# Showing that the image of a curve lies on a surface?

I am looking for an intuitive explanation to a problem in one of my practice tests. I'm given a parameterized curve from $\Bbb R$ to $\Bbb R^3$, called ${\bf r}(t) = (\sin t \cos t, \cos^2 t, \cos t)$. I am then given a surface $S$ represented by the equation $x^2 + y^2 = z^2$. I am told to show that the image of ${\bf r}$ lies in $S$.

The answer to this problem implies that I simply replace $x, y$, and $z$ in the surface equation with the corresponding values in ${\bf r}(t)$. Doing so gives back the equation $\cos^2t(\sin^2t + \cos^2t) = \cos^2 t$.

Based on the above, the problem concludes:

...for all points in the image of ${\bf r}(t)$, the sum of the squares of the $x$ and $y$ coordinates is equal to the square of the $z$ coordinate. We conclude that the image of ${\bf r}$ lies on the surface defined by $S$.

I was wondering if I could have an intuitive explanation for what is happening here -- namely, what does it mean to show that the image lies on a surface, and (more generally) how do we show that an image lies on a surface? In this particular problem, I think that there is something in the fact that the equation can be reduced to $\sin^2 t + \cos^2 t = 1$. However, I am not sure how to reason through these kinds of problems.

The surface $S$ is defined as the set of all points in $\mathbb{R}^3$ whose coordinates satisfy $x^2 + y^2 = z^2$.

What is the image of your curve? Well, the points of the image of your curve are precisely points of the form $r(t) = (\sin t\cos t, \cos^2 t, \cos t)$. As $t$ ranges over all of $\mathbb{R}$, $r(t)$ ranges over all points in the image of your curve.

Thus, you want to ask: Do all points in the image of your curve have coordinates satisfying $x^2 + y^2 = z^2$? The important thing to note is that here $x,y,z$ just refer to the first, second, and third coordinates of your points. Well, the points of the image of your curve are of the points with coordinates $(\sin t\cos t,\cos^2 t,\cos t)$, and thus you want to check if the coordinates satisfy:

(first coordinate squared) + (second coordinate squared) = (third coordinate squared)

How do you check that? Well you simply verify that $$(\sin t\cos t)^2 + (\cos^2 t)^2 = (\cos t)^2$$ for all $t\in\mathbb{R}$.

If this is true, then that's to say that for any $t\in\mathbb{R}$, the point $(\sin t\cos t,\cos^2 t,\cos t)$ satisfies (first coordinate squared) + (second coordinate squared) = (third coordinate squared), and thus lies on your surface $S$.

Since every point on your curve can be described in this way, you're done.

Intuitively, this is your surface $x^2 + y^2 = z^2$, a lovely cone in $\Bbb R^3$.

This is your curve $r(t) = (\cos t \sin t, \cos^2 t, \cos t)$, also a lovely subset of $\Bbb R^3$.

When we plot them at the same time, we see that one literally lies on the other: And another angle, just for fun: Any point on the curve automatically satisfies the equation defining your surface, that's as intuitive as it gets. If we call the collection of points on the curve $C$, and those on the surface $S$, then we have $C \subset S \subset \Bbb R^3$.