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Calculus Volume 3

4.1 Functions of Several Variables

Calculus Volume 34.1 Functions of Several Variables
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  1. Preface
  2. 1 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 1.1 Parametric Equations
    3. 1.2 Calculus of Parametric Curves
    4. 1.3 Polar Coordinates
    5. 1.4 Area and Arc Length in Polar Coordinates
    6. 1.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Vectors in Space
    1. Introduction
    2. 2.1 Vectors in the Plane
    3. 2.2 Vectors in Three Dimensions
    4. 2.3 The Dot Product
    5. 2.4 The Cross Product
    6. 2.5 Equations of Lines and Planes in Space
    7. 2.6 Quadric Surfaces
    8. 2.7 Cylindrical and Spherical Coordinates
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  4. 3 Vector-Valued Functions
    1. Introduction
    2. 3.1 Vector-Valued Functions and Space Curves
    3. 3.2 Calculus of Vector-Valued Functions
    4. 3.3 Arc Length and Curvature
    5. 3.4 Motion in Space
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  5. 4 Differentiation of Functions of Several Variables
    1. Introduction
    2. 4.1 Functions of Several Variables
    3. 4.2 Limits and Continuity
    4. 4.3 Partial Derivatives
    5. 4.4 Tangent Planes and Linear Approximations
    6. 4.5 The Chain Rule
    7. 4.6 Directional Derivatives and the Gradient
    8. 4.7 Maxima/Minima Problems
    9. 4.8 Lagrange Multipliers
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  6. 5 Multiple Integration
    1. Introduction
    2. 5.1 Double Integrals over Rectangular Regions
    3. 5.2 Double Integrals over General Regions
    4. 5.3 Double Integrals in Polar Coordinates
    5. 5.4 Triple Integrals
    6. 5.5 Triple Integrals in Cylindrical and Spherical Coordinates
    7. 5.6 Calculating Centers of Mass and Moments of Inertia
    8. 5.7 Change of Variables in Multiple Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Vector Calculus
    1. Introduction
    2. 6.1 Vector Fields
    3. 6.2 Line Integrals
    4. 6.3 Conservative Vector Fields
    5. 6.4 Green’s Theorem
    6. 6.5 Divergence and Curl
    7. 6.6 Surface Integrals
    8. 6.7 Stokes’ Theorem
    9. 6.8 The Divergence Theorem
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  8. 7 Second-Order Differential Equations
    1. Introduction
    2. 7.1 Second-Order Linear Equations
    3. 7.2 Nonhomogeneous Linear Equations
    4. 7.3 Applications
    5. 7.4 Series Solutions of Differential Equations
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Learning Objectives

  • 4.1.1. Recognize a function of two variables and identify its domain and range.
  • 4.1.2. Sketch a graph of a function of two variables.
  • 4.1.3. Sketch several traces or level curves of a function of two variables.
  • 4.1.4. Recognize a function of three or more variables and identify its level surfaces.

Our first step is to explain what a function of more than one variable is, starting with functions of two independent variables. This step includes identifying the domain and range of such functions and learning how to graph them. We also examine ways to relate the graphs of functions in three dimensions to graphs of more familiar planar functions.

Functions of Two Variables

The definition of a function of two variables is very similar to the definition for a function of one variable. The main difference is that, instead of mapping values of one variable to values of another variable, we map ordered pairs of variables to another variable.

Definition

A function of two variables z=f(x,y)z=f(x,y) maps each ordered pair (x,y)(x,y) in a subset DD of the real plane ℝ2ℝ2 to a unique real number z.z. The set DD is called the domain of the function. The range of ff is the set of all real numbers zz that has at least one ordered pair (x,y)∈D(x,y)∈D such that f(x,y)=zf(x,y)=z as shown in the following figure.

A bulbous shape is marked domain and it contains the point (x, y). From this point, there is an arrow marked f that points to a point z on a straight line marked range.
Figure 4.2 The domain of a function of two variables consists of ordered pairs (x,y).(x,y).

Determining the domain of a function of two variables involves taking into account any domain restrictions that may exist. Let’s take a look.

