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Chapter 14: Problem 45

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 \(f(x, y, z)=L(x, y, z)\) over the region \(R\) $$\begin{aligned}&f(x, y, z)=x z-3 y z+2 \text { at } P_{0}(1,1,2)\\\&R: \quad|x-1| \leq 0.01, \quad|y-1| \leq 0.01, \quad|z-2|\leq 0.02\end{aligned}$$

Short Answer

Expert verified
Linearization is \( L(x, y, z) = 2x - 6y - 2z + 6 \), and error bound is \( 0.0009 \).

Step by step solution

01

Calculate Partial Derivatives

To linearize the function, we first need the partial derivatives of \( f(x, y, z) = xz - 3yz + 2 \). Calculate \( f_x, f_y, \) and \( f_z \):- \( f_x = \frac{\partial}{\partial x}(xz - 3yz + 2) = z \)- \( f_y = \frac{\partial}{\partial y}(xz - 3yz + 2) = -3z \)- \( f_z = \frac{\partial}{\partial z}(xz - 3yz + 2) = x - 3y \)
02

Evaluate Partial Derivatives at Point \(P_0\)

Evaluate each partial derivative at \( P_{0}(1,1,2) \):- \( f_x(1, 1, 2) = 2 \)- \( f_y(1, 1, 2) = -6 \)- \( f_z(1, 1, 2) = 1 - 3 = -2 \)
03

Apply Linearization Formula

The linearization \( L(x, y, z) \) of \( f \) at \( P_0 \) is computed as:\[L(x, y, z) = f(P_0) + f_x(P_0)(x - 1) + f_y(P_0)(y - 1) + f_z(P_0)(z - 2)\] Calculate \( f(P_0) \): \( f(1, 1, 2) = 1 \times 2 - 3 \times 1 \times 2 + 2 = 2 - 6 + 2 = -2 \)Substitute in the values, \[ L(x, y, z) = -2 + 2(x - 1) - 6(y - 1) - 2(z - 2) \]
04

Simplify the Linearization Expression

Expand and simplify the expression:\[L(x, y, z) = -2 + 2x - 2 - 6y + 6 - 2z + 4\]Combine like terms:\[L(x, y, z) = 2x - 6y - 2z + 6\]
05

Determine Maximum Value of Second-order Derivatives

To estimate the error: Find maximum values of second partial derivatives in the region \( R \).The second-order derivatives are:- \( f_{xx} = 0 \)- \( f_{yy} = 0 \)- \( f_{zz} = 0 \)- \( f_{xy} = 0 \)- \( f_{xz} = 1 \)- \( f_{yz} = -3 \)The maximum second derivative in absolute value is \( |f_{yz}| = 3 \).
06

Calculate Upper Bound of Error

The error can be approximated as:\[|E| \leq \frac{1}{2} M(\Delta x^2 + \Delta y^2 + \Delta z^2)\]where \( M = 3 \) is the maximum absolute value of second partial derivatives, and \( \Delta x = 0.01, \Delta y = 0.01, \Delta z = 0.02 \).Compute:\[E \leq \frac{1}{2}(3)(0.01^2 + 0.01^2 + 0.02^2) = \frac{1}{2}(3)(0.0001 + 0.0001 + 0.0004)\]\[E \leq \frac{1}{2} \times 3 \times 0.0006 = 0.0009\]
07

Final Answer

The linearization of the function is \( L(x, y, z) = 2x - 6y - 2z + 6 \), and the upper bound for the error is \( 0.0009 \).

