/*! 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 26 Let \(\mathbf{F}_{1}\) and \(\ma... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(\mathbf{F}_{1}\) and \(\mathbf{F}_{2}\) be differentiable vector fields and let \(a\) and \(b\) be arbitrary real constants. Verify the following identities. $$ \begin{array}{l}{\text { a. } \nabla \cdot\left(a \mathbf{F}_{1}+b \mathbf{F}_{2}\right)=a \nabla \cdot \mathbf{F}_{1}+b \nabla \cdot \mathbf{F}_{2}} \\ {\text { b. } \nabla \times\left(a \mathbf{F}_{1}+b \mathbf{F}_{2}\right)=a \nabla \times \mathbf{F}_{1}+b \nabla \times \mathbf{F}_{2}} \\ {\text { c. } \nabla \cdot\left(\mathbf{F}_{1} \times \mathbf{F}_{2}\right)=\mathbf{F}_{2} \cdot \nabla \times \mathbf{F}_{1}-\mathbf{F}_{1} \cdot \nabla \times \mathbf{F}_{2}}\end{array} $$

Short Answer

Expert verified
All identities are verified using linearity and known vector calculus identities.

Step by step solution

01

Understand the Problem

The task is to verify three vector calculus identities involving divergence \( abla \cdot \) and curl \( abla \times \) operations on linear combinations and cross products of vector fields. We are given two differentiable vector fields \( \mathbf{F}_1 \) and \( \mathbf{F}_2 \) and constants \( a \) and \( b \).
02

Derive the Divergence of a Linear Combination

Using the linearity property of the divergence operator, apply the divergence \( abla \cdot \) to the expression \( a \mathbf{F}_1 + b \mathbf{F}_2 \):\[abla \cdot (a \mathbf{F}_1 + b \mathbf{F}_2) = a (abla \cdot \mathbf{F}_1) + b (abla \cdot \mathbf{F}_2) \]This identity holds as divergence is a linear operator.
03

Derive the Curl of a Linear Combination

Using the linearity property of the curl operator, apply the curl \( abla \times \) to the expression \( a \mathbf{F}_1 + b \mathbf{F}_2 \):\[abla \times (a \mathbf{F}_1 + b \mathbf{F}_2) = a (abla \times \mathbf{F}_1) + b (abla \times \mathbf{F}_2) \]This identity holds due to the linear nature of the curl operation.
04

Compute Divergence of a Cross Product

For the divergence of a cross product, use the vector calculus identity:\[abla \cdot (\mathbf{F}_1 \times \mathbf{F}_2) = \mathbf{F}_2 \cdot (abla \times \mathbf{F}_1) - \mathbf{F}_1 \cdot (abla \times \mathbf{F}_2)\]This can be proven using the Levi-Civita symbol and properties of dot and cross products. The identity holds from known vector calculus results.

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Key Concepts

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

Divergence
In vector calculus, the divergence of a vector field provides a measure of how much a point functions as a source or sink for the field. You can think of it as indicating the 'outflow' of a vector field at a certain point. Mathematically, for a vector field \(\mathbf{F} = \langle F_x, F_y, F_z \rangle\), the divergence is expressed as \(abla \cdot \mathbf{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}\). If the divergence is positive, the point behaves like a source, and if negative, like a sink.

The problem asks us to verify the linearity of the divergence operator. This means when you take the divergence of a linear combination of vector fields, like \(a \mathbf{F}_{1} + b \mathbf{F}_{2}\), it results in the linear combination of their divergences: \(a abla \cdot \mathbf{F}_{1} + b abla \cdot \mathbf{F}_{2}\). The principle of linearity makes operations with vector fields more manageable and predictable.
Curl
The curl of a vector field shows the rotation or the 'twist' of the field around a point. Think of it like how much and which way the field spins at a point in space. It's particularly useful for understanding rotational flow in physics, like how air swirls around in a vortex. For a vector field \(\mathbf{F} = \langle F_x, F_y, F_z \rangle\), the curl is computed as follows:

\[ abla \times \mathbf{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}, \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}, \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \]

