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Chapter 14: Problem 39

Find the linearizations \(L(x, y, z)\) of the functions at the given points. $$f(x, y, z)=x y+y z+x z \text { at }$$ a. (1,1,1) b. (1,0,0) c. (0,0,0)

Short Answer

Expert verified
a. \(L(x, y, z) = 2x + 2y + 2z - 3\); b. \(L(x, y, z) = y + z\); c. \(L(x, y, z) = 0\)."

Step by step solution

01

Understand the concept of Linearization

The linearization of a multivariable function at a given point is the first-order Taylor expansion of the function at that point. It approximates the function as a plane tangent to the surface defined by the function at that point.
02

Compute Partial Derivatives

For the function \(f(x, y, z) = xy + yz + xz\), calculate the partial derivatives: - \(f_x = \frac{\partial}{\partial x}(xy + yz + xz) = y + z\)- \(f_y = \frac{\partial}{\partial y}(xy + yz + xz) = x + z\)- \(f_z = \frac{\partial}{\partial z}(xy + yz + xz) = y + x\).
03

Linearization at Point (1,1,1)

For point \((1, 1, 1)\), evaluate the function and partial derivatives at this point:- \(f(1, 1, 1) = 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 = 3\)- \(f_x(1, 1, 1) = 1 + 1 = 2\)- \(f_y(1, 1, 1) = 1 + 1 = 2\)- \(f_z(1, 1, 1) = 1 + 1 = 2\)The linearization formula is: \[L(x, y, z) = f(a, b, c) + f_x(a, b, c)(x - a) + f_y(a, b, c)(y - b) + f_z(a, b, c)(z - c)\]Substitute the values:\[L(x, y, z) = 3 + 2(x - 1) + 2(y - 1) + 2(z - 1)\]\[L(x, y, z) = 2x + 2y + 2z - 3\].
04

Linearization at Point (1,0,0)

Evaluate the function and partial derivatives at \((1, 0, 0)\):- \(f(1, 0, 0) = 1 \cdot 0 + 0 \cdot 0 + 1 \cdot 0 = 0\)- \(f_x(1, 0, 0) = 0 + 0 = 0\)- \(f_y(1, 0, 0) = 1 + 0 = 1\)- \(f_z(1, 0, 0) = 0 + 1 = 1\)Substitute into the linearization formula:\[L(x, y, z) = 0 + 0(x - 1) + 1(y - 0) + 1(z - 0)\]\[L(x, y, z) = y + z\].
05

Linearization at Point (0,0,0)

Evaluate the function and partial derivatives at \((0, 0, 0)\):- \(f(0, 0, 0) = 0 \cdot 0 + 0 \cdot 0 + 0 \cdot 0 = 0\)- \(f_x(0, 0, 0) = 0 + 0 = 0\)- \(f_y(0, 0, 0) = 0 + 0 = 0\)- \(f_z(0, 0, 0) = 0 + 0 = 0\)Since the partial derivatives are zero at this point, the linearization is:\[L(x, y, z) = 0\].

