Difficulty: Introductory-Intermediate | Prerequisites: Calculus I domain-finding (restrictions on square roots and denominators), basic algebra (factoring quadratics), familiarity with circles and conic sections.
Up to now, your functions have taken one input and returned one output: y = f(x). Section 14.1 extends this to functions of two (or more) variables: z = f(x, y). These functions produce surfaces in 3D space rather than curves in 2D. The first task with any new function is still the same one you learned in Calculus I: find the domain, meaning the set of inputs where the function is defined. The difference is that the domain is now a region in the xy-plane (or in xyz-space) rather than an interval on the number line. You will also learn to read contour maps, which slice a surface at constant heights to produce a family of curves.
Multivariable functions take (x, y) as input and return z. Finding their domain works the same way as in single-variable calculus: set what is under a square root to be greater than or equal to zero, and set any denominator not equal to zero. The domain is now a 2D region, not a 1D interval. Contour maps (level curves) show where the function takes constant values.
Multivariable function (function of two variables)
A function of the form z = f(x, y) that accepts an ordered pair (x, y) as input and returns a single real number z. Think of it as: a height map. For every point on a flat plane, the function tells you how high (or low) the surface sits above that point.
Domain of f(x, y)
The set of all (x, y) pairs for which f(x, y) is defined. This is a region in the xy-plane, not an interval on a number line. In simple terms, this means: find every (x, y) you are allowed to plug in without breaking a mathematical rule (no square roots of negatives, no dividing by zero).
Surface
The graph of z = f(x, y) in three-dimensional space. Every point on the surface has the form (x, y, f(x, y)). Think of it as: the 3D equivalent of a curve. Instead of a line drawn on a flat page, you have a sheet stretched through space.
Contour (level curve)
The set of all (x, y) points where f(x, y) equals a fixed constant c. On a contour map, each curve corresponds to a different value of c. In simple terms, this means: pick a height, then trace out every point on the surface at exactly that height. The result, viewed from above, is one contour line.
Contour map
A collection of level curves drawn in the xy-plane, each labeled with its constant z-value. This is how topographic maps work. Think of it as: looking straight down at the surface and drawing the outlines where the surface crosses certain heights.
Level surface
The 3D analogue of a level curve, used for functions of three variables f(x, y, z). Setting f(x, y, z) = c gives a surface in xyz-space rather than a curve in the xy-plane. In simple terms, this means: for a function with three inputs, the "contour" is a whole surface, not just a curve.
Evaluation works exactly as in single-variable calculus: substitute the given values for x and y, then simplify.
Example: f(x, y) = 3x + e^(x^2 - y^2). Then f(1, 2) = 3(1) + e^(1 - 4) = 3 + e^(-3).
If f(x, y) involves a square root, the expression under the root must be greater than or equal to zero.
Single-variable review: f(x) = sqrt(x^2 - 5x + 6). Factor: (x - 2)(x - 3) >= 0. Both factors positive (x >= 3) or both negative (x <= 2). Domain: (-inf, 2] union [3, inf).
Multivariable example: f(x, y) = sqrt(25 - x^2 - y^2). Restriction: 25 - x^2 - y^2 >= 0, which rearranges to x^2 + y^2 <= 25. This is a closed disk of radius 5 centred at the origin.
If f(x, y) has a fraction, the denominator cannot equal zero.
Example: f(x, y) = 12x / (3x^2 + 5xy - 2y^2). Factor the denominator: (3x - y)(x + 2y). Set each factor not equal to zero: y != 3x and y != -x/2. The domain is all of R^2 except the two lines y = 3x and y = -x/2.
When sketching, draw the excluded lines as dashed and shade or label the rest of the plane as the domain.
To sketch a contour map, set f(x, y) = c for several values of c and solve for the resulting curve in the xy-plane.
Example: f(x, y) = sqrt(25 - x^2 - y^2). Setting z = 0 gives x^2 + y^2 = 25 (circle of radius 5). Setting z = 1 gives x^2 + y^2 = 24. Setting z = 2 gives x^2 + y^2 = 21. The contour map is a family of concentric circles that shrink as z increases.
Contour lines that are close together indicate a steep surface; lines that are far apart indicate a gentle slope.
For a function of three variables, f(x, y, z) = c defines a surface in 3D space (not a curve).
Example: f(x, y, z) = x^2 + y^2 + z^2. Setting c = 25 gives the sphere x^2 + y^2 + z^2 = 25. Setting c = 9 gives a smaller sphere of radius 3. The level surfaces are concentric spheres.
