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.