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Key Concepts

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

Partial Derivatives
Partial derivatives are used to study the behavior of multivariable functions by examining how the function changes as each individual variable is varied, while keeping the others constant. For a function like \( f(x, y, z) = xy + yz + xz \), we compute three partial derivatives, each representing the rate of change of \( f \) with respect to \( x \), \( y \), and \( z \) respectively.
  • \( f_x = \frac{\partial}{\partial x}(xy + yz + xz) = y + z \)
  • \( f_y = \frac{\partial}{\partial y}(xy + yz + xz) = x + z \)
  • \( f_z = \frac{\partial}{\partial z}(xy + yz + xz) = y + x \)
Understanding these derivatives helps us craft a linear approximation of the function near specific points. This is the essence of finding the linearization of the function.
Taylor Expansion
The Taylor expansion is a powerful tool used to approximate functions. Particularly, the first-order Taylor expansion is essentially the linearization of the function. This is achieved by approximating the function with its value at a point, plus adjustments based on its derivatives. When dealing with multivariable functions, the linearization formula looks like this:\[L(x, y, z) = f(a, b, c) + f_x(a, b, c)(x - a) + f_y(a, b, c)(y - b) + f_z(a, b, c)(z - c)\]This equation suggests a simple way to approximate \( f(x, y, z) \) around a point \((a, b, c)\) using the function value and its partial derivatives at that point. This allows us to simplify complex multivariable functions to a manageable linear form.
Multivariable Functions
A multivariable function, such as \( f(x, y, z) = xy + yz + xz \), involves more than one variable and provides a framework for modeling real-world phenomena where multiple factors interact.Multivariable functions can create complex surfaces or shapes. Understanding them involves:
  • Analyzing how changes in one variable affect the output while others remain constant, using partial derivatives.
  • Finding patterns or simple approximations using tools like Taylor expansion.
These functions appear in diverse fields, from physics to economics, making their study and simplification crucial for problem-solving.
Tangent Plane
In multivariable calculus, a tangent plane is similar to a tangent line in single-variable calculus, but extends to functions with more variables. It represents the best linear approximation of the function at a given point. To visualize, imagine a plane just barely touching a surface at one point without cutting through it. This plane uses the concept of linearization:
  • The linear approximation of a multivariable function at a point results in an equation of a plane.
  • For point \((a, b, c)\), the tangent plane equation derives from the first-order Taylor expansion. It takes the form \( L(x, y, z) = f(a, b, c) + f_x(a, b, c)(x - a) + f_y(a, b, c)(y - b) + f_z(a, b, c)(z - c) \).
Understanding the tangent plane helps approximate and analyze the behavior of functions in a localized, linear way, simplifying complex relationships between variables.

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Most popular questions from this chapter

Find the limits by rewriting the fractions first. $$\lim _{(x, y) \rightarrow(0,0)} \frac{1-\cos (x y)}{x y}$$

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)? $$\begin{aligned} &f(x, y)=x^{4}+y^{2}-8 x^{2}-6 y+16, \quad-3 \leq x \leq 3,\\\ &-6 \leq y \leq 6 \end{aligned}$$

Show that each function satisfies a Laplace equation. \(f(x, y)=3 x+2 y-4\)

Show that each function satisfies a Laplace equation. \(f(x, y, z)=\left(x^{2}+y^{2}+z^{2}\right)^{-1 / 2}\)

When we try to fit a line \(y=m x+b\) to a set of numerical data points \(\left(x_{1}, y_{1}\right)\) \(\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right),\) we usually choose the line that minimizes the sum of the squares of the vertical distances from the points to the line. In theory, this means finding the values of \(m\) and \(b\) that minimize the value of the function $$w=\left(m x_{1}+b-y_{1}\right)^{2}+\cdots+\left(m x_{n}+b-y_{n}\right)^{2}. \quad (1)$$ (See the accompanying figure.) Show that the values of \(m\) and \(b\) that do this are $$m=\frac{\left(\sum x_{k}\right)\left(\sum y_{k}\right)-n \sum x_{k} y_{k}}{\left(\sum x_{k}\right)^{2}-n \sum x_{k}^{2}},\quad (2)$$ $$b=\frac{1}{n}\left(\sum y_{k}-m \sum x_{k}\right),\quad(3)$$ with all sums running from \(k=1\) to \(k=n\). Many scientific calculators have these formulas built in, enabling you to find \(m\) and \(b\) with only a few keystrokes after you have entered the data. The line \(y=m x+b\) determined by these values of \(m\) and \(b\) is called the least squares line, regression line, or trend line for the data under study. Finding a least squares line lets you 1\. summarize data with a simple expression, 2\. predict values of \(y\) for other, experimentally untried values of \(x\), 3\. handle data analytically. We demonstrated these ideas with a variety of applications in Section 1.4.
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