/*! 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 45 In Exercises \(43-46,\) find the... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 45

In Exercises \(43-46,\) find the linearization \(L(x, y, z)\) of the function \(f(x, y, z)\) at \(P_{0}\) . Then find an upper bound for the magnitude of the error \(E\) in the approximation \(f(x, y, z) \approx L(x, y, z)\) over the region \(R\) . $$ \begin{array}{l}{f(x, y, z)=x y+2 y z-3 x z \text { at } P_{0}(1,1,0)} \\ {R :|x-1| \leq 0.01, \quad|y-1| \leq 0.01, \quad|z| \leq 0.01}\end{array} $$

Short Answer

Expert verified
The linearization is \( L(x, y, z) = x + y - z - 1 \) with an error bound of \( 0.00009 \).

Step by step solution

01

Identify Partial Derivatives

To find the linearization of the function, first compute the partial derivatives of the given function \( f(x, y, z) = xy + 2yz - 3xz \) with respect to each variable: \( x, y, \) and \( z \).- \( \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (xy + 2yz - 3xz) = y - 3z \)- \( \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (xy + 2yz - 3xz) = x + 2z \)- \( \frac{\partial f}{\partial z} = \frac{\partial}{\partial z} (xy + 2yz - 3xz) = 2y - 3x \)
02

Evaluate Partial Derivatives at \( P_0 \)

Evaluate each partial derivative at the point \( P_0(1, 1, 0) \):- \( \frac{\partial f}{\partial x}(1, 1, 0) = 1 - 0 = 1 \)- \( \frac{\partial f}{\partial y}(1, 1, 0) = 1 + 0 = 1 \)- \( \frac{\partial f}{\partial z}(1, 1, 0) = 2 \cdot 1 - 3 \cdot 1 = -1 \)
03

Compute Linearization \( L(x, y, z) \)

The linearization is found using the formula:\[ L(x, y, z) = f(P_0) + \left( \frac{\partial f}{\partial x}(P_0) \right)(x - x_0) + \left( \frac{\partial f}{\partial y}(P_0) \right)(y - y_0) + \left( \frac{\partial f}{\partial z}(P_0) \right)(z - z_0) \]First, evaluate \( f \) at \( P_0 \):\[ f(1, 1, 0) = 1 \cdot 1 + 2 \cdot 1 \cdot 0 - 3 \cdot 1 \cdot 0 = 1 \]Substituting the values, we get:\[ L(x, y, z) = 1 + 1(x - 1) + 1(y - 1) - 1(z - 0) \]Simplifying:\[ L(x, y, z) = x + y - z - 1 \]
04

Determine Error Bound

To find the error bound for \( R : |x-1| \leq 0.01, |y-1| \leq 0.01, |z| \leq 0.01 \), we must consider second partial derivatives.Compute the second partial derivatives of each first partial derivative:- \( \frac{\partial^2 f}{\partial x^2} = 0, \quad \frac{\partial^2 f}{\partial y^2} = 0, \quad \frac{\partial^2 f}{\partial z^2} = 0 \)- \( \frac{\partial^2 f}{\partial x \partial y} = 1, \quad \frac{\partial^2 f}{\partial x \partial z} = -3, \quad \frac{\partial^2 f}{\partial y \partial z} = 2 \)The maximum value of any second partial derivative in \( R \) is 3 (in absolute terms, because \( -3 \) gives the largest magnitude).The error bound \( E \) can be approximated by:\[ E \leq \frac{1}{2} \cdot (0.01)^2 \cdot 3 \cdot (3) = 0.00009 \]Thus, the upper bound for the error is \( 0.00009 \).

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.

