/*! 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 24 Find \(f_{x}, f_{y},\) and \(f_{... [FREE SOLUTION] | 91Ó°ÊÓ

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Find \(f_{x}, f_{y},\) and \(f_{z}\). $$f(x, y, z)=x y+y z+x z$$

Short Answer

Expert verified
\(f_x = y + z\), \(f_y = x + z\), \(f_z = y + x\).

Step by step solution

01

Differentiate with respect to x

To find the partial derivative of the function with respect to \(x\), treat \(y\) and \(z\) as constants. Differentiating \(f(x, y, z) = x y + y z + x z\) with respect to \(x\) gives: \(\frac{\partial f}{\partial x} = y + z\).
02

Differentiate with respect to y

To find the partial derivative of the function with respect to \(y\), treat \(x\) and \(z\) as constants. Differentiating \(f(x, y, z) = x y + y z + x z\) with respect to \(y\) gives: \(\frac{\partial f}{\partial y} = x + z\).
03

Differentiate with respect to z

To find the partial derivative of the function with respect to \(z\), treat \(x\) and \(y\) as constants. Differentiating \(f(x, y, z) = x y + y z + x z\) with respect to \(z\) gives: \(\frac{\partial f}{\partial z} = y + x\).

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Key Concepts

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

Multivariable Calculus
Multivariable Calculus is an extension of calculus to functions of several variables. Think of it as extending the idea of calculus from one dimension, where you deal with functions of a single variable, to higher dimensions, dealing with functions involving two or more variables. This branch is essential because many real-world problems depend on multiple variables, such as calculating the rate of change in temperature over a space or analyzing profit changes in business across various factors.

Imagine you're analyzing the weather conditions in a city. The temperature could be affected by several variables such as time, latitude, and humidity, making it a challenge that Multivariable Calculus can help solve. At the heart of this subject, you have the concept of partial derivatives, which allow you to understand how a function changes as one specific variable changes, keeping the others constant.

Multivariable Calculus forms the foundation for a deeper understanding of phenomena in different fields like physics, engineering, and economics. It is crucial for tasks such as optimizing outputs and modeling natural occurrences.
Differentiation
Differentiation is a fundamental concept in calculus that focuses on finding the rate at which a function changes at any point. In simpler terms, it helps you understand how a function behaves by examining its slope or gradient.

When dealing with multivariable functions, differentiation takes a new twist because you have partial derivatives. These are essentially the derivatives of a function with respect to one variable while keeping others constant. For instance, in the provided function \( f(x, y, z) = xy + yz + xz \), differentiation with respect to \( x \) means treating \( y \) and \( z \) like constant numbers, resulting in \( \frac{\partial f}{\partial x} = y + z \).

Partial derivatives are instrumental in understanding functions of several variables, giving insight into how each variable contributes to the function's change. This information can help in optimization problems, where the goal is to find maximum or minimum values given certain conditions.
  • Key for modeling changes in multiple-variable systems
  • Helps in finding slopes in multiple directions
  • Vital for solving real-world problems with multivariable data
Function of Several Variables
A function of several variables is simply a function that depends on more than one input to determine its output. Unlike single-variable functions, these functions consider multidimensional input spaces, making them more complex and versatile.

In our example function \( f(x, y, z) = x y + y z + x z \), the output depends on the three variables \( x, y, \text{ and } z \). Adjusting any of these variables would result in a change in the function's output, showcasing the interdependent nature of these variables in real-world scenarios.

Working with functions of several variables involves techniques such as finding partial derivatives, optimizing functions, and even performing more advanced tasks like integrating over regions. These functions are crucial in fields that involve modeling systems with multiple inputs, for instance:
  • Economics, where demand might depend on price, income, and personal preferences
  • Physics, where the trajectory of a particle could rely on initial position, force, and time
  • Engineering, to calculate stress or strain based on multiple environmental factors
By understanding these functions, researchers and professionals can predict behaviors or improve system outcomes by adjusting the input variables accordingly.

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Most popular questions from this chapter

Among all closed rectangular boxes of volume \(27 \mathrm{cm}^{3},\) what is the smallest surface area?