Example 4.1

Domains and Ranges for Functions of Two Variables

Find the domain and range of each of the following functions:

  1. f(x,y)=3x+5y+2f(x,y)=3x+5y+2
  2. g(x,y)=9−x2−y2g(x,y)=9−x2−y2

Solution

  1. This is an example of a linear function in two variables. There are no values or combinations of xx and yy that cause f(x,y)f(x,y) to be undefined, so the domain of ff is ℝ2.ℝ2. To determine the range, first pick a value for z.z. We need to find a solution to the equation f(x,y)=z,f(x,y)=z, or 3x−5y+2=z.3x−5y+2=z. One such solution can be obtained by first setting y=0,y=0, which yields the equation 3x+2=z.3x+2=z. The solution to this equation is x=z−23,x=z−23, which gives the ordered pair (z−23,0)(z−23,0) as a solution to the equation f(x,y)=zf(x,y)=z for any value of z.z. Therefore, the range of the function is all real numbers, or ℝ.ℝ.
  2. For the function g(x,y)g(x,y) to have a real value, the quantity under the square root must be nonnegative:
    9−x2−y2≥0.9−x2−y2≥0.

     This inequality can be written in the form
    x2+y2≤9.x2+y2≤9.

     Therefore, the domain of g(x,y)g(x,y) is {(x,y)∈ℝ2|x2+y2≤9}.{(x,y)∈ℝ2|x2+y2≤9}. The graph of this set of points can be described as a disk of radius 33 centered at the origin. The domain includes the boundary circle as shown in the following graph.
    A circle of radius three with center at the origin. The equation x2 + y2 = 9 is given.
    Figure 4.3 The domain of the function g(x,y)=9−x2−y2g(x,y)=9−x2−y2 is a closed disk of radius 3.

     To determine the range of g(x,y)=9−x2−y2g(x,y)=9−x2−y2 we start with a point (x0,y0)(x0,y0) on the boundary of the domain, which is defined by the relation x2+y2=9.x2+y2=9. It follows that x02+y02=9x02+y02=9 and
    g(x0,y0)=9−x02−y02=9−(x02+y02)=9−9=0.g(x0,y0)=9−x02−y02=9−(x02+y02)=9−9=0.

     If x02+y02=0x02+y02=0 (in other words, x0=y0=0),x0=y0=0), then
    g(x0,y0)=9−x02−y02=9−(x02+y02)=9−0=3.g(x0,y0)=9−x02−y02=9−(x02+y02)=9−0=3.

     This is the maximum value of the function. Given any value c between 0and3,0and3, we can find an entire set of points inside the domain of gg such that g(x,y)=c:g(x,y)=c:
    9−x2−y2=c9−x2−y2=c2x2+y2=9−c2.9−x2−y2=c9−x2−y2=c2x2+y2=9−c2.

     Since 9−c2>0,9−c2>0, this describes a circle of radius 9−c29−c2 centered at the origin. Any point on this circle satisfies the equation g(x,y)=c.g(x,y)=c. Therefore, the range of this function can be written in interval notation as [0,3].[0,3].
Checkpoint 4.1

Find the domain and range of the function f(x,y)=36−9x2−9y2.f(x,y)=36−9x2−9y2.

Graphing Functions of Two Variables

Suppose we wish to graph the function z=(x,y).z=(x,y). This function has two independent variables (xandy)(xandy) and one dependent variable (z).(z). When graphing a function y=f(x)y=f(x) of one variable, we use the Cartesian plane. We are able to graph any ordered pair (x,y)(x,y) in the plane, and every point in the plane has an ordered pair (x,y)(x,y) associated with it. With a function of two variables, each ordered pair (x,y)(x,y) in the domain of the function is mapped to a real number z.z. Therefore, the graph of the function ff consists of ordered triples (x,y,z).(x,y,z). The graph of a function z=(x,y)z=(x,y) of two variables is called a surface.

To understand more completely the concept of plotting a set of ordered triples to obtain a surface in three-dimensional space, imagine the (x,y)(x,y) coordinate system laying flat. Then, every point in the domain of the function ff has a unique z-valuez-value associated with it. If zz is positive, then the graphed point is located above the xy-plane,xy-plane, if zz is negative, then the graphed point is located below the xy-plane.xy-plane. The set of all the graphed points becomes the two-dimensional surface that is the graph of the function f.f.

