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Chapter 13: Problem 26

Find the linearization \(L(x, y)\) of the function at each point. \(f(x, y)=(x+y+2)^{2}\) at a. (0,0) b. (1,2)

Short Answer

Expert verified
a. \(L(x, y) = 4 + 4x + 4y\); b. \(L(x, y) = 10x + 10y - 5\).

Step by step solution

01

Compute the partial derivatives

To linearize a function, we first need the partial derivatives with respect to each variable. The function given is \(f(x, y) = (x+y+2)^2\). - The partial derivative with respect to \(x\) is \(f_x = \frac{d}{dx}(x+y+2)^2 = 2(x+y+2) \cdot 1 = 2(x+y+2)\).- The partial derivative with respect to \(y\) is \(f_y = \frac{d}{dy}(x+y+2)^2 = 2(x+y+2) \cdot 1 = 2(x+y+2)\).
02

Write the formula for linearization

The formula for the linearization \(L(x, y)\) of a function \(f(x, y)\) at a point \((a, b)\) is given by:\[ L(x, y) = f(a, b) + f_x(a, b) (x-a) + f_y(a, b) (y-b) \]
03

Evaluate at point (0, 0)

For point \((0,0)\):- Calculate \(f(0, 0) = (0+0+2)^2 = 4\).- Using \(f_x = 2(x+y+2)\), find \(f_x(0, 0) = 2(0+0+2) = 4\).- Using \(f_y = 2(x+y+2)\), find \(f_y(0, 0) = 4\).Substitute these values into the linearization formula:\[ L(x, y) = 4 + 4(x-0) + 4(y-0) = 4 + 4x + 4y \].
04

Evaluate at point (1, 2)

For point \((1, 2)\):- Calculate \(f(1, 2) = (1+2+2)^2 = 25\).- Using \(f_x = 2(x+y+2)\), find \(f_x(1, 2) = 2(1+2+2) = 10\).- Using \(f_y = 2(x+y+2)\), find \(f_y(1, 2) = 10\).Substitute these values into the linearization formula:\[ L(x, y) = 25 + 10(x-1) + 10(y-2) = 25 + 10x - 10 + 10y - 20 = 10x + 10y - 5 \].

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Key Concepts

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

Partial Derivatives
Understanding partial derivatives is key when working with multivariable functions. A partial derivative represents the rate of change of a function when one of the variables changes, keeping the other variables constant. For the function \(f(x, y) = (x+y+2)^2\), we calculate the partial derivatives \(f_x\) and \(f_y\).
\[ f_x = \frac{d}{dx}(x+y+2)^2 = 2(x+y+2) \]
\[ f_y = \frac{d}{dy}(x+y+2)^2 = 2(x+y+2) \]

Partial derivatives help us understand how a small change in one variable affects the function's output. They form the building blocks for creating the linearization of functions, which gives us straight-line approximations of the function near specific points.
Linearization
Linearization is a technique used to approximate a complex function with a simpler linear function around a specific point. This method is particularly useful for multivariable functions. To linearize a function \(f(x, y)\) at a point \((a, b)\), we use the formula:
\[ L(x, y) = f(a, b) + f_x(a, b) (x-a) + f_y(a, b) (y-b) \]

For our example function at the point \((0, 0)\), the linearization results in:
\[ L(x, y) = 4 + 4x + 4y \]
While at \((1, 2)\), it simplifies to:
\[ L(x, y) = 10x + 10y - 5 \]

These linearized equations provide a way to approximate \(f(x, y)\) near those points, making calculations and predictions more straightforward.
Multivariable Functions
Multivariable functions are functions with more than one variable, such as \(f(x, y) = (x+y+2)^2\). These functions model various real-world phenomena that depend on multiple factors.

Working with multivariable functions requires understanding how changes in each variable affect the overall function. We use concepts like partial derivatives to analyze these changes. Linearization is another approach we use to simplify and approximate the function's behavior near certain points, making it manageable and easy to estimate outcomes.

In this context, knowing how to effectively break down and examine the function can provide tremendous insights and predictive power for complex scenarios.

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Most popular questions from this chapter

During the 1920 s, Charles Cobb and Paul Douglas modeled total production output \(P\) (of a firm, industry, or entire economy) as a function of labor hours involved \(x\) and capital invested \(y\) (which includes the monetary worth of all buildings and equipment). The Cobb-Douglas production function is given by $$P(x, y)=k x^{\alpha} y^{1-\alpha}$$ where \(k\) and \(\alpha\) are constants representative of a particular firm or economy. a. Show that a doubling of both labor and capital results in a doubling of production \(P\) b. Suppose a particular firm has the production function for \(k=\) 120 and \(\alpha=3 / 4 .\) Assume that each unit of labor costs $$ 250$ and each unit of capital costs $$ 400, and that the total expenses for all costs cannot exceed $$ 100,000 . Find the maximum production level for the firm.

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 \(R\) \(f(x, y, z)=x^{2}+x y+y z+(1 / 4) z^{2} \quad\) at \(\quad P_{0}(1,1,2)\) \(R:|x-1| \leq 0.01, \quad|y-1| \leq 0.01, \quad|z-2| \leq 0.08\)

Use a CAS to plot the implicitly defined level surfaces. $$x^{2}+z^{2}=1$$

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)? $$\begin{array}{l}f(x, y)=5 x^{6}+18 x^{5}-30 x^{4}+30 x y^{2}-120 x^{3} \\\\-4 \leq x \leq 3,-2 \leq y \leq 2\end{array}$$

The discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\) is zero at the origin for each of the following functions, so the Second Derivative Test fails there. Determine whether the function has a maximum, a minimum, or neither at the origin by imagining what the surface \(z=f(x, y)\) looks like. Describe your reasoning in each case. $$\text { a. } f(x, y)=x^{2} y^{2}$$ $$\text { b. } f(x, y)=1-x^{2} y^{2}$$ $$\text { c. }f(x, y)=x y^{2}$$ $$\text { d. } f(x, y)=x^{3} y^{2}$$ $$\text { e. } f(x, y)=x^{3} y^{3}$$ $$f(x, y)=x^{4} y^{4}$$
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