \mathbf{F} \) is conservative, and the potential function is \( f(x, y) = x\ln y + x^2y^3 + C \).
Step by step solution
01
Check if the Field is Conservative
A vector field \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} \) is conservative if the partial derivative of \( P \) with respect to \( y \) is equal to the partial derivative of \( Q \) with respect to \( x \). First, identify \( P(x, y) = \ln y + 2xy^3 \) and \( Q(x, y) = 3x^2y^2 + \frac{x}{y} \).
02
Compute Partial Derivatives
Calculate \( \frac{\partial P}{\partial y} \) and \( \frac{\partial Q}{\partial x} \):\[ \frac{\partial P}{\partial y} = \frac{1}{y} + 6xy^2 \]\[ \frac{\partial Q}{\partial x} = 6xy^2 + \frac{1}{y} \]
03
Comparison of Partial Derivatives
Check if \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \): \[ \frac{1}{y} + 6xy^2 = 6xy^2 + \frac{1}{y} \]Since both are equal, \( \mathbf{F} \) is a conservative vector field.
04
Integrate to Find Potential Function f
To find \( f(x, y) \), integrate \( P(x, y) \) with respect to \( x \): \[ f(x, y) = \int (\ln y + 2xy^3) \, dx = x\ln y + x^2y^3 + g(y) \] where \( g(y) \) is an arbitrary function of \( y \).
05
Determine the Function g(y)
Differentiate \( f(x, y) = x\ln y + x^2y^3 + g(y) \) with respect to \( y \) and set it equal to \( Q(x, y) \):\[ \frac{\partial f}{\partial y} = \frac{x}{y} + 3x^2y^2 + g'(y) \]Set this equal to \( Q(x, y) = 3x^2y^2 + \frac{x}{y} \):\[ \frac{x}{y} + 3x^2y^2 + g'(y) = 3x^2y^2 + \frac{x}{y} \]Thus, \( g'(y) = 0 \), which implies \( g(y) = C \), where \( C \) is a constant.
06
Write the Potential Function
Since \( g(y) = C \), we have \( f(x, y) = x\ln y + x^2y^3 + C \). Therefore, the function \( f \) whose gradient is \( \mathbf{F} \) is \( f(x, y) = x\ln y + x^2y^3 + C \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Partial derivatives help us understand how a function changes as we change one of its variables while keeping the other variables constant.
In the context of our exercise, we are working with a vector field \( \mathbf{F}(x, y) = P\mathbf{i} + Q\mathbf{j} \). To determine conservatism, we focus on two functions here: \( P(x, y) = \ln y + 2xy^3 \) and \( Q(x, y) = 3x^2y^2 + \frac{x}{y} \).
We need to compute the partial derivatives \( \frac{\partial P}{\partial y} \) and \( \frac{\partial Q}{\partial x} \) and check if they are equal.
In the context of our exercise, we are working with a vector field \( \mathbf{F}(x, y) = P\mathbf{i} + Q\mathbf{j} \). To determine conservatism, we focus on two functions here: \( P(x, y) = \ln y + 2xy^3 \) and \( Q(x, y) = 3x^2y^2 + \frac{x}{y} \).
We need to compute the partial derivatives \( \frac{\partial P}{\partial y} \) and \( \frac{\partial Q}{\partial x} \) and check if they are equal.
- \( \frac{\partial P}{\partial y} = \frac{1}{y} + 6xy^2 \)
- \( \frac{\partial Q}{\partial x} = 6xy^2 + \frac{1}{y} \)
Potential Function
A potential function \( f(x, y) \) serves as a kind of 'landscape' representation of the vector field.
If a field is conservative, we can find this potential function such that the gradient of \( f \) matches our vector field \( \mathbf{F} \).
In this exercise, finding \( f \) involves integrating the function \( P(x, y) \) with respect to \( x \).
After integrating and realizing that \( Q(x, y) \) provides a check for \( \frac{\partial f}{\partial y} \), we ensure the full alignment of the given field with our derived potential function.
If a field is conservative, we can find this potential function such that the gradient of \( f \) matches our vector field \( \mathbf{F} \).
In this exercise, finding \( f \) involves integrating the function \( P(x, y) \) with respect to \( x \).
- \( f(x, y) = \int (\ln y + 2xy^3) \, dx = x\ln y + x^2y^3 + g(y) \)
After integrating and realizing that \( Q(x, y) \) provides a check for \( \frac{\partial f}{\partial y} \), we ensure the full alignment of the given field with our derived potential function.
Gradient
The gradient of a function, denoted as \( abla f \), is a vector that points in the direction of the greatest rate of increase of the function.
It can be considered as a generalization of a derivative to multiple dimensions.
When calculating \( abla f = \mathbf{F} \), and successfully matching these two, we confirm the conservative nature of the field.
It can be considered as a generalization of a derivative to multiple dimensions.
When calculating \( abla f = \mathbf{F} \), and successfully matching these two, we confirm the conservative nature of the field.
- The components of \( abla f \) are the partial derivatives of \( f(x, y) \) with respect to each variable.
- Achieving \( abla f = \mathbf{F} \) validates that our function \( f \) accurately describes the potential landscape of \( \mathbf{F} \).