We are currently looking at functions of 2 variables and their graphs. Today we looked at cross-sections through a couple of different surfaces to try and figure out what they looked like in three dimensions. We did this by looking at slices in different directions and then worked out how they all fitted together. In the following animation I have taken horizontal slices of the function: $f(x,y)=x^2-y^2$

Remember the graph of this is the set of points $\{(x,y,z)\in{\mathbb R}^3|z=f(x,y) and (x,y)\in D\}$

We can take horizontal slices of the surface by fixing the z-value and seeing how x and y are constrained. For instance, let’s fix the z-value to 0. Then we have: $0=x^2-y^2$

This actually gives us two functions in the $xy$-plane: $y=\pm x$. This of course is just given by two lines of gradient +1 and -1 which pass through the origin in the $xy$-plane. In three dimensions then, if we slice through our surface at $z=0$ we should find these two lines which look like: How about at $z=1$? Well in that case we have: $1=x^2-y^2$

which is again two functions given by $y=\pm\sqrt{x^2-1}$. Here the domain of this function (not the original one) is $|x|>1$. These are functions which look like: How about for $z=-1$

In this case we have: $-1=x^2-y^2$

which is again two functions given by $y=\pm\sqrt{x^2+1}$. This looks like: Of course we can look at different slices at different heights $z$. If we were to first of all work out all the slices, and then put them together at the appropriate height in the $z$ direction, we would find the following. Here the first line you see is that at $z=-5$, then $z=-3$ then $-1$, then $0$ then the same but positive values, going up. Finally we see the shape forming, as we put the slices at the appropriate height, and we overlay the actual function on top of this: How clear is this post?