/*! 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 223 For the following exercises, fin... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 223

For the following exercises, find the divergence of \(\mathbf{F}\) $$\mathbf{F}(x, y, z)=3 x y z^{2} \mathbf{i}+y^{2} \sin z \mathbf{j}+x e^{2 z} \mathbf{k}$$

Short Answer

Expert verified
The divergence is \( \nabla \cdot \mathbf{F} = 3yz^2 + 2y \sin z + 2xe^{2z} \).

Step by step solution

01

Identify the Components of the Vector Field

The vector field given is \( \mathbf{F}(x, y, z) = 3xy z^2 \mathbf{i} + y^2 \sin z \mathbf{j} + xe^{2z} \mathbf{k} \). Identify the components of the vector field as:- \( F_1 = 3xy z^2 \) for the \( \mathbf{i} \) component,- \( F_2 = y^2 \sin z \) for the \( \mathbf{j} \) component,- \( F_3 = xe^{2z} \) for the \( \mathbf{k} \) component.
02

Formula for Divergence

The formula for the divergence of a vector field \( \mathbf{F} = F_1 \mathbf{i} + F_2 \mathbf{j} + F_3 \mathbf{k} \) is given by:\[ abla \cdot \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}. \]
03

Calculate Partial Derivative of \( F_1 \) with respect to \( x \)

Compute the partial derivative:\[ \frac{\partial}{\partial x}(3xy z^2) = 3yz^2. \]
04

Calculate Partial Derivative of \( F_2 \) with respect to \( y \)

Compute the partial derivative:\[ \frac{\partial}{\partial y}(y^2 \sin z) = 2y \sin z. \]
05

Calculate Partial Derivative of \( F_3 \) with respect to \( z \)

Compute the partial derivative:\[ \frac{\partial}{\partial z}(xe^{2z}) = 2xe^{2z}. \]
06

Sum the Partial Derivatives

Add the results from Steps 3, 4, and 5 to find the divergence:\[ abla \cdot \mathbf{F} = 3yz^2 + 2y \sin z + 2xe^{2z}. \]

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 91影视!

Key Concepts

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

Vector Field
A vector field is a fascinating concept where each point in a space is assigned a vector. Think of it as a way to map values that have both magnitude and direction within a certain domain. In the given exercise, the vector field \( \mathbf{F} \) is expressed as a combination of functions that depend on the coordinates \( x, y, \) and \( z \).
The field represents:
  • A component along the direction \( \mathbf{i} \) given by \( 3xyz^2 \).
  • A component along the direction \( \mathbf{j} \) given by \( y^2 \sin z \).
  • A component along the direction \( \mathbf{k} \) given by \( xe^{2z} \).
Each of these is responsible for how the field changes as you move in the respective \( x, y, \) or \( z \) directions. Understanding how these components interact helps in visualizing how the vector field behaves globally.
Partial Derivative
Partial derivatives represent a critical tool in Calculus, especially when dealing with functions of multiple variables. They measure the rate at which a function changes as one of its input variables changes, holding the others constant.
The solution exercise involves finding the partial derivatives of the components of the vector field:
  • For \( F_1 = 3xy z^2 \), the partial derivative with respect to \( x \) is \( 3yz^2 \). This shows how \( F_1 \) varies when you change only \( x \).
  • For \( F_2 = y^2 \sin z \), the partial derivative with respect to \( y \) is \( 2y \sin z \), indicating changes in this component as \( y \) varies.
  • For \( F_3 = xe^{2z} \), the partial derivative with respect to \( z \) is \( 2xe^{2z} \), showing how \( F_3 \) changes with \( z \).
By calculating these, we grasp how each piece of the vector field contributes differently to its overall behavior.
Calculus 3
Calculus 3 refers to multivariable calculus, a branch that extends calculus to functions of several variables. Divergence, a concept in Calculus 3, measures a vector field's tendency to originate from or converge at a point in a region.
Using the divergence formula \(abla \cdot \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}\), we find how the vector field expands or compresses volume at any point. Here's how it applies:
  • The divergence \( abla \cdot \mathbf{F} = 3yz^2 + 2y \sin z + 2xe^{2z} \) quantifies the field's behavior.
  • Each term corresponds to the rate of expansion or contraction in its respective direction.
Grasping these ideas can be incredibly useful, not only in mathematics but in physics and engineering where such fields are commonly applied.

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

Use the divergence theorem to evaluate \(\iint_{S} \mathbf{F} \cdot d S\) \(\mathbf{F}(x, y, z)=(2 x y) \mathbf{i}+\left(-y^{2}\right) \mathbf{j}+\left(2 z^{3}\right) \mathbf{k}\) where \(S\) is bounded by paraboloid \(z=x^{2}+y^{2}\) and plane \(z=2\)

In the following exercises, find the work done by force field \(\mathbf{F}\) on an object moving along the indicated path. Compute the work done by force \(\mathbf{F}(x, y, z)=2 x \mathbf{i}+3 y \mathbf{j}-z \mathbf{k}\) along path \(\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+t^{3} \mathbf{k}, \quad\) where \(0 \leq t \leq 1\)

For the following exercises, evaluate the line integrals. Evaluate \(\int_{C} y z d x+x z d y+x y d z\) over the line segment from \((1,1,1)\) to \((3,2,0)\)

True or False? If \(C\) is given by $$x(t)=t, y(t)=t, 0 \leq \mathrm{t} \leq 1,\( then \)\int_{C} x y d s=\int_{0}^{1} t^{2} d t$$

In the following exercises, find the work done by force field \(\mathbf{F}\) on an object moving along the indicated path. \(\mathbf{F}(x, y)=-x \mathbf{i}-2 y \mathbf{j}\) \(C : y=x^{3}\) from \((0,0)\) to \((2,8)\)
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation
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.