Example 4.2

Graphing Functions of Two Variables

Create a graph of each of the following functions:

  1. g(x,y)=9−x2−y2g(x,y)=9−x2−y2
  2. f(x,y)=x2+y2f(x,y)=x2+y2

Solution

  1. In Example 4.1, we determined that the domain of g(x,y)=9−x2−y2g(x,y)=9−x2−y2 is {(x,y)∈ℝ2|x2+y2≤9}{(x,y)∈ℝ2|x2+y2≤9} and the range is {z∈ℝ2|0≤z≤3}.{z∈ℝ2|0≤z≤3}. When x2+y2=9x2+y2=9 we have g(x,y)=0.g(x,y)=0. Therefore any point on the circle of radius 33 centered at the origin in the x,y-planex,y-plane maps to z=0z=0 in ℝ3.ℝ3. If x2+y2=8,x2+y2=8, then g(x,y)=1,g(x,y)=1, so any point on the circle of radius 2222 centered at the origin in the x,y-planex,y-plane maps to z=1z=1 in ℝ3.ℝ3. As x2+y2x2+y2 gets closer to zero, the value of z approaches 3. When x2+y2=0,x2+y2=0, then g(x,y)=3.g(x,y)=3. This is the origin in the x,y-plane.x,y-plane. If x2+y2x2+y2 is equal to any other value between 0and9,0and9, then g(x,y)g(x,y) equals some other constant between 0and3.0and3. The surface described by this function is a hemisphere centered at the origin with radius 33 as shown in the following graph.
    A hemisphere with center at the origin. The equation z = g(x, y) = the square root of the quantity (9 – x2 – y2) is given.
    Figure 4.4 Graph of the hemisphere represented by the given function of two variables.
  2. This function also contains the expression x2+y2.x2+y2. Setting this expression equal to various values starting at zero, we obtain circles of increasing radius. The minimum value of f(x,y)=x2+y2f(x,y)=x2+y2 is zero (attained when x=y=0.).x=y=0.). When x=0,x=0, the function becomes z=y2,z=y2, and when y=0,y=0, then the function becomes z=x2.z=x2. These are cross-sections of the graph, and are parabolas. Recall from Introduction to Vectors in Space that the name of the graph of f(x,y)=x2+y2f(x,y)=x2+y2 is a paraboloid. The graph of ff appears in the following graph.
    A paraboloid with vertex at the origin. The equation z = f(x, y) = x2 + y2 is given.
    Figure 4.5 A paraboloid is the graph of the given function of two variables.

Example 4.3

Nuts and Bolts

A profit function for a hardware manufacturer is given by

f(x,y)=16−(x−3)2−(y−2)2,f(x,y)=16−(x−3)2−(y−2)2,

where xx is the number of nuts sold per month (measured in thousands) and yy represents the number of bolts sold per month (measured in thousands). Profit is measured in thousands of dollars. Sketch a graph of this function.

Solution

This function is a polynomial function in two variables. The domain of ff consists of (x,y)(x,y) coordinate pairs that yield a nonnegative profit:

16−(x−3)2−(y−2)2≥0(x−3)2+(y−2)2≤16.16−(x−3)2−(y−2)2≥0(x−3)2+(y−2)2≤16.

This is a disk of radius 44 centered at (3,2).(3,2). A further restriction is that both xandyxandy must be nonnegative. When x=3x=3 and y=2,y=2, f(x,y)=16.f(x,y)=16. Note that it is possible for either value to be a noninteger; for example, it is possible to sell 2.52.5 thousand nuts in a month. The domain, therefore, contains thousands of points, so we can consider all points within the disk. For any z<16,z<16, we can solve the equation f(x,y)=16:f(x,y)=16:

16−(x−3)2−(y−2)2=z(x−3)2+(y−2)2=16−z.16−(x−3)2−(y−2)2=z(x−3)2+(y−2)2=16−z.

Since z<16,z<16, we know that 16−z>0,16−z>0, so the previous equation describes a circle with radius 16−z16−z centered at the point (3,2).(3,2). Therefore. the range of f(x,y)f(x,y) is {z∈ℝ|z≤16}.{z∈ℝ|z≤16}. The graph of f(x,y)f(x,y) is also a paraboloid, and this paraboloid points downward as shown.

