Session 08 - Multivariable Functions | Tony's Intertwingled Memex

📅 Micro-Lecture

1. Functions of Several Variables & Higher-Dimensional Input Spaces

Mathematical Foundation: In single-variable calculus, functions $y = f(x)$ map a 1D real input to a 1D output, producing a planar curve. In multivariable calculus, a function of two variables $z = f(x, y)$ assigns to each ordered pair $(x, y) \in \mathbb{R}^2$ in its domain a unique real scalar $z \in \mathbb{R}$ . The graph of $f(x, y)$ forms a two-dimensional surface embedded in 3D space $\mathbb{R}^3$ . For three variables, $w = f(x, y, z)$ maps spatial coordinates in $\mathbb{R}^3$ to a scalar output $w \in \mathbb{R}$ .
Engineering Context: Physical engineering systems inherently depend on multiple independent variables: thermodynamic pressure $P(V, T)$ depends on chamber volume and temperature; microchip power dissipation $P(f, V)$ depends on clock frequency and supply voltage; aerodynamic drag $F_D(\rho, v)$ depends on fluid density and velocity.

2. Multivariable Domains & Boundary Classifications

Mathematical Foundation: The domain of $f(x, y)$ is the subset of $\mathbb{R}^2$ where the function is mathematically defined. As detailed in Domain and Range of Multivariable Functions, restrictions arise from non-zero denominators ( $Q(x,y) \ne 0$ ), non-negative even radicals ( $u(x,y) \ge 0$ ), strictly positive logarithmic arguments ( $u(x,y) > 0$ ), and trigonometric asymptotes. Domains in $\mathbb{R}^2$ form planar regions categorized as open (dashed boundary, excluded points) or closed (solid boundary, included points), and bounded (contained in a finite disk) or unbounded. In $\mathbb{R}^3$ , domains form solid volumes, shells, or open half-spaces.
Engineering Context: Domain calculations define the safe operating envelope of engineering systems, identifying structural boundary limits for gas containment vessels and physical material constraints in load-bearing joints.

3. Visualizing Surfaces via Slices: Level Curves, Level Surfaces, and Traces

Mathematical Foundation: Because 3D surfaces and 4D scalar fields cannot be fully visualized in a single 2D projection, multivariable calculus uses dimensional slicing. Slicing a surface $z = f(x, y)$ horizontally with planes $z = c$ yields Level Curves and Contour Maps $f(x, y) = c$ . Closely spaced contour lines indicate steep gradients, while widely spaced lines indicate gradual slopes. Slicing vertically with coordinate planes yields Traces and Cross-Sections of Surfaces ( $z = f(x, k)$ or $z = f(k, y)$ ). For three variables, setting $f(x, y, z) = c$ creates 3D Level Surfaces (isosurfaces), such as isothermal shells or equipotential surfaces.
Engineering Context: Civil engineers use topographical contour maps to route highway grades; aerospace propulsion engineers inspect 3D thermal isosurfaces inside rocket nozzles to identify hot-spots requiring active regenerative cooling.

Segment 1 of 8

Skill Block 1

Real-Life Engineering Applications

🏗️

Civil Engineering

Terrain Surveying & Road Grade Safety (Civil Engineering)

In civil and transportation engineering, route alignment across mountainous topography requires rigorous analysis of surveyed ground elevation models. Before heavy earthmoving equipment and drilling rigs can be deployed to cut roadway grades, geotechnical engineers must establish the exact spatial boundaries of the mathematical elevation model to ensure equipment is never dispatched into unmapped or physically invalid zones.

A civil engineering survey team at UNLV models the localized terrain elevation $z$ (measured in hundreds of meters above a base datum) for a proposed mountain bypass using the multivariable radical function:

z = f(x, y) = \sqrt{16 - x^2 - y^2}

where

x

and

y

represent surveyed horizontal grid coordinates (in kilometers) measured east and north from a base transit station.

A planned construction access corridor requires heavy excavation machinery to traverse the regional grid point $(3, 2)$ .

State the algebraic domain condition for $f(x, y)$ and determine the mathematical domain $D$ in set-builder notation. Describe its geometric boundary and classify the region as open or closed, and bounded or unbounded.
Determine the mathematical range $R$ of the elevation function in interval notation, identifying the maximum mountain peak elevation and the perimeter base elevation in physical units (meters).
Evaluate whether the proposed construction access point $(3, 2)$ lies within the valid terrain domain. Explain what this result means for project safety and heavy equipment operations.

💡 Show Worked Derivation

In civil and transportation engineering, route alignment across mountainous topography requires rigorous analysis of surveyed ground elevation models. Before heavy earthmoving equipment and drilling rigs can be deployed to cut roadway grades, geotechnical engineers must establish the exact spatial boundaries of the mathematical elevation model to ensure equipment is never dispatched into unmapped or physically invalid zones.

A civil engineering survey team at UNLV models the localized terrain elevation $z$ (measured in hundreds of meters above a base datum) for a proposed mountain bypass using the multivariable radical function:

z = f(x, y) = \sqrt{16 - x^2 - y^2}

where

x

and

y

represent surveyed horizontal grid coordinates (in kilometers) measured east and north from a base transit station.

A planned construction access corridor requires heavy excavation machinery to traverse the regional grid point $(3, 2)$ .

State the algebraic domain condition for $f(x, y)$ and determine the mathematical domain $D$ in set-builder notation. Describe its geometric boundary and classify the region as open or closed, and bounded or unbounded.
Determine the mathematical range $R$ of the elevation function in interval notation, identifying the maximum mountain peak elevation and the perimeter base elevation in physical units (meters).
Evaluate whether the proposed construction access point $(3, 2)$ lies within the valid terrain domain. Explain what this result means for project safety and heavy equipment operations.

Step-by-Step Solution:

Determine the Mathematical Domain and Topological Classification: State the Radicand Non-Negativity Condition for Multivariable Radical Functions:
$16 - x^2 - y^2 \ge 0$
Rearranging terms yields:
$x^2 + y^2 \le 16$
- Set-Builder Notation: $D = \{(x, y) \in \mathbb{R}^2 \mid x^2 + y^2 \le 16\}$
- Geometric Description: The domain is a closed circular disk of radius $r = \sqrt{16} = 4\text{ km}$ centered at the base station $(0, 0)$ .
- Classification: Because the region contains its entire circular boundary curve $x^2 + y^2 = 16$ (non-strict inequality $\le$ ), it is closed. Because the entire disk lies within a finite radius of $4\text{ km}$ , it is bounded. (Defined in Domain and Range of Multivariable Functions.)
Determine the Elevation Range: State the Extremal Value Analysis for Bounded Quadratic Radicands: The quantity $x^2 + y^2$ attains its minimum value of $0$ at the origin $(0, 0)$ :
$z_{\max} = f(0, 0) = \sqrt{16 - 0} = 4\text{ (in hundreds of meters)} \implies 400\text{ meters}$
The quantity $x^2 + y^2$ attains its maximum value of $16$ along the perimeter boundary $x^2 + y^2 = 16$ :
$z_{\min} = \sqrt{16 - 16} = 0\text{ meters}$
- Range in Interval Notation: $R = [0, 4]$ In physical elevation units, the terrain elevation spans from $0\text{ m}$ at the outer survey perimeter to a peak summit of $400\text{ m}$ directly above the base transit station.
Evaluate Proposed Access Point $(3, 2)$ : Substitute $x = 3$ and $y = 2$ into the domain condition:
$x^2 + y^2 = 3^2 + 2^2 = 9 + 4 = 13$ Check against the boundary limit: $13 \le 16 \quad \text{(True)}$ Because $13 < 16$ , the point $(3, 2)$ lies strictly within the interior of the domain disk. Evaluating the actual elevation: $z = f(3, 2) = \sqrt{16 - 3^2 - 2^2} = \sqrt{16 - 13} = \sqrt{3} \approx 1.732\text{ hundreds of meters} \implies 173.2\text{ meters}$

Meaning of the Answer & Real-Life Application:

Because the coordinate $(3, 2)$ satisfies the domain condition ( $x^2 + y^2 = 13 \le 16$ ), it represents a physically real, accessible location on the hillside at an elevation of $173.2\text{ m}$ . Had the coordinate evaluation yielded $x^2 + y^2 > 16$ , the radicand would be negative, indicating a point outside the surveyed topography where the mathematical model breaks down (e.g., dropping off a cliff or extending beyond the project survey boundary). Confirming domain containment ensures that earthmoving contractors can safely grade access roads without navigating off-grid drop-offs.

⚙️

Mechanical Engineering

Gas Storage Tank Pressure Thresholds (Mechanical Engineering)

In industrial mechanical and chemical plant engineering, pressurized gas containment vessels operate under automated digital monitoring to prevent explosive rupture or structural implosion. Under high-temperature thermodynamic regimes, non-ideal gases deviate from linear equations of state, exhibiting asymptotic pressure surges when thermal and volumetric inputs approach acoustic resonance or critical condensation limits.

A UNLV mechanical testing lab monitors the internal pressure $P$ (in megapascals, $\text{MPa}$ ) of an experimental carbon dioxide storage vessel using the rational multivariable model:

P = g(x, y) = \frac{10}{x^2 - y^2}

where

x

represents normalized core gas temperature (in hundreds of Kelvin,

x > 0

) and

y

represents a dimensionless chamber volume expansion ratio (

y > 0

).

During a dynamic test cycle, a control valve malfunction attempts to set the operational parameters to $(x, y) = (2, 2)$ .

State the algebraic domain condition for $g(x, y)$ and write the mathematical domain $D$ in set-builder notation. Describe the geometric curves in the $xy$ -plane that must be excluded from the operational domain.
Determine the mathematical range $R$ of the pressure function across its unrestricted domain and across the physical positive quadrant ( $x > 0, y > 0$ ).
Evaluate the pressure function at the target settings $(x, y) = (2, 2)$ . Explain what happens as operating conditions approach this point, and discuss how automated safety control logic must respond.

💡 Show Worked Derivation

In industrial mechanical and chemical plant engineering, pressurized gas containment vessels operate under automated digital monitoring to prevent explosive rupture or structural implosion. Under high-temperature thermodynamic regimes, non-ideal gases deviate from linear equations of state, exhibiting asymptotic pressure surges when thermal and volumetric inputs approach acoustic resonance or critical condensation limits.

A UNLV mechanical testing lab monitors the internal pressure $P$ (in megapascals, $\text{MPa}$ ) of an experimental carbon dioxide storage vessel using the rational multivariable model:

P = g(x, y) = \frac{10}{x^2 - y^2}

where

x

represents normalized core gas temperature (in hundreds of Kelvin,

x > 0

) and

y

represents a dimensionless chamber volume expansion ratio (

y > 0

).

During a dynamic test cycle, a control valve malfunction attempts to set the operational parameters to $(x, y) = (2, 2)$ .

State the algebraic domain condition for $g(x, y)$ and write the mathematical domain $D$ in set-builder notation. Describe the geometric curves in the $xy$ -plane that must be excluded from the operational domain.
Determine the mathematical range $R$ of the pressure function across its unrestricted domain and across the physical positive quadrant ( $x > 0, y > 0$ ).
Evaluate the pressure function at the target settings $(x, y) = (2, 2)$ . Explain what happens as operating conditions approach this point, and discuss how automated safety control logic must respond.

