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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 \(f(x, y)=L(x, y)\) over the rectangle \(R\) \(f(x, y)=e^{x} \cos y\) at \(P_{0}(0,0)\) \(R:|x| \leq 0.1, \quad|y| \leq 0.1\) (Use \(\left.e^{x} \leq 1.11 \text { and }|\cos y| \leq 1 \text { in estimating } E .\right)\)

Short Answer

Expert verified
The linearization is \(L(x, y) = 1 + x\), with error \(|E| \leq 0.0111\).

Step by step solution

01

Evaluate the Function at P0

First, evaluate the function at the point \(P_0(0,0)\). The function \(f(x,y) = e^x \cos y\). Substituting \((0,0)\) gives \(f(0,0) = e^0 \cos 0 = 1 \times 1 = 1\).
02

Compute Partial Derivatives

Next, find the partial derivatives of \(f(x, y)\) with respect to \(x\) and \(y\). The partial derivative with respect to \(x\) is \(f_x(x, y) = e^x \cos y\), and at \((0, 0)\), it is \(f_x(0,0) = 1\). The partial derivative with respect to \(y\) is \(f_y(x, y) = -e^x \sin y\), and at \((0, 0)\), it is \(f_y(0,0) = 0\).
03

Form the Linearization Expression

The linearization of \(f(x, y)\) at \(P_0(0,0)\) is given by: \[L(x, y) = f(0, 0) + f_x(0, 0) \cdot (x-0) + f_y(0, 0) \cdot (y-0)\]Substitute the values from previous steps: \[L(x, y) = 1 + 1 \cdot x + 0 \cdot y = 1 + x\]
04

Calculate the Error Bound

The error bound for linearization over the rectangle \(R: |x| \leq 0.1, |y| \leq 0.1\) involves the second partial derivatives. Compute: \(f_{xx}(x, y) = e^x \cos y\) and \(f_{yy}(x, y) = -e^x \cos y\). Over \(R\), each is bounded by \(1.11\) since \(e^x \leq 1.11\) and \( |\cos y| \leq 1 \). For the formula \( |E| \approx |f_{xx}|| x|^2/2 + |f_{yy}|| y|^2/2 \), substitute bounds: \(|E| \leq 1.11(0.1^2/2 + 0.1^2/2) = 1.11 \times 0.01 = 0.0111\).

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Key Concepts

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

Error Estimation
Error estimation is an important concept in calculus, especially when approximating functions using linearization techniques. Here, we focus on finding an upper bound for the magnitude of the error, |E|, in the linear approximation of a function. The function given is \(f(x, y) = e^x \cos y\), and we are interested in its behavior around the point \(P_0(0,0)\). This is particularly relevant over a small rectangle \(R\) where \(|x| \leq 0.1\) and \(|y| \leq 0.1\).

To estimate the error in approximation, we consider the second partial derivatives \(f_{xx}\) and \(f_{yy}\), since the error bound is related to the curvature of the function. According to Taylor's theorem for multivariable functions, the error |E| can be estimated using:
  • \(f_{xx}(x, y) = e^x \cos y\)
  • \(f_{yy}(x, y) = -e^x \cos y\)
Given that \(e^x \leq 1.11\) and \(|\cos y| \leq 1\), both second derivatives are bounded by 1.11. Therefore, for small values such as \(|x|, |y| \leq 0.1\), we can approximate \[|E| \approx \left(|f_{xx}|\frac{|x|^2}{2} + |f_{yy}|\frac{|y|^2}{2}\right)\]Thus, the error bound becomes \(|E| \leq 1.11 \times (0.01) = 0.0111\).

This gives us confidence that the linear approximation is accurate within the specified region.
Partial Derivatives
Partial derivatives are fundamental tools in multivariable calculus. They represent the rate of change of a function with respect to one variable while keeping others constant. In our problem, we aim to linearize \(f(x, y) = e^x \cos y\) around the point \(P_0(0,0)\).

To find the linearization, we need the first partial derivatives:
  • The partial derivative of \(f\) with respect to \(x\), denoted by \(f_x(x, y)\), is \(e^x \cos y\).
  • The partial derivative of \(f\) with respect to \(y\), denoted by \(f_y(x, y)\), is \(-e^x \sin y\).
These derivatives help us understand how the function changes as \(x\) and \(y\) vary slightly. At the point \((0, 0)\), these evaluate to:
  • \(f_x(0, 0) = 1\)
  • \(f_y(0, 0) = 0\)
These values are crucial in forming the linear approximation \(L(x, y) = 1 + x\). This shows that near \(P_0(0,0)\), the function changes linearly with \(x\) and is constant with respect to \(y\).

Partial derivatives provide insights into the directional sensitivity of functions, which is essential for tasks such as optimization and finding error bounds in approximations.
Multivariable Calculus
Multivariable calculus extends the concepts of calculus to functions of more than one variable. It enables us to explore and understand mathematical phenomena in higher dimensions. The exercise involves linearizing a two-variable function, \(f(x, y) = e^x \cos y\), which is a classic application in multivariable calculus.

In this realm, key concepts include:
  • Functions of Two Variables: These functions depend on two independent variables, which might represent spatial dimensions in real-world scenarios.
  • Linearization: It simplifies the function to a linear form \(L(x, y)\), serving as an approximation close to a given point. This is particularly useful in complex functions where analytical solutions are challenging.
  • Partial Derivatives: These are critical in determining the function's slope in each direction individually, which aids in forming the linear approximation.
  • Error Estimation: Consideration of second partial derivatives offers insight into how the function's curvature can affect the accuracy of the linear approximation.
Understanding these concepts within multivariable calculus is invaluable for solving real-life problems where variables are interdependent. It allows us to analyze surfaces and apply these methods to fields like physics, engineering, and economics, where modeling with several variables is necessary.

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