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Chapter 16: Problem 3

\(3-10\) Determine whether or not \(\mathbf{F}\) is a conservative vector field. If it is, find a function \(f\) such that \(\mathbf{F}=\nabla f\) $$\mathbf{F}(x, y)=(2 x-3 y) \mathbf{i}+(-3 x+4 y-8) \mathbf{j}$$

Short Answer

Expert verified
The vector field \( \mathbf{F} \) is conservative. The potential function is \( f(x, y) = x^2 - 3xy + 2y^2 - 8y + C \).

Step by step solution

01

Recall the conditions for a vector field to be conservative

A vector field \( \mathbf{F} = M(x, y) \mathbf{i} + N(x, y) \mathbf{j} \) is conservative if its curl is zero, that is \( \frac{\partial N}{\partial x} = \frac{\partial M}{\partial y} \). This means that if the cross partial derivatives are equal, the field is conservative.
02

Identify components of the vector field

The given vector field is \( \mathbf{F}(x, y) = (2x - 3y) \mathbf{i} + (-3x + 4y - 8) \mathbf{j} \). Here, \( M(x, y) = 2x - 3y \) and \( N(x, y) = -3x + 4y - 8 \).
03

Compute partial derivatives

Compute the partial derivative \( \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(-3x + 4y - 8) = -3 \). Likewise, compute the partial derivative \( \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(2x - 3y) = -3 \).
04

Compare the computed partial derivatives

Since \( \frac{\partial N}{\partial x} = -3 \) and \( \frac{\partial M}{\partial y} = -3 \), these derivatives are equal, confirming that \( \mathbf{F} \) is conservative.
05

Find potential function \( f \)

As \( \mathbf{F} \) is conservative, find a potential function \( f \text{ such that } abla f = \mathbf{F} \). We need functions satisfying \( \frac{\partial f}{\partial x} = 2x - 3y \) and \( \frac{\partial f}{\partial y} = -3x + 4y - 8 \).
06

Integrate with respect to \( x \)

Integrate \( \frac{\partial f}{\partial x} = 2x - 3y \) with respect to \( x \). The integration results in \( f(x, y) = x^2 - 3xy + g(y) \), where \( g(y) \) is an arbitrary function of \( y \).
07

Differentiate the result with respect to \( y \)

Differentiate \( f(x, y) = x^2 - 3xy + g(y) \) with respect to \( y \) to find \( \frac{\partial f}{\partial y} = -3x + g'(y) \).
08

Equate \( \frac{\partial f}{\partial y} \) to \( N(x, y) \)

Set \(-3x + g'(y) = -3x + 4y - 8\). Solving for \( g'(y) \) gives \( g'(y) = 4y - 8 \).
09

Integrate \( g'(y) \)

Integrate \( g'(y) = 4y - 8 \) with respect to \( y \) to find \( g(y) = 2y^2 - 8y + C \), where \( C \) is a constant.
10

Write the full potential function \( f \)

Combine results to find \( f(x, y) = x^2 - 3xy + 2y^2 - 8y + C \). This is the function \( f \) such that \( abla f = \mathbf{F} \).

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Key Concepts

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

Potential Function
In the context of vector calculus, a potential function is a scalar function whose gradient is equal to a given vector field. More simply, if a vector field \( \mathbf{F} \) is conservative, there exists a function \( f(x, y) \) such that the gradient \( abla f \) is equal to \( \mathbf{F} \).
  • The vector field is derived from the potential function, which implies that its line integral is path-independent.
  • This characteristic makes solving certain problems more straightforward, as working with a scalar function is often simpler than dealing directly with the vector field.
For example, finding a potential function involves solving for \( f \) such that its partial derivatives with respect to \( x \) and \( y \) match the components of \( \mathbf{F} \). In our original problem, integrating with respect to each variable helped us find the potential function \( f(x, y) = x^2 - 3xy + 2y^2 - 8y + C \).
Curl of a Vector Field
The curl of a vector field measures the tendency of the field to rotate around a point. For a two-dimensional vector field \( \mathbf{F} = M(x, y) \mathbf{i} + N(x, y) \mathbf{j} \), the curl boils down to checking whether the cross partial derivatives of the components are equal.
  • If \( \frac{\partial N}{\partial x} = \frac{\partial M}{\partial y} \), the vector field has zero curl, meaning it is conservative.
  • This property is crucial as it confirms that the vector field can be expressed as the gradient of a potential function, meaning it is irrotational.
In the exercise, we computed \( \frac{\partial N}{\partial x} = -3 \) and \( \frac{\partial M}{\partial y} = -3 \), indicating that the curl is zero and confirming the conservative nature of the vector field.
Partial Derivatives
Partial derivatives represent how a function changes as the variables change. They measure the rate at which the function's value changes as one of the variables is varied while the other remains constant.
  • For a given function \( f(x, y) \), the partial derivative \( \frac{\partial f}{\partial x} \) is calculated by differentiating \( f \) with respect to \( x \), treating \( y \) as a constant.
  • Likewise, \( \frac{\partial f}{\partial y} \) involves differentiating with respect to \( y \) while keeping \( x \) constant.
In our solution, we used partial derivatives to check the curl condition and to find the potential function by integrating the partial derivatives.
Integration with Respect to Variables
Integration with respect to variables is the process of finding a function when given its derivative. It reverses the process of differentiation.
  • When we integrate a function with respect to \( x \), we treat \( y \) as a constant and add a function of \( y \) as an integration constant.
  • Conversely, integrating with respect to \( y \) treats \( x \) as a constant and the integration constant may involve functions of \( x \).
During our problem-solving process, integration was crucial to recovering the potential function from the vector field. Specifically, we integrated \( \frac{\partial f}{\partial x} = 2x - 3y \) to obtain a part of the function, then integrated with respect to \( y \) to solve for the remaining terms and the potential function itself.

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Most popular questions from this chapter

Use Gauss's Law to find the charge contained in the solid hemisphere \(x^{2}+y^{2}+z^{2} \leq a^{2}, z \geqslant 0,\) if the electric field is $$\mathbf{E}(x, y, z)=x \mathbf{i}+y \mathbf{j}+2 z \mathbf{k}$$

Prove each identity, assuming that \(S\) and \(E\) satisfy the conditions of the Divergence Theorem and the scalar functions and components of the vector fields have continuous second-order partial derivatives. $$ \iint_{S}(f \nabla g-g \nabla f) \cdot \mathbf{n} d S=\iiint_{E}\left(f \nabla^{2} g-g \nabla^{2} f\right) d V $$

Use Gaus's Law to find the charge enclosed by the cube with vertices \((\pm 1, \pm 1, \pm 1)\) if the electric field is $$\mathbf{E}(x, y, z)=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}$$

Let \(\mathbf{F}\) be an inverse square field, that is, \(\mathbf{F}(r)=c \mathbf{r} /|\mathbf{r}|^{3}\) for some constant \(c,\) where \(r=x \mathbf{i}+y \mathbf{j}+z\) . . Show that the flux of \(\mathbf{F}\) across a sphere \(S\) with center the origin is independent of the radius of \(S .\)

Verify that Stokes' Theorem is true for the given vector field \(\mathbf{F}\) and surface \(S\). $$\begin{array}{l}{\mathbf{F}(x, y, z)=x \mathbf{i}+y \mathbf{j}+x y z \mathbf{k}} \\ {S \text { is the part of the plane } 2 x+y+z=2 \text { that lies in the first }} \\ {\text { octant, oriented upward }}\end{array}$$
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