Step-by-Step Solution:

Identify the Denominator Restriction and Domain: State the Non-Zero Denominator Condition for Rational Functions:
$x^2 - y^2 \ne 0$
Factoring as a difference of squares:
$(x - y)(x + y) \ne 0 \implies y \ne x \quad \text{and} \quad y \ne -x$
- Set-Builder Notation: $D = \{(x, y) \in \mathbb{R}^2 \mid y \ne \pm x\}$
- Geometric Description: The domain excludes the two perpendicular diagonal lines $y = x$ and $y = -x$ intersecting at the origin. Within the physical operating quadrant where temperature $x > 0$ and expansion ratio $y > 0$ , the single excluded operational boundary is the diagonal line $y = x$ . (Defined in Domain and Range of Multivariable Functions.)
Determine the Pressure Range: State the Range Analysis for Rational Inversion Functions: Let $u = x^2 - y^2$ .
- In the first quadrant ( $x > 0, y > 0$ ), when temperature dominates ( $x > y$ ), $u \in (0, \infty)$ , yielding positive pressures $P = \frac{10}{u} \in (0, \infty)$ .
- When volume expansion dominates ( $y > x$ ), $u \in (-\infty, 0)$ , yielding negative gauge pressures $P \in (-\infty, 0)$ (representing vacuum conditions).
- Because a fraction with a fixed non-zero numerator ( $10$ ) can never equal zero ( $10/u \ne 0$ for any finite $u$ ), zero is excluded from the range.
- Range: $R = (-\infty, 0) \cup (0, \infty)$
Evaluate at $(2, 2)$ and Asymptotic Limit Analysis: Substitute $x = 2$ and $y = 2$ into the denominator:
$x^2 - y^2 = 2^2 - 2^2 = 4 - 4 = 0$
Evaluating $g(2, 2) = \frac{10}{0}$ , which is undefined.
- As operating conditions approach $(2, 2)$ from the region where $x > y$ (temperature higher than expansion ratio), the denominator approaches $0$ through positive values: $\lim_{(x, y) \to (2, 2), x > y} \frac{10}{x^2 - y^2} = +\infty\text{ MPa}$
- As operating conditions approach $(2, 2)$ from the region where $x < y$ , the denominator approaches $0$ through negative values: $\lim_{(x, y) \to (2, 2), x < y} \frac{10}{x^2 - y^2} = -\infty\text{ MPa}$

Meaning of the Answer & Real-Life Application:

The line $y = x$ represents an infinite pressure singularity in the empirical state equation. In physical reality, attempting to operate near $x = y$ drives the containment vessel toward infinite hoop stress, guaranteeing explosive over-pressure rupture if $x \to y^+$ , or severe vacuum collapse if $x \to y^-$ . Control systems in UNLV energy facilities utilize software interlocks that define an exclusion buffer around $y = x$ (e.g., enforcing $|x - y| \ge 0.25$ ), triggering immediate emergency pressure-relief venting and burner cut-offs whenever operational telemetry approaches the singular diagonal line.

📚 1. Core Theory 💡 2. Worked Example ✍️ 3. Active Practice

Functions of Two Variables — Domain, Range, and Planar Regions

General Overview
Skill Block 1 covers evaluating multivariable functions $z = f(x, y)$ , solving for algebraic domains in $\mathbb{R}^2$ , sketching and classifying domain regions (open vs closed, bounded vs unbounded), and determining output ranges.

Conceptual Framework

Definition of Two-Variable Function: $z = f(x, y), \quad f: D \subset \mathbb{R}^2 \to \mathbb{R}$ (Detailed in Multivariable Functions.)
Domain Identification Rules in $\mathbb{R}^2$ :
- Even Radicals ( $\sqrt{u}$ ): Require $u(x, y) \ge 0$ .
- Fractions ( $\frac{P}{Q}$ ): Require $Q(x, y) \ne 0$ .
- Radical Denominator ( $\frac{1}{\sqrt{u}}$ ): Requires $u(x, y) > 0$ .
- Natural Logarithm ( $\ln(u)$ ): Requires $u(x, y) > 0$ .
- Tangent ( $\tan(u)$ ): Requires $u(x, y) \ne \frac{\pi}{2} + n\pi$ for $n \in \mathbb{Z}$ . (Formulated in Domain and Range of Multivariable Functions.)
Topological Boundary Classifications:
- Open Region: Contains none of its boundary points (strict inequality $<, >$ ; dashed boundary).
- Closed Region: Contains all of its boundary points (non-strict inequality $\le, \ge$ ; solid boundary).
- Bounded Region: Enclosed entirely within a circle of finite radius centered at the origin.
- Unbounded Region: Extends infinitely in at least one direction across $\mathbb{R}^2$ .
Range Determination: Analyze boundary values and extremal behavior of the radicand, denominator, or argument across the domain $D$ .

Key Equations & Structural Rules

Circular Disk Domain: $x^2 + y^2 \le r^2 \implies \text{Closed, Bounded Disk of Radius } r$
Open Disk Domain: $x^2 + y^2 < r^2 \implies \text{Open, Bounded Disk of Radius } r$
Linear Half-Plane Domain: $ax + by \le c \implies \text{Closed, Unbounded Half-Plane}$

Worked Example A: Radical Function Domain and Range

Given the two-variable function:

f(x, y) = \sqrt{9 - x^2 - y^2}

Determine the mathematical domain in set-builder notation. Describe its geometric boundary and classify the region as open or closed, and bounded or unbounded.
Determine the mathematical range in interval notation.
Evaluate the function at the origin $(0, 0)$ and at the coordinate point $(1, 2)$ .

💡 Reveal Worked Solution

Analyze Domain Condition: State the Radicand Non-Negativity Condition:
$9 - x^2 - y^2 \ge 0$
Rearrange the terms:
$x^2 + y^2 \le 9$
- Set-Builder Notation: $D = \{(x, y) \in \mathbb{R}^2 \mid x^2 + y^2 \le 9\}$
- Geometric Description: The domain consists of all points lying on or inside a circle centered at the origin with radius $r = \sqrt{9} = 3$ .
- Classification: Because the region contains its entire boundary circle $x^2 + y^2 = 9$ (non-strict inequality $\le$ ), it is closed. Because it is contained within radius $3$ , it is bounded. (Defined in Domain and Range of Multivariable Functions.)
Analyze Range: State the Extremal Value Formulation for Quadratic Radicands: The sum of squares is bounded on the domain: $0 \le x^2 + y^2 \le 9$ .
- The maximum radicand occurs at the origin $(0, 0)$ where $x^2 + y^2 = 0$ : $z_{\max} = f(0, 0) = \sqrt{9 - 0} = 3$
- The minimum radicand occurs along the boundary where $x^2 + y^2 = 9$ : $z_{\min} = \sqrt{9 - 9} = 0$
- Range in Interval Notation: $R = [0, 3]$
Evaluate at Points:
- At $(0, 0)$ : $f(0, 0) = \sqrt{9 - 0^2 - 0^2} = \sqrt{9} = 3$
- At $(1, 2)$ : Check domain: $1^2 + 2^2 = 1 + 4 = 5 \le 9$ (valid interior point). $f(1, 2) = \sqrt{9 - 1^2 - 2^2} = \sqrt{9 - 5} = \sqrt{4} = 2$

Worked Example B: Rational Function with Excluded Boundary Lines

Given the rational function:

g(x, y) = \frac{1}{x^2 - y^2}

Find the mathematical domain and describe the geometric curves excluded from the $xy$ -plane.
Determine the mathematical range in interval notation.
Evaluate $g(3, 1)$ and explain why $g(2, 2)$ is undefined.

💡 Reveal Worked Solution

Analyze Domain Condition: State the Non-Zero Denominator Condition for Rational Functions:
$x^2 - y^2 \ne 0$
Factor the difference of squares:
$(x - y)(x + y) \ne 0 \implies y \ne x \quad \text{and} \quad y \ne -x$
- Set-Builder Notation: $D = \{(x, y) \in \mathbb{R}^2 \mid y \ne \pm x\}$
- Geometric Description: The domain consists of all points in the $xy$ -plane except for the two intersecting diagonal lines $y = x$ and $y = -x$ .
- Classification: The region is open (boundary lines are excluded) and unbounded (spans infinitely across $\mathbb{R}^2$ ). (Defined in Domain and Range of Multivariable Functions.)
Analyze Range: State the Range Formulation for Rational Inversion Functions: Let $u = x^2 - y^2$ . The expression $u$ can take on any non-zero real value: $u \in (-\infty, 0) \cup (0, \infty)$ .
- When $u > 0$ , as $u \to 0^+$ , $g \to +\infty$ , and as $u \to \infty$ , $g \to 0^+$ . Output spans $(0, \infty)$ .
- When $u < 0$ , as $u \to 0^-$ , $g \to -\infty$ , and as $u \to -\infty$ , $g \to 0^-$ . Output spans $(-\infty, 0)$ .
- Because the numerator is constant ( $1 \ne 0$ ), $g(x, y)$ can never equal zero.
- Range in Interval Notation: $R = (-\infty, 0) \cup (0, \infty)$
Evaluate at Points:
- At $(3, 1)$ : $g(3, 1) = \frac{1}{3^2 - 1^2} = \frac{1}{9 - 1} = \frac{1}{8}$
- At $(2, 2)$ : Denominator evaluates to $2^2 - 2^2 = 0$ . Division by zero is undefined; $(2, 2)$ lies on the excluded boundary line $y = x$ .

Practice Problem 1: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Explain the fundamental difference between the domain of a single-variable function $y = f(x)$ and the domain of a two-variable function $z = f(x, y)$ in terms of geometric dimensionality.
(b) In Worked Example A, explain why the range of $f(x, y) = \sqrt{9 - x^2 - y^2}$ is bounded by $[0, 3]$ rather than $(-\infty, 3]$ .
(c) In the linked application Terrain Surveying & Road Grade Safety (Civil Engineering), why is evaluating whether coordinates satisfy the domain condition critical before heavy equipment is dispatched to a mountain site?

Practice Space Solve this problem on paper or tablet.

Review the definition of multivariable domain in Domain and Range of Multivariable Functions and inspect the radicand non-negativity condition $x^2 + y^2 \le 16$ in the Terrain Surveying application.

Answer: * (a) Single-variable domains are 1D subsets (intervals or points) along the real number line $\mathbb{R}$ , whereas two-variable domains are 2D geometric regions (disks, half-planes, or regions bounded by curves) in the coordinate plane $\mathbb{R}^2$ .

(b) The square root radical operator $\sqrt{\cdot}$ natively produces non-negative principal outputs ( $\ge 0$ ), and the radicand $9 - (x^2 + y^2)$ has a minimum value of $0$ on the boundary, preventing negative outputs.
(c) Bounding coordinates within the mathematical domain guarantees that the surveyed elevation model is physically valid and prevents dispatching heavy machinery into unmapped terrain or off-cliff hazards.

Step-by-Step Execution

(a) For a single-variable function $f: \mathbb{R} \to \mathbb{R}$ , the input is a single real scalar $x$ , making the domain a one-dimensional subset of the number line. For a function of two variables $f: \mathbb{R}^2 \to \mathbb{R}$ , an input is an ordered pair $(x, y)$ , meaning the domain represents a two-dimensional geometric region in the $xy$ -plane.
(b) The radical expression $\sqrt{u}$ is defined exclusively as the principal (non-negative) root, establishing an absolute lower bound of $0$ . Within the domain disk $x^2 + y^2 \le 9$ , the maximum radicand is $9 - 0 = 9$ (at the origin), yielding $z = 3$ , while the minimum radicand is $9 - 9 = 0$ (on the boundary circle), yielding $z = 0$ . Thus, $R = [0, 3]$ .
(c) In the civil engineering scenario, the radical model $z = \sqrt{16 - x^2 - y^2}$ models an surveyed hillside. Points where $x^2 + y^2 > 16$ produce imaginary roots, indicating points outside the surveyed physical embankment. Verifying that a coordinate lies in the interior of the domain ensures the road grade exists and can bear heavy construction vehicle loads.

Practice Problem 2

Given the radical function with a linear boundary:

f(x, y) = \sqrt{4 - 2x - y}

Find the mathematical domain in set-builder notation and describe the boundary line in the $xy$ -plane.
Classify the domain as open or closed, and bounded or unbounded.
Evaluate $f(1, 1)$ and determine if the point $(3, 1)$ lies in the domain.

