/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 In Exercises \(7-12,\) find a po... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Exercises \(7-12,\) find a potential function \(f\) for the field \(\mathbf{F}\) $$ \mathbf{F}=(y \sin z) \mathbf{i}+(x \sin z) \mathbf{j}+(x y \cos z) \mathbf{k} $$

Short Answer

Expert verified
The potential function is \( f(x, y, z) = xy \sin z + C \).

Step by step solution

01

Identify Potential Function

A potential function for a vector field \( \mathbf{F} = (y \sin z) \mathbf{i} + (x \sin z) \mathbf{j} + (xy \cos z) \mathbf{k} \) can be determined if it is conservative. A conservative vector field \( \mathbf{F} \) is one where there exists a scalar function \( f(x, y, z) \) such that \( \mathbf{F} = abla f \). Start by observing the components: \( F_x = y \sin z \), \( F_y = x \sin z \), and \( F_z = xy \cos z \).
02

Determine Each Component's Contribution

Since \( \mathbf{F} = abla f \), we use \( f_x = y \sin z \), \( f_y = x \sin z \), and \( f_z = xy \cos z \). The partial derivatives lead us to integrate each component separately, ensuring consistency across variables.
03

Integrate First Component for \( f(x, y, z) \)

Integrate \( f_x = y \sin z \) with respect to \( x \):\[ f(x, y, z) = \int y \sin z \, dx = xy \sin z + g(y, z)\]where \( g(y, z) \) is an arbitrary function of \( y \) and \( z \).
04

Integrate Second Component for Consistency

Next, integrating the second component \( f_y = x \sin z \) with respect to \( y \):\[ xy \sin z + h(x, z) = \int x \sin z \, dy = xy \sin z + h(x, z)\]This implies \( g(y, z) = h(x, z) \).
05

Check Third Component for Completeness

Differentiate \( f(x, y, z) = xy \sin z + g(y, z) \) with respect to \( z \) and set it equal to \( f_z \):\[ \frac{\partial}{\partial z} [xy \sin z + g(y, z)] = xy \cos z + \frac{\partial g}{\partial z} = xy \cos z\]Thus, \( \frac{\partial g}{\partial z} = 0 \), ensuring that \( g(y, z) \) is independent of \( z \), i.e., \( g(y, z) = C \), where \( C \) is a constant.
06

Compose Potential Function

After confirming the components, the potential function \( f(x, y, z) \) is:\[ f(x, y, z) = xy \sin z + C\]

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.

Potential Function
A potential function is a scalar function whose gradient yields a given vector field. In essence, if a vector field \( \mathbf{F} \) is conservative, there is a scalar function \( f(x, y, z) \) such that \( abla f = \mathbf{F} \). This means the function \( f \) describes a field of potential energy from which the vector field can be derived. To find this potential function, the components of the vector field \( \mathbf{F} \) are expressed as the partial derivatives \( f_x, f_y, \) and \( f_z \) of the potential function \( f \). Solving for \( f \) involves integrating these partial derivatives. The challenge is ensuring consistency across different variables so that the discovered \( f \) represents fully the vector field.
Conservative Vector Field
A conservative vector field is characterized by being the gradient of some scalar potential function. This means that the work done by moving along a path in the vector field is independent of the path taken, depending only on the end points. **Key Properties of Conservative Vector Fields:**
  • The line integral over any closed path is zero.
  • Such a field can be described by a scalar potential function \( f \) such that \( \mathbf{F} = abla f \).
  • The vector field is irrotational, meaning its curl is zero: \( abla \times \mathbf{F} = \mathbf{0} \).
When dealing with a conservative vector field like \( \mathbf{F} = (y \sin z) \mathbf{i} + (x \sin z) \mathbf{j} + (xy \cos z) \mathbf{k} \), you can search for a potential function by integrating the components while checking these properties.
Partial Derivatives
Partial derivatives represent how a function changes as one of its input variables changes, while the others are held constant. In the context of potential functions and conservative vector fields, the components of a vector field correspond to the partial derivatives of the potential function:
  • Given \( \mathbf{F} = abla f \), \( F_x \) relates to \( f_x \), the derivative of \( f \) with respect to \( x \).
  • Similarly, \( F_y = f_y \) and \( F_z = f_z \).
These relationships allow us to reconstruct the potential function \( f \) by integrating each partial derivative with respect to its variable, while ensuring all components align consistently to form a coherent function.
Integration
Integration is the process of finding the antiderivative of a function. In vector calculus, integration is used to determine the potential function from its partial derivatives. Let us break down the process: - Start with the component expressions of the vector field, corresponding to \( f_x, f_y, \) and \( f_z \). - Integrate each component to find part of the potential function. - Make sure to consider arbitrary functions that can arise during integration, ensuring these are consistent with the other components. During the integration of \( f_x = y \sin z \), for example, integrating with respect to \( x \) gives \( xy \sin z + g(y, z) \), where \( g(y, z) \) is an arbitrary function that depends on \( y \) and \( z \). Checking subsequent components helps refine this arbitrary function, leading to the full potential function. This intricate process ensures that the reconstructed function truly represents the original vector field consistently.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

