Guessing the graphs of multivariable functions

If you understand the parabola, the straight line, sine and cosine, exponential and logarithm, you are good to go. Most graphs of functions in 3D are combinations of those. So a function can be a wave in one direction and another wave in the other direction. Or a parabola in one direction and another parabola in the other direction. It's not practical to plot such graphs on paper and most teachers won't ever ask you to plot such functions by hand. What we learn in calculus is how to take advantage of analytical geometry and level curves to have a pretty good idea of how some of the easiest two variable functions can be plotted.

If you have a good grasp of how to move the graphs of functions of one variable up and down, left and right, it becomes natural to imagine the same motion for functions of two variables. The easiest cases are those in which one variable is kept constant, while the other is one of the known functions of one variable. To multiply one variable by a constant deforms the graph in the same way a function of one variable is deformed, except that in the case of two variables one variable does not depend on the other, making the deformation asymmetric.

To explain why adding one variable to another can create graphs where one axis displays one behaviour, while the other axis displays another, one is required to understand the concept of linear dependency from linear algebra. When we have plots of parametric curves in 2D, X and Y do not depend on each other. For the 3D space it's the same concept. In calculus we study functions of one or more variables but they do not have any dependency with each other. To plot a function of three variables we would need a 4th axis that is not dependent on the other three. However we do not have the means to do it. If you add a 4th axis it'll inevitably be part of the already existing 3D space, making it linearly dependent with the others. Some teachers of calculus say this at some point during a calculus course, but most of the time they skip it.

If you can follow this example, all other cases are really variations of this very simple visualisation. Notice that it's a parabola parallel to the [math]\displaystyle{ x }[/math] axis. We displaced it following the path of the [math]\displaystyle{ y }[/math] axis.

To check the graphs use wolfram, geogebra, desmos or google. Geogebra may be unable to plot 3D functions though.

Constant function

• [math]\displaystyle{ f(x,y) = k }[/math]

This is the easiest function. For every pair [math]\displaystyle{ (x,y) }[/math] the function is assuming the same constant. It's a plane parallel to the XY plane.

Addition of functions

• [math]\displaystyle{ f(x,y) = x + y }[/math]

This is analogous to the identity function in one variable. Except that now we have a plane, not a single straight line. If we multiply one of the variables by a constant, we are changing the slope in that direction but not in the other direction.

• [math]\displaystyle{ f(x,y) = x^2 + y^2 }[/math]

This is analogous to a parabola. Except that now we have two parabolas. Two parabolas how? One for each axis. Imagine one parabola and then spin | revolve it around its own vertical axis. The name for this shape is "paraboloid". If we mess with one variable multiplying it by a constant, the paraboloid loses its symmetry. If we increase the power of one variable to 4, 6, etc. We "flatten" the parabola in the same way that happens for a single variable function.

• [math]\displaystyle{ f(x,y) = x^2 + 0 }[/math]

With the second variable being a constant we have a parabola that is translated along the [math]\displaystyle{ y }[/math] axis at a constant rate. Essentially, one parabola in front of another in 3D. It's an "U" or half-pipe shape (not really a perfect "U" mind you). Think about snowboarding's or skateboarding's ramps. With this same reasoning we can explain the graphs of log, exp and waves where the second variable is a constant.

• [math]\displaystyle{ f(x,y) = \sin(x) + y }[/math]

This is very similar to [math]\displaystyle{ \sin(x) + x }[/math], except that one variable is not "affecting" the other. Along [math]\displaystyle{ x }[/math] it's a sine wave. Along [math]\displaystyle{ y }[/math] is the identity function. In the previous example all parabolas were at the same "height". Now each wave is not at the same "height" but at some [math]\displaystyle{ y }[/math] units up or down. One way to visualise it is to think on roof tiles. There is a wave pattern but at the same time it's not parallel to the ground, it has a slope.

• [math]\displaystyle{ f(x,y) = \sin(x) + y^2 }[/math]

Very similar to the previous, except that the [math]\displaystyle{ y }[/math] axis is "bent" and is now a parabola. Unlike the case of [math]\displaystyle{ \sin(x) + x^2 }[/math] in which the power of two "wins" over the wave. With two variables the parabola can "coexist" with the wave because they are in different axis. There is no linear dependency between each variable.

