I lead a recitation in multivariable calculus and have a hard time explaining parametric surfaces/curves to my students. Especially, why does $$\vec{r_1}(t) = (x(t), y(t), z(t))$$ trace out a curve while $$\vec{r_2}(u,v) = (x(u,v), y(u,v),z(u,v))$$ gives us a surface?
What is a very intuitive and simple explanation of this to an average student in calculus-$3$ student?
On
Here's one way to think of it. First, think in two dimensions. Consider a parameterized curve $r(t) = (f(t), g(t))$. Assume $f(t)$ is continuous and piecewise injective, which is a fairly reasonable assumption for most functions, and assume $g(t)$ is continuous.
In each "piece", we can write
$$r(t) = (f(t), g(f^{-1}(f(t)))$$
This curve is precisely the graph of the function $h:S \to \mathbb{R}$ where $h(x) = g \circ f^{-1} (x)$ and $S \subset \mathbb{R}$ is the (connected) image of whatever "piece" of $f(t)$ we're working with. Since your students have probably been exposed to graphs of real functions of one-variable, and are convinced that these functions are usually curves of some sort, then they should be convinced that $r(t)$ is a curve.
For three variables, if
$r(t) = (f(t), g(t), j(t))$, then you can approximate $j(t)$ with piecewise constant functions, so we're really working with $r(t) = (f(t), g(t), C)$, and then just refer back to the argument for functions of two variables.
You can do something similar with surfaces $\mathbb{R}^2 \to \mathbb{R}^3$.
Ignoring the details a little bit; the curve is an image of the one-dimensional real line. Thus, its image should have one dimension. The surface is an image of the two-dimensional real plane. Thus, it should be roughly "two-dimensional."