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Chapter 14: Problem 46

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

Short Answer

Expert verified
a. \(L(x, y, z) = 0\), b. \(L(x, y, z) = 0\), c. \(L(x, y, z) = \frac{\pi}{4} + \frac{1}{2}(x-1) + \frac{1}{2}(y-1) + \frac{1}{2}(z-1)\).

Step by step solution

01

Understand the Function and Points

We need to find the linearization of the function \( f(x, y, z) = \tan^{-1}(xyz) \) at the given points: \((1, 0, 0)\), \((1, 1, 0)\), and \((1, 1, 1)\). Linearization is the first-order Taylor expansion of a function at a given point.
02

Linearization Formula

The linearization of a function \( f(x, y, z) \) at \( (a, b, c) \) 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 \( f_x, f_y, f_z \) are partial derivatives of \( f \) with respect to \( x, y, z \), respectively.
03

Compute Partial Derivatives

Calculate \( f_x, f_y, \) and \( f_z \). Derive:- \( f_x(x, y, z) = \frac{d}{dx} \tan^{-1}(xyz) = \frac{1}{1+(xyz)^2} \cdot yz \)- \( f_y(x, y, z) = \frac{d}{dy} \tan^{-1}(xyz) = \frac{1}{1+(xyz)^2} \cdot xz \)- \( f_z(x, y, z) = \frac{d}{dz} \tan^{-1}(xyz) = \frac{1}{1+(xyz)^2} \cdot xy \)
04

Step 4a: Linearize at (1, 0, 0)

At \((1, 0, 0)\), \(xyz = 0\). Therefore, \(f(1,0,0) = \tan^{-1}(0) = 0\). Partial derivatives:- \( f_x(1, 0, 0) = 0 \)- \( f_y(1, 0, 0) = 0 \)- \( f_z(1, 0, 0) = 0 \)Thus, \( L(x, y, z) = 0 \).
05

Step 4b: Linearize at (1, 1, 0)

At \((1, 1, 0)\), again \( xyz = 0\), so \(f(1,1,0) = 0\). Partial derivatives are:- \( f_x(1, 1, 0) = 0 \)- \( f_y(1, 1, 0) = 0 \)- \( f_z(1, 1, 0) = 0 \)Thus, \( L(x, y, z) = 0 \).
06

Step 4c: Linearize at (1, 1, 1)

At \((1, 1, 1)\), \( xyz = 1\), so \(f(1,1,1) = \tan^{-1}(1) = \frac{\pi}{4} \).Partial derivatives:- \( f_x(1, 1, 1) = \frac{1}{2} \cdot 1 = \frac{1}{2} \)- \( f_y(1, 1, 1) = \frac{1}{2} \cdot 1 = \frac{1}{2} \)- \( f_z(1, 1, 1) = \frac{1}{2} \cdot 1 = \frac{1}{2} \)Thus, \( L(x, y, z) = \frac{\pi}{4} + \frac{1}{2}(x-1) + \frac{1}{2}(y-1) + \frac{1}{2}(z-1) \).

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Key Concepts

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

Partial Derivatives
To understand linearization and to find the linear approximation of a multivariable function like \( f(x, y, z) = \tan^{-1}(xyz) \), it's crucial to grasp the concept of partial derivatives. Partial derivatives are similar to derivatives but in a multivariable setting. While derivatives look at the rate of change of a function in one dimension, partial derivatives examine how a function changes with respect to one variable while keeping the other variables constant.

  • Finding Partial Derivatives: The function \( f(x, y, z) = \tan^{-1}(xyz) \) has three partial derivatives: \( f_x \), \( f_y \), and \( f_z \), which are found by differentiating with respect to \( x \), \( y \), and \( z \) respectively.
  • Example Calculation: For \( f_x(x, y, z) \), we use the chain rule: \( f_x(x, y, z) = \frac{1}{1+(xyz)^2} \cdot yz \). This same logic applies to \( f_y \) and \( f_z \).
These derivatives help in constructing the first-order Taylor expansion used in linearization.
First-order Taylor Expansion
The first-order Taylor expansion is a method to approximate a function near a given point using its partial derivatives. For multivariable functions, it expresses the function as a sum of its value at a specific point plus the changes seen in each direction via partial derivatives. This expansion helps simplify complex functions, making them easier to analyze locally.

