/*! 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 7 Prove: If \(f(x, y, z)\) is a co... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 7

Prove: If \(f(x, y, z)\) is a composite function \(F(u)\), where \(u=g(x, y, z)\), then grad \(f=\) \(F^{\prime}(u) \operatorname{grad} g\).

Short Answer

Expert verified
Question: Given that \(f(x, y, z)\) is a composite function of the form \(F(u)\), where \(u = g(x, y, z)\), prove that the gradient of \(f\) can be expressed as the product of the derivative of \(F\) with respect to \(u\) and the gradient of \(g\). Answer: If \(f(x, y, z)\) is a composite function \(F(u)\), where \(u=g(x, y, z)\), then \(\operatorname{grad} f = F^{\prime}(u) \operatorname{grad} g\).

Step by step solution

01

Compute the derivative of \(f\) with respect to \(x\), \(y\), and \(z\).

To find the gradient of \(f(x, y, z)\), we need to compute its partial derivatives with respect to \(x\), \(y\), and \(z\). To do this, we use the chain rule since \(f=F(u)\) and \(u=g(x, y, z)\). For the derivative with respect to \(x\), the chain rule states: \(\frac{\partial f}{\partial x} = \frac{\partial F}{\partial u} \frac{\partial u}{\partial x} = F^{\prime}(u) \frac{\partial g}{\partial x}\). Similarly, for the derivatives with respect to \(y\) and \(z\), we have: \(\frac{\partial f}{\partial y} = F^{\prime}(u) \frac{\partial g}{\partial y}\) and \(\frac{\partial f}{\partial z} = F^{\prime}(u) \frac{\partial g}{\partial z}\).
02

Form the gradient of \(f\) and \(g\)

The gradient of a function is a vector composed of its partial derivatives. So, we have: \(\operatorname{grad} f=\left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right)\) and \(\operatorname{grad} g=\left(\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}, \frac{\partial g}{\partial z}\right)\).
03

Substitute the derivatives from Step 1 into the gradient of f

Now, let's substitute the derivatives from Step 1 into the gradient of \(f\): \(\operatorname{grad} f=\left(F^{\prime}(u) \frac{\partial g}{\partial x}, F^{\prime}(u) \frac{\partial g}{\partial y}, F^{\prime}(u) \frac{\partial g}{\partial z}\right)\).
04

Factor out \(F^{\prime}(u)\) and obtain the final expression for the gradient of \(f\)

We can factor out \(F^{\prime}(u)\) from each component of the gradient of \(f\). Thus, we get: \(\operatorname{grad} f = F^{\prime}(u) \left(\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}, \frac{\partial g}{\partial z}\right) = F^{\prime}(u) \operatorname{grad} g\). This proves that if \(f(x, y, z)\) is a composite function \(F(u)\), where \(u=g(x, y, z)\), then \(\operatorname{grad} f = F^{\prime}(u) \operatorname{grad} g\).

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.

Chain Rule
The chain rule is a fundamental concept in calculus, particularly useful when dealing with composite functions. A composite function is created when one function is applied to the result of another function. In the context of partial derivatives, the chain rule allows us to find the derivative of a composite function with respect to one of its independent variables by linking the derivatives of the inner and outer functions.

For example, if we have a function in the form of f(x, y, z) = F(g(x, y, z)), where f is a composite function of u = g(x, y, z), and we want to compute the derivative of f with respect to x, the chain rule tells us to first take the derivative of the outer function F(u) with respect to u, and then multiply it by the derivative of the inner function g(x, y, z) with respect to x. This property is vital when we are trying to find the gradient of a composite function because it helps us decompose the process into more manageable steps.
Partial Derivatives
Partial derivatives are another cornerstone in the study of multivariable functions and vector calculus. They measure how a function changes as only one of its input variables changes, all other variables staying constant. In essence, a partial derivative gives us the rate of change of the function in the direction of one of the independent variables.

In the given example, grad f is the vector of partial derivatives of the function f with respect to each of its variables, x, y, and z. Here, we see partial derivatives playing a crucial role. By computing each partial derivative individually, using the chain rule, we build the components of the gradient vector. This allows us to understand the behavior and the steepest ascent direction of the function at any point in its domain. Understanding partial derivatives is indispensable for analyzing gradients in vector calculus.
Vector Calculus
Vector calculus extends the concepts of calculus to vector fields and functions of several variables. One of the key objects in vector calculus is the gradient vector. This vector gives the direction and magnitude of the steepest slope of a scalar field at a given point. In the context of our exercise, the gradient of f is found by taking the partial derivatives of f with respect to each variable and bundling them into a vector.

