Visualizing the Derivative of a Surface Parametrization
Suppose we have a differentiable function
from a region in R2 to R3.
We’ve seen that its partial derivatives
fit conveniently in a three-by-two matrix.
But what does this matrix represent?
To explore this question, let’s look at
an example.
Here’s a parametrization of a torus sitting
in space.
This is a function, f, from a square to R3.
You feed it two numbers, s and t, both at
least zero and at most two pi, and it spits
out a point on the torus.
The x coordinate of this point is given by
two plus cosine s, times cosine t.
The y coordinate is two plus cosine s, times
sine t, and the z coordinate is given by sine
s.
This might look like a jumble of symbols,
but let’s think a bit about what this parametrization
is doing to the square.
First, let’s look at what it’s doing to
the horizontal line s=0.
If we set s equal to zero then the function
becomes: 3 times the cosine of t, 3 times
the sine of t, zero.
As t varies from 0 to two pi, this traces
out a circle of radius 3 in the xy plane.
This is the big circle curving around the
torus, as far from the hole as possible.
What if we look at a different horizontal
line instead, like s equals pi over two, for
example?
When we set s equal to pi over two the function
becomes: 2 times the cosine of t, 2 times
the sine of t, one.
Now as t varies this traces out a circle of
radius two in the plane z equals one, hovering
one unit above the xy plane.
In fact, this parametrization turns all horizontal
lines into horizontal circles, parallel to
the xy plane, running around the hole in the
torus.
Now let’s look at a vertical line instead.
If we set t equal to zero the function becomes:
Two plus cosine s, zero, sine s.
This is a circle in the xz plane, with radius
1, centered at the point (2,0,0).
This circle curves through the hole in the
torus.
In fact, our parametrization turns all vertical
lines into vertical circles, running through
the torus’ hole.
If you had a square made out of really squishy
rubber, you could think of this function as
a set of instructions for stretching and squashing
the square, gluing its edges together to make
a torus, and then placing the torus in space.
Now let’s look at the derivative of this
function.
First we treat s as a constant and t as a
variable to find the derivative with respect
to t.
We get that dx/dt is negative the quantity
two plus cosine s, times sine t.
dy/dt is the quantity two plus cosine s, times
cosine t.
And dz/dt is zero.
Next we take the partial derivatives with
respect to s.
dx/ds is negative sine s cosine t, dy/ds is
negative sine s sine t, and dz/ds is cosine
of s.
Now we have a three by two matrix, full of
partial derivatives.
And we can ask again: What does this all mean?
Well, we know that matrices represent linear
transformations.
If we pick values for s and t and plug them
in we’ll get a matrix full of numbers, representing
a linear transformation that turns vectors
in the plane into vectors in space.
For example, if we set s and t both equal
to zero, this matrix becomes: Zero three zero,
zero zero one.
So what does this particular linear transformation
do?
If we feed it the vector one zero, pointing
to the right on the t axis, we’ll get the
vector zero three zero in space.
And if we feed it the vector zero one, pointing
upwards on the s axis, we’ll get the vector
zero zero one.
We could picture both of these vectors based
at the origin in R3, like we usually do.
But we could also imagine them based somewhere
else -- like at the point f(0,0), for example.
This is the point three zero zero in space,
it’s where the parametrization sends the
lower left corner of our square.
And notice that these vectors both fit neatly
on our torus now, both of them tangent to
it.
We can do this for any point in the square.
If we pick a point in the square, call it
(a,b), we get one parametrized path in space
by setting s equal to b.
This path traces out the horizontal circle
which passes through f(a,b).
When we differentiate f with respect to s,
and then plug in a and b for s and t, we get
the velocity vector of this path, right as
it passes through the point in question.
If we instead let s vary and set t equal to
a, we get a different parametrized path, this
time tracing out the vertical circle that
passes through f(a,b).
And when we differentiate f with respect to
t and then plug in a and b, we get the velocity
vector for this second path as it passes through
our point.
So, when we take our derivative matrix and
plug in a and b, we get a linear transformation.
We can think of this linear transformation
as taking vectors in the plane, based at (a,b),
and turning them into tangent vectors on the
torus, based at the point f(a,b).
Here’s one way to think about it.
Suppose you have a particle moving around
in your square, along a differentiable path.
We can imagine another particle moving around
on the torus, simultaneously, following an
analogous path: So when one particle is at
a point p, the other particle is at the point
f(p).
At any given moment these particles both have
velocity vectors -- one in the plane, the
other in space, tangent to the torus.
What’s the relationship between these two
vectors?
The matrix of derivatives at the relevant
point gives us exactly that relationship.
If you feed this matrix the velocity vector
in the plane, you always get the velocity
vector in space.
The matrix tells you what the parametrization
does to vectors based at points in the domain.
It might stretch them, or squash them, or
point them in one direction or another as
it places them on the torus.
But in general, if you want to know what a
parametrization is doing to a certain vector,
just ask the matrix of derivatives.
