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In Exercises \(33-40,\) sketch a typical level surface for the function. $$ f(x, y, z)=z $$

Short Answer

Expert verified
The level surfaces are horizontal planes at different heights given by \( z = c \).

Step by step solution

01

Understanding the Function

The function given is \( f(x, y, z) = z \). This tells us that for any point \((x, y, z)\), the value of the function is simply the \(z\)-coordinate of that point.
02

Defining a Level Surface

A level surface for a function \( f \) is defined by setting \( f(x, y, z) = c \), where \( c \) is a constant. In this exercise, you want to sketch the surface for which \( z = c \).
03

Identifying the Type of Surface

Setting \( f(x, y, z) = c \) results in \( z = c \). This equation represents a plane parallel to the \(xy\)-plane at a height or level \(z = c\).
04

Sketching the Level Surface

To sketch this surface, draw a plane parallel to the \(xy\)-plane. This plane sits above or below the origin at the distance \(c\), depending on the value of \(c\).
05

Considering Various Levels

Since \( c \) can be any real number, there can be infinitely many such planes for different values of \(c\). This means that the level surfaces of \( f(x, y, z) = z \) are horizontal planes at every possible value of \( z = c \).

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Key Concepts

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

Level Surfaces
Level surfaces are fascinating mathematical concepts often encountered in multivariable calculus. In simple terms, a level surface is the set of points \( (x, y, z) \) in space where a function \(f(x, y, z)\) takes on a constant value \( c \). For any function of three variables, such as \(f(x, y, z) = z\), you create a level surface by setting \(f(x, y, z) = c\). In this scenario, you set \(z = c\), forming the all-important plane in three-dimensional space.
This fundamental idea means all points on this plane yield the function value \(c\), providing a geometric viewpoint on the behavior of the function. To visualize these surfaces, you simply draw planes parallel to the \(xy\)-plane at various heights corresponding to different constants \(c\). Each level surface reflects a different slice of the three-dimensional space, helping you to understand not just the height but the distribution of values within the volume defined by the function.
Functions of Several Variables
Functions of several variables extend the concept of functions from high school mathematics, where typically you dealt with one variable mappings. In multivariable calculus, a function can depend on two, three, or even more variables. Consider the function \(f(x, y, z) = z\), a simple yet clear example of a function involving three variables.
  • The function relates points in three-dimensional space \( (x, y, z) \).
  • Each output corresponds directly to one of these coordinates: here, none other than \(z\).

Understanding functions in this advanced context involves moving beyond straight lines or curves, often requiring visualizing surfaces or volumes. They allow you to model real-world phenomena where multiple factors contribute to the outcome, such as temperature, pressure, or distance in physics. As you become comfortable with these concepts, interpreting complex relationships in science and engineering becomes more intuitive. It enhances your toolkit for harnessing mathematics to solve problems.
Coordinate Planes
Coordinate planes are crucial for interpreting functions in multivariable calculus, offering a concrete way to visualize and understand complex relationships. When dealing with three-dimensional space, like in the function \(f(x, y, z) = z\), you encounter three main coordinate planes:
  • The \(xy\)-plane, where \(z = 0\), acts as the base layer of the three-dimensional grid.
  • The \(xz\)-plane, where \(y = 0\), represents a vertical slice through the space.
  • The \(yz\)-plane, where \(x = 0\), represents another orthogonal slice.

Each plane corresponds to zeroing one of the coordinates, simplifying the complexity of analyzing sections of the space. Coordinate planes help you sketch and understand functions by breaking them down into manageable parts. These planes provide a structured framework in which functions like \(f(x, y, z) = z\) manifest as tangible geometric representations, like slices of a multilayer cake, each layer representing different function values.

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Most popular questions from this chapter

In Exercises \(31-38\) , find the absolute maxima and minima of the functions on the given domains. \(T(x, y)=x^{2}+x y+y^{2}-6 x+2\) on the rectangular plate \(0 \leq x \leq 5,-3 \leq y \leq 0\)

Each of Exercises \(59-62\) gives a function \(f(x, y)\) and a positive number \(\epsilon .\) In each exercise, show that there exists a \(\delta>0\) such that for all \((x, y),\) $$ \sqrt{x^{2}+y^{2}}<\delta \Rightarrow|f(x, y)-f(0,0)|<\epsilon $$ $$ f(x, y)=x^{2}+y^{2}, \quad \epsilon=0.01 $$

In Exercises \(43-46,\) 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 \(R\) . $$ \begin{array}{l}{f(x, y, z)=x^{2}+x y+y z+(1 / 4) z^{2} \quad \text { at } \quad P_{0}(1,1,2)} \\ {R : \quad|x-1| \leq 0.01, \quad|y-1| \leq 0.01, \quad|z-2| \leq 0.08}\end{array} $$

Use a CAS to perform the following steps implementing the method of Lagrange multipliers for finding constrained extrema: a. Form the function \(h=f-\lambda_{1} g_{1}-\lambda_{2} g_{2},\) where \(f\) is the function to optimize subject to the constraints \(g_{1}=0\) and \(g_{2}=0 .\) b. Determine all the first partial derivatives of \(h\) , including the partials with respect to \(\lambda_{1}\) and \(\lambda_{2},\) and set them equal to \(0 .\) c. Solve the system of equations found in part (b) for all the unknowns, including \(\lambda_{1}\) and \(\lambda_{2} .\) d. Evaluate \(f\) at each of the solution points found in part (c) and select the extreme value subject to the constraints asked for in the exercise. Minimize \(f(x, y, z)=x y+y z\) subject to the constraints \(x^{2}+y^{2}-2=0\) and \(x^{2}+z^{2}-2=0\)

Can you conclude anything about \(f(a, b)\) if \(f\) and its first and second partial derivatives are continuous throughout a disk centered at \((a, b)\) and \(f_{x x}(a, b)\) and \(f_{y y}(a, b)\) differ in sign? Give reasons for your answer.

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