Practice Space Solve this problem on paper or tablet.

State the Radicand Non-Negativity Condition: $4 - 2x - y \ge 0$ , and solve for $y$ to identify the boundary line equation. (Defined in Domain and Range of Multivariable Functions.)

Answer: 1. $D = \{(x, y) \in \mathbb{R}^2 \mid y \le 4 - 2x\}$ ; boundary line is $y = -2x + 4$ . 2. Closed and unbounded. 3. $f(1, 1) = 1$ ; the point $(3, 1)$ is not in the domain ( $4 - 6 - 1 = -3 < 0$ ).

Step-by-Step Execution

Analyze Domain Condition: State the Radicand Non-Negativity Condition:
$4 - 2x - y \ge 0 \implies y \le 4 - 2x$
- Set-Builder Notation: $D = \{(x, y) \in \mathbb{R}^2 \mid y \le 4 - 2x\}$
- Geometric Description: The domain consists of all points lying on and below the straight line $y = -2x + 4$ in the $xy$ -plane (a closed half-plane).
Classify Domain: Because the inequality is non-strict ( $\le$ ), the boundary line $y = -2x + 4$ is included in the domain, making the region closed. Because the half-plane extends infinitely in the $x$ and negative $y$ directions, it is unbounded.
Evaluate at Points:
- At $(1, 1)$ : $4 - 2(1) - 1 = 4 - 2 - 1 = 1 \ge 0$ $f(1, 1) = \sqrt{1} = 1$
- At $(3, 1)$ : $4 - 2(3) - 1 = 4 - 6 - 1 = -3 < 0$ Because the radicand is negative, $(3, 1)$ is not in the domain ( $f(3, 1) \notin \mathbb{R}$ ).

Practice Problem 3

Given the rational radical function:

g(x, y) = \frac{1}{\sqrt{16 - x^2 - y^2}}

Find the mathematical domain in set-builder notation.
Explain why the boundary curve $x^2 + y^2 = 16$ is excluded, and classify the domain as open or closed, and bounded or unbounded.
Determine the mathematical range in interval notation.

Practice Space Solve this problem on paper or tablet.

State the Combined Denominator and Radical Condition: when an even radical appears in the denominator, the radicand must be strictly positive ( $u > 0$ ). (Defined in Domain and Range of Multivariable Functions.)

Answer: 1. $D = \{(x, y) \in \mathbb{R}^2 \mid x^2 + y^2 < 16\}$ . 2. The boundary circle $x^2 + y^2 = 16$ causes division by zero ( $\sqrt{0} = 0$ in denominator); the region is open and bounded. 3. $R = [1/4, \infty)$ .

Step-by-Step Execution

Analyze Domain Condition: State the Strict Positivity Condition for Denominator Radicals: The radicand must be non-negative ( $\ge 0$ ) for the square root, and the denominator cannot equal zero:
$16 - x^2 - y^2 > 0 \implies x^2 + y^2 < 16$
- Set-Builder Notation: $D = \{(x, y) \in \mathbb{R}^2 \mid x^2 + y^2 < 16\}$
Classify Domain: Setting $x^2 + y^2 = 16$ yields $\sqrt{16 - 16} = \sqrt{0} = 0$ in the denominator, which is undefined. Therefore, the boundary circle is drawn with a dashed line and excluded.
- The domain is an open region (contains none of its boundary points).
- Because the disk is completely contained within radius $4$ , it is bounded.
Analyze Range: State the Range Formulation for Inverted Radicals: Within the open disk $x^2 + y^2 < 16$ , the term $x^2 + y^2$ takes on values in $[0, 16)$ .
- At the origin $(0, 0)$ , $x^2 + y^2 = 0$ , giving the minimum value of $g$ : $g(0, 0) = \frac{1}{\sqrt{16 - 0}} = \frac{1}{4}$
- As points approach the boundary circle ( $x^2 + y^2 \to 16^-$ ), the denominator approaches $0$ from the positive side: $\lim_{x^2+y^2 \to 16^-} \frac{1}{\sqrt{16 - x^2 - y^2}} = +\infty$
- Range in Interval Notation: $R = [1/4, \infty)$

Practice Problem 4

Given the logarithmic multivariable function:

k(x, y) = \ln(x^2 - y^2)

State the algebraic domain condition and express the domain in set-builder notation.
Graphically describe the regions in the $xy$ -plane where the function is defined.
Evaluate $k(3, 1)$ in exact form and round to three decimal places.

Practice Space Solve this problem on paper or tablet.

State the Strict Positivity Condition for Logarithmic Arguments: $\ln(u)$ requires $u > 0$ . Factoring as a difference of squares yields $x^2 > y^2 \implies |x| > |y|$ . (Defined in Domain and Range of Multivariable Functions.)

Answer: 1. $D = \{(x, y) \in \mathbb{R}^2 \mid x^2 > y^2\}$ . 2. The open horizontal wedge regions between the intersecting asymptotes $y = x$ and $y = -x$ containing the positive and negative $x$ -axes ( $|x| > |y|$ ). 3. $k(3, 1) = \ln(8) = 3\ln(2) \approx 2.079$ .

Step-by-Step Execution

Analyze Domain Condition: State the Logarithmic Positivity Rule: The argument of the natural logarithm must be strictly greater than zero:
$x^2 - y^2 > 0 \implies x^2 > y^2 \implies |x| > |y|$
- Set-Builder Notation: $D = \{(x, y) \in \mathbb{R}^2 \mid x^2 > y^2\}$
Geometric Description: The boundary curves are the lines $x^2 - y^2 = 0 \implies y = \pm x$ . Testing sample test points in the four regions divided by these diagonal lines:
- On the $x$ -axis at $(2, 0)$ : $2^2 - 0^2 = 4 > 0$ (Valid $\implies$ Left and Right regions are inside domain).
- On the $y$ -axis at $(0, 2)$ : $0^2 - 2^2 = -4 < 0$ (Invalid $\implies$ Top and Bottom regions are excluded). The domain consists of the two open horizontal conical wedges containing the $x$ -axis where $|x| > |y|$ , with the boundary lines $y = \pm x$ dashed and excluded.
Evaluate at $(3, 1)$ : Substitute $x = 3$ and $y = 1$ :
$k(3, 1) = \ln(3^2 - 1^2) = \ln(9 - 1) = \ln(8)$ In simplified logarithmic form: $k(3, 1) = \ln(2^3) = 3\ln(2) \approx 2.079$

Practice Problem 5

Given the trigonometric multivariable function:

h(x, y) = \tan(x - y)

State the condition required for the tangent function to be defined.
Determine the mathematical domain in set-builder notation and describe the excluded geometry in the $xy$ -plane.
Evaluate $h\left(\frac{\pi}{4}, 0\right)$ and determine if $\left(\frac{\pi}{2}, 0\right)$ lies in the domain.

Practice Space Solve this problem on paper or tablet.

State the Tangent Undefined Condition: $\tan(\theta)$ is undefined when $\theta = \frac{\pi}{2} + n\pi$ for any integer $n \in \mathbb{Z}$ . (Defined in Domain and Range of Multivariable Functions.)

Answer: 1. $x - y \ne \frac{\pi}{2} + n\pi$ for $n \in \mathbb{Z}$ . 2. $D = \{(x, y) \in \mathbb{R}^2 \mid y \ne x - \frac{\pi}{2} - n\pi, \, n \in \mathbb{Z}\}$ ; excludes an infinite set of parallel lines with slope $1$ . 3. $h(\pi/4, 0) = 1$ ; the point $(\pi/2, 0)$ is not in the domain (tangent is undefined).

Step-by-Step Execution

Analyze Domain Condition: State the Trigonometric Tangent Pole Condition:
$\tan(x - y) = \frac{\sin(x - y)}{\cos(x - y)}$ The function is undefined wherever $\cos(x - y) = 0$ , which occurs at odd integer multiples of $\frac{\pi}{2}$ : $x - y \ne \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z}$
Formulate Domain: Solve for $y$ :
$y \ne x - \frac{\pi}{2} - n\pi, \quad n \in \mathbb{Z}$
- Set-Builder Notation: $D = \left\{(x, y) \in \mathbb{R}^2 \;\middle|\; y \ne x - \frac{\pi}{2} - n\pi, \, n \in \mathbb{Z}\right\}$
- Geometric Description: The domain is the entire $xy$ -plane excluding an infinite sequence of evenly spaced parallel lines with slope $m = 1$ , each separated by vertical distance $\pi$ .
Evaluate at Points:
- At $(\pi/4, 0)$ : $x - y = \frac{\pi}{4} - 0 = \frac{\pi}{4}$ $h\left(\frac{\pi}{4}, 0\right) = \tan\left(\frac{\pi}{4}\right) = 1$
- At $(\pi/2, 0)$ : $x - y = \frac{\pi}{2} - 0 = \frac{\pi}{2}$ This matches the condition for $n = 0$ . $\cos(\pi/2) = 0$ , so $\tan(\pi/2)$ is undefined. The point $(\pi/2, 0)$ lies directly on an excluded boundary line and is not in the domain.

Segment 2 of 8

📅 Regroup 1

Emphasize that multivariable domains are 2D geometric regions, not 1D intervals. Always sketch the boundary curve first, then use a test point (like $(0,0)$ ) to decide which side to shade.
Clarify the difference between non-strict inequalities ( $\le$ , closed solid boundary) and strict inequalities ( $<$ , open dashed boundary).

Segment 3 of 8

Skill Block 2

Real-Life Engineering Applications

🏗️

Civil Engineering

Structural Stress Analysis in Spherical Joints (Civil & Mechanical Engineering)

In civil and mechanical structural engineering, large-span space frame roofs and pedestrian bridges rely on solid spherical cast-steel node joints to anchor converging diagonal structural struts. Under combined tension, compression, and shear loads from bridge traffic and wind gusts, a 3D internal stress tensor field develops inside each spherical node.

A structural engineering research team at UNLV models the equivalent Von Mises stress distribution $\sigma$ (in megapascals, $\text{MPa}$ ) inside a solid spherical steel support joint of radius $R = 5\text{ cm}$ using the three-variable radical function:

\sigma = h(x, y, z) = 50\sqrt{25 - (x^2 + y^2 + z^2)}

where

x, y, z

represent spatial coordinates (in centimeters) measured from the geometric center core of the node

(0, 0, 0)

.

An ultrasonic nondestructive evaluation (NDE) transducer inspects internal material properties at the coordinate point $(3, 4, 2)$ .

State the algebraic domain condition for $h(x, y, z)$ and write the mathematical domain $D$ in set-builder notation. Describe the geometric 3D solid region it represents.
Determine the mathematical range $R$ of the stress function in interval notation. Identify the coordinates where maximum stress occurs and where stress drops to zero.
Evaluate whether the inspection point $(3, 4, 2)$ lies within the physical material domain of the steel joint. Explain the engineering significance of this result for structural inspection workflows.

💡 Show Worked Derivation

In civil and mechanical structural engineering, large-span space frame roofs and pedestrian bridges rely on solid spherical cast-steel node joints to anchor converging diagonal structural struts. Under combined tension, compression, and shear loads from bridge traffic and wind gusts, a 3D internal stress tensor field develops inside each spherical node.

A structural engineering research team at UNLV models the equivalent Von Mises stress distribution $\sigma$ (in megapascals, $\text{MPa}$ ) inside a solid spherical steel support joint of radius $R = 5\text{ cm}$ using the three-variable radical function:

\sigma = h(x, y, z) = 50\sqrt{25 - (x^2 + y^2 + z^2)}

where

x, y, z

represent spatial coordinates (in centimeters) measured from the geometric center core of the node

(0, 0, 0)

.

An ultrasonic nondestructive evaluation (NDE) transducer inspects internal material properties at the coordinate point $(3, 4, 2)$ .

