/*! 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 49 Find the gradient \(\nabla f\) f... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the gradient \(\nabla f\) for \(f=x y+z x+y z\)

Short Answer

Expert verified
\( \nabla f = (y + z, x + z, x + y) \)

Step by step solution

01

Understand the Function

The function given is \( f(x, y, z) = xy + zx + yz \). This is a multivariable function with three variables: \( x \), \( y \), and \( z \).
02

Recognize the Gradient

The gradient of a function \( f(x, y, z) \) is a vector that consists of the partial derivatives of \( f \) with respect to each variable. It is represented as \( abla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right) \).
03

Calculate Partial Derivative with Respect to x

To find \( \frac{\partial f}{\partial x} \), differentiate \( f \) with respect to \( x \): \[\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (xy + zx + yz) = y + z\]
04

Calculate Partial Derivative with Respect to y

To find \( \frac{\partial f}{\partial y} \), differentiate \( f \) with respect to \( y \): \[\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (xy + zx + yz) = x + z\]
05

Calculate Partial Derivative with Respect to z

To find \( \frac{\partial f}{\partial z} \), differentiate \( f \) with respect to \( z \): \[\frac{\partial f}{\partial z} = \frac{\partial}{\partial z} (xy + zx + yz) = x + y\]
06

Write the Gradient Vector

Combine the partial derivatives calculated into a gradient vector:\[abla f = (y + z, x + z, x + y)\]

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.

Multivariable Calculus
Welcome to the world of multivariable calculus! This branch of calculus deals with functions that have more than one variable. In the exercise we examined, the function is given as \( f(x, y, z) = xy + zx + yz \). Here, we have three variables: \( x \), \( y \), and \( z \).

Multivariable calculus allows us to explore how changes in these variables affect the entire function. Instead of dealing with single-variable functions like \( f(x) \), we now deal with scenarios where numerous factors play a role,
  • leading to more complex models and calculations.
  • enabling us to address real-world phenomena that involve more than one changing element.
Understanding the landscape of a multivariable function is crucial in predicting outcomes and optimizing results, such as finding the maximum or minimum of a function.
With the tools and techniques from multivariable calculus, you can analyze everything from airflow over a wing to patterns of economic growth.
Partial Derivatives
Partial derivatives are a cornerstone of multivariable calculus. They allow us to study how a function changes as one variable changes, keeping the others constant. In our function \( f(x, y, z) = xy + zx + yz \), we calculated partial derivatives for each variable.

For example, to find \( \frac{\partial f}{\partial x} \), we look at how \( f \) changes if only \( x \) changes:
  • The term \( xy \) changes to \( y \).
  • The term \( zx \) changes to \( z \).
  • The term \( yz \) stays the same since \( x \) doesn’t affect it directly.
Hence, \( \frac{\partial f}{\partial x} = y + z \).

The process is similar for the other variables:
  • \( \frac{\partial f}{\partial y} = x + z \)
  • \( \frac{\partial f}{\partial z} = x + y \)
Having these partial derivatives is critical for understanding the gradient and provides a snapshot of the function's behavior in different directions.
Vector Calculus
Vector calculus extends our understanding of calculus into the realm of vectors. A core element of this field is the concept of the gradient, a vector that encapsulates the rate of change of a multivariable function in all dimensions.

In our exercise, the gradient \( abla f \) was a vector comprising the partial derivatives of \( f \), given by \( abla f = (y + z, x + z, x + y) \).
  • This vector displays the direction and rate of greatest increase of the function.
  • Each component shows how the function changes as we move along a particular axis.
  • The gradient can help us locate points of maximum stress or equilibrium in physical systems.
Vector calculus finds applications in fields like electromagnetics, fluid dynamics, and computer graphics. By understanding the gradient, you grasp how multi-dimensional vectors help navigate complex scenarios and solve intricate problems efficiently.

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

A body of mass \(m\) moves along the curve \(\boldsymbol{r}(t)=x(t) \boldsymbol{i}+y(t) \boldsymbol{j}+z(t) \boldsymbol{k}\), where \(x=2 \cos 3 t\), \(y=2 \sin 3 t\) and \(z=3 t\) at time \(t\). (i) Find the velocity and acceleration at time \(t\). (ii) Find the force acting on the body. Describe the motion of the body (iii) in the \(x\) - and \(y\) - directions, (iv) in the \(x y\)-plane, (v) in the \(z\)-direction, (vi) overall.

\text { Show that } \boldsymbol{a}=(1,2,3), \boldsymbol{b}=(0,-3,2) \text { and } \boldsymbol{c}=(-13,2,3) \text { are orthogonal vectors. }

A body undergoes a displacement from \(\boldsymbol{r}_{1}=(0,0,0)\) to \(\boldsymbol{r}_{2}=(2,3,1)\) under the influence of the conservative force \(\boldsymbol{F}=x \boldsymbol{i}+2 y \boldsymbol{j}+3 z \boldsymbol{k}\). (i) Calculate the work \(W\left(\boldsymbol{r}_{1} \rightarrow \boldsymbol{r}_{2}\right)\) done on the body. (ii) Find the potential-energy function \(V(\boldsymbol{r})\) of which the components of the force are \((-)\) the partial derivatives. (iii) Confirm that \(W\left(\boldsymbol{r}_{1} \rightarrow \boldsymbol{r}_{2}\right)=V\left(\boldsymbol{r}_{1}\right)-V\left(\boldsymbol{r}_{2}\right)\).

\text { Find the value of } \lambda \text { for which } \boldsymbol{a}=(\lambda, 3,1) \text { and } \boldsymbol{b}=(2,1,-1) \text { are orthogonal. }

A charge \(q\) moving with velocity \(v\) in the presence of an electric field \(\boldsymbol{E}\) and a magnetic field \(\mathbf{B}\) experiences a total force \(\boldsymbol{F}=q \boldsymbol{E}+q \boldsymbol{v} \times \boldsymbol{B}\) called the Lorentz force. Calculate the force acting on the charge \(q=3\) moving with velocity \(v=(2,3,1)\) in the presence of the electric field \(\boldsymbol{E}=2 \boldsymbol{i}\) and magnetic field \(\boldsymbol{B}=3 \boldsymbol{j}\).

See all solutions

Recommended explanations on Chemistry 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.