Intuition with graphs of parametarizations

Question

I have trouble intuitively understanding why a certain graph belongs to a parametarization in a certain number of parameters.

When I ask myself why the graph of the function $f(x) = y$ is a curve, it's because if it were a surface it would fail the vertical line test (for any one input there would need to be multiple outputs).

Also, the graph of $f(x, y) = z$ is a surface and not a volume because for any 2 inputs there would need to be multiple outputs.

I can't seem to develop that kind of intuition for parametarizations though. Why is the graph of $\vec{r}(t) = (x(t), y(t), z(t))$ a curve, and the graph of $\vec{r}(u, v) = (x(u, v), y(u, v), z(u, v))$ a surface?

Is there more advanced math involved behind the curtain that prevents the intuition?

Answer 1

What your intuition is very good! You have started feeling what manifolds are!

Manifolds are how we make mathematicaly rigorous the ideas behind curves, surfaces and higher dimensional analogues.

One way to get start you wondering about this stuff is what we call dimension of a set. One naive way is to say that if a set can be parametrized by $n$ variables then it should be of dimension $n$. But this leads to problems as for example the square can be parametrized by $2$ variables trivially but also by $1$ (as Peano's space filling curve demonstrates).

To fix that we ask that every point of the set should have a neighbourhoud homeomorphic (="looks like") some $\mathbb{R^n}.$ What we would call a line or curve cant look localy like any $\mathbb{R^n}$ for $n\geq 2$ since removing a point from a curve breaks it into 2 pieces while any of the higher spaces still remain connected. In fact Brower's famous theorem states that the notion of dimension is good , meaning that $\mathbb{R^n}$ can't be the same as $\mathbb{R^m}$.

The study of manifolds can be motivated by the implicit function theorem which states that localy the level set of a smooth function on a regular value is a graph.

There are many introductory texts about what manifolds are and what we can do with them. I suggest 4:

1. Conlon's Differentiable Manifolds(I think this is one very good start)
2. Tu's An Intoruction to Manifolds
3. Guillemin's and Pollack's Differential Topology
4. Lee's Introduction to smooth manifolds

Answer 2

A curve/surface/volume is defined by the number of "independent" directions you can move within the object. A curve is one-dimensional, since you can only move up and down the curve, while a surface is two-dimensional, since you can move up/down and side-to-side within the surface. Algebraically, we can see the number of independent directions within an object by the number of independent inputs in the parameterization. (By independent I mean linearly independent, which you can think of as directions that are right-angled to each other.)

Answer 3

Why is the graph of $r(t)=(x(t),y(t),z(t))$ a curve, and the graph of$ r(t)=(x(u,v),y(u,v),z(u,v))$ a surface?

Suppose you are sitting in a calculus class and you are asked to draw a curve on a peace of paper.

You start the curve at 8:00 o'clock at a point P=(3,5) where x=3 and y=5 are the coordinate of the starting point.

If t represents the time, we say the starting point is at x(8)=3 and y(8)=5. Now as you trace the curve, time passes from 8:00 to 8:10 and at each time t between 8:00 and 8:10 your pencil is at the point P(t) = (x(t), y(t)) on the curve.

For the three dimensional case consider a fly in the same calculus class.

At each instance of time t, the fly is at some point P(x,y,z) in class room. We say x=x(t), y=y(t), and z=z(t) are coordinates of the point P at time t.

There P=(x(t),y(t),z(t)) describes the curve which fly traces in the class room.

In case of a surface consider the surface of a sphere.

The surface is not a solid, you may think of it as a two dimensional manifold embedded in three dimensional space.

At each point on the sphere you may think of the surface as a curved two dimensional plane.

If you unfold the curved surface to make it flat you have P= (u(t),v(t)) where u(t) and v(t) are coordinates of the point p on the flat surface.

When you return back to the actual three dimensional space, you have P=(x,y,z) where x,y,and z depend on u(t), hence r(t)= (x(u,v),y(u,v),z(u,v)) sounds right for a surface in three dimension.