State the algebraic domain condition for $h(x, y, z)$ and write the mathematical domain $D$ in set-builder notation. Describe the geometric 3D solid region it represents.
Determine the mathematical range $R$ of the stress function in interval notation. Identify the coordinates where maximum stress occurs and where stress drops to zero.
Evaluate whether the inspection point $(3, 4, 2)$ lies within the physical material domain of the steel joint. Explain the engineering significance of this result for structural inspection workflows.

Step-by-Step Solution:

State the Domain Condition and Classify 3D Volume: State the Radicand Non-Negativity Condition for Three-Variable Functions:
$25 - (x^2 + y^2 + z^2) \ge 0$
Rearranging terms:
$x^2 + y^2 + z^2 \le 25$
- Set-Builder Notation: $D = \{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 + z^2 \le 25\}$
- Geometric Description: The domain represents a solid 3D sphere (a closed ball) centered at the origin $(0, 0, 0)$ with a boundary radius of $R = \sqrt{25} = 5\text{ cm}$ .
- Topological Classification: Because the domain includes its outer spherical boundary surface $x^2 + y^2 + z^2 = 25$ , it is a closed and bounded solid volume in $\mathbb{R}^3$ . (Defined in Domain and Range of Multivariable Functions.)
Determine the Stress Range: State the Three-Variable Quadratic Extremum Formulation: The radial distance squared from the center is $\rho^2 = x^2 + y^2 + z^2 \ge 0$ .
- Maximum Stress ( $\rho = 0$ ): At the geometric center core $(0, 0, 0)$ , $\rho^2 = 0$ : $\sigma_{\max} = h(0, 0, 0) = 50\sqrt{25 - 0} = 50(5) = 250\text{ MPa}$
- Minimum Stress ( $\rho = 5$ ): Along the entire outer spherical shell $x^2 + y^2 + z^2 = 25$ : $\sigma_{\min} = 50\sqrt{25 - 25} = 0\text{ MPa}$
- Range in Interval Notation: $R = [0, 250]\text{ MPa}$
Evaluate Inspection Coordinate Point $(3, 4, 2)$ : Compute the distance squared from the origin to the inspection point:
$\rho^2 = x^2 + y^2 + z^2 = 3^2 + 4^2 + 2^2 = 9 + 16 + 4 = 29$ Test the domain inequality: $29 \le 25 \quad \text{(False, since } 29 > 25\text{)}$ Substituting into the mathematical function yields: $\sigma = 50\sqrt{25 - 29} = 50\sqrt{-4} = 100i \notin \mathbb{R}$ The point $(3, 4, 2)$ is located at distance $\rho = \sqrt{29} \approx 5.385\text{ cm} > 5\text{ cm}$ from the center, placing it strictly outside the steel joint.

Meaning of the Answer & Real-Life Application:

The point $(3, 4, 2)$ lies in ambient open air $0.385\text{ cm}$ beyond the spherical metal casing. Evaluating the stress formula generates an imaginary number, signifying that the physical continuum model is non-existent beyond the outer steel surface. In automated NDE robotics, domain-checking algorithms prevent scanner arms from hunting for internal echoes outside the workpiece boundary, preventing false defect alerts and calibration faults during structural inspections.

⚙️

General Engineering

Underground Contaminant Dispersion Plume (Environmental Engineering)

In environmental and hydrological engineering, subterranean chemical spills threaten regional drinking water aquifers. When an industrial solvent leaks through porous underground gravel and fractured limestone, environmental remediation engineers construct 3D spatial dispersion models to estimate pollutant concentrations, design containment grouting walls, and drill monitoring recovery wells.

An environmental engineering firm in Nevada models the chemical pollutant concentration $C$ (in parts per million, $\text{ppm}$ ) migrating from a buried ruptured storage pipe using the 3D rational function:

C(x, y, z) = \frac{100}{x^2 + y^2 + z^2 - 9}

where

x, y, z

represent 3D spatial coordinates (measured in meters) relative to an injection test wellhead

(0, 0, 0)

.

A deep monitoring sensor is installed at the coordinate position $(2, 1, 2)$ .

State the algebraic domain condition for $C(x, y, z)$ and write the mathematical domain $D$ in set-builder notation. Describe the 3D surface that is physically and mathematically excluded from the domain.
Determine the mathematical range $R$ of the concentration function over its valid domain, analyzing the behavior both inside and outside the excluded boundary surface.
Evaluate the contaminant concentration at the sensor location $(2, 1, 2)$ . Explain what this value means for environmental safety zones and groundwater monitoring.

💡 Show Worked Derivation

In environmental and hydrological engineering, subterranean chemical spills threaten regional drinking water aquifers. When an industrial solvent leaks through porous underground gravel and fractured limestone, environmental remediation engineers construct 3D spatial dispersion models to estimate pollutant concentrations, design containment grouting walls, and drill monitoring recovery wells.

An environmental engineering firm in Nevada models the chemical pollutant concentration $C$ (in parts per million, $\text{ppm}$ ) migrating from a buried ruptured storage pipe using the 3D rational function:

C(x, y, z) = \frac{100}{x^2 + y^2 + z^2 - 9}

where

x, y, z

represent 3D spatial coordinates (measured in meters) relative to an injection test wellhead

(0, 0, 0)

.

A deep monitoring sensor is installed at the coordinate position $(2, 1, 2)$ .

State the algebraic domain condition for $C(x, y, z)$ and write the mathematical domain $D$ in set-builder notation. Describe the 3D surface that is physically and mathematically excluded from the domain.
Determine the mathematical range $R$ of the concentration function over its valid domain, analyzing the behavior both inside and outside the excluded boundary surface.
Evaluate the contaminant concentration at the sensor location $(2, 1, 2)$ . Explain what this value means for environmental safety zones and groundwater monitoring.

Step-by-Step Solution:

State the Denominator Condition and Identify Excluded Surface: State the Three-Variable Non-Zero Denominator Condition:
$x^2 + y^2 + z^2 - 9 \ne 0 \implies x^2 + y^2 + z^2 \ne 9$
- Set-Builder Notation: $D = \{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 + z^2 \ne 9\}$
- Geometric Description: The mathematical domain consists of all points in 3D space except for those lying exactly on a hollow spherical shell of radius $\rho = \sqrt{9} = 3\text{ meters}$ centered at the wellhead $(0, 0, 0)$ . (Defined in Domain and Range of Multivariable Functions.)
Determine the Concentration Range: State the Radial Distance Coordinate Substitution: Let $\rho^2 = x^2 + y^2 + z^2$ , where $\rho \ge 0$ . The denominator is $d = \rho^2 - 9$ .
- Inside the Excluded Shell ( $\rho \in [0, 3)$ ): As $\rho^2$ ranges from $0$ to $9$ , the denominator $d \in [-9, 0)$ .
  - At the center wellhead $(0, 0, 0)$ , $C = \frac{100}{-9} \approx -11.11\text{ ppm}$ .
  - As $\rho \to 3^-$ , $d \to 0^- \implies C \to -\infty$ .
  - Inner range component: $(-\infty, -100/9]$ .
- Outside the Excluded Shell ( $\rho \in (3, \infty)$ ): As $\rho^2$ ranges from $9$ to $\infty$ , the denominator $d \in (0, \infty)$ .
  - As $\rho \to 3^+$ , $d \to 0^+ \implies C \to +\infty$ .
  - As $\rho \to \infty$ , $d \to \infty \implies C \to 0^+$ .
  - Outer range component: $(0, \infty)$ .
- Full Mathematical Range: $R = \left(-\infty, -\frac{100}{9}\right] \cup (0, \infty)$
Evaluate at Sensor Coordinate $(2, 1, 2)$ : Compute the squared distance from the wellhead:
$\rho^2 = 2^2 + 1^2 + 2^2 = 4 + 1 + 4 = 9$ Notice that $\rho^2 = 9 \implies \rho = 3\text{ meters}$ ! Substitute into the concentration expression: $C(2, 1, 2) = \frac{100}{9 - 9} = \frac{100}{0} \implies \textbf{Undefined (Singularity)}$

Meaning of the Answer & Real-Life Application:

The sensor at $(2, 1, 2)$ sits precisely on the mathematical boundary sphere $\rho = 3\text{ m}$ . In physical hydrology, this radius corresponds to the outer perimeter of an impermeable compacted grout containment barrier where the fluid flow model exhibits an asymptotic boundary singularity. Because the theoretical equation diverges at $\rho = 3\text{ m}$ , field engineers must reposition monitoring sensors slightly beyond the containment shell (e.g., at $\rho = 4\text{ m}$ , where $C = \frac{100}{16 - 9} = 14.29\text{ ppm}$ ) to obtain reliable, finite sensor readings without digital telemetry overflow.

📚 1. Core Theory 💡 2. Worked Example ✍️ 3. Active Practice

Functions of Three Variables — Spatial Domains and Solid Volumes

General Overview
Skill Block 2 extends multivariable functions to three independent variables $w = f(x, y, z)$ , analyzing 3D input domains as solid volumes or open half-spaces in $\mathbb{R}^3$ , identifying excluded boundary surfaces (spheres, planes, cylinders), and determining output ranges.

Conceptual Framework

Definition of Three-Variable Function: $w = f(x, y, z), \quad f: D \subset \mathbb{R}^3 \to \mathbb{R}$ (Detailed in Multivariable Functions.)
Domain Geometry in $\mathbb{R}^3$ : While 2-variable functions have 2D planar domains, 3-variable functions have 3D volumetric domains in $xyz$ -space.
- Polynomials: Unrestricted domain: all of 3D space $\mathbb{R}^3$ .
- Even Radicals ( $\sqrt{u}$ ): $u(x, y, z) \ge 0 \implies$ Solid 3D volumes (balls, solid cylinders).
- Rational Functions ( $\frac{P}{Q}$ ): $Q(x, y, z) \ne 0 \implies$ 3D space with excluded surfaces (spherical shells, planes).
- Logarithms ( $\ln(u)$ ): $u(x, y, z) > 0 \implies$ Open 3D half-spaces bounded by a plane.
Euclidean Distance Squared in 3D: $\rho^2 = x^2 + y^2 + z^2$ Measures the square of the distance from the origin $(0, 0, 0)$ to $(x, y, z)$ .

Key Equations & Structural Rules

Solid Ball Domain in 3D: $x^2 + y^2 + z^2 \le R^2 \implies \text{Solid Sphere of Radius } R$
Excluded Spherical Shell: $x^2 + y^2 + z^2 \ne R^2 \implies \mathbb{R}^3 \text{ with Shell of Radius } R \text{ Hollowed Out}$
Open Half-Space: $ax + by + cz > d \implies \text{Open Region on One Side of Plane } ax + by + cz = d$

Worked Example A: 3D Radical Function and Solid Ball Domain

Given the three-variable function:

h(x, y, z) = \sqrt{9 - x^2 - y^2 - z^2}

Find the mathematical domain in set-builder notation and describe the 3D solid region.
Determine the mathematical range in interval notation.
Evaluate $h(1, 2, 0)$ and determine if $(2, 2, 2)$ lies in the domain.

