/*! 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 6 If \(\mathbf{v}=x \mathbf{i}+x^{... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 26: Problem 6

If \(\mathbf{v}=x \mathbf{i}+x^{2} y \mathbf{j}-3 x^{3} \mathbf{k}\), and \(\phi=x y z\), find \(\phi \mathbf{v}, \frac{\partial}{\partial x}(\phi \mathbf{v}), \frac{\partial \phi}{\partial x}, \frac{\partial \mathbf{v}}{\partial x}\). Deduce that \(\frac{\partial}{\partial x}(\phi \mathbf{v})=\phi \frac{\partial \mathbf{v}}{\partial x}+\frac{\partial \phi}{\partial x} \mathbf{v}\)

Short Answer

Expert verified
\( \frac{\partial}{\partial x}(\phi \mathbf{v}) = \phi \frac{\partial \mathbf{v}}{\partial x} + \frac{\partial \phi}{\partial x} \mathbf{v} \) is verified.

Step by step solution

01

Calculate \( \phi \mathbf{v} \)

Find the product of the scalar function \( \phi = xyz \) and the vector function \( \mathbf{v} = x \mathbf{i} + x^2 y \mathbf{j} - 3x^3 \mathbf{k} \). Multiply each component of \( \mathbf{v} \) by the scalar \( \phi \).\[ \phi \mathbf{v} = (xyz)(x \mathbf{i}) + (xyz)(x^2 y \mathbf{j}) - (xyz)(3x^3 \mathbf{k}) \]This simplifies to:\[ \phi \mathbf{v} = x^2 yz \mathbf{i} + x^3 y^2 z \mathbf{j} - 3x^4 yz \mathbf{k} \].
02

Compute \( \frac{\partial}{\partial x}(\phi \mathbf{v}) \)

Differentiate each component of \( \phi \mathbf{v} \) with respect to \( x \):- For \( x^2 yz \mathbf{i} \), the derivative is \( 2xyz \mathbf{i} \).- For \( x^3 y^2 z \mathbf{j} \), the derivative is \( 3x^2 y^2 z \mathbf{j} \).- For \( -3x^4 yz \mathbf{k} \), the derivative is \( -12x^3 yz \mathbf{k} \).Thus, \[ \frac{\partial}{\partial x}(\phi \mathbf{v}) = 2xyz \mathbf{i} + 3x^2 y^2 z \mathbf{j} - 12x^3 yz \mathbf{k} \].
03

Find \( \frac{\partial \phi}{\partial x} \)

Differentiate \( \phi = xy z \) with respect to \( x \):\[ \frac{\partial \phi}{\partial x} = yz \].
04

Calculate \( \frac{\partial \mathbf{v}}{\partial x} \)

Differentiate each component of \( \mathbf{v} = x \mathbf{i} + x^2 y \mathbf{j} - 3x^3 \mathbf{k} \) with respect to \( x \):- For \( x \mathbf{i} \), the derivative is \( \mathbf{i} \).- For \( x^2 y \mathbf{j} \), the derivative is \( 2xy \mathbf{j} \).- For \( -3x^3 \mathbf{k} \), the derivative is \( -9x^2 \mathbf{k} \).Thus, \[ \frac{\partial \mathbf{v}}{\partial x} = \mathbf{i} + 2xy \mathbf{j} - 9x^2 \mathbf{k} \].
05

Verify the Product Rule for Vector Functions

Now, verify the identity:\[ \frac{\partial}{\partial x}(\phi \mathbf{v}) = \phi \frac{\partial \mathbf{v}}{\partial x} + \frac{\partial \phi}{\partial x} \mathbf{v} \]Compute each term:- \( \phi \frac{\partial \mathbf{v}}{\partial x} = xyz(\mathbf{i} + 2xy \mathbf{j} - 9x^2 \mathbf{k}) = xyz \mathbf{i} + 2x^2 y^2 z \mathbf{j} - 9x^3 yz \mathbf{k} \)- \( \frac{\partial \phi}{\partial x} \mathbf{v} = yz(x \mathbf{i} + x^2 y \mathbf{j} - 3x^3 \mathbf{k}) = x yz \mathbf{i} + x^2 y^2 z \mathbf{j} - 3x^3 yz \mathbf{k} \)Combine them:\[ xyz \mathbf{i} + 2x^2 y^2 z \mathbf{j} - 9x^3 yz \mathbf{k} + x yz \mathbf{i} + x^2 y^2 z \mathbf{j} - 3x^3 yz \mathbf{k} \]This simplifies to:\[ 2xyz \mathbf{i} + 3x^2 y^2 z \mathbf{j} - 12x^3 yz \mathbf{k} \]which matches the expression for \( \frac{\partial}{\partial x}(\phi \mathbf{v}) \).