If you cannot make any headway with \(\lim _{(x, y) \rightarrow(0,0)} f(x, y)\) in rectangular coordinates, try changing to polar coordinates. Substitute \(x=r \cos \theta, y=r \sin \theta,\) and investigate the limit of the resulting expression as \(r \rightarrow 0 .\) In other words, try to decide whether there exists a number \(L\) satisfying the following criterion: $$|r|<\delta \Rightarrow|f(r, \theta)-L|<\epsilon$$ If such an \(L\) exists, then $$\lim _{(x, y) \rightarrow(0,0)} f(x, y)=\lim _{r \rightarrow 0} f(r \cos \theta, r \sin \theta)=L$$ For instance, $$\lim _{(x, y) \rightarrow(0,0)} \frac{x^{3}}{x^{2}+y^{2}}=\lim _{r \rightarrow 0} \frac{r^{3} \cos ^{3} \theta}{r^{2}}=\lim _{r \rightarrow 0} r \cos ^{3} \theta=0$$ To verify the last of these equalities, we need to show that Equation (1) is satisfied with \(f(r, \theta)=r \cos ^{3} \theta\) and \(L=0 .\) That is, we need to show that given any \(\epsilon>0,\) there exists a \(\delta>0\) such that for all \(r\) and \(\theta\) $$|r|<\delta \Rightarrow\left|r \cos ^{3} \theta-0\right|<\epsilon$$ since $$\left|r \cos ^{3} \theta\right|=|r|\left|\cos ^{3} \theta\right| \leq|r| \cdot 1=|r|$$ the implication holds for all \(r\) and \(\theta\) if we take \(\delta=\epsilon\) In contrast, $$\frac{x^{2}}{x^{2}+y^{2}}=\frac{r^{2} \cos ^{2} \theta}{r^{2}}=\cos ^{2} \theta$$takes on all values from 0 to 1 regardless of how small \(|r|\) is, so that \(\lim _{(x, y) \rightarrow(0,0)} x^{2} /\left(x^{2}+y^{2}\right)\) does not exist. In each of these instances, the existence or nonexistence of the limit as \(r \rightarrow 0\) is fairly clear. Shifting to polar coordinates does not always help, however, and may even tempt us to false conclusions. For example, the limit may exist along every straight line (or ray) \(\theta=\) constant and yet fail to exist in the broader sense. Example 5 illustrates this point. In polar coordinates, \(f(x, y)=\left(2 x^{2} y\right) /\left(x^{4}+y^{2}\right)\) becomes $$f(r \cos \theta, r \sin \theta)=\frac{r \cos \theta \sin 2 \theta}{r^{2} \cos ^{4} \theta+\sin ^{2} \theta}$$for \(r \neq 0 .\) If we hold \(\theta\) constant and let \(r \rightarrow 0,\) the limit is \(0 .\) On the path \(y=x^{2},\) however, we have \(r \sin \theta=r^{2} \cos ^{2} \theta\) and $$\begin{aligned} f(r \cos \theta, r \sin \theta) &=\frac{r \cos \theta \sin 2 \theta}{r^{2} \cos ^{4} \theta+\left(r \cos ^{2} \theta\right)^{2}} \\ &=\frac{2 r \cos ^{2} \theta \sin \theta}{2 r^{2} \cos ^{4} \theta}=\frac{r \sin \theta}{r^{2} \cos ^{2} \theta}=1 \end{aligned}$$ Find the limit of \(f\) as \((x, y) \rightarrow(0,0)\) or show that the limit does not exist. $$f(x, y)=\frac{2 x}{x^{2}+x+y^{2}}$$

Find the linearizations \(L(x, y, z)\) of the functions at the given points. \(f(x, y, z)=\sqrt{x^{2}+y^{2}+z^{2}}\) at a. (1,0,0) b. (1,1,0) c. (1,2,2)

Extrema on a circle of intersection \(\quad\) Find the extreme values of the function \(f(x, y, z)=x y+z^{2}\) on the circle in which the plane \(y-x=0\) intersects the sphere \(x^{2}+y^{2}+z^{2}=4.\)

Find the linearizations \(L(x, y, z)\) of the functions at the given points. \(f(x, y, z)=\tan ^{-1}(x y z)\) at a. (1,0,0) b. (1,1,0) c. (1,1,1)

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