💡 Reveal Worked Solution

Analyze Domain Condition: State the Three-Variable Radicand Non-Negativity Condition:
$9 - x^2 - y^2 - z^2 \ge 0$
Rearrange terms:
$x^2 + y^2 + z^2 \le 9$
- Set-Builder Notation: $D = \{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 + z^2 \le 9\}$
- Geometric Description: The domain is a solid 3D sphere (closed ball) centered at the origin $(0, 0, 0)$ with a radius of $R = \sqrt{9} = 3$ .
- Topological Classification: Closed and bounded in $\mathbb{R}^3$ . (Defined in Domain and Range of Multivariable Functions.)
Analyze Range: State the Radial Extremum Analysis for 3D Radicals: Let $\rho^2 = x^2 + y^2 + z^2 \in [0, 9]$ .
- At the origin $(0, 0, 0)$ , $\rho = 0$ , giving the maximum value: $h_{\max} = \sqrt{9 - 0} = 3$
- On the boundary shell $\rho^2 = 9$ : $h_{\min} = \sqrt{9 - 9} = 0$
- Range in Interval Notation: $R = [0, 3]$
Evaluate at Points:
- At $(1, 2, 0)$ : Compute $\rho^2 = 1^2 + 2^2 + 0^2 = 1 + 4 + 0 = 5 \le 9$ (valid interior point). $h(1, 2, 0) = \sqrt{9 - 5} = \sqrt{4} = 2$
- At $(2, 2, 2)$ : Compute $\rho^2 = 2^2 + 2^2 + 2^2 = 4 + 4 + 4 = 12$ . Since $12 > 9$ , the point lies outside the solid ball; $h(2, 2, 2) = \sqrt{-3} \notin \mathbb{R}$ is not in the domain.

Worked Example B: 3D Rational Function with Excluded Spherical Shell

Given the three-variable rational function:

g(x, y, z) = \frac{1}{x^2 + y^2 + z^2 - 1}

Find the mathematical domain and describe the excluded 3D surface.
Determine the mathematical range in interval notation.
Evaluate $g(0, 0, 0)$ and $g(2, 0, 0)$ , and explain why points with $x^2 + y^2 + z^2 = 1$ are undefined.

💡 Reveal Worked Solution

Analyze Domain Condition: State the Three-Variable Non-Zero Denominator Condition:
$x^2 + y^2 + z^2 - 1 \ne 0 \implies x^2 + y^2 + z^2 \ne 1$
- Set-Builder Notation: $D = \{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 + z^2 \ne 1\}$
- Geometric Description: The domain consists of the entirety of three-dimensional space $\mathbb{R}^3$ with the spherical shell of radius $R = 1$ centered at the origin removed. (Defined in Domain and Range of Multivariable Functions.)
Analyze Range: State the Range Formulation for 3D Inverted Fields: Let $\rho^2 = x^2 + y^2 + z^2 \ge 0$ . Denominator is $d = \rho^2 - 1$ .
- Inside the unit sphere ( $\rho^2 \in [0, 1)$ ): $d \in [-1, 0)$ . At the origin $\rho = 0$ , $g(0,0,0) = \frac{1}{-1} = -1$ . As $\rho \to 1^-$ , $d \to 0^- \implies g \to -\infty$ . This covers $(-\infty, -1]$ .
- Outside the unit sphere ( $\rho^2 \in (1, \infty)$ ): $d \in (0, \infty)$ . As $\rho \to 1^+$ , $d \to 0^+ \implies g \to +\infty$ . As $\rho \to \infty$ , $d \to \infty \implies g \to 0^+$ . This covers $(0, \infty)$ .
- Range in Interval Notation: $R = (-\infty, -1] \cup (0, \infty)$
Evaluate at Points:
- At $(0, 0, 0)$ : $g(0, 0, 0) = \frac{1}{0 - 1} = -1$
- At $(2, 0, 0)$ : $g(2, 0, 0) = \frac{1}{2^2 + 0^2 + 0^2 - 1} = \frac{1}{4 - 1} = \frac{1}{3}$
- Points satisfying $x^2 + y^2 + z^2 = 1$ lie on the unit spherical shell, making the denominator zero and causing an infinite division-by-zero discontinuity.

Practice Problem 6: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Why can we not graph a three-variable function $w = f(x, y, z)$ directly using the standard 3D coordinate system used for two-variable functions $z = f(x, y)$ ?
(b) In Worked Example B, explain why the interval $(-1, 0]$ is excluded from the range of $g(x, y, z) = \frac{1}{x^2 + y^2 + z^2 - 1}$ .
(c) In the linked application Structural Stress Analysis in Spherical Joints (Civil & Mechanical Engineering), why was the inspection point $(3, 4, 2)$ physically invalid for ultrasonic evaluation?

Practice Space Solve this problem on paper or tablet.

Review the definition of input spaces for multivariable functions in Multivariable Functions and inspect the distance calculation $\rho = \sqrt{x^2 + y^2 + z^2}$ in the Structural Stress Joint application.

Answer: * (a) Graphing $w = f(x, y, z)$ requires four perpendicular axes (three input coordinates $x, y, z$ plus one output coordinate $w$ ), creating a 4D hypersurface that cannot be plotted directly in 3D space.

(b) Inside the sphere ( $\rho^2 < 1$ ), the denominator minimum is $-1$ (at the origin), making $-1$ the largest negative output possible; outside the sphere ( $\rho^2 > 1$ ), the fraction is strictly positive ( $> 0$ ), leaving the gap $(-1, 0]$ unreachable.
(c) The squared distance to $(3, 4, 2)$ is $3^2 + 4^2 + 2^2 = 29 > 25$ , placing the point in open air $0.385\text{ cm}$ beyond the outer spherical steel casing.

Step-by-Step Execution

(a) For a function $z = f(x, y)$ , there are two input variables and one output variable, totaling three dimensions $(x, y, z)$ which match physical 3D space. For $w = f(x, y, z)$ , there are three input variables and one output variable, which requires a four-dimensional coordinate space $(x, y, z, w)$ . As a result, functions of three variables are visualized using 3D Level Surfaces rather than direct 4D surface graphs.
(b) For $\rho^2 \in [0, 1)$ , the denominator $\rho^2 - 1$ varies from $-1$ to $0^-$ . Inverting this gives $\frac{1}{\rho^2 - 1} \in (-\infty, -1]$ . The value $-1$ is achieved at the origin, and values closer to zero cannot be formed because $\rho^2 \ge 0$ . For $\rho^2 > 1$ , the denominator is strictly positive, giving outputs in $(0, \infty)$ . Therefore, no inputs can ever produce outputs in the interval $(-1, 0]$ .
(c) The steel joint has a physical radius of $5\text{ cm}$ , defined by the domain ball $x^2 + y^2 + z^2 \le 25$ . Calculating the distance of the probe at $(3, 4, 2)$ yields $\sqrt{9 + 16 + 4} = \sqrt{29} \approx 5.385\text{ cm}$ . Substituting this into the stress formula produces $\sqrt{25 - 29} = \sqrt{-4} = 2i$ , which is non-real. Ultrasonic waves cannot measure steel stress in ambient air outside the joint boundary.

Practice Problem 7

Given the 3D polynomial function:

f(x, y, z) = x^2 + y^2 + z^2

State the mathematical domain and range in appropriate notation.
Evaluate $f(1, 2, 2)$ and $f(-2, 3, -1)$ .
Explain the geometric meaning of the output value in terms of distance from the origin.

Practice Space Solve this problem on paper or tablet.

State the 3D Euclidean Distance Formula: the distance from the origin to $(x, y, z)$ is $\rho = \sqrt{x^2 + y^2 + z^2}$ . Thus, $f(x, y, z) = \rho^2$ . (Defined in Multivariable Functions.)

Answer: 1. Domain: $D = \mathbb{R}^3$ ; Range: $R = [0, \infty)$ . 2. $f(1, 2, 2) = 9$ ; $f(-2, 3, -1) = 14$ . 3. The output represents the square of the straight-line Euclidean distance from the origin $(0, 0, 0)$ to the point $(x, y, z)$ .

Step-by-Step Execution

Analyze Domain and Range:
- The expression is a polynomial with no denominators, radicals, or logarithms. The domain is all of 3D space: $D = \{(x, y, z) \in \mathbb{R}^3\} = \mathbb{R}^3$
- Each squared term is non-negative ( $x^2 \ge 0, y^2 \ge 0, z^2 \ge 0$ ). The minimum output occurs at $(0, 0, 0)$ , where $f(0, 0, 0) = 0$ . As points move arbitrarily far from the origin, $f \to \infty$ . $R = [0, \infty)$
Evaluate at Points:
- At $(1, 2, 2)$ : $f(1, 2, 2) = 1^2 + 2^2 + 2^2 = 1 + 4 + 4 = 9$
- At $(-2, 3, -1)$ : $f(-2, 3, -1) = (-2)^2 + 3^2 + (-1)^2 = 4 + 9 + 1 = 14$
Geometric Meaning: By the 3D Pythagorean theorem, the distance from the origin to $(x, y, z)$ is $\rho = \sqrt{x^2 + y^2 + z^2}$ . Therefore, $f(x, y, z) = \rho^2$ , representing the square of the radial distance from the origin.

Practice Problem 8

Given the 3D rational function:

g(x, y, z) = \frac{1}{x^2 + y^2 + z^2 - 9}

Find the mathematical domain in set-builder notation and describe the excluded boundary surface.
Determine the mathematical range in interval notation.
Evaluate $g(2, 1, 2)$ and explain the result.

Practice Space Solve this problem on paper or tablet.

State the Non-Zero Denominator Condition for 3D Rational Functions: $x^2 + y^2 + z^2 - 9 \ne 0$ . (Defined in Domain and Range of Multivariable Functions.)

Answer: 1. $D = \{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 + z^2 \ne 9\}$ ; excludes the spherical shell of radius $3$ . 2. $R = (-\infty, -1/9] \cup (0, \infty)$ . 3. Undefined; the point $(2, 1, 2)$ lies directly on the excluded spherical shell ( $2^2 + 1^2 + 2^2 = 9$ ).

Step-by-Step Execution

Analyze Domain Condition: State the Three-Variable Non-Zero Denominator Condition:
$x^2 + y^2 + z^2 - 9 \ne 0 \implies x^2 + y^2 + z^2 \ne 9$
- Set-Builder Notation: $D = \{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 + z^2 \ne 9\}$
- Geometric Description: The domain consists of all points in $\mathbb{R}^3$ except those lying on the sphere of radius $R = \sqrt{9} = 3$ centered at the origin.
Analyze Range: Let $\rho^2 = x^2 + y^2 + z^2 \ge 0$ .
- Inside the sphere ( $\rho^2 < 9$ ), $\rho^2 - 9 \in [-9, 0)$ . At the origin $\rho = 0$ , $g = -1/9$ . As $\rho \to 3^-$ , $g \to -\infty$ . Range component: $(-\infty, -1/9]$ .
- Outside the sphere ( $\rho^2 > 9$ ), $\rho^2 - 9 \in (0, \infty)$ . As $\rho \to 3^+$ , $g \to +\infty$ . As $\rho \to \infty$ , $g \to 0^+$ . Range component: $(0, \infty)$ .
- Range in Interval Notation: $R = (-\infty, -1/9] \cup (0, \infty)$
Evaluate at $(2, 1, 2)$ : Compute the denominator:
$x^2 + y^2 + z^2 - 9 = 2^2 + 1^2 + 2^2 - 9 = 4 + 1 + 4 - 9 = 9 - 9 = 0$ $g(2, 1, 2) = \frac{1}{0} \implies \textbf{Undefined}$ The point $(2, 1, 2)$ is at distance $\rho = \sqrt{9} = 3$ from the origin, lying exactly on the excluded boundary sphere.

Practice Problem 9

Given the 3D solid sphere function:

h(x, y, z) = \sqrt{25 - (x^2 + y^2 + z^2)}

Determine the mathematical domain in set-builder notation and state its radius.
Determine the mathematical range in interval notation.
Evaluate $h(3, 4, 0)$ and explain why $h(3, 4, 2)$ is undefined.

Practice Space Solve this problem on paper or tablet.

State the Radicand Non-Negativity Condition: $25 - (x^2 + y^2 + z^2) \ge 0$ . (Defined in Domain and Range of Multivariable Functions.)

Answer: 1. $D = \{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 + z^2 \le 25\}$ ; solid ball of radius $5$ . 2. $R = [0, 5]$ . 3. $h(3, 4, 0) = 0$ (boundary point); $h(3, 4, 2)$ is undefined because $3^2 + 4^2 + 2^2 = 29 > 25$ lies outside the solid sphere.

