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Chapter 16: Problem 7

Determine whether or not \(\mathbf{F}\) is a conservative vector field. If it is, find a function \(f\) such that \(\mathbf{F}=\nabla f\) $$ \mathbf{F}(x, y)=\left(y e^{x}+\sin y\right) \mathbf{i}+\left(e^{x}+x \cos y\right) \mathbf{j} $$

Short Answer

Expert verified
Yes, \( \mathbf{F} \) is conservative. The potential function is \( f(x, y) = y e^{x} + x \sin y + C \).

Step by step solution

01

Check if \( \mathbf{F} \) is conservative

A vector field \( \mathbf{F}(x, y) = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j} \) is conservative if 1) it is defined on a simply connected domain, and 2) its curl is zero. Here, \( P = y e^{x} + \sin y \) and \( Q = e^{x} + x \cos y \). Compute \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \):- \( \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(e^{x} + x \cos y) = e^{x} + \cos y \)- \( \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(y e^{x} + \sin y) = e^{x} + \cos y \)Since \( \frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y} \), \( \mathbf{F} \) is conservative.
02

Find potential function \( f \)

Since \( \mathbf{F} \) is conservative, we find a function \( f(x, y) \) such that \( abla f = \mathbf{F} \). This requires:- \( \frac{\partial f}{\partial x} = P = y e^{x} + \sin y \) - \( \frac{\partial f}{\partial y} = Q = e^{x} + x \cos y \)Start with \( \frac{\partial f}{\partial x} = y e^{x} + \sin y \) and integrate with respect to \( x \):\[ f(x, y) = \int (y e^{x} + \sin y) \, dx = y e^{x} + x \sin y + g(y) \]Here, \( g(y) \) is an unknown function of \( y \).
03

Determine \( g(y) \)

Differentiate the partial solution \( f(x, y) = y e^{x} + x \sin y + g(y) \) with respect to \( y \), and set it equal to \( Q \):- \( \frac{\partial f}{\partial y} = e^{x} + x \cos y + g'(y)\)Since \( \frac{\partial f}{\partial y} = Q = e^{x} + x \cos y \), it follows that:- \( e^{x} + x \cos y + g'(y) = e^{x} + x \cos y \)Thus, \( g'(y) = 0 \), which implies \( g(y) \) is a constant. Assume \( g(y) = C \) for simplicity.
04

Solution expression

Therefore, the scalar potential function \( f \) is:\[ f(x, y) = y e^{x} + x \sin y + C \]where \( C \) is an arbitrary constant. This confirms \( \mathbf{F} \) is indeed a conservative vector field with potential function \( f(x, y) = y e^{x} + x \sin y + C \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Potential Function
A potential function, often denoted by \( f \), is a scalar-valued function whose gradient yields a given vector field \( \mathbf{F} \). This means that if you have a conservative vector field, you can express it as \( \mathbf{F} = abla f \). Understanding this concept is crucial, as it allows us to identify the existence of a potential function, simplifying many problems involving vector fields.
  • If a vector field \( \mathbf{F} \) can be expressed as the gradient of \( f \), then \( \mathbf{F} \) is conservative.

  • The potential function \( f(x, y) \) is found such that its partial derivatives align with the components of \( \mathbf{F} \).

  • This usually involves integrating the components of \( \mathbf{F} \) with respect to their respective variables.
Once the potential function is found, it confirms the relationship \( \mathbf{F} = abla f \). This verification step is important to ensure mathematical correctness.
Curl of a Vector Field
The curl of a vector field in two dimensions is a measure of the rotation or "curling" effect that the field has at a point. A vector field \( \mathbf{F} = (P, Q) \) is conservative if its curl \( abla \times \mathbf{F} = 0 \).For 2D vector fields, this condition simplifies to:\[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0 \]
  • When the curl is zero, it means there is no rotational component to the vector field, suggesting a potential function could exist.

  • The curl helps to check if a vector field is path-independent, which is a key characteristic of conservative fields.
Determining the curl involves taking partial derivatives of the vector field components and checking their equality. This simple calculation gives significant insight into the nature of the vector field.
Partial Derivatives
Partial derivatives are a fundamental part of multivariable calculus, allowing us to analyze how a function changes with respect to one variable at a time, while holding other variables constant.In the context of vector fields and potential functions:
  • Partial derivatives help express vector field components as derivatives of a potential function.

  • They allow us to compute the curl of a vector field to test for conservativeness.

  • By calculating partial derivatives, you can determine if \( \mathbf{F} \) reflects an underlying scalar field \( f \).
For example, understanding the equality \( \frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y} \) is a key step in identifying the conservativeness of \( \mathbf{F} \). This knowledge is pivotal when searching for potential functions.
Simply Connected Domain
A simply connected domain is a crucial concept when dealing with conservative vector fields. A domain is simply connected if it is connected and has no holes, meaning any closed curve within the domain can be continuously contracted to a single point without exiting the domain. For a vector field to be conservative, it needs to be defined on a simply connected domain.
  • This requirement ensures that the potential function can be defined consistently over the entire region, without ambiguity.

  • In practical terms, a vector field defined over a simply connected domain will have no obstacles preventing the existence of a potential function.
Verifying this condition allows us to confidently proceed with determining potential functions and apply the fundamental theorem for gradients, linking vector fields to their respective potential functions with assurance.

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