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Find the linearizations \(L(x, y, z)\) of the functions in Exercises \(37-42\) at the given points. $$ \begin{array}{l}{f(x, y, z)=\tan ^{-1}(x y z) \text { at }} \\\ {\begin{array}{lll}{\text { a. }(1,0,0)} & {\text { b. }(1,1,0)} & {\text { c. }(1,1,1)}\end{array}}\end{array} $$

Short Answer

Expert verified
Linearizations are: a. 0, b. z, c. \( \frac{\pi}{4} + \frac{1}{2}(x+y+z-3)\).

Step by step solution

01

Understand Linearization

The linearization of a function at a point provides an approximation that is linear and tangent to the function at that point. For a multivariable function like \(f(x, y, z)\), the linearization equation is \(L(x, y, z) = f(a, b, c) + f_x(a, b, c)(x-a) + f_y(a, b, c)(y-b) + f_z(a, b, c)(z-c)\), where \(f_x\), \(f_y\), and \(f_z\) are the partial derivatives of the function \(f\).
02

Find Partial Derivatives

First calculate the partial derivatives of \(f(x, y, z) = \tan^{-1}(xyz)\):- \(f_x = \frac{yz}{1+(xyz)^2}\)- \(f_y = \frac{xz}{1+(xyz)^2}\)- \(f_z = \frac{xy}{1+(xyz)^2}\).
03

Calculate Linearization at Point (1, 0, 0)

At \((1, 0, 0)\):- \(f(1, 0, 0) = \tan^{-1}(0) = 0\)- \(f_x(1, 0, 0) = \frac{0 \, \cdot \, 0}{1+(0)^2} = 0\)- \(f_y(1, 0, 0) = \frac{1 \, \cdot \, 0}{1+(0)^2} = 0\)- \(f_z(1, 0, 0) = \frac{1 \, \cdot \, 0}{1+(0)^2} = 0\).Thus, \(L(x, y, z) = 0\).
04

Calculate Linearization at Point (1, 1, 0)

At \((1, 1, 0)\):- \(f(1, 1, 0) = \tan^{-1}(0) = 0\)- \(f_x(1, 1, 0) = \frac{0}{1+0^2} = 0\)- \(f_y(1, 1, 0) = \frac{0}{1+0^2} = 0\)- \(f_z(1, 1, 0) = \frac{1}{1+0^2} = 1\).Thus, \(L(x, y, z) = z\).
05

Calculate Linearization at Point (1, 1, 1)

At \((1, 1, 1)\):- \(f(1, 1, 1) = \tan^{-1}(1) = \frac{\pi}{4}\)- \(f_x(1, 1, 1) = \frac{1}{2}\)- \(f_y(1, 1, 1) = \frac{1}{2}\)- \(f_z(1, 1, 1) = \frac{1}{2}\).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.

Linearization
Linearization is a technique used in multivariable calculus to approximate a function using its tangent line or plane at a particular point. This method simplifies complex functions into linear expressions that are easier to work with, especially for calculations or estimations close to the evaluated point.
For a function of several variables, like \( f(x, y, z) \), the linearization at a point \( (a, b, c) \) is expressed as:
  • \( L(x, y, z) = f(a, b, c) + f_x(a, b, c)(x-a) + f_y(a, b, c)(y-b) + f_z(a, b, c)(z-c) \)
This formula includes the value of the function at the given point plus the contributions from each variable based on their partial derivatives. It's a way to grasp how changes in each variable affect the function in the immediate surrounding area of the point \( (a, b, c) \).
This concept is widely used in various scientific and engineering fields wherever multivariable functions appear. It is particularly useful when precise solutions are difficult to obtain, and approximations suffice.
Partial Derivatives
Partial derivatives are fundamental in finding linearizations as they provide the rate of change of the function concerning each independent variable, while keeping the other variables constant. For the function \( f(x, y, z) = \tan^{-1}(xyz) \), its partial derivatives at a given point reveal how each variable individually influences the function's output.
Specifically:
  • The partial derivative with respect to \( x \):\[ f_x = \frac{yz}{1+(xyz)^2} \]
  • The partial derivative with respect to \( y \):\[ f_y = \frac{xz}{1+(xyz)^2} \]
  • The partial derivative with respect to \( z \):\[ f_z = \frac{xy}{1+(xyz)^2} \]
These derivatives describe the slope of the tangent plane in the direction of the respective variable. The calculations show how sensitive the function is to changes in \( x \), \( y \), or \( z \) at any point.
Using partial derivatives is crucial in constructing the linear approximation since it helps determine the coefficients of each variable in the linear representation.
Tangent Approximation
The tangent approximation, as a component of linearization, offers a simplified, linear form close to the original function when evaluated near the given point. Imagine drawing a tangent plane to the surface defined by the multivariable function; this plane acts as an approximate flat representation of the surface around the point of tangency.
To practically construct the tangent approximation, one calculates the function's value and its partial derivatives at a specific point, like \((1, 1, 1)\) for our function \( f(x, y, z) = \tan^{-1}(xyz) \). The linearization at this point results in the formula:
  • \( L(x, y, z) = \frac{\pi}{4} + \frac{1}{2}(x-1) + \frac{1}{2}(y-1) + \frac{1}{2}(z-1) \)
This linear expression simplifies the complex behavior of the original function in a small neighborhood around the point. The tangent approximation becomes an excellent tool when you need to make quick calculations or predictions about the function's behavior without delving into complex computations.

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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 nonnegative 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.

Extrema on a curve of intersection Find the extreme values of \(f(x, y, z)=x^{2} y z+1\) on the intersection of the plane \(z=1\) with the sphere \(x^{2}+y^{2}+z^{2}=10\) .

Change along the involute of a circle Find the derivative of \(f(x, y)=x^{2}+y^{2}\) in the direction of the unit tangent vector of the curve $$\mathbf{r}(t)=(\cos t+t \sin t) \mathbf{i}+(\sin t-t \cos t) \mathbf{j}, \quad t>0$$

In Exercises \(65-70,\) 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 critical plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. $$ f(x, y)=x^{3}-3 x y^{2}+y^{2}, \quad-2 \leq x \leq 2, \quad-2 \leq y \leq 2 $$

To find the extreme values of a function \(f(x, y)\) on a curve \(x=x(t), y=y(t),\) we treat \(f\) as a function of the single variable \(t\) and use the Chain Rule to find where \(d f / d t\) is zero. As in any other single-variable case, the extreme values of \(f\) are then found among the values at the a. critical points (points where \(d f / d t\) is zero or fails to exist), and b. endpoints of the parameter domain. Find the absolute maximum and minimum values of the following functions on the given curves. Functions: a. \(f(x, y)=x+y\) c. \(h(x, y)=2 x^{2}+y^{2}\) Curves: i. The semicircle \(x^{2}+y^{2}=4, \quad y \geq 0\) ii. The quarter circle \(x^{2}+y^{2}=4, \quad x \geq 0, \quad y \geq 0\) Use the parametric equations \(x=2 \cos t, y=2 \sin t\)

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