Step-by-Step Execution

Analyze Domain Condition: State the Three-Variable Radicand Non-Negativity Condition:
$25 - (x^2 + y^2 + z^2) \ge 0 \implies x^2 + y^2 + z^2 \le 25$
- Set-Builder Notation: $D = \{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 + z^2 \le 25\}$
- Geometric Description: A solid 3D ball of radius $R = \sqrt{25} = 5$ centered at $(0, 0, 0)$ .
Analyze Range: The sum $\rho^2 = x^2 + y^2 + z^2$ is bounded between $0$ (at the origin) and $25$ (on the outer boundary shell).
- Maximum output at $(0, 0, 0)$ : $\sqrt{25 - 0} = 5$ .
- Minimum output on the shell: $\sqrt{25 - 25} = 0$ .
- Range: $R = [0, 5]$ .
Evaluate at Points:
- At $(3, 4, 0)$ : $\rho^2 = 3^2 + 4^2 + 0^2 = 9 + 16 + 0 = 25$ $h(3, 4, 0) = \sqrt{25 - 25} = \sqrt{0} = 0$ This point rests directly on the outer spherical boundary shell.
- At $(3, 4, 2)$ : $\rho^2 = 3^2 + 4^2 + 2^2 = 9 + 16 + 4 = 29$ $h(3, 4, 2) = \sqrt{25 - 29} = \sqrt{-4} = 2i \notin \mathbb{R}$ Because $29 > 25$ , the point lies outside the solid ball, so the function is undefined in the real numbers.

Practice Problem 10

Given the logarithmic function of three variables:

m(x, y, z) = \ln(x + 2y + 3z)

State the domain condition and write the domain in set-builder notation.
Geometrically describe the domain in $\mathbb{R}^3$ and name the boundary surface.
Evaluate $m(1, 1, 1)$ in exact form and round to three decimal places.

Practice Space Solve this problem on paper or tablet.

State the Strict Positivity Condition for 3D Logarithms: $x + 2y + 3z > 0$ . The equation $x + 2y + 3z = 0$ defines a plane passing through the origin with normal vector $\vec{n} = \langle 1, 2, 3 \rangle$ . (Defined in Domain and Range of Multivariable Functions.)

Answer: 1. $D = \{(x, y, z) \in \mathbb{R}^3 \mid x + 2y + 3z > 0\}$ . 2. An open half-space in $\mathbb{R}^3$ bounded by the plane $x + 2y + 3z = 0$ . 3. $m(1, 1, 1) = \ln(6) \approx 1.792$ .

Step-by-Step Execution

Analyze Domain Condition: State the Logarithmic Positivity Condition: The argument of the logarithm must be strictly positive:
$x + 2y + 3z > 0$
- Set-Builder Notation: $D = \{(x, y, z) \in \mathbb{R}^3 \mid x + 2y + 3z > 0\}$
Geometric Description: The boundary surface is the flat plane $x + 2y + 3z = 0$ passing through the origin $(0, 0, 0)$ with normal vector $\vec{n} = \langle 1, 2, 3 \rangle$ . The condition $x + 2y + 3z > 0$ defines an open half-space extending infinitely on the positive side of this boundary plane (in the direction of the normal vector), with the plane itself excluded.
Evaluate at $(1, 1, 1)$ : Substitute $x = 1, y = 1, z = 1$ :
$x + 2y + 3z = 1 + 2(1) + 3(1) = 1 + 2 + 3 = 6 > 0$ $m(1, 1, 1) = \ln(6) \approx 1.792$

Segment 4 of 8

📅 Regroup 2

Review the geometric leap from 2D planar domains to 3D solid volumetric domains. Excluded curves in 2D (like $x^2 + y^2 = 9$ ) become excluded 3D surfaces (like the sphere $x^2 + y^2 + z^2 = 9$ ).
Remind students that three-variable functions $w = f(x, y, z)$ cannot be graphed directly in 3D; their behavior is visualized through 3D level surfaces.

Segment 5 of 8

Skill Block 3

Real-Life Engineering Applications

🏗️

Civil Engineering

Topographical Contour Mapping & Road Grades (Civil & Construction Engineering)

In civil and construction engineering, topographic surveys provide essential contour maps for highway grading, stormwater drainage channels, and earthwork cut-and-fill estimates. Highway safety regulations set by the Department of Transportation limit maximum allowable road grades (slopes) to ensure that loaded commercial haul trucks and transit vehicles can ascend without engine overheating and brake failure.

A civil engineering team is surveying a mountain valley where ground elevation $z$ (measured in dekameters, $1\text{ dam} = 10\text{ m}$ ) is modeled by the multivariable paraboloid function:

z = f(x, y) = x^2 + y^2

where

x

and

y

represent horizontal grid coordinates (in kilometers) measured east and north from a construction benchmark.

Determine the algebraic equations for the level curves (contour lines) at elevations $c = 1\text{ dam}$ ( $10\text{ m}$ ) and $c = 4\text{ dam}$ ( $40\text{ m}$ ). Describe their 2D geometric shape in the $xy$ -plane and state their radii.
Determine the algebraic equation of the vertical cross-sectional trace along the cutting plane $y = 0$ (the $xz$ -plane). Describe the 2D profile curve of this mountain slice.
Meaning & Real-Life Application: Explain what the spacing between successive elevation contours indicates about terrain steepness, and how civil engineers utilize vertical traces to verify that roadway cuts do not exceed legal grade percentage limits.

💡 Show Worked Derivation

In civil and construction engineering, topographic surveys provide essential contour maps for highway grading, stormwater drainage channels, and earthwork cut-and-fill estimates. Highway safety regulations set by the Department of Transportation limit maximum allowable road grades (slopes) to ensure that loaded commercial haul trucks and transit vehicles can ascend without engine overheating and brake failure.

A civil engineering team is surveying a mountain valley where ground elevation $z$ (measured in dekameters, $1\text{ dam} = 10\text{ m}$ ) is modeled by the multivariable paraboloid function:

z = f(x, y) = x^2 + y^2

where

x

and

y

represent horizontal grid coordinates (in kilometers) measured east and north from a construction benchmark.

Determine the algebraic equations for the level curves (contour lines) at elevations $c = 1\text{ dam}$ ( $10\text{ m}$ ) and $c = 4\text{ dam}$ ( $40\text{ m}$ ). Describe their 2D geometric shape in the $xy$ -plane and state their radii.
Determine the algebraic equation of the vertical cross-sectional trace along the cutting plane $y = 0$ (the $xz$ -plane). Describe the 2D profile curve of this mountain slice.
Meaning & Real-Life Application: Explain what the spacing between successive elevation contours indicates about terrain steepness, and how civil engineers utilize vertical traces to verify that roadway cuts do not exceed legal grade percentage limits.

Step-by-Step Solution:

Calculate Level Curves (Horizontal Slices): State the Level Curve Equation for Two-Variable Functions:
$f(x, y) = c$
- For $c = 1\text{ dam}$ ( $10\text{ m}$ ): $x^2 + y^2 = 1$ This is a circle in the $xy$ -plane centered at the origin $(0, 0)$ with radius $r_1 = \sqrt{1} = 1\text{ km}$ .
- For $c = 4\text{ dam}$ ( $40\text{ m}$ ): $x^2 + y^2 = 4$ This is a concentric circle centered at $(0, 0)$ with radius $r_2 = \sqrt{4} = 2\text{ km}$ . (Defined in Level Curves and Contour Maps.)
Determine the Vertical Cross-Sectional Trace ( $y = 0$ ): State the Vertical Trace Formulation along Coordinate Planes: Substitute $y = 0$ into the surface equation $z = f(x, y)$ :
$z = f(x, 0) = x^2 + 0^2 \implies z = x^2$
- Geometric Description: In the vertical $xz$ -plane, the trace is an upward-opening parabola with its vertex (lowest point) at $(x, z) = (0, 0)$ . (Defined in Traces and Cross-Sections of Surfaces.)
Slope and Terrain Grade Interpretation:
- Contour Spacing vs. Gradient: Between $c = 1$ and $c = 4$ , elevation increases by $\Delta z = 3\text{ dam} = 30\text{ m}$ over a radial distance of $\Delta r = 2 - 1 = 1\text{ km} = 1000\text{ m}$ , yielding an average grade of $\frac{30}{1000} = 3\%$ . If we look further out to $c = 9$ ( $r = 3\text{ km}$ ), the elevation increases by another $50\text{ m}$ over $1\text{ km}$ ( $5\%$ grade). Because $r = \sqrt{c}$ , equal vertical elevation contours produce circular contour rings that grow progressively closer together at larger radii, indicating that the valley walls become increasingly steep further from the origin.

Meaning of the Answer & Real-Life Application:

On a civil engineering contour map, concentric circular level curves indicate a symmetrical bowl or depression. When grading a highway route, transportation engineers examine the vertical trace $z = x^2$ along the proposed road centerline. The instantaneous road grade is given by the derivative of the trace: $\frac{dz}{dx} = 2x$ . For example, at $x = 1.5\text{ km}$ , $\frac{dz}{dx} = 3\text{ dam/km} = 3\%$ , well within typical AASHTO interstate standards ( $\le 6\%$ ). However, at $x = 3.5\text{ km}$ , the grade increases to $7\%$ , signaling the need for switchbacks, cut-and-fill retaining walls, or tunnel borings to maintain vehicle safety.

🚀

Aerospace Engineering

Aerospace Structural Thermal Isosurfaces (Aerospace & Mechanical Engineering)

In aerospace and mechanical propulsion engineering, rocket combustion chambers and hypersonic re-entry heat shields must withstand severe thermal gradients exceeding thousands of degrees. Because thermal stress and material yielding are volumetric phenomena occurring inside solid metal structures, aerospace engineers construct 3D thermal isosurfaces (level surfaces of temperature) to map heat flow and optimize regenerative cooling channels.

An aerospace propulsion team at UNLV is testing a prototype liquid rocket engine nozzle injector. During a steady-state hot-fire simulation, the internal temperature field $T$ (measured in hundreds of degrees Celsius, $^{\circ}\text{C}$ ) within the solid copper-alloy injector flange is modeled by the three-variable quadratic function:

T = g(x, y, z) = x^2 + y^2 + z^2

where

x, y, z

represent spatial coordinates (in centimeters) measured from the central ignition core

(0, 0, 0)

.

Determine the mathematical equations for the level surfaces (isothermal shells) at temperature thresholds $c = 1$ ( $100^{\circ}\text{C}$ ) and $c = 4$ ( $400^{\circ}\text{C}$ ). Describe their 3D geometric shape and state their radii.
Determine the mathematical equation for the 2D cross-sectional temperature slice passing through the principal horizontal plane $z = 0$ . Describe the resulting 2D profile.
Meaning & Real-Life Application: Explain what these level surfaces and cross-sections tell propulsion engineers about thermal barrier coating boundaries and structural thermal cracking risks.

💡 Show Worked Derivation

In aerospace and mechanical propulsion engineering, rocket combustion chambers and hypersonic re-entry heat shields must withstand severe thermal gradients exceeding thousands of degrees. Because thermal stress and material yielding are volumetric phenomena occurring inside solid metal structures, aerospace engineers construct 3D thermal isosurfaces (level surfaces of temperature) to map heat flow and optimize regenerative cooling channels.

An aerospace propulsion team at UNLV is testing a prototype liquid rocket engine nozzle injector. During a steady-state hot-fire simulation, the internal temperature field $T$ (measured in hundreds of degrees Celsius, $^{\circ}\text{C}$ ) within the solid copper-alloy injector flange is modeled by the three-variable quadratic function:

T = g(x, y, z) = x^2 + y^2 + z^2

where

x, y, z

represent spatial coordinates (in centimeters) measured from the central ignition core

(0, 0, 0)

.

