Surfaces in Three Dimensions

A third axis

The x- and y-axes lie in a flat plane, the xy-plane, which holds the inputs of a function of two variables. A third axis, the z-axis, stands up out of that plane at right angles to both, and z measures height above it. A point now has three coordinates: (1, 2, 5) is the point 5 units above (1, 2), and (1, 2, −5) is 5 units below it.

The graph of z = f(x, y) is every point (x, y, f(x, y)): above each input, the point at the height the function gives. For $z=x^2+y^2$z = x² + y² the input (1, 2) gives 1 + 4 = 5, so (1, 2, 5) is on the surface.

Read a surface by its slices

A surface is hard to draw from its equation directly, but its slices are easy. Fix one variable and the equation becomes a curve in the other two. These curves are called traces.

For $z=x^2+y^2$z = x² + y², fix x = 0 and the trace is $z=y^2$z = y², a parabola opening upward. Fix y = 0 and it is $z=x^2$z = x², the same parabola along the other axis. Fix y = 1 and it is $z=x^2+1$z = x² + 1, the same parabola raised by 1. Fix the height instead, z = 4, and the trace is $x^2+y^2=4$x² + y² = 4, a circle of radius 2.

Upward parabolas in every vertical slice and circles in every level slice: the surface is a bowl, lowest at the origin, where z = 0, and rising in every direction. It is called a paraboloid.

The bowl $z=x^2+y^2$z = x² + y² over the disc $x^2+y^2\le 16$x² + y² ≤ 16, up to height 16. The gold curve is its trace above the line y = 1, the parabola $z=x^2+1$z = x² + 1; the other wires are traces above lines x = constant and y = constant.

A slice through a different bowl

The instrument below uses $f\left(x,y\right)=x^2+2y^2$f(x, y) = x² + 2y², a bowl that rises twice as steeply along y as along x. On the left are its level curves, ellipses; the dashed line is the slice the held variable cuts. On the right is that slice drawn as a curve of its own. Holding y = 1 leaves $z=x^2+2$z = x² + 2, an upward parabola, and at x = 1.5 its height is 2.25 + 2 = 4.25.

$f=x^2+2y^2$f = x² + 2y² at the point (1.5, 1), held at y = 1: the slice is $z=x^2+2$z = x² + 2, and its gradient at x = 1.5 is 2 × 1.5 = 3. Switch to holding x and the slice through the same point is $z=2.25+2y^2$z = 2.25 + 2y², with gradient 4 × 1 = 4 at y = 1. Drag the point onto the y-axis, where x = 0, and the slice held at y is flat there.

Change one sign: a saddle

Now take $z=x^2-y^2$z = x² − y². Along y = 0 the trace is $z=x^2$z = x², an upward parabola, so the origin is the lowest point of that slice. Along x = 0 it is $z=-y^2$z = −y², a downward parabola, so the origin is the highest point of that one. The surface rises along x and falls along y.

The level traces are no longer circles. At z = 0, $x^2=y^2$x² = y², which is the pair of lines y = x and y = −x. At z = 1 the trace $x^2-y^2=1$x² − y² = 1 is a hyperbola crossing the x-axis at $\pm 1$±1; at z = −1, $y^2-x^2=1$y² − x² = 1 is a hyperbola crossing the y-axis instead.

A point that is a minimum along one slice and a maximum along another is a saddle point, and the surface is a saddle.

$z=x^2-y^2$z = x² − y² over the square with $|x|\le 2$|x| ≤ 2 and $|y|\le 2$|y| ≤ 2. The gold trace above y = 0 is $z=x^2$z = x², rising to 4 at each end; the wires the other way are downward parabolas such as $z=-y^2$z = −y² above x = 0.

Every vertical cut through the saddle

Cut the saddle with a vertical plane through the origin at angle $\theta$θ to the x-axis. A point at distance $\rho$ρ along that cut is $x=\rho \cos \theta$x = ρ cos θ, $y=\rho \sin \theta$y = ρ sin θ, so $z={\rho }^2\left({\cos }^2\theta -{\sin }^2\theta \right)=\left(\cos 2\theta \right){\rho }^2$z = ρ²(cos² θ − sin² θ) = (cos 2θ)ρ². At $\theta =0$θ = 0 the cut is $z={\rho }^2$z = ρ², a bowl; at $\theta =90^{\circ }$θ = 90° it is $z=-{\rho }^2$z = −ρ², a dome; at $\theta =45^{\circ }$θ = 45° it is z = 0, flat, along the line y = x.

The level curves of $z=x^2-y^2$z = x² − y², and a cut through the origin at $\theta =20^{\circ }$θ = 20°: the slice is $z=\left(\cos 40^{\circ }\right){\rho }^2\approx 0.77{\rho }^2$z = (cos 40°)ρ² ≈ 0.77ρ², an upward parabola. Turn the cut to 45° and the slice is flat; past 45° it opens downward, down to $z=-{\rho }^2$z = −ρ² at 90°.

Planes and cones

With no squares at all the surface is a plane. z = 3 is flat, at height 3 above every point. z = x + y tilts: its traces x = 0 and y = 0 are the straight lines z = y and z = x, and its level traces x + y = k are parallel straight lines.

Straight traces do not always mean a plane. $z=\sqrt{x^2+y^2}$z = √(x² + y²) is a cone: its level traces are circles, of radius k at height k, but its traces x = 0 and y = 0 are z = |y| and z = |x|, each a V of two straight half-lines meeting at the origin, where a bowl would be rounded.

So the formula shows the shape. Both squares with a plus sign: a bowl. One square of each sign: a saddle. Both with a minus sign, as in $z=9-x^2-y^2$z = 9 − x² − y²: a dome, highest at the origin. No squares: a plane.

The plane z = x + y over the square with $|x|\le 2$|x| ≤ 2 and $|y|\le 2$|y| ≤ 2: every wire is a straight line, and the plane climbs from −4 at one corner to 4 at the opposite one.

The usual mistakes

Mistaking $z=x^2-y^2$z = x² − y² for a bowl. A bowl rises in every direction; this surface falls along y.

Calling the saddle point a maximum or a minimum. It is the lowest point of one trace and the highest point of another.

Losing the sign in a value. At (1, 2), $x^2-y^2$x² − y² is 1 − 4 = −3, below the xy-plane, not 5.

Taking z = 3 to be a point or a line. It is the whole plane at height 3.

A roof, and a guitar string

In the applications below, a roof over a square hall is a saddle, and its traces along the two center lines are parabolas that open opposite ways. Then the displacement of a guitar string depends on the position along it and on the time: holding the time gives the shape of the string, and holding the position gives the motion of one point.