The beauty of using the gradient in vector calculus is its ability to succinctly express the rate of change in all directions of space. Moreover, when dealing with the gradient of a composite function, as in our problem, the chain rule provides a mechanism to link the gradient of the outer function to the gradient of the inner function, resulting in a product involving these gradients that still abides by the principles of vector calculus. This ability to manipulate and combine gradients is essential in fields ranging from physics to optimization in mathematics.

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

Prove (3.48). [Hint: Use the result of Problem 5 following Section 2.12.]

Normal curvature on \(S\). The results of Problem \(19(\mathrm{c})\) show that all paths \(C\) on \(S\) through a chosen point \(P_{0}\) with fixed tangent vector \(\mathbf{T}\) and principal normal \(\mathbf{N}\) at \(P_{0}\), have the same curvature \(\kappa\) at \(P_{0}\). Hence it suffices to consider plane sections of \(S\). We consider such sections by planes through \(P_{0}\), other than the tangent plane to \(S\) at \(P_{0}\), let \(\theta=\angle(\mathbf{N}, \mathbf{n})\) for each path \(C\) obtained as such a plane section. a) Show that for fixed \(\mathbf{T}\) and variable \(\mathbf{N}\) one has \(\kappa \cos \theta=\) const. at \(P_{0}\). Here the constant has absolute value \(\kappa_{n}\), where \(\kappa_{n}\), called the normal curvature of \(S\), is the curvature at \(P_{0}\) of the section of the surface containing the tangent line to the path at \(P_{0}\) and the normal line to the surface at \(P_{0}\). [Thus \(\kappa=\pm \kappa_{n}\); the \(\pm\) sign follows from the fact that either \(0 \leq \theta<\pi / 2\) or \(\pi / 2<\theta \leq \pi\) for all \(\mathbf{N}\).] b) Show that $$ \kappa_{n}=\left|L\left(\frac{d u}{d s}\right)^{2}+2 M \frac{d u}{d s} \frac{d v}{d s}+N\left(\frac{d v}{d s}\right)^{2}\right| $$

Show that in 3-dimensional space the exterior product \(w_{i j}=u_{i} \wedge v_{j}\) of two covariant vectors \(u_{i}, v_{j}\) can be interpreted in terms of the vector product. In detail, show that, in standard coordinates \(x, y, z\) with usual orientation and \(u_{1}=u_{x}, \ldots, v_{1}=v_{x}, \ldots\), the three components of the vector product \(\left(u_{x} \mathbf{i}+\cdots\right) \times\left(v_{x} \mathbf{i}+\cdots\right)\) appear with \(+\) or \(-\) signs as the nonzero components of \(w_{i j}\).

. Let \(u_{i}, v_{i}, w_{i j}\) be alternating covariant tensors in \(E^{3}\). a) Show that \(u_{i} v_{j}\) fails to be alternating if \(u_{1} v_{1} \neq 0\). b) Show that \(u_{i} w_{j k}\) fails to be alternating if \(u_{1} w_{21} \neq 0\). c) In one coordinate system let \(u_{i}=1\) for \(i=1,2,3\). Show that if \(u_{i} v_{j}\) is alternating, then \(v_{j}=0\) for \(j=1,2,3\).

Prove that the continuity equation (3.17) can be written in the form $$ \frac{\partial \rho}{\partial t}+\operatorname{grad} \rho \cdot \mathbf{u}+\rho \operatorname{div} \mathbf{u}=0 $$ or, in terms of the Stokes derivative (Problem 12 following Section 2.8), thus: $$ \frac{D \rho}{D t}+\rho \operatorname{div} \mathbf{u}=0 $$ Prove that (3.17) reduces to (3.18) when \(\rho \equiv\) const. It will be shown in Chapter 5 that the same simplification can be made when \(\rho\) is variable, provided that the fluid is incompressible. This follows from the fact that \(D \rho / D t\) measures the variation in density at a point moving with the fluid; for an incompressible fluid this local density cannot vary.
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation

Logic and Functions

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.