Determine the mathematical equations for the level surfaces (isothermal shells) at temperature thresholds $c = 1$ ( $100^{\circ}\text{C}$ ) and $c = 4$ ( $400^{\circ}\text{C}$ ). Describe their 3D geometric shape and state their radii.
Determine the mathematical equation for the 2D cross-sectional temperature slice passing through the principal horizontal plane $z = 0$ . Describe the resulting 2D profile.
Meaning & Real-Life Application: Explain what these level surfaces and cross-sections tell propulsion engineers about thermal barrier coating boundaries and structural thermal cracking risks.

Step-by-Step Solution:

Calculate 3D Level Surfaces (Isothermal Shells): State the Level Surface Equation for Three-Variable Functions:
$g(x, y, z) = c$
- For $c = 1$ ( $100^{\circ}\text{C}$ ): $x^2 + y^2 + z^2 = 1$ This is a 3D spherical shell centered at the origin $(0, 0, 0)$ with radius $R_1 = \sqrt{1} = 1\text{ cm}$ . All material points located exactly $1\text{ cm}$ from the central core experience a steady temperature of $100^{\circ}\text{C}$ .
- For $c = 4$ ( $400^{\circ}\text{C}$ ): $x^2 + y^2 + z^2 = 4$ This is an expanded concentric spherical shell centered at $(0, 0, 0)$ with radius $R_2 = \sqrt{4} = 2\text{ cm}$ . All material points $2\text{ cm}$ from the core experience $400^{\circ}\text{C}$ . (Defined in Level Surfaces.)
Determine the Planar Cross-Section ( $z = 0$ ): State the Cross-Sectional Cutting Plane Formulation: Substitute $z = 0$ into the three-variable temperature model:
$T(x, y, 0) = x^2 + y^2 + 0^2 \implies T(x, y) = x^2 + y^2$
- Geometric Description: Slicing the solid 3D temperature field through the midplane $z = 0$ collapses the distribution onto a 2D circular paraboloid field in the $xy$ -plane, where temperature increases radially as the square of the distance from the center: $T(r) = r^2$ . (Defined in Traces and Cross-Sections of Surfaces.)
Engineering Synthesis and Thermal Stress Analysis:
- Volumetric Heat Distribution: Because the function $g(x, y, z) = \rho^2$ depends purely on the radial distance $\rho = \sqrt{x^2 + y^2 + z^2}$ , the thermal field exhibits spherical radial symmetry.
- Thermal Gradients & Material Selection: The radial rate of temperature change is $\frac{dT}{d\rho} = 2\rho$ . At $\rho = 2\text{ cm}$ , the thermal gradient is twice as steep as at $\rho = 1\text{ cm}$ .

Meaning of the Answer & Real-Life Application:

In liquid rocket engines, copper alloys begin to experience rapid thermal creep and reduced tensile strength above $400^{\circ}\text{C}$ . Identifying the $c = 4$ level surface ( $R = 2\text{ cm}$ ) establishes the critical physical boundary beyond which active cooling jackets (such as milled regenerative cooling channels carrying cryogenic liquid methane) must be routed. Furthermore, high temperature differentials between nested spherical shells generate differential thermal expansion, producing localized hoop stresses. Mapping these isosurfaces enables UNLV propulsion engineers to apply ceramic thermal barrier coatings precisely where thermal flux is highest, preventing fatigue micro-cracks and combustion chamber burnout.

📚 1. Core Theory 💡 2. Worked Example ✍️ 3. Active Practice

Visualizing Multivariable Surfaces — Level Curves, Level Surfaces, and Traces

General Overview
Skill Block 3 covers constructing and interpreting 2D level curves $f(x, y) = c$ , analyzing vertical traces ( $x = k, y = k$ ) to reconstruct 3D quadric surfaces (paraboloids, saddles), and extending level sets into 3D as level surfaces $f(x, y, z) = c$ .

Conceptual Framework

Horizontal Slices (Level Curves): For a surface $z = f(x, y)$ , setting $z = c$ produces the 2D curve $f(x, y) = c$ in the $xy$ -plane.
- Tightly packed lines $\implies$ Steep terrain / rapid change.
- Widely spaced lines $\implies$ Gentle slope / slow change. (Detailed in Level Curves and Contour Maps.)
Vertical Slices (Traces): For a surface $z = f(x, y)$ :
- Slicing with the plane $x = k$ yields the 2D curve $z = f(k, y)$ in the $yz$ -plane.
- Slicing with the plane $y = k$ yields the 2D curve $z = f(x, k)$ in the $xz$ -plane. (Detailed in Traces and Cross-Sections of Surfaces.)
3D Level Sets (Level Surfaces): For a function of three variables $w = f(x, y, z)$ , setting $w = c$ produces a 3D surface $f(x, y, z) = c$ in $\mathbb{R}^3$ . (Detailed in Level Surfaces.)

Key Equations & Structural Rules

Circular Paraboloid: $z = x^2 + y^2 \implies \text{Level curves: Circles } x^2 + y^2 = c; \quad \text{Traces: Upward Parabolas } z = x^2, z = y^2$
Hyperbolic Paraboloid (Saddle): $z = x^2 - y^2 \implies \text{Level curves: Hyperbolas } x^2 - y^2 = c; \quad \text{Traces: Opposite Parabolas } z = x^2, z = -y^2$
Spherical Level Surfaces: $x^2 + y^2 + z^2 = c \implies \text{Concentric Spherical Shells of Radius } \sqrt{c}$
Cylindrical Level Surfaces: $x^2 + y^2 = c \implies \text{Circular Cylinders along the } z\text{-axis}$

Worked Example A: Level Curves, Vertical Traces, and Paraboloid Reconstruction

Let $f(x, y) = x^2 + y^2$ .

Find the equations of the level curves for $c = 1, 4, 9$ and describe them in words.
Find the vertical traces in the planes $x = 0$ and $y = 0$ .
Synthesize the 2D slices to name and describe the 3D surface $z = f(x, y)$ .

💡 Reveal Worked Solution

Find Level Curves: State the Level Curve Equation:
$f(x, y) = c \implies x^2 + y^2 = c$
- For $c = 1$ : $x^2 + y^2 = 1$ (circle centered at $(0, 0)$ with radius $r = 1$ ).
- For $c = 4$ : $x^2 + y^2 = 4$ (circle centered at $(0, 0)$ with radius $r = 2$ ).
- For $c = 9$ : $x^2 + y^2 = 9$ (circle centered at $(0, 0)$ with radius $r = 3$ ).
- Geometric Description: Concentric circles centered at the origin whose radii increase as $r = \sqrt{c}$ . (Defined in Level Curves and Contour Maps.)
Find Vertical Traces: State the Vertical Trace Formulation:
- In the plane $x = 0$ ( $yz$ -plane): Substitute $x = 0$ into $z = x^2 + y^2$ : $z = 0^2 + y^2 \implies z = y^2$ This is an upward-opening parabola with its vertex at $(y, z) = (0, 0)$ .
- In the plane $y = 0$ ( $xz$ -plane): Substitute $y = 0$ into $z = x^2 + y^2$ : $z = x^2 + 0^2 \implies z = x^2$ This is an upward-opening parabola with its vertex at $(x, z) = (0, 0)$ . (Defined in Traces and Cross-Sections of Surfaces.)
3D Surface Synthesis: Because the horizontal cross-sections are circles and the orthogonal vertical cross-sections are identical upward-opening parabolas, the 3D surface is a circular paraboloid (a smooth, rotational bowl shape resting at the origin $(0, 0, 0)$ and opening upwards along the positive $z$ -axis).

Worked Example B: Saddle Surface Analysis via Level Curves and Traces

Let $f(x, y) = x^2 - y^2$ .

Find the level curves for $c = -4, -1, 0, 1, 4$ and sketch their qualitative behavior.
Find the vertical traces in the planes $x = 0$ and $y = 0$ .
Describe the 3D surface $z = f(x, y)$ and explain why the origin is called a saddle point.

💡 Reveal Worked Solution

Find Level Curves: State the Level Curve Equation:
$x^2 - y^2 = c$
- For $c = 0$ : $x^2 - y^2 = 0 \implies (x - y)(x + y) = 0 \implies y = x$ and $y = -x$ . The level set consists of two perpendicular straight lines intersecting at the origin.
- For $c = 1$ and $c = 4$ : $x^2 - y^2 = c \implies \frac{x^2}{c} - \frac{y^2}{c} = 1$ . These are horizontal hyperbolas opening to the left and right along the $x$ -axis, with vertices at $(\pm \sqrt{c}, 0)$ .
- For $c = -1$ and $c = -4$ : $x^2 - y^2 = c \implies y^2 - x^2 = |c| \implies \frac{y^2}{|c|} - \frac{x^2}{|c|} = 1$ . These are vertical hyperbolas opening upward and downward along the $y$ -axis, with vertices at $(0, \pm \sqrt{|c|})$ . (Defined in Level Curves and Contour Maps.)
Find Vertical Traces:
- In the plane $x = 0$ ( $yz$ -plane): $z = 0^2 - y^2 \implies z = -y^2$ This is a downward-opening parabola with its maximum at $(y, z) = (0, 0)$ .
- In the plane $y = 0$ ( $xz$ -plane): $z = x^2 - 0^2 \implies z = x^2$ This is an upward-opening parabola with its minimum at $(x, z) = (0, 0)$ . (Defined in Traces and Cross-Sections of Surfaces.)
3D Surface Synthesis: The surface is a hyperbolic paraboloid (commonly called a saddle surface or Pringles chip). Along the $x$ -direction, the origin is a local minimum ( $z$ curves upward like $x^2$ ); along the perpendicular $y$ -direction, the origin is a local maximum ( $z$ curves downward like $-y^2$ ). Because the surface bends up in one direction and down in the other, $(0, 0, 0)$ is a saddle point.

Practice Problem 11: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Explain the geometric distinction between a horizontal level curve $f(x, y) = c$ and a vertical trace $z = f(x, k)$ .
(b) In Worked Example A, explain why the spacing between concentric circular level curves decreases as elevation $c$ increases, and what this reveals about the slope of a paraboloid.
(c) In the linked application Aerospace Structural Thermal Isosurfaces (Aerospace & Mechanical Engineering), how does a 3D level surface help propulsion engineers identify cooling channel placement?

Practice Space Solve this problem on paper or tablet.

Review the definitions in Level Curves and Contour Maps and Traces and Cross-Sections of Surfaces, and inspect the radial temperature model $T = \rho^2$ in the Aerospace Thermal Isosurfaces problem.

Answer: * (a) A level curve is formed by slicing the surface with a horizontal plane $z = c$ (holding the output constant); a vertical trace is formed by slicing the surface with a vertical plane $y = k$ or $x = k$ (holding one input constant).

(b) Since the radius is $r = \sqrt{c}$ , each additional unit of radius produces a quadratic increase in elevation ( $\Delta c = 2r \Delta r$ ). Equal increments in elevation require progressively smaller radial steps, crowding the contours together and signaling that the bowl becomes steeper with distance from the origin.
(c) Level surfaces represent 3D shells of constant temperature ( $T = 400^{\circ}\text{C}$ ); identifying the spatial location of these shells indicates where thermal creep and material failure occur, allowing engineers to place cooling jackets precisely at critical thermal boundaries.

