Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 16

For \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\) and \(r=|\mathbf{r}|\), simplify the following. a. \(\nabla \times(\mathbf{k} \times \mathbf{r})\). b. \(\nabla \cdot\left(\frac{\mathrm{r}}{r}\right)\). c. \(\nabla \times\left(\frac{\mathbf{r}}{r}\right)\). d. \(\nabla \cdot\left(\frac{\mathrm{r}}{r^{3}}\right)\). e. \(\nabla \times\left(\frac{\mathrm{r}}{r^{3}}\right)\).

Short Answer

Expert verified
The answers for the exercise are: (a) \( \mathbf{i} \), (b) 3, (c) 0, (d) 0, and (e) 0.

Step by step solution

01

Solution for (a)

To get the result of \(\nabla \times(\mathbf{k} \times \mathbf{r})\), notice that \( \mathbf{k} \times \mathbf{r} = - y \mathbf{i} + x \mathbf{j} \). Now apply the curl operator \( \nabla \times \) to get the result as \( (0 - (-1)) \mathbf{i} + ((0 - 0) \mathbf{j} + (0 - 0) \mathbf{k} = \mathbf{i} \).
02

Solution for (b)

For \(\nabla \cdot \left( \frac{\mathbf{r}}{r} \right)\), notice that \( \frac{\mathbf{r}}{r} \) is a unit vector. Now apply the divergence operator (\( \nabla \cdot \)) on it, thus \( \nabla \cdot ( \frac{\mathbf{r}}{r} ) = 3 \).
03

Solution for (c)

For \(\nabla \times \left( \frac{\mathbf{r}}{r} \right)\), notice that \( \frac{\mathbf{r}}{r} \) is a unit vector. Considering \(\nabla \times \) is a curl operator, it will yield 0 when applied to an uniform vector field, thus \(\nabla \times \left( \frac{\mathbf{r}}{r} \right) = 0\).
04

Solution for (d)

In \(\nabla \cdot\left( \frac{\mathrm{r}}{r^{3}} \right)\), remember \(\nabla \cdot \) is the divergence operator and \( \frac{r}{r^{3}} \) simplifies to \( \frac{1}{r^{2}} \). Thus, the divergence operator applied to \( \frac{1}{r^{2}} \) yields 0.
05

Solution for (e)

Just like in previous steps, for \(\nabla \times \left( \frac{\mathbf{r}}{r^{3}} \right)\), considering that \(\nabla \times\) is the curl operator and it applied to \( \frac{\mathbf{r}}{r^{3}} \), we will also get the result as 0.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Compute the following determinants using the permutation symbol. Verify your answer. a. \(\left|\begin{array}{ccc}3 & 2 & 0 \\ 1 & 4 & -2 \\ -1 & 4 & 3\end{array}\right|\) b. \(\left|\begin{array}{ccc}1 & 2 & 2 \\ 4 & -6 & 3 \\ 2 & 3 & 1\end{array}\right|\)

Use Stokes' Theorem to evaluate the integral $$ \int_{C}-y^{3} d x+x^{3} d y-z^{3} d z $$ for \(C\) the (positively oriented) curve of intersection between the cylinder \(x^{2}+y^{2}=1\) and the plane \(x+y+z=1\)

Zaphod Beeblebrox was in trouble after the infinite improbability drive caused the Heart of Gold, the spaceship Zaphod had stolen when he was President of the Galaxy, to appear between a small insignificant planet and its hot sun. The temperature of the ship's hull is given by \(T(x, y, z)=\) \(e^{-k\left(x^{2}+y^{2}+z^{2}\right)}\) Nivleks. He is currently at \((1,1,1)\), in units of globs, and \(k=2\) globs \(^{-2}\). (Check the Hitchhikers Guide for the current conversion of globs to kilometers and Nivleks to Kelvin.) a. In what direction should he proceed so as to decrease the temperature the quickest? b. If the Heart of Gold travels at \(e^{6}\) globs per second, then how fast will the temperature decrease in the direction of fastest decline?

Compute the gradient of the following: a. \(f(x, y)=x^{2}-y^{2}\) b. \(f(x, y, z)=y z+x y+x z\). c. \(f(x, y)=\tan ^{-1}\left(\frac{y}{x}\right)\). d. \(f(x, y, z)=2 y^{x} \cos z-5 \sin z \cos y\).

Let \(C\) be a closed curve and \(D\) the enclosed region. Prove the identity $$ \int_{C} \phi \nabla \phi \cdot \mathbf{n} d s=\int_{D}\left(\phi \nabla^{2} \phi+\nabla \phi \cdot \nabla \phi\right) d A. $$
See all solutions

Recommended explanations on Combined Science Textbooks

Synergy

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.

Sign-up for free