/*! 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 13 Calculate the gradient \(\nabla ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Calculate the gradient \(\nabla f\) of the following functions, \(f(x, y, z):(\mathbf{a}) f=\ln (r),\) (b) \(f=r^{n}\), (c) \(f=g(r),\) where \(r=\sqrt{x^{2}+y^{2}+z^{2}}\) and \(g(r)\) is some unspecified function of \(r .\) [Hint: Use the chain rule.]

Short Answer

Expert verified
(a) \(\nabla f = \left(\frac{x}{r^2}, \frac{y}{r^2}, \frac{z}{r^2}\right)\), (b) \(\nabla f = \left(nx r^{n-2}, ny r^{n-2}, nz r^{n-2}\right)\), (c) \(\nabla f = g'(r)\left(\frac{x}{r}, \frac{y}{r}, \frac{z}{r}\right)\).

Step by step solution

01

Understand the Gradient Concept

The gradient of a function is a vector of partial derivatives. For a function \( f(x, y, z) \), the gradient is given by \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \). This vector points in the direction of the greatest rate of increase of the function.
02

Define \( r \) and its Partial Derivatives

Given \( r = \sqrt{x^2 + y^2 + z^2} \), recognize \( r \) as the distance from the origin to a point \( (x, y, z) \). Calculate partial derivatives: \( \frac{\partial r}{\partial x} = \frac{x}{r} \), \( \frac{\partial r}{\partial y} = \frac{y}{r} \), \( \frac{\partial r}{\partial z} = \frac{z}{r} \) using the chain rule.
03

Calculate \( \nabla f \) for \( f = \ln(r) \)

Using the chain rule, \( \frac{\partial f}{\partial x} = \frac{1}{r} \cdot \frac{\partial r}{\partial x} = \frac{x}{r^2} \), \( \frac{\partial f}{\partial y} = \frac{y}{r^2} \), \( \frac{\partial f}{\partial z} = \frac{z}{r^2} \). Thus, \( abla f = \left( \frac{x}{r^2}, \frac{y}{r^2}, \frac{z}{r^2} \right) \).
04

Calculate \( \nabla f \) for \( f = r^n \)

Use the chain rule: \( \frac{\partial f}{\partial x} = n r^{n-1} \cdot \frac{\partial r}{\partial x} = n x r^{n-2} \), \( \frac{\partial f}{\partial y} = n y r^{n-2} \), \( \frac{\partial f}{\partial z} = n z r^{n-2} \). Therefore, \( abla f = \left( n x r^{n-2}, n y r^{n-2}, n z r^{n-2} \right) \).
05

Calculate \( \nabla f \) for \( f = g(r) \)

Using the chain rule: \( \frac{\partial f}{\partial x} = g'(r) \cdot \frac{\partial r}{\partial x} = g'(r) \frac{x}{r} \), \( \frac{\partial f}{\partial y} = g'(r) \frac{y}{r} \), \( \frac{\partial f}{\partial z} = g'(r) \frac{z}{r} \). So, \( abla f = \left( g'(r) \frac{x}{r}, g'(r) \frac{y}{r}, g'(r) \frac{z}{r} \right) \).

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 91Ó°ÊÓ!