Domain restriction for square roots:
\text{If } f(x,y) = \sqrt{g(x,y)}, \quad \text{then } g(x,y) \geq 0Domain restriction for fractions:
\text{If } f(x,y) = \frac{h(x,y)}{g(x,y)}, \quad \text{then } g(x,y) \neq 0Level curve equation:
f(x,y) = c \quad \text{(solve for the curve in the } xy\text{-plane)}Level surface equation:
f(x,y,z) = c \quad \text{(solve for the surface in } xyz\text{-space)}Contour maps are the mathematics behind topographic maps (elevation as a function of latitude and longitude), weather maps (pressure isobars, temperature isotherms), and medical imaging (CT scans use level surfaces of tissue density). Domain analysis matters any time an engineering formula has physical limits on its inputs, such as a stress function that is only valid inside a certain region of a material.
Students often describe the domain of f(x, y) as an interval (e.g. "x is between -5 and 5") instead of as a region in the xy-plane. The domain is a 2D set, not two separate 1D intervals.
When factoring a denominator to find excluded points, students sometimes set the factors equal to zero and exclude only those individual points. If the factors are linear in x and y, the excluded set is entire lines, not isolated points.
Students frequently confuse level curves with cross-sections. A level curve sets f(x, y) = c and lives in the xy-plane. A cross-section fixes one variable (e.g. y = 0) and plots the resulting curve in the xz-plane.
Students sometimes think contour lines can cross. They cannot: a function assigns only one z-value to each (x, y), so two contour lines at different heights can never share a point.
Expect at least one domain-finding problem on the exam, likely involving a combination of square roots and denominators.
Contour map sketching is a common exam question. You may be given f(x, y) and asked to draw level curves for three or four values of c.
Know how to describe a domain both algebraically (as an inequality like x^2 + y^2 <= 25) and geometrically ("a closed disk of radius 5 centred at the origin").
True or False: The domain of f(x, y) = sqrt(9 - x^2 - y^2) is the disk x^2 + y^2 <= 9. True.
Fill in the blank: To find the domain of a rational function f(x, y) = p(x,y)/q(x,y), set ______ not equal to zero. The denominator q(x, y).
True or False: A level curve of f(x, y) is drawn in the xyz-space. False. It is drawn in the xy-plane.
Fill in the blank: The level surfaces of f(x, y, z) = x^2 + y^2 + z^2 are ______. Concentric spheres centred at the origin.
True or False: Two different level curves of the same function can cross each other. False.
Q: Find the domain of f(x, y) = sqrt(16 - 4x^2 - y^2).
A: Require 16 - 4x^2 - y^2 >= 0, i.e. 4x^2 + y^2 <= 16, i.e. x^2/4 + y^2/16 <= 1. The domain is the interior and boundary of an ellipse centred at the origin with semi-axes a = 2 (along x) and b = 4 (along y).
Q: Find and sketch the domain of f(x, y) = 1 / (x^2 - y).
A: Require x^2 - y != 0, so y != x^2. The domain is all of R^2 except the parabola y = x^2. Sketch the parabola as a dashed curve; the domain is everything else.
Q: Sketch the contour map of f(x, y) = x^2 + y^2 for c = 1, 4, 9.
A: Setting x^2 + y^2 = 1 gives a circle of radius 1. Setting = 4 gives radius 2. Setting = 9 gives radius 3. The contour map is concentric circles centred at the origin, growing larger as c increases.
Q: Describe the level surfaces of f(x, y, z) = x^2 + y^2 - z^2 for c = 0, c = 1, and c = -1.
A: c = 0 gives x^2 + y^2 = z^2, a cone. c = 1 gives x^2 + y^2 - z^2 = 1, a hyperboloid of one sheet. c = -1 gives x^2 + y^2 - z^2 = -1, a hyperboloid of two sheets.
Q: Evaluate f(2, -1) for f(x, y) = ln(x + y) + sqrt(x - y).
A: f(2, -1) = ln(2 + (-1)) + sqrt(2 - (-1)) = ln(1) + sqrt(3) = 0 + sqrt(3) = sqrt(3).
Domain-finding here is a direct extension of the single-variable domain work from Calculus I. The factoring techniques are identical; the only new element is that the domain is a region rather than a union of intervals.
Contour maps and level surfaces lay the groundwork for partial derivatives (Section 14.3) and the gradient vector (Section 14.6). The gradient at any point is perpendicular to the level curve through that point, a fact you will use heavily later in the course.
Level surfaces also connect to the topic of quadric surfaces (ellipsoids, hyperboloids, paraboloids) from earlier in MATH 163.
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