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Tony's Intertwingled Memex • Reference Manual

Multivariable Functions

Formulas & Definitions
Core Definition

A function of two variables f:D⊂R2→Rf: D \subset \mathbb{R}^2 \to \mathbb{R}f:D⊂R2→R is a mathematical rule that assigns to each ordered pair of real numbers (x,y)(x, y)(x,y) in a set DDD (the domain) exactly one real number, denoted by z=f(x,y)z = f(x, y)z=f(x,y).

More generally, a real-valued function of nnn variables f:D⊂Rn→Rf: D \subset \mathbb{R}^n \to \mathbb{R}f:D⊂Rn→R maps an ordered nnn-tuple of real numbers (x1,x2,…,xn)(x_1, x_2, \dots, x_n)(x1​,x2​,…,xn​) to a single real scalar output w=f(x1,x2,…,xn)w = f(x_1, x_2, \dots, x_n)w=f(x1​,x2​,…,xn​).

Concept Explanation

From Single-Variable to Multivariable Inputs

In single-variable calculus, functions y=f(x)y = f(x)y=f(x) map a 1D input along the xxx-axis to a 1D output along the yyy-axis, generating a 2D planar curve. Real physical and engineering systems, however, rarely depend on a single parameter. Temperature in a room depends on position (x,y,z)(x, y, z)(x,y,z); pressure in a reaction vessel depends on temperature TTT and volume VVV; aerodynamic lift depends on airspeed vvv, air density ρ\rhoρ, wing area AAA, and angle of attack α\alphaα.

Single-Variable:     x in R  ------[ f ]------>  y in R         (Graph: 2D Curve)
Two-Variable:   (x, y) in R^2 -----[ f ]------>  z in R         (Graph: 3D Surface)
Three-Variable: (x,y,z) in R^3 ----[ f ]------>  w in R         (Graph: 4D Hypersurface)

Geometric Graph of a Function of Two Variables

The graph of a function of two variables f(x,y)f(x, y)f(x,y) is the set of all points (x,y,z)∈R3(x, y, z) \in \mathbb{R}^3(x,y,z)∈R3 such that z=f(x,y)z = f(x, y)z=f(x,y) and (x,y)∈D(x, y) \in D(x,y)∈D. Geometrically, this graph forms a two-dimensional surface hovering over the domain region DDD in the xyxyxy-plane.

                    z ^
                      |          * (x, y, f(x,y))  <-- Point on 3D Surface
                      |         /|
                      |        / |
                      |       /  | z = f(x, y)
                      |      /   |
                      |     /    |
                      +----+-----+-------------> y
                     /     |    /
                    /      |   /
                   /       |  /
                  v x      * (x, y, 0) in Domain D

Evaluating Multivariable Functions

Evaluating a multivariable function requires substituting all input coordinates simultaneously into the expression. Order matters: f(a,b)f(a, b)f(a,b) means x=ax = ax=a and y=by = by=b.


Engineering Applications

  • Thermodynamics & Heat Transfer: The Ideal Gas Law expresses pressure as a multivariable function of temperature and volume: P(V,T)=nRTVP(V, T) = \frac{nRT}{V}P(V,T)=VnRT​.
  • Civil & Structural Engineering: The deflection δ(x,y)\delta(x, y)δ(x,y) across an elastic bridge deck slab subjected to vehicular loads.
  • Electrical & Computer Engineering: Power dissipation in a microchip processor P(f,V)=C⋅f⋅V2P(f, V) = C \cdot f \cdot V^2P(f,V)=C⋅f⋅V2 as a function of clock frequency fff and core voltage VVV.
  • Aerodynamics: Drag force FD(ρ,v)=12CDAρv2F_D(\rho, v) = \frac{1}{2} C_D A \rho v^2FD​(ρ,v)=21​CD​Aρv2 acting on an aircraft fuselage as atmospheric density ρ\rhoρ and flight velocity vvv vary.

Example Problems

Example 1: Function Evaluation on a Thermodynamic Surface

Let T(x,y)=100−2x2−3y2T(x, y) = 100 - 2x^2 - 3y^2T(x,y)=100−2x2−3y2 describe the steady-state temperature (in ∘C^{\circ}\text{C}∘C) on a thin metal cooling plate, where xxx and yyy are measured in centimeters from the plate center.

  1. Evaluate the temperature at the center (0,0)(0,0)(0,0).
  2. Evaluate the temperature at the edge coordinates (2,−1)(2, -1)(2,−1).
  3. Determine the coordinates where T(x,y)=100T(x, y) = 100T(x,y)=100.
Solution
  1. Center (0,0)(0,0)(0,0):
    T(0,0)=100−2(0)2−3(0)2=100∘CT(0,0) = 100 - 2(0)^2 - 3(0)^2 = 100^{\circ}\text{C}T(0,0)=100−2(0)2−3(0)2=100∘C
  2. Edge Point (2,−1)(2, -1)(2,−1):
    T(2,−1)=100−2(2)2−3(−1)2=100−2(4)−3(1)=100−8−3=89∘CT(2,-1) = 100 - 2(2)^2 - 3(-1)^2 = 100 - 2(4) - 3(1) = 100 - 8 - 3 = 89^{\circ}\text{C}T(2,−1)=100−2(2)2−3(−1)2=100−2(4)−3(1)=100−8−3=89∘C
  3. Points where T=100T = 100T=100:
    100−2x2−3y2=100  ⟹  −2x2−3y2=0  ⟹  2x2+3y2=0100 - 2x^2 - 3y^2 = 100 \implies -2x^2 - 3y^2 = 0 \implies 2x^2 + 3y^2 = 0100−2x2−3y2=100⟹−2x2−3y2=0⟹2x2+3y2=0
    Since x2≥0x^2 \ge 0x2≥0 and y2≥0y^2 \ge 0y2≥0, the only real solution is the origin (0,0)(0,0)(0,0).

Example 2: Three-Variable Function Evaluation

A scalar acoustic pressure field inside a reverberation chamber is modeled by:

p(x,y,z)=801+x2+2y2+3z2p(x, y, z) = \frac{80}{1 + x^2 + 2y^2 + 3z^2}p(x,y,z)=1+x2+2y2+3z280​
Evaluate the pressure at the origin (0,0,0)(0, 0, 0)(0,0,0) and at the measurement point (1,2,1)(1, 2, 1)(1,2,1).

Solution
  1. At the origin (0,0,0)(0,0,0)(0,0,0):
    p(0,0,0)=801+0+0+0=80 Pap(0, 0, 0) = \frac{80}{1 + 0 + 0 + 0} = 80\text{ Pa}p(0,0,0)=1+0+0+080​=80 Pa
  2. At point (1,2,1)(1, 2, 1)(1,2,1):
    p(1,2,1)=801+(1)2+2(2)2+3(1)2=801+1+8+3=8013≈6.15 Pap(1, 2, 1) = \frac{80}{1 + (1)^2 + 2(2)^2 + 3(1)^2} = \frac{80}{1 + 1 + 8 + 3} = \frac{80}{13} \approx 6.15\text{ Pa}p(1,2,1)=1+(1)2+2(2)2+3(1)280​=1+1+8+380​=1380​≈6.15 Pa

Connections & References

  • Parent Concepts: Domain of Algebraic Functions, Vectors
  • Sub-Concepts: Domain and Range of Multivariable Functions, Level Curves and Contour Maps, Traces and Cross-Sections of Surfaces, Level Surfaces
  • Course Sessions: Session 08 - Multivariable Functions, Session 09 - Partial Derivatives