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 that helps us find the derivative of composite functions. In simple terms, it's how we deal with functions nested within other functions, kind of like peeling an onion. When we deal with multivariable functions, such as those involving three or more variables, the chain rule provides a pathway to differentiate these functions by connecting their various layers.
For instance, consider a function that depends on another function, like how our exercise involves function values depending on the variable \( r \). Here, \( r = \sqrt{x^2 + y^2 + z^2} \), is an intermediary function that changes the outcome of the final expression \( f(r) \).
Using the chain rule, we determine how our function \( f \) changes as its inner function \( r \) changes, allowing us to find partial derivatives linked with \( r \). These derivatives connect through the chain rule providing:
  • \( \frac{\partial f}{\partial x} = \frac{df}{dr} \cdot \frac{\partial r}{\partial x} \)
  • \( \frac{\partial f}{\partial y} = \frac{df}{dr} \cdot \frac{\partial r}{\partial y} \)
  • \( \frac{\partial f}{\partial z} = \frac{df}{dr} \cdot \frac{\partial r}{\partial z} \)
Mastering the chain rule equips you to systematically tackle complex derivative problems in multivariable calculus.
Gradient Vector
The gradient vector, often denoted as \( abla f \), is an essential tool in vector calculus. It is a vector composed of all the partial derivatives of a function. In layman's terms, it tells you the direction in space where the function increases the most. This concept is crucial in fields like optimization and physics.
For example, with a function \( f(x, y, z) \) dependent on three variables, the gradient \( abla f \) forms the vector:
  • \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \)
This vector shows us how each variable affects the function's change. It's like having a compass pointing towards the steepest ascent of a hill.
When calculating the gradient in our exercise, we make use of \( r = \sqrt{x^2 + y^2 + z^2} \) to find the individual changes along the \( x, y, \) and \( z \) axes. This solution aids in determining how each coordinate direction affects the function's value.
Partial Derivatives
Partial derivatives provide a way to understand how a function changes when altering only one variable while keeping others constant. They are to multivariable calculus what regular derivatives are to single-variable calculus.
To derive these partial derivatives, we treat the function as dependent solely on one variable at a time. For example, from the function \( r = \sqrt{x^2 + y^2 + z^2} \), calculating the partial derivative with respect to \( x \) gives \( \frac{\partial r}{\partial x} = \frac{x}{r} \). The same goes for finding the derivatives with respect to \( y \) and \( z \).
  • \( \frac{\partial r}{\partial y} = \frac{y}{r} \)
  • \( \frac{\partial r}{\partial z} = \frac{z}{r} \)
By conducting this operation, you crack open the behavior of our function from various angles, allowing us to compute things like the gradient effectively.
Partial derivatives are not just abstract notions; they represent real changes in systems and are vital in understanding gradients and applying the chain rule in complex scenarios.

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

If a particle's potential energy is \(U(\mathbf{r})=k\left(x^{2}+y^{2}+z^{2}\right),\) where \(k\) is a constant, what is the force on the particle?

In one dimension, it is obvious that a force obeying Hooke's law is conservative (since \(F=-k x\) depends only on the position \(x,\) and this is sufficient to guarantee that \(F\) is conservative in one dimension). Consider instead a spring that obeys Hooke's law and has one end fixed at the origin, but whose other end is free to move in all three dimensions. (The spring could be fastened to a point in the ceiling and be supporting a bouncing mass \(m\) at its other end, for instance.) Write down the force \(\mathbf{F}(\mathbf{r})\) exerted by the spring in terms of its length \(r\) and its equilibrium length \(r_{\mathrm{o}} .\) Prove that this force is conservative. [Hints: Is the force central? Assume that the spring does not bend.]

Calculate the gradient \(\nabla f\) of the following functions, \(f(x, y, z):\) (a) \(f=x^{2}+z^{3} .\) (b) \(f=k y\), where \(k\) is a constant. (c) \(f=r \equiv \sqrt{x^{2}+y^{2}+z^{2}} .\) [Hint: Use the chain rule.] (d) \(f=1 / r\).

Which of the following forces is conservative? (a) \(\mathbf{F}=k(x, 2 y, 3 z)\) where \(k\) is a constant. (b) \(\mathbf{F}=k(y, x, 0) .\) (c) \(\mathbf{F}=k(-y, x, 0)\). For those which are conservative, find the corresponding potential energy \(U,\) and verify by direct differentiation that \(\mathbf{F}=-\nabla U\).

Near to the point where I am standing on the surface of Planet \(X\), the gravitational force on a mass \(m\) is vertically down but has magnitude \(m \gamma y^{2}\) where \(\gamma\) is a constant and \(y\) is the mass's height above the horizontal ground. (a) Find the work done by gravity on a mass \(m\) moving from \(\mathbf{r}_{1}\) to \(\mathbf{r}_{2}\), and use your answer to show that gravity on Planet \(X,\) although most unusual, is still conservative. Find the corresponding potential energy. (b) Still on the same planet, I thread a bead on a curved, frictionless, rigid wire, which extends from ground level to a height \(h\) above the ground. Show clearly in a picture the forces on the bead when it is somewhere on the wire. (Just name the forces so it's clear what they are; don't worry about their magnitude.) Which of the forces are conservative and which are not? (c) If I release the bead from rest at a height \(h\), how fast will it be going when it reaches the ground?

See all solutions

Recommended explanations on Physics Textbooks

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.