May 17, 2010

Last time we talked about derivatives on surfaces.  We noted that there were two main derivatives that we care about, $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ called the partial derivative in the x-direction and the partial derivative in the y-direction.  To calculate the partial in the x-direction for some function of x and y, we simply make y a constant and derive with respect to x.

“But!…” you might begin, “But in calculus, we had a function’s graph and we had a graph of the derivative!  Both were usually functions!  Don’t we have anything like that for surfaces?”

Well, yes.  We do.  Sort of.  Because, at every point, we have an x and y partial derivative, we can assign to each point $(x_0, y_0)$ of our surface $f(x,y)$ an ordered pair of partial derivatives.  Namely, at the point $(x_0, y_0)$, we have the ordered pair of partial derivatives $(\frac{\partial f}{\partial x}(x_0,y_0), \frac{\partial f}{\partial y}(x_0,y_0))$.  Now, a relatively nice way to represent this kind of thing is as a vector field.  A vector field is sort of like a regular graph, but for each point on the graph we assign a vector.

Here is an image (blatantly stolen from wikipedia) of a surface with its associated vector field below it.

Cool, but how do we do this in practice?  It seems like a real drag to find out this vector thing at every point.  It turns out to not be so bad, actually.  In fact, there’s even a cool little symbol that we can use while we’re doing this.

We define the symbol $\nabla$ by the following application to a function: $\nabla f = (\frac{\partial f}{\partial x},\frac{\partial f}{\partial y})$.  This is sort of obvious: we want the direction derivative vector field, so let’s take the partials at each point.  Let’s do an example or two.

1. Let $f(x,y) = 2x + y^2$.  Then we have $\nabla f(x,y) = (2, 2y)$.  If we want the associated vector at the point $(3,4)$, we need only plug in this point into $\nabla f(x,y)$: we get $(2,8)$.
2. Let $f(x,y) = yx^2 + 2xy$.  Then we have $\nabla f(x,y) = (2xy, x^2 + 2x)$.  If we want the associated vector at the point $(2,-2)$, we need only plug in this point into $\nabla f(x,y)$: we get $(-8,8)$.

We call this taking the gradient of f, and this is a very important part of finding a derivative in any direction!  The gradient of other functions is also defined in the same way: if we are given $g(x,y,z,a,b,c)$ or some other function of some-number-of-variables, we can define the gradient by simply taking the partial derivatives in each of the main directions (for g, above, we’d take it in the x, y, z, a, b, and c direction) and make an ordered pair out of it.

Next post I’ll discuss directional derivatives some more.  For now, here are some exercises!

Exercises:

Let’s let $f(x,y) = 2x + 2xy + 2y$, $g(x,y,z) = 2xz + 2xy + 2yz$, and $h(x,y,z,a) = x + y^2 + z^3 + a^4 + xyza$.

1. Find the gradient of f at the point $(2,-1)$.
2. Find the gradient of g at the point $(3,2,1)$.
3. Find the gradient of h at the point $(2,-1,2,-1)$.
4. Suppose we’re given $f(x) = x^2$.  What is the gradient of this function?  In particular, what is the gradient of a function of one variable?  Is this cool (y/n)?