Just like with divergence, one of the exercises is to verify the linearity of the curl. When computing the curl of a linear combination of vector fields \(a \mathbf{F}_1 + b \mathbf{F}_2\), it separates into the curls of the individual fields weighted by the constants: \(a abla \times \mathbf{F}_{1} + b abla \times \mathbf{F}_{2}\). This property simplifies complex magnetic and fluid dynamic problems.
Cross Product Identities
Cross product identities in vector calculus are essential to relate different vector operations. These identities are particularly useful in solving problems involving torque, angular momentum, and electromagnetism. The cross product of two vectors \(\mathbf{F}_{1}\) and \(\mathbf{F}_{2}\) is a vector perpendicular to both \(\mathbf{F}_{1}\) and \(\mathbf{F}_{2}\), given by:

\( \mathbf{F}_{1} \times \mathbf{F}_{2} \).

The original exercise involves a specific identity called the 'divergence of a cross product'. This identity is \(abla \cdot (\mathbf{F}_{1} \times \mathbf{F}_{2}) = \mathbf{F}_{2} \cdot abla \times \mathbf{F}_{1} - \mathbf{F}_{1} \cdot abla \times \mathbf{F}_{2}\). This can be derived using vector identities and Levi-Civita symbols and provides a relation between the divergence and the curl of vector fields involved in cross multiplication. Understanding these identities can greatly aid in simplifying and solving complex physical problems.

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Most popular questions from this chapter

Use a CAS to perform the following steps for finding the work done by force \(\mathbf{F}\) over the given path: a. Find \(d \mathbf{r}\) for the path \(\mathbf{r}(t)=g(t) \mathbf{i}+h(t) \mathbf{j}+k(t) \mathbf{k}\) b. Evaluate the force \(\mathbf{F}\) along the path. c. Evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) $$ \begin{array}{l}{\mathbf{F}=2 x y \mathbf{i}-y^{2} \mathbf{j}+z e^{x} \mathbf{k} ; \quad \mathbf{r}(t)=-t \mathbf{i}+\sqrt{t} \mathbf{j}+3 t \mathbf{k}} \\ {1 \leq t \leq 4}\end{array} $$

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 \(\mathbf{r}(u, \boldsymbol{v})=f(u, v) \mathbf{i}+g(u, v) \mathbf{j}+h(u, v) \mathbf{k}\) 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 \(P_{0}\) . 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 \(\quad\) The parabolic cylinder surface \(\mathbf{r}(x, y)=\) \(x \mathbf{i}+y \mathbf{j}-x^{2} \mathbf{k},-\infty

Find the surface integral of the field \(\mathbf{F}\) over the portion of the given surface in the specified direction. \(\begin{array}{l}{\mathbf{F}(x, y, z)=-\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}} \\\ {S : \text { rectangular surface } z=0, \quad 0 \leq x \leq 2, \quad 0 \leq y \leq 3} \\ {\quad \text { direction } \mathbf{k}}\end{array}\)

Find the flux of the field \(\mathbf{F}(x, y, z)=z^{2} \mathbf{i}+x \mathbf{j}-3 z \mathbf{k}\) outward through the surface cut from the parabolic cylinder \(z=4-y^{2}\) by the planes \(x=0, x=1,\) and \(z=0\).

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 \(\mathbf{r}(u, \boldsymbol{v})=f(u, v) \mathbf{i}+g(u, v) \mathbf{j}+h(u, v) \mathbf{k}\) 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 \(P_{0}\) . 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. Hemisphere The hemisphere surface \(\mathbf{r}(\phi, \theta)=(4 \sin \phi \cos \theta) \mathbf{i}\) \(+(4 \sin \phi \sin \theta) \mathbf{j}+(4 \cos \phi) \mathbf{k}, 0 \leq \phi \leq \pi / 2,0 \leq \theta \leq 2 \pi\) at the point \(P_{0}(\sqrt{2}, \sqrt{2}, 2 \sqrt{3})\) corresponding to \((\phi, \theta)=\) \((\pi / 6, \pi / 4)\)

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