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.

Product Rule
The product rule is a fundamental concept in calculus that helps us differentiate the product of two functions. In the context of vector calculus, it becomes necessary when dealing with the derivative of a scalar function multiplied by a vector function. The basic idea of the product rule for functions \( u(x) \) and \( v(x) \) is given by:
  • \((uv)' = u'v + uv'\)
This principle can be extended to vector calculus, where we consider a scalar function \( \phi \) and a vector function \( \mathbf{v} \). The derivative of their product is given by:
  • \( \frac{\partial}{\partial x}(\phi \mathbf{v}) = \phi \frac{\partial \mathbf{v}}{\partial x} + \frac{\partial \phi}{\partial x} \mathbf{v} \)
This rule allows us to neatly express the derivative of the vector function's product, separating it into components related to the scalar and vector functions individually.
Vector Function Derivatives
Vector function derivatives are an extension of function derivatives to vector-valued functions. For a vector function \( \mathbf{v}(x) = f_1(x)\mathbf{i} + f_2(x)\mathbf{j} + f_3(x)\mathbf{k} \), taking the derivative with respect to \( x \) involves finding the derivative of each component function:
  • \( \frac{\partial \mathbf{v}}{\partial x} = \frac{\partial f_1}{\partial x}\mathbf{i} + \frac{\partial f_2}{\partial x}\mathbf{j} + \frac{\partial f_3}{\partial x}\mathbf{k} \)
This results in a new vector whose components are the derivatives of the original function's components. In the context of our exercise, this allows us to find \( \frac{\partial \mathbf{v}}{\partial x} \) by differentiating each part of \( \mathbf{v} \) with respect to \( x \). Understanding this, we can better work with derivatives involving vector functions.
Scalar and Vector Multiplication
Multiplying scalar and vector quantities requires a careful application of multiplication principles. A scalar function \( \phi \) can be multiplied to a vector function \( \mathbf{v} \) by scaling each component of the vector function. The scalar multiplication is distributive across the vector components:
  • \( \phi \mathbf{v} = \phi(f_1 \mathbf{i} + f_2 \mathbf{j} + f_3 \mathbf{k}) = \phi f_1 \mathbf{i} + \phi f_2 \mathbf{j} + \phi f_3 \mathbf{k} \)
This principle emphasizes how the scalar uniformly affects all parts of the vector. In our calculation, \( \phi \mathbf{v} \) was computed by multiplying \( \phi = xyz \) with each part of the given vector \( \mathbf{v} \). Resultantly, this process allows us to transform a vector function by a scalar, preparing it for differentiation.
Partial Differentiation
Partial differentiation is a technique used when dealing with functions of multiple variables. It involves taking the derivative of the function with respect to one of the variables while keeping the other variables constant. If you have a function like \( \phi = xyz \), the partial derivative with respect to \( x \) is calculated as if \( y \) and \( z \) are constants:
  • \( \frac{\partial \phi}{\partial x} = yz \)
This allows us to find how the function behaves when one specific variable changes. In our exercise, both scalar and vector functions required partial differentiation to apply the product rule correctly. Understanding partial derivatives is crucial for analyzing functions that depend on more than one variable across calculus and vector calculus.

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

For any vector field \(\mathbf{F}\) is \(\nabla \cdot(-\mathbf{F})\) the same as \(-(\nabla \cdot \mathbf{F}) ?\)

Given \(\phi=3 x^{2} y+x z\) (a) determine \(\nabla \phi\) (b) determine \(\nabla(7 \phi)\) (c) determine \(7 \nabla \phi\) (d) Is \(\nabla(7 \phi)\) the same as \(7 \nabla \phi\) ?

If \(\mathbf{v}=\nabla \phi\), find \(\phi\) when \(\mathbf{v}=\left(2 x-4 y^{2}\right) \mathbf{i}-8 x y \mathbf{j}\)

Suppose \(\mathbf{F}=P(x, y) \mathbf{i}+Q(x, y) \mathbf{j}\) is a two-dimensional vector field. Show that \(\mathbf{F}\) is irrotational if \(\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}\).

Is \(-\nabla \phi\) the same as \(\nabla(-\phi)\) ? Explain your answer.
See all solutions

Recommended explanations on Physics Textbooks

Conservation of Energy and Momentum

Read Explanation

Quantum Physics

Read Explanation

Waves Physics

Read Explanation

Mechanics and Materials

Read Explanation

Wave Optics

Read Explanation

Thermodynamics

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.