• [math]\displaystyle{ f(x,y) = \sin(x) + \sin(y) }[/math]

We have waves with the same period on both directions. Whenever [math]\displaystyle{ \sin(x) }[/math] is minimum, [math]\displaystyle{ \sin(y) }[/math] is too. And the same for maximum.

Product of functions

• [math]\displaystyle{ f(x,y) = xy }[/math]

At first glance this may appear a flat surface, but it's not. It's not a linear function at all. Think about the identity function, each ordered pair is [math]\displaystyle{ x = y }[/math]. What happens if they are equal to each other and we are calculating the product of them? Then we have [math]\displaystyle{ x^2 }[/math] or [math]\displaystyle{ y^2 }[/math]. This means that along the diagonals this function is a parabola. Now comes the question: a parabola upwards or downwards? It depends on the product [math]\displaystyle{ xy }[/math]. As we know from the Cartesian plane, some of the pairs are going to have both variables within the positive range. Some are going to have one positive and the other negative, in which case the parabola is inverted. It's hard to put on words, but in one direction and at a certain angle it's a parabola upwards, in the other direction and at a certain angle it's a parabola downwards. Try this imagination exercise: pick up a square made of a very flexible material. Take two vertexes opposite to each other and pull them up. The other two vertexes, pull them down.

• [math]\displaystyle{ f(x,y) = (xy)^2 }[/math]

This graph is very similar to the previous with one difference: it won't assume any negative values. This means that we have two axis behaving as parabolas upwards. Apply [math]\displaystyle{ x = 1 }[/math] and we have a parabola parallel to the [math]\displaystyle{ y }[/math] axis. Apply [math]\displaystyle{ y = 1 }[/math] and we have a parabola parallel to the [math]\displaystyle{ x }[/math] axis. Where does the function grow faster? When [math]\displaystyle{ |x| = |y| }[/math]. That is, along the diagonals. Along both axis this function is always zero.

Now, can you guess the graph of [math]\displaystyle{ f(x,y) = (2 - x^2)(2 - y^2) \ ? }[/math] It's almost the same as before, but it's not longer zero at [math]\displaystyle{ (0,0) }[/math]. The negative signs can trick us into thinking that the parabolas are downwards. But we have this [math]\displaystyle{ {x^2}{y^2} \gt x^2 }[/math] and [math]\displaystyle{ {x^2}{y^2} \gt y^2 }[/math] for all [math]\displaystyle{ x,y \gt 1 }[/math]. What happens at [math]\displaystyle{ (100, 0) }[/math] and [math]\displaystyle{ (0,100) \ ? }[/math] Along the axes the graph is no longer always zero, it's going down.

• [math]\displaystyle{ f(x,y) = \frac{1}{x^2 + y^2} }[/math]

At first glance this function does resemble [math]\displaystyle{ 1/x^2 }[/math]. We are right here, the shape of this graph does resemble its single variable counterpart. If we choose one variable as a null constant, the function's graph is reduced to a single variable problem. It's symmetric in both directions. We can take an educated guess here and compare it to the paraboloid. In the paraboloid's case we take one parabola and spin it around its own vertical axis. We can do the same here and spin | revolve [math]\displaystyle{ 1/x^2 }[/math] around its own vertical axis to get an idea of this graph. The shape is going to resemble a water vortex flipped upside down.

• [math]\displaystyle{ f(x,y) = \frac{1}{(x + y)^2} }[/math]

It bears some resemblance to the previous, but without spinning | revolving the graph. Every sum [math]\displaystyle{ x + y }[/math] is going to be squared and as such, this function never assumes negative values too. What happens when the sum is [math]\displaystyle{ 0 \lt |x + y| \lt 1 \ ? }[/math] We have fractions such as [math]\displaystyle{ 1/10 }[/math] and to calculate the inverse of that means that [math]\displaystyle{ f(x,y) }[/math] grows towards infinity for such points. When we make [math]\displaystyle{ x = 0 }[/math], [math]\displaystyle{ y }[/math] is close to 1 but not equal to 1. Do the same for the other way around, [math]\displaystyle{ y = 0 }[/math] and [math]\displaystyle{ x }[/math] very close to 1. We can subtract small quantities from one variable and add them to the other while keeping the sum of both under 1 and above 0. This should give us the idea that the division by zero occurs over a straight line, specifically the line that passes through the points [math]\displaystyle{ (0,1) }[/math] and [math]\displaystyle{ 1,0 }[/math] of this function's domain.

