Functions of Two Variables

One input, one output

A function of one variable takes a number and gives back a number: $f\left(x\right)=x^2$f(x) = x² takes 3 and gives 9. Its graph pairs each input with its output, so each point of the graph, (3, 9) for instance, is fixed by a single number, the input 3. The inputs lie along a line, and the graph is a curve above it.

Two inputs

A function of two variables takes a pair of numbers and gives back one number, written f(x, y). Take $f\left(x,y\right)=x^2+y^2$f(x, y) = x² + y². The input (3, 4) gives 9 + 16 = 25, and (2, 3) gives 4 + 9 = 13.

A pair of numbers is a point of a plane, so the inputs now fill a whole plane rather than a line. Different inputs can give the same output: (−3, 4) gives 9 + 16 = 25 too, because squaring removes the sign. In fact every point at distance 5 from the origin gives 25, since $x^2+y^2$x² + y² is the square of that distance.

Three inputs of $f\left(x,y\right)=x^2+y^2$f(x, y) = x² + y², each labeled with its output: (3, 4) and (−3, 4) both give 25, and (2, 3) gives 13. Every input on the dashed circle of radius 5 gives 25.

The order of the inputs

$f\left(x,y\right)=x^2+y^2$f(x, y) = x² + y² gives the same output when the inputs swap, but most functions do not. Take g(x, y) = x − 2y: g(2, 3) = 2 − 6 = −4, while g(3, 2) = 3 − 4 = −1. The first number always goes in for x and the second for y.

Domain and range

The domain is the set of inputs the function accepts, now a region of the plane. $x^2+y^2$x² + y² accepts every point, so its domain is the whole plane. Its outputs are never negative, and every height from 0 up is reached, so its range is $z\ge 0$z ≥ 0.

$\sqrt{x-y}$√(x − y) needs $x-y\ge 0$x − y ≥ 0, that is $x\ge y$x ≥ y: the points on or below the line y = x. It gives $\sqrt{5-1}=2$√(5 − 1) = 2 at (5, 1) and 0 at (3, 3), and nothing at (1, 5). Its range is again $z\ge 0$z ≥ 0.

$\frac{1}{x^2+y^2}$1/(x² + y²) accepts every point except the origin, where it would divide by 0. Its outputs are all positive, and every positive number is reached, so its range is z > 0.

The domain of $\sqrt{x-y}$√(x − y) shaded: the points on or below the line y = x. The input (5, 1) gives 2 and (3, 3), on the line, gives 0; the hollow point (1, 5) lies outside, where x − y is negative.

The output is a height

To draw a function of two variables, stand each output up as a height above its input: above the point (x, y), mark the point at height z = f(x, y). That needs a third axis, z, at right angles to the plane of inputs.

One height above each point of a region joins up into a surface, not a curve. Take $z=4-x^2-y^2$z = 4 − x² − y² over the disc $x^2+y^2\le 4$x² + y² ≤ 4. Above the origin the height is 4; above (1, 1) it is 2; above every point of the circle of radius 2 it is 0. The surface is a hill, highest in the middle and falling to 0 at the edge of the disc.

$z=4-x^2-y^2$z = 4 − x² − y² above the disc of radius 2: the dashed oval is the disc of inputs on the floor, and the surface over it is highest, at 4, above the center. Each wire is the curve the surface makes above one line x = constant or y = constant.

Hold one input still

Fixing one input turns a function of two variables into a function of one, which is what each wire on the surface shows. Hold y = 1 in $z=4-x^2-y^2$z = 4 − x² − y² and what is left is $z=3-x^2$z = 3 − x², a parabola, the curve the surface makes above the line y = 1.

The applications below use this move. The cost of a crate depends on the side of its base and its height; once the height is fixed at 2 meters, the cost is a quadratic in the side alone, and setting it equal to a budget gives the side.

The usual mistakes

Combining the inputs the wrong way. For f(x, y) = x + y, f(2, 3) is 2 + 3 = 5, not 2 × 3 = 6 or 3 − 2 = 1.

Swapping the inputs. g(x, y) = x − 2y gives −4 at (2, 3) and −1 at (3, 2).

Reading the output as a point. f(3, 4) = 25 is a height above the point (3, 4); the point on the surface is (3, 4, 25).

Leaving the domain out. $\sqrt{x-y}$√(x − y) has no value at (1, 5), and $\frac{1}{x^2+y^2}$1/(x² + y²) none at the origin.

A crate, and a kicked ball

In the applications below, the cost of a crate is a function of the side of its base and its height, and two designs that hold the same volume cost different amounts. Then the range of a kicked ball depends on its speed and its angle together, so the faster kick need not go farther.