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Magazine Chapter 15: Problem 27
Find a potential function for \(\mathbf{F}\). $$\mathbf{F}=\frac{2 x}{y} \mathbf{i}+\left(\frac{1-x^{2}}{y^{2}}\right) \mathbf{j}, \quad\\{(x, y): y>0\\}$$
Short Answer
Expert verified
The potential function is \( f(x, y) = \frac{x^2}{y} - \frac{1}{y} + C \), where \( C \) is a constant.
Step by step solution
01
Identify Conditions for a Potential Function
A potential function, \( f(x, y) \), exists if and only if the vector field \( \mathbf{F} \) is conservative. For \( \mathbf{F} = M \mathbf{i} + N \mathbf{j} \), this requires that \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). Let \( M = \frac{2x}{y} \) and \( N = \frac{1-x^2}{y^2} \).
02
Compute Partial Derivatives
Calculate \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \).\[ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}\left(\frac{2x}{y}\right) = -\frac{2x}{y^2} \]\[ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}\left(\frac{1-x^2}{y^2}\right) = -\frac{2x}{y^2} \]
03
Verify Conservativity
Since \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} = -\frac{2x}{y^2} \), \( \mathbf{F} \) is conservative. A potential function \( f(x, y) \) can therefore be found.
04
Find Potential Function from \( M \)
Integrate \( M = \frac{2x}{y} \) with respect to \( x \):\[ f(x, y) = \int \frac{2x}{y} \ dx = \frac{x^2}{y} + g(y) \] where \( g(y) \) is an arbitrary function of \( y \).
05
Determine \( g(y) \) Using \( N \)
Differentiate \( f(x, y) = \frac{x^2}{y} + g(y) \) with respect to \( y \) and set it equal to \( N = \frac{1-x^2}{y^2} \):\[ \frac{\partial}{\partial y} \left( \frac{x^2}{y} + g(y) \right) = -\frac{x^2}{y^2} + g'(y) \]Equate this to \( N(x, y) \):\[ -\frac{x^2}{y^2} + g'(y) = \frac{1-x^2}{y^2} \]Solving gives:\[ g'(y) = \frac{1}{y^2} \]Thus, \( g(y) = -\frac{1}{y} + C \), where \( C \) is a constant.
06
Assemble the Potential Function
Using \( f(x, y) = \frac{x^2}{y} + g(y) \), substitute \( g(y) = -\frac{1}{y} \).Thus, the potential function is:\[ f(x, y) = \frac{x^2}{y} - \frac{1}{y} + C \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Conservative Vector Field
A conservative vector field is a type of vector field where the line integral along any closed path is zero. This property implies that the force exerted by this field is path-independent. In simpler terms, the work done by the field in moving an object from one point to another only depends on the starting and ending points, not the path taken.
- For a vector field to be conservative, there must exist a potential function from which the vector field can be derived as a gradient.
- Mathematically, if \( \mathbf{F} = abla f \), then \( \mathbf{F} \) is conservative.
- The condition for conservativity in two dimensions involves ensuring that the mixed partial derivatives are equal, that is \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).
Gradient Field
A gradient field is another way to describe a conservative vector field. It is a vector field that is the gradient of a potential function. This means it often has a very nice mathematical structure that benefits from the properties of the potential function.
- Given a potential function \( f(x, y) \), the gradient field is represented as \( abla f = \left( \frac{\partial f}{\partial x} \right) \mathbf{i} + \left( \frac{\partial f}{\partial y} \right) \mathbf{j} \).
- In gradient fields, potential functions can be interpreted as physical potential energies or scalar potentials from which the field can "flow."
- To find a potential function for a gradient field, you integrate the components of \( \mathbf{F} \) and find a function \( f(x, y) \) whose gradient gives back the original vector field \( \mathbf{F} \).
Partial Derivatives
Partial derivatives are a fundamental concept in multivariable calculus, used to understand how a function changes when you vary one of its variables, while keeping the others constant. This concept is essential for the study of vector fields.
- If you have a function \( f(x, y) \), the partial derivative of \( f \) with respect to \( x \), denoted \( \frac{\partial f}{\partial x} \), tells you how \( f \) changes as you change \( x \), while keeping \( y \) fixed.
- Similarly, the partial derivative with respect to \( y \), \( \frac{\partial f}{\partial y} \), tells you how \( f \) changes as you change \( y \), while keeping \( x \) fixed.
- Partial derivatives are used to verify the conditions of conservativity in vector fields by ensuring that the mixed derivatives \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).