A paraboloid center seemingly on the positive z axis. The equation z = f(x, y) = 16 – (x – 3)2 – (y – 2)2 is given.
Figure 4.6 The graph of the given function of two variables is also a paraboloid.

Level Curves

If hikers walk along rugged trails, they might use a topographical map that shows how steeply the trails change. A topographical map contains curved lines called contour lines. Each contour line corresponds to the points on the map that have equal elevation (Figure 4.7). A level curve of a function of two variables f(x,y)f(x,y) is completely analogous to a contour line on a topographical map.

This figure consists of two figures marked a and b. Figure a shows a topographic map of Devil’s Tower, which has its lines very close together to indicate the very steep terrain. Figure b shows a picture of Devil’s Tower, which has very steep sides.
Figure 4.7 (a) A topographical map of Devil’s Tower, Wyoming. Lines that are close together indicate very steep terrain. (b) A perspective photo of Devil’s Tower shows just how steep its sides are. Notice the top of the tower has the same shape as the center of the topographical map.

Definition

Given a function f(x,y)f(x,y) and a number cc in the range of f,af,a level curve of a function of two variables for the value cc is defined to be the set of points satisfying the equation f(x,y)=c.f(x,y)=c.

Returning to the function g(x,y)=9−x2−y2,g(x,y)=9−x2−y2, we can determine the level curves of this function. The range of gg is the closed interval [0,3].[0,3]. First, we choose any number in this closed interval—say, c=2.c=2. The level curve corresponding to c=2c=2 is described by the equation

9−x2−y2=2.9−x2−y2=2.

To simplify, square both sides of this equation:

9−x2−y2=4.9−x2−y2=4.

Now, multiply both sides of the equation by −1−1 and add 99 to each side:

x2+y2=5.x2+y2=5.

This equation describes a circle centered at the origin with radius 5.5. Using values of cc between 0and30and3 yields other circles also centered at the origin. If c=3,c=3, then the circle has radius 0,0, so it consists solely of the origin. Figure 4.8 is a graph of the level curves of this function corresponding to c=0,1,2,and3.c=0,1,2,and3. Note that in the previous derivation it may be possible that we introduced extra solutions by squaring both sides. This is not the case here because the range of the square root function is nonnegative.

Three concentric circles with center at the origin. The largest circle marked c = 0 has a radius of 3. The medium circle marked c = 1 has a radius slightly less than 3. The smallest circle marked c = 2 has a radius slightly more than 2.
Figure 4.8 Level curves of the function g(x,y)=9−x2−y2,g(x,y)=9−x2−y2, using c=0,1,2,c=0,1,2, and 33 (c=3(c=3 corresponds to the origin).

A graph of the various level curves of a function is called a contour map.

Example 4.4

Making a Contour Map

Given the function f(x,y)=8+8x−4y−4x2−y2,f(x,y)=8+8x−4y−4x2−y2, find the level curve corresponding to c=0.c=0. Then create a contour map for this function. What are the domain and range of f?f?

Solution

To find the level curve for c=0,c=0, we set f(x,y)=0f(x,y)=0 and solve. This gives

0=8+8x−4y−4x2−y2.0=8+8x−4y−4x2−y2.

We then square both sides and multiply both sides of the equation by −1:−1:

4x2+y2−8x+4y−8=0.4x2+y2−8x+4y−8=0.

Now, we rearrange the terms, putting the xx terms together and the yy terms together, and add 88 to each side:

4x2−8x+y2+4y=8.4x2−8x+y2+4y=8.

Next, we group the pairs of terms containing the same variable in parentheses, and factor 44 from the first pair:

4(x2−2x)+(y2+4y)=8.4(x2−2x)+(y2+4y)=8.

Then we complete the square in each pair of parentheses and add the correct value to the right-hand side:

4(x2−2x+1)+(y2+4y+4)=8+4(1)+4.4(x2−2x+1)+(y2+4y+4)=8+4(1)+4.

Next, we factor the left-hand side and simplify the right-hand side:

4(x−1)2+(y+2)2=16.4(x−1)2+(y+2)2=16.

