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Chapter 14: Problem 35

Find the linearization \(L(x, y)\) of the function \(f(x, y)\) at \(P_{0} .\) Then find an upper bound for the magnitude \(|E|\) the error in the approximation \(f(x, y) \approx L(x, y)\) over the rectangle \(R .\) $$ \begin{array}{l}{f(x, y)=x^{2}-3 x y+5 \text { at } P_{0}(2,1)} \\ {R :|x-2| \leq 0.1, \quad|y-1| \leq 0.1}\end{array} $$

Short Answer

Expert verified
The linearization is \( L(x, y) = x - 6y + 11 \) with error bound \(|E| \leq 0.015\).

Step by step solution

01

Find the Partial Derivatives

To find the linearization, we first need the partial derivatives of the function. The partial derivative of \( f \) with respect to \( x \) is \( f_x = 2x - 3y \). The partial derivative of \( f \) with respect to \( y \) is \( f_y = -3x \).
02

Compute the Gradient at \( P_0 \)

Evaluate the partial derivatives at the point \( P_0(2, 1) \). For \( f_x \), we substitute \( x = 2 \) and \( y = 1 \) to get \( f_x(2, 1) = 2(2) - 3(1) = 4 - 3 = 1 \). For \( f_y \), substitute \( x = 2 \) and \( y = 1 \) to get \( f_y(2, 1) = -3(2) = -6 \).
03

Calculate the Function Value at \( P_0 \)

Substitute \( x = 2 \) and \( y = 1 \) into the original function \( f(x, y) = x^2 - 3xy + 5 \) to find \( f(2, 1) = 2^2 - 3(2)(1) + 5 = 4 - 6 + 5 = 3 \).
04

Write the Linearization Formula

The linearization of \( f(x, y) \) at \( P_0 \) is given by \( L(x, y) = f(P_0) + f_x(P_0)(x - x_0) + f_y(P_0)(y - y_0) \). Substituting the values we calculated, \( L(x, y) = 3 + 1(x - 2) - 6(y - 1) \). Simplify to get \( L(x, y) = x - 6y + 11 \).
05

Determine the Second Partial Derivatives

Find the second partial derivatives to use in determining the error bound. \( f_{xx} = 2 \), \( f_{yy} = 0 \), \( f_{xy} = -3 \), and \( f_{yx} = -3 \).
06

Calculate the Error Bound

The error bound \(|E|\) can be estimated using the formula for the remainder of the Taylor series. A rough bound is related to the maximum of the absolute values of the second derivatives over the region \( R \). The second partials are constants, so the largest term is \(|f_{xy}| = 3\). The bound using second derivatives is \(|E| \leq \frac{1}{2} \times 0.1^2 \times 3 = 0.015\).
07

Conclude the Error Bound Calculation

Summing up, the error in approximating \( f(x, y) \) by \( L(x, y) \) over the rectangle \( R \) is bounded by \( 0.015 \). This is a small error, indicating the linear approximation is quite accurate within the given region.

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Key Concepts

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

Error Estimation
Estimating the error in linear approximation is crucial to understand how well the linearization represents the actual function in a given region. The error, denoted as \(|E|\), indicates the difference between the function \(f(x, y)\) and its linear approximation \(L(x, y)\). For functions of several variables, the error can be approximated using the higher-order partial derivatives within a specified region around a point.
  • The Taylor series gives us a framework to estimate the error.
  • In this exercise, we calculate the error bound using second partial derivatives.
  • The bound helps in understanding the reliability of the approximation.
For the function in the exercise, the error is bound by \(0.015\), indicating the linear approximation is quite accurate within the specified region R.
Partial Derivatives
Partial derivatives are essential tools when working with multivariable functions. They measure the rate of change of the function with respect to each variable while keeping other variables constant. For a function like \(f(x, y) = x^{2} - 3xy + 5\), there are partial derivatives with respect to \(x\) and \(y\). In this exercise:
  • The partial derivative with respect to \(x\) is \(f_x = 2x - 3y\).
  • The partial derivative with respect to \(y\) is \(f_y = -3x\).
These derivatives help form the gradient, which is used to build the linear approximation.
Second Partial Derivatives
Second partial derivatives are the derivatives of the first partial derivatives. They provide information about the curvature of the function. In the context of this exercise, the second partial derivatives are calculated to estimate the error bound.
  • The second derivative with respect to \(x\) twice is \(f_{xx} = 2\).
  • The second derivative with respect to \(y\) twice is \(f_{yy} = 0\).
  • The mixed second partial derivatives are \(f_{xy} = -3\) and \(f_{yx} = -3\).
These values are constant in this scenario and help locate the maximum term that affects the error estimate.
Gradient
The gradient of a function in multiple variables is a vector consisting of its partial derivatives. It points in the direction of the greatest rate of increase of the function and is fundamental in finding the linearization.
  • At the point \(P_0(2, 1)\), the gradient is calculated using the partial derivatives \(f_x\) and \(f_y\).
  • The gradient vector is \((f_x(P_0), f_y(P_0))\), which at this point gives us \((1, -6)\).
The gradient is crucial as it is used in the linearization equation to describe how changes in \(x\) and \(y\) affect \(f(x, y)\).
Taylor Series Approximation
The Taylor series approximation is a powerful method to approximate functions by polynomials. For multivariable functions, this involves expanding the function in terms of its derivatives evaluated at a point. Linearization is essentially a truncation of the Taylor series to the first degree.
  • In the exercise, \(L(x, y)\) is the linear approximation derived by truncating the Taylor series.
  • Taylor's theorem assures that the remainder term, or error, becomes smaller as the point of interest approaches the expansion point.
This approximation provides a simple and quick way to estimate function values near the point \(P_0\), with errors measured as discussed in the error estimation section.

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Most popular questions from this chapter

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