# I'm not enough of a math geek. Need vector calculus help.

I’ve found myself in the unenviable position of trying help teach a class in analytical fluid dynamics. Filling in for a missing instructor is more like it. As I peruse my trusty texts to refresh my memory of some of these god-awful derivations, I’ve run into a bit of a vector calculus shortcut I don’t quite get. It goes like this:

The equation has a number of terms, all of which are gradients of some scalar field. One of them is pressure, and so I’ll just focus on what this author does to the ÑP term. (If it looks messed up on your computer, that’s supposed to be ‘del’ P*)*.

So there’s a term in the equation that is ÑP. Then it says:

In the equation that follows, that pressure term has become simply dP.

I don’t get it. I tried taking the dot product myself, using the definition of the del operator. I ended up with nothing but a tangled mess of curly d’s and regular d’s. I tried some vector derivative identities, but ran into dead end there, too.

Apparently, ÑP · ds = dP. But I don’t really understand why. Can someone spell this out for me?

ÑP, taken in Cartesian coordinates, is (@P/@x)i + (@P/@y)**j ** + (@P/@z)k; the @'s in this equation are partial derivatives (i.e. a ‘d’ with the top limb curled backward); I don’t know how to write them as such in this font. i, j, and k are the unit vectors in the x, y, and z direction.

The line differential in Cartesian coordinates is (dx)i + (dy)j + (dz)k. Note here the d’s are the usual d used in total differentiation. Thus the dot product is given by:

(@P/@x)dx + (@P/@y)dy + (@P/@z)dz

This just happens to be the total derivative of a function P(x,y,z):

dP = (@P/@x)dx + (@P/@y)dy + (@P/@z)dz

dP here is shorthand for a total differential of P; it is a shorthand way of linking a small change in the overall function P to small, related changes in the variables x, y, and z. If you know the path along which P is varying in x, y, and z–say, fix y=f(x) and z=g(x)–you can compute dy = f’(x)dx and dz=g’(x)dx, plug into the formula above, and have dP=Q(x)dx (where Q(x) is the result of combining functions in x from the three added terms).

Any function of several variables has a total derivative. The only conditions are that the function must be differentiable in the ranges of the independent variable.

And there it was. The concept I was missing…the total derivative.

I was able to get that expanded expression with the partials and totals. I just didn’t recognize it instantly as the definition of dP.

Awesome…thanks!