🌱 Tony's Memex Tony's Intertwingled Memex
Digital Garden | OER Curriculum
Home / Digital Garden / Multivariable Functions

Multivariable Functions

🖨️ Printable Reference / PDF
Core Definition

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

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

Concept Explanation

From Single-Variable to Multivariable Inputs

In single-variable calculus, functions y=f(x)y = f(x) map a 1D input along the xx-axis to a 1D output along the yy-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); pressure in a reaction vessel depends on temperature TT and volume VV; aerodynamic lift depends on airspeed vv, air density ρ\rho, wing area AA, and angle of attack α\alpha.

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) is the set of all points (x,y,z)∈R3(x, y, z) \in \mathbb{R}^3 such that z=f(x,y)z = f(x, y) and (x,y)∈D(x, y) \in D. Geometrically, this graph forms a two-dimensional surface hovering over the domain region DD in the xyxy-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) means x=ax = a and y=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}.
  • Civil & Structural Engineering: The deflection δ(x,y)\delta(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^2 as a function of clock frequency ff and core voltage VV.
  • Aerodynamics: Drag force FD(ρ,v)=12CDAρv2F_D(\rho, v) = \frac{1}{2} C_D A \rho v^2 acting on an aircraft fuselage as atmospheric density ρ\rho and flight velocity vv vary.

Example Problems

Example 1: Function Evaluation on a Thermodynamic Surface

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

  1. Evaluate the temperature at the center (0,0)(0,0).
  2. Evaluate the temperature at the edge coordinates (2,−1)(2, -1).
  3. Determine the coordinates where T(x,y)=100T(x, y) = 100.
Solution
  1. Center (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}
  2. Edge Point (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}
  3. Points where T=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 = 0
    Since x2≥0x^2 \ge 0 and y2≥0y^2 \ge 0, the only real solution is the origin (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}
Evaluate the pressure at the origin (0,0,0)(0, 0, 0) and at the measurement point (1,2,1)(1, 2, 1).

Solution
  1. At the origin (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}
  2. At point (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}

Connections & References