Last, we divide both sides by 16:16:

(x−1)24+(y+2)216=1.(x−1)24+(y+2)216=1.
4.1

This equation describes an ellipse centered at (1,−2).(1,−2). The graph of this ellipse appears in the following graph.

An ellipse with center (1, –2), major axis vertical and of length 8, and minor axis horizontal of length 4.
Figure 4.9 Level curve of the function f(x,y)=8+8x−4y−4x2−y2f(x,y)=8+8x−4y−4x2−y2 corresponding to c=0.c=0.

We can repeat the same derivation for values of cc less than 4.4. Then, Equation 4.1 becomes

4(x−1)216−c2+(y+2)216−c2=14(x−1)216−c2+(y+2)216−c2=1

for an arbitrary value of c.c. Figure 4.10 shows a contour map for f(x,y)f(x,y) using the values c=0,1,2,and3.c=0,1,2,and3. When c=4,c=4, the level curve is the point (−1,2).(−1,2).

An series of four concentric ellipses with center (1, –2). The largest one is marked c = 0 and has major axis vertical and of length 8 and minor axis horizontal of length 4. The next smallest one is marked c = 1 and is only slightly smaller. The next two are marked c = 2 and c = 3 and are increasingly smaller. Finally, there is a point marked c = 4 at the center (1, –2).
Figure 4.10 Contour map for the function f(x,y)=8+8x−4y−4x2−y2f(x,y)=8+8x−4y−4x2−y2 using the values c=0,1,2,3,and4.c=0,1,2,3,and4.
Checkpoint 4.2

Find and graph the level curve of the function g(x,y)=x2+y2−6x+2yg(x,y)=x2+y2−6x+2y corresponding to c=15.c=15.

Another useful tool for understanding the graph of a function of two variables is called a vertical trace. Level curves are always graphed in the xy-plane,xy-plane, but as their name implies, vertical traces are graphed in the xzxz- or yz-planes.yz-planes.

Definition

Consider a function z=f(x,y)z=f(x,y) with domain D⊆ℝ2.D⊆ℝ2. A vertical trace of the function can be either the set of points that solves the equation f(a,y)=zf(a,y)=z for a given constant x=ax=a or f(x,b)=zf(x,b)=z for a given constant y=b.y=b.

Example 4.5

Finding Vertical Traces

Find vertical traces for the function f(x,y)=sinxcosyf(x,y)=sinxcosy corresponding to x=−π4,0,andπ4,x=−π4,0,andπ4, and y=−π4,0,andπ4.y=−π4,0,andπ4.

Solution

First set x=−π4x=−π4 in the equation z=sinxcosy:z=sinxcosy:

z=sin(−π4)cosy=−2cosy2≈−0.7071cosy.z=sin(−π4)cosy=−2cosy2≈−0.7071cosy.

This describes a cosine graph in the plane x=−π4.x=−π4. The other values of zz appear in the following table.

cc Vertical Trace for x=cx=c
−π4−π4 z=−2cosy2z=−2cosy2
00 z=0z=0
π4π4 z=2cosy2z=2cosy2
Table 4.1 Vertical Traces Parallel to the xz-Planexz-Plane for the Function f(x,y)=sinxcosyf(x,y)=sinxcosy

In a similar fashion, we can substitute the y-valuesy-values in the equation f(x,y)f(x,y) to obtain the traces in the yz-plane,yz-plane, as listed in the following table.

dd Vertical Trace for y=dy=d
−π4−π4 z=−2sinx2z=−2sinx2
00 z=sinxz=sinx
π4π4 z=2sinx2z=2sinx2
Table 4.2 Vertical Traces Parallel to the yz-Planeyz-Plane for the Function f(x,y)=sinxcosyf(x,y)=sinxcosy

The three traces in the xz-planexz-plane are cosine functions; the three traces in the yz-planeyz-plane are sine functions. These curves appear in the intersections of the surface with the planes x=−π4,x=0,x=π4x=−π4,x=0,x=π4 and y=−π4,y=0,y=π4y=−π4,y=0,y=π4 as shown in the following figure.