The Linearization Formula

For a function \( f(x, y, z) \) around a point \( (a, b, c) \), its linearization is written as:\[L(x, y, z) = f(a, b, c) + f_x(a, b, c)(x-a) + f_y(a, b, c)(y-b) + f_z(a, b, c)(z-c)\]Here's how this is relevant to our original exercise:
  • Base Value: The term \( f(a, b, c) \) gives the value of the function at point \( (a, b, c) \).
  • First-order Terms: The terms with \( f_x \), \( f_y \), and \( f_z \) account for the function's rate of change in their respective directions.
This formula provides a linear approximation that is accurate close to the point of expansion.
Multivariable Calculus
Multivariable calculus, an extension of single-variable calculus, explores functions of more than one variable. Unlike single-variable calculus where functions depend on one variable, multivariable calculus deals with functions that involve several variables and their interactions.

Relevance of Multivariable Calculus

For problems such as finding the linearization of \( f(x, y, z) = \tan^{-1}(xyz) \):
  • Partial Derivatives: By understanding partial derivatives, we can handle how the function shifts along each direction.
  • Taylor Expansion: It allows for accurate approximations of the function near a specific point. This is essential for simplifying complex functions locally.
  • Applications: Such concepts are used in diverse fields like physics, engineering, and economics.
By leveraging multivariable calculus concepts, we can solve and simplify problems that involve multiple dynamic variables.

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Most popular questions from this chapter

a. Maximum on a sphere Show that the maximum value of \(a^{2} b^{2} c^{2}\) on a sphere of radius \(r\) centered at the origin of a Cartesian \(a b c-\) coordinate system is \(\left(r^{2} / 3\right)^{3}\) . b. Geometric and arithmetic means Using part (a), show that for non negative numbers \(a, b,\) and \(c\) $$(a b c)^{1 / 3} \leq \frac{a+b+c}{3}$$ that is, the geometric mean of three nonnegative numbers is less than or equal to their arithmetic mean.

Among all the points on the graph of \(z=10-x^{2}-y^{2}\) that lie above the plane \(x+2 y+3 z=0,\) find the point farthest from the plane.

The linearization of \(f(x, y)\) is a tangent-plane approximation Show that the tangent plane at the point \(P_{0}\left(x_{0}, y_{0}, f\left(x_{0}, y_{0}\right)\right)\) on the surface \(z=f(x, y)\) defined by a differentiable function \(f\) is the plane $$ f_{x}\left(x_{0}, y_{0}\right)\left(x-x_{0}\right)+f_{y}\left(x_{0}, y_{0}\right)\left(y-y_{0}\right)-\left(z-f\left(x_{0}, y_{0}\right)\right)=0 $$ or $$ z=f\left(x_{0}, y_{0}\right)+f_{x}\left(x_{0}, y_{0}\right)\left(x-x_{0}\right)+f_{y}\left(x_{0}, y_{0}\right)\left(y-y_{0}\right) $$ Thus, the tangent plane at \(P_{0}\) is the graph of the linearization of \(f\) at \(P_{0}\)( see accompanying figure).

Show that \(T=\frac{1}{\sqrt{x^{2}+y^{2}}}\) satisfies the equation \(T_{x x}+T_{y y}=T^{3}.\)

In Exercises \(71-74,\) find a function \(z=f(x, y)\) whose partial derivatives are as given, or explain why this is impossible. $$\frac{\partial f}{\partial x}=2 x e^{x y^{2}}+x^{2} y^{2} e^{x y^{2}}+3, \quad \frac{\partial f}{\partial y}=2 x^{3} y e^{x y^{2}}-e^{y}$$
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