Directional Derivatives Part 1: Gradients.

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.

Gradient99

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)?
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