This figure consists of two figures marked a and b. In figure a, a function is given in three dimensions and it is intersected by three parallel x-z planes at y = ±π/4 and 0. In figure b, a function is given in three dimensions and it is intersected by three parallel y-z planes at x = ±π/4 and 0.
Figure 4.11 Vertical traces of the function f(x,y)f(x,y) are cosine curves in the xz-planesxz-planes (a) and sine curves in the yz-planesyz-planes (b).
Checkpoint 4.3

Determine the equation of the vertical trace of the function g(x,y)=−x2−y2+2x+4y−1g(x,y)=−x2−y2+2x+4y−1 corresponding to y=3,y=3, and describe its graph.

Functions of two variables can produce some striking-looking surfaces. The following figure shows two examples.

This figure consists of two figures marked a and b. In figure a, the function f(x, y) = x2 sin y is given; it has some sinusoidal properties by increases as the square along the maximums of the sine function. In figure b, the function f(x, y) = sin(ex) cos(ln y) is given in three dimensions; it decreases gently from the corner nearest (–2, 20) but then seems to bunch up into a series of folds that are parallel to the x and y axes.
Figure 4.12 Examples of surfaces representing functions of two variables: (a) a combination of a power function and a sine function and (b) a combination of trigonometric, exponential, and logarithmic functions.

Functions of More Than Two Variables

So far, we have examined only functions of two variables. However, it is useful to take a brief look at functions of more than two variables. Two such examples are

f(x,y,z)=x2−2xy+y2+3yz−z2+4x−2y+3x−6(a polynomial in three variables)f(x,y,z)=x2−2xy+y2+3yz−z2+4x−2y+3x−6(a polynomial in three variables)

and

g(x,y,t)=(x2−4xy+y2)sint−(3x+5y)cost.g(x,y,t)=(x2−4xy+y2)sint−(3x+5y)cost.

In the first function, (x,y,z)(x,y,z) represents a point in space, and the function ff maps each point in space to a fourth quantity, such as temperature or wind speed. In the second function, (x,y)(x,y) can represent a point in the plane, and tt can represent time. The function might map a point in the plane to a third quantity (for example, pressure) at a given time t.t. The method for finding the domain of a function of more than two variables is analogous to the method for functions of one or two variables.

Example 4.6

Domains for Functions of Three Variables

Find the domain of each of the following functions:

  1. f(x,y,z)=3x−4y+2z9−x2−y2−z2f(x,y,z)=3x−4y+2z9−x2−y2−z2
  2. g(x,y,t)=2t−4x2−y2g(x,y,t)=2t−4x2−y2

Solution

  1. For the function f(x,y,z)=3x−4y+2z9−x2−y2−z2f(x,y,z)=3x−4y+2z9−x2−y2−z2 to be defined (and be a real value), two conditions must hold:
    1. The denominator cannot be zero.
    2. The radicand cannot be negative.
    Combining these conditions leads to the inequality
    9−x2−y2−z2>0.9−x2−y2−z2>0.

     Moving the variables to the other side and reversing the inequality gives the domain as
    domain(f)={(x,y,z)∈ℝ3|x2+y2+z2<9},domain(f)={(x,y,z)∈ℝ3|x2+y2+z2<9},

     which describes a ball of radius 33 centered at the origin. (Note: The surface of the ball is not included in this domain.)
  2. For the function g(x,y,t)=2t−4x2−y2g(x,y,t)=2t−4x2−y2 to be defined (and be a real value), two conditions must hold:
    1. The radicand cannot be negative.
    2. The denominator cannot be zero.
    Since the radicand cannot be negative, this implies 2t−4≥0,2t−4≥0, and therefore that t≥2.t≥2. Since the denominator cannot be zero, x2−y2≠0,x2−y2≠0, or x2≠y2,x2≠y2, Which can be rewritten as y=±x,y=±x, which are the equations of two lines passing through the origin. Therefore, the domain of gg is
    domain(g)={(x,y,t)|y≠±x,t≥2}.domain(g)={(x,y,t)|y≠±x,t≥2}.
Checkpoint 4.4

Find the domain of the function h(x,y,t)=(3t−6)y−4x2+4.h(x,y,t)=(3t−6)y−4x2+4.

Functions of two variables have level curves, which are shown as curves in the xy-plane.xy-plane. However, when the function has three variables, the curves become surfaces, so we can define level surfaces for functions of three variables.