a. Exact differential form How are the constants \(a, b,\) and \(c\) related if the following differential form is exact? \(\left(a y^{2}+2 c z x\right) d x+y(b x+c z) d y+\left(a y^{2}+c x^{2}\right) d z\) b. Gradient field For what values of \(b\) and \(c\) will $$ \mathbf{F}=\left(y^{2}+2 c z x\right) \mathbf{i}+y(b x+c z) \mathbf{j}+\left(y^{2}+c x^{2}\right) \mathbf{k} $$ be a gradient field?

In Exercises \(41-44\) , use a CAS and Green's Theorem to find the counterclockwise circulation of the field \(\mathbf{F}\) around the simple closed curve \(C .\) Perform the following CAS steps. \(\mathbf{F}=x^{-1} e^{y} \mathbf{i}+\left(e^{y} \ln x+2 x\right) \mathbf{j}\) \(C :\) The boundary of the region defined by \(y=1+x^{4}\) (below) \(\quad\) and \(y=2\) (above)

Integral along different paths Evaluate \(\int_{C} 2 x \cos y d x-x^{2}\) \(\sin y d y\) along the following paths \(C\) in the \(x y\) -plane. a. The parabola \(y=(x-1)^{2}\) from \((1,0)\) to \((0,1)\) b. The line segment from \((-1, \pi)\) to \((1,0)\) c. The \(x\) -axis from \((-1,0)\) to \((1,0)\) d. The astroid \(\mathbf{r}(t)=\left(\cos ^{3} t\right) \mathbf{i}+\left(\sin ^{3} t\right) \mathbf{j}, 0 \leq t \leq 2 \pi\) counterclockwise from \((1,0)\) back to \((1,0)\)

Gradient of a line integral Suppose that \(\mathbf{F}=\nabla f\) is a conservative vector field and $$ g(x, y, z)=\int_{(0,0,0)}^{(x, y, z)} \mathbf{F} \cdot d \mathbf{r} $$ Show that \(\nabla g=\mathbf{F}\)

The tangent plane at a point \(P_{0}\left(f\left(u_{0}, v_{0}\right), g\left(u_{0}, v_{0}\right), h\left(u_{0}, v_{0}\right)\right)\) on a parametrized surface \(\mathbf{r}(u, v)=f(u, v) \mathbf{i}+g(u, v) \mathbf{j}+h(u, v) \mathbf{k}\) is the plane through \(P_{0}\) normal to the vector \(\mathbf{r}_{u}\left(u_{0}, v_{0}\right) \times \mathbf{r}_{v}\left(u_{0}, v_{0}\right),\) the cross product of the tangent vectors \(\mathbf{r}_{u}\left(u_{0}, v_{0}\right)\) and \(\mathbf{r}_{v}\left(u_{0}, v_{0}\right)\) at \(P_{0} .\) In Exercises \(49-52,\) find an equation for the plane tangent to the surface at \(P_{0} .\) Then find a Cartesian equation for the surface and sketch the surface and tangent plane together. Parabolic cylinder The parabolic cylinder surface \(\mathbf{r}(x, y)=\) \(x \mathbf{i}+y \mathbf{j}-x^{2} \mathbf{k},-\infty

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.