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Key Concepts

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

Partial Derivatives
When you're dealing with functions of multiple variables, understanding how the function changes with respect to each variable separately is crucial. That's where partial derivatives come in. For a function like \( f(x, y, z) = xz - 3yz + 2 \), you need the partial derivative with respect to each variable independently.
  • To find \( f_x \), treat \( y \) and \( z \) as constants and differentiate with respect to \( x \). Here, \( f_x = z \).
  • Similarly, for \( f_y \), treat \( x \) and \( z \) as constants, leading to \( f_y = -3z \).
  • For \( f_z \), with \( x \) and \( y \) constant, the derivative is \( f_z = x - 3y \).
This gives us a snapshot of how the function behaves in each direction. It's like looking at the slopes of a hill in different directions to understand its shape.
Error Approximation
In mathematics, especially calculus, it's not just about getting a value鈥攊t's about knowing how close that value is to the actual result. Error approximation helps us estimate the difference between the real function and its linear approximation.
To find this error for the function \( f(x, y, z) \), you consider how much the second-order terms could affect the result within a certain region. The error bound formula gives:
  • \( |E| \leq \frac{1}{2} M (abla x^2 + abla y^2 + abla z^2) \),
  • where \( M \) is the maximum absolute value of all second-order derivatives.
  • This formula shows that smaller regions give more accurate approximations.
Understanding this tells us how confident we can be about our approximation.
Second-order Derivatives
Second-order derivatives are about how the rate of change itself changes. These are crucial when estimating error in linear approximations. Higher-order changes mean the function curve isn't a straight line鈥攊t bends.
For \( f(x, y, z) \), you compute second-order derivatives like:
  • \( f_{xx} = 0 \)
  • \( f_{yy} = 0 \)
  • \( f_{zz} = 0 \)
  • \( f_{xy} = 0 \)
  • \( f_{xz} = 1 \)
  • \( f_{yz} = -3 \)
The largest of these, in absolute value, determines the potential error in our linearization. Understanding second-order derivatives means we get a comprehensive view of how small changes in variables can impact the entire function's behavior.
Mathematical Functions
At the core of calculus is understanding how mathematical functions behave. A function is not just an equation鈥攊t's a relationship between variables. For \( f(x, y, z) = xz - 3yz + 2 \), each term has a specific role.
  • \( xz \) and \( -3yz \) show how \( z \) interacts with both \( x \) and \( y \). The constant \( 2 \) is just a vertical shift.
  • By linearizing, you essentially approximate the function with a plane near a point \( P_0 \).
  • This is useful in real-world problems where exact models are too complex.
Understanding functions is key to grasping how different mathematical tools and concepts fit together to approximate, analyze, and build predictions based on real data.

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Most popular questions from this chapter

(A) find the function's domain, (b) find the function's range, (c) describe the function's level curves, (d) find the boundary of the function's domain, (e) determine if the domain is an open region, a closed region, or neither, and (f) decide if the domain is bounded or unbounded. $$f(x, y)=\ln \left(x^{2}+y^{2}-1\right)$$

To find the extreme values of a function \(f(x, y)\) on a curve \(x=x(t), y=y(t),\) we treat \(f\) as a function of the single variable \(t\) and use the Chain Rule to find where \(d f / d t\) is zero. As in any other single-variable case, the extreme values of \(f\) are then found among the values at the a. critical points (points where \(d f / d t\) is zero or fails to exist), and b. endpoints of the parameter domain. Find the absolute maximum and minimum values of the following functions on the given curves. Function: \(f(x, y)=x y\) Curves: i) The line \(x=2 t, \quad y=t+1\) ii) The line segment \(x=2 t, \quad y=t+1, \quad-1 \leq t \leq 0\) iii) The line segment \(x=2 t, \quad y=t+1, \quad 0 \leq t \leq 1\)

Find and sketch the level curves \(f(x, y)=c\) on the same set of coordinate axes for the given values of \(c .\) We refer to these level curves as a contour map. $$f(x, y)=x^{2}+y^{2}, \quad c=0,1,4,9,16,25$$

Find the limits by rewriting the fractions first. $$\lim _{P \rightarrow(\pi, 0,3)} z e^{-2 y} \cos 2 x$$

Find the specific function values. $$f(x, y, z)=\frac{x-y}{y^{2}+z^{2}}$$ a. \(f(3,-1,2)\) b. \(f\left(1, \frac{1}{2},-\frac{1}{4}\right)\) c. \(f\left(0,-\frac{1}{3}, 0\right)\) d. \(f(2,2,100)\)
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