Chapter 6: Problem 26
Is vector field \(\mathbf{F}(x, y)=(P(x, y), Q(x, y))=(\sin x+y) \mathbf{i}+(\cos y+x) \mathbf{j}\) a gradient field?
Short Answer
Expert verified
Yes, the vector field \(\mathbf{F}(x, y)\) is a gradient field.
Step by step solution
01
Define a Gradient Field
A vector field \(\mathbf{F}(x, y) = (P(x, y), Q(x, y))\) is a gradient field if there exists a scalar function \(f(x, y)\) such that \(\mathbf{F} = abla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)\). This implies that \(P = \frac{\partial f}{\partial x}\) and \(Q = \frac{\partial f}{\partial y}\).
02
Check the Condition for a Gradient Field
For \(\mathbf{F}(x, y)\) to be a gradient field, the curl must be zero in a two-dimensional field. This means \(\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}\).
03
Compute Partial Derivatives
Calculate \(\frac{\partial P}{\partial y}\) and \(\frac{\partial Q}{\partial x}\). For \(P(x, y) = \sin x + y\), \(\frac{\partial P}{\partial y} = 1\). For \(Q(x, y) = \cos y + x\), \(\frac{\partial Q}{\partial x} = 1\).
04
Compare Partial Derivatives
Verify if the partial derivatives are equal: \(\frac{\partial P}{\partial y} = 1\) and \(\frac{\partial Q}{\partial x} = 1\). Since they are equal, the curl is zero, which confirms the field is conservative.
05
Conclude the Vector Field Nature
Since the partial derivatives are equal, \(\mathbf{F}(x, y)\) satisfies the condition of being a gradient field, therefore \(\mathbf{F}(x, y)\) is a gradient field.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Partial derivatives are a fundamental concept in calculus used to find the rate of change of a multivariable function with respect to one variable, while keeping other variables constant. In the context of vector fields, they help in determining if a field is conservative—that is, if it has a potential function.
For any vector field \( \mathbf{F}(x, y) = (P(x, y), Q(x, y)) \), partial derivatives are employed to express the functions \( P \) and \( Q \) in relation to \( x \) and \( y \).
The partial derivative of \( P \) with respect to \( y \), denoted as \( \frac{\partial P}{\partial y} \), examines how \( P \) changes as \( y \) varies. Similarly, \( \frac{\partial Q}{\partial x} \) analyzes the change in \( Q \) as \( x \) changes. When these derivatives are equal, it confirms that the vector field is "conservative" and potentially a gradient field.
For any vector field \( \mathbf{F}(x, y) = (P(x, y), Q(x, y)) \), partial derivatives are employed to express the functions \( P \) and \( Q \) in relation to \( x \) and \( y \).
The partial derivative of \( P \) with respect to \( y \), denoted as \( \frac{\partial P}{\partial y} \), examines how \( P \) changes as \( y \) varies. Similarly, \( \frac{\partial Q}{\partial x} \) analyzes the change in \( Q \) as \( x \) changes. When these derivatives are equal, it confirms that the vector field is "conservative" and potentially a gradient field.
Scalar Functions
Scalar functions are functions that return a single value, also known as a scalar, typically represented by \( f(x, y) \) in two-dimensional space. Their importance in analyzing vector fields lies in their relationship with gradient fields.
When determining if a vector field \( \mathbf{F}(x, y) \) is a gradient field, one needs to find a scalar function such that \( \mathbf{F} = abla f \). The gradient of a scalar function, denoted \( abla f \), provides a vector field that outlines the direction and rate of fastest increase of the function in space.
In practical terms, finding a scalar function means locating \( f \) for which the partial derivatives equal the components of \( \mathbf{F} \). For instance, for \( \mathbf{F}(x, y) = (\sin x + y, \cos y + x) \), identify \( f \) satisfying both \( \frac{\partial f}{\partial x} = \sin x + y \) and \( \frac{\partial f}{\partial y} = \cos y + x \).
When determining if a vector field \( \mathbf{F}(x, y) \) is a gradient field, one needs to find a scalar function such that \( \mathbf{F} = abla f \). The gradient of a scalar function, denoted \( abla f \), provides a vector field that outlines the direction and rate of fastest increase of the function in space.
In practical terms, finding a scalar function means locating \( f \) for which the partial derivatives equal the components of \( \mathbf{F} \). For instance, for \( \mathbf{F}(x, y) = (\sin x + y, \cos y + x) \), identify \( f \) satisfying both \( \frac{\partial f}{\partial x} = \sin x + y \) and \( \frac{\partial f}{\partial y} = \cos y + x \).
Curl in Vector Fields
The concept of curl is essential in vector calculus and corresponds to the amount of rotation or the "twisting force" in a vector field. For two-dimensional vector fields, curl provides a simplicity check for conservative fields.
The mathematical condition for a two-dimensional vector field \( \mathbf{F}(x, y) = (P, Q) \) being conservative implies the curl is zero. This translates to verifying that \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \), meaning that the vector field has no rotational component.
The mathematical condition for a two-dimensional vector field \( \mathbf{F}(x, y) = (P, Q) \) being conservative implies the curl is zero. This translates to verifying that \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \), meaning that the vector field has no rotational component.
- If these partial derivatives are equal, there exists a scalar potential function, making \( \mathbf{F} \) a gradient field.
- If not, the field is not conservative. This lack of curl suggests a path independence of the field, making it possible to identify potential functions more easily.
Conservative Fields
Conservative fields are vector fields that can be represented as the gradient of a scalar function. This means they have path-independent properties, allowing potential functions to be defined across the field.
The primary criterion for a vector field to be conservative is the equality of certain partial derivatives, ensuring no rotational component. For example, in a field like \( \mathbf{F} = (\sin x + y, \cos y + x) \), this involves ensuring \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \).
The primary criterion for a vector field to be conservative is the equality of certain partial derivatives, ensuring no rotational component. For example, in a field like \( \mathbf{F} = (\sin x + y, \cos y + x) \), this involves ensuring \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \).
- When this condition holds, \( \mathbf{F} \) is indicative of a gradient field where work done along any path between two points is independent of the path chosen.
- This property is a central tenet of conservative fields, enabling the field to be expressed in terms of a potential function \( f \) that satisfies these derivative conditions.