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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.

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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.

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Intuitively, this is your surface $x^2 + y^2 = z^2$, enter image description here

a lovely cone in $\Bbb R^3$.

This is your curve $r(t) = (\cos t \sin t, \cos^2 t, \cos t)$, enter image description here

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: enter image description here

And another angle, just for fun: enter image description here

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 $C \subset S \subset \Bbb R^3$.

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