Composite functions

• [math]\displaystyle{ f(x,y) = \sin(x + y) }[/math]

Don't be fooled thinking that to sum arguments is the same as to sum functions! For composite functions in two variables it's a little tricky to get the idea of the graph. The first thing to notice is that we have one sine function, not two as in [math]\displaystyle{ \sin(x) + \sin(y) }[/math]. This should give us the idea that there only one wave pattern in a single direction, no interference with waves in any other direction. The second thing to notice is that the angle is going to be a sum of two parts. How many pairs [math]\displaystyle{ (x,y) }[/math] result in, say, [math]\displaystyle{ \pi/2 \ ? }[/math] There are infinitely many combinations that add up to the right angle. This should give us the idea that, for every angle, there are infinitely many pairs. With that idea in mind we can think that each "height" of the sine wave is going to be constant along a straight line over this function's domain. This is exactly the idea of this graph, a series of waves propagating in a single direction without "spreading" over time in circular patterns. Think about waves parallel to the shoreline.

• [math]\displaystyle{ f(x,y) = \sin(x^2 + y^2) }[/math]

If you are quick to notice it we have what seems to be the equation of a circle as the sine's argument. We have that [math]\displaystyle{ \theta = (x^2 + y^2) }[/math]. With this we can already say that this sine wave is propagating in all directions in a circular pattern. The same pattern of throwing a stone on a lake, except that this function's graph keeps its ripples over long distances from the origin. Multiply the function by a decreasing function, in all directions, and we effectively dampen the oscillations.

• [math]\displaystyle{ f(x,y) = \sin(xy) }[/math]

Knowing the graph of [math]\displaystyle{ f(x,y) = xy }[/math] helps here. Along the axes the function is zero and [math]\displaystyle{ \sin(0) = 0 }[/math]. With the product [math]\displaystyle{ xy }[/math] the waves are neither circular nor straight as the previous two examples. To keep the angle a constant what should happen to the product [math]\displaystyle{ xy \ ? }[/math]. One variable should be the inverse of the other. This reminds us of this single variable function [math]\displaystyle{ f(x) = 1/x }[/math]. With that we can take an educated guess and say that the waves are propagating in all directions in a fashion similar to [math]\displaystyle{ 1/x }[/math] and [math]\displaystyle{ -1/x }[/math]. That is, along the curves of the inverse of [math]\displaystyle{ x }[/math] the sine is a constant.

• [math]\displaystyle{ f(x,y) = e^{-x^2 - y^2} }[/math]

If you know the graph of [math]\displaystyle{ e^{-x^2} }[/math] the graph of the two variable version should come naturally. It's symmetric in both directions. Remember that [math]\displaystyle{ e^{-x^2 - y^2} = e^{-x^2} \cdot e^{-y^2} }[/math]. If you look at [math]\displaystyle{ -x^2 - y^2 }[/math] it does resemble the equation of a circle. That means that for every [math]\displaystyle{ e^n }[/math] there are infinitely many pairs [math]\displaystyle{ (x,y) }[/math] that satisfy that number. The number being the distance from the origin. The graph resembles a hill.

• [math]\displaystyle{ f(x,y) = \ln(x^2 + y^2) }[/math]

The first thing to notice is that the argument cannot be zero. Again, [math]\displaystyle{ x^2 + y^2 }[/math] does remind us of the circle. The domain of this function is [math]\displaystyle{ x^2 + y^2 \gt 0 }[/math]. The graph of [math]\displaystyle{ f(x) = \ln(x^2) }[/math] is very similar to the graph of [math]\displaystyle{ f(x) = \ln(x) }[/math] except that the square root allows it to be plotted for negative inputs. With this in mind we can safely assume that the graph is symmetric in both directions. It resembles those illustrations that depict how a black hole distorts space-time.