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

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

Concept of Linearization

The linearization of a function at a given point provides the best linear approximation to the function near that point. For a function \(f(x, y, z)\), the linearization \(L(x, y, z)\) at a point \((a, b, c)\) is given by: \[L(x, y, z) = f(a, b, c) + f_x(a, b, c)(x-a) + f_y(a, b, c)(y-b) + f_z(a, b, c)(z-c)\] where \(f_x, f_y,\) and \(f_z\) are the partial derivatives of \(f\) with respect to \(x, y,\) and \(z\), respectively.
02

Compute Partial Derivatives

Given the function \(f(x, y, z) = xy + yz + xz\), compute 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 (1, 1, 1)

First, evaluate the function and partial derivatives at the point \((1,1,1)\): - \(f(1,1,1) = 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 = 3 \) - \(f_x(1,1,1) = 1 + 1 = 2\) - \(f_y(1,1,1) = 1 + 1 = 2\) - \(f_z(1,1,1) = 1 + 1 = 2\) Substitute these values into the linearization formula: \[L(x, y, z) = 3 + 2(x-1) + 2(y-1) + 2(z-1)\] Simplify: \[L(x, y, z) = 2x + 2y + 2z - 3\]
04

Linearization at (1, 0, 0)

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

Linearization at (0, 0, 0)

Evaluate function and partial derivatives at the point \((0,0,0)\): - \(f(0,0,0) = 0 \cdot 0 + 0 \cdot 0 + 0 \cdot 0 = 0\) - \(f_x(0,0,0) = 0 + 0 = 0\) - \(f_y(0,0,0) = 0 + 0 = 0\) - \(f_z(0,0,0) = 0 + 0 = 0\) Substitute these into the formula:\[L(x, y, z) = 0 + 0(x-0) + 0(y-0) + 0(z-0) = 0\] The linearization at this point is simply: \[L(x, y, z) = 0\]

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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 a fundamental concept in multivariable calculus, especially when dealing with functions of two or more variables, like our function \( f(x, y, z) = xy + yz + xz \). To understand partial derivatives, first recognize that they measure how a function changes as we adjust one variable while keeping others constant.

Here's how you compute them:
  • For \( f_x \), consider \( y \) and \( z \) as constants and differentiate \( f \) with respect to \( x \). This gives \( f_x = y + z \).
  • For \( f_y \), treat \( x \) and \( z \) as constants and differentiate with respect to \( y \), resulting in \( f_y = x + z \).
  • For \( f_z \), assume \( x \) and \( y \) are constants and differentiate with respect to \( z \), yielding \( f_z = x + y \).
Each of these derivatives tells us how the function's value changes when we slightly alter one variable, keeping the others fixed. This is crucial for constructing linear approximations.
Linear Approximation
Linear approximation helps us understand complex, multivariable functions near specific points by simplifying them into linear functions. The core idea is that we can approximate a function by its tangent plane at a given point. This makes calculations much simpler and is a powerful tool when precise functions are unnecessarily complicated.

Given a point \((a, b, c)\) and a function \( f(x, y, z) \), we use the partial derivatives calculated to form the linear function \( L(x, y, z) \). This is done with the formula:
  • \( L(x, y, z) = f(a, b, c) + f_x(a, b, c)(x-a) + f_y(a, b, c)(y-b) + f_z(a, b, c)(z-c) \)
In this way, the terms involve the change in each variable multiplied by how sensitive the function is to changes in that variable. The result is a simple linear equation that closely approximates \( f \) near \((a, b, c)\).

Linearization is particularly useful in engineering, physics, and economics, where predicting slight changes around a known point maximizes efficiency.
Multivariable Calculus
Multivariable calculus extends the principles of calculus to functions of several variables, like \( f(x, y, z) = xy + yz + xz \). In this realm, we explore how functions behave in three-dimensional space, offering insights into problems involving surfaces and volumes.

In multivariable calculus, concepts such as limits, derivatives, integrals, and linear approximations are generalized to accommodate functions with multiple inputs and outputs. These new aspects require considering how variables interact with each other, adding a layer of complexity when performing calculations.

Here’s why it’s important:
  • It allows the examination of gradients and directional changes in multivariable functions, like finding the steepest ascent on a surface.
  • It forms the foundation for fields like vector calculus, used in physics for describing physical phenomena.
  • Techniques from multivariable calculus are indispensable in fields such as economics, engineering, and data science.
Approaching a problem in multivariable calculus means understanding the entire landscape created by the function, not just a single path like in single-variable calculus.

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