/*! 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 11 Use Taylor's formula to find a q... [FREE SOLUTION] | 91影视

91影视

Use Taylor's formula to find a quadratic approximation of \(f(x, y)=\cos x \cos y\) at the origin. Estimate the error in the approximation if \(|x| \leq 0.1\) and \(|y| \leq 0.1\).

Short Answer

Expert verified
The quadratic approximation is \(1 - \frac{x^2}{2} - \frac{y^2}{2}\), with an error of approximately \(10^{-3}\).

Step by step solution

01

Write the Taylor expansion formula

The Taylor expansion of a function of two variables \(f(x, y)\) around the origin (0, 0) up to second order is given by: \[ f(x, y) \approx f(0, 0) + \left.\frac{\partial f}{\partial x}\right|_{(0,0)}x + \left.\frac{\partial f}{\partial y}\right|_{(0,0)}y + \frac{1}{2}\left(\left.\frac{\partial^2 f}{\partial x^2}\right|_{(0,0)}x^2 + 2\left.\frac{\partial^2 f}{\partial x \partial y}\right|_{(0,0)}xy + \left.\frac{\partial^2 f}{\partial y^2}\right|_{(0,0)}y^2\right). \]

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.

Quadratic Approximation
Quadratic approximation is a powerful mathematics tool, especially useful for estimating the behavior of a function near a particular point. In our case, this point is the origin. Imagine a scenario where we want to predict how a function behaves without having to solve it exactly for every value. That's where quadratic approximation steps in. It allows us to simplify a function to
  • its value at the point of interest,
  • the first and second derivatives concerning each variable.
This results in a polynomial, a mathematical expression that鈥檚 more manageable.

For the function given, which is \( f(x,y) = \cos x \cos y \) at the origin (0, 0), we start by noting that \( f(0,0) \) is simply \( \cos 0 \cdot \cos 0 = 1 \).

The approximation then adds terms involving the first and second partial derivatives evaluated at this point, capturing the essence of the function's curve. The end result is much like saying, "here's a gentle curve that quite closely follows what the function does near (0,0).鈥
Partial Derivatives
Partial derivatives are central in finding the coefficients for a quadratic approximation. Think of them as the slope or rate of change of the function with respect to one variable, keeping all others constant. They help in understanding how the function behaves as we slightly change one of the inputs.
For example, with the function \( f(x,y) = \cos x \cos y \), we would find:
  • \( \frac{\partial f}{\partial x} \) measures change with respect to x alone,;
  • \( \frac{\partial f}{\partial y} \) measures change with respect to y alone,;
  • \( \frac{\partial^2 f}{\partial x^2}, \frac{\partial^2 f}{\partial x \partial y}, \) and \(\frac{\partial^2 f}{\partial y^2} \) account for curvature in all possible ways.
Evaluating these at the origin gives us specific values that feed into the quadratic approximation.

For instance, since both cosine functions are involved, each partial derivative should consider changes to the cosine components鈥 products, evaluated at zero to get straightforward results. By using partial derivatives, we are assembling a precise local picture of how the function changes right around the point.
Error Estimation
Error estimation in the context of Taylor's formula helps us understand the accuracy of our quadratic approximation. While approximation simplifies calculations, it鈥檚 always important to know how much we might err or deviate from the actual function.
The error term in a Taylor series considers higher-order derivatives (those beyond what is included in the approximation) and how much they contribute out of the domain range we鈥檙e examining. For this exercise, that means considering \( |x| \leq 0.1 \) and \( |y| \leq 0.1 \), which are small intervals around the origin.

In practical terms, to ensure that the quadratic approximation closely matches \( \cos x \cos y \) near (0,0), we want to keep these variables limited within this range. When bounded properly, the error term is minimized. Thus, with small x and y values as given, the quadratic approximation remains highly reliable for predictions, with any error likely to be minute.

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

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. $$f(x, y)=x^{2}+y^{2}, \quad c=0,1,4,9,16,25$$

Consider the function \(f(x, y)=x^{2}+y^{2}+2 x y-x-y+1\) over the square \(0 \leq x \leq 1\) and \(0 \leq y \leq 1\). a. Show that \(f\) has an absolute minimum along the line segment \(2 x+2 y=1\) in this square. What \(i s\) the absolute minimum value? b. Find the absolute maximum value of \(f\) over the square.

Find the maximum value of \(s=x y+y z+x z\) where \(x+y+z=6\).

By considering different paths of approach, show that the functions have no limit as \((x, y) \rightarrow(0,0)\). $$h(x, y)=\frac{x^{2} y}{x^{4}+y^{2}}$$

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 curves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. d. Calculate the function's second partial derivatives and find the discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\). e. Using the max-min tests, classify the critical points found in part (c). Are your findings consistent with your discussion in part (c)? $$\begin{aligned} &f(x, y)=\left\\{\begin{array}{ll} x^{5} \ln \left(x^{2}+y^{2}\right), & (x, y) \neq(0,0) \\ 0, & (x, y)=(0,0) \end{array}\right.,\\\ &-2 \leq x \leq 2,-2 \leq y \leq 2 \end{aligned}$$

See all solutions

Recommended explanations on Math Textbooks

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.