Welcome to Math 21a! Although we won’t have our first classes until next week, it would be great if you got started thinking about some basic issues involving multi-dimensional space ahead of time. To get ready for our first class please try your best to do the following two sets of problems. Just do your best – you’ll get credit for having tried, even if your answers aren’t completely correct! To help you out, go ahead and get the textbook, Stewart: Multivariable Calculus, Concepts and Contexts, and read section 9.1. Also, as part of the process of doing these problems, find somebody else in the class and compare your answers with theirs. If you have similar answers, great, but if not, try to work with each other to figure out what’s going on.

Be prepared to discuss your answers to these problems at the first class. On your answer sheet, please indicate who was the other person from 21a that you worked with (note, this person doesn’t have to be from your particular section of 21a, and if you’d like to work in a group of three, that’s fine too, just write down both of the other student’s names).

Imagine being a multidimensional fence builder. If you worked in one-dimensional space (you can picture this space as simply a number line), then your fences would just be single points, dividing up the number line into two parts, to the left and right of your point. If you built two different fence-points, then you’d split up the number line universe into three regions.

Now imagine working as a multidimensional fence builder in two-dimensional space (which you can picture as a flat tabletop continuing on forever, with no edges). Now the fences you build are lines, and so if you build a single fence, your fence will again divide the two dimensional universe into two regions. What happens if you build two fences? Assume that the fences are different, but they could be parallel or they could intersect in a single point. How many regions will your fences create? What if you built three distinct fences? Four distinct fences?

Okay, time for 3-D! In three-dimensional space, which you can finally imagine as just, more or less, our universe, your fences will be planes. Building one fence will not only take a long time, but will divide space into two regions. Now, try the same questions – what's the maximum number of regions that will be produced if you build two fences? How about three fences? Four fences?

(As a total beyond-what-is-necessary-at-this-time experience, imagine being a four-dimensional fence builder, working in four dimensions – your fences are three-dimensional, and a single fence will still cut space into two regions. Now what’s the maximum number of regions you can produce if you build two, three, four fences? Five-dimensional fence building...? )

Just as points in two-dimensional space can be described with two coordinates, usually called x and y, then points in three-dimensional space can be described using three coordinates x, y and z (check in the textbook in section 9.1 for details). In two-dimensional space the set of points described by the equation x = 3 forms a line (a vertical line using our usual picture of the xy-plane). In three-dimensional space the set of points described by the equation x = 3 forms a plane, slicing through the x-axis at the point 3 on the x-axis in the same way.

Now, let's try to sketch the surface given by the equation x² + y² = 9 in three-dimensional space.

As a warm-up, let's remember how we would plot the points described by the same equation

x² + y² = 9 in two-dimensional space. Let's call the set of all such plotted points S. For example, the point (1, Ö 8) is in S, because 1² + (Ö 8)² = 9. Similarly, the point (1,2) is not in S because 1² + 2² ¹ 9. In fact, we know from earlier courses that when we plot all the points in S we get a circle of radius 3 centered at the origin.

Now let's go back to three-dimensional space and try to do what we were originally asked to do, which was to visualize the surface given by the equation x² + y² = 9. We need to find points (x,y,z) which satisfy this equation. Using the same reasoning as before, we see that (1, Ö 8, 0) is on our surface, because 1² + (Ö 8)² = 9, just as before. Notice that the value of z really doesn't make any difference, as "z" doesn't show up in the equation x² + y² = 9. So, (1, Ö 8, 10) and (1, Ö 8, 999) are also on the surface, because they satisfy the equation x² + y² = 9, too. So in fact when we find points (x,y,z) which are on the surface defined by this equation, then we can take z to be anything provided that x and y satisfy x² + y² = 9. But we know that the set of points (x,y) satisfying x² + y² = 9 forms a circle in the xy-plane… this is just the set S we were thinking

about a few moments ago! Thus, finally, we know that our surface consists of points (x,y,z) where the (x,y) part lies on the circle of radius 3 about the origin (in the xy-plane) and z can be anything. Thus the final object just looks like a cylinder, like a pipe of radius 3 running vertically up and down around the vertical z axis.

Now, as best you can at this point, try to draw a rough sketch (meaning it doesn’t have to be super accurate, but it should convey the basic sense of the object to someone else) of each of the following surfaces in three-dimensional space. Give it your best try, we realize that you might not have worked with anything like this before now! Credit will be given for your effort, even if the answers are not completely correct!

(a) y² + z² = 1 (b) z = y²

(c) xy = 1 (d) x² + y² = z