Definition

Given a function f(x,y,z)f(x,y,z) and a number cc in the range of f,f, a level surface of a function of three variables is defined to be the set of points satisfying the equation f(x,y,z)=c.f(x,y,z)=c.

Example 4.7

Finding a Level Surface

Find the level surface for the function f(x,y,z)=4x2+9y2−z2f(x,y,z)=4x2+9y2−z2 corresponding to c=1.c=1.

Solution

The level surface is defined by the equation 4x2+9y2−z2=1.4x2+9y2−z2=1. This equation describes a hyperboloid of one sheet as shown in the following figure.

This figure consists of four figures. The first is marked c = 0 and consists of a double cone (that is, two nappes) with their apex at the origin. The second is marked c = 1 and it looks remarkably similar to the first except that there is no apex at which the cones meet: instead, the two nappes are connected. Similarly, the next figure marked c = 2 has the two nappes connect, but this time their connection is larger (that is, the radius of their connection is greater). The final figure marked c = 3 also has the two nappes connect in an even larger fashion.
Figure 4.13 A hyperboloid of one sheet with some of its level surfaces.
Checkpoint 4.5

Find the equation of the level surface of the function

g(x,y,z)=x2+y2+z2−2x+4y−6zg(x,y,z)=x2+y2+z2−2x+4y−6z

corresponding to c=2,c=2, and describe the surface, if possible.

Section 4.1 Exercises

For the following exercises, evaluate each function at the indicated values.

1.

W(x,y)=4x2+y2.W(x,y)=4x2+y2. Find W(2,−1),W(2,−1), W(−3,6).W(−3,6).

2.

W(x,y)=4x2+y2.W(x,y)=4x2+y2. Find W(2+h,3+h).W(2+h,3+h).

3.

The volume of a right circular cylinder is calculated by a function of two variables, V(x,y)=πx2y,V(x,y)=πx2y, where xx is the radius of the right circular cylinder and yy represents the height of the cylinder. Evaluate V(2,5)V(2,5) and explain what this means.

4.

An oxygen tank is constructed of a right cylinder of height yy and radius xx with two hemispheres of radius xx mounted on the top and bottom of the cylinder. Express the volume of the cylinder as a function of two variables, xandy,xandy, find V(10,2),V(10,2), and explain what this means.

For the following exercises, find the domain of the function.

5.

V(x,y)=4x2+y2V(x,y)=4x2+y2

6.

f(x,y)=x2+y2−4f(x,y)=x2+y2−4

7.

f(x,y)=4ln(y2−x)f(x,y)=4ln(y2−x)

8.

g(x,y)=16−4x2−y2g(x,y)=16−4x2−y2

9.

z(x,y)=y2−x2z(x,y)=y2−x2

10.

f(x,y)=y+2x2f(x,y)=y+2x2

Find the range of the functions.

11.

g(x,y)=16−4x2−y2g(x,y)=16−4x2−y2

12.

V(x,y)=4x2+y2V(x,y)=4x2+y2

13.

z=y2−x2z=y2−x2

For the following exercises, find the level curves of each function at the indicated value of cc to visualize the given function.

14.

z(x,y)=y2−x2,z(x,y)=y2−x2, c=1c=1

15.

z(x,y)=y2−x2,z(x,y)=y2−x2, c=4c=4

16.

g(x,y)=x2+y2;c=4,c=9g(x,y)=x2+y2;c=4,c=9

17.

g(x,y)=4−x−y;c=0,4g(x,y)=4−x−y;c=0,4

18.

f(x,y)=xy;c=1;c=−1f(x,y)=xy;c=1;c=−1

19.

h(x,y)=2x−y;c=0,−2,2h(x,y)=2x−y;c=0,−2,2

20.

f(x,y)=x2−y;c=1,2f(x,y)=x2−y;c=1,2

21.

g(x,y)=xx+y;c=−1,0,2g(x,y)=xx+y;c=−1,0,2

22.

g(x,y)=x3−y;c=−1,0,2g(x,y)=x3−y;c=−1,0,2

23.

g(x,y)=exy;c=12,3g(x,y)=exy;c=12,3

24.

f(x,y)=x2;c=4,9f(x,y)=x2;c=4,9

25.

