# I'm not enough of a math geek. Need vector calculus help. **URL:** \ **Category:** Factual Questions\ **Created:** [September 19, 2005, 6:16pm UTC](https://boards.straightdope.com/t/im-not-enough-of-a-math-geek-need-vector-calculus-help/322394 "2005-09-19T18:16:48Z")\ **Posts on this page:** 4\ **Page:** 1
**Author:** ![aerodave](https://avatars.discourse-cdn.com/v4/letter/a/f08c70/32.png) [@aerodave](https://boards.straightdope.com/u/aerodave)\ **Post date:** [September 19, 2005, 6:16pm UTC](https://boards.straightdope.com/t/im-not-enough-of-a-math-geek-need-vector-calculus-help/322394/1 "2005-09-19T18:16:48Z")
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 Ñ_P_ 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 Ñ_P_. Then it says: > [@](#): > > Let us calculate the change in the magnitude of each of these terms along an arbitrary path whose length and direction are defined by the vector _ **ds** _. To do this, we take the dot product of each term in the above equation and the vector _ **ds** _ 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, Ñ_P_ · _ **ds** _ = _dP_. But I don’t really understand why. Can someone spell this out for me? ---
**Author:** ![CJJ](https://avatars.discourse-cdn.com/v4/letter/c/ecc23a/32.png) [@CJJ](https://boards.straightdope.com/u/CJJ)\ **Post date:** [September 19, 2005, 7:15pm UTC](https://boards.straightdope.com/t/im-not-enough-of-a-math-geek-need-vector-calculus-help/322394/2 "2005-09-19T19:15:01Z")
> [@aerodave](#): > > 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 Ñ_P_ 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 Ñ_P_. 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, Ñ_P_ · _ **ds** _ = _dP_. But I don’t really understand why. Can someone spell this out for me? Ñ_P_, 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](http://scholar.chem.nyu.edu/0652/notes/math/node7.html) 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. ---
**Author:** ![aerodave](https://avatars.discourse-cdn.com/v4/letter/a/f08c70/32.png) [@aerodave](https://boards.straightdope.com/u/aerodave)\ **Post date:** [September 19, 2005, 7:41pm UTC](https://boards.straightdope.com/t/im-not-enough-of-a-math-geek-need-vector-calculus-help/322394/3 "2005-09-19T19:41:03Z")
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! ---
**Author:** ![cerberus](https://avatars.discourse-cdn.com/v4/letter/c/71e660/32.png) [@cerberus](https://boards.straightdope.com/u/cerberus)\ **Post date:** [September 19, 2005, 11:33pm UTC](https://boards.straightdope.com/t/im-not-enough-of-a-math-geek-need-vector-calculus-help/322394/4 "2005-09-19T23:33:16Z")
Don’t Stokes’ and Green’s Theorems say soemthing about this?