- So I have here
a very complicated function.
It's got a two-dimensional input,
two different coordinates to its input,
and then a three-dimensional output.
Specifically, it's a three-dimensional vector,
and each one of these is some expression,
its a bunch of cosines and sines
that depends on the two input coordinates.
And in the last video, we talked about how
to visualize functions that have a single input,
a single parameter
and then, a two-dimensional vector output.
So some kind of expression of T
and another expression of T.
And this is sort of the three-dimensional analog of that.
So what we're going to do,
we're just going to visualize things in the output space,
and we're gonna try to think of all
the possible points that could be outputs.
So, for example, let's just start off simple,
let's get a feel for this function
by evaluating it at a simple pair of points.
So, let's say
we evaluate this function F,
at T equals zero, I think will probably be pretty simple,
and then S is equal to pi.
So let's think about what this would be.
We go up and we say,
okay, T of zero, cosine of zero is one,
so this whole thing is gonna be one,
same with this one.
And sine of zero is zero.
So this over here's gonna be zero,
and this is also gonna be zero.
Now cosine of pi is negative one.
So this here's gonna be negative one.
This one here's also gonna be negative one.
And then sine of pi, just like sine of zero,
So this whole thing actually ends up simplifying quite a bit
so that the top is three times one plus negative one,
one times negative one is negative one,
and we get two.
Then we have three times zero plus zero,
so the Y component is just zero,
and then the Z component is also zero.
So what that would mean
is that this output is gonna be the point
that's two along the x-axis,
and, there's nothing else to it,
it's just two along the x-axis.
So go ahead and--whoop,
move the graph about,
add that point there.
So that's what would correspond
to this one particular input, zero and pi.
And, you know you can do this with a whole bunch,
and you might add a couple of other points
based on other inputs that you find.
But this would take forever,
to start to get a feel for the function as a whole.
And another thing you can do, is say,
okay, maybe rather than thinking of
evaluating at a particular point,
imagine one of the inputs was constant.
So let's imagine that S stayed constant at pi.
But then we let T range freely.
So, that means we're gonna have some kind
of different output here.
And, we're gonna let T
just be some kind of variable while the output is pi.
So what that means is we keep all of these,
these negative one, negative one, and zero
for what sine of pi is.
But the output now
is gonna be three cosine of T,
cosine of T,
plus negative one times the cosine of T,
so it's gonna be minus cosine of T.
The next part is gonna still be three sine of T,
this is no longer zero.
I should probably erase those guys actually so...
We're no longer evaluating
(inaudible) when T was zero.
So, three times sine of T,
that's just still the function that we're dealing with.
sine of T,
and then minus one times sine of T.
So minus sine
Keep drawing it in green just to be consistent.
And then the bottom stays at zero.
And this whole thing actually simplifies,
three cosine T minus cosine T,
that's just two cosine T.
And then the same deal for the other one.
It's gonna be two sine of T.
So this whole thing,
actually simplifies down
to this. So this is again
when we're letting S stay constant
and T ranges freely.
And when you do that,
what you're gonna end up getting
is a circle that you draw.
And you can maybe see why it's a circle
'cause you have this cosine/sine pattern.
It's a circle with radius two,
and it should make sense that
it runs through that first point that we evaluated.
So that's what happens if you let
just one of the variables run.
But now let's do the same thing,
but think instead of what happens
is S varies and T stays constant.
I encourage you to work it out for yourself,
I'll go ahead and just kind of draw it,
because I kinda wanna give the intuition here.
So in that case you're gonna get a circle
that looks like this.
So again I encourage you to try
to think through for the same reasons.
Imagine that you let S run freely,
keep T constant at zero.
Why is that you would get a circle
that looks like this?
And in fact, if you let both T and S run freely,
a very nice way to visualize that
is to imagine that this circle,
which represents S running freely,
sweeps throughout space as you let T run freely.
And what you're gonna end up getting when you do that,
is a shape that goes like this.
This is a doughnut. We have a fancy word
for this in mathematics we call it a torus.
But it turns out the function here
is a fancy way of drawing the torus.
And in another video I'm gonna go through in more detail
if you were just given the torus,
how you can find this function,
how you can kind of get the intuitive feel for that.
And in that it'll involve going through,
in a bit more detail,
why when you sweep the circle out
it gets the torus just so.
And what the relationship between
this red circle and the blue circle is.
But here I just kind of want to give
an intuition for what parametric surfaces are all about,
how it's a way of visualizing something
that has a two-dimensional input
and a three-dimensional output.