Step-by-Step Execution

(a) Slicing the graph of $z = f(x, y)$ with a horizontal plane $z = c$ freezes the dependent variable (output), producing a curve in $\mathbb{R}^2$ where every point has the identical elevation. Slicing with a vertical plane $y = k$ freezes an independent input variable, producing a 2D profile curve in the $xz$ -plane showing how elevation changes as only $x$ varies.
(b) For $z = r^2$ , we have $r = \sqrt{z}$ . The derivative is $\frac{dr}{dz} = \frac{1}{2\sqrt{z}}$ . As elevation $z$ increases, the horizontal radial expansion needed to gain another unit of elevation shrinks towards zero. On a contour map with equal elevation intervals ( $\Delta z = 1$ ), the spacing between successive contour rings becomes tighter and tighter, visually demonstrating that the slope ( $\frac{dz}{dr} = 2r$ ) is increasing.
(c) In solid rocket nozzles, temperature cannot be modeled as a flat 2D map because heat conducts three-dimensionally through the metal injector flange. By computing the level surface $T(x, y, z) = 400^{\circ}\text{C}$ (which forms a sphere of radius $R = 2\text{ cm}$ ), propulsion engineers pinpoint the exact physical boundary where copper loses structural tensile strength, directing where regenerative coolant channels must circulate to prevent catastrophic nozzle burnout.

Practice Problem 12

Given the elliptic paraboloid function:

f(x, y) = 4x^2 + y^2

Find the equations of the level curves for $c = 4$ and $c = 16$ . Describe their geometric shape in the $xy$ -plane, identifying the semi-major and semi-minor axes.
Find the vertical traces in the coordinate planes $x = 0$ and $y = 0$ .
Describe the 3D surface $z = f(x, y)$ and explain how it differs from a circular paraboloid.

Practice Space Solve this problem on paper or tablet.

State the Standard Form Equation of an Ellipse: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ . Substitute $f(x, y) = c$ and divide by $c$ . (Defined in Level Curves and Contour Maps.)

Answer: 1. For $c = 4$ : $\frac{x^2}{1} + \frac{y^2}{4} = 1$ (ellipse with $a = 1$ along $x$ , $b = 2$ along $y$ ); for $c = 16$ : $\frac{x^2}{4} + \frac{y^2}{16} = 1$ (ellipse with $a = 2, b = 4$ ). 2. In $x = 0$ : $z = y^2$ (upward parabola); in $y = 0$ : $z = 4x^2$ (steeper upward parabola). 3. Elliptic paraboloid; unlike a circular paraboloid, the bowl is elongated into an oval cross-section because the slope along the $x$ -axis is four times steeper than along the $y$ -axis.

Step-by-Step Execution

Find Level Curves: State the Level Curve Equation:
$4x^2 + y^2 = c$
- For $c = 4$ : $\frac{4x^2}{4} + \frac{y^2}{4} = 1 \implies \frac{x^2}{1^2} + \frac{y^2}{2^2} = 1$ This is an ellipse centered at $(0, 0)$ with semi-axis $1$ along the $x$ -axis and semi-axis $2$ along the $y$ -axis.
- For $c = 16$ : $\frac{4x^2}{16} + \frac{y^2}{16} = 1 \implies \frac{x^2}{2^2} + \frac{y^2}{4^2} = 1$ This is an expanded concentric ellipse with semi-axis $2$ along the $x$ -axis and semi-axis $4$ along the $y$ -axis.
Find Vertical Traces:
- Trace in $x = 0$ ( $yz$ -plane): $z = 4(0)^2 + y^2 \implies z = y^2$ This is a standard upward-opening parabola with vertex at $(0, 0)$ .
- Trace in $y = 0$ ( $xz$ -plane): $z = 4x^2 + 0^2 \implies z = 4x^2$ This is a much steeper upward-opening parabola with vertex at $(0, 0)$ .
3D Surface Synthesis: The surface is an elliptic paraboloid. Because the coefficient of $x^2$ ( $4$ ) is greater than the coefficient of $y^2$ ( $1$ ), the surface rises much faster along the $x$ -direction than the $y$ -direction, creating an oval, trough-like bowl rather than a radially symmetric circular bowl.

Practice Problem 13

Given the linear multivariable function:

f(x, y) = 2x + 3y

Find the equations of the level curves for $c = 0, 6, 12$ . Describe their geometric shape in the $xy$ -plane and state their common slope.
Find the vertical traces in the coordinate planes $x = 0$ and $y = 0$ .
Describe the 3D surface $z = f(x, y)$ .

Practice Space Solve this problem on paper or tablet.

State the Level Curve Equation for Linear Functions: $2x + 3y = c$ , and solve for $y$ in slope-intercept form $y = mx + b$ . (Defined in Level Curves and Contour Maps.)

Answer: 1. $y = -\frac{2}{3}x + \frac{c}{3}$ ; parallel straight lines with slope $m = -2/3$ . 2. In $x = 0$ : $z = 3y$ (straight line); in $y = 0$ : $z = 2x$ (straight line). 3. A flat tilted plane passing through the origin $(0, 0, 0)$ with normal vector $\vec{n} = \langle 2, 3, -1 \rangle$ .

Step-by-Step Execution

Find Level Curves: Set $2x + 3y = c$ and solve for $y$ :
$3y = -2x + c \implies y = -\frac{2}{3}x + \frac{c}{3}$
- For $c = 0$ : $y = -\frac{2}{3}x$ (line through origin).
- For $c = 6$ : $y = -\frac{2}{3}x + 2$ ( $y$ -intercept at $2$ ).
- For $c = 12$ : $y = -\frac{2}{3}x + 4$ ( $y$ -intercept at $4$ ).
- Geometric Description: A family of equally spaced, parallel straight lines across the $xy$ -plane, all possessing a constant slope of $m = -2/3$ .
Find Vertical Traces:
- In $x = 0$ ( $yz$ -plane): $z = 2(0) + 3y \implies z = 3y$ A straight line through the origin with slope $3$ .
- In $y = 0$ ( $xz$ -plane): $z = 2x + 3(0) \implies z = 2x$ A straight line through the origin with slope $2$ .
3D Surface Synthesis: Because all horizontal level curves are parallel straight lines and all vertical traces are straight lines, the graph $z = 2x + 3y \implies 2x + 3y - z = 0$ is a flat tilted plane in $\mathbb{R}^3$ passing through the origin with constant normal vector $\vec{n} = \langle 2, 3, -1 \rangle$ .

Practice Problem 14

Given the 3D spherical field function:

g(x, y, z) = x^2 + y^2 + z^2

Find the equations of the level surfaces for $c = 1, 4, 9$ . Describe their geometric shape in $\mathbb{R}^3$ and state their radii.
Determine the algebraic equation of the planar cross-section along the equatorial plane $z = 0$ .
Describe the physical behavior of this scalar field in space.

Practice Space Solve this problem on paper or tablet.

State the Level Surface Equation for Three-Variable Functions: $g(x, y, z) = c \implies x^2 + y^2 + z^2 = c$ . (Defined in Level Surfaces.)

Answer: 1. Spherical shells centered at $(0, 0, 0)$ with radii $r = 1, 2, 3$ . 2. $x^2 + y^2 = c$ ; concentric circles in the $xy$ -plane. 3. A radially symmetric field whose scalar value depends solely on the squared distance $\rho^2$ from the origin, remaining constant on spherical shell surfaces.

Step-by-Step Execution

Find Level Surfaces: State the Level Surface Formulation:
$x^2 + y^2 + z^2 = c$
- For $c = 1$ : $x^2 + y^2 + z^2 = 1$ (spherical shell of radius $r = 1$ ).
- For $c = 4$ : $x^2 + y^2 + z^2 = 4$ (spherical shell of radius $r = 2$ ).
- For $c = 9$ : $x^2 + y^2 + z^2 = 9$ (spherical shell of radius $r = 3$ ).
- Geometric Description: A family of concentric, nested 3D spherical shells expanding outward from the coordinate origin.
Equatorial Cross-Section ( $z = 0$ ): Substitute $z = 0$ into the level surface equation:
$x^2 + y^2 + 0^2 = c \implies x^2 + y^2 = c$ In the horizontal cutting plane $z = 0$ , the cross-sections are concentric circles centered at the origin of radius $\sqrt{c}$ .
Field Interpretation: The function describes a point-source or centrally symmetric field (such as gravitational potential, electrostatic potential around a point charge, or thermal radiation from a central point heater) where output values depend strictly on spatial distance $\rho = \sqrt{x^2 + y^2 + z^2}$ from the core.

Practice Problem 15

Given the 3D scalar function where the coordinate $z$ is absent:

F(x, y, z) = x^2 + y^2

Determine the equations of the level surfaces for $c = 1, 4, 9$ .
Geometrically describe these level surfaces in $\mathbb{R}^3$ .
Find the vertical cross-sectional trace produced by slicing the field with the plane $x = 0$ .

Practice Space Solve this problem on paper or tablet.

State the Cylindrical Surface Property in 3D: when an equation in $\mathbb{R}^3$ involves only two variables (e.g., $x^2 + y^2 = c$ ), the curve in that coordinate plane is extruded infinitely along the missing coordinate axis ( $z$ -axis). (Defined in Level Surfaces.)

Answer: 1. $x^2 + y^2 = c$ for $c = 1, 4, 9$ . 2. Concentric circular cylinders centered along the $z$ -axis with radii $r = 1, 2, 3$ , extending infinitely in the positive and negative $z$ -directions. 3. $y^2 = c \implies y = \pm \sqrt{c}$ ; two parallel vertical planes parallel to the $xz$ -plane.

Step-by-Step Execution

Find Level Surfaces: State the Level Surface Equation:
$F(x, y, z) = c \implies x^2 + y^2 = c$
- For $c = 1$ : $x^2 + y^2 = 1$ .
- For $c = 4$ : $x^2 + y^2 = 4$ .
- For $c = 9$ : $x^2 + y^2 = 9$ .
Geometric Description in $\mathbb{R}^3$ : In 2D, $x^2 + y^2 = c$ is a circle of radius $\sqrt{c}$ . However, because the coordinate $z$ does not appear in the equation, $z$ can take on any real value: $-\infty < z < \infty$ . Each level surface is a circular cylinder centered coaxial with the $z$ -axis with radius $r = \sqrt{c}$ , extending infinitely along the $z$ -axis. As $c$ increases, the cylinders expand in radius, forming nested concentric pipes.
Cross-Section in $x = 0$ ( $yz$ -plane): Substitute $x = 0$ into the level surface equation:
$0^2 + y^2 = c \implies y^2 = c \implies y = \pm \sqrt{c}$ In 3D space, $y = \sqrt{c}$ and $y = -\sqrt{c}$ are two flat vertical planes parallel to the $xz$ -plane, located at distance $\sqrt{c}$ on either side of the $z$ -axis.

Segment 6 of 8

📅 Regroup 3

Review how 2D contour maps translate into 3D topography. Remember: closely spaced level curves mean steep slopes; parallel linear level curves mean tilted planes; concentric circles mean paraboloids or cones.
Emphasize the distinction between level curves ( $f(x, y) = c$ , horizontal slices in $\mathbb{R}^2$ ) and level surfaces ( $f(x, y, z) = c$ , 2D shells in $\mathbb{R}^3$ ).

Segment 7 of 8

📅 Synthesis Wrap-up

Core Takeaways from Session 08:
1. Multivariable functions $z = f(x, y)$ and $w = f(x, y, z)$ map higher-dimensional coordinates to single real outputs, modeling physical systems with multiple coupled parameters.
2. The domain of a two-variable function is a planar region in $\mathbb{R}^2$ (classified as open/closed, bounded/unbounded); the domain of a three-variable function is a solid volume or open half-space in $\mathbb{R}^3$ .
3. Slicing surfaces horizontally yields level curves (contour lines); slicing vertically yields traces. Combining orthogonal slices allows engineers to reconstruct complex quadric surfaces.
4. In three dimensions, level sets $f(x, y, z) = c$ form level surfaces (isosurfaces), allowing engineers to visualize volumetric fields such as thermal and pressure distributions.
Knewton Alta Homework Alignment:
- Functions of Two Variables
- Functions of More Than Two Variables and Applications
- Level Curves and Traces of Functions of Two Variables

Segment 8 of 8