f(x,y)=xy−x;c=−2,0,2f(x,y)=xy−x;c=−2,0,2

26.

h(x,y)=ln(x2+y2);c=−1,0,1h(x,y)=ln(x2+y2);c=−1,0,1

27.

g(x,y)=ln(yx2);c=−2,0,2g(x,y)=ln(yx2);c=−2,0,2

28.

z=f(x,y)=x2+y2,z=f(x,y)=x2+y2, c=3c=3

29.

f(x,y)=y+2x2,f(x,y)=y+2x2, c=c= any constant

For the following exercises, find the vertical traces of the functions at the indicated values of xx and y, and plot the traces.

30.

z=4−x−y;x=2z=4−x−y;x=2

31.

f(x,y)=3x+y3,x=1f(x,y)=3x+y3,x=1

32.

z=cosx2+y2z=cosx2+y2 x=1x=1

Find the domain of the following functions.

33.

z=100−4x2−25y2z=100−4x2−25y2

34.

z=ln(x−y2)z=ln(x−y2)

35.

f(x,y,z)=136−4x2−9y2−z2f(x,y,z)=136−4x2−9y2−z2

36.

f(x,y,z)=49−x2−y2−z2f(x,y,z)=49−x2−y2−z2

37.

f(x,y,z)=16−x2−y2−z23f(x,y,z)=16−x2−y2−z23

38.

f(x,y)=cosx2+y2f(x,y)=cosx2+y2

For the following exercises, plot a graph of the function.

39.

z=f(x,y)=x2+y2z=f(x,y)=x2+y2

40.

z=x2+y2z=x2+y2

41.

Use technology to graph z=x2y.z=x2y.

Sketch the following by finding the level curves. Verify the graph using technology.

42.

f(x,y)=4−x2−y2f(x,y)=4−x2−y2

43.

f(x,y)=2−x2+y2f(x,y)=2−x2+y2

44.

z=1+e−x2−y2z=1+e−x2−y2

45.

z=cosx2+y2z=cosx2+y2

46.

z=y2−x2z=y2−x2

47.

Describe the contour lines for several values of cc for z=x2+y2−2x−2y.z=x2+y2−2x−2y.

Find the level surface for the functions of three variables and describe it.

48.

w(x,y,z)=x−2y+z,c=4w(x,y,z)=x−2y+z,c=4

49.

w(x,y,z)=x2+y2+z2,c=9w(x,y,z)=x2+y2+z2,c=9

50.

w(x,y,z)=x2+y2−z2,c=−4w(x,y,z)=x2+y2−z2,c=−4

51.

w(x,y,z)=x2+y2−z2,c=4w(x,y,z)=x2+y2−z2,c=4

52.

w(x,y,z)=9x2−4y2+36z2,c=0w(x,y,z)=9x2−4y2+36z2,c=0

For the following exercises, find an equation of the level curve of ff that contains the point P.P.

53.

f(x,y)=1−4x2−y2,P(0,1)f(x,y)=1−4x2−y2,P(0,1)

54.

g(x,y)=y2arctanx,P(1,2)g(x,y)=y2arctanx,P(1,2)

55.

g(x,y)=exy(x2+y2),P(1,0)g(x,y)=exy(x2+y2),P(1,0)

56.

The strength EE of an electric field at point (x,y,z)(x,y,z) resulting from an infinitely long charged wire lying along the y-axisy-axis is given by E(x,y,z)=k/x2+y2,E(x,y,z)=k/x2+y2, where kk is a positive constant. For simplicity, let k=1k=1 and find the equations of the level surfaces for E=10andE=100.E=10andE=100.

57.

A thin plate made of iron is located in the xy-plane.xy-plane. The temperature TT in degrees Celsius at a point P(x,y)P(x,y) is inversely proportional to the square of its distance from the origin. Express TT as a function of xandy.xandy.

58.

Refer to the preceding problem. Using the temperature function found there, determine the proportionality constant if the temperature at point P(1,2)is50°C.P(1,2)is50°C. Use this constant to determine the temperature at point Q(3,4).Q(3,4).

59.

Refer to the preceding problem. Find the level curves for T=40°C andT=100°C,T=40°C andT=100°C, and describe what the level curves represent.

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