Partial Derivatives
Partial derivatives extend the concept of a derivative, which we usually apply to single-variable functions, to functions of multiple variables. In multivariable calculus, when a function depends on more than one variable, like our function \(f(x, y, z) = xy + 2yz - 3xz\), a partial derivative represents the rate of change of the function with respect to one of those variables, while keeping the others constant.
  • The notation \( \frac{\partial f}{\partial x} \) indicates the function's rate of change with respect to \(x\), treating \(y\) and \(z\) as constants. Similarly, \( \frac{\partial f}{\partial y} \) and \( \frac{\partial f}{\partial z} \) focus on \(y\) and \(z\) respectively.
  • The process begins by differentiating the function while focusing on the change of one variable at a time. For example, to find \( \frac{\partial f}{\partial x} \), differentiate \(f(x, y, z) = xy + 2yz - 3xz\) while considering \(y\) and \(z\) constant, resulting in \(y - 3z\).
Getting the partial derivatives correctly lays the foundation for further calculations, such as linearization and finding error bounds.
Linear Approximation
Linear approximation uses the concept of a tangent plane to approximate a multivariable function near a specific point, known as \(P_0\). This method can simplify complex or non-linear functions within a small region around the given point. To create the linearization \(L(x, y, z)\) of the function \(f(x, y, z)\) at \(P_0(1, 1, 0)\), follow these steps:
  • First, calculate the value of the function at \(P_0\). For our function, \(f(1, 1, 0) = 1\).
  • Next, incorporate the evaluated partial derivatives at \(P_0\), which we calculated earlier to be 1 for \( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \) and -1 for \( \frac{\partial f}{\partial z} \).
  • The linearized function \(L(x, y, z)\) is given by
    \[ L(x, y, z) = f(P_0) + \left( \frac{\partial f}{\partial x}(P_0) \right)(x - x_0) + \left( \frac{\partial f}{\partial y}(P_0) \right)(y - y_0) + \left( \frac{\partial f}{\partial z}(P_0) \right)(z - z_0) \] Resulting in
    \(L(x, y, z) = x + y - z - 1\).
This linear function approximates \(f(x, y, z)\) near \(P_0\), converting a potentially complex 3-variable context into a simpler, manageable linear form.
Error Bound
Error bounds are crucial for understanding the accuracy of a linear approximation. They provide a way to estimate how far off your linear function \(L(x, y, z)\) could be from the actual function \(f(x, y, z)\) within a specific region \(R\). The error is influenced by the second partial derivatives of the function. These second derivatives measure how the first derivatives (the slopes) change, giving insights into the function's curvature.
  • In our example, we calculated the second partial derivatives and obtained their maximum magnitudes within region \(R\).
  • The largest absolute value among the second derivatives was found to be 3, coming from the mixed partial derivative \( \frac{\partial^2 f}{\partial x \partial z} \).
  • The formula to estimate the error bound is:
    \[ E \leq \frac{1}{2} \cdot \text{(maximum second derivative's magnitude)} \cdot (\text{length of the interval})^2 \] Inserting our maximum and interval lengths, we get
    \(E \leq 0.00009\).
This upper bound ensures that the linear approximation \(L(x, y, z)\) deviates from the true value \(f(x, y, z)\) by no more than \(0.00009\) within region \(R\). This tight bound reinforces confidence in using \(L(x, y, z)\) for practical applications close to \(P_0\).

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. Maximum on line of intersection Find the maximum value of \(w=x y z\) on the line of intersection of the two planes \(x+y+z=40\) and \(x+y-z=0\) b. Give a geometric argument to support your claim that you have found a maximum, and not a minimum, value of \(w\) .

In Exercises \(65-70,\) you will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level critical plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. $$ \begin{array}{l}{f(x, y)=2 x^{4}+y^{4}-2 x^{2}-2 y^{2}+3, \quad-3 / 2 \leq x \leq 3 / 2} \\ {-3 / 2 \leq y \leq 3 / 2}\end{array} $$

Show that \(f(x, y, z)=x+y-z\) is continuous at every point \(\left(x_{0}, y_{0}, z_{0}\right) .\)

Show that \((0,0)\) is a critical point of \(f(x, y)=x^{2}+k x y+y^{2}\) no matter what value the constant \(k\) has. (Hint: Consider two cases: \(k=0\) and \(k \neq 0 . )\)

In Exercises \(31-36,\) find the linearization \(L(x, y)\) of the function \(f(x, y)\) at \(P_{0} .\) Then find an upper bound for the magnitude \(|E|\) of the error in the approximation \(f(x, y) \approx L(x, y)\) over the rectangle \(R\) . $$ \begin{array}{l}{f(x, y)=\ln x+\ln y \text { at } P_{0}(1,1)} \\ {R :|x-1| \leq 0.2, \quad|y-1| \leq 0.2}\end{array} $$
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Geometry

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.